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General Discussion => Q&A Section => Topic started by: ckiki lwai on 06/23/2008 10:41 am

Title: Basic Rocket Science Q & A
Post by: ckiki lwai on 06/23/2008 10:41 am
 ;D
I hope the mods find this new Q & A allright

Suppose you traveled from Earth to Mars using a Hohmann transfer orbit.
How do you calculate the time you have to wait to get another chance to fly back to Earth again using a Hohmann transfer orbit?

Edit: Sorry didn't read James Lowe1 post in the section intro thread...
I can't remove it anymore, delete it when you get the chance.
Title: Re: Basic Rocket Science Q & A
Post by: Spacenick on 07/18/2008 01:06 pm
It's not strictly Orbital mechanics, but I dind't find a thread that fitted the topic better. Can someone describe to me how a tank pressure fed rocket operates, the thing I don't understand is how the tank pressure can push fuel into the chamber even though the chamber pressure must be many times higher than the tank pressure?
Title: Re: Basic Rocket Science Q & A
Post by: zeke01 on 07/18/2008 01:32 pm
Simple: For a tank pressure fed rocket you increase the pressure in the tanks until it's higher than the combustion chamber pressure.   To obtain high efficiency this would require thick walled tanks and pipes which makes it a very heavy, albeit very simple, propulsion system when compared to other types of rocket cycles.

Most efficient, chemical rocket engines nowadays use pumps to boost fluid pressures from low-pressure tanks to values required for efficient combustion, reducing the amount of high pressure (thick & heavy) plumbing.

zeke
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 07/18/2008 01:57 pm
;D
I hope the mods find this new Q & A allright

Suppose you traveled from Earth to Mars using a Hohmann transfer orbit.
How do you calculate the time you have to wait to get another chance to fly back to Earth again using a Hohmann transfer orbit?


It has to do with knowing the where the planets are in their orbits and how long it takes for them to get into the right positions.  It is every 26 months for earth to mars
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 07/18/2008 01:59 pm
It's not strictly Orbital mechanics, but I dind't find a thread that fitted the topic better. Can someone describe to me how a tank pressure fed rocket operates, the thing I don't understand is how the tank pressure can push fuel into the chamber even though the chamber pressure must be many times higher than the tank pressure?

The tank pressure is always higher than the combustion chamber for a pressure fed system
Title: Re: Basic Rocket Science Q & A
Post by: Spacenick on 07/18/2008 02:09 pm
I guess this means that the heat from the combustion is used to augment the tank pressure, right? Else one wouldn't really need the combustion, or am I wrong with this.
Title: Re: Basic Rocket Science Q & A
Post by: William Barton on 07/18/2008 02:18 pm
I guess this means that the heat from the combustion is used to augment the tank pressure, right? Else one wouldn't really need the combustion, or am I wrong with this.

The combution chamber he's talking about is the rocket engine. The tank pressure has to be higher than the chamber pressure or the engine will starve (or worse, blow back through the valves and feedlines, creating a nice bomb).
Title: Re: Basic Rocket Science Q & A
Post by: Spacenick on 07/18/2008 02:31 pm
Yeah, I'm talking about exactly the same chamber it's the one LH2 and LOX get pumped into to combust kicking the gas out through the nozzle producing the thrust.
The thing is if the tank pressure were higher than the pressure in the chamber WITHOUT using energy from the combustion, one wouldn't need the combustion and could let the fuel flow directly to the nozzle, so my question was how the pressure is augmented so that it is higher than the chamber pressure.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 07/18/2008 02:47 pm
Yeah, I'm talking about exactly the same chamber it's the one LH2 and LOX get pumped into to combust kicking the gas out through the nozzle producing the thrust.
The thing is if the tank pressure were higher than the pressure in the chamber WITHOUT using energy from the combustion, one wouldn't need the combustion and could let the fuel flow directly to the nozzle, so my question was how the pressure is augmented so that it is higher than the chamber pressure.

The pressure isn't augmented.  It is by design higher in the tank than combustion chamber. 

combustion adds energy and it provides the bulk of the  increase the velocity of the exhaust.

If you have trouble grasping this, then I bet a jet engine throws you into a tizzy.   Need to read a basic rocket engine book
Title: Re: Basic Rocket Science Q & A
Post by: Spacenick on 07/18/2008 02:56 pm
Well i think I got it, the important thing in the chamber is not the increase of pressure due to combustion but the high temperature of the exhaust leading to an extreme increase in volume and particle speed (which in itself isn't anything else than temperature). This in turn makes the exhaust exit the nozzle at extremly high speeds (the higher the better). Not really part of the physics we did in school physics though^^
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 07/18/2008 11:07 pm
Rocket Propulsion Elements by Sutton is the standard.  I advise the 6th edition.  There was a 7th updated by Biblarz, but it was full of errors.

Fundamentals of Astrodynamics by Bate, Mueller and White (known in classes as BMW) is an excellent, cheap text on orbital mechanics.
Title: Re: Basic Rocket Science Q & A
Post by: jabe on 07/18/2008 11:33 pm
Rocket Propulsion Elements by Sutton is the standard.  I advise the 6th edition.  There was a 7th updated by Biblarz, but it was full of errors.

Fundamentals of Astrodynamics by Bate, Mueller and White (known in classes as BMW) is an excellent, cheap text on orbital mechanics.
Thanks for the tip...
I recently bought the "BMW" book and love it..Now i need to find the sixth edition by sutton..

cheers
jb
BTW how can the 7th be so full of errors?????
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 07/19/2008 03:13 am
It's not strictly Orbital mechanics, but I dind't find a thread that fitted the topic better. Can someone describe to me how a tank pressure fed rocket operates, the thing I don't understand is how the tank pressure can push fuel into the chamber even though the chamber pressure must be many times higher than the tank pressure?

The pressure feed engines the shuttle uses on orbit have a chamber pressure of about 125 psi and tank pressures of about 300 psi.  I was very surprised when I first learned a rocket engine can operate with such low pressure.  The key to efficiency is high temperature, not high pressure.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 07/19/2008 03:15 am
Well i think I got it, the important thing in the chamber is not the increase of pressure due to combustion but the high temperature of the exhaust leading to an extreme increase in volume and particle speed (which in itself isn't anything else than temperature). This in turn makes the exhaust exit the nozzle at extremly high speeds (the higher the better). Not really part of the physics we did in school physics though^^

You are correct.  It is the temperature that causes the exhaust velocity to be high.
Title: Re: Basic Rocket Science Q & A
Post by: antonioe on 08/31/2008 12:33 am
There were numerous hints at the true role of the chemical reaction (call it "combustion" if you wish) in a rocket engine; let me summarize what was spread over the previous comments:

The true core of a rocket engine is the nozzle, or, more precisely the convergent-divergent nozzle, a frighteningly simple but also frighteningly subtle machine invented by Gustav de Laval (http://en.wikipedia.org/wiki/De_Laval_nozzle) around the turn of the (nineteenth) century.  A Laval nozzle converts gas temperature and pressure into gas velocity in an extremely efficient (lossless) way.  Indeed, the process is very close to "adiabatic", i.e. there is almost no exchange of heat between the gas and anything else during its transit through the nozzle (well, there is some; it looks like a lot to the nozzle, but it looks like very little to the gas).

As heat and pressure are converted into velocity, the temperature (as well as the pressure) of the gas decreases.  If we had injected propellant into the nozzle at tank temperature (either because the tanks are at high pressure, as in a pressure-fed engine, or because we mechanically compressed it with a turbopump, in which case it will be a bit hotter than the tanks due to the compression), it will be mightily cold by the time it expands towards ambient pressure.  Cold also means dense, so the speed will not be very high, since the gas has decreased its volume a lot.

Enter the "combustion"; it takes the (not very hot) compressed gas and heats it up so that, when it cools down through the nozzle, it still remains at a high temperature and low density.  The chain of events could be conceived as follows:

1.- Pump or tank pressure causes the gas to want to leave through the nozzle, towards the area of lower density "out there".

2.- Chemical reaction increases the energy of the gas in the form of thermal energy.

3.- The fluid-dynamics of the nozzle accelerate the gas leaving the nozzle at the expense of pressure and temperature, therefore converting (in a most efficient way) thermal energy into mechanical (gas velocity) energy.

The main effect of the pressure at the beginning of the nozzle is to increase the "thrust density" of the device: high (chamber) pressure engines are small for the amount of propellant that goes through them in a seconds, therefore have high thrust for their size; conversely, low pressure engines have low thrust.  Sometimes (e.g. when you are lifting off vertically against 1 g) you want a substantial amount of thrust.  Other times (e.g., while delicately approaching the ISS) you do NOT want a lot of thrust.  This explains why we have both high pressure AND low pressure engines.

The efficiency of this process, when the expansion is well matched to the outside pressure, is remarkable: it is practically at the limit of the Carnot efficiency between the combustion chamber temperature and the outside conditions.  This leaves the chemistry of the reaction (the energy it adds to a unit mass of propellant) as the physical limit of the performance of any chemical rocket engine.

Society has been so spoiled by Moore's law of technological rate of improvement in electronics that they simply cannot believe that the asymptote of technological improvement in rocketry was achieved in 1970, with no hope of significant improvements...

I check my edition of Sutton; I have two: a third edition (1963 - I graduated in 1972) and a fifth (1986 - the year I left MIT to join Orbital).  According to the bookstore markings, I paid $28.95 for the thrid edition - new -  and $45.95 for the fifth, also new... *SIGH*)
Title: Re: Basic Rocket Science Q & A
Post by: JWag on 09/08/2008 01:44 pm
There were numerous hints at the true role of the chemical reaction (call it "combustion" if you wish) in a rocket engine; let me summarize what was spread over the previous comments:

<snip>

Thank you for writing that.
Title: Re: Basic Rocket Science Q & A
Post by: BriA on 12/31/2008 11:11 pm
;D
I hope the mods find this new Q & A allright

Suppose you traveled from Earth to Mars using a Hohmann transfer orbit.
How do you calculate the time you have to wait to get another chance to fly back to Earth again using a Hohmann transfer orbit?


It has to do with knowing the where the planets are in their orbits and how long it takes for them to get into the right positions.  It is every 26 months for earth to mars

I actually had to solve this exact problem a couple months ago.  Couldn't resist the chance to give a detailed explanation.

1)  The 26 months referred to earlier is the synodic period, which is calculated using the periods of the two planets in question.  It refers to the amount of time it takes for two planets in circular, coplanar orbits to come back to their same position relative to each other.  Granted, neither Earth nor Mars are in exactly circular orbits, nor are they coplanar.  But it's close enough for a first approximation (and for a homework problem).

2)  There is a formula to calculate the angle (I'll call it the beta angle) between the two planets that must exist to start a Hohmann transfer from Earth to Mars, and vice versa.  In order to do a Hohmann from Earth to Mars, Earth must be trailing Mars by about 44 degrees.  To do the return transfer back to Earth, Earth must be trailing Mars by about 75 degrees.

3)  In order to actually solve the problem of how long you have to wait to do a Hohmann transfer back to Earth after arriving at Mars, you must look at the angle between the planets at arrival (found using the period of time for the initial hohmann), the angle between the planets at departure (known), and the angular distance each planet covers during the interim (unknown).  You then set the interim as the variable, find the amount of angular distance each planet has to cover during the time period in question (Earth covers 2*pi plus the unknown angle, Mars covers all three angles but does not make a full revolution), divide it by the angular velocity of each planet (360 degrees/planet period), set the two times equal to each other, and solve for the unknown angle.

As an aside, another book I would recommend to anyone interested in this topic (in addition to the ones that have already been touted, which are all excellent) is "Orbital Mechanics" by Prussing and Conway.  The formulas for all the equations I mentioned above are in the tail end of Chapter 6.
Title: Re: Basic Rocket Science Q & A
Post by: AnimatorRob on 01/06/2009 07:24 pm
This seems like a good place to ask this..
As I understand it, an upper stage is responsible for the final ,precise delta-v to place the payload in is correct orbit, varying it's burn time to correct for variations in first stage performance, atmospheric conditions, etc. My question is how is that accomplished with a solid fueled upper stage, for example,  a Castor 30.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 01/06/2009 08:22 pm
My question is how is that accomplished with a solid fueled upper stage, for example,  a Castor 30.

By the spacecraft, another stage or  requirements that account for the errors.

The mission is planned for "excess" performance then flies t
Title: Re: Basic Rocket Science Q & A
Post by: ckiki lwai on 01/06/2009 10:24 pm
;D
I hope the mods find this new Q & A allright

Suppose you traveled from Earth to Mars using a Hohmann transfer orbit.
How do you calculate the time you have to wait to get another chance to fly back to Earth again using a Hohmann transfer orbit?


It has to do with knowing the where the planets are in their orbits and how long it takes for them to get into the right positions.  It is every 26 months for earth to mars

I actually had to solve this exact problem a couple months ago.  Couldn't resist the chance to give a detailed explanation.

1)  The 26 months referred to earlier is the synodic period, which is calculated using the periods of the two planets in question.  It refers to the amount of time it takes for two planets in circular, coplanar orbits to come back to their same position relative to each other.  Granted, neither Earth nor Mars are in exactly circular orbits, nor are they coplanar.  But it's close enough for a first approximation (and for a homework problem).

2)  There is a formula to calculate the angle (I'll call it the beta angle) between the two planets that must exist to start a Hohmann transfer from Earth to Mars, and vice versa.  In order to do a Hohmann from Earth to Mars, Earth must be trailing Mars by about 44 degrees.  To do the return transfer back to Earth, Earth must be trailing Mars by about 75 degrees.

3)  In order to actually solve the problem of how long you have to wait to do a Hohmann transfer back to Earth after arriving at Mars, you must look at the angle between the planets at arrival (found using the period of time for the initial hohmann), the angle between the planets at departure (known), and the angular distance each planet covers during the interim (unknown).  You then set the interim as the variable, find the amount of angular distance each planet has to cover during the time period in question (Earth covers 2*pi plus the unknown angle, Mars covers all three angles but does not make a full revolution), divide it by the angular velocity of each planet (360 degrees/planet period), set the two times equal to each other, and solve for the unknown angle.

As an aside, another book I would recommend to anyone interested in this topic (in addition to the ones that have already been touted, which are all excellent) is "Orbital Mechanics" by Prussing and Conway.  The formulas for all the equations I mentioned above are in the tail end of Chapter 6.
Well thanks anyway, I had the same question on my exam only with Pluto instead of Mars, but I didn't had the time to solve the question anyway. I did pass the exam tough :P
Title: Re: Basic Rocket Science Q & A
Post by: C5C6 on 01/10/2009 07:28 pm
Since all objects in Earth have the same rotational speed as the Earth itself (around 1700km/h), if I wanted a sattelite in an orbit with the same orbital speed, could I launch a rocket facing totally orthogonal to the earth's surface without providing any additional orbital speed and only using the rocket to reach the needed altitude???

by the way, why arent sattelites injected directly to geosynchronous orbits?? why do they must first go to LEO and then use a transfer orbit???
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 01/10/2009 07:31 pm

by the way, why arent sattelites injected directly to geosynchronous orbits?? why do they must first go to LEO and then use a transfer orbit???

to get to the equator before doing the plane change
Title: Re: Basic Rocket Science Q & A
Post by: C5C6 on 01/10/2009 07:54 pm
thks!!! and about the first question?? would be possible???
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 01/10/2009 08:18 pm
thks!!! and about the first question?? would be possible???

It is possible but would be horribly inefficient. An orbit with a 1700 km/h circular velocity would be so high that you'd be fighting gravity losses the entire way. The current method (thrusting vertically only as much as needed to get above most of the atmosphere, gradually curving horizontally to build up to orbital velocity, then performing impulsive transfers to higher orbits) is far, far more efficient.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 01/11/2009 09:41 pm
My question is how is that accomplished with a solid fueled upper stage, for example,  a Castor 30.

Solids generally have very well predicted impulse.  Therefore, the precision necessary to achieve the target orbit can be controlled by having the excess performance on lower stages and losing it during the coast before the solid ignites.  This method means that the flight computer calculates the variable coast time rather than a variable burn time.  I suppose you could also do it by changing the vector of the solid's burn.  That might be a harder guidance problem.
Title: Re: Basic Rocket Science Q & A
Post by: Nascent Ascent on 01/11/2009 09:53 pm
1) Does it (or would it) ever make sense to launch an orbit-bound rocket at an acute angle with respect to the ground?

2) I know there are different nose cose profiles (ogive, power series, haack, etc) that are optimized for different purposes. Why do the Shuttle ET and the SRB have different profiles?
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 01/12/2009 03:00 am
1) Pegasus does, so I guess it would make sense if the LV is using aerodynamic lift.
Title: Re: Basic Rocket Science Q & A
Post by: edkyle99 on 01/12/2009 04:49 am
1) Does it (or would it) ever make sense to launch an orbit-bound rocket at an acute angle with respect to the ground?

Japan's M-V was launched that way, from a launch rail (see photo).  I believe that NASA's Scout may have also been launched tilted slightly from vertical.  Pegasus, of course, is nearly horizontal when dropped.

 - Ed Kyle

Title: Re: Basic Rocket Science Q & A
Post by: scienceguy on 01/16/2009 03:08 am
You know the rocket equations that say how much of a SSTO would have to be fuel to achieve orbit at Earth? (I think it is about 80%) Anyway, I was just wondering what percentage of a SSTO taking off from Mars would have to be fuel.

Thanks in advance
Title: Re: Basic Rocket Science Q & A
Post by: eeergo on 01/26/2009 10:20 pm
I think this is the best place to post this question:

Regarding GEO perturbations: recently, I read an article in The Space Review (http://www.thespacereview.com/article/1290/1 (http://www.thespacereview.com/article/1290/1)) which talked about DSP23's latest status, and explanied the East-West perturbations are mainly caused because of a pair of bulges at or close to Earth's Equator, quoted to be at 75ºE and 225ºE. I had a faint knowledge of the geoid and its characteristics, but didn't recall its geometry or the key to interpreting it, so I looked for some images, like this one:

(http://cddis.nasa.gov/926/egm96/geoid_050.gif)

I found that what you find at 75ºE is the major Indic Ocean perturbation, but it's a negative perturbation, meaning (and I checked this several times through different sources) it exerts less gravitational pull. The strongest pull is caused by the Indonesian anomaly. At 225ºE, we find something similar, though less obvious: a negative anomaly in the North-East Pacific. There is a positive anomaly nearby, in the Peruvian Andes. The simplest explanation for the geoid is that a ball wouldn't roll in any point in its surface, because it shows the isogravity lines (or whatever you call them)

Intuition (and simple physics) tell the satellites should be accelerated by the positive anomalies, hence causing an oscillatory motion between two points centered in the gravity well: the strongest gravity source. Are people somehow disoriented by the graphical representations of the geoid, thinking the wells are the "lowest" points in the figure? Or am I missing something entirely and being ridiculously stubborn?
Title: Re: Basic Rocket Science Q & A
Post by: simonbp on 01/27/2009 11:08 pm
I think they may have meant that the orbits get bulges at 75 and 225? Or, it may be that the Geoid is low there because the seafloor is higher density than land, and so has a larger effective gravity...

If you think the Earth has a tricky gravity field, here's a low-res grav map for the Moon; there's a reason we're sending GRAIL...

Simon ;)
Title: Re: Basic Rocket Science Q & A
Post by: eeergo on 01/28/2009 10:54 am
I think they may have meant that the orbits get bulges at 75 and 225? Or, it may be that the Geoid is low there because the seafloor is higher density than land, and so has a larger effective gravity...

That's why at first I thought I was reading things incorrectly, but the more I think about it, the more convinced I am they read the geoid incorrectly. Here's the exact quote from The Space Review:

Quote
In fact, there are two “bulges” along the Equator at approximately 75° and 225° East longitude. These gravity “troughs” pull satellites in geostationary orbit east or west towards whichever is closest, giving the satellite an apparent east or west drift.

The gravity vector should always be perpendicular to the surface of the geoid, so the ocean floor, even though higher density, should have lower gravitational pull (the gravity vector gets 'attracted' by the redder, higher mounds in the geoid)

The 'bulges' in the orbits you mention are a good clue as to what they may be trying to say in the article... but that would make a satellite drifting out of its GEO slot over Gabon oscillate between said point and the central Pacific, more or less. Instead of the 8-135ºE oscillation, it would move between 8 and 230-240ºE, almost twice the distance.
Title: Re: Basic Rocket Science Q & A
Post by: Crispy on 01/28/2009 12:53 pm
1) Does it (or would it) ever make sense to launch an orbit-bound rocket at an acute angle with respect to the ground?

Japan's M-V was launched that way, from a launch rail (see photo).  I believe that NASA's Scout may have also been launched tilted slightly from vertical.  Pegasus, of course, is nearly horizontal when dropped.

 - Ed Kyle

And on a semi-related note, the launch system of the Soyuz rotates on its base in order to put the rocket in the right orientation, depending on the launch direction (azimuth). Modern rockets perform this adjustment early in flight using roll control, but this was beyond the avionics of the early R-7.
Title: Re: Basic Rocket Science Q & A
Post by: Heg on 01/28/2009 02:19 pm
You know the rocket equations that say how much of a SSTO would have to be fuel to achieve orbit at Earth? (I think it is about 80%) Anyway, I was just wondering what percentage of a SSTO taking off from Mars would have to be fuel.

Thanks in advance

From my (very rough) calcullations, for LOX / LH2  (isp 460 sec.) it would be about 54.5%. Perhaps more realistic isp 448 sec. gives you about 55.4%. But if you'd like to use liquid methane (isp at ~380 sec.) instead of LH2, your percentage rises to 61.5%. Note that these figures don't take into account the atmosferic drag, trajectory shaping, etc.
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 01/29/2009 01:55 am

From my (very rough) calcullations, for LOX / LH2  (isp 460 sec.) it would be about 54.5%. Perhaps more realistic isp 448 sec. gives you about 55.4%. But if you'd like to use liquid methane (isp at ~380 sec.) instead of LH2, your percentage rises to 61.5%. Note that these figures don't take into account the atmosferic drag, trajectory shaping, etc.


A more fun number would be what is the mass fraction required for SSTER (Single Stage To Earth Return)
Title: Re: Basic Rocket Science Q & A
Post by: Heg on 01/29/2009 07:26 pm

From my (very rough) calcullations, for LOX / LH2  (isp 460 sec.) it would be about 54.5%. Perhaps more realistic isp 448 sec. gives you about 55.4%. But if you'd like to use liquid methane (isp at ~380 sec.) instead of LH2, your percentage rises to 61.5%. Note that these figures don't take into account the atmosferic drag, trajectory shaping, etc.


A more fun number would be what is the mass fraction required for SSTER (Single Stage To Earth Return)

Granted 8) Here are the numbers:

Isp 460 sec. - 67%
Isp 448 sec. - 68%
Isp 380 sec. - 74%

General formula is:

m(p)/m(t) = 1-(1/e^(ΔV/V(e)))

where:

m(p) - mass of the propellant
m(t) - total mass of the vehicle (including propellant)
e = 2.71828...
ΔV - here it's either minimal orbital velocity (for SSTO), or the escape velocity (for SSTER). For Mars, the numbers are: 3556 km/sec and 5029 km/sec, respectively
V(e) - exhaust velocity of the engine(s). Note that V(e) = Isp·g(0), where Isp is the specific impulse and g(0) = 9.81 m/s^2

* * *

[Just have been playing around the basic rocket equation for a while ;):

ΔV = V(e)·ln(m(t)/m(d))

m(d) - dry mass of the vehicle; m(p) = m(t) - m(d)]

Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 02/21/2009 05:23 pm
OK, here's a stupid question:

For some launch vehicles with compartments, the air from these compartments is often replaced with nitrogen, to suppress any possible fires. Why don't they use helium, instead, which would make the launch vehicle lighter?
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 02/21/2009 06:03 pm
Too expensive.  Those compartments vent to atmosphere anyway.  Some materials and electronics are sensitive to helium too.

Helium is actually used in places where the temperature might be so cold (due to the presence of an LH2 tank) that the GN2 would freeze.  There's at least one other application for helium purge, but I'm not feeling the gumption to google it right now to determine if it's public or proprietary.
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 02/21/2009 06:13 pm
Some materials and electronics are sensitive to helium too.

Sensitive to helium in what sense? It's supposed to be an inert gas. Diffusion problems?
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 02/21/2009 06:19 pm
Some materials and electronics are sensitive to helium too.
Sensitive to helium in what sense? It's supposed to be an inert gas. Diffusion problems?

Dunno exactly.  Helium can go through anything, apparently weakening some metals a tiny bit.  Only a problem in very low margin areas.  There are some spacecraft (guidance?) boxes that don't like helium.  I don't know the mechanism, just that the providers ask to make sure that their boxes aren't around any helium sources.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 02/21/2009 09:09 pm
OK, here's a stupid question:

For some launch vehicles with compartments, the air from these compartments is often replaced with nitrogen, to suppress any possible fires. Why don't they use helium, instead, which would make the launch vehicle lighter?


Helium is one of those elements that we have enough of - for now - but that will not always be the case. Mission planners and engine designers are already aware of the decreasing availability of helium. It's not a problem - yet - but it will be, especially if we use it like you suggest and it just gets vented. There just is not an over abundant supply of it on earth and we need to be aware of our usage and not waste it.
Title: Re: Basic Rocket Science Q & A
Post by: AnalogMan on 02/21/2009 11:29 pm
OK, here's a stupid question:

For some launch vehicles with compartments, the air from these compartments is often replaced with nitrogen, to suppress any possible fires. Why don't they use helium, instead, which would make the launch vehicle lighter?


Helium is one of those elements that we have enough of - for now - but that will not always be the case. Mission planners and engine designers are already aware of the decreasing availability of helium. It's not a problem - yet - but it will be, especially if we use it like you suggest and it just gets vented. There just is not an over abundant supply of it on earth and we need to be aware of our usage and not waste it.

Twenty-five years ago I used to work in laboratories that used liquid helium for cooling a multitude of sensors on scientific experiments on a daily basis - even then we were very careful to capture and recover as much of the total boil-off as we could to minimise costs to our facility.
Title: Re: Basic Rocket Science Q & A
Post by: madscientist197 on 02/22/2009 07:51 am
I knew that Hydrogen embrittlement was a real issue, but I didn't realise that there was any issue with Helium.
Title: Re: Basic Rocket Science Q & A
Post by: nomadd22 on 02/22/2009 11:51 am
 The problem with helium is the places it gets into. It diffuses through things that are impermeable to air or nitrogen over time, and when the things are exposed to vacuum in just the few minutes it takes to launch you suddenly have pressurized helium in places you don't want it because it takes time to diffuse back out.
 It also makes transponders transmit in a high squeaky voice.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 02/22/2009 04:28 pm
Some materials and electronics are sensitive to helium too.
Sensitive to helium in what sense? It's supposed to be an inert gas. Diffusion problems?

Dunno exactly.  Helium can go through anything, apparently weakening some metals a tiny bit.  Only a problem in very low margin areas.  There are some spacecraft (guidance?) boxes that don't like helium.  I don't know the mechanism, just that the providers ask to make sure that their boxes aren't around any helium sources.

It was IMU with Hemispherical Resonant Gyro
http://www.es.northropgrumman.com/solutions/hrg/index.html

Helium changes the frequency* and mucks them up.

* not all of them operate in a vacuum
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 02/25/2009 05:01 am
I have a question about using regenerative cooling with rocket engines. I read somewhere that this works best with liquid hydrogen / liquid oxygen rockets. Apparently the thermodynamic qualities of H2 are very good, better than those of hydrocarbons, which have additional problems with coking as well. Oxygen is rarely used as a coolant because it's very reactive, especially when it's very hot.

What I don't understand is why you couldn't use a separate cooling fluid that does not have coking problems and doesn't react with the nozzle and use a heat exchanger. Then you could still use the oxygen as a heat sink. The Skylon people want to do something similar, but not for cooling the nozzle but for use in their precooler in air-breathing mode.

This sounds like a very simple concept, yet I have never read anything about it, so there's probably something wrong with it. Anyone know what?
Title: Re: Basic Rocket Science Q & A
Post by: yinzer on 02/25/2009 05:54 am
How does having a separate heat exchanger help with oxygen reactivity?  When you're done, you still have a bunch of hot, high pressure oxygen.

There is also the added weight of the second exchanger and the power required to move the coolant through it (which is much higher than you might imagine - thousands of horsepower for a mid-to-large rocket engine).
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 02/25/2009 09:21 am
How does having a separate heat exchanger help with oxygen reactivity?  When you're done, you still have a bunch of hot, high pressure oxygen.

I was thinking that the heat capacity and conductivity of the oxygen tank would be so enormous that the oxygen wouldn't get very hot at all, even locally.

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There is also the added weight of the second exchanger

Ah, I can see how that would be a problem. Would the heat exchanger be more massive than an ablative nozzle?

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and the power required to move the coolant through it (which is much higher than you might imagine - thousands of horsepower for a mid-to-large rocket engine).

But is that any different from a normal regeneratively cooled engine?
Title: Re: Basic Rocket Science Q & A
Post by: DMeader on 02/25/2009 12:34 pm
How about the weight of the fluid itself?

Oxidizer tank would have to be built stronger to handle the increase in pressure, as the oxidixer absorbs heat the pressure would begin to rise immediately.

Also adding complexity. Another fluid system, more pipes, more valves, tankage on the LV, infrastructure for servicing. Something else to fail.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 02/25/2009 12:47 pm
How about the weight of the fluid itself?

Oxidizer tank would have to be built stronger to handle the increase in pressure, as the oxidixer absorbs heat the pressure would begin to rise immediately.

Also adding complexity. Another fluid system, more pipes, more valves, tankage on the LV, infrastructure for servicing. Something else to fail.

I see, thanks guys.
Title: Re: Basic Rocket Science Q & A
Post by: Eerie on 02/25/2009 06:21 pm
Can I find somewhere a total mass of everything artificial in orbit around Earth?
Title: Re: Basic Rocket Science Q & A
Post by: yinzer on 02/26/2009 02:05 am
How does having a separate heat exchanger help with oxygen reactivity?  When you're done, you still have a bunch of hot, high pressure oxygen.

I was thinking that the heat capacity and conductivity of the oxygen tank would be so enormous that the oxygen wouldn't get very hot at all, even locally.

The amount of oxygen available for cooling doesn't change.
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There is also the added weight of the second exchanger
Ah, I can see how that would be a problem. Would the heat exchanger be more massive than an ablative nozzle?
Depends, but probably.
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and the power required to move the coolant through it (which is much higher than you might imagine - thousands of horsepower for a mid-to-large rocket engine).
But is that any different from a normal regeneratively cooled engine?

In a normal regeneratively cooled engine, you have one heat exchanger operating at a large temperature differential.  With your scheme, you have two heat exchangers each operating at a smaller temperature differential, which means more pressure drop and thus power.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 02/26/2009 02:24 am
In a normal regeneratively cooled engine, you have one heat exchanger operating at a large temperature differential.  With your scheme, you have two heat exchangers each operating at a smaller temperature differential, which means more pressure drop and thus power.

Thanks! Continuing with my naive ideas, why couldn't you construct a rocket in such a way that either the fuel or oxygen tank surrounds the combustion chamber and nozzle for cooling? I'm not suggesting this is a good idea and maybe this would be an excellent way to build a bomb, I'm just trying to understand.
Title: Re: Basic Rocket Science Q & A
Post by: usn_skwerl on 02/27/2009 05:42 am
Just an observation I had and wanted to pass along because I didn't quite hear it worded correctly and/or dumbed down until very recently.

The way I heard it described (for those of us unsure about specific impulse (measured in seconds)), it's defined as how long a rocket will fire using a pound of prop/oxidizer to provide a pound of thrust. If this is inaccurate, please correct me.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 02/27/2009 12:15 pm
Just an observation I had and wanted to pass along because I didn't quite hear it worded correctly and/or dumbed down until very recently.

The way I heard it described (for those of us unsure about specific impulse (measured in seconds)), it's defined as how long a rocket will fire using a pound of prop/oxidizer to provide a pound of thrust. If this is inaccurate, please correct me.

That is incorrect

It is thrust per propellant flow rate or  impulse (change in momentum) per unit of propellant.
Title: Re: Basic Rocket Science Q & A
Post by: William Barton on 02/27/2009 12:28 pm
Just an observation I had and wanted to pass along because I didn't quite hear it worded correctly and/or dumbed down until very recently.

The way I heard it described (for those of us unsure about specific impulse (measured in seconds)), it's defined as how long a rocket will fire using a pound of prop/oxidizer to provide a pound of thrust. If this is inaccurate, please correct me.

That is incorrect

It is thrust per propellant flow rate or  impulse (change in momentum) per unit of propellant.

Isn't propellant flow rate a compound factor? Lb-thrust per lb-prop per second? (I will not be surprised if I got that wrong.)
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 02/28/2009 01:07 pm
Occasionally I see people mention air-started SSME's. I get the impression that this is either hard or dangerous. Somehow the J2X is supposed to be better in this regard. Is there a simple explanation of what the problem is and how J2X is better?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 02/28/2009 01:22 pm
Occasionally I see people mention air-started SSME's. I get the impression that this is either hard or dangerous. Somehow the J2X is supposed to be better in this regard. Is there a simple explanation of what the problem is and how J2X is better?

it is only hard for the SSME since it wasn't designed for it.   It is head started which means the start box is small, the initial conditions are very critical.  Also it depends on ground GSE for start and purges
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 03/02/2009 02:40 pm
From the SpaceX thread:
Now i need to ask... I guess I'm missing a key point on how rocket engines are started..is there a summary some where on how they start up rocket engines..I didn't realize a charge is needed to start the engine..
jb

There's plenty of information on this.  Use Sutton Rocket Propulsion Elements or Huzel and Huang Modern Engineering for Design of Liquid-propellant Rocket Engines.  An older version can be found here (http://www.spl.ch/publication/sp125.html) for free.
Title: Re: Basic Rocket Science Q & A
Post by: kkattula on 03/02/2009 02:46 pm
Occasionally I see people mention air-started SSME's. I get the impression that this is either hard or dangerous. Somehow the J2X is supposed to be better in this regard. Is there a simple explanation of what the problem is and how J2X is better?

it is only hard for the SSME since it wasn't designed for it.   It is head started which means the start box is small, the initial conditions are very critical.  Also it depends on ground GSE for start and purges

There's non-ground GSE?   ;)
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/02/2009 02:59 pm
it is only hard for the SSME since it wasn't designed for it.   It is head started which means the start box is small, the initial conditions are very critical.  Also it depends on ground GSE for start and purges

Thanks!
Title: Re: Basic Rocket Science Q & A
Post by: Spacenick on 03/02/2009 05:17 pm
How much energy/m^2 does a tile on the Space Shuttle have to dissipate?
Wouldn't it be possible to construct a metallic heat shield cooled by some material say wax being molten during reentry? I mean a phase transition should take a lot of energy off the metallic heat shield.
Title: Re: Basic Rocket Science Q & A
Post by: LegendCJS on 03/02/2009 06:50 pm
How much energy/m^2 does a tile on the Space Shuttle have to dissipate?
Wouldn't it be possible to construct a metallic heat shield cooled by some material say wax being molten during reentry? I mean a phase transition should take a lot of energy off the metallic heat shield.

You can do back of the envelope calculations for things like this by assuming that all the kenetic energy in an orbiting vehicle is disipated as heat.  Thus:

1/2 *mass*orbital_velocity^2 = dissipated_energy_per_m^2 * area_of_vehicle

Where mass is the vehicles mass,
orbital velocity is 175000 mph or close to that,
dissipated_energy_per_m^2 is the quantity you want to know about
area_of_vehicle is the cross-sectional area profile of the vehicle during re-entry.

Now you need the mass of wax able to undergo a phase change and take up the same amount of energy- if this is any significant amount of the vehicle mass then you have to account for the added mass in the first equation.

However, I'm not aware of the phase of wax in zero pressure at orbital ambient temperatures, it might already want to liquefy on you every time the vehicle is in sunlight...
Title: Re: Basic Rocket Science Q & A
Post by: renclod on 03/02/2009 07:01 pm
How much energy/m^2 does a tile on the Space Shuttle have to dissipate?

7 giga joules per m^2, give or take

Title: Re: Basic Rocket Science Q & A
Post by: kneecaps on 03/02/2009 07:16 pm
Wouldn't it be possible to construct a metallic heat shield cooled by some material say wax being molten during reentry? I mean a phase transition should take a lot of energy off the metallic heat shield.

You are really just talking about an ablative heat shield using a novel material :) The key point with the Shuttles TPS is that it is not an ablator and you can dust it off and fly it again (on paper!). It is simply unfortunate that such a great heatshield is so susceptible to phsyical damage.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/02/2009 07:23 pm
I'm trying to understand how a rocket nozzle works and I have some questions. I only have a very limited understanding of thermodynamics and fluid flow. I worked on the software design of a multiphase flow simulator with experts on multiphase flow and thermodynamics for a couple of years and picked up a tiny little bit.

So here are my questions and my own possibly highly incorrect guesses what the answers are:

- How essential is choked flow through the throat? Would a de Laval nozzle still have an effect if the flow were not choked? My guess is yes, but it would be less effective.

- How essential is it that the fluid is compressible? Would it work with a liquid? My guess is absolutely essential and no, it wouldn't work.

- How essential is it that the gas is hot? My guess is very important, though I wonder what the reference point is.

- What if you left off the nozzle? My guess is you would still have thrust just less and your gas would be very hot.

- What if you left off both the throat and the nozzle? My guess is you'd get a nice flame but very little thrust.

- What if you kept the nozzle but left off the throat? My guess is that this is the same as my first question, you wouldn't get choked flow.
Title: Re: Basic Rocket Science Q & A
Post by: Nascent Ascent on 03/02/2009 07:29 pm
Quote
How essential is choked flow through the throat? Would a de Laval nozzle still have an effect if the flow were not choked? My guess is yes, but it would be less effective.

It wouldn't BE a deLaval nozzle if it weren't "choked".  The convergent-divergent aspect of the deLaval nozzle is the basis of the design.

It is also essential that the working medium is compressible (i.e. a gas).
Title: Re: Basic Rocket Science Q & A
Post by: kneecaps on 03/02/2009 07:30 pm
http://www.grc.nasa.gov/WWW/K-12/airplane/ienzl.html

Have a play with this applet, put it in 'nozzle' mode, it starts as a turbine otherwise, see if you can answer some of your questions by experimentation, always more educational :)
Title: Re: Basic Rocket Science Q & A
Post by: LegendCJS on 03/02/2009 07:31 pm
Wouldn't it be possible to construct a metallic heat shield cooled by some material say wax being molten during reentry? I mean a phase transition should take a lot of energy off the metallic heat shield.

You are really just talking about an ablative heat shield using a novel material :) The key point with the Shuttles TPS is that it is not an ablator and you can dust it off and fly it again (on paper!). It is simply unfortunate that such a great heatshield is so susceptible to phsyical damage.

I'm pretty sure he was talking about a system where the metal was on the "outside" and the interior of the metal shell was packed with wax.  The wax would melt during re-entry, but a couple of hours later on the ground the wax would have re-solidified, and thus his idea was for a re-usable heat shield. 

Not to say it has any chance of being practical:  at a latent heat of fusion of 200 J per gram, you would need 35,000kg of parafin to melt in order to absorb the quoted 7 giga-joules per meter squared for every square meter of heat shield.  Plus wax expands in volume quite a bit when it melts.
Title: Re: Basic Rocket Science Q & A
Post by: kneecaps on 03/02/2009 07:37 pm
Wouldn't it be possible to construct a metallic heat shield cooled by some material say wax being molten during reentry? I mean a phase transition should take a lot of energy off the metallic heat shield.

You are really just talking about an ablative heat shield using a novel material :) The key point with the Shuttles TPS is that it is not an ablator and you can dust it off and fly it again (on paper!). It is simply unfortunate that such a great heatshield is so susceptible to phsyical damage.

I'm pretty sure he was talking about a system where the metal was on the "outside" and the interior of the metal shell was packed with wax.  The wax would melt during re-entry, but a couple of hours later on the ground the wax would have re-solidified, and thus his idea was for a re-usable heat shield. 

Not to say it has any chance of being practical:  at a latent heat of fusion of 200 J per gram, you would need 35,000kg of parafin to melt in order to absorb the quoted 7 giga-joules per meter squared for every square meter of heat shield.  Plus wax expands in volume quite a bit when it melts.

Ah...you saw it the other way around to me :) I see what was intended now :)
Title: Re: Basic Rocket Science Q & A
Post by: Spacenick on 03/02/2009 07:55 pm
@LegendCJS: Yeah, that was kind of the idea, seems like most materials will be molten too easily.
 Maybe someone could expand how the real concepts for metallic actively cooled heat shields looked like?
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/02/2009 07:58 pm
Thanks!

It wouldn't BE a deLaval nozzle if it weren't "choked".

So does that mean it stops being a de Laval nozzle when it's turned off? ;-)
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 03/03/2009 05:45 am
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- How essential is choked flow through the throat?

Choked means Mach 1.  If it doesn't get to Mach 1, it won't be supersonic in the divergent section.  Low thrust, very low Isp.

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How essential is it that the fluid is compressible? Would it work with a liquid?

Force is rate of change of momentum, right?  M-dot v in this case.  So for high thrust you need to maximize each one.  A liquid is going to be denser and more viscous, requiring more energy to get both a higher m-dot and higher v.  Low thrust, very low Isp.

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- How essential is it that the gas is hot? My guess is very important, though I wonder what the reference point is.

It needs to be energetic - so it can become speedy.  When it's nearly static near the injector face or unburned solid propellant, it's REALLY hot.  Adiabatically, in the nozzle, it's also REALLY hot - if you say stick your finger in the flow.  If, instead, you put your finger on the edge of the flow in a big enough nozzle, it might FEEL cold because so much of the static enthalpy has become velocity (this is not wind chill factor).

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What if you left off the nozzle? My guess is you would still have thrust just less and your gas would be very hot.

You would have an uncontrolled expansion from the throat, full of shocks too.  Lots of entropy and other losses.  Temperature would vary depending on where in the shock-expansion field you were.

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What if you left off both the throat and the nozzle?

Remember the scene in Hot Shots where the aircraft tech is roasting weinies or marshmallows at the nozzle of the fighter?

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What if you kept the nozzle but left off the throat?

No, it would still choke at the throat, unless the mass flow were so low that it didn't need to accelerate to sonic to get out of the combustion chamber.  Flow in the nozzle would be indeterminate (subsonic or supersonic) without more information like chamber pressure and ambient pressure.

Huzel & Huang; Sutton; J.D. Anderson are your friends.
Title: Re: Basic Rocket Science Q & A
Post by: renclod on 03/03/2009 07:20 am
Maybe someone could expand how the real concepts for metallic actively cooled heat shields looked like?

I would suggest you read this piece online at spacedaily under "Rocket Science" :
" Cult spacecraft Part One: The Little Spaceplane That Couldn't "
Dec.04, 2008
by Jeffrey F. Bell

Title: Re: Basic Rocket Science Q & A
Post by: William Barton on 03/03/2009 08:06 am
Am I correcting in thinking the big problem with active TPS is heat rejection? In one of my old stories, I had a liquid-metal (sodium) cooled TPS that used the collected heat to drive a laser to dump the energy overboard. It being a story, I kind of glossed over the conversion efficiency issues, "in the future" someone would figure it out.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/03/2009 08:27 am
Hi Antares,

Thanks for the lengthy reply. And thanks 'kneecaps' for the link to that applet. Based on your replies I have some gedanken experiments I'd like to run by you if that's ok.

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- How essential is choked flow through the throat?
Choked means Mach 1.  If it doesn't get to Mach 1, it won't be supersonic in the divergent section.  Low thrust, very low Isp.

And compared to the situation without the nozzle? Would it still convert some (possibly tiny) amount of heat to macroscopic kinetic energy?

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How essential is it that the fluid is compressible? Would it work with a liquid?
Force is rate of change of momentum, right?  M-dot v in this case.  So for high thrust you need to maximize each one.  A liquid is going to be denser and more viscous, requiring more energy to get both a higher m-dot and higher v.  Low thrust, very low Isp.

I'm imagining attaching a garden hose to a de Laval nozzle and turning on the tap. I'd expect water to come out at a higher velocity than without the nozzle. I'd also be worried about my hose or nozzle bursting. I'd expect the same thrust as without the nozzle, just with a meaner jet. I wouldn't expect a hot water tap to produce better results than a cold water tap.

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- How essential is it that the gas is hot? My guess is very important, though I wonder what the reference point is.
It needs to be energetic - so it can become speedy.  When it's nearly static near the injector face or unburned solid propellant, it's REALLY hot.  Adiabatically, in the nozzle, it's also REALLY hot - if you say stick your finger in the flow.  If, instead, you put your finger on the edge of the flow in a big enough nozzle, it might FEEL cold because so much of the static enthalpy has become velocity (this is not wind chill factor).

So a leaf blower would produce the same 'thrust' with a nozzle as without it? And a hair dryer slightly more with a nozzle than without it?

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Quote
What if you left off both the throat and the nozzle?
Remember the scene in Hot Shots where the aircraft tech is roasting weinies or marshmallows at the nozzle of the fighter?

Lol!

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Quote
What if you kept the nozzle but left off the throat?
No, it would still choke at the throat, unless the mass flow were so low that it didn't need to accelerate to sonic to get out of the combustion chamber.  Flow in the nozzle would be indeterminate (subsonic or supersonic) without more information like chamber pressure and ambient pressure.

Let me see if I understand.

Suppose we have a fixed pressure at the inlet and a (lower) fixed pressure at the outlet. Is it the case that for every positive pressure differential the flow through a de Laval nozzle would choke in steady state?

And the only way it could not choke would be if instead you specified mass flow at the inlet and that mass flow were sufficiently low to prevent choking at the throat? And even then it would depend on the other conditions.

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Huzel & Huang; Sutton; J.D. Anderson are your friends.

And having some experts around doesn't hurt either. Thanks again guys!
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 03/03/2009 05:48 pm
Quote
Quote
Quote
How essential is it that the fluid is compressible? Would it work with a liquid?
Force is rate of change of momentum, right?  M-dot v in this case.  So for high thrust you need to maximize each one.  A liquid is going to be denser and more viscous, requiring more energy to get both a higher m-dot and higher v.  Low thrust, very low Isp.
I'm imagining attaching a garden hose to a de Laval nozzle and turning on the tap. I'd expect water to come out at a higher velocity than without the nozzle. I'd also be worried about my hose or nozzle bursting. I'd expect the same thrust as without the nozzle, just with a meaner jet. I wouldn't expect a hot water tap to produce better results than a cold water tap.

No, the water would come out slower because it did not reach Mach 1 at the throat.  In all of these cases, m-dot is constant and m-dot equals density times area times velocity.  So if your area goes up in an incompressible flow or a compressible, subsonic flow, your velocity goes down.

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- How essential is it that the gas is hot? My guess is very important, though I wonder what the reference point is.
It needs to be energetic - so it can become speedy.  When it's nearly static near the injector face or unburned solid propellant, it's REALLY hot.  Adiabatically, in the nozzle, it's also REALLY hot - if you say stick your finger in the flow.  If, instead, you put your finger on the edge of the flow in a big enough nozzle, it might FEEL cold because so much of the static enthalpy has become velocity (this is not wind chill factor).
So a leaf blower would produce the same 'thrust' with a nozzle as without it? And a hair dryer slightly more with a nozzle than without it?

Both of those would be less thrust because the flow is subsonic and the nozzle would decelerate the flow.

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Quote
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What if you kept the nozzle but left off the throat?
No, it would still choke at the throat, unless the mass flow were so low that it didn't need to accelerate to sonic to get out of the combustion chamber.  Flow in the nozzle would be indeterminate (subsonic or supersonic) without more information like chamber pressure and ambient pressure.
Let me see if I understand.

1) Suppose we have a fixed pressure at the inlet and a (lower) fixed pressure at the outlet. Is it the case that for every positive pressure differential the flow through a de Laval nozzle would choke in steady state?

2) And the only way it could not choke would be if instead you specified mass flow at the inlet and that mass flow were sufficiently low to prevent choking at the throat? And even then it would depend on the other conditions.

1) No.  There's a diagram in Anderson's Fundamentals of Aerodynamics that you really need to look at showing the varieties of nozzle flow depending on pressure ratio.  Area ratio is the other factor obviously.
2) "It would depend on the other conditions." Yes.  If the pressure or area ratio were low, the flow wouldn't accelerate sufficiently to be sonic at the throat.

Read Anderson.  Ignore the equations if you're not good at math.  The explanations and pictures are really good.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/03/2009 05:54 pm
Read Anderson. 

I will. Thanks again.

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Ignore the equations if you're not good at math.  The explanations and pictures are really good.

I studied maths, I should be OK :-)
Title: Re: Basic Rocket Science Q & A
Post by: Spacenick on 03/03/2009 06:40 pm
I thought about this equation again:
"1/2 *mass*orbital_velocity^2 = dissipated_energy_per_m^2 * area_of_vehicle"

and it seems to me that it can't be correct. It' uses the correct equation for the energy dissipated over the whole reentry, but because this energy is is not only converted into space craft heating, but also in kinetic energy of air molecules (as in pushing the air out of the way), heating and compressing of air, creation of plasma from air and electric discharges.
I'm pretty sure that only a tiny percentage of the orbital energy is actually converted into heating up of the spacecraft itself, especially since the blunt shape of reentry vehicles keeps the shock wave off the vehicle hull and therefor reduces energy transfer extremely.

I foudn the wikipedia article for atmospheric reentry to be very imformative
http://forum.nasaspaceflight.com/index.php?topic=13543.75
Title: Re: Basic Rocket Science Q & A
Post by: vt_hokie on 03/03/2009 11:45 pm
I found that what you find at 75ºE is the major Indic Ocean perturbation, but it's a negative perturbation, meaning (and I checked this several times through different sources) it exerts less gravitational pull.

Yes, this is one of the gravity wells, the other being near 105 deg W longitude.  A geostationary satellite that fails on-orbit will oscillate around the nearest gravity well with an amplitude roughly equal to the original distance from that well.  So, the further from a well a failed satellite is, the more of a problem it becomes for other satellites when it fails!

Of course, such sats will be above GEO when drifting west and below when drifting east.  (And meanwhile they'll have higher inclinations, up to ~15 deg depending on where they are in the inc cycle.)  But they have to be watched for potential intercepts with operational spacecraft nonetheless!

Title: Re: Basic Rocket Science Q & A
Post by: vt_hokie on 03/04/2009 04:22 pm
Digging up old notes on the subject:

The mass of the Earth is not uniformly distributed, and the resultant non-circular shape of the Earth’s equator is described by the term ‘triaxial Earth’ or triaxiality.  The planet’s mass distribution can be roughly approximated by a spherically symmetric Earth with additional masses at 15° West and 165° East longitude.   This is referred to as a ‘triaxial Earth’ because the Earth would have one polar and two equatorial axes.  The additional masses correspond to mid-ocean regions, where the oceanic tectonic plates are heavier than continental plates.

Question for the real experts: So, if you were to place a satellite in geostationary orbit directly over one of the gravity wells, would its natural eccentricity be determined primarily by solar pressure and the satellite's area to mass ratio?  I believe lunar, solar, and other gravitational perturbations are negligible in comparison when it comes to drift.
Title: Re: Basic Rocket Science Q & A
Post by: DMeader on 03/04/2009 05:13 pm
Concerning the discussion up the thread a bit about nozzle designs... do any of the factors mentioned apply to cold-gas thrusters like some attitude control jets? How about thrusters where a monopropellant like hydrazine or hydrogen peroxide is decomposed across a catalyst bed? Do those concerns not apply so much to units that tend to be that small?
Title: Re: Basic Rocket Science Q & A
Post by: Herb Schaltegger on 03/05/2009 02:27 am
Concerning the discussion up the thread a bit about nozzle designs... do any of the factors mentioned apply to cold-gas thrusters like some attitude control jets? How about thrusters where a monopropellant like hydrazine or hydrogen peroxide is decomposed across a catalyst bed? Do those concerns not apply so much to units that tend to be that small?

They apply to any gas thruster if the mass flow is more than a mere trickle.  Regardless of whether the gas is produced via combustion or not, it will expand into the area of low pressure (e.g., the vacuum at one end of the nozzle).  The flow will still choke at the narrowest point.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/05/2009 09:53 am
They apply to any gas thruster if the mass flow is more than a mere trickle.  Regardless of whether the gas is produced via combustion or not, it will expand into the area of low pressure (e.g., the vacuum at one end of the nozzle).  The flow will still choke at the narrowest point.

Is the cold gas in a cold gas thruster cold enough that the nozzle cannot convert much thermal energy to (macroscopic) kinetic energy, or does the gas get appreciably colder and faster as it expands through the nozzle?
Title: Re: Basic Rocket Science Q & A
Post by: DMeader on 03/05/2009 11:13 am
They apply to any gas thruster if the mass flow is more than a mere trickle.  Regardless of whether the gas is produced via combustion or not, it will expand into the area of low pressure (e.g., the vacuum at one end of the nozzle).  The flow will still choke at the narrowest point.

Is the cold gas in a cold gas thruster cold enough that the nozzle cannot convert much thermal energy to (macroscopic) kinetic energy, or does the gas get appreciably colder and faster as it expands through the nozzle?

By "cold gas" I meant just compressed gas like nitrogen. For example the VDU roll thrusters on MIR or the thrusters on the MMU. Not necessarily cryogenic gas.
Title: Re: Basic Rocket Science Q & A
Post by: Herb Schaltegger on 03/05/2009 12:47 pm
Basic physics - when gas expands, it cools.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/05/2009 01:22 pm
Basic physics - when gas expands, it cools.

I know, I was wondering how much and how much use the Laval nozzle is in that situation. I am not as dumb as I look. I studied mathematics and computer science. In my first year I also studied physics and passed all exams for that.
Title: Re: Basic Rocket Science Q & A
Post by: nomadd22 on 03/05/2009 01:49 pm
 I'm exactly as dumb as I look. I'm still trying to figure out why a Geo sat that's a half a degree off the equator (like MSAT1) needs less fuel for station keeping. It's Sband, made for low gain ground antennas so it doesn't have to stay perfectly still.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/05/2009 01:52 pm
Alas, while I'm not quite as dumb as I look, I am sadly not smart enough to answer that question...
Title: Re: Basic Rocket Science Q & A
Post by: vt_hokie on 03/05/2009 08:09 pm
I'm exactly as dumb as I look. I'm still trying to figure out why a Geo sat that's a half a degree off the equator (like MSAT1) needs less fuel for station keeping. It's Sband, made for low gain ground antennas so it doesn't have to stay perfectly still.

That inclination is 10 times what GEO sats are typically held to.  I don't see how that would save much propellant though.  Inclination grows at close to 1 degree per year (I think it might be 0.86 deg/yr, though the rate will decrease as inclination gets greater, and the lunar perturbation varies cyclically with the moon's inclination.  But I'm too lazy to look it up right now).  So, if you started near zero inclination, you could let it go for 6 months or so with no inclination burns.  But then you'd have to prevent the satellite from exceeding 0.5 deg.  So, while with a larger "box" maybe you could do longer burns and reduce the frequency of maneuvers (though that also depends on propulsion limitations), you still need to perform inclination maneuvers.

Edit: Okay, I looked it up...

The moon has a larger effect on inclination than the sun, with the rate of inclination change ranging from 0.48˚ per year to 0.68˚ per year over an 18.6 year period.  This period corresponds to the variation in the moon’s orbital inclination (from 18˚ to 28˚) over the same time period.  The rate of inclination change due to solar effects is approximately 0.27˚ per year.  The solar perturbation is strongest at the solstice, when the sun is furthest from the Earth’s equatorial plane.  In addition to affecting inclination, both the lunar and solar perturbations affect the right ascension of the ascending node (Ω), the angle between the longitude where the satellite crosses the equator in a northerly direction and the vernal equinox direction – i.e. the longitude where the sun appears on the first day of spring.
Title: Re: Basic Rocket Science Q & A
Post by: nomadd22 on 03/05/2009 08:31 pm
MSAT 1 isn't drifting. MSV took it off the equatorial plane on purpose because they were running out of fuel and they say it takes less to stay in place if you're out of place......uh....sort of. Now that I bothered to look it up, it's actually 2 1/2 degrees off.
 The result is that if you track the sat it appears to make a daily figure eight around the point you'd expect it to be.
 It's also one of the Boeing busses that has solar array woes, is several years past it's planned life and has a failed transponder.
 But the little sucker kept our fleet of ships going after Katrina when public utilities were down for months and Iridium had 200 users trying to make a call for every available channel.
 It's replacement, MSV-1 is riding a Proton Briz-M later this year. It'll be a big one at 5600kg.
Title: Re: Basic Rocket Science Q & A
Post by: vt_hokie on 03/05/2009 08:45 pm
MSAT 1 isn't drifting. MSV took it off the equatorial plane on purpose because they were running out of fuel and they say it takes less to stay in place if you're out of place......uh....sort of. Now that I bothered to look it up, it's actually 2 1/2 degrees off.
 

Okay, so it's nearing end of life and they're just letting inclination go, it would seem.  That's pretty typical.  The satellite might still generate some revenue for a while, or sometimes the "inclined" sats are simply used as placeholders to hold an orbital slot until the operator can get a replacement launched.
Title: Re: Basic Rocket Science Q & A
Post by: nomadd22 on 03/06/2009 12:23 pm
MSAT 1 isn't drifting. MSV took it off the equatorial plane on purpose because they were running out of fuel and they say it takes less to stay in place if you're out of place......uh....sort of. Now that I bothered to look it up, it's actually 2 1/2 degrees off.
 

Okay, so it's nearing end of life and they're just letting inclination go, it would seem.  That's pretty typical.  The satellite might still generate some revenue for a while, or sometimes the "inclined" sats are simply used as placeholders to hold an orbital slot until the operator can get a replacement launched.
OK. It might be simpler than I thought. Maybe the fuel savings comes from maintaining longitude but letting latitude get sloppy. I'd thought they put it where it was on purpose. Wouldn't be the first time I made things more complicated than they were.
 But, it's more than just a placeholder. There are many, many customers using that thing, from sailboats to large ships. (Like mine) We just bought a mess of replacement radios on MSV's promise that the new sats would be compatible. I can even get internet access with them at a blistering 4800bps.
Title: Re: Basic Rocket Science Q & A
Post by: vt_hokie on 03/07/2009 04:34 pm
OK. It might be simpler than I thought. Maybe the fuel savings comes from maintaining longitude but letting latitude get sloppy.

Yep, that's exactly it!

Quote
I'd thought they put it where it was on purpose.

If only it were possible to center it at a latitude other than zero.  Stupid physics! ;)

Quote
But, it's more than just a placeholder. There are many, many customers using that thing, from sailboats to large ships. (Like mine) We just bought a mess of replacement radios on MSV's promise that the new sats would be compatible. I can even get internet access with them at a blistering 4800bps.

That's cool!  I wonder how much inclination they can tolerate while still providing continuous coverage.
Title: Re: Basic Rocket Science Q & A
Post by: nomadd22 on 03/07/2009 08:26 pm


That's cool!  I wonder how much inclination they can tolerate while still providing continuous coverage.
Most MSAT remotes are autotracking marine or mobile antennas or low gain antennas that you just point in the general direction, so the sat can afford to move around some. It's mostly Ku band birds who might have neighbors on the same frequency and remote, fixed high gain antennas that have to be really precise. We're just starting to get marine Ku band antennas now. Suckers cost a fortune because they have to maintain half a degree precision in rough seas.
Title: Re: Basic Rocket Science Q & A
Post by: vt_hokie on 03/09/2009 04:22 pm
Most MSAT remotes are autotracking marine or mobile antennas or low gain antennas that you just point in the general direction, so the sat can afford to move around some. It's mostly Ku band birds who might have neighbors on the same frequency and remote, fixed high gain antennas that have to be really precise. We're just starting to get marine Ku band antennas now. Suckers cost a fortune because they have to maintain half a degree precision in rough seas.

Thanks for the explanation, makes sense.  I can see why that would be expensive!

Thought I'd mention that when you do inclination maneuvers, usually the goal isn't to reduce magnitude to zero, but rather to control growth.  As I mentioned, GEO sats are typically maintained to within 0.05 deg of the equator.  To fully reset the inc cycle, you actually want to drive the inclination out to ~0.05 deg.  This is how I explained it a while back...hopefully it makes sense.  (And hopefully I got it right!  I basically just re-worded what was in orbital dynamics texts, but corrections are always welcome. :) )

Ideally, the inclination cycle is controlled such that the inclination vector is set to begin near the inclination deadband (typically 0.05˚) with the right ascension, Ω, near 270˚.  The magnitude of inclination will decrease toward zero and then begin increasing as Ω moves toward 90˚.  North/south stationkeeping maneuvers are required to reset the cycle when the inclination magnitude has increased back to the deadband limit and Ω is near 90˚.  The maneuvers occur at the intersection of the existing orbital plane and the target plane, within 30˚ of the ascending or descending node. 
 
So, in that sense, you do want to purposely take the satellite out of a truly equatorial orbit by a small amount.  Also, you can save some propellant by only correcting for secular perturbations, but that would be a whole new topic that I'm guessing is more than you want to know! :D

It should be easy, but throw in some component malfunctions, unpredictable east/west coupling due to thruster plume impingement and attitude control firings, etc. and it ends up being a real headache! ;) Also, I'm not a huge fan of arcjets!  I don't care how efficient they are, they're a pain from an operational standpoint!  :)
Title: Re: Basic Rocket Science Q & A
Post by: vt_hokie on 03/09/2009 10:56 pm

Intuition (and simple physics) tell the satellites should be accelerated by the positive anomalies, hence causing an oscillatory motion between two points centered in the gravity well: the strongest gravity source. Are people somehow disoriented by the graphical representations of the geoid, thinking the wells are the "lowest" points in the figure? Or am I missing something entirely and being ridiculously stubborn?

Revisiting this question, I think I see what you're saying here.  You're wondering why the stable longitudes are 90 deg off of the "bulges" rather than directly over them, right?  You're making me think for the first time in months...my brain hurts!  ;)  I'm not sure how to best describe it other than to say that at the low point, the effects of the two additional mass regions cancel each other out and no longer change the shape of the orbit.  (But then that's true at the unstable equilibrium points as well.  The key is how does the shape of the orbit change at all other longitudes.)  Hopefully someone here can explain it better, but I managed to find a figure which might illustrate it better than I'm saying it.  Check out page 260:

Stable longitudes (http://books.google.com/books?id=HMM1YRGVsLIC&pg=PA259&lpg=PA259&dq=geostationary+stable+longitude&source=bl&ots=yymfKx-gyq&sig=_XEDL75A1ZNSaOtHVgWZTpTh1F4&hl=en&ei=-KO1Sa6dIo-yMcrTjbgH&sa=X&oi=book_result&resnum=3&ct=result#PPA260,M1)
 
And here's a better explanation:

Pages 137, 138 (http://books.google.com/books?id=_KCNcxYPUZgC&pg=PA137&lpg=PA137&dq=geostationary+triaxial+oscillation&source=bl&ots=gvanu27S7L&sig=CWc9VtrM10UG3OcHxy1MX_03HEw&hl=en&ei=5se1SdP5MYL0NIHLgNwK&sa=X&oi=book_result&resnum=2&ct=result#PPA138,M1)

Figure 5.18 makes perfect sense...I feel dumb now for getting confused over something so obvious!

So, of course you're correct about satellites being accelerated by the positive anomalies.  But it's the old "slow down to speed up, speed up to slow down" paradox of orbital mechanics!  :)   
Title: Re: Basic Rocket Science Q & A
Post by: eeergo on 03/10/2009 07:20 pm
That's a great explanation, very clear! Stupid orbital mechanics... :)

Thank you very much for the outstanding reference, I hadn't found this resource whey trying to understand the effect.
Title: Re: Basic Rocket Science Q & A
Post by: Spacenick on 03/10/2009 07:40 pm
I have a question about heat transfer and cooling/heating of sattelites.
In many sources, especially in main stream media, it sounds like an unbelievably big problem, that in space, the sun side is so much hotter than the shadow. And they talk about huge figures hundreds of degrees celsius of temperature difference.
On the other hand, my common sense tells me, that because of the low pressure there is probably very little convection and therefor there is only radiation for heat transfer.
Therefor, i'd think that though the air moluecules in LEO are technically really hot, they transfer very little energy to a spacecraft.
So how is the cooling/heating acomplished? I'd guess that making the spacecraft reflective on the outside would block most of the energy coming from the sun, an because of little convection the spacecraft should cool very little when provided with basic insulation against heat transfer to the outer hull (which would remove the energy by way of infrared radiation).
Then I'd guess the electronics or other systems (e.g. a human in a spacecraft) would provide enough heat to keep the sacecraft from freezing.
So what, would be left would be to provide a way of radiating exactly as much energy out of the spacecraft as needed o keep constant temperature, so how would I do that?
Title: Re: Basic Rocket Science Q & A
Post by: cgrunska on 03/10/2009 08:54 pm
I was reading a blurb about the new Merlin Vaccum engine and it showed it on earth with a big flame plume coming out the backend, and an artist rendering of the engine in space with "flares" around the nozzle.

Basically my question is thus, how does an engine, turned on, look in space? There's no fire coming out the backside. Is there anything besides maybe a blue ring around the inner nozzle? Is there even that? Any 'movement field' that can be seen?

Pretty basic question. Thanks for the answer!
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 03/10/2009 09:39 pm
It depends on the propellants. 
Title: Re: Basic Rocket Science Q & A
Post by: cgrunska on 03/10/2009 10:27 pm
a good point

chemical combustion engine vs electrical engine then?
I assume a nuclear propelled spacecraft is an electrical engine and not an explosive one.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 03/10/2009 10:36 pm
I was referring to differences in chemical propellants.  LO2/LH2, RP1/LOX, N2O4/UDMH
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/10/2009 11:03 pm
I have a question about the corrosiveness of N2O4. I have read that adding a few percent of NO reduces the corrosiveness and that the resulting mixture is called MON, which stands for mixed oxides of nitrogen. Various mixture ratios are in use.

So how much of a difference does this make and how much of a problem is this corrosiveness? I'm particularly interested in the effect on the near-term feasibility of orbital hypergolic propellant depots. How long can you realistically store MMH/MON in space before corrosion renders your depot inoperable?
Title: Re: Basic Rocket Science Q & A
Post by: nomadd22 on 03/10/2009 11:06 pm
 Not sure why you think there wouldn't be fire coming out the backside in space. Near the engine it would look about the same as in the air. Yellow/white flame for most kerosene type and flame so blue you can hardly see it for H2/O2.
 I can't say i know what the water trail for a hydrogen engine looks like in vacuum.
 If you go to Spacex.com you can watch the film of the 4th Falcon 1 launch. They had a camera on the second stage picking up it's flight through the vacuum.
Title: Re: Basic Rocket Science Q & A
Post by: nomadd22 on 03/10/2009 11:22 pm
I have a question about heat transfer and cooling/heating of sattelites.
In many sources, especially in main stream media, it sounds like an unbelievably big problem, that in space, the sun side is so much hotter than the shadow. And they talk about huge figures hundreds of degrees celsius of temperature difference.
On the other hand, my common sense tells me, that because of the low pressure there is probably very little convection and therefor there is only radiation for heat transfer.
Therefor, i'd think that though the air moluecules in LEO are technically really hot, they transfer very little energy to a spacecraft.
So how is the cooling/heating acomplished? I'd guess that making the spacecraft reflective on the outside would block most of the energy coming from the sun, an because of little convection the spacecraft should cool very little when provided with basic insulation against heat transfer to the outer hull (which would remove the energy by way of infrared radiation).
Then I'd guess the electronics or other systems (e.g. a human in a spacecraft) would provide enough heat to keep the sacecraft from freezing.
So what, would be left would be to provide a way of radiating exactly as much energy out of the spacecraft as needed o keep constant temperature, so how would I do that?
Temperature in space doesn't mean much because there's not much there to be hot or cold, so you're right, and convection isn't a factor.
 You just have to worry about direct solar radiation and heat produced by spacecraft systems.
 Satellites in equatorial or low inclination orbits stay evenly baked since they have to rotate once an orbit to keep the same face toward the earth. They can keep the electronics just the right temp by adjusting the design of the satellite to control how much heat it picks up and how much it radiates back out.
 The small lower powered ones radiate enough heat on the dark sides to make up for what the sunlit sides pick up.
 Big ones, like the space station can have big honkin heat pump systems that usually use ammonia for refrigerant and big radiators kept edge on to the sun.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 03/10/2009 11:28 pm
The flame becomes very diffuse and thin in the vacuum.  Look at the rocketcams on a Delta first stage.  Youtube has many
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 03/11/2009 12:22 am
I have a question about heat transfer and cooling/heating of sattelites.
In many sources, especially in main stream media, it sounds like an unbelievably big problem, that in space, the sun side is so much hotter than the shadow. And they talk about huge figures hundreds of degrees celsius of temperature difference.
On the other hand, my common sense tells me, that because of the low pressure there is probably very little convection and therefor there is only radiation for heat transfer.
Therefor, i'd think that though the air moluecules in LEO are technically really hot, they transfer very little energy to a spacecraft.
So how is the cooling/heating acomplished? I'd guess that making the spacecraft reflective on the outside would block most of the energy coming from the sun, an because of little convection the spacecraft should cool very little when provided with basic insulation against heat transfer to the outer hull (which would remove the energy by way of infrared radiation).
Then I'd guess the electronics or other systems (e.g. a human in a spacecraft) would provide enough heat to keep the sacecraft from freezing.
So what, would be left would be to provide a way of radiating exactly as much energy out of the spacecraft as needed o keep constant temperature, so how would I do that?

There is no air pressure and few to none air molecules where most spacecraft operate, hence no convection.  Sunward side of the spacecraft experiences high radiative heat input.  The sides facing  away would radiate heat into the black body of deep space.   MLI (multi layered insulation) is used to cover the spacecraft and comes in many types, with many different outer layers (white beta cloth, aluminized mylar, kapton, etc).  MLI both retains and reflect heat.  Since there is no convention, electronics would overheat and fluids would freeze. The electronics are usually mounted on radiators or connected to them by heatpipes.  Heaters are employed to keep parts like propellant tanks warm. 
Title: Re: Basic Rocket Science Q & A
Post by: charlieb on 03/11/2009 12:42 am
I have a question about heat transfer and cooling/heating of sattelites.
In many sources, especially in main stream media, it sounds like an unbelievably big problem, that in space, the sun side is so much hotter than the shadow. And they talk about huge figures hundreds of degrees celsius of temperature difference.
On the other hand, my common sense tells me, that because of the low pressure there is probably very little convection and therefor there is only radiation for heat transfer.
Therefor, i'd think that though the air moluecules in LEO are technically really hot, they transfer very little energy to a spacecraft.
So how is the cooling/heating acomplished? I'd guess that making the spacecraft reflective on the outside would block most of the energy coming from the sun, an because of little convection the spacecraft should cool very little when provided with basic insulation against heat transfer to the outer hull (which would remove the energy by way of infrared radiation).
Then I'd guess the electronics or other systems (e.g. a human in a spacecraft) would provide enough heat to keep the sacecraft from freezing.
So what, would be left would be to provide a way of radiating exactly as much energy out of the spacecraft as needed o keep constant temperature, so how would I do that?

The following concerns GEO Comm type-satellites, but generically covers any spacecraft to a degree:

Satellite thermal configurations are designed to maintain all spacecraft equipment within allowable temperature limits during transfer orbit and throughout their life on orbit. Thermal control is achieved using both passive and active elements. Passive elements include thermal blankets, coatings, shields, radiators, and heat pipes. Active elements consist of ground and autonomously controlled heaters. Thermistors provide temperature data that are used for thermal control and health monitoring.

Blankets, coatings, and solar reflectors are used to control radiative heat transfer to and from the spacecraft, as well as among units within the spacecraft bus. Multi-Layer Insulation (MLI) blankets are used externally on the east, west, anti-Earth, and Earth spacecraft surfaces to reduce diurnal temperature variations. Optical Solar Reflectors (OSRs), mounted on the north and south communication panels, subsystem module  radiator panels, battery panel radiators, and the Earth sensor radiator panel, minimize solar absorption during periods when there is significant solar flux on these surfaces. The OSRs also increase the spacecraft thermal emissivity, thereby cooling the radiator panels, which contain the majority of high-heat dissipation units.

Internal surface coatings help control radiative heat transfer within the satellite and aid in maintaining elements within limits.

The solar array temperature is controlled entirely by passive means. Absorption of solar energy on the front (cell) side of the array is emitted by the highly emissive, black graphite back surface of the array.

The antenna reflectors are subjected to a diurnal solar flux. MLI blankets covering the reflector backup structures help reduce temperature variations and thermally isolate the antennas from the spacecraft mainbody. MLI blankets are also used to maintain a constant temperature of the antenna tower structure(s).

Heat pipes are embedded in the spacecraft radiator panels (north and south communication panels, subsystem module panels, and battery panels) to minimize temperature variations and to convect heat from high-dissipation units to cooler areas of the panel. Heat pipes consist of a pair of ammonia-filled tubes that convect heat, and thus provide additional paths for heat transfer to take full advantage of the OSR-covered radiator panels.

Thermistors are used to sense the temperatures of all major and critical subsystems, indicating the thermal state of health of the spacecraft and providing performance data and diagnostic capability. Thermistors also allow positive verification of the proper operation of the spacecraft during all phases of operations and test and allow timely detection of anomalies
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 03/11/2009 08:25 am
From what I gather, LOX/LH2 exhaust is invisible in vacuum. If you looked into the engine nozzle and chamber you'd see a brilliant glow, but past the nozzle water vapor is invisible. Even in darkness (nighttime launches), the exhaust from the engine is invisible.

LOX/RP1 has apparently a more particulate exhaust (sooty particles/unburned fuel) that's visible even in vacuum, although it doesn't "burn", it's a dark color as it expands and cools rapidly. In nighttime operation the exhaust near the nozzle is presumably illuminated by combustion chamber glow diffusion - see Falcon 1 flight 4 2nd stage restart footage.

As for N2O4/UDMH - Delta II 2nd stage footage suggests it's also invisible in vacuum.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 03/11/2009 11:47 am
I have a question about the corrosiveness of N2O4. I have read that adding a few percent of NO reduces the corrosiveness and that the resulting mixture is called MON, which stands for mixed oxides of nitrogen. Various mixture ratios are in use.

So how much of a difference does this make and how much of a problem is this corrosiveness? I'm particularly interested in the effect on the near-term feasibility of orbital hypergolic propellant depots. How long can you realistically store MMH/MON in space before corrosion renders your depot inoperable?

It isn't a problem.  Comsats have serviceable lives of more than 15 years with MON.  Cassini is now over ten.  MGS was over ten
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/11/2009 11:51 am
It isn't a problem.  Comsats have serviceable lives of more than 15 years with MON.  Cassini is now over ten.  MGS was over ten

Thanks again Jim. Also: long live the internet, I've just found a scanned copy of a Rocketdyne nitrogen tetroxide handling manual from 1961 :-)
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 03/19/2009 02:37 am
Was wondering if anyone could help me find reference material on mass ratio scaling laws for first stages based on propellant density. Specifically, I'm curious as to how CH4-LOX stacks up against LH2-LOX and Kerolox for a first stage. It looks like the average propellant density nearly evens out with Kerolox versus Methlox, so I just wonder.


I've just found a scanned copy of a Rocketdyne nitrogen tetroxide handling manual from 1961 :-)

Ironic it's called a "handling" manual. Somehow, I think a strictly literal interpretation would be bad :-0 .
Title: Re: Basic Rocket Science Q & A
Post by: Steven Pietrobon on 03/19/2009 04:10 am
strangequark, I wrote two papers on this subject which are attached. In a first stage performance is limited by propellant volume. I show in my paper that the criteria for choosing a fuel for a first stage is its impulse density Id, equal to the product of the propellant density (kg/L) and the exhuast speed (m/s). A list of propellants is given below

Propellants  MR   dp (kg/L)  ve (m/s) Id (Ns/L)
O2/H2        5.0  0.3251     4455     1448
O2/H2        6.0  0.3622     4444     1610
O2/H2        7.5  0.4120     4365     1798
O2/CH4       3.6  0.8376     3656     3062
O2/C2H6      3.2  0.9252     3634     3362
O2/C3H8      3.1  0.9304     3613     3362
O2/C3H4      2.4  0.9666     3696     3573
O2/RP–1      2.8  1.0307     3554     3663
O2/C7H8      2.4  1.0954     3628     3974
HTP/C3H4     6.5  1.2553     3319     4166
HTP/RP–1     7.3  1.3059     3223     4209
HTP/C7H8     6.6  1.3496     3288     4437

HTP is 98% hydrogen peroxide, RP-1 is rocket grade kerosene and C7H9 is quadricyclene or RP-X2 (an exotic hydrocarbon fuel). O2/H2 (liquid oxygen and liguid hydrogen) has the best exhaust speed, but a very poor density, which makes it a bad choice as a first stage propellant (that's why the Delta-IV is so huge). A good combination is O2/RP-1, but there are better combinations, such as HTP/RP-1 which will give 15% more performance, plus it has the advantage of being non-cryogenic at the disadvantage of being unstable in the presence of impurities.

So in answer to your specific question about O2/CH4, that performs worse than O2/RP-1 in a first stage. This means your fuel tanks will need to be about 20% larger in order to have the same performance as O2/RP-1. Against O2/H2 it performs much better. Your fuel tanks will be about 47% smaller (the actual percentage depends on the required delta-v for the first stage, larger values will decrease this amount, but even to orbital speeds, O2/CH4 will still perform better).

For an upper stage, mass is all important, so the high exhaust speed of O2/H2 means that's the best propellant to use.

By the way, for single stage to orbit, my second paper shows that any combination O2 or HTP with any other hydrocarbon fuel will outperform O2/H2. A good choice is O2/RP-1, but O2/C7H8 gives 13% more performance.
Title: Re: Basic Rocket Science Q & A
Post by: Eerie on 03/20/2009 01:40 pm
Then why all SSTO projects seems to use O2/H2?
Title: Re: Basic Rocket Science Q & A
Post by: gospacex on 03/20/2009 10:40 pm
strangequark, I wrote two papers on this subject which are attached. In a first stage performance is limited by propellant volume. I show in my paper that the criteria for choosing a fuel for a first stage is its impulse density Id, equal to the product of the propellant density (kg/L) and the exhuast speed (m/s). A list of propellants is given below

Propellants  MR   dp (kg/L)  ve (m/s) Id (Ns/L)
O2/H2        5.0  0.3251     4455     1448
O2/H2        6.0  0.3622     4444     1610
O2/H2        7.5  0.4120     4365     1798
O2/CH4       3.6  0.8376     3656     3062
O2/C2H6      3.2  0.9252     3634     3362
O2/C3H8      3.1  0.9304     3613     3362
O2/C3H4      2.4  0.9666     3696     3573
O2/RP–1      2.8  1.0307     3554     3663
O2/C7H8      2.4  1.0954     3628     3974
HTP/C3H4     6.5  1.2553     3319     4166
HTP/RP–1     7.3  1.3059     3223     4209
HTP/C7H8     6.6  1.3496     3288     4437


HTP is 98% hydrogen peroxide

It is an explosion hazard, and is more expensive than LOX.

I think O2/RP–1 looks good (no wonder it is such a widely used combination).

O2/C3H8 (that's LOX/propane) has some advantages non-obvious from this table: O2 and C3H8 are cryocompatible (easy to build common bulkhead tank for them), and C3H8 is somewhat easier on engines than RP-1: less sooting, and no nasty kerosene residue. I don't know for sure, but maybe it's possible to build an engine in RD-180 class without oxygen-rich preburner for this combination (with RP-1 it's not possible). Also both are cheap.
Title: Re: Basic Rocket Science Q & A
Post by: Steven Pietrobon on 03/23/2009 04:02 am
HTP is 98% hydrogen peroxide

It is an explosion hazard, and is more expensive than LOX.

So are solid rocket motors, which are widely used. Provided that proper precautions are followed, HTP is safe to use, as shown by prior British and US experience. Also, a HTP stage only needs to be filled just prior to launch, unlike solids where an explosion hazard is always present from the time the stage is filled to its use on the launch pad.

Nitrous Oxide (N2O) is also an explosion hazard and is going to be used on a crewed spacecraft. HTP is much less sensitive to shock (unlike N2O), practically requiring a small explosion in order to set it off.

Propellant cost is a small fraction of launch costs. The 15% performance increase more than makes up for any propellant cost increase.
Title: Re: Basic Rocket Science Q & A
Post by: Steven Pietrobon on 03/23/2009 04:18 am
Then why all SSTO projects seems to use O2/H2?

That's a very good question. I believe this is because people only look at exhuast speed ve (divide by g = 9.8065 m/s^2 to get specific impulse) as the performance criteria. O2/H2 has one of the highest ve and is thus thought to be the best propellant. However, it also has the poorest density. O2/RP-1 has a density which is 185% greater than O2/H2, while having an exhaust speed that is only 20% less. This means that in a first stage, you can reduce propellant volume by up to 56%. If you do the sums, a vertical takeoff SSTO vehicle will always have much better performance with O2/RP-1 than with O2/H2.

For horizontal takeoff, then O2/H2 may have better performance due to its much lower total liftoff weight.
Title: Re: Basic Rocket Science Q & A
Post by: gospacex on 03/23/2009 04:33 am
HTP is 98% hydrogen peroxide

It is an explosion hazard, and is more expensive than LOX.

So are solid rocket motors, which are widely used.

I don't like solids either. It doesn't make me like HTP. Imagine a manned rocket at the pad fueled by HTP. Unlike LOX one, which can only burn, HTP *can* explode, LAS wouldn't be able to save the crew.

Quote
Provided that proper precautions are followed, HTP is safe to use, as shown by prior British and US experience.

With enough care, *anything* is safe to use. Like solids or rocketplanes with fragile TPS...

Quote
Propellant cost is a small fraction of launch costs. The 15% performance increase more than makes up for any propellant cost increase.

BTW, do you have data on HTP/C3H8 pair?
Title: Re: Basic Rocket Science Q & A
Post by: sbt on 03/23/2009 07:07 am

To add a data point:

The British Gamma decomposed the HTP to Water + O2 by passing it over a catalyst (which had the disadvantage in Gammas case of needing insertion 2 hrs before launch due to a short 'shelf' life). Steam plus O2 at 500 Celsius is Hypergolic with RP1 so no igniter was needed.

By choosing a Silver-plated Nickel Wire catalyst on cost grounds the UK limited its HTP to 85% H2O2. Higher concentrations would have required something like Platinum.

As well as being creating a Hypergolic engine the decomposition of HTP can also be used to power the fuel and oxidiser turbo-pumps, as with Gamma, giving you a Closed Cycle engine for much less fuss than Staged Combustion.

Finally, although you might not think it, HTP can be used in a Regenerative design – as in Gamma.

The reason cost was more of a factor in selecting the catalyst than might be expected in a launcher program was Gamma was a derivative of one of, and related to many of, motors intended for JATO, Rocket and/or Rocket Assisted fighters and the Blue Steel Stand Off Missile.

In addition to performance HTP was chosen for its relatively benign handling on airfields. It didn't need Cryogenic storage or handling and if spilled it had no dangerous off gassing and could be safely washed away with copious water.

The big question in my mind is how the Catalytic Decomposition approach and Regenerative cooling would scale up beyond Gamma and the larger Stentor Large Chamber (24,000 lb Thrust).

There is also the issue of Freezing at relatively high temperatures (It does, after all, contain water).)

On the Explosion Hazard and LAS. If the Ares I LAS can pull a crew away from an exploding SRM then I wouldn't be so sure that one couldn't be devised to protect a crew from a HTP stage. It might be more difficult for a Single Stage system though, as you loose the separation and buffer provided by the 2nd Stage.

As to why its not proposed more often? One big factor is, I suspect, that LOX/LH and LOX/RP1 are regarded as 'developed technologies' whereas HTP based bi-propellant engines haven't been demonstrated much beyond the British program.

Finally the UK Bloodhound SSC record breaking car project will use a HTP/HTPB Hybrid for main propulsion – again using decomposition and no igniter.

Rick
Title: Re: Basic Rocket Science Q & A
Post by: madscientist197 on 03/23/2009 08:19 am
There's no way that a crew could survive either a solid stage or a HTP stage detonating without another stage as a buffer in between -- no LAS could relocate the capsule fast enough.

Using HTP for stage combustion is an interesting idea, it certainly would reduce the temperatures involved a lot and make production cheaper. Not sure if it would give the best performance though.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 03/23/2009 09:42 am

On the Explosion Hazard and LAS. If the Ares I LAS can pull a crew away from an exploding SRM then I wouldn't be so sure that one couldn't be devised to protect a crew from a HTP stage.


SRM's don't detonate
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 03/23/2009 10:56 am
Unlike LOX one, which can only burn ...

gospacex, this is a general comment and not aimed specifically at you.

I keep seeing this statement ("Unlike LOX one, which can only burn") and others like it all over the place, and while I know it's a fine distinction, it is *not* true that LOX (oxygen) burns. Oxygen does not burn. It is not a fuel. It is an oxidizer. Fuel cannot burn without the presence of an oxidizer to support combustion, and oxygen, in this case, is that oxidizer. In fact the term "burn" is common speech for combustion and actually means "oxidizing" the fuel, such as hydrogen, kerosene, RP1, gasoline, paper, wood, etc. Combustion requires at least 2 things, a fuel to burn and an oxidizer to support the burn. The fuel burns, the oxidizer does not. It molecularly combines with the fuel to create the "oxidation process", called combustion. There are other oxidizers as well besides oxygen. All of them are oxidizers, not fuel, and none of them "burn". They all support combustion and allow the fuel, whatever it is, to combust, or “burn”. Chemistry 101.

Likewise it is incorrect to describe LOX as explosive. It is not. What is generally being referred to when this term is misapplied is that the ET, as a "pressure vessel" can rupture under the high internal pressure. That does not take a spark or ignition or combustion of any kind, just a structural imperfection in a weld or some other less than satisfactory "mechanical" property of the structure of the tank. Granted, the end result is often the same, especially if the rupture also releases large quantities of fuel to mix with the released oxygen (Challenger) which then supports the "fuel" exploding, but the oxygen did not explode, the structural pressure vessel ruptured.

Can we use the correct terminology please?
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 03/23/2009 11:10 am
Can we use the correct terminology please?

From where I stand, gospacex did use the correct term - a "LOX one" as in engine means it has a fuel in addition to LOX. If the two are not well mixed, no detonation is possible. Any by definition, keeping both fuel and oxidizer (in this case LOX) in separate tanks prevents them from mixing. I don't see that statement as implying it's LOX itself that burns.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 03/23/2009 11:20 am
Can we use the correct terminology please?

From where I stand, gospacex did use the correct term - a "LOX one" as in engine means it has a fuel in addition to LOX. If the two are not well mixed, no detonation is possible. Any by definition, keeping both fuel and oxidizer (in this case LOX) in separate tanks prevents them from mixing. I don't see that statement as implying it's LOX itself that burns.

Oxygen does *not* burn.
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 03/23/2009 11:30 am
What did I just say above? He did NOT say "Unlike LOX, which can only burn" he said "Unlike LOX one [i.e. fuel/LOX combo], which can only burn".

I'd think the vast majority of people here know oxygen doesn't actually burn, but maybe it's just my impression.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 03/23/2009 12:07 pm
What did I just say above? He did NOT say "Unlike LOX, which can only burn" he said "Unlike LOX one [i.e. fuel/LOX combo], which can only burn".

I'd think the vast majority of people here know oxygen doesn't actually burn, but maybe it's just my impression.

Google "LOX one" and you won't get the reference you refer to. You will gets hundreds of hits something like this:
"Lyrics for One Two Three Four by The Lox. Album DJ Clue "
So that's just another example of what I was trying to get at and afaik, that term isn't found in a technical manual or reputable dictionary either.

My post was in regards to the use of proper terminology, especially when discussing things of a technical nature. Introducing coloqual terms into a technical discussion is not appropriate given the nature of the subjects. It can and has led to misinterpretations.

I have seen professors ask students to leave the class when they did that, and my request was for folks to take pains to use the proper terminology and use it correctly. This isn't a rap around a campfire. Perhaps I'm old school, but I like to know that the terms I see used in a technical discussion are actually there in the dictionary and continue to mean what they have always meant.

But that's just me.

Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 03/23/2009 12:10 pm
I have seen professors ask students to leave the class when they did that, and my request was for folks to take pains to use the proper terminology and use it correctly. This isn't a rap around a campfire. Perhaps I'm old school, but I like to know that the terms I see used in a technical discussion are actually there in the dictionary and continue to mean what they have always meant.

Fair enough. I'll drop this subject now.
Title: Re: Basic Rocket Science Q & A
Post by: gospacex on 03/23/2009 12:23 pm
Unlike LOX one, which can only burn ...

gospacex, this is a general comment and not aimed specifically at you.

I keep seeing this statement ("Unlike LOX one, which can only burn") and others like it all over the place, and while I know it's a fine distinction, it is *not* true that LOX (oxygen) burns.

You are right. My phrase was "Imagine a manned rocket at the pad fueled by HTP. Unlike LOX one..."

Correct phrase should be "Imagine a manned rocket at the pad fueled by HTP. Unlike LOX fueled one..."

(I'm not sure whether it's 100% correct to say "fueled by [some oxidizer]", but nothing better comes to mind)
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 03/23/2009 12:42 pm
Unlike LOX one, which can only burn ...

gospacex, this is a general comment and not aimed specifically at you.

I keep seeing this statement ("Unlike LOX one, which can only burn") and others like it all over the place, and while I know it's a fine distinction, it is *not* true that LOX (oxygen) burns.

You are right. My phrase was "Imagine a manned rocket at the pad fueled by HTP. Unlike LOX one..."

Correct phrase should be "Imagine a manned rocket at the pad fueled by HTP. Unlike LOX fueled one..."

(I'm not sure whether it's 100% correct to say "fueled by [some oxidizer]", but nothing better comes to mind)

That’s fine. I take your meaning and I understood what you were getting at in the first place. That wasn’t the point, and is why I started by saying it was a general comment, not specific to you.

I see so much these days of folks allowing their English to slip into word usages that are not correct, being linguistically lazy. Folks like me who speak English as a native language often forget that English is *THE* most difficult language on earth to learn as a foreign language. That’s why it is so important, especially in a technical conversation, to take pains to use the correct terminology and to apply it correctly. There are many people on this forum from non-English speaking nations and what they are reading here is a foreign language to them. We English-speakers too often forget that. When we allow ourselves to become casual with how we use our terms, it can make it difficult for them. Hell, it is even difficult for English speakers sometimes to figure out what someone has said.

So like I said, my point was to encourage folks to think about what they want to say before they post it, not only so that it is conceived correctly, but also that it is spoken correctly. Set your language bar high and then reach for it. That’s all I was getting at. I used “burning oxygen” as the example, because I see that incorrect usage so often.
Title: Re: Basic Rocket Science Q & A
Post by: sbt on 03/23/2009 06:24 pm

On the Explosion Hazard and LAS. If the Ares I LAS can pull a crew away from an exploding SRM then I wouldn't be so sure that one couldn't be devised to protect a crew from a HTP stage.


SRM's don't detonate

Ok – I picked that up a while ago (and didn't say they did).

Just to fill in the gaps, does anyone know:

a) Whether HTP actually Detonates in the true sense (i.e. reacts along a shock front) or if it deflagrates (or something else)?

b) If the behaviour of bulk HTP in a stage is understood – does the 'whole lot go up' in one instantaneous bang or does the oxidiser disperse so that only part goes up?

Mind you, this is for info only, not because I actually believe that anybody is going to build a large HTP stage nor am I particularly advocating one – just doing thought experiments.

Rick
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 03/23/2009 06:41 pm

a) Whether HTP actually Detonates in the true sense (i.e. reacts along a shock front) or if it deflagrates (or something else)?

b) If the behaviour of bulk HTP in a stage is understood – does the 'whole lot go up' in one instantaneous bang or does the oxidiser disperse so that only part goes up?


It rapidly decomposes (explodes)  and hence its use as a monopropellant.
Title: Re: Basic Rocket Science Q & A
Post by: A_M_Swallow on 03/24/2009 01:44 am

Mind you, this is for info only, not because I actually believe that anybody is going to build a large HTP stage nor am I particularly advocating one – just doing thought experiments.


There is water on Mars so ISRU H2O2 could be used in a Mars ascent stage.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 03/24/2009 02:56 am
Another thing that needs to be pointed out: solids can only ignite under a very hot ignition source to begin with.  I don't think even a typical match already burning would do it.  Please correct my impression if I'm wrong.
Title: Re: Basic Rocket Science Q & A
Post by: Steven Pietrobon on 03/24/2009 05:17 am
From HTP safety manual

"HTP Explosion Hazard: The action of detonators on HTP has shown that it is possible to partially explode 90% material if it closely confined, and under severe conditions of shock and confinement it has been known to detonate at 80%-85% strength."

100% HTP has the same freezing point as water, 0 C. So if you can keep large volumes of water from freezing, you can do the same with HTP.

If anybody wants a scan of various paper and manuals relating to HTP, email me and I will send them to you.
Title: Re: Basic Rocket Science Q & A
Post by: Steven Pietrobon on 03/24/2009 05:48 am
BTW, do you have data on HTP/C3H8 pair?

I don't have it on hand, but the program I use gives a density of 1.2284 kg/L, exhaust speed of 3242 m/s, impulse density of 3982 Ns/L and a mixture ratio of 7.8. This is better than O2/RP-1, but worse than HTP/RP-1 as a first stage propellant.
Propellants  MR   dp (kg/L)  ve (m/s) Id (Ns/L)
O2/H2        5.0  0.3251     4455     1448
O2/H2        6.0  0.3622     4444     1610
O2/H2        7.5  0.4120     4365     1798
O2/CH4       3.6  0.8376     3656     3062
F2/H2       14.6  0.6553     4704     3083
O2/C2H6      3.2  0.9252     3634     3362
O2/C3H8      3.1  0.9304     3613     3362
O2/C3H4      2.4  0.9666     3696     3573
O2/RP–1      2.8  1.0307     3554     3663
O2/C7H8      2.4  1.0954     3628     3974
HTP/C3H8     7.8  1.2284     3242     3982
HTP/C3H4     6.5  1.2553     3319     4166
HTP/RP–1     7.3  1.3059     3223     4209
HTP/C7H8     6.6  1.3496     3288     4437
F2/NH3       3.4  1.1770     4115     4843

By the way, the highest impulse density propellant I know of is F2/NH3 (liquid fluorine and ammonia). F2 has a density of 1.505 kg/L and NH3 0.676 kg/L. With such high density propellants, at a MR of 3.4 you get dp =  1.1770 kg/L, ve = 4115 m/s and Id = 4843 Ns/L, 32% better than O2/RP-1 if you don't mind a hydrofluoric acid exhaust. :-) I've also added data for F2/H2 which has an Id more twice that of O2/H2 with a better ve.

Any other propellant combinations the readers here would like me to try?
Title: Re: Basic Rocket Science Q & A
Post by: duane on 03/25/2009 02:26 am
I have a simple question. How do you destroy a solid rocket booster that is out of control ?

More specifically how are the shuttle SRB's destroyed by the range safety system ?

I read a Wikipedia article on the SRB's and it quoted it using a linear shaped charge, but where exactly is the charge located, and what/how does it destroy the SRB ?

Cut it in half, cut the nozzle off etc.... ??

Thanks for any info
Duane
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 03/25/2009 02:31 am
Along the side of the case.  It's a zipper.

There's occasionally talk of being able to shut down a solid without destroying it by jettisoning the nozzle with ordnance, but that joint itself would be unsafe.
Title: Re: Basic Rocket Science Q & A
Post by: duane on 03/25/2009 02:54 am
Along the side of the case.  It's a zipper.

There's occasionally talk of being able to shut down a solid without destroying it by jettisoning the nozzle with ordnance, but that joint itself would be unsafe.

Is this charge down a significant length of the SRB ?
Does this zipper cause it to break apart, and does this propellant continue to burn as it falls to splashdown ?

What the heck happens if they abort Ares I before, or shortly after clearing the tower ? Would you have a mass of solid rocket fuel burning on/near the pad til it was exhausted ? Even if the SRB was zipped open  ?
 
Thanks a bunch!
Duane
Title: Re: Basic Rocket Science Q & A
Post by: TrueGrit on 03/25/2009 02:54 am
I've also heard of a concept of blowing off the top of an "out of control" solid...  But the fact is that the nature are solids such that they don't go "out of control" in a standard sense.  They quickly transistion to and overpressure situation and blowup.  So while NASA claims the RSRBs don't blowup...  I'm having trouble believing them.
Title: Re: Basic Rocket Science Q & A
Post by: TrueGrit on 03/25/2009 02:59 am
The shaped charge splits the case, the rocket motor takes care of the rest.

Ever see the Delta 241 failure?  It blew up 1500 ft off the pad.  It was due to a solid rocket motor failing, then triggering the destruct system.  Best explaination I've heard used is a "war zone"...  And yes the launch crew was int he middle of it inside a bunker when it happened.

http://www.youtube.com/watch?v=_R4en3IOQB8&feature=related
Title: Re: Basic Rocket Science Q & A
Post by: TrueGrit on 03/25/2009 03:02 am
http://www.youtube.com/watch?v=appMDzLeT_Q&feature=related
Title: Re: Basic Rocket Science Q & A
Post by: duane on 03/25/2009 03:16 am
Yea I saw that video a few times. Pretty impressive. Just wonder what a Ares-I SRB  zippered open that  close to the ground would look like. A falling flaming hunk of molten slag ?  :-)

Thanks a bunch, heading to bed
Duane
Title: Re: Basic Rocket Science Q & A
Post by: AnalogMan on 03/25/2009 10:38 am
I read a Wikipedia article on the SRB's and it quoted it using a linear shaped charge, but where exactly is the charge located, and what/how does it destroy the SRB ?

Cut it in half, cut the nozzle off etc.... ??

"The Linear Shaped Charge, installed in the systems tunnel over 70% of the motor case, cuts through the motor case to allow rapid combustion of all propellant, destroying the SRB’s. the ET, and the Orbiter"
Title: Re: Basic Rocket Science Q & A
Post by: Heg on 03/25/2009 11:09 am

(...)

Any other propellant combinations the readers here would like me to try?

Given the perfect impulse density which F2/NH3 offers, not to mention nice exhaust velocity, practically in the O2/H2 ballpark (see RS-68), I'd like to ask you how would the ammonia perform combined with other oxidizers, like O2, HTP, or N2O4? Would any of these combinations ever work?
Title: Re: Basic Rocket Science Q & A
Post by: duane on 03/25/2009 10:16 pm
Thanks to everyone for the RSS info on the shuttle SRB's
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 03/26/2009 12:49 am
strangequark, I wrote two papers on this subject which are attached. In a first stage performance is limited by propellant volume. I show in my paper that the criteria for choosing a fuel for a first stage is its impulse density Id, equal to the product of the propellant density (kg/L) and the exhuast speed (m/s). A list of propellants is given below

Propellants  MR   dp (kg/L)  ve (m/s) Id (Ns/L)
O2/H2        5.0  0.3251     4455     1448
O2/H2        6.0  0.3622     4444     1610
O2/H2        7.5  0.4120     4365     1798
O2/CH4       3.6  0.8376     3656     3062
O2/C2H6      3.2  0.9252     3634     3362
O2/C3H8      3.1  0.9304     3613     3362
O2/C3H4      2.4  0.9666     3696     3573
O2/RP–1      2.8  1.0307     3554     3663
O2/C7H8      2.4  1.0954     3628     3974
HTP/C3H4     6.5  1.2553     3319     4166
HTP/RP–1     7.3  1.3059     3223     4209
HTP/C7H8     6.6  1.3496     3288     4437

HTP is 98% hydrogen peroxide, RP-1 is rocket grade kerosene and C7H9 is quadricyclene or RP-X2 (an exotic hydrocarbon fuel). O2/H2 (liquid oxygen and liguid hydrogen) has the best exhaust speed, but a very poor density, which makes it a bad choice as a first stage propellant (that's why the Delta-IV is so huge). A good combination is O2/RP-1, but there are better combinations, such as HTP/RP-1 which will give 15% more performance, plus it has the advantage of being non-cryogenic at the disadvantage of being unstable in the presence of impurities.

So in answer to your specific question about O2/CH4, that performs worse than O2/RP-1 in a first stage. This means your fuel tanks will need to be about 20% larger in order to have the same performance as O2/RP-1. Against O2/H2 it performs much better. Your fuel tanks will be about 47% smaller (the actual percentage depends on the required delta-v for the first stage, larger values will decrease this amount, but even to orbital speeds, O2/CH4 will still perform better).

For an upper stage, mass is all important, so the high exhaust speed of O2/H2 means that's the best propellant to use.

By the way, for single stage to orbit, my second paper shows that any combination O2 or HTP with any other hydrocarbon fuel will outperform O2/H2. A good choice is O2/RP-1, but O2/C7H8 gives 13% more performance.

Hi Steven,

Thanks for the papers, just got the chance to really read through them. I do have a few nits to pick, however. This "density impulse" you use, it looks like it arbitrarily assumes that mass ratio scales linearly with propellant density. Tank mass seems like it roughly would (assuming: constant pressure, spherical tanks or constant height/radius cylinders), but I was curious if there's anything empirical covering the other systems as well. Or am I looking too much into it, and first stage tank mass so dwarfs the other components that a linear scaling is fair? Secondly, I was a bit thrown off by the assumption used in defining density impulse. You basically use the ln(1+x)=x assumption, but that's only valid for pretty small x, say about 0.1. So that assumption would only work for mass ratio of 1.1, at best (assuming ~450 Isp, to high-side ballpark) a first stage delta-V of just 421 m/s.
Title: Re: Basic Rocket Science Q & A
Post by: Steven Pietrobon on 03/26/2009 03:02 am
This "density impulse" you use, it looks like it arbitrarily assumes that mass ratio scales linearly with propellant density. Tank mass seems like it roughly would (assuming: constant pressure, spherical tanks or constant height/radius cylinders), but I was curious if there's anything empirical covering the other systems as well. Or am I looking too much into it, and first stage tank mass so dwarfs the other components that a linear scaling is fair?

Yes, tank mass is roughly proportional to tank volume. In the near single stage to orbit (NSTO) paper I plot engine mass (kg) to propellant flow rate (kg/s) against thrust. I found a variation from about 2 to 4 kgs/L with thrusts from 0 to 8 MN. So, there is a constant term, plus some linear term dependent on the thrust.

I believe the constant term comes about from pipe and turbo pump sizes being inversely proportional to propellant density. So for the same thrust, a low density propellant will require a bigger engine than a high density propellant, simply because the pipes and turbo pumps will be bigger. However, thrust structure mass will be proportional to thrust which is where I think the linear term comes from.

Quote
Secondly, I was a bit thrown off by the assumption used in defining density impulse. You basically use the ln(1+x)=x assumption, but that's only valid for pretty small x, say about 0.1. So that assumption would only work for mass ratio of 1.1, at best (assuming ~450 Isp, to high-side ballpark) a first stage delta-V of just 421 m/s.

Yes, for a very close approximation, the delta-V's need to be quite small. However, if we are just interested in seeing which propellant is better as a first stage, then Figure 3 in the NSTO paper (attached below) quite clearly shows that up to about a delta-V of 3 km/s, the delta-V approximation using Id is quite valid. At higher delta-V's there is some cross over in the curves. For example, above 6 km/s O2/C7H8 performs better than HTP/C7H8. That is below dv = 6 km/s, the high Id of HTP/C7H8 allows it to perform better, while above dv = 6 km/s the better exhaust speed ve of O2/C7H8 allows it to perform better. This implies that for a traditional first stage, it is better to use HTP/C7H8. For an SSTO vehicle, O2/C7H8 is better to use, which is why I chose that propellant in this paper for my NSTO vehicle.
Title: Re: Basic Rocket Science Q & A
Post by: Steven Pietrobon on 03/26/2009 04:58 am
Updated table. We see that NH3 does not work so well with O2, N2O4 or HTP. I've also added some hydrazine N2H4 results. This gives a new record for impulse density with F2, a staggering value of 5506 Ns/L, 50% greater than O2/RP-1.

You might also note that F2 burns with water H2O quite well! F2 also burns with any hydrocarbon, literally sucking the hydrogen off and leaving a carbon soot. With H2O it leaves O2 in the exhaust! F2 also burns with HTP quite nicely, having the highest density combination, which gives it a high impulse density.

Propellants  MR   dp (kg/L)  ve (m/s) Id (Ns/L)
O2/H2        5.0  0.3251     4455     1448
O2/H2        6.0  0.3622     4444     1610
O2/H2        7.5  0.4120     4365     1798
O2/NH3       1.4  0.8896     3399     3024
O2/CH4       3.6  0.8376     3656     3062
O2/C2H6      3.2  0.9252     3634     3362
O2/C3H8      3.1  0.9304     3613     3362
O2/C3H4      2.4  0.9666     3696     3573
O2/RP–1      2.8  1.0307     3554     3663
O2/C7H8      2.4  1.0954     3628     3974

N2O4/NH3     2.0  1.0428     3097     3230
N2O4/UDMH    2.9  1.1823     3350     3961
N2O4/MMH     2.4  1.2051     3366     4056
N2O4/N2H4    1.4  1.2156     3371     4097

HTP/NH3      3.1  1.1216     3068     3441
HTP/C3H8     7.8  1.2286     3240     3981
HTP/N2H4     2.2  1.2608     3283     4139
HTP/C3H4     6.5  1.2553     3319     4166
HTP/RP–1     7.3  1.3059     3223     4209
HTP/C7H8     6.6  1.3496     3288     4437

F2/H2       14.6  0.6553     4704     3083
F2/H2O       2.1  1.2942     2876     3722
F2/HTP       0.88 1.4689     2966     4357
F2/NH3       3.4  1.1770     4115     4843
F2/N2H4      2.3  1.3073     4212     5506
Title: Re: Basic Rocket Science Q & A
Post by: gospacex on 03/26/2009 03:45 pm

(...)
Any other propellant combinations the readers here would like me to try?

Given the perfect impulse density which F2/NH3 offers, not to mention nice exhaust velocity, practically in the O2/H2 ballpark (see RS-68), I'd like to ask you how would the ammonia perform combined with other oxidizers, like O2, HTP, or N2O4? Would any of these combinations ever work?

I don't see much reason to mess with potentially explosive and/or toxic fuels or oxidizers, performance increase would not be worth it.
Additional complexity of handling them securely is adding some costs, and still, can't be made 100.00% safe.

Would you like to read reports about poisonous spills from US rockets?

Would you like to put astronauts atop a rocket with 700 tons of concentrated hydrogen peroxide (as opposed to "a rocket with 700 tons of LOX")?
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 03/27/2009 12:25 am

Would you like to read reports about poisonous spills from US rockets?


Cough ... Titan ... Cough ... Proton ... Cough Cough ... Delta II Second Stage ... Cough Cough ...

Maybe not as toxic as some of the fuels in the table, but pretty nasty and toxic none the less...
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 03/27/2009 03:15 am
Steven,

Thanks for reminding me about HTP. I might have to think about acquiring some for my garage project. The catalytic decomposition eases ignition, and it doesn't have any nitrogen in it to murder your Isp.
Title: Re: Basic Rocket Science Q & A
Post by: kkattula on 03/27/2009 07:44 am
Steven,

Thanks for reminding me about HTP. I might have to think about acquiring some for my garage project. The catalytic decomposition eases ignition, and it doesn't have any nitrogen in it to murder your Isp.

Be very, very careful!   

HTP is catalyzed by a lot of organic sustances, including leather and human skin. Have plenty of water around to dilute any spills and a shower for any one exposed.  Never work alone as the pain from exposure can prevent someone from treating themself.

IIRC, an amateur rocketry enthusiast died a couple of years ago working with HTP.

Do plenty of research.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/28/2009 05:42 pm
Does anyone know a good reference text for ascent trajectories? I found some old Pascal code by our very own Steven Pietrobon and some papers that didn't go into a lot of depth. I'd like to write some code to get decent payload estimates to L1 for various Jupiter variants. I have a couple of years of experience with numerical simulation, but I know next to nothing about aerodynamics.
Title: Re: Basic Rocket Science Q & A
Post by: Nikola on 03/29/2009 12:08 am
I was always puzzled by the fact that smaller rockets can put smaller proportion of their weight into orbit. For example Falcon 1 can put 500kg, or 1.3% of launch mass, while Falcon 9 can put 10t or 3.1%. Why is that? They use same components, same manufacturing, same engines (apart from merlin<--->kestrel switch) etc.

Also Elon Musk said that they choose F1 size so that is as small as possible and capable of reaching orbit (can't remember where I read this, I'll search for it later). Does that mean that there is some lower cutoff mass for a launch vehicle?

I don't see any theoretical reason for this. For delta-v only important thing is mass fraction. I mean, what are the practical problems to make, for example, rocket with a mass of 100kg that can put like 1kg into orbit?
Title: Re: Basic Rocket Science Q & A
Post by: hop on 03/29/2009 12:25 am
The cube/square relationship of volume to surface area means that smaller LVs have higher aero losses. You get similar effects with the minimum gauge of materials.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 03/29/2009 02:21 am
Does anyone know a good reference text for ascent trajectories? I found some old Pascal code by our very own Steven Pietrobon and some papers that didn't go into a lot of depth. I'd like to write some code to get decent payload estimates to L1 for various Jupiter variants. I have a couple of years of experience with numerical simulation, but I know next to nothing about aerodynamics.

Attached is a small simulation I wrote in Quick Basic.  It has a very efficient numerical integration routine and the basics of aerodynamics built in.  You should be able to use the aero stuff in what you are doing.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/29/2009 02:26 am
Attached is a small simulation I wrote in Quick Basic.  It has a very efficient numerical integration routine and the basics of aerodynamics built in.  You should be able to use the aero stuff in what you are doing.

Thanks! Any recommendations for books if I want to learn more?
Title: Re: Basic Rocket Science Q & A
Post by: Oberon_Command on 03/29/2009 02:51 am
Danny, if you or anyone else would like, I could port that code to C, C++, or Java, since some people might find it easier to read or use that way. I was thinking of doing that anyway, since I'm one of those who, despite having started with QB, finds C-like languages more readable, but it occurred to me that others might find such a translation useful...
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 03/29/2009 03:11 am
Nikola, also for your question there are some things that every rocket has to have.  A 3x rocket doesn't need a 3x flight computer.  The closer one starts to that minimum function and payload rocket, the more payload one gets for a slight growth in the overall size.  Look at Athena I vs Athena II and Falcon 1 vs 1e.
Title: Re: Basic Rocket Science Q & A
Post by: Nikola on 03/29/2009 03:15 am
The cube/square relationship of volume to surface area means that smaller LVs have higher aero losses. You get similar effects with the minimum gauge of materials.

That's true - smaller LV will have higher relative air drag. Can you, please, give some example of the effects that appear "with the minimum gauge of materials"? That's what interest me the most.

For example if you reduce the dimensions of a fuel tank by half and reduce the thickens of walls by half, tank will still have same mass fraction. I would guess this is example where scaling is possible. I don't have any experience in building rocket hardware so I could be wrong.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 03/29/2009 03:16 am
Danny, if you or anyone else would like, I could port that code to C, C++, or Java, since some people might find it easier to read or use that way. I was thinking of doing that anyway, since I'm one of those who, despite having started with QB, finds C-like languages more readable, but it occurred to me that others might find such a translation useful...

c might be useful to many out there.   Lot is c code in the world today.  By the way, you must be a really strange dude if you think c is inherently more readable than QB ;-)
Title: Re: Basic Rocket Science Q & A
Post by: Oberon_Command on 03/29/2009 04:00 am
Slightly off-topic:
Once upon a time, I thought the same as you... then I actually switched to C, used it for a few years, and I came to see the beauty and simplicity in single-character braces as opposed to constantly capitalized potential polysyllabic keywords (woo, alliteration!), shorter type-names, and always declaring your variables before using them. Then again, I do find myself working with regular expressions from time to time, which are of a whole order less readable than BASIC or C... :P

This is not to say that beautiful BASIC code is impossible, or that ugly C-syntax code is impossible. Idiomatic C++ is particularly hard on the eyes at times, for sure.

I'll do that port tomorrow, then, when I have more caffeine in my system.
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 03/29/2009 06:40 am
Steven,

Thanks for reminding me about HTP. I might have to think about acquiring some for my garage project. The catalytic decomposition eases ignition, and it doesn't have any nitrogen in it to murder your Isp.

Be very, very careful!   

HTP is catalyzed by a lot of organic sustances, including leather and human skin. Have plenty of water around to dilute any spills and a shower for any one exposed.  Never work alone as the pain from exposure can prevent someone from treating themself.

IIRC, an amateur rocketry enthusiast died a couple of years ago working with HTP.

Do plenty of research.

Definitely planning on it. I have a few brains at KSC I will likely pick, regarding general safety practices with strong oxidizers. Might invest in something like a butyl coverall if I do go through with it. The "hypergolic with test engineers" issue is definitely not to be taken lightly.  ;)
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 03/29/2009 12:05 pm
The "hypergolic with test engineers" issue is definitely not to be taken lightly.  ;)

"Peroxides kill chemists" is another warning I've heard.
Title: Re: Basic Rocket Science Q & A
Post by: hop on 03/29/2009 10:34 pm
That's true - smaller LV will have higher relative air drag. Can you, please, give some example of the effects that appear "with the minimum gauge of materials"? That's what interest me the most.
Described here (http://www.jerrypournelle.com/slowchange/SSX.html) as:
Quote
Note also that to get high mass ratios you need BIG rocket ships. That follows because there are minimum weights to many rocket parts. This is known as 'minimum gauge', meaning that you can't make it any thinner or lighter. Combustion chambers have to be rugged. Pipes must be thick enough to hold pressures that don't get smaller just because you're trying to make a small sized rocket.
But this is sort of specialized case of a larger issue: A lot of things just don't scale arbitrarily. See 3-5 here http://www.av8n.com/physics/scaling.htm#htoc14

For rockets, engine cooling is a good example: You need to cool the surface area of your combustion chamber, but propellant flow scales by volume. Small engines are harder to cool.
Quote
For example if you reduce the dimensions of a fuel tank by half and reduce the thickens of walls by half, tank will still have same mass fraction.
Tank mass does scale roughly with volume over a fairly large range (see http://yarchive.net/space/launchers/fuel_tank_scaling_laws.html ), but at extremes other things would start to take over.

strangequark:
If you haven't already, I suggest you read armadillos update archives (http://www.armadilloaerospace.com/n.x/Armadillo/Home). They did quite a lot with H2O2.
Title: Re: Basic Rocket Science Q & A
Post by: kkattula on 03/30/2009 05:48 am

...
strangequark:
If you haven't already, I suggest you read armadillos update archives (http://www.armadilloaerospace.com/n.x/Armadillo/Home). They did quite a lot with H2O2.


NB:  Armadillo started with HTP then moved to a 'mixed mono-prop' of 50% conc. H2O2 mixed with Methanol. (in a 92%:8% mixture IIRC).

50% conc. H2O2 mixed with Methanol is fairly stable, but tricky to decompose.

HTP mixed with Methanol tends to DETONATE (not just decompose) if you look at it wrong.
Title: Re: Basic Rocket Science Q & A
Post by: sbt on 03/30/2009 07:10 pm
Danny, if you or anyone else would like, I could port that code to C, C++, or Java, since some people might find it easier to read or use that way. I was thinking of doing that anyway, since I'm one of those who, despite having started with QB, finds C-like languages more readable, but it occurred to me that others might find such a translation useful...

c might be useful to many out there.   Lot is c code in the world today.  By the way, you must be a really strange dude if you think c is inherently more readable than QB ;-)

And, if you don't object, I'll have a bash at putting it into Python.

Any particular wishes concerning a licence to be attached? Otherwise I'll tag it as being GPL which will keep your name attached and prevent somebody rolling it into their commercial product yet allow others to make improvements.

Rick
Title: Re: Basic Rocket Science Q & A
Post by: Steven Pietrobon on 03/31/2009 04:03 am
Checkout Peroxide Propulsion (http://www.peroxidepropulsion.com/) which has rocket grade HTP at 80%, 85% and 90% concentration.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 03/31/2009 07:21 am
Danny, if you or anyone else would like, I could port that code to C, C++, or Java, since some people might find it easier to read or use that way. I was thinking of doing that anyway, since I'm one of those who, despite having started with QB, finds C-like languages more readable, but it occurred to me that others might find such a translation useful...

c might be useful to many out there.   Lot is c code in the world today.  By the way, you must be a really strange dude if you think c is inherently more readable than QB ;-)

And, if you don't object, I'll have a bash at putting it into Python.

Any particular wishes concerning a licence to be attached? Otherwise I'll tag it as being GPL which will keep your name attached and prevent somebody rolling it into their commercial product yet allow others to make improvements.

Rick

Thanks for asking.  Sounds like a good idea.  Is that what GNU does to their code?

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: vt_hokie on 03/31/2009 04:06 pm
c might be useful to many out there.   Lot is c code in the world today.  By the way, you must be a really strange dude if you think c is inherently more readable than QB ;-)

I never liked C, but then I never really used it beyond an intro course in college.  Hate to say it, but I haven't really done any programming since my FORTRAN 77 days in school! 
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 03/31/2009 04:16 pm
c might be useful to many out there.   Lot is c code in the world today.  By the way, you must be a really strange dude if you think c is inherently more readable than QB ;-)

I never liked C, but then I never really used it beyond an intro course in college.  Hate to say it, but I haven't really done any programming since my FORTRAN 77 days in school! 

So you should do fairly well with both POST and OTIS code.
Title: Re: Basic Rocket Science Q & A
Post by: William Barton on 03/31/2009 04:25 pm
c might be useful to many out there.   Lot is c code in the world today.  By the way, you must be a really strange dude if you think c is inherently more readable than QB ;-)

I never liked C, but then I never really used it beyond an intro course in college.  Hate to say it, but I haven't really done any programming since my FORTRAN 77 days in school! 

I saved my Fortran text for years, though what box it's in now... probably the same box as the Prolog, Pascal, and LISP books. I do most of my work with Visual Studio, these days, so that means C#, which is charitably described as "C-flavored."
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 03/31/2009 04:43 pm
c might be useful to many out there.   Lot is c code in the world today.  By the way, you must be a really strange dude if you think c is inherently more readable than QB ;-)

I never liked C, but then I never really used it beyond an intro course in college.  Hate to say it, but I haven't really done any programming since my FORTRAN 77 days in school! 

I saved my Fortran text for years, though what box it's in now... probably the same box as the Prolog, Pascal, and LISP books. I do most of my work with Visual Studio, these days, so that means C#, which is charitably described as "C-flavored."

Oh - now I 'C'. (grin)

You may find this a little strange, but I actually prefer VB. It's very conversational and sufficiently structured that I can do just about anything I want. Of course C is more powerful; it's just not as much fun. Plus most Macro applications for Window-based software is executed by VB. Makes it convenient to be conversant with it.
Title: Re: Basic Rocket Science Q & A
Post by: sbt on 03/31/2009 07:05 pm

And, if you don't object, I'll have a bash at putting it into Python.

Any particular wishes concerning a licence to be attached? Otherwise I'll tag it as being GPL which will keep your name attached and prevent somebody rolling it into their commercial product yet allow others to make improvements.

Rick

Thanks for asking.  Sounds like a good idea.  Is that what GNU does to their code?

Danny Deger

Yes. There are a few variations on a theme. The LGPL (also GNU) is the main one - which allows inclusion into commercial products as a stand alone library.

There are lots of small variations, if you really want to get deep into things.

Two reasonably simple comparisons (assuming you want to be Open Source)

http://itmanagement.earthweb.com/osrc/article.php/12068_3803101_2/Bruce-Perens-How-Many-Open-Source-Licenses-Do-You-Need.htm

http://developer.kde.org/documentation/licensing/licenses_summary.html

Really deep and 'political':

http://www.gnu.org/philosophy/license-list.html

Personally I would have to have a very good reason to licence something that was my creation (and mine alone) as anything other than GPL. But this is a port of your code so you get to choose.

Anyway - back to the Rockets...

Rick
Title: Re: Basic Rocket Science Q & A
Post by: butters on 05/04/2009 05:36 pm
Are there any general guidelines for sizing diameter versus length for a cylindrical ELV with a given LEO payload capacity, specific impulse, and propellant density?


Suppose you wish to launch a payload with a relatively large diameter for its mass.  Would it be better to launch it in a fat PLF on a skinnier LV or to launch it on a relatively short and squat LV that matches the PLF?


In other words, is there a "golden aspect ratio" for earth-launched rockets, and to what extent does the diameter of the payload affect the optimal diameter of the rocket?
Title: Re: Basic Rocket Science Q & A
Post by: mdo on 05/06/2009 10:53 am
Are there any general guidelines for sizing diameter versus length for a cylindrical ELV with a given LEO payload capacity, specific impulse, and propellant density?


Suppose you wish to launch a payload with a relatively large diameter for its mass.  Would it be better to launch it in a fat PLF on a skinnier LV or to launch it on a relatively short and squat LV that matches the PLF?

- drag is proportional to frontal area per lift-off mass

- this ratio improves with rocket mass; for Shuttle sized vehicles it becomes negligible - Orbiter, Tank, SRBs fully exposed; only small sounding rockets need to be tall

- spherical tanks have the best volume/weight ratio

- constant cylindrical LV diameter has two advantages
1. PLF is lighter; it need not reach inwards to skinny rocket body
2. less turbulent air flow

- payload dimensions vary; so adapt the LV to a certain mass and adapt the fairing to individual volume requirement

Quote
In other words, is there a "golden aspect ratio" for earth-launched rockets, and to what extent does the diameter of the payload affect the optimal diameter of the rocket?

Don't know about golden rules:
Suggest to let the denser one of the two fuel/oxidizer 2nd (upper) stage components drive the diameter of its spherical tank. The other three tanks of upper and core stage will be more cylindrically elongated corresponding to their larger volumes. The 4 tank stack gives a rough estimate of length to diameter ratio for a given propellant.

Mind you, I do not build these things myself; others please correct or expand on solid propellants.

Apropos, some say the SRB was designed by a horse's ass:
http://greyhorsematters.blogspot.com/2009/04/interesting-history-lesson.html (http://greyhorsematters.blogspot.com/2009/04/interesting-history-lesson.html)

edit: typo
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/07/2009 07:13 pm
Rule of thumb for solid motor length:diameter seems to be 12:1.
Title: Re: Basic Rocket Science Q & A
Post by: mdo on 05/08/2009 06:28 am
What is the desired angular change to the flight path achieved by the pitch maneuver performed after clearing the launch tower?

What are typical pitch angles for current LVs such as Atlas V, Ariane 5 etc.?
Title: Re: Basic Rocket Science Q & A
Post by: William Barton on 05/08/2009 07:18 am
c might be useful to many out there.   Lot is c code in the world today.  By the way, you must be a really strange dude if you think c is inherently more readable than QB ;-)

I never liked C, but then I never really used it beyond an intro course in college.  Hate to say it, but I haven't really done any programming since my FORTRAN 77 days in school! 

I saved my Fortran text for years, though what box it's in now... probably the same box as the Prolog, Pascal, and LISP books. I do most of my work with Visual Studio, these days, so that means C#, which is charitably described as "C-flavored."

Oh - now I 'C'. (grin)

You may find this a little strange, but I actually prefer VB. It's very conversational and sufficiently structured that I can do just about anything I want. Of course C is more powerful; it's just not as much fun. Plus most Macro applications for Window-based software is executed by VB. Makes it convenient to be conversant with it.



I don't find it strange at all. When you're working in a tool like Visual Studio, the difference between the languages is a matter of style more than anything else, and if inaccurate Microsoft would rationalize the .Net object model, the differences would be negligible. I sell VB code to my clients along with their projects, based on the theory they can have it maintained by bottom feeders if need be, and will never have trouble finding a programmer if they need one and I'm not available anymore (or they've decided I'm too annoying for the money I charge). There's really nothing you can do in C# you can't do in VB, so the main advantage of C# is, I get to charge more.

I would love to get my hands on some of the flight software for modern LVs and spacecraft. I'm sure I could have no end of fun with it.

Footnote: another posting intervention! When I type i-d-i-o-t (without the spaces," it rendered as "inaccurate." (Idiot?)
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/08/2009 11:25 am
I would love to get my hands on some of the flight software for modern LVs and spacecraft. I'm sure I could have no end of fun with it.

One interesting experiment would be to have a look at the code and then decide whether you still think launching people with that is a good idea. :)

On a related note, I think the Apollo LM software is out there somewhere on the Internet. There's even an emulator for it.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/08/2009 11:35 am
What is the desired angular change to the flight path achieved by the pitch maneuver performed after clearing the launch tower?

What are typical pitch angles for current LVs such as Atlas V, Ariane 5 etc.?

Why don't you go ask what the shuttle pitch profile is?  Try the shuttle Q&A thread.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: AnalogMan on 05/08/2009 12:22 pm

Footnote: another posting intervention! When I type i-d-i-o-t (without the spaces," it rendered as "inaccurate." (inaccurate?)

Heaven help us if we ever need to talk about a clueless idiot  ;)
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/08/2009 12:37 pm
Heaven help us if we ever need to talk about a clueless idiot  ;)

I was about to ask how on earth you did that, but if you look at the wiki markup it's obvious. :) Well done sir!
Title: Re: Basic Rocket Science Q & A
Post by: William Barton on 05/08/2009 12:56 pm
I would love to get my hands on some of the flight software for modern LVs and spacecraft. I'm sure I could have no end of fun with it.

One interesting experiment would be to have a look at the code and then decide whether you still think launching people with that is a good idea. :)

On a related note, I think the Apollo LM software is out there somewhere on the Internet. There's even an emulator for it.

There's a part of me that doesn't want to see. I spent a part of the 1990s as a "rescue programmer," and I shudder to think...
Title: Re: Basic Rocket Science Q & A
Post by: William Barton on 05/08/2009 12:58 pm
Heaven help us if we ever need to talk about a clueless idiot  ;)

I was about to ask how on earth you did that, but if you look at the wiki markup it's obvious. :) Well done sir!

Clever! Now the question of why *those* words...
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/08/2009 03:20 pm
What is the desired angular change to the flight path achieved by the pitch maneuver performed after clearing the launch tower?

What are typical pitch angles for current LVs such as Atlas V, Ariane 5 etc.?

No guideline.  It depends on other requirements.  Usually (1) controllability, (2) structural limits, (3) performance (drag) in that order of importance.

Very small compared to airplanes (quantification could involve eye tar so don't ask).  While q is still high, much of an alpha or beta will cause tumble or tearing off the fairing.  Which is why the constraint is often listed in terms of q*alpha or q*beta.
Title: Re: Basic Rocket Science Q & A
Post by: mdo on 05/08/2009 05:51 pm
Why don't you go ask what the shuttle pitch profile is?  Try the shuttle Q&A thread.

...because access to emergency landing sites, aerodynamic surfaces and the lack of fairing separation (altitude) make the Shuttle a special case muddying the water when trying to understand the basics as in Basic Rocket Science Q&A.

Title: Re: Basic Rocket Science Q & A
Post by: mdo on 05/08/2009 05:56 pm
No guideline.  It depends on other requirements.  Usually (1) controllability, (2) structural limits, (3) performance (drag) in that order of importance.

Very small compared to airplanes (quantification could involve eye tar so don't ask).  While q is still high, much of an alpha or beta will cause tumble or tearing off the fairing.  Which is why the constraint is often listed in terms of q*alpha or q*beta.

Thanks Antares.

The Ariane 5 press kit of the upcoming Herschel/Planck launch, for instance,
http://www.arianespace.com/images/launch-kits/launch-kit-pdf-eng/HERSCHEL-PLANCK-GB.pdf (http://www.arianespace.com/images/launch-kits/launch-kit-pdf-eng/HERSCHEL-PLANCK-GB.pdf)
mentions a 10 s pitch rotation. So, no itar issue. My question is not so much about structural limits but about how much the LV is put off the initial 90 deg climb angle? Maybe it rotates to 70 or 80 deg during those 10 s?

My guess for its purpose is to gain just enough vertical speed during the 1st stage burn so that the LV does not drop back below fairing separation altitude. It is an issue when the upper stage thrust/weight ratio is below 1. Since the angle of attack during atmospheric flight is rather limited I am speculating that the initial pitch maneuver is supposed to put the vehicle on a trajectory that limits vertical v to exactly meet this requirement, but again, I'm not sure this is the case.


Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/08/2009 06:05 pm

1. mentions a 10 s pitch rotation. ....... My question is not so much about structural limits but about how much the LV is put off the initial 90 deg climb angle? Maybe it rotates to 70 or 80 deg during those 10 s?

2. My guess for its purpose is to gain just enough vertical speed during the 1st stage burn so that the LV does not drop back below fairing separation altitude. It is an issue when the upper stage thrust/weight ratio is below 1. Since the angle of attack during atmospheric flight is rather limited I am speculating that the initial pitch maneuver is supposed to put the vehicle on a trajectory that limits vertical v to exactly meet this requirement, but again, I'm not sure this is the case.


1.  There are other "pitch rotations" in the flight.  This is not the only one.  Also a "pitch rotation" is a rate and not a final position.

2.  Since there are other "pitch rotations", and the vehicle attitude is not fixed, therefore the rest of your post is not applicable. 

The initial pitch rates and attitudes are about structural limitations and aerodynamic loads.   Pitch rates and attitudes to deal with gravity losses and second stage start conditions occur later.
Title: Re: Basic Rocket Science Q & A
Post by: mdo on 05/08/2009 07:40 pm
1.  There are other "pitch rotations" in the flight.  This is not the only one.  Also a "pitch rotation" is a rate and not a final position.

Let me phrase the question differently then:

Do LVs actively aim for non zero angle of attack at any time during supersonic flight?

If not, then it needs to be tipped off initially, i.e., after clearing the tower and by some angle that I'm interested in. Thereafter, it does a gravity turn. Well, I hope gravity turn is the proper term for aligning thrust and velocity vectors to achieve zero AOA.

Quote
2.  Since there are other "pitch rotations"...

... would mean my assumption is wrong which is LVs strictly stick to a gravity turn during atmospheric flight and after the first few seconds of ascent - UNLESS those other pitch rotations are performed for the very reason of keeping the AOA continuously close to zero?
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/08/2009 08:27 pm
Why don't you go ask what the shuttle pitch profile is?  Try the shuttle Q&A thread.

...because access to emergency landing sites, aerodynamic surfaces and the lack of fairing separation (altitude) make the Shuttle a special case muddying the water when trying to understand the basics as in Basic Rocket Science Q&A.



You are correct the shuttle is a little different, but it still must not fall apart during first stage and fly an optimum pitch profile in second stage.  Someone in the shuttle world can get you a copy of the pitch profile cue card used by the crew.  Try googling "SODB".  You might try the Apollo forum to get a copy of the Apollo cue card.  The is a Saturn V users manual on the web that I think has a copy.

Have you tried the Atlas and Delta User's guides?  I would help find links, but I am on my Treo.

The launch vehicles often don't shoot for zero angle of attack.  But must always keep angle of attack in limits.

Pitch is pretty much contantly changing.  Constant pitch rate gives the fuel optimum profile.

I have never heard the term "pitch manuever" because pitch is always decreasing (with some exceptions).  There is a roll maneuver that puts the LV in the proper roll attitude and it then stay there.

Hope this helps.  Happy hunting.

Danny Deger

Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/08/2009 08:50 pm

1.  Do LVs actively aim for non zero angle of attack at any time during supersonic flight?

2.  If not, then it needs to be tipped off initially, i.e., after clearing the tower and by some angle that I'm interested in. Thereafter, it does a gravity turn.

3.  Well, I hope gravity turn is the proper term for aligning thrust and velocity vectors to achieve zero AOA.

4.  ... would mean my assumption is wrong which is LVs strictly stick to a gravity turn during atmospheric flight and after the first few seconds of ascent - UNLESS those other pitch rotations are performed for the very reason of keeping the AOA continuously close to zero?


1.  LV's don't sense AOA or even airspeed.  Everything is based on simulations and balloon data

2.  That "angle" is different for every launch vehicle

3.   That is a benefit of a gravity turn vs the reason for it

4.  they may be for load relief
Title: Re: Basic Rocket Science Q & A
Post by: mdo on 05/08/2009 10:20 pm
The launch vehicles often don't shoot for zero angle of attack.  But must always keep angle of attack in limits.

Pitch is pretty much contantly changing.  Constant pitch rate gives the fuel optimum profile.

Great! That's exactly the kind of hints I was looking for; will look up the specific references for some sample values. Cheers.

Quote from: Jim
1.  LV's don't sense AOA or even airspeed...

Points taken. It is apparently a bit more sophisticated than anticipated. Thanks.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/08/2009 10:32 pm
The launch vehicles often don't shoot for zero angle of attack.  But must always keep angle of attack in limits.

Pitch is pretty much contantly changing.  Constant pitch rate gives the fuel optimum profile.

Great! That's exactly the kind of hints I was looking for; will look up the specific references for some sample values. Cheers.

Quote from: Jim
1.  LV's don't sense AOA or even airspeed...

Points taken. It is apparently a bit more sophisticated than anticipated. Thanks.

You are correct.  Ascent profile design is very complicated.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: mdo on 05/09/2009 12:46 pm
FYI, found some reference material in the meantime. Depicted below are the diagrams for pitch and AOA taken from the Saturn V Flight Manual. It shows that AOA peaks at 4 deg shortly after clearing the tower. Then it stays below 1 deg during atmospheric flight. So, the Saturn V first stage is so kind to behave following my intuition: That is, actively pitch over to some precomputed attitude early on and then just duck the oncoming airflow.

The same with Atlas V; the payload user manual says:
At an altitude of 244 m (800 ft) and time from liftoff greater than 10 seconds, the vehicle begins its initial pitch-over phase. At approximately 2,438 m (8,000 ft), the vehicle enters into a nominal zero-pitch and zero-yaw angle-of-attack phase to minimize aerodynamic loads.

It also mentions an optional alpha-bias angle-of-attack steering mode between 80-120 Kft to account for wind.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/09/2009 01:23 pm
FYI, found some reference material in the meantime. Depicted below are the diagrams for pitch and AOA taken from the Saturn V Flight Manual. It shows that AOA peaks at 4 deg shortly after clearing the tower. Then it stays below 1 deg during atmospheric flight. So, the Saturn V first stage is so kind to behave following my intuition: That is, actively pitch over to some precomputed attitude early on and then just duck the oncoming airflow.

The same with Atlas V; the payload user manual says:
At an altitude of 244 m (800 ft) and time from liftoff greater than 10 seconds, the vehicle begins its initial pitch-over phase. At approximately 2,438 m (8,000 ft), the vehicle enters into a nominal zero-pitch and zero-yaw angle-of-attack phase to minimize aerodynamic loads.


No, you are still getting it wrong.   Read the top chart, attitude goes from 0 to 60 degree

The pitch rate doesn't go to zero. there still is one (the actual attitude is changing), it is just at a slow rate as to not generate an angle of attack.  This is not "ducking".   The incoming flow is the velocity vector.

A vehicle could fly straight up and hence would not generate an angle of attach.   But because a more efficient trajectory requires the vehicle to go horizontally eventually, the vehicle has to pitch over to turn the velocity vector.  The rate at which it does is a function of structural limitations and the propulsion system.

Notice for Atlas is says "phase".  The phase is short
Title: Re: Basic Rocket Science Q & A
Post by: mdo on 05/09/2009 02:46 pm
...The pitch rate doesn't go to zero.

Sure, it was never claimed that it's zero. My initial point in reply 178 was about initiating the gravity turn and how much it typically takes to do so. It was not well phrased though because the term pitch over phase was missing from my vocabulary. AOA vs. pitch rate is clarified. Thanks for your patience.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/09/2009 02:48 pm
Anything you're going to find publicly is also going to be very generic.  In the real world, every single one of these is going to be mission unique.

It's like the cone vs the black line in hurricane forecasting.  You seem to be trying to calculate a black line that's the same for every mission.  In reality, there are just bounds on what the rocket can do (the cone).  How it flies on launch day, the black line withing the cone, is subject to a long list of variables, requirements and optimizations of same.  Where one rocket might be flying a steady zero or non-zero alpha, another one would be pitching.  It's truly unique to each mission.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 05/09/2009 07:40 pm
Anything you're going to find publicly is also going to be very generic.  In the real world, every single one of these is going to be mission unique.

It's like the cone vs the black line in hurricane forecasting.  You seem to be trying to calculate a black line that's the same for every mission.  In reality, there are just bounds on what the rocket can do (the cone).  How it flies on launch day, the black line withing the cone, is subject to a long list of variables, requirements and optimizations of same.  Where one rocket might be flying a steady zero or non-zero alpha, another one would be pitching.  It's truly unique to each mission.

Antares
That is perhaps the very best example I have ever seen for any rocket's actual performance vs. its predicted performance.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/09/2009 08:03 pm
Anything you're going to find publicly is also going to be very generic.  In the real world, every single one of these is going to be mission unique.

It's like the cone vs the black line in hurricane forecasting.  You seem to be trying to calculate a black line that's the same for every mission.  In reality, there are just bounds on what the rocket can do (the cone).  How it flies on launch day, the black line withing the cone, is subject to a long list of variables, requirements and optimizations of same.  Where one rocket might be flying a steady zero or non-zero alpha, another one would be pitching.  It's truly unique to each mission.

Antares
That is perhaps the very best example I have ever seen for any rocket's actual performance vs. its predicted performance.

One of the jokes about trajectories and COLA's is that the one trajectory that the LV is not going to fly is the predicted one
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/12/2009 02:16 pm
How much easier would it be to make a battleship inline version of the ET than the real thing? Assuming it would still have enough payload to launch an Orion to the ISS. Since J-130 payload is predicted to be around 60mT, you'd have more than 30mT to play with.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/12/2009 02:18 pm
How much easier would it be to make a battleship inline version of the ET than the real thing? Assuming it would still have enough payload to launch an Orion to the ISS. Since J-130 payload is predicted to be around 60mT, you'd have more than 30mT to play with.

That is a Direct question
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/12/2009 02:25 pm
I was hoping for an unbiased answer.
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 05/12/2009 02:31 pm
It's still a Direct J-130 question and there's no logic in expecting an unbiased answer here only because it's a general thread.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/12/2009 02:32 pm
Very well, I'll ask it there.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/13/2009 07:37 pm
Would a kerosene/LOX upper stage be easier to reuse on orbit than a LH2/LOX stage?
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 05/14/2009 02:07 pm
Would a kerosene/LOX upper stage be easier to reuse on orbit than a LH2/LOX stage?

I think you have coking issues (carbon buildups) with hydrocarbons that will limit how many times the engine can be reused. LH/LOX does not have "that" issue. With time and money you can design arround any issue.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/14/2009 02:18 pm
And for a wet workshop?
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 05/14/2009 02:40 pm
And for a wet workshop?

Good luck getting that Kerosene smell out of the workshop. On the plus side, the Kero tank is much smaller than the LOX tank, so if you only use the LOX tank it is not an issue. With an LH/LOX stage, you will have a much bigger "workshop" due to the low density of LH and the fact you can use both tanks.

Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/16/2009 02:31 am
Do RP engines have to be cleaned after the acceptance hot fire test?  If not, then I fail to see how coking would be an issue on a properly designed and operated engine.
Title: Re: Basic Rocket Science Q & A
Post by: Notriverse on 05/17/2009 03:42 pm
Hello to everyone, I ve got 2 questions... 1. How fast is a Satelit? 2. Why cant a satelit collide with another one?
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/17/2009 06:21 pm
Hello to everyone, I ve got 2 questions... 1. How fast is a Satelit? 2. Why cant a satelit collide with another one?

About 17,000 miles/hours or 25,000 feet/second -- for the ones in low earth orbit.  And they can and do collide with each other.  When they do they turn into many, many smaller pieces and it makes a bit mess.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: MKremer on 05/17/2009 06:25 pm
Hello to everyone, I ve got 2 questions... 1. How fast is a Satelit? 2. Why cant a satelit collide with another one?

1) http://en.wikipedia.org/wiki/Orbital_speed

2) Satellites can collide - 2 of them did just a few months ago with stories about them all over the media and online.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/22/2009 12:24 am
I've been reading a bit about augmented hydrazine thrusters and the discussion seems to focus on monopropellant thrusters. Is there a fundamental reason why bipropellant hydrazine thrusters couldn't benefit from thrust augmentation through electrical heating?
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/22/2009 04:04 am
What kind of power requirements are we talking about?  Most spacecraft are pretty power-limited.  It would be a trade of battery mass, I assume, against propellant mass.  Or if the power was drawn straight off the panels, would it be an appreciable gain in satellite life?  Is that the application envisioned?
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/22/2009 02:15 pm
Well, the background was I was trying to see if there was anything you could do to increase the Isp of storables if you had plentiful power, say next to a solar power satellite. This would only make sense if you had medium thrust. You can already do 500-600 s Isp with arcjet thrusters and ammonia or hydrazine, but with very little thrust. And of course you can do high thrust with storables, but only with mediocre Isp.

I did some more googling and found that people have tried bipropellant arcjet thrusters, but that there are problems with soot deposition. This wouldn't be a problem with hydrazine as a fuel, just with MMH or UDMH, but there may be other reasons why it's not done with hydrazine.
Title: Re: Basic Rocket Science Q & A
Post by: duane on 05/24/2009 02:59 am
Got a crazy question that may have been answered already.

It's regarding the technology and manufacturing capability of private rockets capable of placing objects in orbit. Namely SpaceX's Falcon 1

My question is how much of a step up in technology or avionics etc would need to go into the rocket system to convert it to a small ICBM ?  I understand the guidance and rentry system would need to built/upgraded.

Just musing about how dangerous the technology, plans etc of these private rockets are if they fall into the hand of states such as Iran, Syria etc.. ?

Or is it a order of magnitude more difficult to get a ICBM working ?

Thanks for any high level insight on this.

Thanks a bunch
Duane
Title: Re: Basic Rocket Science Q & A
Post by: StarStuff on 05/24/2009 09:11 am
The first ICBMs were built using '40s and '50s technology, then it was a step up to man-rated rockets.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/24/2009 01:52 pm

My question is how much of a step up in technology or avionics etc would need to go into the rocket system to convert it to a small ICBM ?

None.  A space launch vehicle is just a ICBM with longer range.
Title: Re: Basic Rocket Science Q & A
Post by: Eerie on 05/24/2009 04:07 pm
Just musing about how dangerous the technology, plans etc of these private rockets are if they fall into the hand of states such as Iran, Syria etc.. ?

SpaceX will have to use Falcon-9 with Dragon to nuke someone. ICBM needs a reentry vehicle.

Hmm, I wonder how many megatons you can stuff into Dragon...
Title: Re: Basic Rocket Science Q & A
Post by: duane on 05/24/2009 05:14 pm

My question is how much of a step up in technology or avionics etc would need to go into the rocket system to convert it to a small ICBM ?

None.  A space launch vehicle is just a ICBM with longer range.

Well that kinda sounds frightening!

Could something like a Falcon be accurate enough with just a dumb reentry vehicle (no manuevering) to hit a county (30X30 miles) ?

Do modern launch vehicles still use Intertial Navigation Units or do they use GPS systems, or a combination of both ?

Thanks a bunch!
Duane
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/24/2009 05:31 pm

My question is how much of a step up in technology or avionics etc would need to go into the rocket system to convert it to a small ICBM ?

None.  A space launch vehicle is just a ICBM with longer range.

Well that kinda sounds frightening!

snip

Duane

Thus all the fuss about ITAR.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/24/2009 05:36 pm

1. Could something like a Falcon be accurate enough with just a dumb reentry vehicle (no manuevering) to hit a county (30X30 miles) ?

2.  Do modern launch vehicles still use Intertial Navigation Units or do they use GPS systems, or a combination of both ?


1.  yes

2. INS
Title: Re: Basic Rocket Science Q & A
Post by: duane on 05/24/2009 06:34 pm
Thanks Jim and Danny

I wont bother another question about accuracy. I have a feeling its probably a order or two of magnitude better than 30X30 miles, which is even more scary a thought.

Duane

Title: Re: Basic Rocket Science Q & A
Post by: ginahoy on 05/25/2009 03:13 am
Does anyone know the typical g forces astronauts were/are subjected to in the various vehicles during ascent and reentry? Seems like I recall during Project Mercury, the centrifuge trained up to about 10 g's. Soyuz ballistic reentry is probably the highest, but I don't think it's anywhere close to 10.
David
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/25/2009 03:24 am
Does anyone know the typical g forces astronauts were/are subjected to in the various vehicles during ascent and reentry? Seems like I recall during Project Mercury, the centrifuge trained up to about 10 g's. Soyuz ballistic reentry is probably the highest, but I don't think it's anywhere close to 10.
David

8 with Gemini and 6 with Apollo during ascent
Title: Re: Basic Rocket Science Q & A
Post by: hop on 05/25/2009 03:33 am
Does anyone know the typical g forces astronauts were/are subjected to in the various vehicles during ascent and reentry? Seems like I recall during Project Mercury, the centrifuge trained up to about 10 g's. Soyuz ballistic reentry is probably the highest, but I don't think it's anywhere close to 10.
David
8-9 for ballistic Soyuz reentries. This shouldn't really qualify as "typical" since it only happens in the event of a malfunction. ISTR a nominal lifting Soyuz reentry peaks around 4. Launch escape is 15+.

Apollo lunar reentries were in the 6-7 range.
Title: Re: Basic Rocket Science Q & A
Post by: ginahoy on 05/25/2009 04:14 am
8-9 for ballistic Soyuz reentries. This shouldn't really qualify as "typical" since it only happens in the event of a malfunction. ISTR a nominal lifting Soyuz reentry peaks around 4.

Apollo lunar reentries were in the 6-7 range.


What's the nominal Shuttle reentry force?

Here's an interesting interview Whitson gave to CBS News, in which she discusses the BE at length:

http://www.shroomery.org/forums/showflat.php/Number/8356943

Quote
Launch escape is 15+.

Holy Moses!! Talk about a kick in the pants!
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 05/25/2009 04:27 am
8-9 for ballistic Soyuz reentries. This shouldn't really qualify as "typical" since it only happens in the event of a malfunction. ISTR a nominal lifting Soyuz reentry peaks around 4.

Apollo lunar reentries were in the 6-7 range.


What's the nominal Shuttle reentry force?

About 1.5 during entry. Gets a little higher during TAEM but never above 2.0.

http://spaceflight.nasa.gov/shuttle/reference/green/entare.pdf
Title: Re: Basic Rocket Science Q & A
Post by: lmikers on 05/25/2009 04:55 am
Also, if I may note, the (de)(ac)celeration figure should also be considered together with the *duration* of that peak load (same goes for thermal loads)  Also whether it's negative or positive.  A very short (milliseconds) even a 15+G peak may be more survivable than a longer 8G peak (same for thermal loads) 

[edit] obviously, after a certain threshold of 'neutral' levels of G's, like 4 perhaps, which can be benign for a while 

[edit2] And *up to* a threshold, of course, at which no short duration will help as things will simply be ripped out/pushed in instanteneously
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/25/2009 03:36 pm
Does anyone know the typical g forces astronauts were/are subjected to in the various vehicles during ascent and reentry? Seems like I recall during Project Mercury, the centrifuge trained up to about 10 g's. Soyuz ballistic reentry is probably the highest, but I don't think it's anywhere close to 10.
David

The "winner" is a ballistic ascent abort entry and ballistic lunar entry.  These go up to about 20G.  The Russian have done this at least once and the US tested them in centrifuges in the 60s.  Believe it or not, they are OK.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: ginahoy on 05/25/2009 04:21 pm
... and ballistic lunar entry.

Elaborate please? Sounds like no retros, therefore surviving the g forces would be a moot point.

Edit - Sorry, I took 'lunar entry' literally. I'm guessing you meant reentry. (I need another cup of java)
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/25/2009 04:25 pm
The "winner" is a ballistic ascent abort entry and ballistic lunar entry. 

I read that a normal lunar reentry is mostly ballistic anyway, with only the RCS being used for attitude control. Is this true? And if so, how much difference does the attitude control make?
Title: Re: Basic Rocket Science Q & A
Post by: StarStuff on 05/25/2009 05:39 pm
Apollo attitude during entry was determined by the offset of CG from the axis of symmetry. It could not be changed in flight. The guidance and control were accomplished with roll that moved the direction of lift.

Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/25/2009 05:57 pm
Apollo attitude during entry was determined by the offset of CG from the axis of symmetry. It could not be changed in flight. The guidance and control were accomplished with roll that moved the direction of lift.

Is roll rotation around the axis of symmetry? If so, how can it influence lift?
Title: Re: Basic Rocket Science Q & A
Post by: Lee Jay on 05/25/2009 06:07 pm
Apollo attitude during entry was determined by the offset of CG from the axis of symmetry. It could not be changed in flight. The guidance and control were accomplished with roll that moved the direction of lift.

Is roll rotation around the axis of symmetry? If so, how can it influence lift?

If the CG is not located on the geometric axis the craft will come in "crooked" with aerodynamics compensating for the CG offset.  Roll would then control in which direction the "crookedness" (don't know if they called it skew, skid, slip or just AoA) was pointing thereby controlling the direction of lift.
Title: Re: Basic Rocket Science Q & A
Post by: ginahoy on 05/25/2009 06:17 pm
Is roll rotation around the axis of symmetry? If so, how can it influence lift?

Yaw is rotation around about the 'vertical' axis, which in the case of a capsule, is the axis of symmetry. I get confused because with a capsule, the velocity vector is more or less the same as the vertical axis, whereas with an airplane, they are at right angles. 'Vertical' can be ambiguous because it's referential.
Title: Re: Basic Rocket Science Q & A
Post by: StarStuff on 05/25/2009 06:42 pm
Think of airplane that has no control of pitch with the nose always  trimmed to a positive attitude.

You would have to descend by rolling upside down. That's what Apollo did.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/25/2009 09:27 pm
The "winner" is a ballistic ascent abort entry and ballistic lunar entry. 

I read that a normal lunar reentry is mostly ballistic anyway, with only the RCS being used for attitude control. Is this true? And if so, how much difference does the attitude control make?

A guided entry uses roll to put the lift vector largely up.  In a ballistic entry with a capsule, roll rate is set to a constant value and the end result is no effective lift.

In a controlled entry, the capsule ends up flying in much thinner air for a longer period of time.  I think the guided lunar entries were about 9 Gs, while a ballistic is about 20.  I don't think Apollo had a backup lunar ballistic mode.  I saw one short blurb about the Service module putting a dead Command Module heatshield forward and giving it a roll rate, but never saw enough info in the old Apollo documents to know if this procedure was taken to the point it was ready to use.  This would have been an option in Apollo 13, but the LM would have had to have done the maneuvering and I don't know if there would have been enough time to get back into the command module, close the hatch, and jettison the LM in time.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 05/25/2009 09:39 pm
The "winner" is a ballistic ascent abort entry and ballistic lunar entry. 

I read that a normal lunar reentry is mostly ballistic anyway, with only the RCS being used for attitude control. Is this true? And if so, how much difference does the attitude control make?

A guided entry uses roll to put the lift vector largely up.  In a ballistic entry with a capsule, roll is set to a constant value and the end result is no effective lift.

Nit: A capsule with an offset CG performs a ballistic entry by setting roll *rate* to a constant, so that the direction of the lift vector averages out.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/25/2009 09:44 pm
Nit: A capsule with an offset CG performs a ballistic entry by setting roll *rate* to a constant, so that the direction of the lift vector averages out.

Ah, so does ballistic mean "as if there were no atmosphere and no propulsion" instead of "without propulsion" in this context?
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 05/25/2009 09:46 pm
Nit: A capsule with an offset CG performs a ballistic entry by setting roll *rate* to a constant, so that the direction of the lift vector averages out.

Ah, so does ballistic mean "as if there were no atmosphere and no propulsion" instead of "without propulsion" in this context?

Don't think so; "ballistic" is still reliant on drag for deceleration. I think of it as "drag only, no lift".
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/25/2009 09:49 pm
Is there a special name for a trajectory when you have no propulsion because something's wrong with your RCS?
Title: Re: Basic Rocket Science Q & A
Post by: StarStuff on 05/26/2009 02:23 am
A ballistic entry has zero lift. The only thing you can control is the angle of entry into the atmosphere. The astronauts would experience at least 17g's. NASA considered that to be not survivable.

The Apollo entry was close to ballistic but with a small change to generate some lift: the CG was offset resulting in a small trim attitude. The lift kept the CM in the upper atmosphere long enough so the astronauts would not experience much over 7g. The lift vector could only be controlled by rotating the CM.

Source: http://www.mae.usu.edu/aerospace/publications/Reno_rev_whitmore.pdf

http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20060045567_2006144506.pdf
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/26/2009 03:31 am
The "winner" is a ballistic ascent abort entry and ballistic lunar entry. 

I read that a normal lunar reentry is mostly ballistic anyway, with only the RCS being used for attitude control. Is this true? And if so, how much difference does the attitude control make?

A guided entry uses roll to put the lift vector largely up.  In a ballistic entry with a capsule, roll is set to a constant value and the end result is no effective lift.

Nit: A capsule with an offset CG performs a ballistic entry by setting roll *rate* to a constant, so that the direction of the lift vector averages out.


Not a nit.  A huge difference.  Thanks for the correction.  I edited my post.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/26/2009 03:33 am
A ballistic entry has zero lift. The only thing you can control is the angle of entry into the atmosphere. The astronauts would experience at least 17g's. NASA considered that to be not survivable.

The Apollo entry was close to ballistic but with a small change to generate some lift: the CG was offset resulting in a small trim attitude. The lift kept the CM in the upper atmosphere long enough so the astronauts would not experience much over 7g. The lift vector could only be controlled by rotating the CM.

Source: http://www.mae.usu.edu/aerospace/publications/Reno_rev_whitmore.pdf

http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20060045567_2006144506.pdf


The NASA doctors approved 20 G entries for aborts in about 2004.  We looked very seriously at ballistic entries, but I don't think they made the final cut into Orion.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: ginahoy on 05/26/2009 03:55 am
I would imagine a 20 g force on a launch abort would be very short in duration -- only in the moments after the initial firing. Sorta like an ejection seat on some fighter jets. Correct?
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 05/26/2009 04:13 am
I would imagine a 20 g force on a launch abort would be very short in duration -- only in the moments after the initial firing. Sorta like an ejection seat on some fighter jets. Correct?

You do get a big g-pulse at LAS firing, but don't forget about the subsequent re-entry, which will be steeper than normal due to the lower horizontal velocity.
Title: Re: Basic Rocket Science Q & A
Post by: ginahoy on 05/26/2009 04:18 am
You do get a big g-pulse at LAS firing, but don't forget about the subsequent re-entry, which will be steeper than normal due to the lower horizontal velocity.

I was only thinking about an escape at launch. I guess the system could be used all the way up to orbit.
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 05/26/2009 04:25 am
You do get a big g-pulse at LAS firing, but don't forget about the subsequent re-entry, which will be steeper than normal due to the lower horizontal velocity.

I was only thinking about an escape at launch.

For that particular case you're correct; there would be one g-pulse at LAS firing but the spacecraft would not get high enough to experience a re-entry.

Quote
I guess the system could be used all the way up to orbit.

Systems with a LAS generally would use it until shortly after staging, then jettison it. After that, the stack acceleration is generally low enough (and the atmosphere thin enough) that the spacecraft can get away from the LV without it, so it is jettisoned to save mass. That is how Apollo/Saturn worked and how Ares/Direct would work.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/26/2009 11:45 am
I would imagine a 20 g force on a launch abort would be very short in duration -- only in the moments after the initial firing. Sorta like an ejection seat on some fighter jets. Correct?

Longer 5 seconds or so and it is 15g's.
Title: Re: Basic Rocket Science Q & A
Post by: hop on 05/27/2009 05:23 am
You do get a big g-pulse at LAS firing, but don't forget about the subsequent re-entry, which will be steeper than normal due to the lower horizontal velocity.
Especially if things go wrong like Soyuz 18a (http://en.wikipedia.org/wiki/Soyuz_18a). They actually aborted after LAS jettison, but a similar effect.

A Zond ballistic return (i.e. if the guidance failed in a skip re-entry, which it did on a couple of tests) was 20+ as well AFAIK.
Title: Re: Basic Rocket Science Q & A
Post by: veryrelaxed on 05/27/2009 05:55 am
20+G for how long.  The final objective upon re-entry is to dissipate the TREMENDEOUS both potential (altitude) AND kinetic energy (angular speed) components into other forms of energy (heat mostly + electro magnetic+air friction<-heat again)  You can do it at lower (+/-)G levels lasting longer, or at higher (+/-)G lasting short time.  And that will determine your vehicle's descent profile.  Biologically speaking a high absolute velocity change in as short period as possible is preferable to extended Gs over entire descent.  (there is much more for mititgation as to the body orientation to the deceleartion vector, and more...)
Title: Re: Basic Rocket Science Q & A
Post by: C5C6 on 05/27/2009 09:14 pm
I'm embarassed to ask this but I never knew, are ELV stages recovered for burial or recycle or any other use (excepting reuse of course)???
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/27/2009 10:13 pm
I'm embarassed to ask this but I never knew, are ELV stages recovered for burial or recycle or any other use (excepting reuse of course)???

The only ones I know of are some solids on the Ariane that land on land and sometimes they take one down on chutes that land on water.  Not to reuse, but to look at for problems in the design or manufacturing process.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/27/2009 10:15 pm
20+G for how long. 

snip


For about 5 to 10 seconds.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: C5C6 on 05/27/2009 10:57 pm
I'm embarassed to ask this but I never knew, are ELV stages recovered for burial or recycle or any other use (excepting reuse of course)???

The only ones I know of are some solids on the Ariane that land on land and sometimes they take one down on chutes that land on water.  Not to reuse, but to look at for problems in the design or manufacturing process.

Danny Deger
so the ones that don't have chutes and crash on land or water are not recovered not even for cleanup??? I really find it hard to believe they keep throwing rocket stages to some places and are not removed....you'd have a pile of rocket stages in the atlantic only because of KSC launches.....

another embarassing question: most of second or third stages burn up on reentry don't they?? first stages do not acquire enough speed for a destructive reentry do they???
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 05/27/2009 11:22 pm
I'm embarassed to ask this but I never knew, are ELV stages recovered for burial or recycle or any other use (excepting reuse of course)???

The only ones I know of are some solids on the Ariane that land on land and sometimes they take one down on chutes that land on water.  Not to reuse, but to look at for problems in the design or manufacturing process.

Danny Deger
so the ones that don't have chutes and crash on land or water are not recovered not even for cleanup??? I really find it hard to believe they keep throwing rocket stages to some places and are not removed....you'd have a pile of rocket stages in the atlantic only because of KSC launches.....

another embarassing question: most of second or third stages burn up on reentry don't they?? first stages do not acquire enough speed for a destructive reentry do they???

Almost.  As far as I know the Ariane 4 solids land on land in South America and they are collected and put in a junk pile.  The ones that land in water just stay there.

And you are correct that upper stages for the most part burn up on entry, but some parts make it to the ground.  In the past, upper stages that make orbit often had their tanks over pressure and they blew up making a bit mess on orbit.  Today's upperstages relieve the pressure in the tanks so they stay in one easy to track and avoid piece.

I am not an expert on this, so I might be wrong in a couple of places.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/28/2009 03:49 am
If it made economic sense, they would recover them.  It doesn't.  ELVs also provide a continuous opportunity to improve designs.  It's the state of the art, every time, not 30 year old technology (ahem).
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 05/28/2009 12:42 pm
Well there was that partial Titan II stage that recovered back in the 1960's. Discussion of it and pictures are somewhere in the historical threads. Russian first stages land on land and are cut up and sold as scrap. Again there is a cool set of photos linked somewhere in the historical section.

But rockets in the ocean are a drop in the bucket. I once saw (and don't have a source) a claim that there are over a million ships at the bottom of the ocean. Though it raises my eyebrow, they definately number in the thousands...
Title: Re: Basic Rocket Science Q & A
Post by: DMeader on 05/28/2009 01:37 pm
I've seen some of those photos from Kazakhstan, downrange from Baikonur... sections of rocket bodies re-purposed into shelters, storage buildings, etc.

Would be interesting to see what remains of spent S1-C stages and whatnot on the bottom of the Atlantic. I wonder how much of it would be recognizable?
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/28/2009 09:32 pm
One of the downrange security forces actually shows up at the impact points on Russian rockets to ensure the first group that gets to the hardware has first dibs and that there is not a generic melee over it.

There are beach bars in the Caribbean and Bahamas that have pieces of fairings and tanks that are used as tables and chairs.  I'm not sure if I've seen pictures or if I have a good imagination.
Title: Re: Basic Rocket Science Q & A
Post by: mars sts-107 on 05/30/2009 05:27 pm
what if you could have the rocket engines on the launch pad so it would not a big first stage ???
Title: Re: Basic Rocket Science Q & A
Post by: Spacenick on 05/30/2009 05:35 pm
rockets are not blown into space by a stream of air. They push them selves upwards by accelerating mass downwards (this works in any direction by the way) they work by the newtonian principle of f=ma where a is the acceleration (it's a vector meaning it implies a direction) m is the mass and f is the force working on the rocket.

Therefor you can't put the engines anywhere else then on the rocket itself.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/31/2009 04:40 am
I have a question related to Ares I being used by NASA to relearn rocket development skills.

Let me first explain the background for my question and then get back to the Basic Rocket Science aspect.

Ares I has been criticised for many reasons and one of those reasons is that even if it ever flies, it would duplicate an existing capability. Other alternatives would be inline or side-mount SDLVs. These could not be dismissed as simply duplicating existing capability, the problem being that they exceed existing capabilities by so much that there are no payloads for it. The SDLV would not be useful until an upper stage were developed for it.

One interim solution I can think of is to use an existing upper stage (Centaur or DHCSS) to move payloads beyond LEO. Centaur would be good enough to get an unmanned and partially fueled Orion to L1 on a very slow but energy-efficient trajectory. DHCSS would be good enough for manned missions. NSF poster Jim has repeatedly expressed grave doubts as to whether MSFC would be capable of integrating this existing upper stage with the shuttle stack. The problem doesn't seem to be that the concept itself is flawed, but more that MSFC lacks recent rocket development skills.

So I've thinking about an alternative, one that fits with my continuing fascination with hypergolics. It may be without merit, but here goes anyway:

As I understand it, historically hypergolic rockets have preceded the more difficult cryogenic ones. Given that MSFC needs to relearn rocket development skills, would it help if they started with a hypergolic upper stage? And what if they built a flying battleship version first? Say no fancy Al-Li friction stir welding etc, but plain old stainless steel with plenty of margins? The thing is, an SDLV has such an enormous LEO payload that you could easily absorb the inefficiency of hypergolics. You could still do a one-launch Orion mission to L1. You might not win any prizes in the high-tech department, but if it flies, who cares?

So after this lengthy introduction, let me get to my question:

Would such a hypergolic "training stage" be substantially easier/cheaper/quicker/less risky to develop than a non-battleship cryogenic upper stage? Would it help if they started from an existing design? Or could they perhaps scale up the Orion SM drastically to turn it into a hypergolic EDS? If the answer is 'no', could you explain why?
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/31/2009 05:02 am
Another question regarding hypergolics: the other day Danny Deger mentioned that a hypergolic stage can act as a sort of shield against an exploding first stage, because it would burn, not explode. Is this something that holds for all hypergolic propellant combinations or something that's special to MMH/NTO?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/31/2009 11:51 am
There isn't a mission/need for another upperstage nor the money to 'train" MSFC.

My beef with MSFC is that they should be doing any LV development. Civil service performing such tasks is not needed.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/31/2009 03:14 pm
There isn't a mission/need for another upperstage nor the money to 'train" MSFC.

My beef with MSFC is that they should be doing any LV development. Civil service performing such tasks is not needed.

Agreed, but I'm working from the assumption that politicians will want to preserve the shuttle stack anyway and trying to combine this with incrementalism to reduce risk and move IOC forward. This works very well in my own field of software development.

So

- take a shiny fully-featured Altair as currently envisaged
- now downgrade it to all-hypergolic
- now downgrade it to all metal structures (perhaps even stainless steel) and no composites
- now downgrade it to a single-stage lander
- now remove the landing gear, remove the need for a deeply throttleable engine and remove all the complexities related to landing from the flight software and see if you can now use the Orion engine
- now downgrade the crew compartment to a pressurised cargo carrier
- now downgrade the pressurised cargo carrier to an unpressurised cargo carrier
- now remove the cargo carrier entirely
- now downgrade to a pressure fed system if you hadn't done so already
- now see if you can merge it with the Orion SM

Would you agree that each step above would be a simplification?

Now focusing on the basic rocket science aspect again:

When you are designing a rocket stage, what fraction of the work is taken up by upgrading from a battleship version to an optimised version? Would using stainless steel allow for earlier initial operational capability, at the expense of lower performance and later full operational capability? How much of the vibrational analysis could you eliminate? Are hypergolics substantially easier than cryogenics? If not, how about LOX/RP-1? Is pressure-fed substantially easier than pump-fed?

A fool can ask more questions than a wise man can answer...
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/31/2009 03:46 pm

- take a shiny fully-featured Altair as currently envisaged
- now downgrade it to all-hypergolic
- now downgrade it to all metal structures (perhaps even stainless steel) and no composites
- now downgrade it to a single-stage lander
- now remove the landing gear, remove the need for a deeply throttleable engine and remove all the complexities related to landing from the flight software and see if you can now use the Orion engine
- now downgrade the crew compartment to a pressurised cargo carrier
- now downgrade the pressurised cargo carrier to an unpressurised cargo carrier
- now remove the cargo carrier entirely
- now downgrade to a pressure fed system if you hadn't done so already
- now see if you can merge it with the Orion SM

Would you agree that each step above would be a simplification?

Now focusing on the basic rocket science aspect again:

When you are designing a rocket stage, what fraction of the work is taken up by upgrading from a battleship version to an optimised version? Would using stainless steel allow for earlier initial operational capability, at the expense of lower performance and later full operational capability? How much of the vibrational analysis could you eliminate? Are hypergolics substantially easier than cryogenics? If not, how about LOX/RP-1? Is pressure-fed substantially easier than pump-fed?


That is not feasible, too many changes in between.  An upperstage has different design constraints than an lander
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/31/2009 04:12 pm
That is not feasible, too many changes in between.

I don't believe you, you can turn a word processor into a spreadsheet this way. It may take a while of course. Feel free to take a year for each step. Personally I'd rather have them undertake ten one-year projects than one ten-year project. I'd expect the ten one-year projects to take twenty years and deliver working hardware every two years (=twice in a presidential election cycle). I'd expect the ten-year project to take 7 years after which it would be cancelled without ever delivering working hardware.

Efficient? No. But a lot more efficient and less risky than what we have now. And less risk and earlier returns mean higher risk adjusted net present value.

Quote
  An upperstage has different design constraints than an lander

Can you elaborate? And why couldn't you do those changes incrementally?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/31/2009 04:54 pm
But a lot more efficient and less risky than what we have now.

It is not efficient at all. There is no use for the intermediate steps.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/31/2009 04:57 pm
It is not efficient at all. There is no use for the intermediate steps.

What do you mean no use? It's a working spacecraft every step of the way.

I know you want to get MSFC out of it and I agree, but what if that's not possible?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/31/2009 05:03 pm
It is not efficient at all. There is no use for the intermediate steps.

What do you mean no use? It's a working spacecraft every step of the way.


It isn't a working spacecraft.   The Orion SM has no guidance system.
The SM is not like upper stage nor its a spacecraft.  It is purpose built for the Orion and not very adaptable. The avionic in the CM are not made for vacuum.

But more so there are no missions for the intermediate steps.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/31/2009 05:05 pm
I know you want to get MSFC out of it and I agree, but what if that's not possible?

Keep them away from spacecraft.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/31/2009 05:10 pm
Keep them away from spacecraft.

OK, not just launch vehicles, but spacecraft as well. Should spacecraft be left to KSC and JPL? How about fundamental research into engine technology then? Those metallised gel propellants sound interesting, would that be a valid activity for MSFC? How about ISRU propellants?
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 05/31/2009 05:39 pm
Jim, let me start by expressing my appreciation you're taking the time to answer my persistent questions.

For those of you who might be wondering, the topic of this discussion is to what degree incrementalism is applicable to rocket development. For software development incrementalism absolutely rocks. There are those who dispute this, but they are, shall we say, less correct than they might be.

Jim, who is an expert on all things rocket seems unimpressed by my attempts at an incremental plan. It might be that the problem is with my feeble attempts, but perhaps incrementalism just doesn't apply to rockets. In favour of incrementalism I offer the example of the Russian and European space programs and less confidently the way von Braun worked on the V-2.

It isn't a working spacecraft.   The Orion SM has no guidance system.
The SM is not like upper stage nor its a spacecraft.  It is purpose built for the Orion and not very adaptable. The avionic in the CM are not made for vacuum.

OK, then let's take the required bits from the CM. Or forget about the initial steps as Orion is probably too far along anyway. Assume we start from a working Orion SM + CM.

Quote
But more so there are no missions for the intermediate steps.

Ah, but the trick is to find a convincing mission. This is hard work. I say it can be done, especially if you have an expert doing it, instead of an interested amateur like yours truly.

So here's the new and improved plan with intermediate missions added:

- Orion on EELV - crew rotation to ISS
- Orion on battleship J-130/Aquila - backup crew rotation
- Orion + battleship SDLV + SSPDM - LEO servicing missions
- stretch limo version of Orion + battleship SDLV + SSPDM - MEO servicing missions
- (stretch limo Orion + optimised SDLV) /  (Orion + optimised SDLV + upper stage) + SSPDM - GEO servicing missions
- stretch limo SM + Altair crew compartment - GEO station pressurised resupply
- extend to Altair shuttle - L1 based missions to SEL-2, GEO, LLO and NEOs
- switch to deeply throttleable engine, add landing gear, extend flight software for landings - lunar surface missions
- switch to cryogenic engines - improved performance
- use composites - improved performance
Title: Re: Basic Rocket Science Q & A
Post by: DMeader on 05/31/2009 11:11 pm

So here's the new and improved plan with intermediate missions added:

- Orion on EELV - crew rotation to ISS
- Orion on battleship J-130/Aquila - backup crew rotation
- Orion + battleship SDLV + SSPDM - LEO servicing missions
- stretch limo version of Orion + battleship SDLV + SSPDM - MEO servicing missions
- (stretch limo Orion + optimised SDLV) /  (Orion + optimised SDLV + upper stage) + SSPDM - GEO servicing missions
- stretch limo SM + Altair crew compartment - GEO station pressurised resupply
- extend to Altair shuttle - L1 based missions to SEL-2, GEO, LLO and NEOs
- switch to deeply throttleable engine, add landing gear, extend flight software for landings - lunar surface missions
- switch to cryogenic engines - improved performance
- use composites - improved performance

Seems to me that making all those incremental steps and designing for missions that do not exist would be awfully, awfully expensive. Some of those steps (for instance, "use composites") should come at the very beginning, and others ("- switch to deeply throttleable engine, add landing gear, extend flight software for landings - lunar surface missions") would be better left to a vehicle designed for that mission when the requirement arose, rather than try to cut-and-paste existing into a frankenship.
Title: Re: Basic Rocket Science Q & A
Post by: kraisee on 05/31/2009 11:52 pm
DMeader,
I agree.   There is a point where incremental steps just make the systems more expensive in the long term.   The right balance needs to be struck.

Generally speaking, Martijn is on a good track and the approach simply needs refinement.

Purely from the DIRECT perspective, I'll raise question marks over the following items, either because there are no requirements at this time or because they aren't actually needed to be successful:-

* Stretch limo Orion -- not required, Orion has sufficient propellant for the job already
* GEO servicing -- Are there any satellites in GEO which were ever designed for human servicing?
* GEO station -- Humans and the Van Allen Belt do not mix well. An Earth Orbiting station is safer in LEO, plus how do any other nations get a crew to a station no longer in LEO?
* Given the need to develop new engines for Lunar Landings anyway, why duplicate that effort & cost by making non-cryo engines first and then planning to upgrade them to cryo later?   It would be cheaper to just bite the bullet at the start and make a single generic system -- and the schedule currently allows us to do that
* Composites -- is there any real *need* for them?   Planned or Proposed?   With DIRECT, we currently don't need them.

Ross.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 06/01/2009 01:12 am
I agree.   There is a point where incremental steps just make the systems more expensive in the long term.   The right balance needs to be struck.

It was more of a risk management thing and not wanting to wait ages before anything new flies.

Quote
Purely from the DIRECT perspective, I'll raise question marks over the following items, either because there are no requirements at this time or because they aren't actually needed to be successful:-

Interesting, because I came up with this with the need to preserve the shuttle stack in mind.

Quote
* Stretch limo Orion -- not required, Orion has sufficient propellant for the job already

The idea was to move Orion out of LEO and out of the way of commercial competitors asap, while making use of the enormous margins provided by any SDLV including a derivative of the ET. This allows the SDLV to demonstrate an immediate or at least early technical feat that had not been accomplished before. No need to fly MPLMs or ISS modules as filler missions, although the SDLV would remain available for that as a backup if necessary.

The stretch limo Orion was also meant to evolve into the Altair propulsion system slowly, meaning KSC had something to do even if a full Altair could not be funded.

Quote
* GEO servicing -- Are there any satellites in GEO which were ever designed for human servicing?

Not likely, but I suspect it would go down really well with the public. Inspecting historic satellites in a graveyard orbit would also be interesting. Or servicing training satellites, though that might be better done in MEO. These are example of "filler" mission that help you demonstrate new capabilities (as opposed to just systems) while more powerful and more difficult hardware is under development. On their own, these are probably not very good missions, and it would make sense to rack the old brain to see if you can come up with better ones. My old favourite of an Orion as a makeshift radiation research lab is one possibility. Test runs of Orion + Altair shuttles in LEO/MEO/GEO are also possible, but only after you had the shuttle.

Quote
* GEO station -- Humans and the Van Allen Belt do not mix well. An Earth Orbiting station is safer in LEO, plus how do any other nations get a crew to a station no longer in LEO?

Both are actually features, not bugs. :) Meant as a step towards an L1 station, where similar conditions apply. Being beyond the van Allens gives you an excellent location for radiation shielding research. This needs to be done anyway, better start now than wait for a NEO/moon mission. Prestige, no other nation can do this. Direct LOS communications allow for better telerobotics than even the ISS, since you could bypass TDRS. PR-aspect, the 'Obama-star', a permanently visible beacon of hope and change in the sky. Also meant to provide a permanent justification for the Shuttle stack, once commercial players reach LEO/L1. Excellent staging location for an Altair. It's still EOR, just a bit higher up :). Back and forth to L1 for 1.4 km/s, propulsive braking is a serious possibility, even with hypergolics, even with a battleship Altair. Phasing is much easier than in LEO. Single orbit rendez-vous, cargo/crew transfer, deorbit. Aborts are much quicker than at L1. I don't believe NASA's objections to L1R are serious. GEO staging takes away the excuses of not having a lifeboat and not being able to return quickly and provides them with a new excuse to preserve the shuttle stack indefinitely, meaning noone has to lose face. Eventually use an SDLV with an ACES upper stage to launch a properly modified Dream Chaser to GEO. Should ULA ever develop Atlas Phase 2 on its own dime, it would mean there would be a commercial market for an SDLV and it could be privatised - I don't see this happening in at least 25 years. If it does not, the shuttle stack remains firmly in government hands for a long time.

Quote
* Given the need to develop new engines for Lunar Landings anyway, why duplicate that effort & cost by making non-cryo engines first and then planning to upgrade them to cryo later?

Orion already has hypergolic engines as does the Altair ascent stage. No new engine development that wasn't already planned would be necessary and some (CECE) could be delayed. A single stage design is simpler than a two stage design. Hypergolics allow low-risk propellant depots, which enables an Altair precursor to act as a near-Earth shuttle, capable of performing NEO missions. And of course I desperately want those depots to enable commercial and international participation and I just don't believe DIRECT would be able to deliver on cryo depots - much as the team might want to. GEO staging saves the shuttle stack's bacon anyway, noone needs to worry.

And a refuelable Orion + Altair shuttle allows for an early unmanned Apollo 8 to Mars.

All the cryogenics other than the ET-derivative could be moved off the critical path - including Centaur/DHCSS integration.

Quote
* Composites -- is there any real *need* for them?   Planned or Proposed?   With DIRECT, we currently don't need them.

I saw that in a late 2008 NASA Altair presentation I found on the web. It was one of about 5-7 bullet items of critical technologies NASA needed to master or mature. I can dig up the details if you like. If it's not necessary, great. I moved it to the end of the list because it didn't seem critical once you accepted depots and gateway stations.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 06/01/2009 01:30 am
Seems to me that making all those incremental steps and designing for missions that do not exist would be awfully, awfully expensive.

Likely expensive, don't know about awfully expensive. Constantly flight-certifying new designs would certainly take a lot of effort. I suspect this is a major difference with software development, where most of the equivalent of that can be automated. On the plus side: you'd get really good at it and the guys doing it would have plenty of job security. :)

But the real point is that the individual steps would be much smaller than the existing steps. One or more steps could happen every election cycle. Is there anyone here who believes Ares I + Orion will fly before Obama leaves office? Each step could be undertaken in much less time, with less risk and with a much lower yearly budget. It's yearly budgets and near-term milestones politicians and the public care about. Total amounts, who cares?

Quote
Some of those steps (for instance, "use composites") should come at the very beginning, and others ("- switch to deeply throttleable engine, add landing gear, extend flight software for landings - lunar surface missions") would be better left to a vehicle designed for that mission when the requirement arose, rather than try to cut-and-paste existing into a frankenship.

Yeah, the idea was to have the single ship evolve into a separate Altair, Orion and perhaps a hypergolic (and later cryogenic) small (and later large) EDS. Much like different species evolve from a common ancestor. The stretch limo was an intermediate step, likely destined to be preserved in the fossil record only. The idea is to get from an SDLV + Orion but without an upper stage to a lunar architecture in small steps, with every step being a working system and leading to increased capability or performance. The NEO mission for instance is there in order to be able to do something useful with only a subset of the systems required for a lunar landing.
Title: Re: Basic Rocket Science Q & A
Post by: mrskyking737 on 06/03/2009 02:47 am
I have a question about reentry heating that can't seem to find easily, and have no rocketry background.  Forgive me if the question turns out to be stupidly easy. 

I was thinking about how orbital vehicles need to withstand several thousand degrees during reentry, but suborbital vehicles do not.  I don't have any way to quantify how fast is too fast to not need TPS protection. 

I understand that the White Knight reached 367k and mach 3.09 on it's first flight without anything special in regard to it's construction.  Could the same craft with increased duration been able to handle mach 4 and 367k?  Mach 6 and 450k?  How much is too much for simple aircraft materials to be able to handle the heat? 

How can we figure out in simple fashion where any given speed & corresponding given altitude will experience a certain calculated amount of heat? 

Title: Re: Basic Rocket Science Q & A
Post by: Jim on 06/03/2009 03:56 am
Heat is a function of shape, altitude and speed.  Too complex for be reduced to a table.
Title: Re: Basic Rocket Science Q & A
Post by: veryrelaxed on 06/03/2009 04:35 am
Jim's response is correct.  Thermal loads, as well as the G-loads are complex...  Here *duration* is also important as in any G-load analysis.  The question should be framed "this many Kelvins/Centigrade" *for how long*, and *how even a distribution across the heat absorbing components*?  For example STS due to the higher Lift/Drag ratio provides lower Kelvins but higher durations (especially on the leading edges of the structure) , while lower L/D shapes* provide higher absolute heat loads but for shorter durations... What is 'better'.  It is a complex analysis of trade-offs.

*like capsules or lifting bodies
Title: Re: Basic Rocket Science Q & A
Post by: DMeader on 06/03/2009 01:01 pm
I understand that the White Knight reached 367k and mach 3.09 on it's first flight without anything special in regard to it's construction. 

SS1, not White Knight.
Title: Re: Basic Rocket Science Q & A
Post by: fredm6463 on 06/05/2009 05:41 pm
Does any one know how NASA obtains the hydrogen and oxygen it uses to  fill the Space Shuttle's External Tank.

What process is involved and specifically what devices are used to separate the two elements? And does NASA do it or is it done by a contractor?

Thanks.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 06/05/2009 06:01 pm
The hydrogen is produced by Air Products, I think, near New Orleans, by cracking hydrocarbons IIRC, then trucked in.  Fun fact: NASA couldn't get hydrogen after Katrina because Homeland Security took over distribution of LH2 since it's used in food production (look for partially hydrogenated whatever in your ingredients).

The oxygen is produced in Mims, FL, in a typical air separation unit.  Praxair, I think.

Suggest moving this to the Shuttle or basic rocket science Q&A thread.
Title: Re: Basic Rocket Science Q & A
Post by: fredm6463 on 06/05/2009 06:39 pm
thank you
Title: Re: Basic Rocket Science Q & A
Post by: padrat on 06/05/2009 06:51 pm
The hydrogen comes from steam-reforming natural gas. Actually, after Katrina, we gave most of our hydrogen back to Air Products because Katrina shut down their New Orleans plant for a while at the same time their Canadian plant was down for maintenance, so they didn't have any for their other customers.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 06/05/2009 07:30 pm
Just thinking outside the box here but wouldn't it be *forward looking* if there were a THORIUM (not uranium) nuclear power plant nearby that generated electricity, drew in seawater, and then used some of that electricty to distill and crack the purified water into hydrogen and oxygen gasses? Send the remaining electricity into the grid and send the gasses off to be liquified, maybe even on site at KSC.

Thorium generators (molten salt coolant) can't "melt down" and do not produce any byproducts that can be reprocessed into weapons grade material. And thorium byproduct half lives are extremely short, unlike uranium.

Just musing. Returning you now to your regular programming.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 06/26/2009 04:28 am
All things being equal, where is the aerodynamic center of a rocket?
Title: Re: Basic Rocket Science Q & A
Post by: whitewatcher on 06/30/2009 12:39 pm
1.) I would like to do a WSB transfer from LEO to LLO.

To which extent is the delta-v and the travel time dependant on inclination of departure/destination orbit?

2.) Are there WSB transfers from LLO to LEO? Where can I find information on that subject? (delta-v's, travel times)

3.) Which transfer options are there for travel between EML-1 andd EML-2?
Title: Re: Basic Rocket Science Q & A
Post by: Bob the Avenger on 06/30/2009 05:55 pm
WSB meaning Weak Stability Boundary?
If so I'd suggest you start with "Fly Me to the Moon" by Edward Belbruno. Its an excellent introduction to how ballistic capture orbits work, but for more indepth information, he has written "Capture Dynamics and Chaotic Motions in Celestial Mechanics: With Applications to the Construction of Low Energy Transfers" which I haven't read so I dont know how much thats got in it that you're looking for.
If I remember the basics of it all correctly (big if), then the chaotic nature of the orbits that it utilises means you need to simulate the transfer in reverse, i.e. you start with the object in orbit around the moon(-ish) and you 'run time backwards' in the simulation, and see where it ends up.

I hope that helps, if not, then sorry.
Title: Re: Basic Rocket Science Q & A
Post by: Archibald on 06/30/2009 06:04 pm
Here's some Pdfs I have found

http://www.agi.com/downloads/support/productSupport/literature/pdfs/whitePapers/0800_wsb.pdf

http://esto.nasa.gov/conferences/nstc2007/papers/Belbruno_Edward_C6P1_NSTC-07-0156.pdf





Title: Re: Basic Rocket Science Q & A
Post by: irex31i on 06/30/2009 07:10 pm
Thanks a lot guys.. I have been reading this forum for a long while now. i hold a degree in electronics and am working in a small research company.. I just wanted to appear for the interview at indian space research org. Can someone please guide me about the questions.. i was reading some interview questions here (http://latestexams.com/2009/06/rf-and-radio-engineering-sample-technical-interview-questions/) about radio and rf. Pls i need more.. can you suggest some books.. thanks in advance..
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 06/30/2009 07:19 pm
All things being equal, where is the aerodynamic center of a rocket?

Is the rocket in or out of control?   ::)

On an aircraft wing, aerodynamic center is the point on the wing that the pitching moment is NOT a function of the angle of attack.  It needs to be aft of the center of gravity for the "aircraft" to be stable.

I'm not sure it is even used on rockets, because they don't have wings.  They mostly use the point that the pitching moment is zero.  This is called the center of pressure. 

For a layman, mixing the two terms up is probably OK.  Might even be OK for a pilot.  But an engineer MUST know the difference to do his or her job.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 06/30/2009 07:21 pm
All things being equal, where is the aerodynamic center of a rocket?

In the center of its longitudinal area projection.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 06/30/2009 07:26 pm
All things being equal, where is the aerodynamic center of a rocket?

In the center of its longitudinal area projection.

I think that is center of pressure.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 06/30/2009 10:17 pm
All things being equal, where is the aerodynamic center of a rocket?
In the center of its longitudinal area projection.
I think that is center of pressure.

You're both right.  For an airfoil, the rule of thumb is quarter-chord subsonic, half-chord supersonic.  Just wondering if there was an analogue for rockets, without airfoils.

Which would be like a plane with no wings, which would be a fuselage.  Except this would be a variable area fuselage... I think I answered my own question.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 07/01/2009 12:27 am
All things being equal, where is the aerodynamic center of a rocket?
In the center of its longitudinal area projection.
I think that is center of pressure.

You're both right.  For an airfoil, the rule of thumb is quarter-chord subsonic, half-chord supersonic.  Just wondering if there was an analogue for rockets, without airfoils.

Which would be like a plane with no wings, which would be a fuselage.  Except this would be a variable area fuselage... I think I answered my own question.

With no fins the center of pressure would be in the middle.  Thus the need to add fins to shift it aft.  But most modern rockets are not aerodynamically stable, by are stabilized by moving the thrust.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: Lab Lemming on 07/01/2009 02:26 am
And Ares I can't suffer a first stage turbopump failure.

Saying that non-existent technology X is better than well-used technology Y is disingenuous.

Just thinking outside the box here but wouldn't it be *forward looking* if there were a THORIUM (not uranium) nuclear power plant nearby that generated electricity, drew in seawater, and then used some of that electricty to distill and crack the purified water into hydrogen and oxygen gasses? Send the remaining electricity into the grid and send the gasses off to be liquified, maybe even on site at KSC.

Thorium generators (molten salt coolant) can't "melt down" and do not produce any byproducts that can be reprocessed into weapons grade material. And thorium byproduct half lives are extremely short, unlike uranium.

Just musing. Returning you now to your regular programming.
Title: Re: Basic Rocket Science Q & A
Post by: Lab Lemming on 07/01/2009 02:30 am
At high temperature and low pressure, CH4 + H2O => CO + 3H2.

Obviously, O2 cannot be present.

The hydrogen comes from steam-reforming natural gas. Actually, after Katrina, we gave most of our hydrogen back to Air Products because Katrina shut down their New Orleans plant for a while at the same time their Canadian plant was down for maintenance, so they didn't have any for their other customers.
Title: Re: Basic Rocket Science Q & A
Post by: whitewatcher on 07/01/2009 09:56 am
If so I'd suggest you start with "Fly Me to the Moon" by Edward Belbruno.

This book has a bad amazon comment, but if you recommend it, i'll buy!
Thank you for the recommendation!

Quote
Here's some Pdfs I have found

http://www.agi.com/downloads/support/productSupport/literature/pdfs/whitePapers/0800_wsb.pdf

http://esto.nasa.gov/conferences/nstc2007/papers/Belbruno_Edward_C6P1_NSTC-07-0156.pdf

I've already read the first one, but the second is new to me. Thank you!
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 07/01/2009 03:09 pm
And Ares I can't suffer a first stage turbopump failure.
Ares-I 2nd stage can

Quote
Saying that non-existent technology X is better than well-used technology Y is disingenuous.

What's your beef? I was musing about a potential future efficient application of nuclear technology to ground support equipment for launch vehicles. And fyi, the technology DOES exist. There's a ton of information available if you look for it.
Title: Re: Basic Rocket Science Q & A
Post by: whitewatcher on 07/01/2009 04:29 pm
I just had the opportunity to talk to an export. Just in case somebody's interested:

- The departure inclination can be at least between 0 and 51.6°. Higher inclinations are probably possible, but still need analysis. The farther off-plane the ejection is, the more delta-v is required (but in the order of magnitude of 50, maybe 60 m/s).

- The destination inclination can almost be chosen for free.

- WSB transfers from LLO/EML1/EML2 to earth orbit are possible but end up in a high eccentric orbit (apogee 1,4 mio km or so). Perigee can be chosen at almost no additional delta-v, so aerocapture could be an option.

- Transfer between EML1 and EML2 can be done as follows: Go into an unstable manifold which brings you to a random lunar orbit. After 14 days, the EML's have switched and your orbital plane now intersects the destination EML region. Bring the S/C on a stable manifold to your destination EML. That's the trick!
Title: Re: Basic Rocket Science Q & A
Post by: Archibald on 07/01/2009 05:30 pm
somewhere in these documents it is said that some WSB trajectories might double your payload to the Moon.
WSB trajectories could be of interest when refueling a EML-1 /EML-2 depot...
Title: Re: Basic Rocket Science Q & A
Post by: whitewatcher on 07/01/2009 08:12 pm
somewhere in these documents it is said that some WSB trajectories might double your payload to the Moon.
WSB trajectories could be of interest when refueling a EML-1 /EML-2 depot...

Yeah, that's Belbruno. He is known to be somewhat optimistic.

But I think the combination of some new technologies could fly your cargo to the moon (from LEO) with less than 10% of the mass being fuel.

You could use a low thrust trajectory from LEO to EML-1, use an unstable manifold for transfer to a high lunar orbit, transfer to EML-2 on a stable manifold and then use  a lunar space elevator to get things to the surface. Of couse: you need the elevator and you get your cargo delivered to the far side's equator. ;-) You get the idea ....

But for a fuel mass of 50-60 percent you should be able to bring your cargo to any place on the moon. Maybe staging in EML-1 would be a good idea in this case. (Avoids shutteling high thrust engines of a lander between LEO and EML-1.)

I'm currently writing a summary on lunar logistics and it's VERY interesting stuff.
Title: Re: Basic Rocket Science Q & A
Post by: Bob the Avenger on 07/01/2009 08:23 pm
I really think this stuff can revolutionise (storable) cargo mission to the moon. Also I think I remember Belbruno writing about using these trajectories to place fuel depots around the moon for return journeys or trips onward. Moving things around the Earth-Moon system for comparitively tiney deltaVs should provide us with lots of opportunities untill a Star Trek type impluse drive can be developed/build but thats going off topic somewhat
Title: Re: Basic Rocket Science Q & A
Post by: Lab Lemming on 07/02/2009 12:52 am
My beef is that much Thorium advocacy is based on geological misconceptions:
http://lablemminglounge.blogspot.com/2008/05/thorium-uranium-ratios-and-atomic-power.html

And Ares I can't suffer a first stage turbopump failure.
Ares-I 2nd stage can

Quote
Saying that non-existent technology X is better than well-used technology Y is disingenuous.

What's your beef? I was musing about a potential future efficient application of nuclear technology to ground support equipment for launch vehicles. And fyi, the technology DOES exist. There's a ton of information available if you look for it.
Title: Re: Basic Rocket Science Q & A
Post by: yinzer on 07/02/2009 03:04 am
And Ares I can't suffer a first stage turbopump failure.
Ares-I 2nd stage can

Quote
Saying that non-existent technology X is better than well-used technology Y is disingenuous.

What's your beef? I was musing about a potential future efficient application of nuclear technology to ground support equipment for launch vehicles. And fyi, the technology DOES exist. There's a ton of information available if you look for it.

How is it efficient?  Natural gas is used to make electricity and reformed into hydrogen.  The reformation is more thermodynamically efficient than the generation.  If you add a source of nuclear generated electricity, the efficient thing to do would to be the nuclear power to run refrigerators, light bulbs, and air compressors while using the natural gas to make hydrogen.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 07/02/2009 03:38 pm
And Ares I can't suffer a first stage turbopump failure.
Ares-I 2nd stage can

Quote
Saying that non-existent technology X is better than well-used technology Y is disingenuous.

What's your beef? I was musing about a potential future efficient application of nuclear technology to ground support equipment for launch vehicles. And fyi, the technology DOES exist. There's a ton of information available if you look for it.

How is it efficient?  Natural gas is used to make electricity and reformed into hydrogen.  The reformation is more thermodynamically efficient than the generation.  If you add a source of nuclear generated electricity, the efficient thing to do would to be the nuclear power to run refrigerators, light bulbs, and air compressors while using the natural gas to make hydrogen.

Yes, natural gas can make hydrogen. But I talked about hydrogen and oxygen for launch vehicle propellants. One plant producing all the propellant. Natural gas plants can't do that. Plus there is a *lot* more seawater available, without drilling, than there is natural gas. It can be difficult to find appropriate places to put the natural gas plant, but there are tens of thousands of miles of sea coast that can be accessed much easier. Thorium nuclear energy is also a lot cleaner than the natural gas process. It's non-polluting.
Title: Re: Basic Rocket Science Q & A
Post by: Hobbs on 07/22/2009 05:05 pm
Ok this is a reeeeaaally basic question but just thinking of it now i cant think of a satisfactory answer..

How did the saturn 5 (and for that matter any "inline" rocket) keep upright during launch? , obviously its massively top heavy about where the force is being applied at the bottom which would make it really unstable, it isn't spinning for gyrostabilisation and I cant see the balsa-wood stick, the only way i can think is that the engines were constantly vectored to stay dead on centre of mass but surely if they were just millimetres off then the rocket would tilt.
so how do they manage this?
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 07/22/2009 05:13 pm
the only way i can think is that the engines were constantly vectored to stay dead on centre of mass but surely if they were just millimetres off then the rocket would tilt.

Bingo. The engines gimbal keeping not only the thrust vector through the center of mass, but also by nudging the vector slightly they provide pitch/yaw (and if more than one nozzle) roll control. The rocket tends to "fall over", it's not inherently stable in flight (just like trying to keep a pencil upright on your fingertip) so this gimballing is continuous. The angles involved are really small - a couple of degrees in any direction, typically up to 8 degrees, but for the most part it's enough to have good control authority. Also a factor in controlability is how fast the engine can gimbal to a desired position.

If a rocket is really long (e.g. Saturn V), at the other end you can really feel this kicking around - which is what all Apollo astronauts reported.
Title: Re: Basic Rocket Science Q & A
Post by: cixelsyD on 07/22/2009 05:19 pm
Maybe a simpler question, how did they support the Saturn V before launch, were there platforms near the base that supported it, or did the arms that retracted before launch hold it?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 07/22/2009 05:35 pm
Maybe a simpler question, how did they support the Saturn V before launch, were there platforms near the base that supported it, or did the arms that retracted before launch hold it?

It sat on this

http://www.hq.nasa.gov/pao/History/SP-4204/ch13-4.html
Title: Re: Basic Rocket Science Q & A
Post by: engstudent on 08/13/2009 12:49 pm
How do you calculate the escape velocity of a rocket and how is it dependent on the density of or choice of propellant?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/13/2009 01:02 pm
How do you calculate the escape velocity of a rocket and how is it dependent on the density of or choice of propellant?

Escape velocity or exhaust velocity?
Title: Re: Basic Rocket Science Q & A
Post by: engstudent on 08/13/2009 01:11 pm
How do you calculate the escape velocity of a rocket and how is it dependent on the density of or choice of propellant?

Escape velocity or exhaust velocity?

Escape velocity - exhaust velocity is C = 9.8*Isp in the basic rocket equation relating deltaV, dry mass and propellant mass.  I'm familiar with this [ dV = C*ln((M+P)/M) ] but not the relationship between the deltaV required of a vehicle depending on its... whats the word density impulse?  How do you lower the dV required when C, exhaust velocity is fixed, or how do you compare the benefits of a propellant choice with slightly different C and grossly different density of P?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/13/2009 01:28 pm
Escape velocity is dependent on the planetary body that the vehicle is near.

Ve = sqrt(2GM/r)
Title: Re: Basic Rocket Science Q & A
Post by: engstudent on 08/13/2009 01:43 pm
then this equation may not help me in its current form

Can you assume a vehicle can have a dV equal to the Ve?  So that maybe C*ln( m+P/m ) = sqrt(2GM/r) ?
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 08/13/2009 01:50 pm
Yes, you could assume that if your burn delivering that deltaV is short enough so it doesn't incur additional gravity losses of you flying significantly uphill while accelerating. In that case it would be dV > Ve, though if you already start in orbit and for a high thrust vehicle (chemical propulsion) the gravity losses would not be high.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 08/13/2009 02:08 pm
How do you calculate the escape velocity of a rocket and how is it dependent on the density of or choice of propellant?

Escape velocity or exhaust velocity?


Escape velocity - exhaust velocity is C = 9.8*Isp in the basic rocket equation relating deltaV, dry mass and propellant mass.  I'm familiar with this [ dV = C*ln((M+P)/M) ] but not the relationship between the deltaV required of a vehicle depending on its... whats the word density impulse?  How do you lower the dV required when C, exhaust velocity is fixed, or how do you compare the benefits of a propellant choice with slightly different C and grossly different density of P?

I always thought is was 32.2 times ISP   ::)

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: engstudent on 08/13/2009 02:23 pm
How do you calculate the escape velocity of a rocket and how is it dependent on the density of or choice of propellant?

Escape velocity or exhaust velocity?

I always thought is was 32.2 times ISP   ::)

Danny Deger

Escape velocity - exhaust velocity is C = 9.8*Isp in the basic rocket equation relating deltaV, dry mass and propellant mass.  I'm familiar with this [ dV = C*ln((M+P)/M) ] but not the relationship between the deltaV required of a vehicle depending on its... whats the word density impulse?  How do you lower the dV required when C, exhaust velocity is fixed, or how do you compare the benefits of a propellant choice with slightly different C and grossly different density of P?

could be - I'd just have to change almost everything else before I try to get a number, kg - lbs, etc

I was curious about how the choice of propellant effects a theoretical SSTO rocket in terms of density and energy of the propellants and I'm not sure this helps me get to the effect of propellant choice on a launch vehicle - maybe escape velocity was the wrong concederation.  But Ive seen it used in a paper comparing the theoretical performance of SSTOs, I suppose you cant talk about a launch vehicles performance without concedering, drag, resistance, gravity loss - but I don't know how to do that either yet. {maybe I should start with a simpler question}  Maybe how is orbital velocity related to escape velocity?  :P

 
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 08/13/2009 05:32 pm
Well, circular orbital velocity is the same formula with the 2 taken out.  So dVe is (sqrt(2)-1)*sqrt(GM/r)
Title: Re: Basic Rocket Science Q & A
Post by: Tnarg on 08/24/2009 08:36 am
I'm looking for a program to work out orbits and delta-v budgets for moving around the solar system.  The main questions I have is 'if I increase my delta-v budget my this much how long would it take to get into a stable orbit around mars.

How do people work this out and is there any software I can use to help me work it out.

Thanks.

PS another question I've asked before but never got an answer but I want to know so I'm going to keep asking is:  If I had a 25 tonnes fuel tank that is made to lift into orbit and picked up by my space ship how much would be tank and how much would be fuel?
Title: Re: Basic Rocket Science Q & A
Post by: Mr Natural on 08/24/2009 09:24 am
If I had a 25 tonnes fuel tank that is made to lift into orbit and picked up by my space ship how much would be tank and how much would be fuel?

 The rocket Equation I use is as follows:
m_prop = m_pay*{[e^(delta_v/I_sp*g)]-1}*(1-m_inert)/{1-[f_inert*e^(delta_v/I_sp*g)]}
 where
 m_prop = propellant mass
 m_pay  = payload mass
 m_inert = empty mass, including propulsion system
 delta_v = the velocity requirements of the tank
 I_sp     = Specific Impulse of the propellant
 g         = gravity constant of the body in question (Earth = 9.80665)

This is a very difficult question to answer, as there are many more data elements needed to calculate that.For Example,
1/ What is the delta_v trajectory requirements for the tank
2/ What type of propellant are you using - IE what is the Specific Impulse
3/ A tank cannot by itself get into orbit, it requires a propulsion system
...So the mass of the propulsion system is needed to be known as well, is this included in the 25 tonnes or is it 'extra' mass.

There are many vatiables that need to be known before a constructive answer can be tabulated. 
Title: Re: Basic Rocket Science Q & A
Post by: Tnarg on 08/24/2009 10:01 am
"m_inert = empty mass, including propulsion system"

This is my unknown.  I'm putting together this ship in LEO and I can lift 25 tonnes at a time and If I know every thing else including the empty mass of my space ship and want to work out how many fuel tanks I need, I need to know the 'm_inert mass' of my fuel tanks.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 08/26/2009 05:55 pm
Anyone have a good reference for rocket plume electrical (and RF?) characteristics?
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 08/26/2009 09:45 pm
Just set SCE to Aux and you'll be fine. ;)
Title: Re: Basic Rocket Science Q & A
Post by: C5C6 on 08/27/2009 11:59 pm
Would a capsule in a retrograde orbit need extra shielding for a safe reentry??
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 08/28/2009 12:10 am
Would a capsule in a retrograde orbit need extra shielding for a safe reentry??

Does retrograde mean going West?  If it does the answer is yes.  Probably a 10% increase in mass to the heat shield.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: ginahoy on 09/09/2009 06:13 am
When flying in a supersonic jet (or the shuttle), does the sound level in the cockpit (or crew module) change after the vehicle goes supersonic? I would guess that the answer is no since the crew is moving at the same speed as the source of the noise.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 09/09/2009 02:31 pm
When flying in a supersonic jet (or the shuttle), does the sound level in the cockpit (or crew module) change after the vehicle goes supersonic? I would guess that the answer is no since the crew is moving at the same speed as the source of the noise.

I the F-4, if my back seater was talking too much, I would just go supersonic so I couldn't hear him anymore  ;D

Not really.  I don't remember any difference at all in the sound in the cockpit.  I have a good description of a "fun" ride I had over the Kissimee river on what it is like to go supersonic.  Look in www.lulu.com/dannydeger for a free copy. 

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: ANTIcarrot on 10/07/2009 11:25 pm
As a matter of special effects...

1) In space no one can hear you scream. But what if you stood in the exhaust of a rocket. Ignoring material considerations, what would you be likely to hear as you moved 'upsteam' along the exhaust closer and closer to the engine? Dull roar? High pitched whistle?

2) If you were looking sideways through a hydrogen rocket exhaust in vacuum (at close range, and moving in the same direction) and there were no other bright objects around, would you still see the same faint blue glow and shock cones you see inside the atmosphere? Would sub-optimal altitude compensation produce any interesting effects? What would be the length of the vissible plume?

Any true or suffficiently interesting answers appreciated.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 10/08/2009 04:22 pm
1) In space no one can hear you scream. But what if you stood in the exhaust of a rocket. Ignoring material considerations, what would you be likely to hear as you moved 'upsteam' along the exhaust closer and closer to the engine? Dull roar? High pitched whistle?

2) If you were looking sideways through a hydrogen rocket exhaust in vacuum (at close range, and moving in the same direction) and there were no other bright objects around,
a) would you still see the same faint blue glow and shock cones you see inside the atmosphere? Would sub-optimal altitude compensation produce any interesting effects?
b) What would be the length of the vissible plume?

1) Sound is oscillating gas pressure, sensed on a diaphragm of some sort (ear drum, microphone).  So in a textbook engine, there would be none.  In real life, however, I can't imagine it being anything other than a dull roar.  The turbulence in the flow would be broadband in frequency (chaos theory applies IIRC).

2a) By definition, no shock cones on a plume into vacuum.
2a&b) This is fairly easily solvable using known emission lines of combustion products (which depend on propellant) and equations of black body radiation and flow expansion.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 10/08/2009 04:33 pm
snip
2a) By definition, no shock cones on a plume into vacuum.
snip

This is certainly true after the gasses have dispersed into the vacuum, but close to the nozzle there might be enough gas in the exhaust to form a visible shock wave.  Next time I am next to a nozzle in the vacuum of space, I will try and get a picture for us.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 10/08/2009 06:16 pm
snip
2a) By definition, no shock cones on a plume into vacuum.
snip

This is certainly true after the gasses have dispersed into the vacuum, but close to the nozzle there might be enough gas in the exhaust to form a visible shock wave.

I don't get it. If there is no back pressure on the nozzle exit, what would the (supersonic) exhaust interact with? Indeed, there would have to be something with respect to what the flow would be supersonic in the first place, no?
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 10/08/2009 06:19 pm
snip
2a) By definition, no shock cones on a plume into vacuum.
snip

This is certainly true after the gasses have dispersed into the vacuum, but close to the nozzle there might be enough gas in the exhaust to form a visible shock wave.

I don't get it. If there is no back pressure on the nozzle exit, what would the (supersonic) exhaust interact with? Indeed, there would have to be something with respect to what the flow would be supersonic in the first place, no?

Any time you turn a gas flowing at faster than the speed of sound it generates a shock wave.  The exhaust is a super sonic flow of a gas. 

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 10/08/2009 07:10 pm
Indeed, there would have to be something with respect to what the flow would be supersonic in the first place, no?

The sonic speed of a flow is determined by sqrt(gamma*R*T), all of the flow itself.
gamma = Cp/Cv ratio of specific heats
R = universal gas constant
T = temperature of the flow
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 10/08/2009 11:42 pm
Minor nitpick on the gas constant. For a speed of sound calculation, it is the gas constant for the gas, not the universal constant. For instance:

R.air=287 J/kg-K
R.universal=8314 J/kmol-K
Title: Re: Basic Rocket Science Q & A
Post by: gloomygod on 10/13/2009 07:42 am
Hi there,

I'm looking some sites, and it seems that the space shuttle travels at aprox 17500 mph while orbiting the earth.
That would be roughly 8km/sec.

I guess other non geostationary satellites would travel at similar speeds in order to maintain orbit.

But, which is really the maximum speed at which one of those bodies could travel in open space (while traveling to other planets)? Would it be possible to increase speed to let's say, 1000km/s?

In the vacuum, increasing speed should be "cheap", and just a matter of time. Would these speeds be achievable by a common shuttle/satellite?

Thanks a lot for your answers.

Update: Edited some mistakes. Sorry for my poor english.
Title: Re: Basic Rocket Science Q & A
Post by: khallow on 10/13/2009 04:17 pm
The theoretical upper limit is the speed of light. But in current practice, you'll be moving at the orbital speed for moving around the Sun at that radius plus a little extra. At Earth, that'd be 36 km/s plus a few km/s. For current proposed interstellar missions using electric propulsion (like VASIMR) and radiothermal generators (RTGs), one might be able to obtain speeds of 100-200 km/s.

If we can obtain a high thrust, high isp nuclear powered propulsion system, we should be able to achieve much higher speeds. But I'm not knowledgeable enough to guess how fast that would be.
Title: Re: Basic Rocket Science Q & A
Post by: Analyst on 10/13/2009 05:35 pm
One question to ask is speed relative to what? The Galileo entry probe for example has been very fast relative to Jupiter, just by being pulled by Jupiters gravity.

Analyst
Title: Re: Basic Rocket Science Q & A
Post by: AlexInOklahoma on 10/13/2009 07:11 pm
One question to ask is speed relative to what?
Analyst

And does this thing need to stop (or enter an orbit) at its destination?  Or just blaze on by it meaninglessly?  Can really make a difference overall (form -v- function)   ;)

Alex
Title: Re: Basic Rocket Science Q & A
Post by: Nomadd on 10/13/2009 07:40 pm
Not counting Galileo at Jupiter entry (108,000mph) I believe the fastest travelling away from the sun is Voyager I at 38,600 mph. Dawn might have the greatest change in velocity. Just not all in one direction, since it will have to orbit two different bodies.
Title: Re: Basic Rocket Science Q & A
Post by: edkyle99 on 10/13/2009 07:58 pm
Hi there,

I'm looking some sites, and it seems that the space shuttle travels at aprox 17500 mph while orbiting the earth.
That would be roughly 8km/sec.

I guess other non geostationary satellites would travel at similar speeds in order to maintain orbit.

But, which is really the maximum speed at which one of those bodies could travel in open space (while traveling to other planets)? Would it be possible to increase speed to let's say, 1000km/s?

I believe that NASA's Pluto New Horizons, launched by an Atlas V 551 on January 19, 2006, was given the largest initial velocity of any human-made object to date.  It left Earth at 16.26 km/sec (Earth-relative), or about 36,373 mph.  It took only 9 hours to pass lunar distance, 13 months to fly past Jupiter, and less than 2.5 years to pass Saturn's orbit.  Even at that blistering pace it won't reach Pluto until 2015.

 - Ed Kyle
Title: Re: Basic Rocket Science Q & A
Post by: Pheogh on 10/13/2009 08:09 pm
So hopefully this isn't to far out but under the following conditions (below) what fraction of the speed of light would you guess we could achieve?

- NASA's "entire" current exploration budget to 2020
- Nuclear restrictions removed
- Using known technology or technology that appears likely to mature within that time scale
-Spacecraft in the New Horizon's weight class

*Oh and describe it please!

Basically what I am asking is if for one reason or another we HAD to accelerate a spacecraft in the New Horizon's Class how fast could we get it going by 2020?
Title: Re: Basic Rocket Science Q & A
Post by: Nomadd on 10/13/2009 08:14 pm
 I think one of the Helios probes will hit close to 150,000 mph at it's closest approach to the sun.
 New Horizons didn't have time for the normal inner planet dosey-do since it was trying to get to Pluto before the atmosphere froze out.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 10/13/2009 09:16 pm
So hopefully this isn't to far out but under the following conditions (below) what fraction of the speed of light would you guess we could achieve?

- NASA's "entire" current exploration budget to 2020
- Nuclear restrictions removed
- Using known technology or technology that appears likely to mature within that time scale
-Spacecraft in the New Horizon's weight class

*Oh and describe it please!

Basically what I am asking is if for one reason or another we HAD to accelerate a spacecraft in the New Horizon's Class how fast could we get it going by 2020?

Need more constraints, like a reliability number.  You could pick the lowest cost per kilo existing stage (I'll assume Atlas CCB) and start stamping them out like mad and strap them together asparagus cluster and wedding cake style.  You'd end up with ~150 of them.
Title: Re: Basic Rocket Science Q & A
Post by: gloomygod on 10/13/2009 09:20 pm
Hi there,

Thx moderator for moving the post to the right section.

One question to ask is speed relative to what?
Analyst

And does this thing need to stop (or enter an orbit) at its destination?  Or just blaze on by it meaninglessly?  Can really make a difference overall (form -v- function)   ;)

Alex

That's exactly the question. Yes, it needs to stop, and the speed relatively faster than the target moon/planet/body. Let me ellaborate, although this would almost certainly absurd and impossible.

I was wondering, if it's possible to accelerate a body(satellite) to a speed faster to the rotation speed for example of our moon (~1000 kms/s). If that was possible, would it be feasible to intercept the orbit of the moon at such speed, that the gravity of the moon would decelerate the satellite, until they collide at really slow speeds?

Imagine that we put our satellite going very fast right into the moon's path. The moon continues it's path, and eventually its gravity starts "braking" out satellite. The moon would get closer and closer, and eventually it will catch our satellite.

So I was wondering if it would be possible to calculate and reach such speed, so the collision would transate into a light landing:D

I haven't done the equations, and probably such speed would be orders of magnitude greater than the target moon speed, but I was just curious if that would be possible theoretically.

Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 10/13/2009 10:04 pm
Hi there,

Thx moderator for moving the post to the right section.

One question to ask is speed relative to what?
Analyst

And does this thing need to stop (or enter an orbit) at its destination?  Or just blaze on by it meaninglessly?  Can really make a difference overall (form -v- function)   ;)

Alex

That's exactly the question. Yes, it needs to stop, and the speed relatively faster than the target moon/planet/body. Let me ellaborate, although this would almost certainly absurd and impossible.

I was wondering, if it's possible to accelerate a body(satellite) to a speed faster to the rotation speed for example of our moon (~1000 kms/s). If that was possible, would it be feasible to intercept the orbit of the moon at such speed, that the gravity of the moon would decelerate the satellite, until they collide at really slow speeds?

Imagine that we put our satellite going very fast right into the moon's path. The moon continues it's path, and eventually its gravity starts "braking" out satellite. The moon would get closer and closer, and eventually it will catch our satellite.

So I was wondering if it would be possible to calculate and reach such speed, so the collision would transate into a light landing:D

I haven't done the equations, and probably such speed would be orders of magnitude greater than the target moon speed, but I was just curious if that would be possible theoretically.



It's not possible. The proper way to work the problem is to take the initial states of the spacecraft and the moon and compute the *relative* state between the spacecraft and the moon. Then work the problem in moon-centered coordinates.

You'll see that, because the problem started out with the spacecraft outside the moon's sphere of influence, it will always approach the moon along a hyperbolic path, and will *always* accelerate *toward* the moon as distance decreases. So when the spacecraft strikes the moon, it'll always be going faster than when it started out. That rules out this method of attempting a "free" soft-landing.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 10/13/2009 10:39 pm
Hi there,

I'm looking some sites, and it seems that the space shuttle travels at aprox 17500 mph while orbiting the earth.
That would be roughly 8km/sec.

I guess other non geostationary satellites would travel at similar speeds in order to maintain orbit.

But, which is really the maximum speed at which one of those bodies could travel in open space (while traveling to other planets)? Would it be possible to increase speed to let's say, 1000km/s?

I believe that NASA's Pluto New Horizons, launched by an Atlas V 551 on January 19, 2006, was given the largest initial velocity of any human-made object to date.  It left Earth at 16.26 km/sec (Earth-relative), or about 36,373 mph.  It took only 9 hours to pass lunar distance, 13 months to fly past Jupiter, and less than 2.5 years to pass Saturn's orbit.  Even at that blistering pace it won't reach Pluto until 2015.

 - Ed Kyle

Did I do my math right?  I come up with a whopping 53,300 ft/sec  :o

Any gravity assists on this mission?

Are they going to orbit or fly-by?

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: zeke01 on 10/13/2009 10:59 pm
There was an assist from Jupiter.  New Horizons spacecraft will not be captured by Pluto.  It will go onward to investigate any additional Kuiper objects within its small delta-v capability after visiting the Pluto system.

zeke
Title: Re: Basic Rocket Science Q & A
Post by: hop on 10/14/2009 01:03 am
I believe that NASA's Pluto New Horizons, launched by an Atlas V 551 on January 19, 2006, was given the largest initial velocity of any human-made object to date.  It left Earth at 16.26 km/sec (Earth-relative), or about 36,373 mph.
But thanks to gravity assists, not the fastest spacecraft. According to http://www.heavens-above.com/solar-escape.asp that honor belongs to Voyager 1
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 10/31/2009 04:10 pm
Quote
Quote
Quote
Quote
Hypothesis:  There was no recontact, but the first stage disrupted the upper stage simulator *aerodynamically* and caused it to tumble when it's upper end exited the slipstream of the upper stage simulator.
Lee Jay, I suspect you might be right, especially after checking out klausd's video above.
I like the attempt at a higher-order physical explanation, but this is literally impossible at supersonic speeds.  Such disturbances can only flow upstream at sonic speed (exactly) no matter the speed of the vehicle.  So the USS could not have been disturbed by the flow of the FS since they were somewhere M~4.
Aren't bow shocks generated in front of an object traveling supersonically? I.e. a higher pressure region *some distance* in front of the front end of the SRB? If so, what if one side of the USS aft end dipped into this region? Don't know much about fluid dynamics, just wondering if that's possible here.

Not enough to matter.  The blunt forward end of the separated SRM would have a small bow shock, probably detached by just a few inches from the solid surface.  It would be a minimal detachment distance due to the high supersonic (i.e. not hypersonic) flow regime.
Title: Re: Basic Rocket Science Q & A
Post by: littlepagan on 11/01/2009 01:21 am
hey, what senseations do astornauts experience when in outer space and on takeoff, mostly on takoff
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 11/04/2009 05:47 am
Read any of the astronaut autobiographies for really good descriptions of this.  I'm not copping out.  I just like to see more off-line reading.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 11/08/2009 05:01 am
Experts on this forum have said Titan IIs used for launching early space missions did not need an escape tower because the propellants are hypergolic and will deflagrate instead of detonating. Further googling reveals that this 'barely' enabled the use of ejection seats.

In another thread I wondered if this would apply to catalysed kerosene/peroxide launchers such as the ones Beal Aerospace intended to develop. On reflection this seemed unlikely to me since I imagined the peroxide could decompose if heated. But I was surprised to learn hydrogen peroxide could be used for regenerative cooling, so maybe things are not as bad as I imagined.

So, would you need an escape tower for a kerosene/peroxide launcher or would ejection seats be enough?
Title: Re: Basic Rocket Science Q & A
Post by: MKremer on 11/08/2009 07:24 am
IMO, the Gemini ejection seat was a Dumb Idea, and was mostly a result of mass restrictions, plus R&D time and costs to find an effective LAS tower.

Using ejection seats to me is a Dumb Idea in general, if for nothing else than having multiple flammable/explosive ordinance devices as permanent parts within a pressurized crew cabin.

IMO, if your booster can't handle the additional mass of an effective crew LAS, then you need a bigger rocket.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 11/08/2009 11:44 am
Quote
Using ejection seats to me is a Dumb Idea in general, if for nothing else than having multiple flammable/explosive ordinance devices as permanent parts within a pressurized crew cabin.

*ALL* our military fighter aircraft, indeed *all* modern military fighter aircraft around the entire planet, use this system. To the best of my knowledge in all those years there has never been an accidental ignition of one. It takes a very deliberate, conscious action on the part of the crewman to activate. Don't forget that those astronauts were on loan to NASA from the military. They were all military pilots and they were very, very comfortable with them, as are ALL military pilots to this day.

Gemini was very popular with the pilots in part because it "felt" familiar. The astronauts asked for the ejection seats in the design phase. The spacecraft has been described as a "pilot's dream". It was designed around the pilots and what they wanted in it and everyone who flew in them loved the design. The ejection seat was part of that.

Ejection seats are not a dumb idea. They are a very good idea, are very safe inside the pressurized cabin and are an extremely efficient way to quickly exit a destructing vehicle. While the ignition process has been simplified to facilitate quick activation, it has been designed to be exceptionally safe; they cannot be accidentally activated; it has to be deliberately done. After Challenger, they were the top runner for a safety change to Shuttle but couldn't be incorporated into the STS design without a MAJOR complete redesign of the crew cabin. That's the main reason they are not on Shuttle today.

Ejection seats would likely have saved most of Challenger's crew as it has been reliably reported that they were alive, and probably conscious, inside the descending crew cabin which had survived the explosion intact, all the way down to ocean impact. If I recall correctly, in spite of additional wounds they may have received caused by the breakup of the Shuttle's airframe, the cause of death for the crewmembers was blunt force trauma caused by the cabin's impact with the ocean surface. If they had had ejection seats they could have ejected and most of them would likely have survived. But there was, sadly, no way to incorporate ejection seats into Shuttle after-the-fact.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 11/08/2009 03:20 pm
Sit in your car.  If it's modern, there's a fairly good chance your surrounded by ordnance devices that drive the airbags.  It's been a major growth segment for suppliers who previously had only government business.
Title: Re: Basic Rocket Science Q & A
Post by: Art LeBrun on 11/08/2009 04:01 pm
Shown is the integration of a Delta first stage and the second stage/shroud structure. Note the "many registration pins" ready to be inserted in the matching holes in the first stage. Is this a common matching procedure? Why are so many pins required? And how can I improve my vernacular on these questions?
Title: Re: Basic Rocket Science Q & A
Post by: Art LeBrun on 11/08/2009 06:34 pm
Experts on this forum have said Titan IIs used for launching early space missions did not need an escape tower because the propellants are hypergolic and will deflagrate instead of detonating. Further googling reveals that this 'barely' enabled the use of ejection seats.


So, would you need an escape tower for a kerosene/peroxide launcher or would ejection seats be enough?

Isn't there an altitude restriction for ejection seats? How often has lox/kerosene "detonated"?
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 11/08/2009 09:55 pm
On first order, I don't find detonation credible as the root cause of an LOC event.  There would have to be fuel/LOX or fuel/air mixing BEFORE the application of an ignition source.  More likely, the propellants would deflagrate with the existing ignition sources (exhaust or firing ordnance).  The crew compartment would be able to get away from the cloud before mixing and ignition could occur.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 11/08/2009 10:44 pm
So are you saying detonation is unlikely to begin with, regardless of whether the propellants are hypergolic? What about Challenger?
Title: Re: Basic Rocket Science Q & A
Post by: Art LeBrun on 11/09/2009 12:53 am
So are you saying detonation is unlikely to begin with, regardless of whether the propellants are hypergolic? What about Challenger?

Challenger broke up with the propellants spilling and igniting in the air. Not much of a detonation...............
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 11/09/2009 01:01 am
We're getting close to a very sensitive subject here, so I'll try to be careful. Suppose the ejection seats for the pilot and commander had not been disabled, would they have been able to survive the disaster? I realise it is unacceptable to give some crew members an escape option others do not have, but I'm wondering how this relates to potential survivability in a future spaceplane.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 11/09/2009 01:14 am
To the best of my knowledge in all those years there has never been an accidental ignition of one.

Here's such a story, but it was a passenger who accidentally triggered an eject, not a pilot or a malfunction:

Plane passenger accidentally activates ejector seat - and survives (http://www.telegraph.co.uk/news/newstopics/howaboutthat/6480818/Plane-passenger-accidentally-activates-ejector-seat---and-survives.html)

Quote
Ejection seats would likely have saved most of Challenger's crew as it has been reliably reported that they were alive, and probably conscious, inside the descending crew cabin which had survived the explosion intact, all the way down to ocean impact.

I suppose that answers my previous question.
Title: Re: Basic Rocket Science Q & A
Post by: kch on 11/09/2009 01:47 am
We're getting close to a very sensitive subject here, so I'll try to be careful. Suppose the ejection seats for the pilot and commander had not been disabled, would they have been able to survive the disaster? I realise it is unacceptable to give some crew members an escape option others do not have, but I'm wondering how this relates to potential survivability in a future spaceplane.

To the best of my knowledge, the only flight with disabled ejection seats was STS-5 (first non-Orbital-Flight-Test flight, crew of 4) in Columbia.  After that flight, the ejection seats were removed to save weight and make room for 4 crew on the flight deck.  IIRC, none of the other orbiters (with the possible exception of Enterprise) had ejection seats.
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 11/09/2009 05:04 am

Ejection seats would likely have saved most of Challenger's crew as it has been reliably reported that they were alive, and probably conscious, inside the descending crew cabin which had survived the explosion intact, all the way down to ocean impact.

It's been reliably reported that they were alive but probably *unconscious*.
Title: Re: Basic Rocket Science Q & A
Post by: MKremer on 11/09/2009 05:21 pm
Shown is the integration of a Delta first stage and the second stage/shroud structure. Note the "many registration pins" ready to be inserted in the matching holes in the first stage. Is this a common matching procedure? Why are so many pins required? And how can I improve my vernacular on these questions?

Those are the bolts that attach the interstage adapter to the first stage. If you look closely, there are threads at the end of each one.

A few more views, showing the bolt holes from inside and adapters being mated both with and without the bolts in place:

http://mediaarchive.ksc.nasa.gov/detail.cfm?mediaid=24565
http://mediaarchive.ksc.nasa.gov/detail.cfm?mediaid=24594
http://mediaarchive.ksc.nasa.gov/detail.cfm?mediaid=43889


Title: Re: Basic Rocket Science Q & A
Post by: Art LeBrun on 11/10/2009 03:41 am
Shown is the integration of a Delta first stage and the second stage/shroud structure. Note the "many registration pins" ready to be inserted in the matching holes in the first stage. Is this a common matching procedure? Why are so many pins required? And how can I improve my vernacular on these questions?

Those are the bolts that attach the interstage adapter to the first stage. If you look closely, there are threads at the end of each one.

A few more views, showing the bolt holes from inside and adapters being mated both with and without the bolts in place:

http://mediaarchive.ksc.nasa.gov/detail.cfm?mediaid=24565
http://mediaarchive.ksc.nasa.gov/detail.cfm?mediaid=24594
http://mediaarchive.ksc.nasa.gov/detail.cfm?mediaid=43889




Thank you for the images and explanation. Now for additional questions.

1) why some with and some without bolts?
2) how are bolts secured? threaded insert on first stage or nuts attached later? access panel obviously available........I seem to see counterbores for bolts on interstage.
Title: Re: Basic Rocket Science Q & A
Post by: sbt on 11/10/2009 11:50 pm
To the best of my knowledge in all those years there has never been an accidental ignition of one.

Here's such a story, but it was a passenger who accidentally triggered an eject, not a pilot or a malfunction:

We also suffered one during the introduction of the Harrier GR5. A Wander Lamp fell to the floor. The pilot motored the seat down to adjust his eyeline to avoid sun in his eyes. The lamp was under the linkage to part the parachute system. The slug used to drag the cute out of its container was fired, simultaneously triggering the seat straps to release, broke the canopy and pulled the cute into the airstream. The pilot did not survive the decent on a shredded parachute. The lamp was eliminated and a FOD shield added to the bottom of the seat.

Most Ejector Seat accidents, however, are during ground handeling. Personnel MUST check that the pins are in before working in the cockpit area.

Rick
Title: Re: Basic Rocket Science Q & A
Post by: tva on 11/26/2009 08:55 am
Chris has just reported (http://forum.nasaspaceflight.com/index.php?topic=19633.msg510235#msg510235) that Atlantis has raised her orbit with a burn wich was radial down (nose to the Earth).

I tought that a nose-down burn would rather alter apogee/perigee then raise orbit.

Could someone cast some light on this  ??? ?
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 12/12/2009 08:57 pm
Chris has just reported (http://forum.nasaspaceflight.com/index.php?topic=19633.msg510235#msg510235) that Atlantis has raised her orbit with a burn wich was radial down (nose to the Earth).

I tought that a nose-down burn would rather alter apogee/perigee then raise orbit.

Could someone cast some light on this  ??? ?

That was a live mission thread, most likely Chris was quoting NASA PAO from NASA TV and something got garbled in the chain.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 12/28/2009 02:29 am
Moved from SpaceX thread:
IIRC, if LOX soaks into a fuel like boot polish or coal or asphalt, it can form a shock sensitive explosive. So the boot wouldn't explode immediately, it would probably wait until you stamped to attention.

Coal & LOX used to a common mining explosive, but there were too many unexpected detonations.

Google Atlas 71F and oxyliquit.
Title: Re: Basic Rocket Science Q & A
Post by: HMXHMX on 12/28/2009 04:33 am
Moved from SpaceX thread:
IIRC, if LOX soaks into a fuel like boot polish or coal or asphalt, it can form a shock sensitive explosive. So the boot wouldn't explode immediately, it would probably wait until you stamped to attention.

Coal & LOX used to a common mining explosive, but there were too many unexpected detonations.

Google Atlas 71F and oxyliquit.

And you get:  "Access to this server is forbidden from your client"
Title: Re: Basic Rocket Science Q & A
Post by: Art LeBrun on 12/28/2009 04:41 am
Moved from SpaceX thread:
IIRC, if LOX soaks into a fuel like boot polish or coal or asphalt, it can form a shock sensitive explosive. So the boot wouldn't explode immediately, it would probably wait until you stamped to attention.

Coal & LOX used to a common mining explosive, but there were too many unexpected detonations.

Google Atlas 71F and oxyliquit.

And you get:  "Access to this server is forbidden from your client"

click on "view as html" on second line. I wonder what the recent SLC-2W event was?
Title: Re: Basic Rocket Science Q & A
Post by: yinzer on 12/28/2009 08:26 pm
Summary: LOX and RP-1 coming out an overflow line pooled in the flame deflector and exploded prior to launch.  Damage to the vehicle caused premature engine shut down followed by range safety pushing the button at T+303.

Given that the vehicle was able to take off and fly more or less normally for five seconds after a substantial leak of fuel and oxidizer on the launch pad, I'm not convinced that this shows LOX/RP-1 to be any more dangerous than your typical hypergolics.
Title: Re: Basic Rocket Science Q & A
Post by: Art LeBrun on 12/28/2009 09:01 pm
I would like to know the actual effect on Atlas 71F from the explosion at ignition. It appears to have functioned through staging so we had about 130 seconds of flight at least and what caused the sustainer cutoff at what time?

Reminds me of two Atlas E at the Cape when the hydraulics rise off valve failed to close allowing fluid to drain and after staging the sustainer engine had no flight control.
Title: Re: Basic Rocket Science Q & A
Post by: yinzer on 12/28/2009 10:51 pm
From the report:
Quote
Pre-launch instrumentation staging disconnected prematurely resulting in loss of all booster data measurement.
Sustainer and vernier engine shutdown occurred prematurely during flight.
Range Safety destroyed vehicle at 303 seconds.

So... who knows.  Could have damaged one of the valves that seals off a line that separates during staging, resulting in a leak of fuel or oxidizer or pressurant or something.
Title: Re: Basic Rocket Science Q & A
Post by: tminus9 on 12/29/2009 02:18 pm
To convert earth-centered inertial cartesian coordinates (e.g., J2000) to geodetic latitude and longitude, I know that the rotation, precession and nutation of the earth since the epoch must be accounted for.

Rotation seems easy enough, but I've yet to find an accessible (i.e., assuming enough physics and math to understand, but not assuming any previous knowledge of astronomy) resource about calculating precession and nutation matrices.

1. Does anyone know of a good (preferably online) description of how to calculate the precession and nutation matrices, given an arbitrary epoch? Included examples would be even better to verify my results. I know the calculation gets pretty hairy, so even an approximation is fine for my purposes.

2. If I neglect precession and nutation, what is the ballpark error in the resulting geodetic calculations? (I assume it depends on the epoch.)
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 12/29/2009 06:54 pm
To convert earth-centered inertial cartesian coordinates (e.g., J2000) to geodetic latitude and longitude, I know that the rotation, precession and nutation of the earth since the epoch must be accounted for.

Rotation seems easy enough, but I've yet to find an accessible (i.e., assuming enough physics and math to understand, but not assuming any previous knowledge of astronomy) resource about calculating precession and nutation matrices.

1. Does anyone know of a good (preferably online) description of how to calculate the precession and nutation matrices, given an arbitrary epoch? Included examples would be even better to verify my results. I know the calculation gets pretty hairy, so even an approximation is fine for my purposes.

Vallado's book Fundamentals of Astrodynamics and Applications contains a good discussion of this, and he has graciously put his sample code online, in several languages:

http://www.smad.com/vallado/

The code can be understood without reference to the book, but I highly recommend the book anyway.

Quote
2. If I neglect precession and nutation, what is the ballpark error in the resulting geodetic calculations? (I assume it depends on the epoch.)

Yes, it does depend on the epoch. Generally speaking, the results are good enough for backyard astronomy (that is, you can look up and find the object in question even though the computed position is off) but you wouldn't want to use it for professional applications.
Title: Re: Basic Rocket Science Q & A
Post by: butters on 01/05/2010 01:38 pm
Basic orbital mechanics question:


Is it possible to put a satellite in a geostationary orbit over a pole, or in a sort of halo orbit that stays within, let's say, 25 degrees latitude of a (rotational) pole?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 01/05/2010 01:42 pm
Basic orbital mechanics question:


Is it possible to put a satellite in a geostationary orbit over a pole, or in a sort of halo orbit that stays within, let's say, 25 degrees latitude of a (rotational) pole?

No
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 01/05/2010 02:15 pm
But there are other useful orbits that make communications at the poles easier. They require a min of two birds for 24-7 coverage.

google:

molniya orbits ( http://en.wikipedia.org/wiki/Molniya_orbit )

and

tundra orbits ( http://en.wikipedia.org/wiki/Tundra_orbit )
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 01/05/2010 10:52 pm
Quick question about C3.

I understand that this relates the energy of a trajectory to escape velocity (C3=0).

If a mission is described as having a positive C3 (say 1.5km2/s2), is there an easy way to equate that to a delta-V, over and above the delta-V necessary to achieve escape velocity?

Thanks, Martin
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 01/05/2010 11:05 pm
If a mission is described as having a positive C3 (say 1.5km2/s2), is there an easy way to equate that to a delta-V, over and above the delta-V necessary to achieve escape velocity?

Try this wiki page (http://en.wikipedia.org/wiki/Specific_orbital_energy#Voyager_1), in particular the Voyager 1 example. Since you can deduce V-inf from C3 for a hyperbolic orbit, you should be able to calculate V at periapsis and compare that to circular orbital velocity at that point (and also to escape velocity at that point, it being SQRT(2) greater than orbital)
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 01/05/2010 11:54 pm
Many thanks.

So a C3 of 1.5km2/s2, equates to V-inf of sqrt (2 * 1.5) = 1.73 km/s.

Would it be an over-simplification to say that an EDS would need to burn to escape velocity + 1.73 km/s to achieve that C3?

cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 01/06/2010 12:17 am
So a C3 of 1.5km2/s2, equates to V-inf of sqrt (2 * 1.5) = 1.73 km/s.

No, V-inf equals sqrt(2 * epsilon). Since C3 already is 2*epsilon (equation further up that page), V-inf becomes sqrt(C3), in your case that's V-inf of 1.22 km/s

Quote
Would it be an over-simplification to say that an EDS would need to burn to escape velocity + 1.73 km/s to achieve that C3?

Yes, and a gross inaccuracy. An EDS would need a smaller burn than that because of the Oberth effect. In effect having a small boost above V-esc you get out of the gravity well a lot quicker and that translates into a larger velocity at infinity.

That's why you need the equation in that Voyager 1 example to find the actual velocity at periapsis of that hyperbola. If you do the calculation, you'll find that an EDS needs substantially less than 1.22 km/s to reach that specific velocity at infinity. The deeper into the gravity well you are when burning to escape, the more efficient this effect is. It's one of the reasons parking orbits are just low enough not to decay over a period until the 2nd burn.
Title: Re: Basic Rocket Science Q & A
Post by: toddbronco2 on 01/06/2010 04:10 am
Basic orbital mechanics question:


Is it possible to put a satellite in a geostationary orbit over a pole, or in a sort of halo orbit that stays within, let's say, 25 degrees latitude of a (rotational) pole?

Jim is right that it's not possible to be geostationary above a rotational pole, but there are orbits that are loosely associated with the L3 family of Halo orbits that are similar to your description, in that they dwell for long periods of time above the north-pole.  Notice the nearly vertical orange colored orbits that pass closest to 'Earth' (the black dot).  Source: D. Grebow (https://engineering.purdue.edu/people/kathleen.howell.1/Publications/publications.html)
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 01/07/2010 04:41 pm
Constellation assumes that TLI will take advantage of the lowest delta-V opportunities. I've seen contradictory comments that these opportunities occur either once or twice per month. Which is correct?

How does the delta-V vary between these opportunities - say if you wanted to just pick the best opportunity on any particular day?

Many thanks, Martin
Title: Re: Basic Rocket Science Q & A
Post by: Namechange User on 01/07/2010 04:47 pm
Constellation assumes that TLI will take advantage of the lowest delta-V opportunities. I've seen contradictory comments that these opportunities occur either once or twice per month. Which is correct?


Who cares.  Either is more frequent than actual launch rates will be. 
Title: Re: Basic Rocket Science Q & A
Post by: jongoff on 01/07/2010 05:08 pm
Constellation assumes that TLI will take advantage of the lowest delta-V opportunities. I've seen contradictory comments that these opportunities occur either once or twice per month. Which is correct?


Who cares.  Either is more frequent than actual launch rates will be. 

MP99, it varies depending on if you're launching from the surface or from LEO.  If you're launching from *the surface*, you get daily opportunities, because you can pick your launch time such that when your assembly is ready to depart LEO, your stack's orbital plane intersects the point where the moon will be when your stack arrives in its vicinity.  Once you're in LEO though, you've got a plane, and planes precess at a rate that gives you launch windows on the order of 2-3 times per month.  Depends a lot on inclination and altitude.  So, if you're doing an EOR mission (say one with a depot for instance), you only get 2-3 good departure dates per month. 

But OV-106 is right.  Having launch window frequency being a real issue would be a nice problem to have.  If it ever becomes a problem (say with a depot-based architecture), you always have the option of putting up 1-2 more depots in additional orbital planes.  Then you can send missions every 2-3 days....but we're a *long* way from that being important.

~Jon

[edit: corrected a dumb error in line two...sigh]
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 01/07/2010 07:30 pm
Constellation assumes that TLI will take advantage of the lowest delta-V opportunities. I've seen contradictory comments that these opportunities occur either once or twice per month. Which is correct?


Who cares.  Either is more frequent than actual launch rates will be. 

MP99, it varies depending on if you're launching from the surface or from LEO.  If you're launching from LEO, you get daily opportunities, because you can pick your launch time such that when your assembly is ready to depart LEO, your stack's orbital plane intersects the point where the moon will be when your stack arrives in its vicinity.  Once you're in LEO though, you've got a plane, and planes precess at a rate that gives you launch windows on the order of 2-3 times per month.  Depends a lot on inclination and altitude.  So, if you're doing an EOR mission (say one with a depot for instance), you only get 2-3 good departure dates per month. 

But OV-106 is right.  Having launch window frequency being a real issue would be a nice problem to have.  If it ever becomes a problem (say with a depot-based architecture), you always have the option of putting up 1-2 more depots in additional orbital planes.  Then you can send missions every 2-3 days....but we're a *long* way from that being important.

~Jon

I'm thinking of CxP modified for dual launch.

If the lander launches to take advantage of a minimum delta-V window (max payload), could the crew launch a few days later, even though it's not optimal?

How much extra delta-V would be required? How does it vary as the month progresses?

cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: jongoff on 01/08/2010 04:39 am
Constellation assumes that TLI will take advantage of the lowest delta-V opportunities. I've seen contradictory comments that these opportunities occur either once or twice per month. Which is correct?


Who cares.  Either is more frequent than actual launch rates will be. 

MP99, it varies depending on if you're launching from the surface or from LEO.  If you're launching from LEO, you get daily opportunities, because you can pick your launch time such that when your assembly is ready to depart LEO, your stack's orbital plane intersects the point where the moon will be when your stack arrives in its vicinity.  Once you're in LEO though, you've got a plane, and planes precess at a rate that gives you launch windows on the order of 2-3 times per month.  Depends a lot on inclination and altitude.  So, if you're doing an EOR mission (say one with a depot for instance), you only get 2-3 good departure dates per month. 

But OV-106 is right.  Having launch window frequency being a real issue would be a nice problem to have.  If it ever becomes a problem (say with a depot-based architecture), you always have the option of putting up 1-2 more depots in additional orbital planes.  Then you can send missions every 2-3 days....but we're a *long* way from that being important.

~Jon

I'm thinking of CxP modified for dual launch.

If the lander launches to take advantage of a minimum delta-V window (max payload), could the crew launch a few days later, even though it's not optimal?

How much extra delta-V would be required? How does it vary as the month progresses?

cheers, Martin

Well, my orbital dynamics fu is pretty weak, but AIUI, launching at a time that is more than a day or so away from a good launch window makes the delta-V costs prohibitive.  As in several times higher dV requirements.  There *may* be some clever trick, but I'm not smart enough (and don't have good enough tools) to know for sure.

~Jon
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 01/08/2010 02:07 pm
Going back to my earlier question re C3 - let's see if I've got it now.

C3 is numerically twice the kinetic energy per Kg that an escaping object would attain if it was allowed to continue to infinity. Measured in J/Kg, which is also equivalent to m2/s2 (meters per second, squared).



Specific orbital energy is the total orbital energy of the object per Kg of mass, composed of kinetic and potential terms. Kinetic is always positive. Potential is negative, tending to zero at inifinity. (For a circular orbit, kinetic energy is half the potential, but of opposite sign.)

An object on the Earth's surface has potential energy but no kinetic (baring that from the rotational speed for the Earth's latitude).

To lift an object to LEO it needs to gain a small increase in potential energy, and a large amount of kinetic. The net delta-V can be equated to the specific energy gain via 1/2mv2, where m=1, ie 1/2v2. The energy of this orbit is -GM/2a, where a = semi-major axis. AFAICT, this is the delta-V to reach an orbit with the same inclination as the latitude of the launch site, further delta-V will be  required for plane changes.

For injection out of Earth's gravity from LEO, add the energy required to escape (+GM/2a) to the additional energy required to reach the destination (half of the C3 value). Again, assumes that all orbital parameters at start of the injection are ideal for the required orbit. Again, delta-V can be equated to this total specific energy via 1/2v2

For TMI via a hohmann orbit, I presume this would equate to the difference in energy between the orbits of Earth & Mars, within the Sun's gravitational field.



OK, how did I do?

cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 01/09/2010 11:51 am
Let's see how well (or not) this amateur can do...

To relate delta-v to energy you can't just say 0.5 * delta_v ^2 = delta-E. The efficiency of the burn depends strongly on your current velocity.

If we idealise things and consider only infinitely short, impulsive burns then we obtain instantaneous changes in velocity while positions remain unchanged. In real life situations with finite duration burns of significant thrust we get small changes to position and potentially large changes to velocity. If positions remain unchanged, then potential energy remains unchanged too. Kinetic energy does change of course. The total change in energy due to an impulsive burn therefore consists entirely of a change in kinetic energy. And since kinetic energy is quadratic in velocity, the increase in energy for a given delta-v increases linearly with the starting velocity. The faster you go already, the more benefit you get from your burn. This is the famous Oberth effect. For true orbits (i.e. not suborbital trajectories) this means the most efficient place to increase your energy is at perigee. On suborbital trajectories you can hardly expect to be able to execute a subterranean perigee burn...

Of course there's more to it than just energy. To begin with velocity is a vector. You may also want to change the orientation of your orbital plane and the orientation of your major axis within that plane. It's no use flying in the wrong direction with the right speed!

You typically need more than one burn to get into the desired orbit. In theory you could get into a sea level Earth orbit with a single impulsive burn that raises your perigee from the center to the surface of the Earth. In practice your burn cannot be impulsive and even if it could be you would burn up in the atmosphere and smash into mountains and other obstacles. To get into circular LEO you would need at least two impulsive burns: one burn to raise either apogee or perigee to the desired altitude, and once at the new apogee another circularisation burn to raise the new perigee to the desired altitude. In the real world you need to raise your current apogee by enough to escape the effects of drag and heating and to give your engines enough time to raise your perigee.

All in all you'd need more delta-v than you'd expect from just comparing energies.
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 01/09/2010 05:55 pm
Let's see how well (or not) this amateur can do...

To relate delta-v to energy you can't just say 0.5 * delta_v ^2 = delta-E. The efficiency of the burn depends strongly on your current velocity.

Yeah, I'd found that by working through examples on the wiki page that ugordan pointed to.

It's applicable to the C3 part of the calculation, but not the escape element.


Quote
If we idealise things and consider only infinitely short, impulsive burns then we obtain instantaneous changes in velocity while positions remain unchanged. In real life situations with finite duration burns of significant thrust we get small changes to position and potentially large changes to velocity. If positions remain unchanged, then potential energy remains unchanged too. Kinetic energy does change of course. The total change in energy due to an impulsive burn therefore consists entirely of a change in kinetic energy. And since kinetic energy is quadratic in velocity, the increase in energy for a given delta-v increases linearly with the starting velocity. The faster you go already, the more benefit you get from your burn. This is the famous Oberth effect. For true orbits (i.e. not suborbital trajectories) this means the most efficient place to increase your energy is at perigee. On suborbital trajectories you can hardly expect to be able to execute a subterranean perigee burn...

Of course there's more to it than just energy. To begin with velocity is a vector. You may also want to change the orientation of your orbital plane and the orientation of your major axis within that plane. It's no use flying in the wrong direction with the right speed!

You typically need more than one burn to get into the desired orbit. In theory you could get into a sea level Earth orbit with a single impulsive burn that raises your perigee from the center to the surface of the Earth. In practice your burn cannot be impulsive and even if it could be you would burn up in the atmosphere and smash into mountains and other obstacles. To get into circular LEO you would need at least two impulsive burns: one burn to raise either apogee or perigee to the desired altitude, and once at the new apogee another circularisation burn to raise the new perigee to the desired altitude. In the real world you need to raise your current apogee by enough to escape the effects of drag and heating and to give your engines enough time to raise your perigee.

All in all you'd need more delta-v than you'd expect from just comparing energies.

Taking the 100km circ LEO example where SOE = −30.8 MJ/kg and v = 7.85km/s.

To escape, another total ∆v of 7.85km/s would be required by 1/2(∆v)2.

Instead, Oberth gives ∆SEO = (v * ∆v) + 1/2(∆v)2 and ∆v of 3.25 km/s is required for escape, as you'd expect.



Starting from the same LEO, a ∆v of 4.25 km/s gives me a ∆SEO of 42.4 MJ/Kg. Presumably, I have this right that this results in a C3 = (42.4 - 30.8) * 2 = 23.2 km2/s2?

cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: yinzer on 01/09/2010 06:30 pm
It all boils down to the equation for specific orbital energy:
Quote
C3/2 = E = (v^2/2) - µ/r

Trying to relate delta-energy and delta-velocity is going to require calculus, so is probably best avoided.  Just figure out what terms of the specific orbital energy equation you know and solve for the remaining one.

For escape, C3 = 0 so you can solve the above equation with appropriate values of µ and r.  100 km from Earth gives 0 = v^2/2 - (3.99e6 / 6.46e3), or v = 11.1 km/sec.  If your orbit has you going 7.85 km/sec at 100 km, you indeed need a delta-V of 3.25 km/sec.

For your other example, after adding 4.25 km/sec from your LEO, you are at 100km altitude and travelling 12.1 km/sec.  v^2/2 - µ/r, or (12.1^2)/2 - 3.99e6/6.46e3, or 11.47 km^2/sec^2.  This is energy, so double it for C3, and you end up with roughly the same number (22.9 vs. 23.2, but that might be rounding errors).
Title: Re: Basic Rocket Science Q & A
Post by: tnphysics on 01/12/2010 12:33 am
Can one use a lunar gravity assist to aid TMI?
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 01/12/2010 10:27 am
Yes, Farquhar proposed a double gravity assist from both the Earth and the moon from SEL-2. If you want to launch straight from LEO it might be different.
Title: Re: Basic Rocket Science Q & A
Post by: tnphysics on 01/14/2010 01:38 am
From LEO
Title: Re: Basic Rocket Science Q & A
Post by: yinzer on 01/14/2010 03:01 am
Theoretically you could, but the moon's gravity is not that strong so it won't help that much.  There may not be many times when the Moon is in a useful place relative to the outbound trajectory to Mars, either.
Title: Re: Basic Rocket Science Q & A
Post by: Cbased on 01/18/2010 10:25 am
When do you usually stop using the term "altitude" and start using "distance from/to a planet"?
Thank you.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 01/18/2010 11:28 am
When do you usually stop using the term "altitude" and start using "distance from/to a planet"?
Thank you.


Generally, "alitude" is used to identify an orbital path around a planet or moon and is repeating with each succeeding orbit while "distance" generally describes, for spacecraft anyway, how far it is from the planet/moon as it is approaching or departing.
Title: Re: Basic Rocket Science Q & A
Post by: thomson on 01/19/2010 05:17 pm
Very basic question.

In one message (http://forum.nasaspaceflight.com/index.php?topic=13206.msg526703#msg526703 (http://forum.nasaspaceflight.com/index.php?topic=13206.msg526703#msg526703)) Robotbeat stated that 195 second hover (a recent feat accomplished by Masten team) is enough for a trip from L2 to high Mars orbit. My question is how does one convert hover time to delta-v? Is it a matter of simple multiplying hover time by G? So you get:

dV= t*G = 195s*9.81m/s^2 = 1912m/s

Is this right?

My second question is built on top of the first one. If this is correct, then one could calculate required hover time for a vehicle capable of reaching orbital velocity. I've assumed v0=9300m/s, so required hover time is 948s. Ignoring atmosphere drag, it is valid to say that if Masten vehicle would be able to achieve ~20% (195s/948s) of orbital velocity?

Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 01/19/2010 05:29 pm
Generally, "alitude" is used to identify an orbital path around a planet or moon and is repeating with each succeeding orbit while "distance" generally describes, for spacecraft anyway, how far it is from the planet/moon as it is approaching or departing.

There's also the difference between distance to the surface (altitude) and distance to the center of mass of the planet/moon. This is the terminology the Orbiter (http://orbit.medphys.ucl.ac.uk/) flight simulator uses. Altitude is more useful close to the surface, distance from the center of mass is more useful when you're farther away and when integrating the equations of motion.
Title: Re: Basic Rocket Science Q & A
Post by: sdsds on 01/19/2010 08:18 pm
Very basic question.

In one message (http://forum.nasaspaceflight.com/index.php?topic=13206.msg526703#msg526703 (http://forum.nasaspaceflight.com/index.php?topic=13206.msg526703#msg526703)) Robotbeat stated that 195 second hover (a recent feat accomplished by Masten team) is enough for a trip from L2 to high Mars orbit. My question is how does one convert hover time to delta-v? Is it a matter of simple multiplying hover time by G? So you get:

dV= t*G = 195s*9.81m/s^2 = 1912m/s

Is this right?

My second question is built on top of the first one. If this is correct, then one could calculate required hover time for a vehicle capable of reaching orbital velocity. I've assumed v0=9300m/s, so required hover time is 948s. Ignoring atmosphere drag, it is valid to say that if Masten vehicle would be able to achieve ~20% (195s/948s) of orbital velocity?

The assumptions underlying your second question probably don't lead to accurate results.  For a first approximation, try thinking about it another way.  If a vehicle was launched straight up, and the top of its ascent was at an altitude where the velocity required to reach orbit was 7500 m/s, and if the vehicle could then instantaneously propel itself sideways to gain 7500 m/s, it would be in orbit.  If instead that propulsion were distributed over 948 seconds, it would likely fail to make orbit because during those 15+ minutes the vehicle would have fallen to an altitude where its trajectory enters the atmosphere.  Does that way of thinking about it help at all?  (There are better ways to think about it involving the rocket equation and thrust/weight ratios, but those ways are "rocket science."  They require <shiver> logarithms!)

P.S.:  It would be interesting to calculate the altitude to which Xoie would have to be lifted so that she could make orbit under her own propulsion!
Title: Re: Basic Rocket Science Q & A
Post by: jongoff on 01/19/2010 10:54 pm
Very basic question.

In one message (http://forum.nasaspaceflight.com/index.php?topic=13206.msg526703#msg526703 (http://forum.nasaspaceflight.com/index.php?topic=13206.msg526703#msg526703)) Robotbeat stated that 195 second hover (a recent feat accomplished by Masten team) is enough for a trip from L2 to high Mars orbit. My question is how does one convert hover time to delta-v? Is it a matter of simple multiplying hover time by G? So you get:

dV= t*G = 195s*9.81m/s^2 = 1912m/s

Is this right?

That's basically right.  You're providing a constant 1g=9.81m/s^2 worth of acceleration for most of the flight (except for brief accelerations and decelerations at the ascent/descent/translation tipping), so if you multiply that by time that is how much velocity change you could have done *assuming that throttling profile, and those atmospheric conditions*.  Our mission averaged Isp is actually a lot worse than our full-throttle Isp, which is less than our vacuum Isp, which is a lot less than our Vacuum Isp with a high expansion ratio nozzle extension (I can't remember what our mission-averaged Isp was, but I think it was in the 180-190s range, our full throttle launch altitude Isp is around 215-220s range, our vacuum Isp with the short nozzle is probably in the 250s range, and with a full altitude-optimized nozzle is in the 300-310s range).

Quote
My second question is built on top of the first one. If this is correct, then one could calculate required hover time for a vehicle capable of reaching orbital velocity. I've assumed v0=9300m/s, so required hover time is 948s. Ignoring atmosphere drag, it is valid to say that if Masten vehicle would be able to achieve ~20% (195s/948s) of orbital velocity?

Only sort of.  One of the big challenges for long hovers is that you have to have a very wide throttling range, and your backpressure losses at low-throttle start eating your lunch.  Designing for a long hover that has to get down to low throttles really killed our mission-averaged Isp, and it only gets worse the longer the duration you have to do.

Without some sort of altitude compensation, multiple engines (that you can shutoff in series), or flow-separation control, we'd be hard pressed to get anywhere near even half of that.

~Jon
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/23/2010 09:56 pm
I'm guessing for stability, but are there other reasons? 
What would be the result in air pressure, if an SRB were placed between 2 fins?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 01/23/2010 10:04 pm
I'm guessing for stability, but are there other reasons? 
What would be the result in air pressure, if an SRB were placed between 2 fins?

No other reason.  Too many variables to answer the last question.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 01/23/2010 11:21 pm
Assuming their angle of incidence is zero, fins always move the center of pressure toward the fin.  Assuming the center of pressure is aft of the center of gravity, they also act like dampers in a spring, mass, damper system.
Title: Re: Basic Rocket Science Q & A
Post by: Fequalsma on 01/24/2010 01:45 am
Yep, positive static stability is a GOOD thing!  Gotta
keep CG in front of the AC, or bad things will happen...
F=ma

Assuming their angle of incidence is zero, fins always move the center of pressure toward the fin.  Assuming the center of pressure is aft of the center of gravity, they also act like dampers in a spring, mass, damper system.
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/24/2010 02:45 pm
I'm guessing for stability, but are there other reasons? 
What would be the result in air pressure, if an SRB were placed between 2 fins?

  Too many variables to answer the last question.

Oh, I don't know about that.  Let's think this thing through.  The fin is parallel with the SRB (roughly speaking).  And the SRB skirt would be at a 45 deg. angle to the fin (roughly speaking).  Wouldn't a column of air comming straight down from the SRB get caught between the fin and skirt (45 deg. pointing down), and get thrusted beneath the base of the rocket?  Might be another base heating variable.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 01/24/2010 02:54 pm
I'm guessing for stability, but are there other reasons? 
What would be the result in air pressure, if an SRB were placed between 2 fins?

  Too many variables to answer the last question.

Oh, I don't know about that.  Let's think this thing through.  The fin is parallel with the SRB (roughly speaking).  And the SRB skirt would be at a 45 deg. angle to the fin (roughly speaking).  Wouldn't a column of air comming straight down from the SRB get caught between the fin and skirt (45 deg. pointing down), and get thrusted beneath the base of the rocket?  Might be another base heating variable.

What skirt?  Whose says the skirt and fin are on the same level? 

No, too many variables and you even say it.
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/24/2010 03:56 pm
I'm guessing for stability, but are there other reasons? 
What would be the result in air pressure, if an SRB were placed between 2 fins?



  Too many variables to answer the last question.

Oh, I don't know about that.  Let's think this thing through.  The fin is parallel with the SRB (roughly speaking).  And the SRB skirt would be at a 45 deg. angle to the fin (roughly speaking).  Wouldn't a column of air comming straight down from the SRB get caught between the fin and skirt (45 deg. pointing down), and get thrusted beneath the base of the rocket?  Might be another base heating variable.

What skirt?  Whose says the skirt and fin are on the same level? 



The SRB skirt.  Of all the pictures I've seen of Ares V (or Direct) the SRB skirt is at the base of the rocket.  Fins could also be placed at the base of the rocket, unless you can think of a reason why they shouldn't be.  All the rocket nozzles are located beneath the base of the rocket.  Now that I think of it, the air flow between the fin and SRB skirt, MIGHT flow directly on to the nozzles!  And with 2 SRB's, that would be at 4 locations, near the perimeter of the base of the rocket.  Thanks, Jim.

Not all of the flow would go beneath the base, to the nozzles.  Maybe 1/2 of the flow, nearest the 1st stage core.  This is because the flow goes between the skirt (which is round) and the fin section, which is flat, and the base core (which is round, but we can consider it flat).
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 01/24/2010 04:37 pm
Fins could also be placed at the base of the rocket, unless you can think of a reason why they shouldn't be.

No need, only extra weight and drag.

This is what is used.  The scoops are on right and left of the base of the stage
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/24/2010 04:50 pm
Fins could also be placed at the base of the rocket, unless you can think of a reason why they shouldn't be.

No need, only extra weight and drag.

This is what is used.  The scoops are on right and left of the base of the stage

Thank you, I see the scoops with the holes in them.  Now the question is, are the fins more aerodynamic than the scoops?  IIRC, the fins on Apollo, only had an internal tubing (conduit) structure.

The scoops are a pretty clever idea.  But another question, would be the quantity of air flow.  And which method would give more.
Title: Re: Basic Rocket Science Q & A
Post by: Art LeBrun on 01/25/2010 01:49 am
Fins could also be placed at the base of the rocket, unless you can think of a reason why they shouldn't be.

No need, only extra weight and drag.

This is what is used.  The scoops are on right and left of the base of the stage

Jim's image is launch of GT-9.

Attached is an image of GT-2 12557. Notice scoops are not in place but a hole pattern seems to be part of the mounting. Any comments that will help? Did these scoops fly on all Gemini-Titans?
Title: Re: Basic Rocket Science Q & A
Post by: Integrator on 01/25/2010 02:08 am
I understood on Saturn V the fins were to add stability in case of an abort.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 01/25/2010 03:00 am
Yep, positive static stability is a GOOD thing!  Gotta
keep CG in front of the AC, or bad things will happen...

Not with a digital FCS.  B-2s fly just fine.... usually.
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 01/25/2010 05:58 am
I understood on Saturn V the fins were to add stability in case of an abort.

That is my understanding as well. They did not add much stability to a nominal ascent, but in a first-stage loss of control case, they would delay the onset of high rates enough to allow the LES to pull the CM away.
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/25/2010 01:36 pm
Fins could also be placed at the base of the rocket, unless you can think of a reason why they shouldn't be.

No need, only extra weight and drag.

This is what is used.  The scoops are on right and left of the base of the stage

I congradulate Jim for comming up with a simple, yet practical solution for the base heating problem, concerning the RS-68 engine!  Well done.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 01/25/2010 01:48 pm

I congradulate Jim for comming up with a simple, yet practical solution for the base heating problem, concerning the RS-68 engine!  Well done.

That isn't a solution.  The base heating problem is radiant heat from SRB.  The scoops are for base recirculation issues.
Title: Re: Basic Rocket Science Q & A
Post by: Downix on 01/25/2010 01:53 pm

I congradulate Jim for comming up with a simple, yet practical solution for the base heating problem, concerning the RS-68 engine!  Well done.

That isn't a solution.  The base heating problem is radiant heat from SRB.  The scoops are for base recirculation issues.
If he'd read the thread he claimed was solved, he'd know this too.
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/25/2010 03:53 pm

I congradulate Jim for comming up with a simple, yet practical solution for the base heating problem, concerning the RS-68 engine!  Well done.

That isn't a solution.  The base heating problem is radiant heat from SRB.  The scoops are for base recirculation issues.

It may not be the total solution, but it certainly is a partial solution in keeping the base heat from rising any higher than it otherwise would.  Maybe in combination with heat dissipating fins on the RS-68 nozzles, to keep the ablative from disintigrating faster.  As you stated, the SRB heat radiation problem isn't likely to be solved, short of extending the perimeter of the base to cover the RS-68 nozzles, that are nearest to the SRB exhaust.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 01/25/2010 04:00 pm
Maybe in combination with heat dissipating fins on the RS-68 nozzles,

Not feasible.  The fins wouldn't dissipate heat, they would absorb it from the SRB's.  Plus the construction of the nozzle (composite overlap) is not conducive to fins
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 01/25/2010 04:00 pm
This problem is not going to be solved on this forum.  It is too complex
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/28/2010 02:21 pm
  As you stated, the SRB heat radiation problem isn't likely to be solved, short of extending the perimeter of the base to cover the RS-68 nozzles, that are nearest to the SRB exhaust.

What is this "Boat Tail" that I read about in other threads?  Is it similar to an inverted, tapered, interstage at the base of the rocket?
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/28/2010 07:51 pm
On page 3, I found a picture of a Boat Tail on a Model Rocket:

http://www.2020vertical.com/nar_edu_cd_dev/lessons/apogee/Reports/Rocket_parts.pdf

Boattail
Boattail is a drag reducing part on the back of the rocket.
It helps direct airflow around the base of the rocket. In effect,
it keeps the flow smooth, which reduces the aerodynamic drag
and allows the rocket to fly higher into the air.


Hmmmm.... Combine a Boat Tail with air scoops, and what do you have?  A possible solution to base heating, for the RS-68 engine?
Title: Re: Basic Rocket Science Q & A
Post by: MarsInMyLifetime on 01/28/2010 10:55 pm
That model rocketry magazine also describes shock cords, which don't apply to SRBs, so you have to watch how you extrapolate some information you come across.  ;)

I come back to what Jim said: the problem is radiant heat, and the problem exists beyond the atmosphere where scoops and boattails no longer matter. Not to discourage your thinking, but Jim mentioned radiant heat specifically because redirected air is a solution to the wrong problem.

Model rocketry does have some good things to offer on subsonic problems of smaller rockets, though. This NAR test report is a nice layman's introduction to the subject of base drag and boattails:
http://www.interactiveinstruments.com/pdfs/28.pdf
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/29/2010 03:20 pm
That model rocketry magazine also describes shock cords, which don't apply to SRBs, so you have to watch how you extrapolate some information you come across.  ;)

Understood.  I needed a picture and description of a boat tail on a rocket, and this was the best I could find.

I come back to what Jim said: the problem is radiant heat, and the problem exists beyond the atmosphere where scoops and boattails no longer matter.

The air traveling thru the scoops must be traveling at sonic speeds with in the boat tail.  And could be directed downward to the RS-68 nozzles by short ducts.  I disagree about the problem existing beyond the atmosphere, because we're talking about the 1st stage only.  And the SRB's fall away after 2 min. 

 
Not to discourage your thinking, but Jim mentioned radiant heat specifically because redirected air is a solution to the wrong problem.

The RS-68 nozzles are completly covered, and are protected directly from the radiant heat of the SRB's by a (carbon fiber) boat tail.  Unless you're saying that the radiant heat (from the SRB) is comming directly thru the RS-68 plume.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 01/29/2010 03:30 pm
The air traveling thru the scoops must be traveling at sonic speeds with in the boat tail.  And could be directed downward to the RS-68 nozzles by short ducts.  I disagree about the problem existing beyond the atmosphere, because we're talking about the 1st stage only. 


First stage does leave the sensible atmosphere
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/29/2010 03:41 pm
The air traveling thru the scoops must be traveling at sonic speeds with in the boat tail.  And could be directed downward to the RS-68 nozzles by short ducts.  I disagree about the problem existing beyond the atmosphere, because we're talking about the 1st stage only. 


First stage does leave the sensible atmosphere

Does a sensible atmosphere exist, before the SRB's fall away?  I agree that there is a "point of diminishing returns" for the air scoops, as the rocket gains altitude.
Title: Re: Basic Rocket Science Q & A
Post by: Downix on 01/29/2010 03:51 pm
On page 3, I found a picture of a Boat Tail on a Model Rocket:

http://www.2020vertical.com/nar_edu_cd_dev/lessons/apogee/Reports/Rocket_parts.pdf

Boattail
Boattail is a drag reducing part on the back of the rocket.
It helps direct airflow around the base of the rocket. In effect,
it keeps the flow smooth, which reduces the aerodynamic drag
and allows the rocket to fly higher into the air.


Hmmmm.... Combine a Boat Tail with air scoops, and what do you have?  A possible solution to base heating, for the RS-68 engine?

You do realize this is why the DIRECT proposals using the RS-68 had a boattail, yes?

The issue with Ares V having one is due to the sheer # of engines.  DIRECT had, at most, 3 engines on an 8.4m core.  A 10m core would not add enough width, and the engines would burn out before they had used that much fuel, and would be unable to lift that size anyways.

Review DIRECT 2.0 if you want to see how they worked on solving these issues.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 01/29/2010 03:52 pm
The air traveling thru the scoops must be traveling at sonic speeds with in the boat tail.  And could be directed downward to the RS-68 nozzles by short ducts.  I disagree about the problem existing beyond the atmosphere, because we're talking about the 1st stage only. 


First stage does leave the sensible atmosphere

Does a sensible atmosphere exist, before the SRB's fall away?  I agree that there is a "point of diminishing returns" for the air scoops, as the rocket gains altitude.

it is gone maybe 30 seconds before that
Title: Re: Basic Rocket Science Q & A
Post by: MarsInMyLifetime on 01/29/2010 05:29 pm
Kyle, I'll note that scoops have their place--they make great sense for cruise missiles, which fly at one speed at one height, so the design variables are well controlled for that case. The SRB phase has the complexity of speeds ranging from 0 to Mach 4 and air pressures from sea level to 25 nautical miles, so the conditions are extremely dynamic during the period where their radiant heating is the concern.
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/29/2010 08:21 pm



You do realize this is why the DIRECT proposals using the RS-68 had a boattail, yes?

IIRC, Direct had a smaller boattail that didn't completly cover the nozzles of the RS-68.  I guess the reason is, that the engines were lined up perpendicular to the SRB's and were a good distance away.


The issue with Ares V having one is due to the sheer # of engines.  DIRECT had, at most, 3 engines on an 8.4m core.  A 10m core would not add enough width, and the engines would burn out before they had used that much fuel, and would be unable to lift that size anyways.

Any version of Direct that Nasa may adopt for HLV, will most likely have a streched tank, and one or more engines than Direct (in a square pattern)
Title: Re: Basic Rocket Science Q & A
Post by: edkyle99 on 01/30/2010 12:44 am
I'm guessing for stability, but are there other reasons? 
What would be the result in air pressure, if an SRB were placed between 2 fins?

Early missiles like V-2 and Corporal and Redstone had fins in part because the fins had movable surfaces that were part of the missile control system (along with movable vanes that extended into the exhaust).  The rocket nozzles didn't move in these rockets. 

Later missiles, like Jupiter (the real Jupiter) and Atlas and Titan (and later Thors) did not have any fins.  (Early R&D Thors had small fins, but on one flight some of the fins were ripped off of the missile, and the missile kept flying stable, so Douglas deleted them from the design.)

Saturn I Block I did not have fins.  They were added to the Block II Saturn for two reasons.  First, to provide stability for a projected winged Air Force payload.  Second, to provide stability for a few seconds in case of engine shutdown and activation of an Apollo launch escape system (Saturn I Block II was slated for several manned Apollo flights when it was designed).  Fins appeared on Saturn IB and Saturn V for the same reasons. 

 - Ed Kyle
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/30/2010 03:16 pm
Kyle, I'll note that scoops have their place--they make great sense for cruise missiles, which fly at one speed at one height, so the design variables are well controlled for that case. The SRB phase has the complexity of speeds ranging from 0 to Mach 4 and air pressures from sea level to 25 nautical miles, so the conditions are extremely dynamic during the period where their radiant heating is the concern.

I'm not advocating a perfect solution for the base heating problem, concerning the RS-68 engines.  I appreciate all the black and white analysis, concerning this problem.  But the final solution, is probably closer to a dark grey.  Meaning, a combination of good (not perfect) and practical solutions.  Remember, the problem is keeping the ablative nozzles from disintigratiing too quickly BEFORE the SRB's fall away.
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 01/30/2010 03:26 pm
I just noticed something on page 3 of the illustration of the rocket.  It's called a "Boattail shoulder".  Is this a necessary part of the boattail, or can it be ignored? 

http://www.2020vertical.com/nar_edu_cd_dev/lessons/apogee/Reports/Rocket_parts.pdf
Title: Re: Basic Rocket Science Q & A
Post by: Downix on 02/01/2010 12:10 pm
I just noticed something on page 3 of the illustration of the rocket.  It's called a "Boattail shoulder".  Is this a necessary part of the boattail, or can it be ignored? 

http://www.2020vertical.com/nar_edu_cd_dev/lessons/apogee/Reports/Rocket_parts.pdf
It's what holds the boattail in place.
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 02/03/2010 03:25 pm
I don't see how there would be any savings in money or propellent.  You still need a large (commercial) rocket to get the propellent into space.  And there aren't any on the drawing boards as far as I can tell.  Then you launch your deep space rocket with an empty EDS and "filler up" in orbit?  I see this as costing significantly more than having one large rocket.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 02/03/2010 04:54 pm
I don't see how there would be any savings in money or propellent.  You still need a large (commercial) rocket to get the propellent into space.  And there aren't any on the drawing boards as far as I can tell.  Then you launch your deep space rocket with an empty EDS and "filler up" in orbit?  I see this as costing significantly more than having one large rocket.

Because the rocket with the empty EDS can be much smaller. and fly more often
Title: Re: Basic Rocket Science Q & A
Post by: bad_astra on 02/03/2010 05:00 pm
I don't see how there would be any savings in money or propellent.  You still need a large (commercial) rocket to get the propellent into space.  And there aren't any on the drawing boards as far as I can tell.  Then you launch your deep space rocket with an empty EDS and "filler up" in orbit?  I see this as costing significantly more than having one large rocket.
Aquarius LV by Loral is designed soley for tanker duty. There's a good explanation of it on youtube somewhere.

http://homepage.mac.com/fcrossman/NorCalSAMPE/Comp_WS_papers/Turner_012204.pdf
Title: Re: Basic Rocket Science Q & A
Post by: Downix on 02/03/2010 09:35 pm
I don't see how there would be any savings in money or propellent.  You still need a large (commercial) rocket to get the propellent into space.  And there aren't any on the drawing boards as far as I can tell.  Then you launch your deep space rocket with an empty EDS and "filler up" in orbit?  I see this as costing significantly more than having one large rocket.

Because the rocket with the empty EDS can be much smaller. and fly more often
This reminds me, I've often thought that refueling an EDS might not be the most efficient method.  Could not the depot *be* the EDS fuel tank, the individual pods?  So, you fly up, dock with the fuel tank, already in orbit waiting for you.  End result, you take the fuel tank with you when you go.  The Depot then would be a central maintenance system to keep the fuel tanks cryo before usage.
Title: Re: Basic Rocket Science Q & A
Post by: Patchouli on 02/03/2010 10:11 pm
I don't see how there would be any savings in money or propellent.  You still need a large (commercial) rocket to get the propellent into space.  And there aren't any on the drawing boards as far as I can tell.  Then you launch your deep space rocket with an empty EDS and "filler up" in orbit?  I see this as costing significantly more than having one large rocket.

Because the rocket with the empty EDS can be much smaller. and fly more often
This reminds me, I've often thought that refueling an EDS might not be the most efficient method.  Could not the depot *be* the EDS fuel tank, the individual pods?  So, you fly up, dock with the fuel tank, already in orbit waiting for you.  End result, you take the fuel tank with you when you go.  The Depot then would be a central maintenance system to keep the fuel tanks cryo before usage.

One idea I played around with was launching the LOX and fuel tanks of the EDS separately.
Or just having a LOX depot since it is the bulk of the propellant mass.
Title: Re: Basic Rocket Science Q & A
Post by: A_M_Swallow on 02/03/2010 11:23 pm
This reminds me, I've often thought that refueling an EDS might not be the most efficient method.  Could not the depot *be* the EDS fuel tank, the individual pods?  So, you fly up, dock with the fuel tank, already in orbit waiting for you.  End result, you take the fuel tank with you when you go.  The Depot then would be a central maintenance system to keep the fuel tanks cryo before usage.

The propellant needs to be taken to the depot in some sort of tank so you are a third of the way there.  The spacecraft and depot would need systems able to remove the old tanks and fit the new tanks.  The tanks would need to have some kind of docking system that the liquids can flow through.
Title: Re: Basic Rocket Science Q & A
Post by: jongoff on 02/04/2010 01:17 am
I don't see how there would be any savings in money or propellent.  You still need a large (commercial) rocket to get the propellent into space.  And there aren't any on the drawing boards as far as I can tell.  Then you launch your deep space rocket with an empty EDS and "filler up" in orbit?  I see this as costing significantly more than having one large rocket.

The key savings comes from not needing to build and pay the fixed costs an HLV (or even necessarily an "EDS") in the first place.  While having a big LOX/LH2 upper stage on the size scale of an ACES or Raptor stage would help a lot, you can actually do ESAS-class missions using existing launch vehicles if you use depots in LEO and L2.  The Centaur stages would need a Long-Duration Mission Kit (like ULA has suggested in several papers and proposals), but would otherwise be basically the same upper stage that has been flying with a good overall reliability record for decades.  So again, you save the cost of developing all that new hardware and paying all the infrastructure costs for it.  A depot, which can be a fairly simple single-launch spacecraft is likely going to be a lot less than an HLV to develop and operate.  That savings directly translates into more missions and sooner.

The other big potential savings comes down the road as the demand for lots of smaller flights provides economic incentives for the development of high flight-rate, low-cost transportation like fully-reusable LVs, gun launch schemes of various sorts, various beamed-propulsion concepts, etc.  There are methods that could drive the cost of propellant to orbit by at least an order of magnitude or more.  With a non-depot, HLV-centric architecture, even if someone invented a earth-to-LEO matter teleporter that could put propellant up for free, they wouldn't be able to benefit from it.

HLVs only win economically if you assume that a) Depots are going to be so super hard to develop, b) that they'll be super expensive to operate, c) make really pessimistic assumptions about how tanker trips are done, and d) assume that it's impossible to improve over current costs of access to orbit.

~Jon
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 02/04/2010 11:41 am
HLVs only win economically if you assume that a) Depots are going to be so super hard to develop, b) that they'll be super expensive to operate, c) make really pessimistic assumptions about how tanker trips are done, and d) assume that it's impossible to improve over current costs of access to orbit.

~Jon

There is a lot more to HLV's than just economics. There are also mission requirements to consider. 5m plf's do not support a *lot* of things. Explain, for example, how depots enable large volume plf's, which will be necessary for any planetary EDL system. The *size* of the EDL system is just one example of the need for very large plf's. Only the HLV provides large plf's.

Please don't take this as being anti-depot; because I'm not. I'm very pro-depot, as you know. It's just that so many people want to occupy the polar opposites and so few take the position that a healthy program makes effective use of *both*. Stop demonizing the HLV because we need one, as the Augustine Commission clearly stated ("don't skimp on the heavy lift").

People who insist that either the depot must "win" or the HLV must "win" are either uninformed or just not being honest with themselves. As you know, the Jupiter is a HLV, but every proposal we made to Congress, NASA and industry lobbied hard, *very* hard, for the depot-based architecture. Why? Because, as I stated above, "a healthy program makes effective use of *both*".
Title: Re: Basic Rocket Science Q & A
Post by: cro-magnon gramps on 02/04/2010 08:59 pm
this may have been asked before, so forgive my ignorance

can a vasimr engine be scaled up to launch a HLV (80 to 100mt)
what are the restrictions,

is there an air breathing engine (ie Skylon) that could be scaled up to HLV capabilities

is internal combustion still the most efficient method of launching a rocket of this size, and all we can do is tweek it

this is not a question of do we need a HLV, but a real question of Advanced Rocket Propulsion
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 02/04/2010 09:03 pm
People who insist that either the depot must "win" or the HLV must "win" are either uninformed or just not being honest with themselves.

Careful there. I don't think Team Direct is in a position to lecture others on honesty or truthfulness. The truth of the matter is that indeed it is not either/or if you look at it exclusively from the point of view of exploration. Depots add flexibility, but that's something you do not strictly need if you have an HLV, at least not in the early stages. On the other hand, if you look at it from the point of commercial development of space and seeking maximum synergies with exploration, then absence of an HLV is probably crucial, at least in the short term.

There's an asymmetry here: presence of depots is not crucial to exploration in the short term, but absence of HLV is crucial for commercial development of space in the short term. Presence of depots is not harmful to exploration, even in the long term and absence of HLV is not harmful to exploration, even in the long term.

That is the best argument for looking at it as either/or. Deliberately leaving out that point of view would be, well, dishonest.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 02/05/2010 12:01 pm
People who insist that either the depot must "win" or the HLV must "win" are either uninformed or just not being honest with themselves.
Careful there. I don't think Team Direct is in a position to lecture others on honesty or truthfulness.

Are you actually accusing us of being dishonest and untruthful?
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 02/05/2010 06:24 pm
1) can a vasimr engine be scaled up to launch a HLV (80 to 100mt)
what are the restrictions,

2) is internal combustion still the most efficient method of launching a rocket of this size, and all we can do is tweek it

1) Power is a far bigger problem for an electric engine than thrust, so no.  You'd need gigawatts or high hundreds of megawatts.

2) Thrust is the need on a booster engine.  A reusable (and of course retrievable) air-breather is the way to go so it needs half as much tank.  I'm of the opinion that NACA was progressing nicely in building us up from the ground, and then the Moon Race got in the way.  Start where X-15 left off.  It probably ends up at rocket-based combined cycle, if one insists on a single stage.
Title: Re: Basic Rocket Science Q & A
Post by: ginahoy on 02/07/2010 05:35 am
I've always been curious about the 2 days it takes for the shuttle to reach ISS. Aside from the inspections that must be done, presumably the reason for a slow approach is to conserve fuel and thus maximize payload capacity.

Just out of curiosity, is it even possible for shuttle to reach ISS in its first orbit, even an empty shuttle?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 02/07/2010 10:18 am
I've always been curious about the 2 days it takes for the shuttle to reach ISS. Aside from the inspections that must be done, presumably the reason for a slow approach is to conserve fuel and thus maximize payload capacity.

Just out of curiosity, is it even possible for shuttle to reach ISS in its first orbit, even an empty shuttle?

No, it does have the propellant or thrust to weight ratio like Gemini.  If it did, the ISS couldn't take the plume impingement

The payload mass is not significant, the orbiter still weighs around 200k lb.
Title: Re: Basic Rocket Science Q & A
Post by: ginahoy on 02/07/2010 04:42 pm
I was talking about *launching* into ISS orbit. Without 50k lbs of payload, and with maximum propellant load, presumably the SSME's with OMS assist would be capable of reaching a higher orbit at liftoff. But I have no clue how much higher.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 02/07/2010 04:59 pm
I was talking about *launching* into ISS orbit. Without 50k lbs of payload, and with maximum propellant load, presumably the SSME's with OMS assist would be capable of reaching a higher orbit at liftoff. But I have no clue how much higher.

The Shuttle has always been able to reach the ISS with payload.  The "slow" approach is not for saving ET propellant but OMS and RCS.  The slow approach allows for the orbiter to make small delta v changes vs the large ones that Gemini and Apollo did. 
Title: Re: Basic Rocket Science Q & A
Post by: ginahoy on 02/07/2010 05:14 pm
OK. Not sure if you answered my question. What I had in mind is shuttle reaching ISS orbit at MECO. If this is possible, then it seems possible that shuttle could arrive at station vicinity within first orbit (retaining enough OMS/RCS prop for nominal docking/undocking maneuvers & landing). Just wondering if possible, not whether it's desirable.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 02/07/2010 05:52 pm
OK. Not sure if you answered my question. What I had in mind is shuttle reaching ISS orbit at MECO. If this is possible, then it seems possible that shuttle could arrive at station vicinity within first orbit (retaining enough OMS/RCS prop for nominal docking/undocking maneuvers & landing). Just wondering if possible, not whether it's desirable.

That is my point, it can't, it would be approaching the station too fast  and OMS-2 is performed. There will be dispersion, which will require too large of delta V  and the plumes would have a major affect on the station.

You don't approach your car parking spot at 50 mph.  You slow down way before it.

Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 02/07/2010 06:05 pm
I've always been curious about the 2 days it takes for the shuttle to reach ISS. Aside from the inspections that must be done, presumably the reason for a slow approach is to conserve fuel and thus maximize payload capacity.

No, to first order, propellant consumption is proportional to delta-H so for ISS at a given altitude, the propellant required to reach it is constant.

First orbit rendezvous has narrow and infrequent launch windows. That is why it is not done. Flight day 3 rendezvous allows every-day launch windows while standardizing the timeline. Soyuz performs rendezvous on flight day 3 just like the shuttle, and for the same reason.

Quote
Just out of curiosity, is it even possible for shuttle to reach ISS in its first orbit, even an empty shuttle?

As Jim wrote, the shuttle has the propellant to do so, but this would require the shuttle to perform a large amount of RCS braking and the plume impingement from the braking burns would destroy ISS.
Title: Re: Basic Rocket Science Q & A
Post by: ginahoy on 02/07/2010 06:48 pm
Ok, thanks. I understand now.
Title: Re: Basic Rocket Science Q & A
Post by: helloworld on 02/08/2010 08:05 pm
I couldn't find any info on google.....
Could anyone tell me what it means please?
Title: Re: Basic Rocket Science Q & A
Post by: Lee Jay on 02/08/2010 08:09 pm
When you launch, you're typically in an elliptical (oval-shaped) orbit.  If you wait until the point where you are highest (apogee) and do a little burn there, you can change your orbit to circular, thus "circularizing" your orbit.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 02/08/2010 08:14 pm
If you were to launch a spacecraft from the Earth with an instantaneous boost (say with a launch cannon or something), you would end up in a highly elliptical orbit. Its highest point could be high above the atmosphere, but its lowest point would be deep beneath the surface, close to the center of the Earth. This means your spacecraft would crash into the ground very soon. A circularisation burn at the highest point would lift the lowest point to equal height, turning the trajectory into a true orbit.

Real launches are more complicated than this because the period of acceleration is stretched out over a period of several minutes, but the situation is similar. You typically first need to gain height to get above the atmosphere and then you want to raise your lowest point as soon as possible. The circularisation burn can then be small because your lowest point is then already above the atmosphere.
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 02/12/2010 04:05 pm
The air traveling thru the scoops must be traveling at sonic speeds with in the boat tail.  And could be directed downward to the RS-68 nozzles by short ducts.  I disagree about the problem existing beyond the atmosphere, because we're talking about the 1st stage only. 


First stage does leave the sensible atmosphere

Does a sensible atmosphere exist, before the SRB's fall away?  I agree that there is a "point of diminishing returns" for the air scoops, as the rocket gains altitude.

it is gone maybe 30 seconds before that

For a quick review:

1.  Air scoops located at the top perimeter of the boat tail.

2.  The boat tail completly covers the RS-68 nozzles (except at the bottom, of course).

My question is, if the sensible atmosphere is gone at 1:30 sec. after liftoff, inside the boattail would be a vacuum.  How does this vacuum affect the radiant heat comming off the RS-68 nozzles?  I would guess, because there is no air left in the boatttail, the heat would still radiate inside the boattail wall, but the temperature couldn't climb any higher than the outside temperature of the RS-68 nozzle.  Is that a fair analysis? 

And once the boattail has served its purpose after SRB seperation at 2:00 min. it could be removed with line charges.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 02/12/2010 04:29 pm

2.  The boat tail completly covers the RS-68 nozzles (except at the bottom, of course).


Not that easy. The engine exhausts hot hydrogen from the heat exchanger and roll control nozzle (which is from the turbopump exhaust).  Also the nozzles have to be able to move.

http://www.tallgeorge.com/images/projectconstellation/Boeing%20Rocketdyne%20RS-68%20Engine.jpg

We aren't going to find a solution here.  It is too complex
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 02/12/2010 08:14 pm

2.  The boat tail completly covers the RS-68 nozzles (except at the bottom, of course).


Not that easy. The engine exhausts hot hydrogen from the heat exchanger and roll control nozzle (which is from the turbopump exhaust).  Also the nozzles have to be able to move.

http://www.tallgeorge.com/images/projectconstellation/Boeing%20Rocketdyne%20RS-68%20Engine.jpg


Your points are valid, if Nasa choses the Ares V (Max) with 6 engines, or the Ares V (Super Max) with 7 engines.  However, we both know that that's never going to happen.  The most likely scenario is a 8.4m core that is stretched with 4 engines (maybe 5).  In this case, the nozzles have plenty of room to gimbal, and the enviornment with in the boattail should be acceptable (IMO) to the hot hydrogen exhausts from the heat exchanger and roll control nozzles.

Remember, that you said, the main problem was the radiant heat comming from the SRB plume.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 02/13/2010 02:51 pm
2.  The boat tail completly covers the RS-68 nozzles (except at the bottom, of course).

Uh, no.  Almost the whole nozzle is outside the heat shield.
Title: Re: Basic Rocket Science Q & A
Post by: kyle_baron on 02/13/2010 03:14 pm
2.  The boat tail completly covers the RS-68 nozzles (except at the bottom, of course).

Uh, no.  Almost the whole nozzle is outside the heat shield.

That's the way Direct, uses a boattail.  I'm looking at the boattail from a model rocket point of view, as shown on page 3:

http://www.2020vertical.com/nar_edu_cd_dev/lessons/apogee/Reports/Rocket_parts.pdf

Boattail
Boattail is a drag reducing part on the back of the rocket.
It helps direct airflow around the base of the rocket. In effect,
it keeps the flow smooth, which reduces the aerodynamic drag
and allows the rocket to fly higher into the air.
Title: Re: Basic Rocket Science Q & A
Post by: Warren Platts on 04/11/2010 03:13 am
I have a question: How is the mass of a spacecraft supposed to scale with respect to the payload it is required to carry, assuming the overall volume of the spacecraft doesn't change much?

I'm thinking specifically of lunar SSTO landers where the maximum g-force wouldn't exceed 0.5 g.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 04/11/2010 04:53 am
I have a question: How is the mass of a spacecraft supposed to scale with respect to the payload it is required to carry, assuming the overall volume of the spacecraft doesn't change much?

I'm thinking specifically of lunar SSTO landers where the maximum g-force wouldn't exceed 0.5 g.

Volume has no bearing for a lunar SSTO, where there is no air.

Title: Re: Basic Rocket Science Q & A
Post by: Warren Platts on 04/11/2010 05:41 am
Quote from: The Jim
Volume has no bearing for a lunar SSTO, where there is no air.
True, except for the fact that it's got to fit within the faring of an EELV (unless of course we want to contemplate manufacturing the SSTO right there on the Moon! ;)). I was thinking of a modified and beefed up ULA DTAL SSTO (using the ACES-71 tank volume) with the RL-10 motors modified to run on ALLOX (powdered aluminum and LO2 monopropellant). The Isp is low, but the thrust is high. A BOTE calculation reveals the following:

Outward Leg (mass in mt)
 30 Empty Mass
200 Payload
 31 Return Propellant
261 Total 1st Leg "dry" mass
272 Launch propellant
304 Total propellant (tank capacity = 320 mt)
534 GLOW

Return Leg
30 Empty Mass
31 Return propellant
61 Gross Return Weight


Obviously, the 200 ton payload is overly optimistic since gravity losses would be a lot under this configuration. But even if the payload had to be slashed in half, 100 tons to orbit for a reusable SSTO would be pretty good--that's the holy grail of rocket science, is it not?

However, the dry mass is based on the mass of an ordinary DTAL (stretched out a little, but still able to fit within a standard EELV faring). Since the GLOW is massively higher, though, how much would you have to increase the dry weight of the SSTO to handle the extra weight of the ALLOX propellant and still be able to fit into a standard EELV (heavy) faring?
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 04/23/2010 04:46 pm
It is often said that the velocity gain achieved by airlaunch is fairly trivial, and could be matched simply by a slightly larger first stage. It is also said that the velocity generated by a simple hop to 100 km is about 3 percent of that required for orbital.

While watching some recent launches, I noticed that its not uncommon for the vehicle to achieve Mach One at about 50,000 feet, which is in the ball park of the velocity gain generated by air launch. So, my question becomes a little hypothetical, but:

If a Falcon 1 were carried aloft by a Lockheed 1011 (or comparable aircraft), and air dropped at .85 Mach and 40,000 feet, how much smaller could the first stage be to take advantage of the altitude gain and velocity increase from air launch, and still meet the nominal Falcon 1 payload capability?

I am thinking that the first stage could be reduced by close to 50% and still do the job. If so, the claims in my first paragraph may be true, but misleading.
Title: Re: Basic Rocket Science Q & A
Post by: gospacex on 04/23/2010 11:36 pm
It is often said that the velocity gain achieved by airlaunch is fairly trivial, and could be matched simply by a slightly larger first stage. It is also said that the velocity generated by a simple hop to 100 km is about 3 percent of that required for orbital.

Additional pros:
* air launch eliminates lower atmosphere portion of the flight, where ground-launched rockets lose about 0.5 km/s to drag
* engine bells can be optimized to vacuum operation
* you have a mobile launch platform. Very mobile one. Want to launch from North Pole? Can do.

Additional cons:
* size limits imposed by carrier aircraft
* all engines need to be air-startable (no GSE for you!)
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 04/23/2010 11:42 pm
More potential advantages:

- you can fly to an equatorial launch point
- more flexible launch windows
- fewer azimuth restrictions
- you are above the weather (though his has been disputed)
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 04/24/2010 05:05 pm
Yeah, but I still want to know how much could be chopped off the Falcon I first stage if the vehicle were airlaunched.  I suspect that it would be a lot more than 3%.

Title: Re: Basic Rocket Science Q & A
Post by: notsorandom on 04/25/2010 08:13 pm
I remember reading an interview with Elon Musk where he said that SpaceX had looked into air launching. I really wish I remember where that was that I saw it! However, If I remember correctly he stated that early on SpaceX had studied it and found in terms of cost that air launch was more expensive for what they wanted to do. That's cost though not performance.
Title: Re: Basic Rocket Science Q & A
Post by: HMXHMX on 04/25/2010 10:09 pm
I remember reading an interview with Elon Musk where he said that SpaceX had looked into air launching. I really wish I remember where that was that I saw it! However, If I remember correctly he stated that early on SpaceX had studied it and found in terms of cost that air launch was more expensive for what they wanted to do. That's cost though not performance.

I recall the statement, too.  He was focused – wrongly – on pure performance enhancement.  While there are undeniable performance improvements, the real value of air-launching comes from operational flexibility and regulatory advantages.  Over-ocean launches at sufficient remove from the coastline will be treated like SeaLaunch (no destruct required, only engine shutdown).  And one can hit a first orbit rendezvous on any day, from anywhere.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 04/26/2010 12:36 am
I remember reading an interview with Elon Musk where he said that SpaceX had looked into air launching. I really wish I remember where that was that I saw it! However, If I remember correctly he stated that early on SpaceX had studied it and found in terms of cost that air launch was more expensive for what they wanted to do. That's cost though not performance.

He wants to go to Mars
Title: Re: Basic Rocket Science Q & A
Post by: notsorandom on 04/26/2010 12:55 am
I recall the statement, too.  He was focused – wrongly – on pure performance enhancement.  While there are undeniable performance improvements, the real value of air-launching comes from operational flexibility and regulatory advantages.  Over-ocean launches at sufficient remove from the coastline will be treated like SeaLaunch (no destruct required, only engine shutdown).  And one can hit a first orbit rendezvous on any day, from anywhere.

There are other advantages as well. The plane can fly towards the equator and launch at a lower latitude to increase payload. Not so much of an advantage for SpaceX since they have Kwajalein near the equator. Also, a plane can fly around or above weather.

He wants to go to Mars

One would need quite a large airplane to launch a rocket the size of the Falcon 9 let alone the Falcon 9 Heavy. For example, an AN-225 can lift 250,000 kg, that's quite a bit less then rockets in that class.
Title: Re: Basic Rocket Science Q & A
Post by: drbobguy on 04/26/2010 02:38 pm
I have a question about scaling effects...

I know that you can't scale a rocket or rocket engine without effect (that is, a rocket twice as big in every respect won't behave the same as a smaller rocket).  That's partly because the Earth's gravity doesn't scale with the rocket, and partly because of fluid effects (viscosity changes with scale).

But I was wondering what the smallest possible orbital rocket might be.  Is it conceivable to build a small rocket with staging, cryogenic engines, etc. that could take a small payload to orbit?  What are the limits on this?  Large hobbyist rockets are obviously limited by solid-fuel ISP's, but is there any way you could build a small rocket that could reach orbit?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 04/26/2010 02:51 pm
I have a question about scaling effects...

I know that you can't scale a rocket or rocket engine without effect (that is, a rocket twice as big in every respect won't behave the same as a smaller rocket).  That's partly because the Earth's gravity doesn't scale with the rocket, and partly because of fluid effects (viscosity changes with scale).

But I was wondering what the smallest possible orbital rocket might be.  Is it conceivable to build a small rocket with staging, cryogenic engines, etc. that could take a small payload to orbit?  What are the limits on this?  Large hobbyist rockets are obviously limited by solid-fuel ISP's, but is there any way you could build a small rocket that could reach orbit?

See Vanguard
Title: Re: Basic Rocket Science Q & A
Post by: TyMoore on 04/26/2010 03:07 pm
Yep---Vanguard is just about as small as you can go. Liftoff weight was around 11 tons if I remember right.
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 04/26/2010 03:48 pm
Also see Scout for what you can do with a solid. About the same height, but twice the launch mass. Has there been a "smaller" solid?

Edit: could look at this ref: http://www.astronautix.com/lvs/propilot.htm

Several attempts at a min vehicle, "may have" had a success... Big * next to that "may".
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 04/26/2010 06:15 pm


If a Falcon 1 were carried aloft by a Lockheed 1011 (or comparable aircraft), and air dropped at .85 Mach and 40,000 feet, how much smaller could the first stage be to take advantage of the altitude gain and velocity increase from air launch, and still meet the nominal Falcon 1 payload capability?

I am thinking that the first stage could be reduced by close to 50% and still do the job. If so, the claims in my first paragraph may be true, but misleading.


I get a 22% reduction in first stage mass, while keeping the 1010kg payload of the Falcon 1e. Had to make some assumptions/tweaks, but I got a reasonable set of assumptions (for pmf, and total dV) that gives a gross liftoff weight of 35,200 kg, which is pretty close to the published 35,180. Then, I held second stage values constant, dropped the dV by 290 m/s, and ramped the Isp on the Merlin to 310 s (from an averaged value of 285). This reoptimization gave a new gross liftoff weight of 29,300 kg. This is only a step above BOTE, but probably not too bad an estimate.
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 04/26/2010 07:29 pm
One thing to consider is a ground-launched F1 still flies fairly straight up when it hits Mach 1, while an air-launched one would be flying level. It would then need to pitch up to lessen the drag losses so the gain from going Mach 0.85 level in the air wouldn't be a completely useful and "free" boost. I think.
Title: Re: Basic Rocket Science Q & A
Post by: drbobguy on 04/26/2010 11:29 pm
Well I was thinking quite a bit smaller, like say a cryogenic rocket less than one ton.
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 05/01/2010 01:30 am


If a Falcon 1 were carried aloft by a Lockheed 1011 (or comparable aircraft), and air dropped at .85 Mach and 40,000 feet, how much smaller could the first stage be to take advantage of the altitude gain and velocity increase from air launch, and still meet the nominal Falcon 1 payload capability?

I am thinking that the first stage could be reduced by close to 50% and still do the job. If so, the claims in my first paragraph may be true, but misleading.


I get a 22% reduction in first stage mass, while keeping the 1010kg payload of the Falcon 1e. Had to make some assumptions/tweaks, but I got a reasonable set of assumptions (for pmf, and total dV) that gives a gross liftoff weight of 35,200 kg, which is pretty close to the published 35,180. Then, I held second stage values constant, dropped the dV by 290 m/s, and ramped the Isp on the Merlin to 310 s (from an averaged value of 285). This reoptimization gave a new gross liftoff weight of 29,300 kg. This is only a step above BOTE, but probably not too bad an estimate.

Is dropping the dV by 290 m/s purely from the velocity gained by the aircraft, or does that take into account the additional altitude from the airlaunch, as well? Or for that matter, the gravity losses that were NOT incurred on a rocket propelled climb to 40,000 feet?
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 05/01/2010 01:32 am
I have a question about scaling effects...

I know that you can't scale a rocket or rocket engine without effect (that is, a rocket twice as big in every respect won't behave the same as a smaller rocket).  That's partly because the Earth's gravity doesn't scale with the rocket, and partly because of fluid effects (viscosity changes with scale).

But I was wondering what the smallest possible orbital rocket might be.  Is it conceivable to build a small rocket with staging, cryogenic engines, etc. that could take a small payload to orbit?  What are the limits on this?  Large hobbyist rockets are obviously limited by solid-fuel ISP's, but is there any way you could build a small rocket that could reach orbit?


http://garvspace.com/NLV.htm

"The NLV is a two-stage vehicle that is being designed to have the capability to launch nanosat-class (up to 10 kg) payloads to low Earth orbit.  GSC is developing this concept in cooperation with CSULB for the academic and small payload users market.  Recent flight demonstrations involving the reusable P-7 test vehicle have focused on operationally responsive spacelift (ORS) issues for the Air Force, while the next generation of prototypes is addressing vehicle design and performance objectives."
Title: Re: Basic Rocket Science Q & A
Post by: jongoff on 05/01/2010 03:17 am
I remember reading an interview with Elon Musk where he said that SpaceX had looked into air launching. I really wish I remember where that was that I saw it! However, If I remember correctly he stated that early on SpaceX had studied it and found in terms of cost that air launch was more expensive for what they wanted to do. That's cost though not performance.

I recall the statement, too.  He was focused – wrongly – on pure performance enhancement.  While there are undeniable performance improvements, the real value of air-launching comes from operational flexibility and regulatory advantages.  Over-ocean launches at sufficient remove from the coastline will be treated like SeaLaunch (no destruct required, only engine shutdown).  And one can hit a first orbit rendezvous on any day, from anywhere.

Even then, depending on how you do the air drop, and what sort of flight path angle you can release the vehicle into, it can make a pretty respectable performance difference.  Especially if you want to do a two stage RLV.

~Jon
Title: Re: Basic Rocket Science Q & A
Post by: jongoff on 05/01/2010 03:21 am


If a Falcon 1 were carried aloft by a Lockheed 1011 (or comparable aircraft), and air dropped at .85 Mach and 40,000 feet, how much smaller could the first stage be to take advantage of the altitude gain and velocity increase from air launch, and still meet the nominal Falcon 1 payload capability?

I am thinking that the first stage could be reduced by close to 50% and still do the job. If so, the claims in my first paragraph may be true, but misleading.


I get a 22% reduction in first stage mass, while keeping the 1010kg payload of the Falcon 1e. Had to make some assumptions/tweaks, but I got a reasonable set of assumptions (for pmf, and total dV) that gives a gross liftoff weight of 35,200 kg, which is pretty close to the published 35,180. Then, I held second stage values constant, dropped the dV by 290 m/s, and ramped the Isp on the Merlin to 310 s (from an averaged value of 285). This reoptimization gave a new gross liftoff weight of 29,300 kg. This is only a step above BOTE, but probably not too bad an estimate.

Is dropping the dV by 290 m/s purely from the velocity gained by the aircraft, or does that take into account the additional altitude from the airlaunch, as well? Or for that matter, the gravity losses that were NOT incurred on a rocket propelled climb to 40,000 feet?

A lot depends on the flight path angle you release the rocket into.  Done correctly, I've seen analyses that indicate that you can avoid as much as 1000m/s of delta-V between altitude, velocity at staging, and the drag and gravity losses you've avoided.  Admittedly that was for a maneuver where you light the rocket engine before dropping the stage (and then using lift from the carrier plane to do a zoom climb of sorts), which is...kind of sporty.  But if you can make that work, that's a huge performance benefit, on top of all the other benefits of air launch.

~Jon
Title: Re: Basic Rocket Science Q & A
Post by: tnphysics on 05/01/2010 02:44 pm
If you have wings on your stage, you can do it the ordinary way. Use "wet wings".

Also, what kind of Isp would you get for B2H6/LOX? Use just enough LOX to burn the B (about 1.73 to 1 O/F). Assume 20MPa chamber pressure and altitude compensating nozzle. Engine start @ 40,000 ft.

The intended application is an SSTO RLV. B2H6 is much denser than hydrogen, and should have a (at least in theory) better Isp, IF your expansion ratio is sufficiently large-enough to deal with the B2O3 particles. Perhaps add a little Al to form Al2O3 particles as condensation nuclei for the B2O3.
Title: Re: Basic Rocket Science Q & A
Post by: jongoff on 05/01/2010 07:51 pm
Also, what kind of Isp would you get for B2H6/LOX? Use just enough LOX to burn the B (about 1.73 to 1 O/F). Assume 20MPa chamber pressure and altitude compensating nozzle. Engine start @ 40,000 ft.

Boronated fuels have, at least from what I've heard, always been a big disappointment compared to theoretical calculations, and a real pain in the neck.

~Jon
Title: Re: Basic Rocket Science Q & A
Post by: tnphysics on 05/01/2010 08:28 pm
If only LiF was not toxic. Imagine 542 sec Isp!

IMO Li, Be, and F2 are all far too toxic to be practical (except in deep space, but I suspect that's a job for an NTR), but boron oxides are far less toxic, and the theoretical Isps are incredible. What about adding a small amount of B to LH2 and burning it with LOX?

Even if you get no Isp improvement, you DO get a density improvement, and we all know that that must help. In fact, you would need to have a worse Isp than hydrolox to not have a performance boost.
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 05/03/2010 03:55 pm

A lot depends on the flight path angle you release the rocket into.  Done correctly, I've seen analyses that indicate that you can avoid as much as 1000m/s of delta-V between altitude, velocity at staging, and the drag and gravity losses you've avoided.  Admittedly that was for a maneuver where you light the rocket engine before dropping the stage (and then using lift from the carrier plane to do a zoom climb of sorts), which is...kind of sporty.  But if you can make that work, that's a huge performance benefit, on top of all the other benefits of air launch.

~Jon

Is there a theoretical possibility of using a lighter payload fairing if the rocket is carried to altitude? Wouldn't it be the case that the dynamic stresses on the payload fairing would be lessened and therefore, lighter materials could be used?
Title: Re: Basic Rocket Science Q & A
Post by: kch on 05/03/2010 04:24 pm
If only LiF was not toxic. Imagine 542 sec Isp!

IMO Li, Be, and F2 are all far too toxic to be practical (except in deep space, but I suspect that's a job for an NTR), but boron oxides are far less toxic, and the theoretical Isps are incredible. What about adding a small amount of B to LH2 and burning it with LOX?

Even if you get no Isp improvement, you DO get a density improvement, and we all know that that must help. In fact, you would need to have a worse Isp than hydrolox to not have a performance boost.

Just had to see what that last bit was ... :)
Title: Re: Basic Rocket Science Q & A
Post by: tnphysics on 05/03/2010 08:29 pm
Also, what kind of Isp would you get for B2H6/LOX? Use just enough LOX to burn the B (about 1.73 to 1 O/F). Assume 20MPa chamber pressure and altitude compensating nozzle. Engine start @ 40,000 ft.

Boronated fuels have, at least from what I've heard, always been a big disappointment compared to theoretical calculations, and a real pain in the neck.

~Jon
Why is it so difficult? What are the problems?
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 05/03/2010 09:04 pm
Why is it so difficult? What are the problems?

Boron trioxide (the stable oxide) is like sand. At O/F 1.73, Pc of 1000psi, I get that 40% of the mass flow is Boron trioxide once you've expanded to atmospheric pressure. For a sea-level nozzle, the B2O3 would be a bunch of entrained liquid particles, they'd be little sand grains in a vacuum engine, which do not accelerate as well in a nozzle, and will bring down your Isp. With 40% of the flow being like that, it sounds like it could bring down your Isp a whole lot. That sound reasonable, Mr. Goff?
Title: Re: Basic Rocket Science Q & A
Post by: tnphysics on 05/04/2010 12:05 am
I thought more would be B2O3. Does it decompose at some high temperature?

I was thinking 2B2H6+3O2->2B2O3+6H2

Also, is that by mass or by moles?
Title: Re: Basic Rocket Science Q & A
Post by: jongoff on 05/04/2010 03:21 am
Why is it so difficult? What are the problems?

Boron trioxide (the stable oxide) is like sand. At O/F 1.73, Pc of 1000psi, I get that 40% of the mass flow is Boron trioxide once you've expanded to atmospheric pressure. For a sea-level nozzle, the B2O3 would be a bunch of entrained liquid particles, they'd be little sand grains in a vacuum engine, which do not accelerate as well in a nozzle, and will bring down your Isp. With 40% of the flow being like that, it sounds like it could bring down your Isp a whole lot. That sound reasonable, Mr. Goff?

Yeah, most boronated fuels are:

1-extremely toxic/nasty
2-their exhausts are extremely toxic/nasty/corrosive/sticky/abrasive
3-the large amount of solids in the exhaust kills the Isp compared to its theoretical potential
4-most of them are pyrophoric, and can produce explosive mixtures with air quite readily.

Both the US and the USSR sunk tons of money into Boronated fuels about 50 years ago.  Both of them abandoned them in disgust after trying to make them work.  Clark has a whole chapter on the topic in "Ignition".

This is one of those areas where while in theory there's no difference between theory and practice, but in practice there is.

~Jon
Title: Re: Basic Rocket Science Q & A
Post by: jongoff on 05/04/2010 03:23 am
I thought more would be B2O3. Does it decompose at some high temperature?

I was thinking 2B2H6+3O2->2B2O3+6H2

Also, is that by mass or by moles?

Not to be rude Tnphysics, but it's often important to remember that almost every rocket propellant you can even sanely imagine (and several that you'd have to be insane to imagine) have been tested over the past 60 years.  While there's a large range of workable propellants that haven't made it over the TRL valley of death, almost everything you can imagine has at least been tried in a lab somewhere.

~Jon
Title: Re: Basic Rocket Science Q & A
Post by: jongoff on 05/04/2010 03:24 am
Also, what kind of Isp would you get for B2H6/LOX? Use just enough LOX to burn the B (about 1.73 to 1 O/F). Assume 20MPa chamber pressure and altitude compensating nozzle. Engine start @ 40,000 ft.

Boronated fuels have, at least from what I've heard, always been a big disappointment compared to theoretical calculations, and a real pain in the neck.

~Jon
Why is it so difficult? What are the problems?

http://lmgtfy.com/?q=boronated+rocket+fuel  :-)

~Jon
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 05/04/2010 12:47 pm

http://lmgtfy.com/?q=boronated+rocket+fuel  :-)

~Jon

Okay, the geek in me says, that was a way cool flame... ISP in the five hundreds?
Title: Re: Basic Rocket Science Q & A
Post by: jongoff on 05/04/2010 03:48 pm

http://lmgtfy.com/?q=boronated+rocket+fuel  :-)

~Jon

Okay, the geek in me says, that was a way cool flame... ISP in the five hundreds?

Sorry.  I've been wanting to use that somewhere ever since my coworker got me with it.  :-)

~Jon
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 05/04/2010 11:28 pm
Quadricyclane? Any luck with that? How much does it cost per kilogram? Even if it doesn't make sense for use on a first stage, would it make sense at the tip of the rocket, like for a lander or upper stage?

Also, these higher energy propellants may require exotic high-temperature alloys for the chamber/throat/nozzle.
Title: Re: Basic Rocket Science Q & A
Post by: tnphysics on 05/05/2010 01:26 am
I do not know about quadricyclane. I do know that one does not need exotic allows. Al (!) can be used for even the nozzle throat, provided it is adequately cooled. The latter is not a problem for any sufficiently large engine.
Title: Re: Basic Rocket Science Q & A
Post by: toddbronco2 on 05/07/2010 03:53 am
Does anybody know where I can find a video of a rocket engine experiencing thrust instability?  Specifically, I'd like to find a video and audio example of a rocket engine that is "chugging."  I've looked around a bit on the web but I haven't had much luck so far
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/28/2010 03:21 am
Who/when/where was the first to discover the significance of maximum dynamic pressure as a driving environment?

I think this is going to be a hard question.
Title: Re: Basic Rocket Science Q & A
Post by: dks13827 on 05/28/2010 12:44 pm
Who/when/where was the first to discover the significance of maximum dynamic pressure as a driving environment?

I think this is going to be a hard question.
The NACA guys perhaps ?  Faget comes to mind as
one of the early guys who learned fast !!!
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/28/2010 01:33 pm
Who/when/where was the first to discover the significance of maximum dynamic pressure as a driving environment?

I think this is going to be a hard question.
The NACA guys perhaps ?  Faget comes to mind as
one of the early guys who learned fast !!!

No, because NACA didn't do launch vehicles and Faget only did Mercury and not other earlier spacecraft or launch vehicles.

Maybe one of the NRL White Sands (V-2/Viking) people, Thor designers or Bossart who designed a minimalist vehicle that had to depend on internal pressure to survive this regime.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 06/14/2010 04:54 am
After looking at the Amro ad: what are the pros and cons of orthogrid vs isogrid?
Title: Re: Basic Rocket Science Q & A
Post by: DavisSTS on 07/24/2010 10:48 am
Why is it that when you re-enter, you have all the heat and friction, but not the same problem when you launch through that same area?
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 07/24/2010 10:50 am
Because you reenter at much higher velocities. Airbreathing launch vehicles would have heating problems on ascent too, but those do not exist yet.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 07/24/2010 12:05 pm
There is some heating but during ascent, most of the velocity gain is outside the atmosphere to avoid drag (friction).  During entry, the vehicle reaches the lower parts of the atmosphere at higher velocities.
Title: Re: Basic Rocket Science Q & A
Post by: Ratty_2 on 08/07/2010 06:15 pm
Hi,

I am having a small problem with a launcher design I am doing at the moment and wondered if any one could help. I have designed a 3 stage winged launch vehicle. The first stage uses a liquid engine while the upper two stages uses hybrids. The second stage uses a cluster of hybrids, 4, so the diameter has to be 1.86 m. So the diameter of the first and second stage is 1.86 m, but to save mass, the third stage is 1.28 m. The fairing is released during the second stage burn. Now my problem is, is that I am unsure whether the fairing will collide with the second stage during separation (due to it being a larger diameter)? I cannot find any dynamic information on typical fairing separations, so I am unsure on typical clearance values. Does anyone know the general capabilities of fairing separation mechanisms and have an inclination whether this will be a problem?

At the moment I have three courses of action,

1) it will be ok, so I dont modify,
2) add a flared skirt to increase the fairing diameter closer to the second stage, but I am worried about the aerodynamic consequences,
3) have a hinge point of the fairing via rails down to the top of the second stage, so essentially the fairing is hinged about the larger diameter, but the strength of this rail is the issue.

Any ideas would be greatly appreciated:)
Title: Re: Basic Rocket Science Q & A
Post by: Hoonte on 08/09/2010 02:15 pm
Hi,
I don't understand a bit about orbital mechanics but I do know that the higher you get, the slower your orbital time will be.

So just for the fun of it. Does anyone knows what the orbital time will be if the earth was a vacuum and the apogeum/perigeum would be around 9 kilometres (so it won't hit the everest).

I gues that this time would be the fastest possible time to travel around earth even with an atmosphere.

Forgive my bizarre mind.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/09/2010 02:25 pm
http://www.stuegli.com/phyzx/calculators/calc-orbitvel.htm


84.660365294035 minutes at 17673.549882780328 mph
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 08/09/2010 02:29 pm
That's some fantastic accuracy in those numbers  ;D
Title: Re: Basic Rocket Science Q & A
Post by: Lee Jay on 08/09/2010 02:30 pm
You could go faster than what Jim said if you could apply continuous delta-V directed radially inward.  For example, if you had a hypothetical aircraft that flew in the upper atmosphere upside down generating lift toward the center of the Earth instead of away from it like a usual airplane, you could in theory go faster.  Practice and theory in this regard are still quite a ways apart.
Title: Re: Basic Rocket Science Q & A
Post by: JohnFornaro on 08/09/2010 03:16 pm
Bizzarre minds need great accuracy.  I'm there with that.  Great answer.  Two more points for Jim!  Hufflepuff?
Title: Re: Basic Rocket Science Q & A
Post by: Lee Jay on 08/10/2010 07:58 pm
Orbital mechanics question.

Let's say you're in a perfectly circular orbit around a gravitation point source (no gravitational gradients or anomalies).  Let's move our frame of reference into your frame and call your now-fixed location "A".  If you were to apply a small delta-V in either direction along the velocity vector, what would be the shape of your flight path relative to point "A"?
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 08/11/2010 01:49 am
It could be even faster if you built a craft that could orbit inside the lithosphere!

Again, given enough time and money someone would say could turn the theory in practice. 
Title: Re: Basic Rocket Science Q & A
Post by: gospacex on 08/11/2010 03:04 am
It could be even faster if you built a craft that could orbit inside the lithosphere!

No problem, as long as you are ok with very small craft: LHC already pushed the speed pretty much to the limit.
Title: Re: Basic Rocket Science Q & A
Post by: Urvabara on 08/11/2010 06:16 am
So, is it easy to vary the thrust force in chemical rockets? How does that change the propellant flow rate? If I halve the thrust, does the propellant flow rate halve too?
Title: Re: Basic Rocket Science Q & A
Post by: simonbp on 08/11/2010 06:31 am
Interestingly enough, you don't find many natural orbital systems faster than a few hours. The closest binary stars have periods of greater than ~3 hours, as do the closest asteroid satellites...

Now, for very massive systems, that's still pretty fast; e.g. PSR B1913+16, with a period of 7.5 hours and orbital velocity of the secondary of 450 km/s...

http://en.wikipedia.org/wiki/PSR_B1913%2B16
Title: Re: Basic Rocket Science Q & A
Post by: mdo on 08/11/2010 07:42 am
Interestingly enough, you don't find many natural orbital systems faster than a few hours. The closest binary stars have periods of greater than ~3 hours, as do the closest asteroid satellites...

Although slightly OT, a quick search in the ADS database reveals that some exotic star systems manage an even faster pace. For instance, Stella et al. claimed the discovery of a so called low mass X-ray binary with an 685 s orbital period.
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 08/11/2010 07:56 am
Interestingly enough, you don't find many natural orbital systems faster than a few hours.

I don't find that very surprising. Such systems tend to decay fast (on cosmic timescales) so statistically we don't see many of them.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 08/11/2010 03:20 pm
So, is it easy to vary the thrust force in chemical rockets? How does that change the propellant flow rate? If I halve the thrust, does the propellant flow rate halve too?

No.  You have to look at combustion stability and turbodynamics.  The motor may not be able to operate at that point without tearing itself up.  Even if those two items are still healthy, you have to do the detailed aerothermal analyses to see how the nozzle flow works to determine what mass flow effects half thrust.  It would probably be within a few percent of half the mass flow, assuming the nozzle was still choked.
Title: Re: Basic Rocket Science Q & A
Post by: Citabria on 08/11/2010 03:54 pm
Bizzarre minds need great accuracy.  I'm there with that.  Great answer.  Two more points for Jim!  Hufflepuff?

At LEO velocity, relativistic effects are about 3 in 10 billion. I doubt that calculator includes relativity, so any digit after the tenth in Jim's answers is wrong. One point deducted! Slytherin.

(Seriously, one must also consider the accuracy of the constants used in the calculation, as well as the precision of the variables.)
Title: Re: Basic Rocket Science Q & A
Post by: KelvinZero on 09/25/2010 12:38 am
Laymans' Orbital mechanics question. I hope this is the right thread.

I have heard that there are low energy but slow orbital transfers for example to get from the earth to the moon, or to get back. I looked on line but mainly found papers with very scary titles. I could not find a good summary of what savings in fuel are plausible, at what cost in increased travel time. Does anyone know of such a summary or just some interesting specific examples? Im just trying to get an idea of whether taking 6 months saves 25% of your propellant or is an order of magnitude improvement etc.

Title: Re: Basic Rocket Science Q & A
Post by: Antares on 09/25/2010 02:15 am
Google interplanetary superhighway and/or low energy transfers.  That'll get you fairly layman, or at least technical generalist, answers.  Here's a Discover article that says Lunar Observer could have saved 25% of its fuel and 30% for a generic mission.

http://discovermagazine.com/1994/sep/gravitysrim419/article_view?b_start:int=1&-C=
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 09/25/2010 09:21 am
For L1/L2 it's the difference between 3.8km/s and 3.2km/s delta-v, which is substantial. Useful for cargo and (well-insulated or space storable) propellant that don't mind taking a long detour.
Title: Re: Basic Rocket Science Q & A
Post by: KelvinZero on 09/25/2010 11:02 am
Ah ok.. yes those were the sort of numbers I found online. For some reason I thought there were bigger eg x10 potential savings for a 6 month trip. I guess I was mistaken.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 09/25/2010 11:08 am
That may be true for interplanetary routes.
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 09/25/2010 02:29 pm
Google interplanetary superhighway and/or low energy transfers.  That'll get you fairly layman, or at least technical generalist, answers.  Here's a Discover article that says Lunar Observer could have saved 25% of its fuel and 30% for a generic mission.

http://discovermagazine.com/1994/sep/gravitysrim419/article_view?b_start:int=1&-C=

Quote
“You have to lose 600 miles per second of velocity to get off the Earth orbit and allow the spacecraft to be captured by the moon’s gravity,”

That should be metres per second, I reckon. Defo not "miles per second".

cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: KelvinZero on 09/26/2010 04:57 am
Thanks for that link.
It occurs to me that perhaps whatever I read initially was also assuming some more efficient but lower thrust form of propulsion that becomes practical if the timeframe is six months?
Title: Re: Basic Rocket Science Q & A
Post by: ycs86 on 10/05/2010 01:41 am
What exactly will happen when a thrust, that is perpendicular to the orbital plane, is applied to an orbiting spacecraft? It would no doubt change the inclination of the orbital plane, but is that all that's gonna happen? Will it somehow affect the altitude?
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 10/05/2010 01:52 am
The application of thrust perpendicular to the orbital plane will increase the speed, although if the magnitude of the burn is small compared to orbital speed (about 7800 m/s in low earth orbit), then this effect will be very small.  It's just the Pythagorean theorem.  If the initial orbit is circular, then the new orbit will be in a different plane, as you surmise, and elliptical, with perigee at the point where the burn was performed.
Title: Re: Basic Rocket Science Q & A
Post by: Malderi on 10/05/2010 07:48 pm
I read that plane change maneuvers use less fuel in higher orbits compared to lower ones. This is why, for example, GEO birds launched from Canaveral will do their GTO burn in a 28.5 degree orbit, and make it equatorial only when they're all the way up there.

Why is this the case? I don't have any understanding of the equations (or the reasoning behind them).
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 10/05/2010 07:53 pm
Orbital velocity is much lower in high energy orbits, since much of the kinetic energy has been converted into potential energy. You can also combine the plane change with the circularisation burn which is inevitable and relatively large.
Title: Re: Basic Rocket Science Q & A
Post by: DavisSTS on 10/06/2010 12:20 pm
Having an argument with a friend. "200g acceleration on a human body wouldn't necessarily kill you, the time over which this acceleration occurs has to be taken into account."

He's wrong, yes?
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 10/06/2010 12:33 pm
My 2c is that your friend's probably right. What's the deceleration for example if you bang your head against a wall or jump from a table and the force your feet experience the instant you hit the floor? Human body is a good shock absorber so can withstand some punishment if it's short enough.
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 10/06/2010 12:41 pm
Colonel John Stapp, did the rocket sled research that determined the limits. His record as listed by wiki is 45g  ( http://en.wikipedia.org/wiki/John_Stapp ).
Title: Re: Basic Rocket Science Q & A
Post by: DavisSTS on 10/06/2010 12:46 pm
The argument started when I said there was no way the crew of Challenger could have drowned as they would have died in the 200G impact with the water if they were still alive.
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 10/06/2010 03:45 pm
Well, the Columbia report did point a flaw in the seat design that resulted in the head not being properly restrained (broken neck) during tumbling of the crew compartment. (Don't know if that would apply to Challenger, since they where dressed differently).
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 10/07/2010 03:24 am
Well, the Columbia report did point a flaw in the seat design that resulted in the head not being properly restrained (broken neck) during tumbling of the crew compartment. (Don't know if that would apply to Challenger, since they where dressed differently).

I suspect the friend is right, in that a 200-g deceleration can be tolerated for a very short time.  That time, however, is probably shorter than the time it takes for a human body or crew cabin falling from great height to decelerate to zero speed on impact with the water, putting survivability of Challenger's crew in grave doubt.

If I remember correctly, by the way, there were signs that emergency oxygen supplies aboard Challenger had been manually activated following disintegration of the orbiter.
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 10/07/2010 03:28 am
Well, the Columbia report did point a flaw in the seat design that resulted in the head not being properly restrained (broken neck) during tumbling of the crew compartment. (Don't know if that would apply to Challenger, since they where dressed differently).

I suspect the friend is right, in that a 200-g deceleration can be tolerated for a very short time.  That time, however, is probably shorter than the time it takes for a human body or crew cabin falling from great height to decelerate to zero speed on impact with the water, putting survivability of Challenger's crew in grave doubt.

If I remember correctly, by the way, there were signs that emergency oxygen supplies aboard Challenger had been manually activated following disintegration of the orbiter.

Emergency *air* supplies, not oxygen. The distinction is non-trivial, since air would not have kept the Challenger crew conscious in the event of a cabin depressurization.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 10/07/2010 05:34 am
CxP 70024 requires less than 500 G/s of jerk.
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 11/05/2010 01:17 am
Having an argument with a friend. "200g acceleration on a human body wouldn't necessarily kill you, the time over which this acceleration occurs has to be taken into account."

He's wrong, yes?

Attached is a figure from NASA STD 3000 on shock Gs allowed.  It shows 200 Gs as an upper limit to short G loads.  20 Gs is a good rule of thumb for ascent and entry Gs that last for several seconds.

Danny Deger
Title: Re: Basic Rocket Science Q & A
Post by: tnphysics on 11/24/2010 03:29 pm
Here is an idea for a high (SSME-esque, at least in vac) performance GG engine.

The engine runs on LH2/LOX. The GG exhaust, after expansion in the turbine, is  used to cool nozzle parts. This heats it hot enough that it has an exhaust velocity equal to that of the main chamber gasses. The extremely low molecular weight should make this possible despite the low temp.

Is this possible? If so, is it practical?
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 11/24/2010 05:26 pm
If you've got 1500F H2 and H2O coming out of the turbines, it's probably not enough of a heat sink vs -425F H2.

Also, you'd need a lot of pressure left after the turbine to get it through cooling passages.  That's not as efficient if you're not going to burn it the rest of the way.

I could imagine, though, basically a shell over the inner nozzle with many wide channels instead of hundreds of tubes.  Structural design would be a lot different than we're accustomed to, but it's doable.

Still, though, I think the much weaker heat sink makes it a non-starter.  Interesting academic exercise, though.
Title: Re: Basic Rocket Science Q & A
Post by: drbobguy on 11/24/2010 10:18 pm
Here is an idea for a high (SSME-esque, at least in vac) performance GG engine.

The engine runs on LH2/LOX. The GG exhaust, after expansion in the turbine, is  used to cool nozzle parts. This heats it hot enough that it has an exhaust velocity equal to that of the main chamber gasses. The extremely low molecular weight should make this possible despite the low temp.

Is this possible? If so, is it practical?

I'm no expert so can'd do the math, but it seems very unlikely and you might as well do the easy thing and just use it for film-cooling with an annular injector midway down the nozzle, as in the J-2.
Title: Re: Basic Rocket Science Q & A
Post by: Malderi on 12/10/2010 05:10 pm
What is the relationship between delta-V (in m/s, for example) and degrees of plane change capability, depending on altitude? I'm just curious for a rough rule of thumb, especially for LEO. So, when a satellite has 500 m/s of delta-V, how many degrees of plane change could you do in an ISS-like orbit?
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 12/10/2010 05:53 pm
http://en.wikipedia.org/wiki/Orbital_inclination_change

It's about halfway down the page.
Title: Re: Basic Rocket Science Q & A
Post by: Malderi on 12/10/2010 06:23 pm
Ah, I should remember to look through Wikipedia more thoroughly. Thanks.
Title: Re: Basic Rocket Science Q & A
Post by: scienceguy on 12/31/2010 12:37 am
I just wanted to know, what is a good power to weight ratio for power systems for spacecraft? What is it for say a small nuclear reactor? What is it for a fusion reactor?
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 12/31/2010 04:49 pm
1) I just wanted to know, what is a good power to weight ratio for power systems for spacecraft?
2) What is it for say a small nuclear reactor?
3) What is it for a fusion reactor?
1) Modern conventional solar arrays typically do about 50-80W/kg, though the UltraFlex arrays (being built for Orion, was baselined for ST-8 before ST-8 was canceled, and is baselined for many new missions in the latest planetary science decadal survey and already used on the Phoenix lander) will be able to do at least 150-175W/kg, at the solar array-level (not the cell-level, which could be much higher), and could likely be extended to 300 or 500W/kg with new ultra-high efficiency (~35%) IMM cells which are reportedly one fifteenth the thickness of typical multijunction cells (and thus much lighter, allowing the structure to also be lighter). So, I would say anything above 75W/kg is "good," but in a decade, that will probably seem like very low performance.

RTGs have much lower specific power, though don't need sunlight, so work well even in the deep outer solar system, though their power decreases considerably over time as the fuel decays and the thermocouples degrade. MMRTGs (the modern RTG design used for MSL) have about 2.8W/kg, which is, to be honest, pretty bad (which is part of the reason for the push to ASRGs which have about 4W/kg, which also are more efficient, though they haven't been proven in space). Older RTGs, like used on Cassini, have a tad over 5W/kg specific power, but the last one is on its way to Pluto right now.

2) The specific power for the only nuclear reactor ever orbited by the US is a little less than 1.75W/kg unshielded (and even worse when shielded). So, pretty horrible... in fact, at 1 AU from the Sun, that's one hundredth of the performance of the latest Ultraflex arrays. And the prototype fission reactor only produced about 500W electric power. Not exactly fair, since it was pretty much just a proof-of-concept, but nuclear fission power in space has a lot left to prove before it competes with solar power (or RTGs). However, for a lot of power in the outer solar system, there are not really many other options than nuclear fission, though the amount of development left to do is pretty enormous (and billions of dollars need to be spent). In the inner solar system, solar power is awesome.

There are some notional concepts for very high specific power fission space reactors (~200W/kg), but the cost of development would be... um... out of this world. Solar power could much easier reach this specific power level in the inner solar system (and has actually been demonstrated at more than this performance on the IKAROS solar sail spacecraft with thin film solar cells).

3) Has never been demonstrated to produce electricity, even on Earth (unless you count solar power...). Unless there is some kind of breakthrough, it's not an option. And if fusion power only will work with ITER-style tokamaks for the next century, I highly doubt fusion would be better than fission power for space propulsion in the next century.

Regarding solar versus nuclear:
Power on the surface of celestial bodies is a little different story. Lunar night is a long time to store electricity in some kind of battery, so nuclear power has an edge, here (also, it would be easier to shield on the Moon, where you can put the reactor behind a hill or use regolith). A similar situation is faced on Mars, where, although night isn't nearly as long as on the Moon, there are some pretty intense and long-lasting dust storms. Also, the Martian atmosphere can be used as a heat sink instead of just having to rely radiators. Also, the Martian winds can cause problems for the very lightweight solar arrays that can be used in microgravity and vacuum. So although solar power is a pretty good solution, advanced "RTGs" (like improved, high power ASRGs) and new nuclear fission reactors are probably superior for surface power, if you have the money for them.

EDIT:For reference, solar power specific power figures given above are for 1AU distance from the sun when fully lit. Solar power available per area decreases as you get further from the Sun (and increases as you get closer to the Sun).

Power at Mars = ~50% of power at 1AU
asteroid belt   = ~10-20% of power at 1AU
Jupiter           = ~4% of power at 1AU
Saturn           = ~1%
Uranus           = ~0.25%
Neptune         = ~0.1%

Power at Venus =~200% of power at 1AU (if you can handle the high temperatures)

(there are some design issues with operating a solar array with very low sunlight, but they can be addressed)
Title: Re: Basic Rocket Science Q & A
Post by: hop on 12/31/2010 08:54 pm
Not exactly fair, since it was pretty much just a proof-of-concept
Fairer examples:
http://www.svengrahn.pp.se/trackind/RORSAT/RORSAT.html
http://en.wikipedia.org/wiki/TOPAZ_nuclear_reactor
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 01/21/2011 06:20 am
I'm looking for a comparison of the operational and safety difficulties of nitrous oxide, hydrogen peroxide and liquid oxygen as oxidizers. If performance (e.g. density and specific impulse) isn't a distinguishing factor which would you rather deal with? Here are the principle safety issues that I'm aware of:
* Liquid oxygen is cryogenic.
* The decomposition of hydrogen peroxide is catalyzed by an inconveniently wide range of substances including various contaminants and human skin.
* An ignition source in gaseous nitrous oxide will cause catastrophic deflagration. Organics dissolved in liquid nitrogen may make it shock sensitive.

Hydrogen peroxide engines can be ignited easy by simply catalytically decomposing the peroxide before injection. Liquid oxygen doesn't allow any ignition shortcuts. Nitrous oxide is in between, with catalytic decomposition possible but only at high temperatures.

There's some discussion of hydrogen peroxide in the following post and in a few pages of posts before and after:
http://forum.nasaspaceflight.com/index.php?topic=13543.msg380864#msg380864

Webpages on hydrogen peroxide and nitrous oxide safety respectively:
http://www.solvaychemicals.us/static/wma/pdf/6/6/0/4/HH-2323.pdf
http://www.spg-corp.com/nitrous-oxide-safety.html
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 01/21/2011 09:47 am
Heh, I've just been thinking about a related topic, triggered by a discussion on Paul Breed's blog.

Way out of the box presure system...  (http://unreasonablerocket.blogspot.com/2010/12/way-out-of-box-presure-system.html)
Is nitrous oxide safe? (http://www.spl.ch/publication/SPL_Papers/N2O_safety_e.pdf)
Hybrid safety (http://myweb.tiscali.co.uk/aspirespace/TechPapers_files)

I was thinking that using nitrous oxide as a pressurant for a storable oxidiser instead of as a (self pressurising) storable oxidiser itself might be a nice way to get the best of both worlds. And by using it as an oxidiser for a (continuously running?) torch igniter instead of a cat pack you might be able to add lots of stabilisers to the peroxide without getting into trouble with poisoning now unneeded peroxide cat packs. You could similarly use propane or acetylene to pressurise a storable fuel (although the vapour pressure of propane is a bit low), or use a diaphragm and pressurise it with the oxidiser or perhaps with a diaphragm and nitrous.

But the safety problems associated with nitrous oxide may make this uninteresting. On the other hand, peroxide has its own safety issues, and "there are no nice oxidisers". Which would be safer, liquid high test peroxide or gaseous nitrous oxide? Most of the time your nitrous would be liquid, but at engine cut-off it would be mostly gaseous. Still not a good time to go boom. On the other hand isn't nitrous used safely with acetylene welding torches? Some googling suggests that flashback arresters are standard practice for acetylene, but I haven't been able to find a nitrous compatible one. The ones I've seen explicitly warn they are not suitable for nitrous (and silane, both for unexplained reasons). On the other hand, the hybrid safety document I linked to above suggests a flashback arrester for nitrous should be easier than for most gases, because it has a larger quenching distance than typical fuel/air mixtures. One potential reason is that even the pure nitrous upstream of the arrester, being a monopropellant, is an explosion hazard whereas a pure fuel wouldn't be. But that reason doesn't apply to silane and doesn't take away from the larger quenching distance.
Title: Re: Basic Rocket Science Q & A
Post by: Downix on 01/26/2011 07:21 am
Satellites
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 01/26/2011 07:50 am
Deep Space 1.

Dawn.

cheers, Martin  :)
Title: Re: Basic Rocket Science Q & A
Post by: pathfinder_01 on 01/26/2011 08:56 am
One more probe to add. ESA's Smart 1.
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 01/26/2011 02:59 pm
Hayabusa.
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 01/26/2011 03:14 pm
Many GEO comsats for station keeping
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 01/26/2011 11:51 pm
GOCE, which is a pretty cool application of it - to overcome aero drag.
Title: Re: Basic Rocket Science Q & A
Post by: blue sky on 01/28/2011 03:46 pm
how can i find the detail about the current satellite that use electric thrusters? i need sources and i want to do a research about them.
Title: Re: Basic Rocket Science Q & A
Post by: blue sky on 01/28/2011 04:37 pm
One more probe to add. ESA's Smart 1.
thanks. but i want the detail about these satellites.
Title: Re: Basic Rocket Science Q & A
Post by: Jason1701 on 01/29/2011 01:44 am
Hello,

In my research I'm trying to optimize the path of my theoretical Centaur-based craft. This has led me to wonder how pitch programs are optimized for real rockets. Do the engineers just try a whole bunch of pitch kick time and delta-angle values and see which are the best? Would this need to be repeated if a mission required a new set of orbital parameters? Or is there a more analytic way to find a good pitch program?

Your help is much appreciated.
Jason
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 01/29/2011 02:23 am
Hello,

In my research I'm trying to optimize the path of my theoretical Centaur-based craft. This has led me to wonder how pitch programs are optimized for real rockets. Do the engineers just try a whole bunch of pitch kick time and delta-angle values and see which are the best? Would this need to be repeated if a mission required a new set of orbital parameters? Or is there a more analytic way to find a good pitch program?

Google "powered explicit guidance".
Title: Re: Basic Rocket Science Q & A
Post by: scienceguy on 02/06/2011 10:37 pm
Hi there,

How thin does aluminum have to be for H2 boiloff to occur? How thick does it have to be to prevent N2 or O2 boiloff?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 02/06/2011 10:41 pm
Hi there,

How thin does aluminum have to be for H2 boiloff to occur? How thick does it have to be to prevent N2 or O2 boiloff?

You mean how thick to prevent boil off?

feet of aluminum
Title: Re: Basic Rocket Science Q & A
Post by: scienceguy on 02/06/2011 10:43 pm
No, actually I wanted boiloff of hydrogen in this case. I just figured there would be lots of people on this forum who knew about this.

So it takes feet of aluminum, you say, to prevent N2 and O2 boiloff?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 02/06/2011 10:58 pm
Actually, there is always boil off.   Only active cooling would prevent boil off
Title: Re: Basic Rocket Science Q & A
Post by: Scotty on 02/06/2011 11:43 pm
LH2 boiloff is a result of heat leaking into the tankage.
Bare aluminum will have a high boiloff, as aluminum is a very good thermal conductor.
To reduce boiloff, you have to reduce heat leakage into the tankage.
That is where foam insulation comes into play.
Title: Re: Basic Rocket Science Q & A
Post by: sitharus on 02/07/2011 01:48 am
Hi there,

How thin does aluminum have to be for H2 boiloff to occur? How thick does it have to be to prevent N2 or O2 boiloff?

As long as ambient temperature is above the boiling temperature of the liquid there will be boiloff. Insulation will help but insulation is never 100% effective, even in a vacuum flask there's still a path for heat conduction.

The only way to prevent boiloff is to alter the environment so the ambient temperature is below the boiling point. Generally this means increasing the pressure, which raises the boiling point..
Title: Re: Basic Rocket Science Q & A
Post by: Nomadd on 02/07/2011 02:13 am
No, actually I wanted boiloff of hydrogen in this case. I just figured there would be lots of people on this forum who knew about this.

So it takes feet of aluminum, you say, to prevent N2 and O2 boiloff?

 Not sure why you think thick aluminum would help. Aluminum is a pretty lousy insulator. That's why they make radiators out of it. Thicker would just give you more mass to warm the H2 and make boiloff worse to start.
Title: Re: Basic Rocket Science Q & A
Post by: Propforce on 02/07/2011 02:40 am
No, actually I wanted boiloff of hydrogen in this case. I just figured there would be lots of people on this forum who knew about this.

So it takes feet of aluminum, you say, to prevent N2 and O2 boiloff?

 Not sure why you think thick aluminum would help. Aluminum is a pretty lousy insulator. That's why they make radiators out of it. Thicker would just give you more mass to warm the H2 and make boiloff worse to start.

You guys are not approaching this problem correctly.  You'll need to know what the boiling point of each gases you 1) want to boil off, and 2) to prevent from boiling off.  Then you'll have to ask what is the "heat source" coming into this "system"? Is it the sun radiation? on the ground? in space?  Finally, you'll need to ask what is my insuation system?  Is it a vaccuum-jacked dewar? MLI blankets?

In your case, LH2 boils off at -423 def. F, while the LO2 boils off at -297 deg. F and LN2 at -320 deg. F, So if you want to boil off LH2 while keeping LO2 & LN2 liquid, then depending on your 'system' and 'environment, you'll just need to keep the temperature in between -423 and -320 deg. F. You'll not only need to cool it down bt also able to draw heat away afterward.

The above numbers assume you're doing this at 1 atmosphere.  Different numbers if you are dong this at lower pressure.

Title: Re: Basic Rocket Science Q & A
Post by: Patchouli on 02/07/2011 02:45 am
No, actually I wanted boiloff of hydrogen in this case. I just figured there would be lots of people on this forum who knew about this.

So it takes feet of aluminum, you say, to prevent N2 and O2 boiloff?

 Not sure why you think thick aluminum would help. Aluminum is a pretty lousy insulator. That's why they make radiators out of it. Thicker would just give you more mass to warm the H2 and make boiloff worse to start.

Aluminum actually is one of the best thermal conductors there is though there are better ones which is why it makes pretty good cookware and engine heads.
http://en.wikipedia.org/wiki/Thermal_conductivity
Very thick aluminum would act like a heat pipe which is why they make CPU heat sinks from it.
Title: Re: Basic Rocket Science Q & A
Post by: scienceguy on 02/07/2011 04:40 am
Thanks for all the responses! Actually what I will have is an almost vacuum compartment with cold hydrogen in it, the hydrogen of which I want to pass into another vacuum compartment with minimal spillage back into the first compartment. That's why I was wondering if I could just use an aluminum barrier. Also, this second compartment will be exposed to air on the outside of another barrier, and I want to keep O2 and N2 out while I build up cold H2 in it, for later use in a third compartment.

(please see attached diagram)
Title: Re: Basic Rocket Science Q & A
Post by: JayP on 02/07/2011 04:03 pm
It looks like you want some sort of molecular sieve as your barrier. That won't work in a one way application like you have laid out. You need a line between the two compartments with a check valve to prevent back flow or you could incrrease the preasure in the first compartment to something higher that the second compartment to ensure constant flow.
Title: Re: Basic Rocket Science Q & A
Post by: Jason1701 on 02/09/2011 01:09 am
Thanks Jorge.

That got me wondering, would there be any need for a pitch maneuver if the rocket had some initial velocity at less than a 90-degree angle? In other words, could one select an initial velocity and angle that (with an engine burn) could place the vehicle in orbit following a gravity turn from the very start?
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 02/09/2011 03:28 am
Thanks Jorge.

That got me wondering, would there be any need for a pitch maneuver if the rocket had some initial velocity at less than a 90-degree angle? In other words, could one select an initial velocity and angle that (with an engine burn) could place the vehicle in orbit following a gravity turn from the very start?

Probably wouldn't be a good idea on Earth. The vertical rise is necessary to get above some of the Earth's thick atmosphere before starting the gravity turn. Otherwise max-Q is too high. You could do it if you beefed up the rocket's structure to withstand higher max-Q, but what would be the point? To avoid an extra guidance phase in the software? That's not a good trade; bits are cheaper than metal.
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 03/11/2011 02:49 pm
What is the difference between the usual S-Band transponder that launch vehicles use, and the usual S-Band transponder used by satellites? Are these indeed different models, and if so, why?

I am guessing that rocket transponders are very high power, whereas satellite transponders are low power.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 04/19/2011 09:04 pm
This might have been answered hundreds of times. But I couldn't find my exact question. The issue is hammerheads. I understand that it all depends on the aerodynamics and rockets are not legos. But my question is. Let's suppose we have two TSLV. The first is 3.8m wide all the way. The second is 3.8m on the first stage and 5.2m on the second stage. Would it be easier, on the general case, to put a (just for the sake of it) 6.6m on the one with the 5.2m? In other words, the closer you get the general profile to a drop, the easier to increase the widest diameter at the top?
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 04/20/2011 02:01 am
Probably, but aero is highly non-linear.  Look up stuff about area ruling (transonic drag) and center of pressure.  Acoustics are an issue too.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 05/05/2011 06:23 pm
I've read that the RD-0120 development used more than 105 prototypes and test engines. Is that a normal number? I thought the plan of the J-2X has something like 10 to 20.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/05/2011 07:25 pm
No, it's very much in the single digits in a modern engine program.  Frequently those are rebuilt as development goes on and improvements are made.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 05/05/2011 07:32 pm
So, they might have count each rebuilt and new piece put together as a different engine? I do recall that they had something close to 106,000s of test by that time.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/06/2011 02:11 am
Hard to say for certain.  Usually the engine serial number follows a certain major component of the engine (every engine I can think of does this).  Then, some part of the serial number is incremented when a rebuild occurs (2 engines I can think of DON'T do this).  If they just change out minor components or remachine something, maybe neither increment or maybe they count it as a whole new engine.  Hard to say without seeing records and the source of 105 for the 0120.
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 05/12/2011 07:37 pm
For rockets where the 2nd stage engine is housed inside the 1st stage prop tanks, how exactly does the engine leave the tank during staging? Is the prop tank blown up?
Title: Re: Basic Rocket Science Q & A
Post by: LegendCJS on 05/12/2011 07:59 pm
For rockets where the 2nd stage engine is housed inside the 1st stage prop tanks, how exactly does the engine leave the tank during staging? Is the prop tank blown up?

Do you know of a rocket where the second stage engine is immersed in the fuel of the first stage tanks before staging?  Don't you mean inter-stage volume?
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 05/13/2011 01:48 am
I believe it was one of the concepts tspace was looking at.
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 05/13/2011 04:55 am
For rockets where the 2nd stage engine is housed inside the 1st stage prop tanks, how exactly does the engine leave the tank during staging? Is the prop tank blown up?

Do you know of a rocket where the second stage engine is immersed in the fuel of the first stage tanks before staging?  Don't you mean inter-stage volume?

Sorry, I forgot to add the word "nozzle" in there, so the nozzle is in the lower stage prop tank.

Its quite common.
Title: Re: Basic Rocket Science Q & A
Post by: sitharus on 05/13/2011 05:02 am
For rockets where the 2nd stage engine is housed inside the 1st stage prop tanks, how exactly does the engine leave the tank during staging? Is the prop tank blown up?

Do you know of a rocket where the second stage engine is immersed in the fuel of the first stage tanks before staging?  Don't you mean inter-stage volume?

Sorry, I forgot to add the word "nozzle" in there, so the nozzle is in the lower stage prop tank.

Its quite common.


So the nozzle of the second stage forms part of the pressure structure of the first stage? I don't think I've ever seen such a design...
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 05/13/2011 06:46 am
For rockets where the 2nd stage engine is housed inside the 1st stage prop tanks, how exactly does the engine leave the tank during staging? Is the prop tank blown up?

Do you know of a rocket where the second stage engine is immersed in the fuel of the first stage tanks before staging?  Don't you mean inter-stage volume?

Sorry, I forgot to add the word "nozzle" in there, so the nozzle is in the lower stage prop tank.

Its quite common.


So the nozzle of the second stage forms part of the pressure structure of the first stage? I don't think I've ever seen such a design...

No, the nozzle extends into the first stage prop tank, saving loads of interstage area.
Title: Re: Basic Rocket Science Q & A
Post by: yinzer on 05/13/2011 07:56 am
Linear shaped charge around the circumference of the tank just below the upper dome, I'd imagine.  Not much different than cutting an interstage with an LSC, just stiffened sheet metal. Might be a pain to not have a pressurant diffuser or pressure relief valve at the very top of the tank, but that should be workable.

Maybe a plug in the throat of the upper stage engine that gets blown out during ignition.

I doubt anyone who actually knows will tell you given that the biggest use of this technique is making a missile short enough to fit in a silo or launch tube, thus ITAR.
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 05/13/2011 01:15 pm
I believe it was one of the concepts tspace was looking at.

Went and looked it up, got the wrong new space company. It was Airlaunch llc.

http://www.airlaunchllc.com/

It was part of the proposed "falcon". A two stage pressure fed Lox/Propane rocket dropped out of a C-17.

Look at the January 13th press release to see a nozzle extraction test. http://www.airlaunchllc.com/News.htm

A collage with a picture of the test.
http://www.airlaunchllc.com/AirLaunch%20Phase%202B%20Milestones%201&2%20Photos.pdf

My memory is fading, but I thought they also released a video of the test at the time.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/13/2011 03:16 pm
Sorry, I forgot to add the word "nozzle" in there, so the nozzle is in the lower stage prop tank.

Its quite common.

Name one.
Title: Re: Basic Rocket Science Q & A
Post by: douglas100 on 05/13/2011 04:16 pm
Shtil'
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 05/13/2011 05:03 pm
Dnepr either has the 2nd stage nozzle in the lower stage prop tank, or the entire Dnepr 2nd stage engine is in the 2nd stage prop tank.

Title: Re: Basic Rocket Science Q & A
Post by: LegendCJS on 05/13/2011 06:06 pm
For rockets where the 2nd stage engine is housed inside the 1st stage prop tanks, how exactly does the engine leave the tank during staging? Is the prop tank blown up?

Do you know of a rocket where the second stage engine is immersed in the fuel of the first stage tanks before staging?  Don't you mean inter-stage volume?

Sorry, I forgot to add the word "nozzle" in there, so the nozzle is in the lower stage prop tank.

Its quite common.


So the nozzle of the second stage forms part of the pressure structure of the first stage? I don't think I've ever seen such a design...

No, the nozzle extends into the first stage prop tank, saving loads of interstage area.

The second stage nozzle is immersed in the fluid propellent of the upper first stage tank?  Let me see a diagram of that!  Some part of the nozzle throat or combustion chamber or engine would have to be part of the pressure containing boundary of the first stage tank in that case.

Or do you mean that the first stage tank's top is not a dome, but contains a concave region sized for the nozzle of the second stage engine, and there is no fluid communication between the second stage nozzle and the first stage propellant?

A simple diagram (even in paint) would save a lot of confusion for me and other people here it seems.
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 05/13/2011 06:26 pm


The second stage nozzle is immersed in the fluid propellent of the upper first stage tank?  Let me see a diagram of that!  Some part of the nozzle throat or combustion chamber or engine would have to be part of the pressure containing boundary of the first stage tank in that case.

Or do you mean that the first stage tank's top is not a dome, but contains a concave region sized for the nozzle of the second stage engine, and there is no fluid communication between the second stage nozzle and the first stage propellant?

A simple diagram (even in paint) would save a lot of confusion for me and other people here it seems.

Its not so easy to show in a diagram, but here is a try, a Rokot diagram.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/13/2011 11:42 pm
.This is one of the coolest things I've learned in the last few years:

Why did they do that? The nuclear arms treaties limited the length of the silos.  So having the upper tank extend up around the second stage nozzle got them a teeny bit more kilotons or a few more miles of range.

There are launch films where the missile is ejected from the silo and THEN the fairing springs up around the warhead, just so they could fit 3 more feet of rocket into the silo.

Brilliant engineering. A very interesting choice in prioritization of design requirements.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 05/13/2011 11:43 pm
For the rest of us, the build and operational complexity of that doesn't make sense for a commercially viable rocket. Just sacrifice some performance for simplicity.
Title: Re: Basic Rocket Science Q & A
Post by: douglas100 on 05/14/2011 11:10 am
From Antares:

Quote
Why did they do that? The nuclear arms treaties limited the length of the silos.  So having the upper tank extend up around the second stage nozzle got them a teeny bit more kilotons or a few more miles of range.

And of course the same technical argument applies to Shtil' and other submarine launched liquid propellant missiles.
Title: Re: Basic Rocket Science Q & A
Post by: Danderman on 05/14/2011 03:22 pm
For the rest of us, the build and operational complexity of that doesn't make sense for a commercially viable rocket. Just sacrifice some performance for simplicity.

Unless the rocket is being carried inside an aircraft.
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 05/16/2011 01:54 pm

For the rest of us, the build and operational complexity of that doesn't make sense for a commercially viable rocket. Just sacrifice some performance for simplicity.

Does it?

1. You eliminate the weight of the interstage.
2. The pressure in the empty tank when severed from the upper stage provides a separation impulse, eliminating the need and weight of ullage motors and separation pneumatics.
3. (Assuming LOX) Keeping the upper engine immersed in LOX tank will precondition the thermal environment.

It has some merit, If you look at Airlaunch LLC's design you'll notice that the Second stage nozzle (Vac Opt I'm sure) is almost as long as the first stage LOX tank. That is a fair amount of interstage kit you are saving, plus you don't need a stretched C-17 ;)

Of course I can think of more downsides than the three upsides I just mentioned...
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 06/10/2011 10:53 pm
D. L. Jensen states in this paper (http://www.aerowebspace.com/AdvancedPerformanceRocketEngines.pdf) that most current US H2 engines are develop with the "wrong", mixture ratios. Given that H2 needs such a huge volumes, it would only seem natural to try to use as little as possible, specially if you get better isp.
But since I know that there's a lot of intelligent people behind current engines, and I've learned better than to blindly trust random papers from the internet, specially when written as angry as this one, I would like to know what he's not saying. In particular, it doesn't makes any sense to me dual clone turbopumps against low pressure and high pressure.
From what I could grasp, the higher the mixture ration, the higher the pressure and temperature that the engine has to handle. And so it becomes more expensive and heavy. A reduction in the proportion of H2 would lower the tank's sizes, so it should offset a bit of the extra weight. And the reduced use of H2 should reduce the turbopump sizes. But there's so much that I guess I'm missing that I would love someone knowledgeable to tell me.
Title: Re: Basic Rocket Science Q & A
Post by: DMeader on 06/10/2011 11:14 pm
I'm not a rocket scientist myself (and I didn't stay at a Holiday Inn Express last night either) but even so some things jump out at me in Mr. Jensen's paper. Pardon me if I make some elementary errors.

Regarding his turbopump arrangement. I was under the impression that the SSME used the low pressure fuel and oxidizer turbopumps to bring the propellants to sufficient pressure to eliminate cavitation in the high-pressure turbopumps. Otherwise the external tank would have to supply propellants at much higher pressure itself and would need to be much stronger and heavier.

Regarding his contentions on "wasteful" mixture ratios. Doesn't H2 beyond stoichiometry contribute to thrust thru greater mass flow through the engine, the extra hydrogen being heated and expanded by combustion? Also, isn't a fuel-rich ratio desirable to protect the engine hardware?

I'm amused by people like Mr. Jensen who come out and say that the people that do this for a living (not me of course, I'm just an interested amateur) are obviously wrong about something so fundamental.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 06/11/2011 04:13 am
Doesn't H2 beyond stoichiometry contribute to thrust thru greater mass flow through the engine, the extra hydrogen being heated and expanded by combustion?

Excess hydrogen improves specific impulse but not by providing additional mass flow.  Rather it increases specific impulse by raising  the efficiency with which the nozzle can convert heat into kinetic  energy.  That in turn is because hydrogen molecules, being simpler than  water molecules, have fewer ways of soaking up energy, so a larger  fraction of the total energy can go into the desired form, kinetic.

Adding mass can improve propulsive efficiency if the added mass need not be carried by the vehicle -- that's a part of the reason that jet engines have much higher specific impulses than rocket engines.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 06/11/2011 05:49 pm
I understand that running the engine fuel rich gives better isp in the H2 case, than a lean mixture. I could also understand that it could get better isp than a stoichiometric ratio, but with an incomplete mixture (i.e. if the engine is run below its optimum pressure and temperature). The graph that he show, is suggestive.
Title: Re: Basic Rocket Science Q & A
Post by: TyMoore on 06/11/2011 05:54 pm
D. L. Jensen states in this paper (http://www.aerowebspace.com/AdvancedPerformanceRocketEngines.pdf) that most current US H2 engines are develop with the "wrong", mixture ratios. Given that H2 needs such a huge volumes, it would only seem natural to try to use as little as possible, specially if you get better isp.
But since I know that there's a lot of intelligent people behind current engines, and I've learned better than to blindly trust random papers from the internet, specially when written as angry as this one, I would like to know what he's not saying. In particular, it doesn't makes any sense to me dual clone turbopumps against low pressure and high pressure.
From what I could grasp, the higher the mixture ration, the higher the pressure and temperature that the engine has to handle. And so it becomes more expensive and heavy. A reduction in the proportion of H2 would lower the tank's sizes, so it should offset a bit of the extra weight. And the reduced use of H2 should reduce the turbopump sizes. But there's so much that I guess I'm missing that I would love someone knowledgeable to tell me.

I read the paper, and one thing that I saw that was wrong right off the bat, was that he was using 43,000 BTU/lb as the enthalpy of combustion for hydrogen. This is wrong: the lower heat of combustion for hydrogen is about 51,000 Btu per pound (non condensing.) [61,000 BTU/lb for higher heat of combustion (condensing)]

The performance of a rocket engine is not just Isp; it is also thrust. You have to weigh both and optimize the engine calculations to perform the mission. This is why there are different mixture ratios. And burning slightly hydrogen rich reduces the average molecular weight of the exhaust, which improves Isp but reduces the overall thrust slightly. Some nuclear thermal rocket engines would actually use a liquid oxygen injection system to boost total thrust by combusting some of the hot hydrogen exiting the nozzle throat. This improves thrust, but reduces overall Isp slightly.

A rocket engine is more than the sum of its "Thrust and Isp." It has to achieve performance goals necessary to achieve the mission.
Title: Re: Basic Rocket Science Q & A
Post by: Gregori on 06/24/2011 03:52 am
I've some basic questions i wanted to know about Mars/Moon missions using chemical propulsion.

1) Does chemical propulsion have an inherent speed limited that prevents it from doing a much faster transit to Mars? Is it that you need a large amount of propellant to slow down and circularize the orbit? Could a huge craft with multiple stages assembled in space speed up the mission?

I've heard of NTRs being able to allow faster transits, but I was never sure why they allowed this. I also heard that New Horizons reached the orbit of Mars in two months with chemical. Would this speed be just too fast to be captured by mars without melting the craft or a crash landing?

2) Does the gravity when closely approaching Mars or the Moon help slow down craft significantly and capture them or does this require lots of propellant and using the atmosphere to slow the craft?

Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 06/24/2011 05:20 am
1) Does chemical propulsion have an inherent speed limited that prevents it from doing a much faster transit to Mars? Is it that you need a large amount of propellant to slow down and circularize the orbit? Could a huge craft with multiple stages assembled in space speed up the mission?

In principal there's no reason that fast transits can't be done with chemical propulsion, it's just a matter of cost.  Multiple in-space stages could be used with chemical or nuclear rockets.  The Daedalus stellar probe design from the 1970s, for example, calls for a two-stage nuclear-fusion rocket.

Quote
I've heard of NTRs being able to allow faster transits, but I was never sure why they allowed this. I also heard that New Horizons reached the orbit of Mars in two months with chemical. Would this speed be just too fast to be captured by mars without melting the craft or a crash landing?

Yeah, to fast to capture.  Could have been slowed down with rockets -- chemical, nuclear, or otherwise -- but now were talking about hauling lots and lots of propellant all the way out to Mars, so things get really expensive.  Aerocapture could help.

Quote
2) Does the gravity when closely approaching Mars or the Moon help slow down craft significantly and capture them or does this require lots of propellant and using the atmosphere to slow the craft?

Basically, if the spacecraft arrives at the planet above the escape velocity, then without braking of some sort it will not be captured.  Essentially, as seen from the planet, it will speed up as it approaches the planet and then will slow back down again after it swings past.  So, no, the gravity of the planet alone will not permit capture.
Title: Re: Basic Rocket Science Q & A
Post by: hop on 06/24/2011 05:26 am
1) Does chemical propulsion have an inherent speed limited that prevents it from doing a much faster transit to Mars?
http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

Equally valid for nuclear thermal, ion, chemical, throwing rocks off the back...
Title: Re: Basic Rocket Science Q & A
Post by: IsaacKuo on 06/24/2011 07:24 am
I also heard that New Horizons reached the orbit of Mars in two months with chemical. Would this speed be just too fast to be captured by mars without melting the craft or a crash landing?

Mars would not "capture" a space probe on its own regardless of speed.  The space probe would enter at greater than escape speed and leave at the same speed, relative to Mars, on a hyperbolic trajectory.  The probe might use a rocket thrust or the as yet untried technique of aerocapture to slow down into Mars orbit, but Mars's gravity isn't going to do the job by itself.

Alternatively, the space probe could simply directly land on Mars after atmospheric entry to slow it down.  This requires a heavy heat shield and it requires the probe to be designed to survive high accelerations.  But this atmospheric entry would be a cakewalk compared to the Galileo atmospheric entry probe.  That probe slammed into Jupiter's atmosphere at a blistering 48km/s and survived 230 gees of deceleration!!!

Quote
2) Does the gravity when closely approaching Mars or the Moon help slow down craft significantly and capture them or does this require lots of propellant and using the atmosphere to slow the craft?

The gravity actually accelerates the probes, but ironically they also help thanks to the Oberth effect and the fact that you only need to slow down to orbital speed (rather than zero speed).  In fact, deeper stronger gravity wells are better.  The Moon's gravity is almost more of a hindrance than a help, but Mars's gravity well is certainly helpful.
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 06/24/2011 09:21 am
D. L. Jensen states in this paper (http://www.aerowebspace.com/AdvancedPerformanceRocketEngines.pdf) that most current US H2 engines are develop with the "wrong", mixture ratios. Given that H2 needs such a huge volumes, it would only seem natural to try to use as little as possible, specially if you get better isp.
But since I know that there's a lot of intelligent people behind current engines, and I've learned better than to blindly trust random papers from the internet, specially when written as angry as this one, I would like to know what he's not saying. In particular, it doesn't makes any sense to me dual clone turbopumps against low pressure and high pressure.
From what I could grasp, the higher the mixture ration, the higher the pressure and temperature that the engine has to handle. And so it becomes more expensive and heavy. A reduction in the proportion of H2 would lower the tank's sizes, so it should offset a bit of the extra weight. And the reduced use of H2 should reduce the turbopump sizes. But there's so much that I guess I'm missing that I would love someone knowledgeable to tell me.

I have a case of insomnia (my wife goes in for surgery tomorrow, and I am 2400 miles away), so I felt like tackling this.

D.L. Jensen, P.E.'s problem is that he doesn't know what stagnation enthalpy is. I have attached a PDF I worked up with the correct way to get exit velocity from chamber enthalpy (hm in his paper).

Basically though, he gives you an equation for specific impulse (exit velocity, but basically the same thing):

Ve2=2Jghm

It should read:

Ve2=2Jg(hm-he)

Where he is the energy you couldn't recover from the flow. This value is smaller for a fuel-rich hydrogen engine.

This is a derivation that an engineering sophomore should understand, someone with a P.E. (that's "Professional Engineer") after their name has no excuse, particularly when it is in every book that covers the basics of rocket engines. He especially had no right to be a sanctimonious ass and insult an entire industry of people with a vastly better grip on the subject.

The guy also gets some basic numbers wrong. His chart showing a monotonic increase of Isp with O/F ratio has RS-68 as O/F=2.5. It is actually at 6. The RL-10 is 5.88, with Isp of 465. Plot those correctly, and you just have a scattered cluster between 5 and 6, with no correlation.

Also, Proponent, I realize you're dead right from our earlier discussion. Redoing this derivation made it painfully obvious. Static enthalpy goes to zero for an infinite nozzle (because temp goes to zero), so you obviously want to just maximize chamber temp in that case. That's basically what this guy did is assumed an infinite expansion ratio, and then somehow figured that the same logic would hold for a real nozzle.

EDIT: I should note that she is going in for gallbladder removal, which is very minor as surgeries go, before anyone gets too concerned.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 06/24/2011 10:10 am
I have a case of insomnia (my wife goes in for surgery tomorrow, and I am 2400 miles away), so I felt like tackling this.

Let us know how things go.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 06/24/2011 03:57 pm
Best wished, no matter how "small" the surgery is.
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 06/24/2011 08:02 pm
Best wished, no matter how "small" the surgery is.


Let us know how things go.

Everything was a success, no issues. Thank you both for your concern, it means a lot.

As for my late night rocket science, I hope I adequately answered where Jensen went astray. High temperature is good, but so is low molecular weight, and it's a trade between the two.

Incidentally, nuclear rockets are governed by the same equation. The nice thing with nuclear is not so much that you can have higher temperatures. Your materials limits mean that they are about the same chamber temp as a chemical rocket.

Where nuclear pulls ahead is your molecular weight and temperature are decoupled. You can use pure hydrogen, which is how you get 1000s of Isp. That's for solid core, of course. If you're talking about more exotic nukes, you can bump up temp too (I'm a fan of the nuclear lightbulb concept, myself).
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 06/24/2011 08:13 pm
Best wished, no matter how "small" the surgery is.


Let us know how things go.

Everything was a success, no issues. Thank you both for your concern, it means a lot.

As for my late night rocket science, I hope I adequately answered where Jensen went astray. High temperature is good, but so is low molecular weight, and it's a trade between the two.

Incidentally, nuclear rockets are governed by the same equation. The nice thing with nuclear is not so much that you can have higher temperatures. Your materials limits mean that they are about the same chamber temp as a chemical rocket.

Where nuclear pulls ahead is your molecular weight and temperature are decoupled. You can use pure hydrogen, which is how you get 1000s of Isp. That's for solid core, of course. If you're talking about more exotic nukes, you can bump up temp too (I'm a fan of the nuclear lightbulb concept, myself).

Glad to hear that.

I too am a complete fan of the nuclear lightbulb. That's why I said "a process to create pure silica" over in the thread about what do we want to see developed. That material is completely transparent to the thermal energy of a nuclear reaction so whatever thermal energy is left over from dealing with the pure hydrogen just goes out thru the glass engine into space. The isp would climb into the tens or even hundreds of thousands of seconds, and the thrust would climb right along with it.
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 06/24/2011 08:31 pm

Glad to hear that.

I too am a complete fan of the nuclear lightbulb. That's why I said "a process to create pure silica" over in the thread about what do we want to see developed. That material is completely transparent to the thermal energy of a nuclear reaction so whatever thermal energy is left over from dealing with the pure hydrogen just goes out thru the glass engine into space. The isp would climb into the tens or even hundreds of thousands of seconds, and the thrust would climb right along with it.

Indeed, I've also wondered about that design using synthetic diamond. If you're not familiar with the CVD process for creating diamond, it's interesting. You make a low pressure, low temp carbon gas, and allow it to deposit on a preexisting diamond. The upshot is the technology has the potential to create relatively large, relatively arbitrary shapes out of diamond. Aside from temp resistance (as long as it's not in an oxidizer...), diamond has phenomenal thermal conductivity.

Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 06/24/2011 08:43 pm

Glad to hear that.

I too am a complete fan of the nuclear lightbulb. That's why I said "a process to create pure silica" over in the thread about what do we want to see developed. That material is completely transparent to the thermal energy of a nuclear reaction so whatever thermal energy is left over from dealing with the pure hydrogen just goes out thru the glass engine into space. The isp would climb into the tens or even hundreds of thousands of seconds, and the thrust would climb right along with it.

Indeed, I've also wondered about that design using synthetic diamond. If you're not familiar with the CVD process for creating diamond, it's interesting. You make a low pressure, low temp carbon gas, and allow it to deposit on a preexisting diamond. The upshot is the technology has the potential to create relatively large, relatively arbitrary shapes out of diamond. Aside from temp resistance (as long as it's not in an oxidizer...), diamond has phenomenal thermal conductivity.



What would it's transparency be to hard ultraviolet?
Title: Re: Basic Rocket Science Q & A
Post by: IsaacKuo on 06/24/2011 08:51 pm
I too am a complete fan of the nuclear lightbulb. That's why I said "a process to create pure silica" over in the thread about what do we want to see developed. That material is completely transparent to the thermal energy of a nuclear reaction so whatever thermal energy is left over from dealing with the pure hydrogen just goes out thru the glass engine into space. The isp would climb into the tens or even hundreds of thousands of seconds, and the thrust would climb right along with it.

Even if the bulb walls are completely transparent, they will still absorb incredible amounts of heat from direct contact with the reactor fuel.  If you had some magical method of preventing direct contact with the reactor fuel, then the bulb walls would be superfluous.  (Unfortunately, magnetic fields aren't strong enough, which is the reason why transparent bulb walls are considered necessary in the first place.)

The bulb walls need to be kept actively cooled to below melting point, which means a high rate of heat transfer if the reactor gas is hot.  If you try to use your propellant to do this cooling, then you need a very high mass rate.  Your performance ends up limited to solid core NTP Isp--which makes the whole thing pretty pointless.

So instead, you need a closed loop cooling system with large radiators to keep the bulb walls from melting while also maintaining a propellant consumption rate low enough to provide high Isp.  This radiator system limits your potential power/weight ratio.  As such, you're never going to have the same sort of high thrust potential as solid core NTP.  And since you're power limited, the potential thrust is at best inversely proportional to the Isp.

All in all, the nuclear lightbulb concept is technologically challenging with numerous potential pitfalls.  It represents a high technology risk compared to nuclear electric, and the potential performance benefit over nuclear electric is questionable.  While it might have an order of magnitude better power/weight ratio than nuclear electric, it might not.  And if not, then it's just a massive R&D investment wasted on something which is inferior to nuclear electric (which is far less expensive to develop, can use a variety of dense storable propellants, provides plentiful electrical power for other mission hardware, and can have much higher Isp).

My bet is that the inherent technology risk of nuclear lightbulb is high enough that we never find out how hard or easy it would have been for us to develop.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 06/26/2011 03:44 am
Nuclear lightbulb /= basic rocket science
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 07/07/2011 09:20 pm
I guess is a silly question, but with tanks and such structures, is the isogrid first machined and then formed, or first formed and then machined?
A related question it, is formed and then welded, or welded and then formed?
All this was from seeing the Dragon's interior. It would seem that they formed the parts, machined them, and then friction welded. Because you can see the rims of each part. I didn't knew you could do friction welding in a curved surface like the Falcon's, though.
I guess that from a structural side the ideal process would be to weld->form->machine. But it's exactly opposite of how is easier to machine.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 07/08/2011 12:07 am
Someone will correct me if I'm wrong.  Machined in a flat sheet then formed to a curved tank.  Usually formed then welded, but not always - depends on other factors.
Title: Re: Basic Rocket Science Q & A
Post by: neutrino78x on 07/08/2011 03:03 am
Hey guys (men and women)... if you were in a stanford torus (http://en.wikipedia.org/wiki/Stanford_torus) of the right dimensions to provide an acceleration equal to that of Earth gravity (9.81 m/s), but you were in an evacuated compartment, that is to say, one in which is there is no matter whatsoever except you and a spacesuit, would you still feel the acceleration pushing you against the outside edge of the torus (the "floor")? Or does there have to be air in the compartment to transfer the force? I think the answer is that you would just float, but I wanted to check what others think, especially if they have a college degree!

btw we assume that you start out in the middle of the compartment, like maybe you were hooked up to a tether or something, and then they evacuated all the air.

--Brian
Title: Re: Basic Rocket Science Q & A
Post by: Sparky on 07/08/2011 03:41 am
Assuming you weren't touching the "floor", and there were no obstacles within your path to bump into, you should be able to float, but only if the entire torus were evacuated (or at least a single corridor along its length). If you were floating inside a compartment, you would be able to experience Microgravity only until the wall bumped into you.

I imagine that even with atmosphere, if a person were to jump at the right speed, in the right direction, they could get very impressive hang time until just the air resistance slowed them enough to fall to the floor.
Title: Re: Basic Rocket Science Q & A
Post by: lcs on 07/08/2011 04:08 am
I imagine that even with atmosphere, if a person were to jump at the right speed, in the right direction, they could get very impressive hang time until just the air resistance slowed them enough to fall to the floor.

In other words, orbital velocity inside the torus is just the v=omega x R.  For the Stanford torus v=pi*1.6 km /60 sec = 94 meters/sec.  I don't think jumping into orbit will be an Olympic event. 
Title: Re: Basic Rocket Science Q & A
Post by: STS-134 on 07/08/2011 04:16 am
I imagine that even with atmosphere, if a person were to jump at the right speed, in the right direction, they could get very impressive hang time until just the air resistance slowed them enough to fall to the floor.

In other words, orbital velocity inside the torus is just the v=omega x R.  For the Stanford torus v=pi*1.6 km /60 sec = 94 meters/sec.  I don't think jumping into orbit will be an Olympic event. 

94 m/s = about 210.7 mph.  At that speed, air resistance will VERY QUICKLY cause you to begin rotating again with the torus, and get pushed to the "floor".
Title: Re: Basic Rocket Science Q & A
Post by: lcs on 07/08/2011 04:21 am
I imagine that even with atmosphere, if a person were to jump at the right speed, in the right direction, they could get very impressive hang time until just the air resistance slowed them enough to fall to the floor.

In other words, orbital velocity inside the torus is just the v=omega x R.  For the Stanford torus v=pi*1.6 km /60 sec = 94 meters/sec.  I don't think jumping into orbit will be an Olympic event. 

94 m/s = about 210.7 mph.  At that speed, air resistance will VERY QUICKLY cause you to begin rotating again with the torus, and get pushed to the "floor".

In any case, the answer to neutrino's question is that even in a vacuum, if you were suspended in the middle of the torus and suddenly released you would fall to the floor, because you would be moving in a straight line at the velocity of the torus, while the torus floor rushed up to meet you as it curved away from your straight line trajectory. 
Title: Re: Basic Rocket Science Q & A
Post by: STS-134 on 07/08/2011 04:42 am
I imagine that even with atmosphere, if a person were to jump at the right speed, in the right direction, they could get very impressive hang time until just the air resistance slowed them enough to fall to the floor.

In other words, orbital velocity inside the torus is just the v=omega x R.  For the Stanford torus v=pi*1.6 km /60 sec = 94 meters/sec.  I don't think jumping into orbit will be an Olympic event. 

94 m/s = about 210.7 mph.  At that speed, air resistance will VERY QUICKLY cause you to begin rotating again with the torus, and get pushed to the "floor".

In any case, the answer to neutrino's question is that even in a vacuum, if you were suspended in the middle of the torus and suddenly released you would fall to the floor, because you would be moving in a straight line at the velocity of the torus, while the torus floor rushed up to meet you as it curved away from your straight line trajectory. 

Well, it depends on the initial condition (velocity).  If your velocity is such that you are initially rotating along with the torus, you will NOT feel any acceleration (because you won't be accelerating, you'll just be moving along a straight line in the non rotating reference frame of the torus hub).  You will be weightless, but as time goes on you'll see the "floor" accelerate toward you at something less than 1.0g, until you eventually hit the floor.  The rate at which you acclerate toward the floor depends on your height above the "floor" when you are released from the tether.  The higher up you are (the closer you are to the "hub"), the slower the "floor" will "accelerate" toward you.  This is basically equivalent to being held above the floor on Earth, and released in an evacuated chamber.  You will accelerate toward the floor at 1.0g, and you will experience weightlessness until you hit the floor.

If, on the other hand, you are inside an evacuated torus, AND stationary with respect to the hub at the center of the torus, you'll just float.
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 07/15/2011 08:05 am
...until you drift into the floor that's moving past at 210MPH.

I suspect the initial contact wouldn't be too painful if the closure rate is very low, but it would set you spinning, and add a small component of lateral velocity.

This will ensure another contact, increasing lateral velocity and spin rate.

Each time the process repeats the increasing lateral velocity increases the effective g forces of the "fall" towards the floor. By the time you're travelling at 105MPH, you've having 1/2g impacts with the floor at a relative velocity of 105MPH while tumbling violently. Suspect that's not survivable.

cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: Warren Platts on 07/25/2011 05:03 am
Q1: Roughly, what is the maximum bonus factor one can expect from the Oberth effect when injecting into Low Mars orbit (or Sun-Mars Lagrange point) fully propulsively with high thrust chemical rockets?

Q2: Given a chemical, single-stage MTV with 11 km/sec nominal delta v, if staged from EML1 or EML2, could one get an effective 44 km/sec total delta v using the Oberth effect when slingshotting around Earth and then braking into Mars orbit?

Q3: When departing Mars, can one get a significant Oberth bonus if the ERV was staged at Mars's Sun-Mars L-point?
Title: Re: Basic Rocket Science Q & A
Post by: IsaacKuo on 07/25/2011 02:39 pm
Q1: Roughly, what is the maximum bonus factor one can expect from the Oberth effect when injecting into Low Mars orbit (or Sun-Mars Lagrange point) fully propulsively with high thrust chemical rockets?

The maximum bonus factor is infinite, in the case of an infinitesmal delta-v at perigee compared to delta-v at apogee (or v_inf, in the case of a parabolic orbit).  It's not a matter of a clean "bonus factor", because the multiplier goes down with higher delta-v.

Quote
Q2: Given a chemical, single-stage MTV with 11 km/sec nominal delta v, if staged from EML1 or EML2, could one get an effective 44 km/sec total delta v using the Oberth effect when slingshotting around Earth and then braking into Mars orbit?

No.  Even if you start from a parabolic escape trajectory and perform an 11km/s burn at an Earth grazing perigee, the resulting v_inf is still only sqrt((v_esc+delta_v)^2 -v_esc^2) = sqrt((11.1+11)^2-11.1^2) = 19.11km/s.  This is a "bonus factor" of only 19.11/11 = 1.73.

The "bonus factor" is higher for smaller delta-v.  For example, a delta-v of .45km/s gives a v_inf of 3.19km/s.  This is a "bonus factor" of 3.19/.45 = 7.09.

A way to calculate the effect geometrically is to use a right triangle.  The base is escape velocity; the hypotenuse is escape velocity plus delta-v.  The other side of the triangle is velocity at infinity.  When delta-v is small compared to escape velocity, the "bonus factor" is high.  When delta-v is large compared to escape velocity, the "bonus factor" is low.

Quote
Q3: When departing Mars, can one get a significant Oberth bonus if the ERV was staged at Mars's Sun-Mars L-point?

I think so, if you're willing to accept very long trip times (on the order of many years).  However, it would be better to stage at a highly elliptical Mars orbit.  This gives you all of the benefit without any of the wait time.  Similarly, staging at a highly elliptical Earth orbit gives you less of a wait time than EML1 or EML2.
Title: Re: Basic Rocket Science Q & A
Post by: Warren Platts on 07/25/2011 05:47 pm
Q1: Roughly, what is the maximum bonus factor one can expect from the Oberth effect when injecting into Low Mars orbit (or Sun-Mars Lagrange point) fully propulsively with high thrust chemical rockets?

The maximum bonus factor is infinite, in the case of an infinitesmal delta-v at perigee compared to delta-v at apogee (or v_inf, in the case of a parabolic orbit).  It's not a matter of a clean "bonus factor", because the multiplier goes down with higher delta-v.

Quote
Q2: Given a chemical, single-stage MTV with 11 km/sec nominal delta v, if staged from EML1 or EML2, could one get an effective 44 km/sec total delta v using the Oberth effect when slingshotting around Earth and then braking into Mars orbit?

No.  Even if you start from a parabolic escape trajectory and perform an 11km/s burn at an Earth grazing perigee, the resulting v_inf is still only sqrt((v_esc+delta_v)^2 -v_esc^2) = sqrt((11.1+11)^2-11.1^2) = 19.11km/s.  This is a "bonus factor" of only 19.11/11 = 1.73.

The "bonus factor" is higher for smaller delta-v.  For example, a delta-v of .45km/s gives a v_inf of 3.19km/s.  This is a "bonus factor" of 3.19/.45 = 7.09.

A way to calculate the effect geometrically is to use a right triangle.  The base is escape velocity; the hypotenuse is escape velocity plus delta-v.  The other side of the triangle is velocity at infinity.  When delta-v is small compared to escape velocity, the "bonus factor" is high.  When delta-v is large compared to escape velocity, the "bonus factor" is low.

Quote
Q3: When departing Mars, can one get a significant Oberth bonus if the ERV was staged at Mars's Sun-Mars L-point?

I think so, if you're willing to accept very long trip times (on the order of many years).  However, it would be better to stage at a highly elliptical Mars orbit.  This gives you all of the benefit without any of the wait time.  Similarly, staging at a highly elliptical Earth orbit gives you less of a wait time than EML1 or EML2.

Thanks for the thoughtful remarks Isaac. The way you express the math is very clear.

Of course the delta v would be split up into two separate burns, but using Oberth's equation, I can see even that won't make that much of a difference:

sqrt((v_esc+delta_v)^2 -v_esc^2) = sqrt((11.1+5.5)^2-11.1^2) = 12.3km/s

Bonus factor: 12.3 / 5.5 = 2.2

The maximum warp for a single-stage craft with a usable payload isn't going to be much more than 11 km/sec nominally. With an Oberth bonus, we can't expect much more than to double that to maybe 20-22 km/sec. Still, that's not bad for a chemical rocket.

As for elliptical orbits versus LaGrange points: I don't see why you think the "wait time" is going to be a big deal--after all, EML1 or 2 is only a week or so from Earth. Since Mars gravity is so weak compared to the Sun, the Sun-Mars L1 point is going to be fairly close to Mars. Also, Lagrange points are ideal locations for propellant depots. (Not to mention that EML1/2 is close to Lunar propellant sources ;) ).
Title: Re: Basic Rocket Science Q & A
Post by: IsaacKuo on 07/25/2011 06:37 pm
Thanks for the thoughtful remarks Isaac. The way you express the math is very clear.

You're welcome!

Quote
As for elliptical orbits versus LaGrange points: I don't see why you think the "wait time" is going to be a big deal--after all, EML1 or 2 is only a week or so from Earth. Since Mars gravity is so weak compared to the Sun, the Sun-Mars L1 point is going to be fairly close to Mars. Also, Lagrange points are ideal locations for propellant depots. (Not to mention that EML1/2 is close to Lunar propellant sources ;) ).

If you want to minimize delta-v, the wait period will be on the same order of magnitude as the orbital period.  You use multiple orbits to reduce the delta-v and take advantage of the Lagrange point instability.

For EML1 and EML2 the wait isn't such a big deal because the orbital period is only a month.  For Sun-Mars L1 and L2, the orbital period is almost 2 years.  Yes, you can get to/from Sun-Mars L1 and L2 faster by using more delta-v--but this defeats the purpose.  You might as well have staged at any random location outside Mars's Hill sphere.

Also, the direction of the hyperbolic escape trajectory after leaving Mars won't line up well with the desired direction to get back to Earth.  This isn't a problem for EML1/EML2 because of the way the Moon orbits Earth.  Within a month, it will be in the correct alignment for the desired Earth escape trajectory.  But the Sun-Mars lagrange point retain the same orientation with respect to Mars's orbit at all times.

As for the ideal location for propellant depots--I favor placing tankers in highly elliptical orbits rather than lagrange points.  It maximizes the delta-v advantage while reducing the wait times.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 08/01/2011 02:29 pm
For a given Isp (I don't care if it's 50 or 5000) and in-space trajectory change (orbit change, maneuvers, escapes), does high thrust or low thrust end up with less propellant required?
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 08/01/2011 02:58 pm
For a given Isp (I don't care if it's 50 or 5000) and in-space trajectory change (orbit change, maneuvers, escapes), does high thrust or low thrust end up with less propellant required?
Here's a quick math-like "proof" that high thrust always uses at most the delta vee of low thrust: a high thrust engine can simulate a low thrust engine by operating in a series of short pulses.

You can sometimes also go the other way simulate a high thrust engine with a low thrust engine, at least in theory. For example you can presumably simulate a high-thrust Hoffmann transfer using a low thrust engine by firing the engine for a short pulse at perigee over many orbits until apogee is raised, and then short pulses at apogee. This trick is of course limited to circumstances where you have patience and repeated orbits.

More practically high thrust enables better use of the Oberth effect, which can reduce delta vee. For example a low-thrust spiral from low earth orbit to escape uses 100% of orbital speed worth of delta vee whereas an infinite-thrust impulse uses only 41%. (See http://en.wikipedia.org/wiki/Hohmann_transfer_orbit .)

Of course propellant usage is more complicated than just delta vee, but surely you can't expect someone answering your short question to incorporate tradeoffs between dry mass and delta vee!
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 08/01/2011 04:46 pm
Nope, no expectations.  Thanks.  Physical intuition is faster than a numerical analysis.
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 08/01/2011 11:23 pm
For a given Isp (I don't care if it's 50 or 5000) and in-space trajectory change (orbit change, maneuvers, escapes), does high thrust or low thrust end up with less propellant required?
It really depends... Given that rocket engines are limited in their T/W, a "really high"-thrust stage will take more propellant than a "moderately high"-thrust, since the dry mass that needs to be put through your delta-v will be higher for the high-thrust stage. Also, higher thrust (if higher than launch accelerations, etc) can also mean higher structural mass requirements in both the rocket stage and payload, thus increasing total propellant needed.

And, of course, if you have too low thrust, your long burn means you can't burn all the necessary propellant deep in the gravity well, thus you need more propellant for a given escape velocity, etc.

But with multiple passes and multiple burns (raising apogee, not perigee, say for escape), it can be possible to "simulate" a higher thrust rocket engine with a much lower thrust rocket engine...

Akin's Law #8:
"8. In nature, the optimum is almost always in the middle somewhere. Distrust assertions that the optimum is at an extreme point."
http://spacecraft.ssl.umd.edu/akins_laws.html
Title: Re: Basic Rocket Science Q & A
Post by: Patchouli on 08/02/2011 05:02 am
For a given Isp (I don't care if it's 50 or 5000) and in-space trajectory change (orbit change, maneuvers, escapes), does high thrust or low thrust end up with less propellant required?
It really depends... Given that rocket engines are limited in their T/W, a "really high"-thrust stage will take more propellant than a "moderately high"-thrust, since the dry mass that needs to be put through your delta-v will be higher for the high-thrust stage. Also, higher thrust (if higher than launch accelerations, etc) can also mean higher structural mass requirements in both the rocket stage and payload, thus increasing total propellant needed.

And, of course, if you have too low thrust, your long burn means you can't burn all the necessary propellant deep in the gravity well, thus you need more propellant for a given escape velocity, etc.

But with multiple passes and multiple burns (raising apogee, not perigee, say for escape), it can be possible to "simulate" a higher thrust rocket engine with a much lower thrust rocket engine...

Akin's Law #8:
"8. In nature, the optimum is almost always in the middle somewhere. Distrust assertions that the optimum is at an extreme point."
http://spacecraft.ssl.umd.edu/akins_laws.html

I believe this actually was done to save a com sat using the RCS because the apogee engine failed but it's also an advantage VASIMR has over ion engines.
A VASIMR rocket is not bothered as much by multiple restarts as a gridded ion engine is.

Might add a another law to the list if a change only nets an extra 3% performance gain but increases the cost by 15% it's best not to implement it.
I'll call it Korolev's law as Russians LVs often tend just accept a lower payload mass fraction if a given design element is lower cost or more robust.

Russian LVs tend to be simple and robust while western LVs tend to be highly tuned hotrods.
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 08/02/2011 05:07 am
...
A VASIMR rocket is not bothered as much by multiple restarts as a gridded ion engine is.
...
...And the main reason is because VASIMR has never flown. ;)

I actually kind of doubt this. Have you seen the flight profile of Dawn (which uses a gridded ion engine)? It's all chopped up... Very many starts and stops.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 08/02/2011 05:08 am
Touche. Usually I'm the one citing Akin.

Stage size is fixed in this thought exercise. A spacecraft or upper stage is going to be structurally designed to live through boost phase g loads anyway. T/W trade space between 0.1 and 1.0.
Title: Re: Basic Rocket Science Q & A
Post by: Patchouli on 08/02/2011 05:34 am
...
A VASIMR rocket is not bothered as much by multiple restarts as a gridded ion engine is.
...
...And the main reason is because VASIMR has never flown. ;)

I actually kind of doubt this. Have you seen the flight profile of Dawn (which uses a gridded ion engine)? It's all chopped up... Very many starts and stops.

No grids to short out which makes it more forgiving plus keep in mind even though it has not flown there has been a lot of ground tests but the Hall thruster also lacks the troublesome grids.

I think gridded ion engines likely will fall out of favor for VASIMR and Hall thrusters at least for larger applications such as high ISP tugs since they can scale better.

Title: Re: Basic Rocket Science Q & A
Post by: Urvabara on 08/08/2011 06:52 am
Let's try to assume a realistic rocket that uses methane as a propellant. No oxygen for this rocket.

How much methane is needed to raise 100 kilograms of cargo onto LEO (400 kilometers above the surface)?

How much power and energy is needed to heat up that methane and produce enough thrust to raise that mass of 100 kilograms onto LEO (400 km)?
Title: Re: Basic Rocket Science Q & A
Post by: Urvabara on 08/08/2011 09:19 am
Is it possible to build a Solar Thermal Rocket with thrust-to-weight ratio > 1?
Title: Re: Basic Rocket Science Q & A
Post by: Galactic Penguin SST on 08/19/2011 04:17 am
One rookie question: Do rocket stages using hypergolic fuel have the danger of fuel/oxidizer sloshing? Why?
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 08/19/2011 05:12 am

I believe this actually was done to save a com sat using the RCS because the apogee engine failed but it's also an advantage VASIMR has over ion engines.
A VASIMR rocket is not bothered as much by multiple restarts as a gridded ion engine is.

Might add a another law to the list if a change only nets an extra 3% performance gain but increases the cost by 15% it's best not to implement it.
I'll call it Korolev's law as Russians LVs often tend just accept a lower payload mass fraction if a given design element is lower cost or more robust.

Russian LVs tend to be simple and robust while western LVs tend to be highly tuned hotrods.

Yeah, it was on Advanced EHF (http://spaceflightnow.com/atlas/av019/100902inquiry.html), a military comsat. Apogee failed, so they had to use the RCS to put it in a stable orbit, otherwise it would have deorbited very quickly. It's been spiraling for a year on Hall thrusters since, which also don't care about restarts, and actually are flying ;).

One rookie question: Do rocket stages using hypergolic fuel have the danger of fuel/oxidizer sloshing? Why?

Yes they do. If you launched a tank with water, it would too. Slosh is nothing exotic, it can just lead to hard to predict effects depending on how it couples with the vehicle. If you have a liquid, it can slosh.

Let's try to assume a realistic rocket that uses methane as a propellant. No oxygen for this rocket.

How much methane is needed to raise 100 kilograms of cargo onto LEO (400 kilometers above the surface)?

How much power and energy is needed to heat up that methane and produce enough thrust to raise that mass of 100 kilograms onto LEO (400 km)?

Well, this is a very strange rocket, and I'm not totally sure of what you're asking, but I gave it a shot. I've attached a file. Basically, we're limited by materials to a temp around combustion temp anyway. So, I've limited to 3000K. Hot methane has a specific heat ratio of about 1.2, molecular weight of 16. Assumed a nozzle with a pressure ratio of 100, which is reasonable (and going to 10 or 1000 doesn't change it that much). So, the Isp is only 314s under these assumptions.

To get a delta-V of 9600 m/s (Low Earth Orbit w/ g losses, drag), and assuming a weightless structure (any reasonable structure, and it won't get to LEO on an SSTO), you need 2160kg of methane.

The energy required is a function of the total propellant mass, and the exit velocity (Isp), I get 10.2 GJ, or about the energy consumed by a typical American car in 2 months.

The max power will occur at max thrust (if Isp is constant). Thrust should be sized by thrust to weight ratio, I chose 1.1 at liftoff. Power will be 88.4 MW at that time, or about the power of a GE90 jet engine (don't get any ideas, different tech).

Is it possible to build a Solar Thermal Rocket with thrust-to-weight ratio > 1?

Well, let's look at the solar collector first. The sun provides about 1000W/m^2 of energy flux near Earth. Power is related to thrust as follows:

P=0.5*T*Isp*g0

Isp is limited to about 1000s, with hydrogen and temps that don't melt the engine.
So, thrust is:

T=2P/(Isp*g0)

Therefore, T/W>1 requires:

2P/(Isp*g0W)>1

Rearranging we have this limit on the weight of the solar collector, assuming it is the bulk of the structure:

W<2DA/(Isp*g0)

Where D is the 1000 W/m^2 solar energy flux, and A is the solar collector area.

Then, rewriting in terms of density:

rho*V*g0<2DA/(Isp*g0)

For a thin shell, the volume, V, of material (not enclosed volume, FYI) is the product of surface area and thickness:

rho*A*t*g0<2DA/(Isp*g0)

Therefore:

rho*t<2/g02
rho*t<0.02

If we're looking at Mylar, with a thickness of say 20 micrometers, then:

rho<1000 kg/m^3

Mylar is heavier than that, and most practical polymers are close. So, your solar collector mass almost breaks the budget. I would say it is very unlikely.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 08/19/2011 02:23 pm
Might add a another law to the list if a change only nets an extra 3% performance gain but increases the cost by 15% it's best not to implement it.
I'll call it Korolev's law as Russians LVs often tend just accept a lower payload mass fraction if a given design element is lower cost or more robust.

Russian LVs tend to be simple and robust while western LVs tend to be highly tuned hotrods.

My uncle was a F-86 pilot in the Korean War and tells a story about the MIG-15's. They were really rugged, built like tanks and not swiss watches. He talks about one that he shot down. It was the last of his rounds so he was starting to head home (he thought he had a kill) when he noticed that the pilot was actually landing the plane in a mushy field. The pilot got out and did a quick field repair to the rudder, where my uncle had hit him, and got back in the cockpit and took off again from a waterlogged field, actually a rice paddy. Positionally, my uncle had the tactical advantage so he was able to get right behind the MIG pilot. That pilot knew he was dead (but he didn't know my uncle was out of shells). My uncle was so impressed with the aircraft and what that pilot did that he wagged him and then turned for home. He took a hell of a chance that the MIG wouldn't turn to engage but it didn't. The MIG pilot must have respected my uncle as much as my uncle respected him.

Just last year I was at an air show in Cheyenne, Wyoming and watched a MIG-15 perform. Afterward, I got to sit in the MIG and the pilot explained all the instrumentation and controls to me. Pretty sparse, only what the pilot really needed. The MIG was privately owned. The owner-pilot belongs to a flying club of vintage fighter aircraft and tours the country in the summer with his MIG. That MIG actually saw combat in Korea, which makes it 60 years old.

I relay that story to make the point that the Russians have been building things like that for decades. They know where the point of diminishing returns is and they don't cross it – they can't afford to. What they do do is make their machines as rugged (not sophisticated) as possible within the budget available. NASA could learn a few things from them in that regard.
Title: Re: Basic Rocket Science Q & A
Post by: Lee Jay on 08/19/2011 02:36 pm
The sun provides about 1000W/m^2 of energy flux near Earth.

1366W/m^2, actually.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 08/19/2011 03:23 pm
One rookie question: Do rocket stages using hypergolic fuel have the danger of fuel/oxidizer sloshing? Why?

Yes.  Slosh is a problem independent of what the liquid is.  It's because the CG of the stage is moving around because the liquid inside it is moving around.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 08/19/2011 04:26 pm
Another rookie question. If you'd have a tank pressurized over the triple point of a liquid (let's use H2 for the example), what would happen if you start with a full liquid charge?
Would it allow better density? (disregarding the tank's mass, of course)
And if you went beyond the critical point?
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 08/19/2011 04:34 pm
The sun provides about 1000W/m^2 of energy flux near Earth.

1366W/m^2, actually.

Meh, call it an efficiency factor, I was working off the top of my head. The fundamental point still stands.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 08/19/2011 07:10 pm
If you'd have a tank pressurized over the triple point of a liquid (let's use H2 for the example), what would happen if you start with a full liquid charge?
Would it allow better density? (disregarding the tank's mass, of course)
And if you went beyond the critical point?

I assume you're familiar with slush hydrogen and densified LOX, but those are more from lower temperature than higher pressure.  The former is advantageous because higher pressure requires heavier tanks.
Title: Re: Basic Rocket Science Q & A
Post by: DarkenedOne on 08/22/2011 03:12 pm
A modern ICBM is able to hit multiple targets using multiple independently targetable reentry vehicles with an accuracy of 150 meters and very high reliability.  Most importantly these ICBMs are fire and forget.  They only require a destination.  On top of that the launch team for these rockets are only a few people.

My question is why do modern rockets require such large launch teams?  The Shuttle has a launch team of 500 people.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/22/2011 03:22 pm
A modern ICBM is able to hit multiple targets using multiple independently targetable reentry vehicles with an accuracy of 150 meters and very high reliability. 

Weapon systems reliability is not same as for space launch.  The make up for failures with numbers.   Also the weapon system payloads are passive for launch.
Title: Re: Basic Rocket Science Q & A
Post by: DarkenedOne on 08/22/2011 05:21 pm
A modern ICBM is able to hit multiple targets using multiple independently targetable reentry vehicles with an accuracy of 150 meters and very high reliability. 

Weapon systems reliability is not same as for space launch.  The make up for failures with numbers.   Also the weapon system payloads are passive for launch.

The Trident missile system has had 135 consecutive successful tests.  That is pretty much the best space launch systems to date.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/22/2011 05:30 pm

The Trident missile system has had 135 consecutive successful tests.  That is pretty much the best space launch systems to date.

When it can put 50klb spacecraft into LEO then use it as example.
Title: Re: Basic Rocket Science Q & A
Post by: Tcommon on 08/22/2011 05:57 pm

The Trident missile system has had 135 consecutive successful tests.  That is pretty much the best space launch systems to date.

When it can put 50klb spacecraft into LEO then use it as example.

DarkenedOne, that's an excellent example. Launching with fewer people reliably is clearly possible, as shown with your example. Its the kind of new technology and designs that NASA should be working on.

It's a clear tip-off when Jim slams questions like this without offering any significant  technical rationale.
Title: Re: Basic Rocket Science Q & A
Post by: AS-503 on 08/22/2011 06:13 pm
ICBM does not equate to manned space launch.

Apples to Oranges.

Almost no commonality. Trajectory, mission type, ground support equiptment, facilities...etc.etc.

I stand by Jim's assertion.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/22/2011 06:16 pm

The Trident missile system has had 135 consecutive successful tests.  That is pretty much the best space launch systems to date.

When it can put 50klb spacecraft into LEO then use it as example.

DarkenedOne, that's an excellent example. Launching with fewer people reliably is clearly possible, as shown with your example. Its the kind of new technology and designs that NASA should be working on.

It's a clear tip-off when Jim slams questions like this without offering any significant  technical rationale.

"50klb spacecraft into LEO" is significant technical rationale.
Trident SLBM are not space launch vehicles.  And RV's are not spacecraft.  or even manned spacecraft.

It has nothing to with "new technology".  It has to do with complexity of the task and the complexity of the payload.  That is the difference between the vehicle.

Sticking with SRM's.

You have excluded the team to build up the trident, that is not part of the submarine crew nor the ICBM launch teams for Minuteman.
Look at a Minotaur launch team, it is in the 40 to 50 range (including the techs to build it up)

Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 08/22/2011 06:24 pm
Also the weapon system payloads are passive for launch.

Can you explain what you mean by that? I sometimes hear "spacecraft on internal power" or words to that effect in launch audio feeds, but what exactly does a spacecraft actively do before and during ascent?
Title: Re: Basic Rocket Science Q & A
Post by: Tcommon on 08/22/2011 06:50 pm
"50klb spacecraft into LEO" is significant technical rationale.
Trident SLBM are not space launch vehicles.  And RV's are not spacecraft.  or even manned spacecraft.

It has nothing to with "new technology".  It has to do with complexity of the task and the complexity of the payload.  That is the difference between the vehicle.,

And yet SLBMs have independently targeted payloads and so are more complicated, at least in one respect

Quote
Sticking with SRM's.

You have excluded the team to build up the trident, that is not part of the submarine crew nor the ICBM launch teams for Minuteman.
Look at a Minotaur launch team, it is in the 40 to 50 range (including the techs to build it up)

I wouldn't suggest launching from a submarine as a way to reduce launch personnel.

Jim you're saying what is, I'm asking what's possible by way of examples that show it can be done, reliably. But you don't get something for nothing. Reducing launch personnel would certainly entail a redesign of the carrier rocket, facilities and operating procedures, if not the entire company that launches it.  And would probably require less flexible standards that the payload must meet.

Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/22/2011 06:56 pm
I'm asking what's possible by way of examples that show it can be done, reliably.


They aren't valid examples.  launching a small warhead is not the same as launching a large spacecraft.

An RV is rugged and self contained.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/22/2011 06:58 pm

And yet SLBMs have independently targeted payloads and so are more complicated, at least in one respect



That isn't that complicated.
a.  The warhead does some of the maneuvering.
b. the target is stationary.
Title: Re: Basic Rocket Science Q & A
Post by: agman25 on 08/22/2011 07:31 pm
I have a question about lattice structures on interstages. It has been my understanding that these are designed to dissipate any fuel or oxidizer that may leak from the upperstage.

I see a simillar structure on the Agni-II, which is a solid fuelled IRBM. Can anybody explain why this design was used when there is no chance of fuel leakage. Does it have any other benifits.
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 08/22/2011 07:33 pm
I have a question about lattice structures on interstages. It has been my understanding that these are designed to dissipate any fuel or oxidizer that may leak from the upperstage.

I see a simillar structure on the Agni-II, which is a solid fuelled IRBM. Can anybody explain why this design was used when there is no chance of fuel leakage. Does it have any other benifits.

The primary purpose of such structures is to allow "fire in the hole" staging, as opposed to separating first and then firing the upper stage.
Title: Re: Basic Rocket Science Q & A
Post by: agman25 on 08/22/2011 07:37 pm
Thanks.
Title: Re: Basic Rocket Science Q & A
Post by: AS-503 on 08/22/2011 07:41 pm
The Apollo LM had "fire in the hole" for lunar ascent and descent abort capability.

I think the S-1b that launched the LM for the "fire in the hole" unmanned test flight was the booster from the Apollo 1 fire.

I wonder what the gas pressure on various portions of the Apollo LM was like given the lack of venting with respect to the previously mentioned lattice structures as an interface. 
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/22/2011 07:45 pm
It went around the descent engine
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 08/22/2011 11:11 pm
An ICBM is little more than a sophisticated gigantic artillery shell whose properties are so well known that its trajectory can be calculated to within a few meters of accuracy for ground impact. The main difference is that the solid propellant for an artillery shell is explosive while the solid propellant for an ICBM is not. The mass of all 10 of those MRVs combined don't come anywhere near the mass of a manned spacecraft. Plus once you launch an ICBM it will go where it is sent, provided something doesn't act on it from outside. If it goes off trajectory it cannot be controlled and is destroyed. Not so for a manned spacecraft. If it starts to go off trajectory you correct the trajectory if you can. The LV has that capability while the ICBM does not. If it can't be corrected you abort the spacecraft to a safe landing. Try doing that with any one of the MRVs.

The comparison is completely off the wall insane.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 08/22/2011 11:21 pm
If that were true then Atlas and Titan could never have been converted to space launch systems.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 08/22/2011 11:34 pm
If that were true then Atlas and Titan could never have been converted to space launch systems.

We were comparing modern solid fuel ICBM's to a manned launcher. Atlas and Titan were liquid fuel.
Title: Re: Basic Rocket Science Q & A
Post by: Tcommon on 08/22/2011 11:38 pm
If that were true then Atlas and Titan could never have been converted to space launch systems.

We were comparing modern solid fuel ICBM's to a manned launcher. Atlas and Titan were liquid fuel.

No, not solid ICBMs specifically. No, not manned launchers specifically.
http://forum.nasaspaceflight.com/index.php?topic=13543.msg796429#msg796429
Title: Re: Basic Rocket Science Q & A
Post by: yinzer on 08/23/2011 03:48 am
The ICBM to launch vehicle comparison is not completely off-the-wall insane, as should be obvious when you consider that there are current launch vehicles that are barely modified ICBMs, eg: Dnepr.

The size of the launch crew and the length of the countdown are essentially design requirements, and the people buying ICBMs want both to be very small while the people buying space launches largely don't care.

In order to meet these requirements ICBM designers do things like include extensive automated test equipment, use solid or storable fuels, house their rockets in silos, buy and build all their rockets ahead of time and in significant quantity, and do regular test flights.  All of these things costs money, both on the launch vehicle side and the payload side.
Title: Re: Basic Rocket Science Q & A
Post by: AS-503 on 08/23/2011 04:14 am

Atlas (Mercury) and Titan (Gemini) were last gasp measures to catch up with the Soviet program in a *big hurry*.

Atlas needed thicker tankage/skin and used a "belly band" on John Glenn's first orbital Mercury. This flight also had engines that were not baffled and prone to 1 in 5 combustion instability (i.e.LOM or LOC).

The Titan had an extremely steep trajectory with 7g load on the crew and NO abort option. Steep trajectory was from underpowered booster for the given task so you can't just put an abort tower in the mix there.

John Young mentions the Gemini/Titan ascent in the special features of the In The Shadow of the Moon DVD, by saying, "I don't think comfort is what your looking for going uphill".

Neither one of these was a good manned launcher (much less optimal) so the comparison is not that valid.

These boosters were *only* "manrated" for the extremes of the Cold War/Space Race.

Each one of these crash programs were never intended to last more than a couple of years. Recall the "gap" from the end of Gemini in November of 1966 until Apollo 7 in October of 1968.
Title: Re: Basic Rocket Science Q & A
Post by: just-nick on 08/24/2011 10:07 pm
I don't think I've seen this explained here before...so here goes:

Sometimes specific impulse (isp) is footnoted as being "shifting".  What does this mean?  Is this during a throttling or ratio transient?   What is it contrasted with?  Fixed?  Static?

Thanks,

  --Nick
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 08/24/2011 10:26 pm
It has to do with idealised models of the expansion process in the nozzle. Shifting equilibrium assumes chemical reactions continue to take place during the expansion process, while frozen assumes that mole fractions remain constant during the expansion process. If the reaction is very fast then Isp may be closer to shifting equilibrium Isp, otherwise it will be closer to the frozen value.
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 08/25/2011 03:49 am
It has to do with idealised models of the expansion process in the nozzle. Shifting equilibrium assumes chemical reactions continue to take place during the expansion process, while frozen assumes that mole fractions remain constant during the expansion process. If the reaction is very fast then Isp may be closer to shifting equilibrium Isp, otherwise it will be closer to the frozen value.

Adding to that, frozen underpredicts. For BOE estimates, I will use frozen. Shifting will always overpredict to an extent, because you are getting back energy from dissociated species recombining (H monatomic, rejoins to become H2, etc).

Also, if you want to try playing with the chemistry, CEA (http://www.grc.nasa.gov/WWW/CEAWeb/ceaHome.htm) is an excellent resource. They have an online version there, but it is kind of buggy. The download is  here (http://www.grc.nasa.gov/WWW/CEAWeb/ceaguiDownload-win.htm). It's not a tool that can be used completely blind, but it is pretty user friendly, as these things go.
Title: Re: Basic Rocket Science Q & A
Post by: mgfitter on 09/14/2011 05:24 pm
Not sure if this is the correct location for this question, but can anyone point me towards information about the thermal environment on the side of a cylindrical rocket during ascent?

I know that the cone of the payload fairing will get heated and I understand that the base of the vehicle will. I also note that the plume can climb back up the side of the vehicle due to low pressures in that region, but I'm really just trying to get a feel for the heating involved above that, say on the cylindrical body of the payload fairing or the upper stage tanking. Does the airflow in that area cause noticeable heating?

Can anyone help?

-MG
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 09/14/2011 08:41 pm
Main sources are radiation from the plume (especially when the plume expands at altitude) and friction/stagnation from aeroheating.  Recirc convective heating is more vehicle-dependent.
Title: Re: Basic Rocket Science Q & A
Post by: mmeijeri on 09/24/2011 05:47 pm
On a SpaceX thread the possibility of using gaseous hydrogen dissolved in liquid methane came up. It looks as if this is impractical since the solubility is far too low. Would it be possible to use slush methane in liquid hydrogen and if so would this be more practical than slush hydrogen?
Title: Re: Basic Rocket Science Q & A
Post by: cambrianera on 09/24/2011 06:22 pm
On a SpaceX thread the possibility of using gaseous hydrogen dissolved in liquid methane came up. It looks as if this is impractical since the solubility is far too low. Would it be possible to use slush methane in liquid hydrogen and if so would this be more practical than slush hydrogen?
Could be possible, a good way to increase density.
I see a problem pumping the liquid/solid mixture (methane is rock solid at LH temperatures).
Some way to avoid damage to impellers of pumps must be studied.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 09/26/2011 09:05 am
On a SpaceX thread the possibility of using gaseous hydrogen dissolved in liquid methane came up. It looks as if this is impractical since the solubility is far too low. Would it be possible to use slush methane in liquid hydrogen and if so would this be more practical than slush hydrogen?

I'm struggling a bit to understand the advantage of doing this.  Is the hope that dissolved methane would add little to the volume, thereby increasing volume-specific impulse (at the cost of weight-specific impulse)?
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 09/26/2011 09:13 pm
I'm struggling a bit to understand the advantage of doing this.  Is the hope that dissolved methane would add little to the volume, thereby increasing volume-specific impulse (at the cost of weight-specific impulse)?

I believe the desire is for a higher, and tailorable density specific impulse.
Title: Re: Basic Rocket Science Q & A
Post by: cambrianera on 09/26/2011 09:20 pm
On a SpaceX thread the possibility of using gaseous hydrogen dissolved in liquid methane came up. It looks as if this is impractical since the solubility is far too low. Would it be possible to use slush methane in liquid hydrogen and if so would this be more practical than slush hydrogen?

I'm struggling a bit to understand the advantage of doing this.  Is the hope that dissolved methane would add little to the volume, thereby increasing volume-specific impulse (at the cost of weight-specific impulse)?
Coming from Elon's Reddit AMA video responses:
"We can certainly improve on the chemical propulsion that has been done thus far, and I think probably a very high efficiency light hydrocarbon that uses predominantly methane is probably a good way to go, and I think that's something SpaceX will end up working on."

I was wondering what Elon was meaning and my idea was:
Obtaining a (still unknown) mixture of liquid hydrocarbons that enables storing hydrogen at mild cryo temperature and low pressure would be a great advantage.
Obviously adding solid methane to LH2 would be smaller advantage.
As you said, trading density to ISP.
Title: Re: Basic Rocket Science Q & A
Post by: aussie invader on 09/27/2011 01:35 am
Hi guys, after reading your blogs about the use of different blow down gases..eg helium as oppossed to nitrogen can someone out there please further comment on this subject. We are building a LSR car using a LR-105 lox/kero injector and would like to use nitrogen as our blow down gas. Our total pressurization time is around 25 seconds for a run with a max 30 second pre pressurization time ( total 55 seconds propellant pressure time) From reading your comments i considered the main difference between using He or N20 was the time that the gas was putting the lox under pressure, am i correct ? Would N20 do the same job on a short duration blow down ? We are going to operate this vehicle in a very remote location where He is hard to source and is very expensive.
Thank you
Rosco McGlashan Western Australia www.aussieinvader.com
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 09/27/2011 02:38 am
On a SpaceX thread the possibility of using gaseous hydrogen dissolved in liquid methane came up. It looks as if this is impractical since the solubility is far too low. Would it be possible to use slush methane in liquid hydrogen and if so would this be more practical than slush hydrogen?

I'm struggling a bit to understand the advantage of doing this.  Is the hope that dissolved methane would add little to the volume, thereby increasing volume-specific impulse (at the cost of weight-specific impulse)?
Coming from Elon's Reddit AMA video responses:
"We can certainly improve on the chemical propulsion that has been done thus far, and I think probably a very high efficiency light hydrocarbon that uses predominantly methane is probably a good way to go, and I think that's something SpaceX will end up working on."

I was wondering what Elon was meaning and my idea was:
Obtaining a (still unknown) mixture of liquid hydrocarbons that enables storing hydrogen at mild cryo temperature and low pressure would be a great advantage.
Obviously adding solid methane to LH2 would be smaller advantage.
As you said, trading density to ISP.

It would be fantastic if hydrogen were significantly soluble in some combination of light hydrocarbons, but I think that's unlikely, given the low solubility of hydrogen in methane.  I would guess that Musk's remark refers to an engine running on a mixture of methane and either propane or proplylene or possibly a more exotic hydrocarbon.

Re dissolving or suspending solid methane in hydrogen, if feasible it certainly permits a higher volume-specific impulse, at a cost.  I remain skeptical that it's worth trouble, when there are easier ways to boost volume-specific impulse.
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 10/05/2011 06:49 pm
I wasn't sure where to put this, and thought it might be useful to someone else.

This is a spreadsheet I built to answer a couple of questions of mine. It allows you to determine payload mass fraction for a given rocket based on a set of general performance metrics. You do not need to know any masses beforehand, but can calculate them afterward based on your desired payload mass.

It is currently set up to show why NTRs are not terribly useful for ground launch, which is the question I was curious about.

Most of the parameters are pretty self explanatory, but a couple notes:

Vehicle T/W and Engine T/W need to be consistent with each other. You can use lunar, Earth, Mars, Europa, or whatever the heck you want, because the gravity constant will cancel out. Just make sure you are using the same gravity constant for both T/W ratios.

Aux Structure Mass Fraction is a fudge factor to assign a percentage of vehicle mass for things other than engines, tanks, payload, or propellant.

Tank Density is the mass of a tank, per unit volume enclosed, in kg/m^3. As a few points of reference, the Shuttle SLWT is about 12 kg/m^3. A modern Centaur is about 25 kg/m^3. The Shuttle SRBs (as much as they were tanks...) were about 160 kg/m^3.
Title: Re: Basic Rocket Science Q & A
Post by: dcporter on 01/19/2012 05:33 pm
Hopefully basic orbital mechanics is like basic rocket science. Apologies if not.

So the first thing you do when you're launching is loft yourself above the atmosphere, so it doesn't rob your v and crash you back down to Earth. Then you build up a bunch of v. (These are obviously happening concurrently e.g. with less-lofted launches.) I've seen the awesome Earth -> LEO chart circulating around here recently, and it lists that as costing 9.5 m/s. Does that include lofting costs? If so, how much of it? If not, how much extra dv-equivalent does it cost to get out of atmo? If I'm not asking the question quite right from my armchair, what is the right question (and answer)?

Thanks!
Title: Re: Basic Rocket Science Q & A
Post by: Jorge on 01/19/2012 05:40 pm
Hopefully basic orbital mechanics is like basic rocket science. Apologies if not.

So the first thing you do when you're launching is loft yourself above the atmosphere, so it doesn't rob your v and crash you back down to Earth. Then you build up a bunch of v. (These are obviously happening concurrently e.g. with less-lofted launches.) I've seen the awesome Earth -> LEO chart circulating around here recently, and it lists that as costing 9.5 m/s. Does that include lofting costs?

Yes. In these parts we call that "gravity losses".
Title: Re: Basic Rocket Science Q & A
Post by: dcporter on 01/19/2012 06:28 pm
Oh true, gravity hits you while you're trying to get up to altitude. Thx for the nomenclature. I meant to ask about losses to atmo friction + gravity loss, is that meaningfully more than gravity loss alone? How much are the two together? (approximately is cool. I'm not launching rockets here.)

Edit: or does gravity loss refer to atmo loss too? Since I guess you wouldn't need it if you were sans atmo.
Title: Re: Basic Rocket Science Q & A
Post by: Lee Jay on 02/10/2012 01:51 pm
What's the advantage of having multiple combustion chambers and nozzles on a given turbo machinery assembly instead of one larger chamber and nozzle?  I know of one possible advantage on regenerative nozzles, but I'm thinking of other types such as on the Soyuz.
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 02/10/2012 02:14 pm
I thought the pro of multiple combustion chambers was it made the whole combustion stability an easier problem to tackle. Hence the use at the dawn of the space age on the R-7.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 02/10/2012 02:17 pm
What's the advantage of having multiple combustion chambers and nozzles on a given turbo machinery assembly instead of one larger chamber and nozzle?  I know of one possible advantage on regenerative nozzles, but I'm thinking of other types such as on the Soyuz.
For what I know, the bigger nozzles have bigger instabilities problems. Thus, it's easier to make more smaller ones, than a single bigger one. A second issue is that it's also shorter (smaller interstage). And a third issue is that since total thrust is just pressure times throat surface, but pressure vessels require proportionally more material for bigger diameters, you have thrust scaling quadratically, and weight cubically. Of course I suspect that the Russians do it because they design for low cost manufacturing. I.e. they have to tolerate very sloppy quality control. For a space application, obviously.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 02/10/2012 04:14 pm
Agree with combustion stability and weight items.  Also a way to get some economy of scale in producing more of the same thing.
Title: Re: Basic Rocket Science Q & A
Post by: Lee Jay on 02/10/2012 09:02 pm
Agree with combustion stability and weight items.  Also a way to get some economy of scale in producing more of the same thing.

Okay, much appreciated, all three of you!
Title: Re: Basic Rocket Science Q & A
Post by: e of pi on 02/10/2012 11:28 pm
Is it possible to inject into a polar lunar orbit directly from Earth, as opposed to entering an equatorial LLO, then plane-changing into polar, and what would the associated delta-v be compared to going to an equatorial orbit? Additionally, if such an injection is possible, is it still able to offer a free-return trajectory? Additionally, in terms of offering total-surface access, would this be any better than staging from L1/L2, given the lower delta-v required of a descent stage going directly down from orbit instead of getting there from a Lagrange point? Thanks for any insight into this.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 02/11/2012 12:37 am
1st question: yes, look at Lunar Prospector and LRO.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 02/11/2012 07:28 am
Getting into lunar polar orbit from earth is hardly any more difficult than getting into lunar equatorial orbit.

The moon is moving around the earth at about 1 km/s.  When the spacecraft arrives in the vicinity of the moon, it is near apogee of a very highly elliptical orbit and is moving very slowly with respect to the earth.  Looking at it from an earth-centered perspective, the spacecraft is nearly motionless and is swept up by the moon.

Entering polar orbit rather than equatorial orbit is principally a matter of arriving in the moon's vicinity above or below the plane of the moon's orbit by somewhat more than a lunar radius.  On arrival, the spacecraft will go into an orbit around the moon such that the earth will initially be in view at all times -- it won't loop behind the moon.  Thus, I don't think a free-return trajectory is possible, at least not without following a much higher-delta-V trajectory to the moon in the first place.

The delta-V requirements for getting to polar orbit directly from earth are basically similar to those for reaching equatorial orbit.  However, the orientation with respect to earth of a polar orbit, unlike that of an equatorial one, will change as the moon revolves around the earth.  If abort-anytime capability is desired, then a large delta-V reserve will be needed for aborts when the orbital orientation is unfavorable.
Title: Re: Basic Rocket Science Q & A
Post by: nyrath on 02/13/2012 08:03 pm
Hi! This is a stupid newbie with a stupid newbie question, so feel free to throw popcorn at me.

I have a question about optimal acceleration profiles for a rocket lifting off from Earth's surface into low Earth orbit.

Say the rocket has a thrust-to-weight ratio way above 1 and an unreasonably high specific impulse of 20,000 seconds or so (NSWR, don't ask) so it is a single stage to orbit rocket. The engine can be throttled so the acceleration can be varied as the rocket loses mass.

Given that it is undesirable for the acceleration to rise above 4g or so in order to prevent damage to the astronauts.

Also given that every second spent in transit to orbit imposes a "gravity tax" of a 9.81 m/s reduction in speed so it is desirable to minimize the transit duration.

So the question is: with these givens, what sort of acceleration profile over the transit duration would be optimal?
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 02/13/2012 09:55 pm
Hi! This is a stupid newbie with a stupid newbie question, so feel free to throw popcorn at me.

I have a question about optimal acceleration profiles for a rocket lifting off from Earth's surface into low Earth orbit.

Say the rocket has a thrust-to-weight ratio way above 1 and an unreasonably high specific impulse of 20,000 seconds or so (NSWR, don't ask) so it is a single stage to orbit rocket. The engine can be throttled so the acceleration can be varied as the rocket loses mass.

Given that it is undesirable for the acceleration to rise above 4g or so in order to prevent damage to the astronauts.

Also given that every second spent in transit to orbit imposes a "gravity tax" of a 9.81 m/s reduction in speed so it is desirable to minimize the transit duration.

So the question is: with these givens, what sort of acceleration profile over the transit duration would be optimal?

Well, for the very unreasonable case where you have an infinitely throttleable rocket, and a 4 g upper limit, you would do something like what’s in the attached. Basically, you keep your accel as high as you can, while staying under a maximum dynamic pressure limit (0.5*air density*velocity^2). Once you’re through the thick part of the atmosphere, density drops off, and you can punch it again. I chose 800 psf for the attached graph, because it is a reasonable upper limit on maximum dynamic pressure (or “Max Q”).
Title: Re: Basic Rocket Science Q & A
Post by: nyrath on 02/14/2012 12:13 am
Thank you very much!
Title: Re: Basic Rocket Science Q & A
Post by: IsaacKuo on 02/14/2012 12:24 am
Also given that every second spent in transit to orbit imposes a "gravity tax" of a 9.81 m/s reduction in speed so it is desirable to minimize the transit duration.
This is not true.  The "gravity tax" is actually much less than that, depending on your thrust angle.

For example, if your acceleration is 4 gees, and you want horizontal acceleration, you only need to tilt your thrust upward by ~15 degrees in order to counteract a downward acceleration of 1 gee.  Your horizontal thrust is reduced, but only down to 3.87 gees.  In other words, the gravity drag is only 0.13 gees.
Title: Re: Basic Rocket Science Q & A
Post by: nyrath on 02/14/2012 02:04 pm
Thank you, I was wondering about that.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 02/14/2012 02:16 pm
Isn't the gravity loss the integrations of g(distance to Earth).sin(gravity normal)? You'd have two components, one is the angle towards the center of the Earth, and the other would be the distance to the point source, right?
Title: Re: Basic Rocket Science Q & A
Post by: Danny Dot on 02/14/2012 02:47 pm
There is no reason to limit to 4 gs.  NASA Standard 3000 allows for much higher g levels.
Title: Re: Basic Rocket Science Q & A
Post by: joertexas on 02/24/2012 05:56 pm
How is C3 calculated? I understand Dv, but C3 is beyond me..

Thanks,

JR
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 02/24/2012 06:32 pm
How is C3 calculated? I understand Dv, but C3 is beyond me..

Thanks,

JR

Does http://en.wikipedia.org/wiki/Specific_orbital_energy help?
Title: Re: Basic Rocket Science Q & A
Post by: IsaacKuo on 02/24/2012 06:47 pm
How is C3 calculated? I understand Dv, but C3 is beyond me..

One way to calculate it is:

v^2 - v_escape^2

Where v is your current velocity and v_escape is the relevant escape velocity.  Note that v_escape depends on your current altitude.

The utility of this value is that it remains constant if you don't perform any more thrusts, as you escape the local gravity well.  It tells you what your eventual velocity will slow down to as you escape, by taking the square root.

Does http://en.wikipedia.org/wiki/Specific_orbital_energy help?

It's important to realize that C3 is actually TWICE the specific orbital energy, since it is missing a factor of 1/2.

http://en.wikipedia.org/wiki/Characteristic_energy
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 02/24/2012 07:44 pm
How is C3 calculated? I understand Dv, but C3 is beyond me..

I got a few answers re C3 here (and subsequent pages):-

http://forum.nasaspaceflight.com/index.php?topic=13543.msg521874#msg521874 (http://forum.nasaspaceflight.com/index.php?topic=13543.msg521874#msg521874)

cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: DarkenedOne on 02/25/2012 05:16 pm
Hey this is just a quick question.

Why do rockets not use the same expanding nozzle technology used in fighter jet aircraft?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 02/25/2012 05:51 pm
Hey this is just a quick question.

Why do rockets not use the same expanding nozzle technology used in fighter jet aircraft?

Exhaust is hotter and at higher pressure
Title: Re: Basic Rocket Science Q & A
Post by: joertexas on 02/27/2012 06:54 pm
How is C3 calculated? I understand Dv, but C3 is beyond me..

I got a few answers re C3 here (and subsequent pages):-

http://forum.nasaspaceflight.com/index.php?topic=13543.msg521874#msg521874 (http://forum.nasaspaceflight.com/index.php?topic=13543.msg521874#msg521874)

cheers, Martin

My math skills just aren't up to this..

I needed to calculate what the approximate C3 would be for a translunar injection from LEO, say, 180km.

Thanks for the responses, though. I am trying to make sense of them.

JR
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 02/27/2012 07:08 pm
Hey this is just a quick question.

Why do rockets not use the same expanding nozzle technology used in fighter jet aircraft?

Exhaust is hotter and at higher pressure

Jet nozzles are strictly converging as well (the throat is the exit), and the geometry of a converging nozzle doesn’t really matter to the flow. You have much more leeway because you won’t trigger shocks, and you can have unsmooth surfaces. Even neglecting the smoothness, if you tried to do that with a converging diverging rocket nozzle, you would probably be restricted to a conical nozzle (otherwise you would have to adjust the entire contour of the nozzle). With that, you either suffer huge cosine losses at high expansion ratio (which is the opposite of what you want), or your nozzle is obscenely long at low expansion ratio, which also makes it impractical.
Title: Re: Basic Rocket Science Q & A
Post by: joertexas on 02/28/2012 06:09 pm
Let me restate the question so I can understand the process:

I'm trying to calculate the approximate payload for a Falcon 9 launching a payload on a translunar injection trajectory. The company's table references C3 and compares it to the payload capacity. How do I calculate C3 from the available data on the TLI trajectory?

Thanks,

JR
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 02/28/2012 06:24 pm
For a first approximation you can assume a TLI burn has C3=0.
Title: Re: Basic Rocket Science Q & A
Post by: joertexas on 02/28/2012 08:28 pm
For a first approximation you can assume a TLI burn has C3=0.

Okay, that makes sense to me :-)

Thanks!

JR
Title: Re: Basic Rocket Science Q & A
Post by: sdsds on 02/28/2012 08:41 pm
For a first approximation you can assume a TLI burn has C3=0.

For a second-order approximation
  TLI from a parking orbit of 130 x 130 nmi @29 deg inclination; C3 = -0.4 km2/sec2
Boeing says so.
Title: Re: Basic Rocket Science Q & A
Post by: joertexas on 02/29/2012 05:53 pm
For a first approximation you can assume a TLI burn has C3=0.

For a second-order approximation
  TLI from a parking orbit of 130 x 130 nmi @29 deg inclination; C3 = -0.4 km2/sec2
Boeing says so.

That's even better, considering the payload capacity increases with negative C3 values.

Thanks! :-)

JR
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 02/29/2012 06:51 pm
For a first approximation you can assume a TLI burn has C3=0.

For a second-order approximation
  TLI from a parking orbit of 130 x 130 nmi @29 deg inclination; C3 = -0.4 km2/sec2
Boeing says so.

That's even better, considering the payload capacity increases with negative C3 values.

Thanks! :-)

JR
The Apollo 11 TLI was -1.8km²/s²
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 03/02/2012 03:30 am
Entering polar orbit rather than equatorial orbit is principally a matter of arriving in the moon's vicinity above or below the plane of the moon's orbit by somewhat more than a lunar radius.  On arrival, the spacecraft will go into an orbit around the moon such that the earth will initially be in view at all times -- it won't loop behind the moon.  Thus, I don't think a free-return trajectory is possible, at least not without following a much higher-delta-V trajectory to the moon in the first place.

I was wrong.  Arthur Schwaniger of MSFC figured it all out in 1963 (paper attached -- see p. 9 of the PDF for the summary).  A free-return trajectory to a lunar polar orbit is possible without a large delta-V at injection.  The price to be paid is that closest approach to the moon is about 20,000 km.
Title: Re: Basic Rocket Science Q & A
Post by: e of pi on 03/02/2012 01:07 pm
Proponent:

Thank you for contnuing to look into this, I'll need to take time to read that paper this weekend. I'm guessing that the reason such an approach wasn't used was the extreme distance, even at close approach? On the whole, for full-surface access, how would injecting into polar lunar orbit compare to staging out of L1/L2?
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 03/03/2012 06:13 pm
What's the optimum tank top and bottom profiles? I would guess a catenary, if weight was the pure consideration, hemisphere if pressure. May be parabolic is a solution for an intermediate case? If minimum tank mass was hemispherical, it would require more intertank, so it might not be an optimum solution unless you have common bulkhead.
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 03/04/2012 12:21 am
Elliptical, hemispherical being a special case if so desired.
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 03/04/2012 03:29 pm
What's the optimum tank top and bottom profiles? I would guess a catenary, if weight was the pure consideration, hemisphere if pressure. May be parabolic is a solution for an intermediate case? If minimum tank mass was hemispherical, it would require more intertank, so it might not be an optimum solution unless you have common bulkhead.
Of course, the optimum tank is spherical. Anything else kind of depends on what you're optimizing. If you're ONLY optimizing mass-per-volume, then a spherical tank still wins, of course. If you're also optimizing usable-volume-to-available-space-for-tank, then it's going to depend on your exact application (in the extreme case, you have a basically completely conformal tank). That's a complicated question and really depends on where you draw the line.

So, often you end up just picking an industry standard tank end shape and call it a day. (Oh, and you also may want to figure out how to transfer loads to the tank, especially if the tank is a load-bearing member of your structure, which it usually is for rockets.)
Title: Re: Basic Rocket Science Q & A
Post by: spacecane on 04/21/2012 11:56 pm
What is the smallest rocket you could make (using practical propellants) to get something like the size of a digital camera into earth orbit?  On some of those discovery channel type shows I've seen some pretty big "model" rockets and they can't get into orbit.

Ignoring the fact that I wouldn't be able to get approval to launch it, what would the minimum size rocket I would need to build in my garage to get lets say 2 lbs into orbit?
Title: Re: Basic Rocket Science Q & A
Post by: Antares on 04/22/2012 01:40 am
Is cost a constraint? This determines which propellant. Propellant (all solid, peroxide, some cryo, all cryo) determines which hardware and then which cost. Could minimize GLOW by the choice of propellant, but that wouldn't optimize cost and performance.
Title: Re: Basic Rocket Science Q & A
Post by: spacecane on 04/22/2012 02:32 am
Is cost a constraint? This determines which propellant. Propellant (all solid, peroxide, some cryo, all cryo) determines which hardware and then which cost. Could minimize GLOW by the choice of propellant, but that wouldn't optimize cost and performance.

I would say to keep the cost and practicality in the reasonable realm of hobbyists.  I'd say no cryo as I'd assume this would make it pretty impractical.  Maybe LO2 as an oxidizer but certainly no LH2.
Title: Re: Basic Rocket Science Q & A
Post by: sbt on 04/22/2012 07:03 am
Is cost a constraint? This determines which propellant. Propellant (all solid, peroxide, some cryo, all cryo) determines which hardware and then which cost. Could minimize GLOW by the choice of propellant, but that wouldn't optimize cost and performance.

I would say to keep the cost and practicality in the reasonable realm of hobbyists.  I'd say no cryo as I'd assume this would make it pretty impractical.  Maybe LO2 as an oxidizer but certainly no LH2.

Legislation is also a constraint - unless you factor that out as part of ignoring the 'no permission to launch' bit.

This is why I understand Hybrids are relatively popular in large-scale UK Amateur Rocketry. Our explosives regulations with respect to private citizens are more strict than those in place in most of the US.

Even in the US your workshop location will affect things. People tend to be unhappy about the preparation of bulk Solid Rocket Fuel in Manhattan Apartments.
Title: Re: Basic Rocket Science Q & A
Post by: spacecane on 04/22/2012 05:12 pm
To be clear I'm not actually intending to do this.  Just trying to get an idea of what the smallest rocket you could make would be to get a small light payload into earth orbit.  The question popped up when I was watching a model rocket show on tv and saw that the "big" model rockets they were building weren't even close to being able to reach orbit.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 04/22/2012 08:45 pm
Just trying to get an idea of what the smallest rocket you could make would be to get a small light payload into earth orbit.

John Whitehead wrote a really neat little paper pretty much on that topic in 2005.  It's the first attachment.  You might also find Jordin Kare's 1994 design for a small SSTO interesting.

Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 04/22/2012 09:44 pm
Proponent:

Thank you for contnuing to look into this, I'll need to take time to read that paper this weekend. I'm guessing that the reason such an approach wasn't used was the extreme distance, even at close approach? On the whole, for full-surface access, how would injecting into polar lunar orbit compare to staging out of L1/L2?

If you start from L1/L2, then you can head to any point anywhere on the moon anytime you want (unless local lighting is a constraint), and you can also leave from the moon to L1/L2 anytime, though the trip between L1/L2 and the surface will take quite a while if done at minimum delta-V.  From L1/L2 you can return to Earth anytime.

From a polar orbit, you can economically descend to a given non-polar landing site only when the orbit passes over the site.  If you're launching from Earth specifically to go to that site, then presumably you'll time your arrival in lunar polar orbit so that you won't have to wait long.  Otherwise, you could have to wait half a month for the orbit to pass over the landing site (give or take -- I'm not sure about the rate of nodal regression).  And if there were additional constraints (like lighting or local topography) that dictated that the approach had to be from the north or from the south, then you might have to wait a full month.

The real problem with a polar orbit, though, comes when you consider the return trip.  Once the lander's on the moon, it could have to wait half a month for the orbiting mother ship to pass overhead again:  only with an overhead pass is a minimum-delta-V rendezvous possible.  If you want to begin the trip home at some other time (say, you need to abort), then enough propellant will have to be held in reserve to change the inclination of the mother ship's orbit by 90 degrees to allow rendezvous.

If a high polar orbit, say 20,000 km, is used, orbital speed is low and plane changes require much less propellant.  On the other, the orbital period at that altitude is measured in days, so opportunities for trans-Earth injection are days apart.  That problem can be solved at the expense of delta-V, but then maybe it would be best to just stick with a low polar orbit in the first place.
Title: Re: Basic Rocket Science Q & A
Post by: fatjohn1408 on 08/30/2012 05:23 pm
I was wondering why heat flux is often seen as a constraint during launch ascent. Can't a thin layer of ablative material around the fairing increase the constraint to unobtainable levels for launchers?

Heat flux scales with velocity to the power of three, therefore the heat flux of ascent vehicles is rather low (Pegasus is limited to around 10,000 - 15,000 Watt per m^2 (this can be seen in the pdf where a graph up to 1.2 BTU per Ft^2 sec is posted). All the while heat of ablation of some materials can be as high as 2500000 Joules/kg. Requiring maximum Pegasus heat flux for 200 seconds in order to ablate 1 kg for every square meter. This seems way more beneficial than adjusting the launch profile and obtaining more gravity losses in the process.

What am I missing? 
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/31/2012 12:33 am
Where do you see it as a constraint?  As free molecular heating?
For Pegasus, it is not just the fairing.  The wings char.
Also, there is interference heating. 
Title: Re: Basic Rocket Science Q & A
Post by: fatjohn1408 on 09/02/2012 07:08 pm
Where do you see it as a constraint?  As free molecular heating?
For Pegasus, it is not just the fairing.  The wings char.
Also, there is interference heating. 

Well I just had one study that claimed that heat flux is an important constraint when simulating launcher trajectories. However, I cannot find a second source, so perhaps my assumption is wrong?
Title: Re: Basic Rocket Science Q & A
Post by: Moe Grills on 12/04/2012 08:18 pm
 Ignoring clear differences in thrust forces, specific impulses
and chemical combinations (combustion) absent in "plasma" rocket engines); doesn't the fact that combustion temperatures in
"chemical" rocket engines, which reach intensities of 3000-4000 Kelvin
in certain cases, generate a plasma?

  I find it difficult to accept the idea that at 3000-4000 Kelvin superheated
CO2, NO2 and or H20 do not behave like a plasma. I would think that highly agitated electrons and nuclei would not tend to recombine immediately inside rocket motor combustion chambers; but I could be wrong.
 
Title: Re: Basic Rocket Science Q & A
Post by: beb on 12/04/2012 08:38 pm
And your question is... what?

The chief difference between chemical engines and plasma engines us that chemical engines do not interact with the exhaust gases. All its thrust comes from the expansion of gases. Plasma engines use electric and magnetic fields to accelerate the plasma exhaust gases.
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 12/04/2012 08:38 pm
The energy source is different. Plasma rockets use electricity, chemical uses the chemical energy.

Flame is, of course, a weak plasma.
Title: Re: Basic Rocket Science Q & A
Post by: Moe Grills on 12/05/2012 10:42 pm
And your question is... what?

The chief difference between chemical engines and plasma engines us that chemical engines do not interact with the exhaust gases. All its thrust comes from the expansion of gases. Plasma engines use electric and magnetic fields to accelerate the plasma exhaust gases.

  I think we're not on the same page; but that's my fault.
My question was sort of ambivalent...my regrets.
But I got the answer I was looking for below.
Title: Re: Basic Rocket Science Q & A
Post by: Moe Grills on 12/05/2012 10:46 pm
The energy source is different. Plasma rockets use electricity, chemical uses the chemical energy.

Flame is, of course, a weak plasma.

Weak plasma or strong.
A combustion generated plasma rather than an electrically generated one
is what I would call it.
Thanks for your answer. Cheers.
BTW...a 'weak' combustion generated plasma still presents unique possibilities if one thinks outside the box....But I'll take that topic to
"advanced concepts".   ;)
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 12/06/2012 07:24 am
The energy source is different. Plasma rockets use electricity, chemical uses the chemical energy.

Flame is, of course, a weak plasma.

And JP Aerospace suggests a plasma engine in which the plasma is the exhaust of a chemical engine, which is then accelerated by electromagnetic means.  The advantage of this weird hybrid approach is that it's not necessary to generate a large amount of electricity to create the plasma in the first place.
Title: Re: Basic Rocket Science Q & A
Post by: cambrianera on 12/06/2012 08:46 am
FYI, a low temperature plasma is normally used to heat treat components in Plasma nitriding; the combination of low pressure and DC high voltage generates a Hydrogen- Nitrogen plasma acceleration layer (the glowing layer in pic).
Concerning rockets engines, beb's explanation is the best, plasma sensibility to EM fields is used to accelerate it (MHD generators do the inverse job, picking energy from the plasma content of a very high temperature combustion flame).
Title: Re: Basic Rocket Science Q & A
Post by: BobCarver on 12/06/2012 06:29 pm
For an in-depth discussion of this topic, see AIAA 95-4079 Rocket-Induced Magnetohydrodynamic Ejector —A Single-Stage-to-Orbit Advanced Propulsion Concept, by J. Cole, J. Campbell, and A. Robertson NASA Marshall Space Flight Center Huntsville, AL at http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19960021025_1996043871.pdf

Title: Re: Basic Rocket Science Q & A
Post by: Warren Platts on 12/20/2012 06:05 pm
Is static electricity buildup ever a problem on ordinary spacecraft operations? If so, are there any engineering solutions that work in the vacuum of space that are commonly applied?
Title: Re: Basic Rocket Science Q & A
Post by: deepseaskydiver on 12/20/2012 06:50 pm
Where do you see it as a constraint?  As free molecular heating?
For Pegasus, it is not just the fairing.  The wings char.
Also, there is interference heating. 

Well I just had one study that claimed that heat flux is an important constraint when simulating launcher trajectories. However, I cannot find a second source, so perhaps my assumption is wrong?

A second source would be the Atlas V User Guide:
"For Atlas V 500 series missions, the PLF is jettisoned during the booster phase of flight. Before PLF jettison,
the RD-180 engine is throttled down to maintain 2.5 g acceleration. Typically, the PLF is jettisoned when the
3-sigma free molecular heat flux falls below 1,135 W/m2 (360 Btu/ft2-hr). For sensitive SC, PLF jettison can
be delayed to reduce the heat flux with minor performance loss. After PLF jettison, the RD-180 is throttled
up."
So it can be a constraint that really originates from the payload and not necessarily from the launch vehicle.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 12/20/2012 07:05 pm
Where do you see it as a constraint?  As free molecular heating?
For Pegasus, it is not just the fairing.  The wings char.
Also, there is interference heating. 

Well I just had one study that claimed that heat flux is an important constraint when simulating launcher trajectories. However, I cannot find a second source, so perhaps my assumption is wrong?

A second source would be the Atlas V User Guide:
"For Atlas V 500 series missions, the PLF is jettisoned during the booster phase of flight. Before PLF jettison,
the RD-180 engine is throttled down to maintain 2.5 g acceleration. Typically, the PLF is jettisoned when the
3-sigma free molecular heat flux falls below 1,135 W/m2 (360 Btu/ft2-hr). For sensitive SC, PLF jettison can
be delayed to reduce the heat flux with minor performance loss. After PLF jettison, the RD-180 is throttled
up."
So it can be a constraint that really originates from the payload and not necessarily from the launch vehicle.

The assumption is wrong and therefore that isn't the second source.
It only affects fairing jettison time.  The trajectory  isn't really affected. 
Title: Re: Basic Rocket Science Q & A
Post by: deepseaskydiver on 12/20/2012 08:19 pm
Where do you see it as a constraint?  As free molecular heating?
For Pegasus, it is not just the fairing.  The wings char.
Also, there is interference heating. 

Well I just had one study that claimed that heat flux is an important constraint when simulating launcher trajectories. However, I cannot find a second source, so perhaps my assumption is wrong?

A second source would be the Atlas V User Guide:
"For Atlas V 500 series missions, the PLF is jettisoned during the booster phase of flight. Before PLF jettison,
the RD-180 engine is throttled down to maintain 2.5 g acceleration. Typically, the PLF is jettisoned when the
3-sigma free molecular heat flux falls below 1,135 W/m2 (360 Btu/ft2-hr). For sensitive SC, PLF jettison can
be delayed to reduce the heat flux with minor performance loss. After PLF jettison, the RD-180 is throttled
up."
So it can be a constraint that really originates from the payload and not necessarily from the launch vehicle.

The assumption is wrong and therefore that isn't the second source.
It only affects fairing jettison time.  The trajectory  isn't really affected. 

Jettison time controls when the throttle down to 2.5 G occurs, which is a pretty significant event during booster flight. I guess you can split hairs about what the conversation is about and what is meant by "constraint", but I thought it would be helpful to point out a case where significant events in a trajectory are based off of heating. After all, it seemed to me the question was looking for more sources as to why "heat flux is an important constraint when simulating launcher trajectories".
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 12/20/2012 09:10 pm
Where do you see it as a constraint?  As free molecular heating?
For Pegasus, it is not just the fairing.  The wings char.
Also, there is interference heating. 

Well I just had one study that claimed that heat flux is an important constraint when simulating launcher trajectories. However, I cannot find a second source, so perhaps my assumption is wrong?

A second source would be the Atlas V User Guide:
"For Atlas V 500 series missions, the PLF is jettisoned during the booster phase of flight. Before PLF jettison,
the RD-180 engine is throttled down to maintain 2.5 g acceleration. Typically, the PLF is jettisoned when the
3-sigma free molecular heat flux falls below 1,135 W/m2 (360 Btu/ft2-hr). For sensitive SC, PLF jettison can
be delayed to reduce the heat flux with minor performance loss. After PLF jettison, the RD-180 is throttled
up."
So it can be a constraint that really originates from the payload and not necessarily from the launch vehicle.

The assumption is wrong and therefore that isn't the second source.
It only affects fairing jettison time.  The trajectory  isn't really affected. 

Jettison time controls when the throttle down to 2.5 G occurs, which is a pretty significant event during booster flight. I guess you can split hairs about what the conversation is about and what is meant by "constraint", but I thought it would be helpful to point out a case where significant events in a trajectory are based off of heating. After all, it seemed to me the question was looking for more sources as to why "heat flux is an important constraint when simulating launcher trajectories".

That only applies to the 5m version of Atlas.  The 4m fairing is jettison during second stage flight and there is no throttling of the second stage engine.  Same applies for the Delta IV and II.   
Title: Re: Basic Rocket Science Q & A
Post by: Warren Platts on 12/26/2012 07:29 pm
1. Is static electricity buildup ever a problem on ordinary spacecraft operations?

2. If so, are there any engineering solutions that work in the vacuum of space that are commonly applied?

1. Electric propulsion that relies on accelerating ions carries away a positive charge, thus charging the spacecraft with a negative charge that must be dissipated.

2. There are "spacecraft neutralizers" (e.g., the LISA probes have them) that consist of field emitter arrays (http://en.wikipedia.org/wiki/Spindt_tip)--arrays of tiny nanospikes that use field emission of electrons to emit electrons directly into the vacuum to neutralize the spacecraft.

http://www.sciencedirect.com/science/article/pii/S0094576508003664

http://erps.spacegrant.org/uploads/images/images/iepc_articledownload_1988-2007/2011index/IEPC-2011-083.pdf



Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 12/28/2012 03:14 pm
1. Electric propulsion that relies on accelerating ions carries away a positive charge, thus charging the spacecraft with a negative charge that must be dissipated.

The exhaust beam of electric thrusters is neutralized, otherwise the positive ions would eventually halt, reverse course, and start accelerating back toward the spacecraft.
Title: Re: Basic Rocket Science Q & A
Post by: Warren Platts on 12/28/2012 10:30 pm
1. Electric propulsion that relies on accelerating ions carries away a positive charge, thus charging the spacecraft with a negative charge that must be dissipated.

The exhaust beam of electric thrusters is neutralized, otherwise the positive ions would eventually halt, reverse course, and start accelerating back toward the spacecraft.

The solar wind plasma serves as a ground; thus, I would guess that any positive ions would be neutralized by solar wind electrons, leaving the spacecraft with a net negative charge--hence the need for spacecraft neutralizers. However, I'm not an expert, which is the reason I asked the question in the first place.
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 01/05/2013 06:25 am
1. Electric propulsion that relies on accelerating ions carries away a positive charge, thus charging the spacecraft with a negative charge that must be dissipated.

The exhaust beam of electric thrusters is neutralized, otherwise the positive ions would eventually halt, reverse course, and start accelerating back toward the spacecraft.

The solar wind plasma serves as a ground; thus, I would guess that any positive ions would be neutralized by solar wind electrons, leaving the spacecraft with a net negative charge--hence the need for spacecraft neutralizers. However, I'm not an expert, which is the reason I asked the question in the first place.

I should have been more clear. I was making a statement of fact, based on previous work at a company which designs and manufactures gridded ion and Hall effect thrusters. A hot cathode, integral to the thruster, serves as an electron source, neutralizes the ion beam, and completes the virtual circuit. There is no reliance on in-space electron sources, and this is why the thrusters generate thrust within a ground-based vacuum chamber.

The cathodes are like the Hollow Cathodes depicted here (http://www.cathode.com/). These can be seen in the Hall Thrusters here (http://upload.wikimedia.org/wikipedia/commons/6/6b/Russian_stationary_plasma_thrusters.jpg), at the top of the devices (often dual-redundant in the Russian designs, as shown).
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 01/05/2013 08:55 pm
I've been wondering about a full stage CH4/LOX engine. I always thought that on a FRSC version, you could use an expander cycle to get as many starts as you'd want. But that made me think, can you make a FSC with a single shaft turbopump? Can the turbopump have two turbines on the same shaft?
So you can start it up on the expander cycle and then transition to the power of both gas generators. Is doing two separated turbopumps more efficient?
I understand that in the H2/LOX case it might get different given the widely different volume, density and specific heat, right?
Title: Re: Basic Rocket Science Q & A
Post by: Christian La Fleur on 03/04/2013 05:40 am
Our planet rotates on itself and revolves around the sun.  If we follow the idea that our galaxy is also in motion, are we not in reality moving at tremendous amounts of speed?.......If we were to create a model of our galaxy and present it in a classroom (for the naked eye)...and we were to move the model over even a fraction of a sliver every minute or hour, this movement would represent millions of kilometers per minute or per hour.......

Thumbs up or Thumbs down........
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 03/04/2013 11:21 am
Thumb up (http://en.wikipedia.org/wiki/Galactic_year)

Quote
The galactic year, also known as a cosmic year, is the duration of time required for the Solar System to orbit once around the center of the Milky Way Galaxy.[1] Estimates of the length of one orbit range from 225 to 250 million "terrestrial" years.[2] According to NASA, the Solar System is traveling at an average speed of 828,000 km/h (230 km/s) or 514,000 mph (143 mi/s),[3] which is about one 1300th of the speed of light.
Title: Re: Basic Rocket Science Q & A
Post by: spacecane on 03/06/2013 12:42 pm
What is the tremendous speed relative to?  If all the Galaxies are rotating "in sync" then maybe we aren't moving at all???  Hmmmm...

It's interesting to think about.  We are certainly moving at a pretty good clip relative to a given point on the sun's surface!
Title: Re: Basic Rocket Science Q & A
Post by: QuantumG on 05/01/2013 09:01 pm
Great answer.

Welcome to the forum!
Title: Re: Basic Rocket Science Q & A
Post by: stevetaylor on 05/11/2013 03:14 pm
a Question?
is the recovery time from a prolonged stay in zero-g affected by recovering in a reduced gravitational environment, say one-third-g?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/11/2013 04:22 pm
a Question?
is the recovery time from a prolonged stay in zero-g affected by recovering in a reduced gravitational environment, say one-third-g?

Unknown
Title: Re: Basic Rocket Science Q & A
Post by: scienceguy on 05/15/2013 03:34 am
When something has a specific impulse (Isp) of 5000 s, what does that mean? What lasts for 5000 s?
Title: Re: Basic Rocket Science Q & A
Post by: QuantumG on 05/15/2013 03:59 am
When something has a specific impulse (Isp) of 5000 s, what does that mean? What lasts for 5000 s?

Heh. I like the phrasing.

Specific impulse is a measure of engine efficiency. Like most measurements of efficiency, that means it is something divided by something. For a rocket, we're concerned with how much thrust it produces for each unit of propellant, per second. That is, we want to divide the thrust by the change in propellant per second, which is often called "mdot".

Say we take the thrust and divide it by the mdot, what do we get? In the civilized world we measure thrust in newtons and we measure mdot in kg/s. We prefer to say newtons, because Newton was awesome, but they are just kg*m/s^2. If we divide kg*m/s^2 by kg/s, we get m/s (it's the same as multiplying kg*m/s^2 by s/kg, see how the kg and one of the s's cancel?) which is a measurement of velocity. What's this velocity measuring? Newton's third law tells us: it's measuring the exhaust velocity of the rocket!

Hang on, we're almost there. In many parts of the world, rocket scientists and engineers don't bother with isp.. they just talk about the exhaust velocity. For various reasons, not the least of which that smaller numbers are easier to remember, it's convenient to divide the exhaust velocity by g, the acceleration due to gravity at the Earth's surface. Like many measurements, this is entirely arbitrary.

So, we have m/s and we divide by m/s^2, what do we get? Well, the m and one of the s's cancel, so you get just seconds! That's why isp is measured in seconds.

To answer your specific question, an isp of 5000 s is telling us that the exhaust velocity of the rocket is 49033.25 m/s, which is very fast indeed. If we can put 1 kg of propellant through the rocket per second, it will produce 49033.25 newtons of thrust!
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 05/15/2013 04:46 am
When something has a specific impulse (Isp) of 5000 s, what does that mean? What lasts for 5000 s?

QuantumG is correct, but if you're looking for a physical meaning to latch onto, it is the number of seconds that 1 pound of propellant can be used to produce 1 pound of thrust.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/15/2013 06:53 am
[quote author=QuantumG link=topic=13543.msg1052304#msg1052304 date=136859038

Say we take the thrust and divide it by the mdot, what do we get? In the civilized world we measure thrust in newtons and we measure mdot in kg/s. We prefer to say newtons, because Newton was awesome, but they are just kg*m/s^2. If we divide kg*m/s^2 by kg/s, we get m/s (it's the same as multiplying kg*m/s^2 by s/kg, see how the kg and one of the s's cancel?) which is a measurement of velocity. What's this velocity measuring? Newton's third law tells us: it's measuring the exhaust velocity of the rocket!

Hang on, we're almost there. In many parts of the world, rocket scientists and engineers don't bother with isp.. they just talk about the exhaust velocity. For various reasons, not the least of which that smaller numbers are easier to remember, it's convenient to divide the exhaust velocity by g, the acceleration due to gravity at the Earth's surface. Like many measurements, this is entirely arbitrary.
[/quote]

In the uncivilized world, we measure thrust in lb and we measure mdot in lb/s and when you divide them, you get s

too late to continue.  Will reconcile lb and g tomorrow day.
Title: Re: Basic Rocket Science Q & A
Post by: QuantumG on 05/15/2013 07:20 am
lbf and lb are not the same thing, and yes, we know you're not civilized ;)
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/15/2013 12:42 pm
lbf and lb are not the same thing, and yes, we know you're not civilized ;)


That is why I was going to reconcile lb an g
Title: Re: Basic Rocket Science Q & A
Post by: scienceguy on 05/15/2013 05:14 pm
Thanks for the responses. I knew it had something to do with fuel, but I was trained in metric and I knew we wouldn't be measuring thrust in kilograms.
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 05/15/2013 07:58 pm
I knew it had something to do with fuel, but I was trained in metric and I knew we wouldn't be measuring thrust in kilograms.

Didn't the Russians use kilogram-force for a long time for their engine specs?
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 05/15/2013 08:22 pm
I knew it had something to do with fuel, but I was trained in metric and I knew we wouldn't be measuring thrust in kilograms.

Didn't the Russians use kilogram-force for a long time for their engine specs?

Yes. The Soviets used metric, but not SI. They still used seconds for Isp, and pressures were something weird too (kgf/cm^2 maybe).
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 05/16/2013 04:41 pm
I knew it had something to do with fuel, but I was trained in metric and I knew we wouldn't be measuring thrust in kilograms.

Didn't the Russians use kilogram-force for a long time for their engine specs?

Yes. The Soviets used metric, but not SI. They still used seconds for Isp, and pressures were something weird too (kgf/cm^2 maybe).
Actually, all the metric pressure units are intimately related. And the SI are the culprits of the pressure measure being so diverse (even as metric).
The correct SI unit is the Pascal (N/m²). Which is very nice as a definition of two units, but it's simply "too small" for normal use.
The traditional measure was the Bar, which is 10N/cm², but also 1.013atmospheres. Thus, for second order approach to atm is an excellent unit and one very easy to relate to. But since 10N/cm² is not unit/unit, SI decided standardize on Pa, but then 1atm=0.1013MPa, a lot of unnecessary numbers.
Now enter the kgf/cm², which is 0.97atm, 0.98Bar and gives you the thrust straight to tonnes force, which is what you usually want to know the T/W at liftoff. Also, when working with pneumatic systems, it gives you your thrust in kg, with is extremely useful when you use kg as the unit of weight. Once you are in orbit, N is the correct unit for acceleration. And for high isp engines like ion drive, N is a more comfortable unit to work with. But those only work in orbit, where you can forget about gravity losses.
Thus, kgf/cm² or Bar would be the units that you'd actually use for pressure systems if the SI snobs hadn't polluted the measures because Bar was not "beautiful enough".
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 05/17/2013 11:51 am
To nit-pick, the SI unit of pressure is the pascal, not the Pascal.  The SI symbols for units derived from people's names are capitalized, but not the units themselves.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 05/17/2013 03:39 pm
To nit-pick, the SI unit of pressure is the pascal, not the Pascal.  The SI symbols for units derived from people's names are capitalized, but not the units themselves.
Yeah, another of those "logic" that come out of an international committee. The full name is not capitalized, to "avoid" confusion the the actual person (when the whole purpose of naming it after someone is to immortalize its name), but it's degrees Celsius (the "d" is not capitalized!). And, to add insult to injury, the symbol is capitalized (Pa). I love metric, but once you allow a bunch of bureaucrats to make decisions nothing practical can come out of it.
That's why the Russians use the metric units the way you actually work with them. When you want pressure and use US customary, you want the force to be in lbf. Same for metric, you usually need it in kgf. So, don't let the SI bureaucrats get in the way of using the most practical units for actual work.
Again, it does makes sense to use Pa for things like Martian atmosphere, and N for the thrust of an orbital vehicle. But plumbers think in terms of the force units of mass.
Title: Re: Basic Rocket Science Q & A
Post by: crickmaster on 05/17/2013 06:08 pm
Got a question regarding interplanetary transfers.  I'm trying do a 2D simulation of a hohmann orbit between Earth and Mars. 

Suppose I have all the orbital elements of Earth and Mars at departure and arrival, and if I had the magnitudes of the velocities in orbits, how do I convert/transform the magnitude into it's xyz components?

I know we're supposed to use trig and some of the angles from the COE, but I'm not entirely sure how to go about it.  Anyone can link me/tell me what these transformations are?
Title: Re: Basic Rocket Science Q & A
Post by: 93143 on 05/17/2013 06:59 pm
But plumbers think in terms of the force units of mass.

Plumbers use the Hazen-Williams formula too.  What's your point?

Any experienced SI user can switch between Pa (or kPa, or MPa; there's a reason metric prefixes exist) and bar almost subconsciously.  He might have to whip out his calculator to switch between bar and atm, if the answer is particularly important...  As for N vs. kgf, the use of separate units for force and mass maintains generality and keeps you from doing stupid things.  So what if it's a little more work to multiply or divide by 9.80665 (or whatever the local value of g happens to be)?  Consistency of units is a valuable thing, as I've found in my CFD research.

Measuring rocket performance in s seems to be a sort of grandfathered artifact that persists partly because it gives you the same number in all measurement systems (and will until someone comes up with an alternative unit of time).  For actual calculations, everyone multiplies it by g.

...

I don't see how you can maintain that a unit is "too small" for a given application, when metric provides such a simple way to scale up or down in multiples of 1000...

I personally think it's great that metric produces integer powers of ten that correspond almost exactly to physical measures like the density of water (998.2 kg/m³ at 20°C) or the Earth's surface gravity (9.80665 N/kg ± 0.5% or so) and atmospheric pressure (101.325 kPa on a standard day at sea level).  It means you can do quick calculations really easily in your head, but when it comes time to do it accurately, it doesn't fool you into thinking you're dealing with a physical identity when it's really only a coincidence.  (Water is the worst for this, because at 4°C it really is almost exactly 1000 kg/m³; this is by design, but it's no longer the standard...)

...

Also, a bar is 100 kPa, or 0.986923 atm.  Your number was backwards.

...

I use bar all the time.  Then I move the decimal five places to the right, because my code is in straight MKS...

...on the other hand, cgs confuses the hell out of me (I don't actually know what an erg is off the top of my head).  Especially in electromagnetics.  I suppose it's mostly to do with what you learned first...
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 05/20/2013 07:09 am
To nit-pick, the SI unit of pressure is the pascal, not the Pascal.  The SI symbols for units derived from people's names are capitalized, but not the units themselves.
Yeah, another of those "logic" that come out of an international committee. The full name is not capitalized, to "avoid" confusion the the actual person (when the whole purpose of naming it after someone is to immortalize its name), but it's degrees Celsius (the "d" is not capitalized!). And, to add insult to injury, the symbol is capitalized (Pa). I love metric, but once you allow a bunch of bureaucrats to make decisions nothing practical can come out of it.
That's why the Russians use the metric units the way you actually work with them. When you want pressure and use US customary, you want the force to be in lbf. Same for metric, you usually need it in kgf. So, don't let the SI bureaucrats get in the way of using the most practical units for actual work.
Again, it does makes sense to use Pa for things like Martian atmosphere, and N for the thrust of an orbital vehicle. But plumbers think in terms of the force units of mass.

I'd like to add to 93143's post above by pointing out that using kgf/cm^2 throws a wrench into things when you're working with dynamic and stagnation pressure. You'd be stuck with cm/s for velocity, and hyls (http://en.wikipedia.org/wiki/Hyl_(unit))/cm^3 for density to keep a consistent set of units. I'll take SI please and thank you.
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 05/21/2013 12:54 pm
From Mars distance at average Mars orbital velocity of 24.1 km and doing standard Hohmann transfer to average Mercury distance [with average velocity of 47.9 km/sec] how much delta-v is needed. And how long does take to get to Mercury distance. And how long does it take to cross Earth's average orbital distance?

And is it true, that if instead you started from Mercury distance and did a standard Hohmann transfer to Mars, it takes the same amount of delta-v.
Or if wanted to circularize orbit at either Mercury or Mars distance is also same amount of delta-v?

For time of travel from Mars to Mercury can just add the length of each of their years and divide by 2. And divide by 2 again for duration of each leg?
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 05/21/2013 04:41 pm
gbaikie: your question is answered in http://en.wikipedia.org/wiki/Hohmann_transfer_orbit (see also http://en.wikipedia.org/wiki/Standard_gravitational_parameter).
Title: Re: Basic Rocket Science Q & A
Post by: IsaacKuo on 05/21/2013 05:15 pm
Also, it's not so simple because of the Oberth effect.  It takes much less delta-v to go between an interplanetary transfer orbit and orbit around a planet than it does to go between that transfer orbit and a point in open space.

Think of a planetary orbit as zooming around a racetrack.  It doesn't take as much delta-v to slow down to it or speed up from it because there's already some velocity just going around and around.  The true orbital mechanics are a bit more complicated than that, but it should give you an intuitive idea.
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 05/21/2013 11:27 pm
Also, it's not so simple because of the Oberth effect.  It takes much less delta-v to go between an interplanetary transfer orbit and orbit around a planet than it does to go between that transfer orbit and a point in open space.
It seems that the Oberth effect does complicate it. When one gets nearer to the sun one get more Oberth effect.

Mars and Mercury are similar planetary masses. The Oberth effect depends upon velocity. Because at Mercury distance one has a higher velocity, one gets less Oberth effect when one "involves" the planet Mercury.
Or because one has lower orbital velocity at Mars, by using the planet Mars one get more significant difference in the Oberth effect.

Quote
Think of a planetary orbit as zooming around a racetrack.  It doesn't take as much delta-v to slow down to it or speed up from it because there's already some velocity just going around and around.  The true orbital mechanics are a bit more complicated than that, but it should give you an intuitive idea.

My question is related to a fast trajectory from Earth to Mars. If you have a highly elliptical orbit, such as mercury distance at perihelion and Mars distance at aphelion. From Earth distance to Mars is fairly short travel time in terms distance and velocity. And you arrive at Mars distance at low velocity difference- between Mars distance orbital speed and the aphelion velocity of the highly elliptical orbit. So fast transit and least braking at Mars. And could have some kind of "free return" back to Earth.

And for a faster transit, one has the elliptical orbit have the perihelion closer to the sun than Mercury, and has slightly less total energy orbital energy if the aphelion of orbit is slightly less than Mars distance- but more importantly less travel distance in the Earth to Mars earth segment of trajectory [though requires much more delta-v to match this trajectory starting from Earth].
But generally it seems the more important variable is how close the perihelion of the elliptical orbit is to the Sun.
In terms of Mars cycler, perhaps the orbital period would a Earth year, some resonant orbit, or some precession orbit.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 05/22/2013 03:38 am
Remember plane change costs. Mercury's orbit is the most inclined of the planets.
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 05/22/2013 06:39 am
Remember plane change costs. Mercury's orbit is the most inclined of the planets.

Hmm. That's kind of neat. It's an added element in terms of getting to Mercury.
But also seems like it gives more options in terms destination, which may be at inclination [a lot space rocks]. You can leave [or arrive] at Mercury at a zero inclination, but you leave [or arrive] at a 7 inclination to solar planetary disk.
Title: Re: Basic Rocket Science Q & A
Post by: torzek on 05/22/2013 11:20 am
I'm a student of Aerospace engineering currently working on my thesis.
Assumption is to put rocket of certain mass and payload into 600 km LEO . how to determine if 2 or 3 stages should be used? Is Tsiolkovsky equation enough? How to calculated velocities of each stage and total velocity taking into consideration earth rotation.

I would greatly appreciate your help as well as proper book suggestion.
Title: Re: Basic Rocket Science Q & A
Post by: jabe on 05/22/2013 12:06 pm
I'm a student of Aerospace engineering currently working on my thesis.
Assumption is to put rocket of certain mass and payload into 600 km LEO . how to determine if 2 or 3 stages should be used? Is Tsiolkovsky equation enough? How to calculated velocities of each stage and total velocity taking into consideration earth rotation.

I would greatly appreciate your help as well as proper book suggestion.

wow..now that is an interesting question.. not so "basic" though. :)
Use whatever they have taught you in class and apply it... ;)
my newbie thoughts..pick a fuel, engine and play with numbers...Elon did that and got the Falcon series of rockets..
my 2 cents
jb
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 05/22/2013 12:37 pm
I'm a student of Aerospace engineering currently working on my thesis.
Assumption is to put rocket of certain mass and payload into 600 km LEO . how to determine if 2 or 3 stages should be used? Is Tsiolkovsky equation enough? How to calculated velocities of each stage and total velocity taking into consideration earth rotation.

I would greatly appreciate your help as well as proper book suggestion.


Three stages will always have performance benefits over two stage rockets. However, cost and risk goes up with more stages. Check out the document below for optimal mass ratios for each stage.

Welcome to the forum!
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 05/22/2013 07:36 pm
Remember plane change costs. Mercury's orbit is the most inclined of the planets.

NASA's GMAT software looks like it can be used to answer the question with high fidelity, but it isn't particularly easy to use.
Title: Re: Basic Rocket Science Q & A
Post by: torzek on 05/23/2013 05:41 pm
thanks for these suggestions,  but still strugglingto find the answer...
Title: Re: Basic Rocket Science Q & A
Post by: Remes on 05/23/2013 09:53 pm
The good news is, you are doing a 2D simulation, so no need for the Z-components.  ;)

In a first step you could ignore the orientation of the spacecraft. Simply treat the spacecraft as a mass point and apply the forces as you need them.

Typically vectors are used to do that sort of calculation.
https://en.wikipedia.org/wiki/Vector_%28mathematics_and_physics%29

You would have a vector for Earth and Mars position. They move on a elipsis independently. The spacecraft has a position and velocity vector vector.

Calculate the distance between the spacecraft and the earth/mars/sun (vector subtraction, then pythagoras) in order to calculate the gravitation forces. Calculate the forces you apply by means of rockets. Based on the sum of forces and the spacecraft mass calculate acceleration, from acceleration velocity, from velocity the new position. That would be one timestep. This procedure is repeated as long as you don't enter the perfect orbit or burn up in the mars atmosphere.

If you need an angle between two vectors you use the scalar vector product. If you need to rotate a vector, use the rotation matrix (https://en.wikipedia.org/wiki/Rotation_matrix).

Hope this helps.
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 05/23/2013 09:54 pm
thanks for these suggestions,  but still strugglingto find the answer...

Well, a one stage rocket could deliver a payload to 600 km orbit.

In terms of costs, a 2 stage rocket could be the cheapest and most reliable- and generally costs are the most important factor, in terms
of engineering anything. But if it's a larger payload, then one might consider 3 stages.
A two stage could considered the minimal because, rockets perform
differently at sea level pressure as compared to vacuum of space- so first stage can have rocket nozzle optimized for in atmosphere, and second stage made to function best in vacuum of space environment.
Also one needs a lot of thrust at the beginning of launch up to the space environment- and higher energy density per volume, such as kerosene and LOX is commonly used in first stage. And one can also use kerosene
with the second stage [see SpaceX]. One could also use solid rocket fuel
in first stage- in some aspect solids are better than Kerosene or one could use strap on solid rockets.
Generally, most efficient chemical rocket as second stage or in space environment is liquid hydrogen & LOX. If you had an existing rocket that used kerosense in both stages, you could probably increase the amount payload delivered to orbit by switching to LH&LOX for second stage.
Solids rockets can used in upper stages, though they not as efficient but are easily stored and are reliable- and many are available "off the shelf".
Etc.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 05/23/2013 11:23 pm
... rockets perform
differently at sea level pressure as compared to vacuum of space- so first stage can have rocket nozzle optimized for in atmosphere, and second stage made to function best in vacuum of space environment.

Only because of the nozzle design. Bell nozzles must be optimize for a specific barometric pressure. Aerospikes otoh are self regulating, dynamically changing their characteristics to match the changing air pressure as the vehicle ascends.
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 05/24/2013 10:15 am
Quote
NASA's GMAT software looks like it can be used to answer the question with high fidelity, but it isn't particularly easy to use.

I think patience might be useful with GMAT, but I lack it.
This is simplier graphic in terms of a broad idea:
http://www.jgiesen.de/hohmann/index.html

So, using a hohmann transfer the planet which takes the shortest period of time to travel to from Earth is Mercury.
Next is Venus and then finally Mars.
Earth to Mercury: 105.5 days
Venus: 146 days
Mars: 259 days

If you are on Mercury, then shortest travel time is Venus, Earth, then Mars.
And from Mercury or Venus it faster to get to Mars than from Earth [Mercury has shortest travel time to Mars].

But if from Earth one could match the Hohmann trajectory going from Mercury to Mars [or Mars to Mercury] then you get to Mars quicker.
So if Mercury to Mars hohmann transfer was passing near Earth [say at distance 1 million km, so Earth gravity didn't affect the trajectory [by much] and launch rocket from Earth to catch up and dock with it. Then from Earth you get to Mars faster than from Mercury. And such a trajectory from Earth would not be a Hohmann trajectory. Though if you didn't stop at Mars you would return to Mercury's orbit [it's a Mercury to Mars Hohmann trajectory]. Such a non-Hohmann trajectory from Earth could get you to Mars in somewhere around 3 months [90 days] or less.

If use simple method [which isn't accurate] and half the years of Mercury and Earth.
So 365 + 88 = 226.5 and half again for 1/2 a year you get 113 days. As compared to 105 days.

So 88 day [Mercury year] and 687 day [Mars year] is 193.75 days for Hohmann trajectory [one leg to or from Mars].
So if using this rough reckoning and took 190 days [6 months] to do Hohmann trajectory from Mercury to Mars, part of the section of trajectory from Earth orbital distance to Mars orbital distance is a small portion of the trajectory, but one traveling at much lower velocity as compared most of the total distance which is traveled. So less than 1/2 but not a 1/4 or less of the total travel time.
Said differently the "year" of Mercury to Mars trajectory is about 387 days. At Mercury distance you going somewhere around +50 km/sec [Mercury average orbital speed is 47.9 km] and at Mars distance going less than Mars orbital average speed of 24.1 km/sec [if were going at Mars orbital speed you stay at Mars distance instead of falling back towards Mercury's orbit].

This fast way to Mars is not what most people expect or mention- they think if you go to Mars so one gets there in less than 3 month you will be going much faster than Mars orbital speed and at an sharp angle to Mars's orbit. And this trajectory you can go to Mars faster than any Hohmann type trajectory and one reach Mars at slower orbital speed than Mars and have no angle to Mars- going slower but exact same direction of Mars orbit.

Of course, we don't always take 8 1/2 month to reach Mars as would with a "true" Hohmann trajectory- you tweak it so it's a Hohmann trajectory with patched conic.

So if leaving from Mercury to Mars, you could also have Hohmann trajectory with patched conic and get to Mars quicker.
And from Earth one could match this faster trajectory- and so arrive at Mars faster and at small angle [which you do a course correction near Mars {patched conic}].

Doing what I am talking about requires more delta-v. But getting to Mars faster than 7 months require more delta-v.
Much is said about using nuclear propulsion to get to Mars fast and getting somewhere around say 20 km/sec of delta-v being applied from low Earth orbit.
So that is more delta-v and it's less efficient and takes longer- in comparison.
What talking about is done with chemical rockets [and can't be done with low thrust rockets] and uses much less total delta-v. And my rough guess it's instead about 10 km/sec of chemical high thrust delta-v from LEO or about 7 km/sec from high Earth orbit- a elliptical orbit with burn near Earth at perigee.

This diagram of space rock is vaguely similar trajectory:
http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2013+JL22&orb=1

This rock would be harder to get to from Earth than what I am talking about. Though if one wanted to get to it with robotic mission, instead going directly from Earth to Rock, one probably go to inner planets and use gravity assist from Venus or Mercury [or both].
But assuming the orbit reached instead nearer to Mars, and assuming one wanted to go directly from Earth to rock, how delta-v would it take?
Title: Re: Basic Rocket Science Q & A
Post by: torzek on 05/25/2013 02:25 pm
Fisrt of all, my question was about earth to LEO transfer.
Secondly, i found that calculating velocieties of various stages is possible by using Tsiolkovski formula but how do i calculate the component of velocity comming from earth rotation????

I would realy appreciate if someone  could post some real calculations insted of basic theoretical description.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 05/25/2013 02:40 pm
how do i calculate the component of velocity comming from earth rotation????

465m/s * cosine ( latitude )


Quote
I would realy appreciate if someone  could post some real calculations insted of basic theoretical description.

Buy a book (http://www.amazon.com/Fundamentals-Astrodynamics-Dover-Aeronautical-Engineering/dp/0486600610/ref=pd_bxgy_b_img_y), a bargain for 11 bucks.

(http://img2.imagesbn.com/p/9780486600611_p0_v1_s260x420.jpg)
Title: Re: Basic Rocket Science Q & A
Post by: torzek on 05/29/2013 04:34 pm
Thanks a lot for all answers. Qeastion solved
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 05/30/2013 05:30 am
Remember plane change costs. Mercury's orbit is the most inclined of the planets.

Hmm. That's kind of neat. It's an added element in terms of getting to Mercury.
But also seems like it gives more options in terms destination, which may be at inclination [a lot space rocks]. You can leave [or arrive] at Mercury at a zero inclination, but you leave [or arrive] at a 7 inclination to solar planetary disk.

I was looking for trajectories to space rocks to see if any trajectory to a rock would be similar to fast Mars. And this rock:
http://ssd.jpl.nasa.gov/sbdb.cgi?sstr=2002+NW16&orb=1
costs 10.995 delta-v and  346 days to get to:(2002 NW16)
http://neo.jpl.nasa.gov/cgi-bin/nhats?dv=11&dur=360&stay=8&launch=2015-2040&H=18&occ=7&sort=n_via_traj&sdir=DESC&action=Display+Table#top
[edit- link appears to work]
That link probably will not work, so
http://neo.jpl.nasa.gov/cgi-bin/nhats
And put in variable of a high delta-v [<=11 km/sec] and H<=18
Anyhow, it reminded me of high delta-v costs to change orbital plane of inclination.
And basically if you want a lower cost to make larger change inclination, go out to Jupiter. But for same reason Mercury is bad place to change inclination.
Anyways got me wondering how they get anything to Mercury. So upshot is I think they wouldn't arrive at Mercury when it's orbit crosses
the solar plane, but change the inclination prior to getting to Mercury.

So changing inclination at Earth requires a lot of delta-v and changing it at Mercury requires even more delta-v.
And I should check Mercury Messenger to see how they did this exactly.*

But anyhow, changing inclination is similar to getting a fast transit to Mars from Earth. The faster the orbital velocity the more delta-v is required to alter it
And I believe both are better done when in a gravity well.

And the difficulty of getting to Mercury has a lot to do with inclination rather than the nearness to the sun.
Of course, fast transit to Mars doesn't need to be same inclination as Mercury. Or it requires less delta-v to get to Mercury distance without the need to change inclination. In other words, if  you send a spacecraft to Mercury's distance from the sun and do Hohmann trajectory to Mars and from the point of Mercury distance, it gets to Mars faster than Hohmann trajectory from Earth distance.
And finally I am not suggesting one would send crew to Mercury distance to get to Mars, rather start from Earth distance and match the trajectory of a Mars to Mercury Hohmann trajectory. [Mercury's distance from the sun- not a trajectory with Mercury's inclination]

And the delta-v to do this, is similar to making a inclination change at Earth orbital distance.

A question is how much inclination change is it similar to?
So, roughly, is it like a 5 percent inclination change or 10, 15, 20%, etc?

* Edit: "This pass increases the inclination of MESSENGER's orbit to match that of Mercury; Venus is mostly sunlit during the spacecraft's approach. "
http://spacex.9f.com/messenger.htm
So used Venus gravity assist to change inclination and also used the burn at Mercury to change it:
http://messenger.jhuapl.edu/the_mission/orbit.html
Title: Re: Basic Rocket Science Q & A
Post by: spacejeff on 06/03/2013 12:14 pm
Hi all.  I am not an aerospace engineer but I have followed the space program with interest all my life and am trying to understand some of the fundamental principles of rocketry.  I apologize in advance for any stupid mistakes or misunderstandings ;-)

I'm trying to understand the rocket equation and how a delta-V budget relates to the energy it takes to get into orbit.  I'm also confused by the results of a paper called "High Altitude Launch for a Practical SSTO" by Landis and Denis from 2003, probably because I don't understand delta-V.

First off, reading Don Petit's "The Tyranny of the Rocket Equation" it seemed like there was a relationship between orbital velocity and the delta-V needed to get there.  Don talks about needing 8km/s to get into LEO (not including drag, etc), and I understand the orbital velocity of LEO is about 8km/s.  But according to Wikipedia (Delta-v_budget) it takes an additional 4km/s of delta-V to get to GEO.  However, higher orbits have a lower velocity, yes?  According to Wikipedia, velocity at GEO is about 3km/s, though it takes 8km/s + 4km/s delta-V to get there.  Is that right?  So is it just a coincidence that the delta-V needed for LEO happens to be close to the actual velocity of LEO?

Now about the delta-V budget needed for a given orbit: I tried to use the rocket equation to help me understand the following (simple?) example: give a single stage rocket a super-powerful first booster stage - something that in a few seconds would zoom it up to 100m/s.  I imagined this would have a significant impact on payload since rockets seem to take forever to get going, but then pick up speed faster later on.

Using Isp = 310sec and g=9.8m/s^2, I solve for mass ratio:

delta-V = Isp * g * ln( MR )
delta-V = 3038 * ln( MR )
delta-V / 3038 = ln( MR)
e^(delta-V / 3038) = MR

Now I use the delta-V needed for LEO (approximately including drag and gravity) = 9km/s and I get a mass ratio of 19.34.  Then, to see the effect of the super-booster, I add the 100m/s from the super-booster to delta-V and solve again.  This gives a mass ratio of 19.99 (3.3% more than without the super-booster).  Is it ok to do this?  To just give the rocket an initial upward velocity and add that to the total delta-V budget?

Can I conclude that the super-booster is worth a 3.3% change in mass ratio?  This would translate to a 3.3% change in mass to orbit, yes?

Once I understand this example I'll ask the question about the High Altitude Launch paper :-)

Thanks,

Jeff

Title: Re: Basic Rocket Science Q & A
Post by: e of pi on 06/03/2013 12:52 pm
There is a relationship between delta-v and orbital speed, but there's another factor: The delta-v to reach an orbit doesn't just include the velocity of that orbit, but also the potential energy required. Given that geosynchronous orbit is 35,785 km up there, the potential energy is enough that in spite of the lower final velocity, it still takes about 13 km/s to get there from Earth's surface. This gravitational potential energy also has to be accounted for in launches to LEO, but because you're only going up about 200 km or so, the effects are smaller--a couple km/s (a total of more like 9 km/s would be closer to correct than 8 km/s). This is why the delta-v to LEO is fairly close to the orbital velocity, but the delta-v to get to a geosynchronous orbit is not close to the velocity of a geosynch orbit--there's a lot more potential energy you have to store up to get to GEO.

Also, to answer your specific question, you need to subtract the delta-v performed by any lower stage (which is essentially what your "super-booster" is, a low-performance first stage). For instance, if the payload needs 9.4 km/s to reach LEO, and it's already going 400 m/s, then the rocket doesn't need to do 9.8 m/s, but rather the remaining 9 km/s. This results in a lower mass ratio required for the upper stage, which implies that more of the vehicle's fueled mass can be structure and payload as opposed to fuel.

Does this help you understand what you need to in order to ask your question?
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 06/03/2013 01:01 pm
You may find the attached paper elucidating.  Despite its seemingly unrelated title, it starts with a calculation of the delta-Vs needed to reach a given circular orbit along two different trajectories.  This shows that, even in the absence of drag, the required delta-V differs from the orbital velocity and depends on the trajectory flown.

Both of the example trajectories in the paper are lossless:  the delta-V actually delivered by the rocket equals its ideal delta-V (that given by the rocket equation).  In reality, however, losses will arise in three ways (gravity, drag and steering) and will increase the ideal delta-V needed to achieve a given actual delta-V.
Title: Re: Basic Rocket Science Q & A
Post by: spacejeff on 06/03/2013 03:41 pm
There is a relationship between delta-v and orbital speed, but there's another factor: The delta-v to reach an orbit doesn't just include the velocity of that orbit, but also the potential energy required. Given that geosynchronous orbit is 35,785 km up there, the potential energy is enough that in spite of the lower final velocity, it still takes about 13 km/s to get there from Earth's surface. This gravitational potential energy also has to be accounted for in launches to LEO, but because you're only going up about 200 km or so, the effects are smaller--a couple km/s (a total of more like 9 km/s would be closer to correct than 8 km/s). This is why the delta-v to LEO is fairly close to the orbital velocity, but the delta-v to get to a geosynchronous orbit is not close to the velocity of a geosynch orbit--there's a lot more potential energy you have to store up to get to GEO.

Ah ha, ok, that makes sense.

Quote
Also, to answer your specific question, you need to subtract the delta-v performed by any lower stage (which is essentially what your "super-booster" is, a low-performance first stage). For instance, if the payload needs 9.4 km/s to reach LEO, and it's already going 400 m/s, then the rocket doesn't need to do 9.8 m/s, but rather the remaining 9 km/s. This results in a lower mass ratio required for the upper stage, which implies that more of the vehicle's fueled mass can be structure and payload as opposed to fuel.
Quote

Ok.  To be sure I understand, you are saying that a certain delta-V invested at launch counts the same as the same delta-V achieved at any other point in flight, yes?  Even though the energy required to accelerate the fully loaded rocket is more than the energy to accelerate the almost empty rocket?

Does this help you understand what you need to in order to ask your question?

Yes.  Let me correct my math: starting with a rocket with Isp = 310sec and 9=9.8m/s^2, I solve for mass ratio:

delta-V = Isp * g * ln( MR )
delta-V = 3038 * ln( MR )
delta-V / 3038 = ln( MR)
e^(delta-V / 3038) = MR

Now I compare two rockets trying to reach 9km/s:

Rocket 1 (no super-booster):

delta-V = 9000m/s
MR1 = 19.35

Rocket 2 (initial upward velocity of 100m/s):

delta-V = 9000m/s - 100m/s = 8900m/s
MR2 = 18.72

So we get a mass ratio 3.3% lower for rocket 2 given an initial delta-V of 100m/s.  This means a 3.3% increase in mass to orbit over rocket 1, yes?

Now to my second question.  In the paper "High Altitude Launch for a Practical SSTO" by Landis and Denis in 2003, they explore the performance increase for a rocket launched from a tall tower (up to 25km tall).  They graph the results of their research in Figure 2 in terms of "Total required delta-V" as they increase launch altitude.  Their graph shows that if a sea level launch requires ~9.39km/sec, a launch from a tower 25km tall requires only ~9.29km/sec (note that this is a difference of 100m/s).  They also conclude that this corresponds to a decrease in propellant mass from ~87.5% to ~85% and an increase in mass to orbit of 19.68%.

How are they seeing a ~20% increase in mass to orbit for 100m/s delta-V if I calculate only 3.3% for my rocket with the super-booster?  What am I missing?

Their paper can be found here:

http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20030022661_2003025516.pdf
Title: Re: Basic Rocket Science Q & A
Post by: e of pi on 06/05/2013 12:14 pm
Spacejeff,

It looks like the math is correct. :) There's two factors they're looking at that you're not yet.

First, you need to understand that the dry mass used in the calculation of a mass ratio has two parts: of course, there's the payload, but there's also the structural mass of the stage itself. So, for instance, continuing with your numbers, the 9 km/s case works out that the sum of these can be 5.22% of the mass at ignition, while the 8.9 km/s case can be 5.39%. Like you say, a 3.33% increase in the dry mass allowed. However, say that the structure of the tanks we can build is such that they must mass 5% of the gross liftoff weight (GLOW). In the former case, then, we have 0.22% of GLOW as the remaining mass that can be pushed up to 9 km/s--the payload. In the second case, we have 0.39% of GLOW. That's 1.79x the original payload! So this is one lesson--when you're doing single-stage and running very close to the maximum achievable delta-v (note that with those tanks and engine performance, the maximum MR with no payload at all would be 20, for a delta-v of 9.13 km/s), the payload increase from any performance improvements can be magnified compared to the actual change in mass ratio.

The second factor is the effects of air pressure on rocket engine performance--basically, since you're expanding the exhaust to ambient, at sea level you get lower performance than at altitude (annoyingly, this also means less thrust right when you need it most--at liftoff. D'oh!). For instance, look at the SSME: at altitude, it produces 2.3 MN of thrust with a specific impulse of 452s, but at sea level down in thicker, higher pressure air, it only makes 1.9 MN from 363s. This specific impulse increase again decreases the required mass ratio (note that by 3km up, air density is down by a third).
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 06/18/2013 08:50 pm
I have a question that may be is for advanced concepts, but let's see if I got the general concepts right.
Let's say that you want to do a sun escape mission. And let's say that your last stage is a SEP stage. Could you launch towards the sun without delta-v penalty (i.e. passing closer to the Sun than Earth's orbit?
This would give you plenty of power, but the main argument is how would the Oberth Effect work there? I understand that the SEP would be running for a long time (months), but it would be closer to the Sun than Earth. And since it is a Sun escape mission, it should give some extra boost.
Of course the thermal requirements might be ugly, but the P/W of the solar system would improve. But I don't want to digress. The important part are the delta-v questions.
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 06/22/2013 09:17 pm
I have a question that may be is for advanced concepts, but let's see if I got the general concepts right.
Let's say that you want to do a sun escape mission. And let's say that your last stage is a SEP stage. Could you launch towards the sun without delta-v penalty (i.e. passing closer to the Sun than Earth's orbit?
This would give you plenty of power, but the main argument is how would the Oberth Effect work there? I understand that the SEP would be running for a long time (months), but it would be closer to the Sun than Earth. And since it is a Sun escape mission, it should give some extra boost.
Of course the thermal requirements might be ugly, but the P/W of the solar system would improve. But I don't want to digress. The important part are the delta-v questions.

Oberth effect only works with high thrust.
Title: Re: Basic Rocket Science Q & A
Post by: Hoonte on 07/08/2013 12:42 pm
I was watching a small clip about the sprint missile as being the fastest Accelerating rocket ever build. (as far as I know)

This info transformed in my mind to the following question.

I know that S-V generated obout 4-5 G on lift off and shuttle even less. But what is the slowest accelerating orbital reaching rocket ever build?

Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 07/08/2013 01:05 pm
I would bet the Delta IV with no solids and large gravity losses would rank high on the list.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 07/09/2013 04:01 am
I know that S-V generated obout 4-5 G on lift off

Its take-off acceleration was much lower than that:  about 7.6 Mlb of thrust divided by about 6 Mlb of weight gives an acceleration of 1.3-ish Gs, which is typical for launch vehicles.

EDIT:  "It's" -> "Its"
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 07/09/2013 05:08 am
I was watching a small clip about the sprint missile as being the fastest Accelerating rocket ever build. (as far as I know)

Wiki says it's 100 gees [which is amazing]:
"The Sprint accelerated at 100 g, reaching a speed of Mach 10 in 5 seconds."
http://en.wikipedia.org/wiki/Sprint_%28missile%29

Quote
This info transformed in my mind to the following question.

I know that S-V generated obout 4-5 G on lift off and shuttle even less. But what is the slowest accelerating orbital reaching rocket ever build?
Successful orbital rocket?
Starting with too low acceleration is dangerous.
So I would say the orbital rocket which blew up the most.

Generally orbital rockets are around 1/3 to 1/2 gee over gravity.
Don't know any which have as much a 2 gees of thrust at point of launch.

Title: Re: Basic Rocket Science Q & A
Post by: R7 on 07/09/2013 05:44 am
Don't know any which have as much a 2 gees of thrust at point of launch.

Solid LVs like Scout (https://en.wikipedia.org/wiki/Scout_%28rocket_family%29) and M-V (http://en.wikipedia.org/wiki/M-V).
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 07/09/2013 06:36 am
Don't know any which have as much a 2 gees of thrust at point of launch.

Solid LVs like Scout (https://en.wikipedia.org/wiki/Scout_%28rocket_family%29) and M-V (http://en.wikipedia.org/wiki/M-V).

Yes. Interesting.
So solid fuel rockets.
Solid rockets are good at providing a lot of thrust.
By design they can provide more thrust at launch by having more surface
area which can be burnt.

Edit: I don't know much about the Japanese launcher; I wonder what the gravity loss is on the M-V rocket?
So it's has mass of 303,100 - 306,000 lb and thrust of 849,855 lb.
That should get going pretty fast.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 07/09/2013 09:22 am
Solid rockets are good at providing a lot of thrust.

Indeed. The Sprint ABM is just crazy, see the video how friction makes missile skin go white hot within seconds from launching.

Solid LVs compensate for lower Isp with higher T/W. Less time boosting = less gravity loss, obviously drag losses increase to some extent, but solid LVs can cope with much higher drag because structurally they are much sturdier than liquid LVs.

http://www.youtube.com/watch?v=msXtgTVMcuA (http://www.youtube.com/watch?v=msXtgTVMcuA)
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 07/10/2013 08:44 am
Solid rockets are good at providing a lot of thrust.

Indeed. The Sprint ABM is just crazy, see the video how friction makes missile skin go white hot within seconds from launching.

Solid LVs compensate for lower Isp with higher T/W. Less time boosting = less gravity loss, obviously drag losses increase to some extent, but solid LVs can cope with much higher drag because structurally they are much sturdier than liquid LVs.

But what if one could have just the beginning of the launch start off fast and not have the disadvantages of solid boosters?
So get less time of the higher gee acceleration, and therefore not have significant increases drags losses [have lower dynamic pressure at Max Q]?

Or suppose another similar idea is one could use shorter burn times of solid booster, so instead 2 mins or more of solid rocket burn times on zero stage booster as was used on Shuttle, so say less the 1 min.
Say about 30 seconds of burn time.
And say, give it something like the Japanese M-V launch acceleration, but only for 30 second.
So about 2 gees thrust or about 10 meter/sec acceleration for 30 second  so at 300 m/s [around 670 mph at 13500' elevation] the zero stage are burn out.
And then the liquid fuel engine continue [or begin] and they are at this start point providing 1/3 or 1/2 a gee of acceleration [or similar acceleration as typical liquids rockets leave a launch pad].

The advantage of shorter burn out, is one uses less total mass of solid rocket- and solid rocket are a more expensive rocket fuel.
And one could design so as to recover the used solid stages [not that they have much value].
But main thing is you don't need this much acceleration at this point in rocket trajectory and because the stages are at low speed and low elevation it should should make fairly easy to recover it.

Or you paying penalty in terms of lower ISP of solid booster,
and this penalty is lessen if done over shorter time frame before getting rid of this stage.
I would say the excellent aspect of solid booster is mostly achieved in first 30 seconds. Also since this zero stage only going a small portion of rocket's total delta-v, you afford a more robust/stronger design. One can afford to carry more dead weight- things for recovery, like parachutes; and rocket self destruct features, etc are carried for less of rocket's total delta-v.
Title: Re: Basic Rocket Science Q & A
Post by: pberrett on 07/14/2013 08:42 am

Hi all

This question is a bit out there but you'll soon see the relevance to space.

When you make popcorn you heat up the corn kernal and as I understand it the water in the kernal turns to steam, expands and because the internal pressure exceeds the external pressure, it explodes.

So let's say you had an amateur radio balloon project and you sent up some corn kernals to the edge of space. There would be little external air pressure due to being on the edge of the atmosphere therefore any internal pressure would be more than the external air pressure.

Would the kernals therefore pop?

If not would I be correct in concluding that they would pop with less heating applied than on earth because of the diminished external air pressure?   
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 07/14/2013 08:55 am
I would wager that they just fracture instead of puff. The temperature actually matters because it causes chemical reactions to soften the starch.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 07/14/2013 09:46 am
I'd wager that seemingly nothing would happen.

http://en.wikipedia.org/wiki/Popcorn

Quote
Each kernel of popcorn contains a certain amount of moisture and oil. Unlike most other grains, the outer hull of the popcorn kernel is both strong and impervious to moisture, and the starch inside consists almost entirely of a hard, dense type.

...

The pressure continues to increase until the breaking point of the hull is reached: a pressure of about 135 psi (930 kPa)[8] and a temperature of 180 °C (356 °F).

It seems to be a very sturdy pressure vessel, just vacuum outside would be way below it's breaking point. The kernel would only slowly dry up.
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 07/14/2013 12:22 pm
I'd wager that seemingly nothing would happen.

http://en.wikipedia.org/wiki/Popcorn

Quote
Each kernel of popcorn contains a certain amount of moisture and oil. Unlike most other grains, the outer hull of the popcorn kernel is both strong and impervious to moisture, and the starch inside consists almost entirely of a hard, dense type.

...

The pressure continues to increase until the breaking point of the hull is reached: a pressure of about 135 psi (930 kPa)[8] and a temperature of 180 °C (356 °F).

It seems to be a very sturdy pressure vessel, just vacuum outside would be way below it's breaking point. The kernel would only slowly dry up.

So in Earth low orbit, you throw a dried cob popcorn so it de-orbits to Earth surface. And you toss [superman tosses] another cob of popcorn
so it's at hohmann transfer and intersects Mars and hits Mars surface.

What happens to the popcorn on the cob?

It seems to me the popcorn would arrive to Earth and Mars, without popping and cob which hits Mars surface is going near supersonic [not a lot kernels would survive intact] and cob hitting earth surface would be much slower and more would survive intact.
And those intact kernels could be popped or grown.
if in suitable environment [Mars is not suitable environment].
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 07/14/2013 12:28 pm
So in Earth low orbit, you throw a dried cob popcorn so it de-orbits to Earth surface. And you toss [superman tosses] another cob of popcorn
so it's at hohmann transfer and intersects Mars and hits Mars surface.

What happens to the popcorn on the cob?

It seems to me the popcorn would arrive to Earth and Mars, without popping and cob which hits Mars surface is going near supersonic [not a lot kernels would survive intact] and cob hitting earth surface would be much slower and more would survive intact.
And those intact kernels could be popped or grown.
if in suitable environment [Mars is not suitable environment].

Depends on how fast Superman throws it.   ;) ;D

Seriously gbaikie, there'd be nothing left of the corn, it would completely disintegrated upon impact.
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 07/14/2013 02:07 pm
So in Earth low orbit, you throw a dried cob popcorn so it de-orbits to Earth surface. And you toss [superman tosses] another cob of popcorn
so it's at hohmann transfer and intersects Mars and hits Mars surface.

What happens to the popcorn on the cob?

It seems to me the popcorn would arrive to Earth and Mars, without popping and cob which hits Mars surface is going near supersonic [not a lot kernels would survive intact] and cob hitting earth surface would be much slower and more would survive intact.
And those intact kernels could be popped or grown.
if in suitable environment [Mars is not suitable environment].

Depends on how fast Superman throws it.   ;) ;D

Seriously gbaikie, there'd be nothing left of the corn on the cob, it would be completed disintegrated upon impact.

Ok.
Now this would be quite light.
It's not pop corn which is green. Or even if it was green and wet, the cob would dry out [freeze dry] in space.
So, it's low density.

I would say impact on surface Earth would be somewhere around baseball pitcher's fast ball [100 mph].
But Mars would much faster.

So can I assume from your comments that the pop corn on cob, survives re-entry?
Or are saying *even if it somehow were to survived re-entry* it would hit the ground at very high velocity [both Earth and Mars]?
 
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 07/14/2013 02:32 pm
Ok.
Now this would be quite light.
It's not pop corn which is green. Or even if it was green and wet, the cob would dry out [freeze dry] in space.
So, it's low density.

I would say impact on surface Earth would be somewhere around baseball pitcher's fast ball [100 mph].
But Mars would much faster.

So can I assume from your comments that the pop corn on cob, survives re-entry?
Or are saying *even if it somehow were to survived re-entry* it would hit the ground at very high velocity [both Earth and Mars]?
 

I'm starting to get confused...

IMO popcorn is a kernel that has popped, a kernel is a pod on a corn on the cob, and a corn on the cob is many kernels on the cob.

This is off topic and should not be here. If you want, you can open your own thread about popcorn reentry different atmospheres...
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 07/29/2013 07:13 pm
Alright I got a few questions... (Keep in mind, I'm not a rocket scientist  ;) )

How do you convert from km2s2 into km/s?

Also, how do you calculate the "slowing down" of a probe as it exits the solar system? (Like how the Voyager's velocity are slowing down relative to the Sun)

Thanks!
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 07/29/2013 07:51 pm
You can't, those are different concepts. Km²/s² is potential energy (and it's wrt a certain body). km/s is an impulse concept. Plus, you might need a lot of km/s to stay at the same km²/s². Say that you want to change your orbital plane, keeping everything else equal. You need km/s but km²/s² is the same.
As someone explained on a thread, you could thing of an orbit like a skater on a halfpipe. The higher that he gets the more km²/s² he has. But if he wants to go higher, lower or simply sideways, he'll need some impulse (km/s).
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 07/29/2013 11:18 pm
Sure you can convert it. Just take the square root of the C3 to get the asymptotic velocity (i.e. the velocity it would have at infinity, i.e. the excess above escape velocity). Take the square root of the whole quantity including the units to make sure it gives you the right units.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 07/30/2013 06:20 am
What Robotbeat said. But make sure the value you got is C3, the same unit is also used for specific orbital energy (of which C3 is a special case) in which case you first need to multiply the value by two before taking the square root. Wiki explains these quite well;

http://en.wikipedia.org/wiki/Specific_orbital_energy
http://en.wikipedia.org/wiki/Characteristic_energy

(but buy the books ;))
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 07/30/2013 03:40 pm
(but buy the books ;))

What books? :)

So what I've been trying to figure out is, if a Delta IV Heavy has a C3 performance of 60 km²s−2: 2,521 kg, that would make the velocity relative to Earth...

60 * 2 = 120

Square root of 120 km2s2 = 10.95 km/s?

Correct?
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 07/30/2013 03:41 pm
No, read carefully! Take the square root of the C3. sqrt(60km^2/s^2)=7.75km/s
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 07/30/2013 03:49 pm
No, read carefully! Take the square root of the C3. sqrt(60km^2/s^2)=7.75km/s

Ah, I multiplied by 2 thinking it was specific orbital energy, but it wasn't.

So if the object is escaping Earth at 7.75 km/s, the object would have a relative velocity to the Sun of 26.75 km/s?
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 07/30/2013 07:45 pm
What books? :)

The one's recommended to you in the book thread.

So if the object is escaping Earth at 7.75 km/s, the object would have a relative velocity to the Sun of 26.75 km/s?

Depends on to what direction the object escapes.
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 07/30/2013 08:35 pm
What Robotbeat said. But make sure the value you got is C3, the same unit is also used for specific orbital energy (of which C3 is a special case) in which case you first need to multiply the value by two before taking the square root. Wiki explains these quite well;

http://en.wikipedia.org/wiki/Specific_orbital_energy
http://en.wikipedia.org/wiki/Characteristic_energy

(but buy the books ;))

I'd guess x2 applies for a circular orbit?

Cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 07/31/2013 09:42 am
Circular orbit simplifies calculations a lot.

r = radius of circular orbit
μ = gravitational parameter (http://en.wikipedia.org/wiki/Gravitational_parameter) of the celestial body you are orbiting

specific orbital energy e = -μ/2*r
characteristic energy C3 = 2*e = -μ/r
orbital velocity v = sqrt(C3) = sqrt(μ/r)

the minuses may seem confusing but it's because potential energy at infinite distance is agreed to be zero to simplify rest of the math. The result is that objects on circular/elliptical orbits always have negative specific orbital energy (thus C3 too).
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 07/31/2013 09:40 pm
Thanks.

I believe I was confusing specific orbital energy with it's gravitational potential energy component.

cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 08/01/2013 02:42 am
No, read carefully! Take the square root of the C3. sqrt(60km^2/s^2)=7.75km/s

Ah, I multiplied by 2 thinking it was specific orbital energy, but it wasn't.
easy mistake, but that's why we use ye ole fudge factor, or fixer coefficient.
Title: Re: Basic Rocket Science Q & A
Post by: Hershey on 09/17/2013 01:18 am
Whenever I read about bimodal NTR's, the brayton power cycle always uses a helium-xenon working fluid. I understand He, but why is Xe used?
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 09/18/2013 06:56 am
I understand He, but why is Xe used?

http://proceedings.aip.org/resource/2/apcpcs/880/1/559_1?isAuthorized=no

Quote
The selected configuration for the Project Prometheus Space Nuclear Power Plant was a direct coupling of Brayton energy conversion loop(s) to a single reactor heat source through the gas coolant/working fluid. A mixture of helium (He) and xenon (Xe) gas was assumed as the coolant/working fluid. Helium has superior thermal conductivity while xenon is added to increase the gas atomic weight to benefit turbomachinery design.

Heavier gas = smaller turbines. Hmm, if turbines were replaced with piston driven generator there would be no need for xenon?
Title: Re: Basic Rocket Science Q & A
Post by: M129K on 09/29/2013 01:45 pm
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 09/29/2013 06:06 pm
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.
You mean gravity losses? That's the integral of  g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.
Title: Re: Basic Rocket Science Q & A
Post by: M129K on 09/29/2013 06:58 pm
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.
You mean gravity losses? That's the integral of  g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.
If I understand it correctly, t is total time? If so, thank you! Seems very accurate.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 09/30/2013 01:05 am
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.
You mean gravity losses? That's the integral of  g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.
If I understand it correctly, t is total time? If so, thank you! Seems very accurate.
Nope, h(t) and rho(t) are functions of t. Thus, you have to integrate from t=0 to T  the function over dt.
Title: Re: Basic Rocket Science Q & A
Post by: baddux on 10/11/2013 09:18 pm
How do you calculate the maximum altitude if you know the speed for a suborbital rocket? You would go straight up and omit all propellant and rocket weight kind of things.

How high could you go with an orbital speed rocket?
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 10/11/2013 10:42 pm
um physics and calc.

rise time = Initial V / acceleration

In this case acceleration would be a constant 9.81 meters/second

Height = acceleration * time^2 / 2

Gets a bit more complicated for rockets that go to very high altitudes


Going straight up at an orbital velocity of 8 km/s using a constant G of 9.81 (it will drop off a little) you would get a height ~3260 km.
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 10/11/2013 10:48 pm
How do you calculate the maximum altitude if you know the speed for a suborbital rocket? You would go straight up and omit all propellant and rocket weight kind of things.

If your max altitude is relatively small (at most a few hundred km) you can get a good approximation using little-g as described by kevin. Equation (4) in http://physics.bu.edu/~redner/211-sp06/class01/equations.html makes things easier since it doesn't involve time.

For higher altitude apogees or more accurate calculations you can use conservation of energy. The specific orbital energy GM/r + v^2/2 is invariant during free fall, where G is the universal gravitational constant, M is the mass of Earth, r is the distance from the center of the Earth and v is the speed. You know r and v at launch and v is zero at apogee (when firing straight up) so you can solve for radius at apogee.

Quote
How high could you go with an orbital speed rocket?

If you shoot a rocket capable of LEO (and no more) straight up it'll reach apogee at a distance from the center of the Earth equal to twice the radius of the Earth. (This is an approximation ignoring complications such as the rotation of the Earth and gravity drag.)

Derivation: the vis-via equation (http://en.wikipedia.org/wiki/Vis-viva_equation) is: v^2 = GM(2/r - 1/a) where v is the speed of a satellite at some point in time, r is the satellite's distance to the center of the Earth at that time, and a is the semi-major axis of its orbit. We'll use this equation a couple of times. Note that a rocket launched straight up and one launched downrange (for LEO) end up with the same v and r (ignoring rotation of Earth and assuming impulsive transfer) so by the vis-via equation the semi-major axis of the resulting orbit is independent of the direction of launch. It's obvious from the definition of semi-major axis that a low earth orbit has semi-major axis equal to the radius of the Earth (plus a small altitude that we can ignore). Since we launch straight up at apogee speed equals zero so we can solve the vis-viva equation for the apogee altitude:
0 = v_app^2 = G M (2/r_app - 1/r_LEO)   ==>   r_app = 2 r_LEO.    Q.E.D.

The vis-via equation can also be used to answer questions such as how much harder (in terms of initial speed) it is to escape Earth's gravity than to orbit it.
Title: Re: Basic Rocket Science Q & A
Post by: baddux on 10/12/2013 10:24 am
Thanks  for answers! The purpose of my question was to figure if you have an orbital tourist vehicle and instead of going to orbit you'd like to do as high suborbital trip as possible and see the Earth as a globe.

(http://s17.postimg.org/leyqokwof/earth_3400km.jpg)
Title: Re: Basic Rocket Science Q & A
Post by: M129K on 10/12/2013 10:32 am
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.
You mean gravity losses? That's the integral of  g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.
If I understand it correctly, t is total time? If so, thank you! Seems very accurate.
Nope, h(t) and rho(t) are functions of t. Thus, you have to integrate from t=0 to T  the function over dt.
Thanks... I guess I'll have to open my math books again.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 10/13/2013 12:24 am
I've looked for this question here and couldn't find it, thought I might have skimmed over it because 56 pages is a bit much to read very well.
How can you estimate the gravitational drag of a rocket? The only thing I could find was "Divide orbital dV by acceleration in planetary G's" but doing that for a few real life rockets has given me ridiculous amounts of gravitational drag.
You mean gravity losses? That's the integral of  g(h(t)).sin(rho(t)), where rho is the angle to the normal of the gravity source (or cos() if you take to the tangent to the Earth's surface) and h it the distance to the gravity source. You can assume that in Earth's case, at 6,371km distance the gravity is 9.8m/s.
If I understand it correctly, t is total time? If so, thank you! Seems very accurate.
Nope, h(t) and rho(t) are functions of t. Thus, you have to integrate from t=0 to T  the function over dt.
Thanks... I guess I'll have to open my math books again.
Throttle and rho are your variables for a launch trajectory (if done in 2D, i.e. no plane changes).
g: 9.8*(R^2)/[(R+h)^2]
R: 6371 (in km, the average radius of the Earth)
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 10/13/2013 03:46 pm
Throttle and rho are your variables for a launch trajectory (if done in 2D, i.e. no plane changes).
g: 9.8*sqrt(R)/sqrt(R+h)
R: 6371 (in km, the average diameter of the Earth)

AIUI the symbol "g" is usually used for the standard acceleration due to gravity near the surface (about 9.8 m/s/s), not the acceleration anywhere else. More importantly the acceleration at altitude is (assuming spherical Earth): (9.8 m/s/s)*(R/(R+h))^2 --- i.e. square, not square-root. Gravity is proportional to distance raised to the -2 power (i.e. inverse square law): http://en.wikipedia.org/wiki/Newtonian_gravity .

Edit: also that's the radius of Earth, not the diameter.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 10/14/2013 04:25 am
Throttle and rho are your variables for a launch trajectory (if done in 2D, i.e. no plane changes).
g: 9.8*sqrt(R)/sqrt(R+h)
R: 6371 (in km, the average diameter of the Earth)

AIUI the symbol "g" is usually used for the standard acceleration due to gravity near the surface (about 9.8 m/s/s), not the acceleration anywhere else. More importantly the acceleration at altitude is (assuming spherical Earth): (9.8 m/s/s)*(R/(R+h))^2 --- i.e. square, not square-root. Gravity is proportional to distance raised to the -2 power (i.e. inverse square law): http://en.wikipedia.org/wiki/Newtonian_gravity .

Edit: also that's the radius of Earth, not the diameter.
Correct in all numerical issues. I simply can't do math on an iPad. So difficult to type that I confuse my own writing. Will correct my post. I thought that Earth surface gravity was g0, since G is the gravitational constant, and I'm use to single letters for functions.
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 10/14/2013 02:07 pm
Your corrected radius of the Earth (3185 km) is wrong. You had the right radius the first time, you just labeled it wrong.
Title: Re: Basic Rocket Science Q & A
Post by: aero on 12/07/2013 05:58 am
If I had a suitable canon and a totally frictionless shell, no drag of any sort on the shell, say it behaved as a point mass then I shot the shell straight up with an initial velocity of 1.908 km/s (delta-V), the shell would come to a stop at 200 km altitude. Is this 1.908 delta-V the same thing that is called "gravity drag?"
Title: Re: Basic Rocket Science Q & A
Post by: sdsds on 12/07/2013 07:41 am
If I had a suitable canon and a totally frictionless shell, no drag of any sort on the shell, say it behaved as a point mass then I shot the shell straight up with an initial velocity of 1.908 km/s (delta-V), the shell would come to a stop at 200 km altitude. Is this 1.908 delta-V the same thing that is called "gravity drag?"

My short answer, "No."

My longer answer: I would not call that "gravity drag" because gravity drag happens over the interval of a propulsive maneuver, and your canon provides its shell with an impulsive (i.e. instantaneous) maneuver. Gravity drag happens when, for example, a vehicle hovers at a fixed altitude. In that case, all of the thrust is lost to gravity drag. In general, gravity drag is delta-v spent fighting gravity.

Note also your shell is going to get back again as it falls all the velocity it lost on ascent. Gravity drag is lost forever.
Title: Re: Basic Rocket Science Q & A
Post by: Hop_David on 12/07/2013 12:48 pm
If I had a suitable canon and a totally frictionless shell, no drag of any sort on the shell, say it behaved as a point mass then I shot the shell straight up with an initial velocity of 1.908 km/s (delta-V), the shell would come to a stop at 200 km altitude. Is this 1.908 delta-V the same thing that is called "gravity drag?"

Given those conditions, it would be better to send the shell along a horizontal path. Accelerating it to 7.96 km/s would result in an orbit with an apogee at 200 km and a perigee at 0 km.

At apogee the circularization burn is .06 km/s. So total delta v from surface to a circular 200 km altitude orbit would be 8.02 km/s.

At 0 km altitude, orbit speed is 7.9 km/s. Going from a 0 km altitude orbit to a 200 km altitude takes .12 km/s.

------

But earth isn't frictionless. Before accelerating to those velocities we must first climb above the atmosphere. During vertical ascent the rocket's thrust has a vertical component. Let's say vertical acceleration is 19.6 meters/sec^2. Subtract gravity's 9.8 meters/sec^2 from that and your net acceleration is 9.8 meters/sec^2. This would be a thrust to weight ratio of 2.

I like to round 9.8 to 10. Accelerating vertically for 3 minutes would incur a gravity loss of about 180 seconds * 10 meter/sec^2. -- about 1.8 km/s.

It's easy to see longer vertical ascents result in higher gravity loss. To make ascent brief, a lower stage should have a high thrust to weight ratio. Which encourages lower ISP propellants and also encourages more rocket engines.
Title: Re: Basic Rocket Science Q & A
Post by: aero on 12/07/2013 04:36 pm
Quote
I like to round 9.8 to 10. Accelerating vertically for 3 minutes would incur a gravity loss of about 180 seconds * 10 meter/sec^2. -- about 1.8 km/s.

Yes, but you'd only reach 162 km in altitude. To reach 200 km you need to accelerate for 200 seconds which is 2 km/s delta-V, using 10 m/s^2 acceleration.

I think the answer boils down to this: Yes, some of the gravity loss goes toward raising the vehicle to orbital altitude, or staging altitude for stage 1, and some of the gravity loss results from holding the vehicle up against the pull of gravity as it accelerates to final horizontal velocity where centrifugal force counterbalances gravity. The more time this takes, the greater this gravity loss. We lump all of this loss together and name it gravity drag.

But no matter how gravity drag is sliced up, the US and payload at 200 km altitude has a gravitational potential energy sufficient to impart 1.9 km/s velocity if it should somehow fall straight down. Looking at conservation of energy, that same delta V is required to raise it up, no matter the path.

My point and the reason for this question in the first place is: Doesn't physics require a minimum gravity drag of 1.9 km/s delta-V, no matter what rocket or canon you use? It follows that people who bandy about a gravity drag of less than 1.9 km/s for their rocket are wrong or more charitably, are being overly optimistic.

Edit add - But I don't understand your example of shooting the shell horizontally. Raising the perigee I got, but getting around the need to add the potential energy of altitude went right over my head. And LEO orbital velocity does not seem to be a fixed number even for a fixed 200 km altitude. I mean, different sources give different values, with 7.8 km/s being common. I guess I'll have to calculate it from first principles to get my own value.
Title: Re: Basic Rocket Science Q & A
Post by: aero on 12/07/2013 07:26 pm
I did that, calculate orbital velocity and got the value of 7783.09281 m/s at 200 km altitude. I used this:

centrifugal acceleration equals the acceleration of gravity in orbit.
centrifugal acceleration equals v^2/r, where r = re +h, re = 6371000 meters, h=200000 meters, and v is tangential velocity.
gravitational acceleration at altitude h  is equal to g0 *(re/(re + h))^2, g0 = 9.80665 m/s^2

equating and simplifying/rearranging, v=sqrt (g0*re^2/(re + h)) = 7783.1 m/s, or 7.7831 km/s at 200 km altitude. Note. At the surface, h=0 and this simplifies to v=sqrt(g0*re) = 7904.313199 = 7.9043 km/s

So now I have a handy dandy little formula to tell me orbital velocity around Earth at my altitude of choice without asking or looking it up.




Title: Re: Basic Rocket Science Q & A
Post by: Hop_David on 12/07/2013 10:07 pm
My point and the reason for this question in the first place is: Doesn't physics require a minimum gravity drag of 1.9 km/s delta-V, no matter what rocket or canon you use? It follows that people who bandy about a gravity drag of less than 1.9 km/s for their rocket are wrong or more charitably, are being overly optimistic.

Edit add - But I don't understand your example of shooting the shell horizontally. Raising the perigee I got, but getting around the need to add the potential energy of altitude went right over my head.

Orbital energy is a combination of kinetic and potential energy. (http://en.wikipedia.org/wiki/Orbital_energy)

Potential energy is -GMm/r (http://en.wikipedia.org/wiki/Potential_energy#General_formula). In earth's case the difference in potential energy between 6378 and 6578 km from center is about 1900 joules per kilogram.

To go from a 0 km circular orbit (7.9 km/s) to a 0 x 200 elliptical orbit takes an acceleration of .06 km/s. Kinetic energy is 1/2 mv2 (http://en.wikipedia.org/wiki/Kinetic_energy#Kinetic_energy_of_rigid_bodies). The difference between 1/2 (7.9 km/s)2 and 1/2 (7.96 km/s)2 is 480 joules per kilogram. A .6 circularization burn at apogee takes velocity from 7.72 to 7.78 km/s, another 470 joules per kilogram.

The difference in kinetic energy between a 0 km orbit (7.9 km/s) and a 200 km orbit (7.78 km/s) is 950 joules per kg. Spending 950 joules to get a kilogram from an orbit where it's kinetic energy differs by 950 joules, the total energy difference is 1900 joules per kilogram.

If you're already going fast, a little increase in speed gives you a lot of extra energy. Also known as the Oberth benefit (http://hopsblog-hop.blogspot.com/2013/10/what-about-mr-oberth.htmlmr-oberth.html).
Title: Re: Basic Rocket Science Q & A
Post by: aero on 12/08/2013 02:02 am
Yes, I know about the Oberth benefit. It has to do with work done by the force of rocket thrust. Since work equals force times distance, applying that force while moving at high speed means the force will act through a longer distance hence doing a lot more work and adding a lot more energy than when that same force (engine burn) is applied at a slow speed. And the speed of a vehicle in orbit is highest at periapsis so that is the place to apply the force for maximum work hence orbital energy added.

What is truly counter intuitive is the result of adding 950 J kinetic energy to reap 1900 J of potential energy. That potential energy would convert to 1.91 km/s if it could respond to the gravity field and fall, (your 1900 J). Atmosphere does not enter because of course you could repeat the same steps while raising the 200 km orbit to a 400 km orbit with very little difference in the delta-V’s and arrive at a very similar result.

So, back to finding the minimum possible gravity drag for a launch vehicle, does your example above mean that gravity drag must always exceed 0.48 km/s? Even though it is not a very achievable minimum, it is good to know that gravity drag numbers larger than that are subject to reduction by “What,” more powerful boosters and wiser choice of launch trajectory profile. Traded against atmospheric drag of course.

Title: Re: Basic Rocket Science Q & A
Post by: Hop_David on 12/08/2013 03:17 am
What is truly counter intuitive is the result of adding 950 J kinetic energy to reap 1900 J of potential energy.

I made an arithmetic error. My spreadsheet (http://clowder.net/hop/railroad/Hohmann.xls) uses units of kilograms, kilometers and seconds. In converting kg * km2/sec2 to joules I multiplied by 1,000 when I should have multiplied by 1,000,000.

But since I made the same error in potential as well as kinetic energies, the two sides of the ledger still came out.

That potential energy would convert to 1.91 km/s if it could respond to the gravity field and fall, (your 1900 J).

As it's falling potential converts to kinetic energy. and kinetic energy  is 1/2 mv2

1900000 joules = 1/2 kg v2
3800000 joules =  kg v2
sqrt(3800000 joules/kg) = v
v = 1949 meters/sec

So your 1.91 km/s sounds in the right ball park.

Atmosphere does not enter because of course you could repeat the same steps while raising the 200 km orbit to a 400 km orbit with very little difference in the delta-V’s and arrive at a very similar result.

Periapsis and apoapsis altitudes can be the entered in the pink cells (F38 and F39) of my spreadsheet (http://clowder.net/hop/railroad/Hohmann.xls). Periapsis and apoapsis circularization burns can be seen at cells J38 and J39. Entering 200 and 400 into periapsis and apoapsis altitudes gives me circularization burns of .0576 and .0581. So I'd agree with you, the burns are nearly the same.

So, back to finding the minimum possible gravity drag for a launch vehicle, does your example above mean that gravity drag must always exceed 0.48 km/s? Even though it is not a very achievable minimum, it is good to know that gravity drag numbers larger than that are subject to reduction by “What,” more powerful boosters and wiser choice of launch trajectory profile. Traded against atmospheric drag of course.

We're attaching different meanings to the term. What I call gravity drag (or gravity loss) is loss incurred while doing an acceleration with a vertical component.

On worlds with an atmosphere, a vertical ascent must be made to get above the atmosphere.

Even on our airless moon, a departing rocket's thrust must have a vertical component or the rocket will fall to the surface. Unless the rocket is on a nearly frictionless track that allows horizontal acceleration over a long distance. In this case, gravity loss would be virtually zero.
Title: Re: Basic Rocket Science Q & A
Post by: aero on 12/08/2013 07:06 am
Quote
We're attaching different meanings to the term.

Are you sure? My position was that when the vertical component of displacement was considered, ~1.9 km/s delta V is needed to reach 200 km. Of course, to do that with a real rocket takes time and gravity will act during that time causing additional losses but my canon example eliminated that time to accelerate. During the 210 second freefalling coast upward, the canon shell would need to accelerate horizontally to orbital velocity but that could be done by most boosters, if only horizontal acceleration were needed. (I'd have to check the available fuel on the booster.)

You countered that by using my canon, the altitude could be gained more cheaply by firing horizontally at orbital velocity so that centrifugal acceleration countered gravity then using orbital mechanics to gain the altitude.

Quote
What I call gravity drag (or gravity loss) is loss incurred while doing an acceleration with a vertical component.

On worlds with an atmosphere, a vertical ascent must be made to get above the atmosphere.

I agree, and since no one has that magic canon the rocket's vertical thrust component must be used for support. When considering real rockets, calculating a minimum gravity loss is complicated by the cosine factor causing the sum of the vertical and horizontal components of the thrust to be greater than the total thrust of the engine. But their must be a minimum gravity drag for a particular engine because their certainly can be more or less.

Maybe I should have asked for a minimum gravity drag trajectory. I bet it has something to do with the vertical and horizontal component of engine thrust during ascent to orbit. And perhaps atmospheric drag although that seems to be much less than gravity drag so that if minimizing gravity drag comes at the expense of a little aero drag, it is still a benefit overall. So, is there a simple control rule like, always maintain the vehicle vertical acceleration positive, or maybe, point the engine down at 45 degrees and blast away? I'm looking for a way that I can know that my rocket simulation is doing the best that it can do. Minimizing gravity drag looks like an area of improvement.

Quote
... moon, ... snip
yes, ok .

Title: Re: Basic Rocket Science Q & A
Post by: Hop_David on 12/08/2013 04:00 pm
Quote
We're attaching different meanings to the term.

Are you sure? My position was that when the vertical component of displacement was considered, ~1.9 km/s delta V is needed to reach 200 km.

My position is that it can be done with less if you have horizontal velocity. The most extreme example would be a 7.9 km/s circular orbit at 0 km altitude. (if the shell is frictionless, as you stipulated, this would be possible). From this orbit it only takes .06 km/s to achieve a 200 km vertical displacement. 

In the real world we can't enjoy the Oberth savings conferred by high speeds from earth's surface. Orbital speeds in the troposphere aren't an option. But neither do we go straight up, turn a hard right angle and then  do the burn for horizontal velocity. The rocket typically starts vertically, then gradually tips towards the horizontal. During vertical ascent it gains horizontal velocity. This is a path somewhere between your vertical cannon shot and my horizontal departure.

You want to drop your shell 200 kilometers and call the final speed your delta V for gravity loss. It just isn't so.

Exploiting the Oberth effect can get you the energy needed with less than 1.9 km/s delta V.

On the other hand, if thrust to weight ratio is bad, you can consume much more than 1.9 km/s. If T/W ≤ 1, no amount of propellant will get you off the ground.

I will say it again. Potential energy difference between altitudes and gravity loss are two different animals.


Maybe I should have asked for a minimum gravity drag trajectory.

Optimum ascent path is complicated and the subject of much study. Above my pay grade, sorry to admit.

You might try Googling "Apollo ascent profile" or "shuttle ascent profile". Looking at various ascent profiles might give you something to copy.
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 12/22/2013 08:35 pm
Alright, I'm having some trouble calculating the ISP of certain rocket engines.

As I understand, specific impulse can be calculated by dividing the total impulse of a rocket motor by its weight. I'm particularly interested in calculating the specific impulse of a rocket motor built by Richard Nakka.

It's properties are:

Motor mass: 2.735 kg
Total impulse: 2008 N * s
ISP: 126 sec

So if I am correct (which I'm obviously not):

2008 N * s / (2.735 kg * 9.8 m/s^2) = 74.9 sec

What did I do wrong?

(Screenshot via Richard Nakka's website (http://www.nakka-rocketry.net/pvcmot11.html).)
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 12/22/2013 09:38 pm
Alright, I'm having some trouble calculating the ISP of certain rocket engines.

As I understand, specific impulse can be calculated by dividing the total impulse of a rocket motor by its weight.

You need to divide the total impulse by the weight of the propellant, not the weight of the entire motor.

2008 N * s / (1.618 kg * 9.8 m/s^2) = 127 s.
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 12/22/2013 11:31 pm
Alright, I'm having some trouble calculating the ISP of certain rocket engines.

As I understand, specific impulse can be calculated by dividing the total impulse of a rocket motor by its weight.

You need to divide the total impulse by the weight of the propellant, not the weight of the entire motor.

2008 N * s / (1.618 kg * 9.8 m/s^2) = 127 s.

Oh, thank you.  ::)
Title: Re: Basic Rocket Science Q & A
Post by: AJA on 12/24/2013 10:07 pm
I've read in quite a few places now, that the Isp of different propellant mixes scales with the density of the fuel. So, for example, RP-1 offers more ISP than methane etc. (All other things being equal.. consider some hypothetical engine that burns Metholox as well as it burns Kerolox).

Why is this? I would've thought that larger molecules would have more degrees of freedom to soak up the thermal energy liberated from combustion, and thereby impart a lesser fraction to translational velocity  - thereby leading to a lower Vexhaust. Same reason why γ (=Cp/Cv) takes on different values in the cases of monoatomic, and polyatomic gases (where Cp = Specific heat at constant pressure, and correspondingly volume)

So what changes in these rocket engines? Like I said before, the Isp seems to be quoted for fuel mixtures, as opposed to specific power plants.

(Yes, I'm assuming there's going to be incomplete combustion, which will bring the energy distribution amongst different modes into play. Otherwise - if an engine burns through the fuel-ox mixture completely, considering the case of Methane vs RP1 - you're only going to have a difference in the percentage of the exhaust that is water and the percentage that is carbon di-oxide. Isn't the momentum carried by these the same?)
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 12/31/2013 09:56 pm
Specific impulse scales with density?  That's a new one on me.  Hydrolox has a notoriously low density yet a very high specific impulse.

Specific impulses are usually quoted for a particular expansion ratio.  In effect, specific impulses given for propellant combinations are actually for particular idealized engines.

More complex molecules (in the exhaust products) will have more non-translational degrees of freedom.  In principle this does make complex exhaust products less attractive, but for a reasonably large expansion ratio, the effect is pretty small.
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 01/02/2014 01:11 am
Here's a high-level view AIUI. (Lots of details omitted.)

For a 100% efficient rocket we have
 H = 1/2 m v^2
where H is the heat of combustion of a parcel of mass m of propellants and v is the exhaust velocity. Solving for v we have v = sqrt(2 H/m). One should therefore expect propellants with a higher heat of combustion (per mass) to produce greater specific impulse.

Water has a molar mass of 18 amu and a heat of formation of -242 kJ/mol (as gas). Carbon dioxide has a molar mass of 44 amu and a heat of formation of -394 kJ/mol (as gas). Observe that a carbon dioxide molecule masses over twice water but provides less than twice the heat of formation, so producing H2O releases more heat per kilogram than producing CO2 does. The mainstream hydrocarbon fuels have heat of formation that can be approximated by zero for present purposes so you can think of e.g. a methane atom as just a convenient way to store a carbon atom and 4 hydrogen atoms. To maximize H/m you want to produce as much water as possible. That's why methane has better ISP than RP-1.
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 01/12/2014 02:46 pm
Is there a number that describes the efficiency of stages as a whole? ISP is for propellant combination and engine performance but does not factor in the dry mass of the stage needed to support the propellant and its associated engine...

(Not sure if that made any sense... everything I said sounded weird ::) )
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 01/12/2014 02:52 pm
Propellant mass fraction is another relevant number.  In some cases, (initial) thrust-to-weight ratio matters a lot.
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 01/12/2014 03:19 pm
Is there a number that describes the efficiency of stages as a whole? ISP is for propellant combination and engine performance but does not factor in the dry mass of the stage needed to support the propellant and its associated engine...

(Not sure if that made any sense... everything I said sounded weird ::) )

One figure I like (which is non-standard as far as I know) is the ISP of theoretical stage that has the same wet mass, zero dry mass, and provides the same delta-vee as the real stage we're evaluating. Call this concept the equivalent specific impulse. Here are the formulae:
 deltaV = I * ln((M_w+M_p)/(M_d+M_p))
 I_equiv = deltaV / ln((M_w+M_p)/(M_p))
where I is specific impulse (in velocity units), M_w is the wet mass of the stage, M_d is the dry mass of the stage, M_p is the mass of the payload (including any later stages), deltaV is the delta vee of the stage, and I_equiv is the equivalent specific impulse described above. The formulae for deltaV is only valid for a single stage but the definition of I_equiv is meaningful both for individual stages and for a multi-stage rocket as a whole.

One nice thing about this equivalent specific impulse is that the equivalent specific impulse of a multi-stage rocket is the ln(mass ratio) weighted average of the equivalent specific impulses of its constituent stages. For example a two-stage rocket with equivalent specific impulses of 350 s and 370s has an overall equivalent specific impulse somewhere between 350 and 370s. This two-stage rocket is better (in terms of mass ratios) than an alternative single stage rocket design with equivalent specific impulse 340s.

One annoying thing about this concept is the equivalent specific impulse of a stage depends on the mass of its payload.
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 01/12/2014 05:40 pm
Thank you both Proponent and DeltaV.

Would total impulse divided by mass of the stage be a a good evaluation of the performance of the stage?
Title: Re: Basic Rocket Science Q & A
Post by: M129K on 01/15/2014 07:38 pm
How can you calculate the ∆V needed to escape Earth and get on a Mars transfer if using low-thrust propulsion? I recently asked this somewhere else, and why it was useful for something like reaching GEO, I still don't know how to calculate escape velocity ∆V.
Title: Re: Basic Rocket Science Q & A
Post by: mheney on 01/15/2014 08:33 pm
Escape velocity is sqrt(2) times your circular orbital velocity - which makes your delta V (sqrt(2) -1) times orbital velocity.  That holds if you only burn at or near perigee.  (you keep the perigee, and raise apogee to infinity.)  If you do continuous burn, then you're essentially trying to raise both apogee and perigee to infinity.  That sees like you'd be doubling the total dV needed that way...
Title: Re: Basic Rocket Science Q & A
Post by: M129K on 01/16/2014 07:06 am
Escape velocity is sqrt(2) times your circular orbital velocity - which makes your delta V (sqrt(2) -1) times orbital velocity.  That holds if you only burn at or near perigee.  (you keep the perigee, and raise apogee to infinity.)  If you do continuous burn, then you're essentially trying to raise both apogee and perigee to infinity.  That sees like you'd be doubling the total dV needed that way...

So... Just double the delta V from 3.22 to 6.44 km/s to escape?
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 01/16/2014 11:57 am
Escape velocity is sqrt(2) times your circular orbital velocity - which makes your delta V (sqrt(2) -1) times orbital velocity.  That holds if you only burn at or near perigee.  (you keep the perigee, and raise apogee to infinity.)  If you do continuous burn, then you're essentially trying to raise both apogee and perigee to infinity.  That sees like you'd be doubling the total dV needed that way...

The delta-V for a very-low-thrust trajectory between two circular orbits is simply the difference in the orbital speeds of the two orbits.  A "circular orbit at infinity" has an orbital speed of zero.  If you're in a parking at, say, 7 km/s, then the low-thrust delta-V to escape is 7 km/s.  This of course ignores the perturbing influences of the moon and sun, but I think it's a pretty good first approximation.

To get to Mars, of course, you need to do more than just escape: you need a positive asymptotic speed.  It's not obvious to me that there's a good simple approximation available here.  If the acceleration is high enough that the delta-V for a Hohmann-like transfer from Earth's orbit to Mars's can be delivered in a couple of months, then I suppose you could just add the delta-V, recognizing that it will be a bit of an underestimate.  If, on the other hand, thrust is so low that several orbits of the sun will be completed on the way to Mars, then low-thrust approximation works well, it's just that this time the relevant circular speeds are those of Earth and Mars about the sun.  Most likely, though, neither of these limiting cases applies.  At least they give you upper and lower bounds on the delta-V required.
Title: Re: Basic Rocket Science Q & A
Post by: mheney on 01/16/2014 02:40 pm
It seems , though, that you only need to fire your engine near perigee.  The target orbit really isn't a circular orbit at infinity - it's a parabola (Actually, a hyperbola, so you've got some added heliocentric velocity once you escape.)

It seems that you'd be "wasting" any delta-v on the 'upper' half of the orbit, which only serves to raise the perigee.
If you set up your orbit so that perigee is at 6 pm solar time (over the terminator, "behind" the Earth looking down from the North), you'd want to do your burn for some fraction of the orbit around perigee.  The effect would be for the apogee to climb out ahead/away from the earth along Earth's orbit, while the perigee remained (roughly) fixed.  (Note that if you wanted to drop into the lower solar system, you'd reverse this, and have your perigee at 6am solar, with the apogee climbing out behind the Earth...)


The closer you get to instantaneous perigee burns, the closer your total delta-V gets to the theoretical minimum - which is sqrt(2) -1 times your orbital velocity.   For a 7 km/s LEO, that'd be about 2.9 km/s delta-V to escape - much less than the 7km/sec for a constant burn.  You'd be trading reaction mass for time-to-escape.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 01/16/2014 04:38 pm
To minimize delta-V, one would could indeed burn only near perigee and accept much longer trips.  With electric propulsion offering exhaust velocities in the neighborhood of 20-30 km/s, however, I would think that a mission planner would typically prefer a higher delta-V.  The high exhaust velocity limits the additional propellant mass needed.
Title: Re: Basic Rocket Science Q & A
Post by: Hop_David on 01/18/2014 04:23 pm
The delta-V for a very-low-thrust trajectory between two circular orbits is simply the difference in the orbital speeds of the two orbits.  A "circular orbit at infinity" has an orbital speed of zero.  If you're in a parking at, say, 7 km/s, then the low-thrust delta-V to escape is 7 km/s.  ...
To get to Mars, of course, you need to do more than just escape: you need a positive asymptotic speed.  It's not obvious to me that there's a good simple approximation available here.  If the acceleration is high enough that the delta-V for a Hohmann-like transfer from Earth's orbit to Mars's can be delivered in a couple of months, then I suppose you could just add the delta-V, recognizing that it will be a bit of an underestimate.  If, on the other hand, thrust is so low that several orbits of the sun will be completed on the way to Mars, then low-thrust approximation works well, it's just that this time the relevant circular speeds are those of Earth and Mars about the sun.  Most likely, though, neither of these limiting cases applies.  At least they give you upper and lower bounds on the delta-V required.

Indeed. Low earth orbit has a big angular velocity, about 4 degrees a minute (360 degrees/90 minutes).

To enjoy an Oberth benefit (http://hopsblog-hop.blogspot.com/2013/10/what-about-mr-oberth.html), we'd need to do the acceleration in the perigee's neighborhood, let's say over 60 degrees or about 15 minutes. If the ion rocket accelerates at 100 micro-gees, 100 micro-gees * 15 minutes = ~ 1 meter/second. A neglible burn.

Instead of an impulsive burn at perigee sending the rocket to escape, the ship's path would be a gradual spiral out of earth's gravity well. From LEO to escape would take about 7 km/s as you say. Although the delta V budget is more than twice that of an impulsive burn, ion exhaust speeds can be 4 or 5 times that of chemical. So the exponent in the rocket equation (delta V/V exhaust) is smaller in spite of the extra delta V.

However earth's orbit about the sun is a much more leisurely degree per day. (360 degrees/365 days). In this case we'd remain in a 60 degree neighborhood of perihelion for two months. 100 micro-gees * 2 months = 5 km/s.

For heliocentric orbits, low thrust, high ISP engines might be able to enjoy some Oberth benefit. Once out of planetary wells, it seems to me we can have our ion cake and eat it too.
Title: Re: Basic Rocket Science Q & A
Post by: sdsds on 01/18/2014 10:45 pm
However earth's orbit about the sun is a much more leisurely degree per day. (360 degrees/365 days). In this case [of a spacecraft in heliocentric orbit] we'd remain in a 60 degree neighborhood of perihelion for two months. 100 micro-gees * 2 months = 5 km/s.

For heliocentric orbits, low thrust, high ISP engines might be able to enjoy some Oberth benefit. Once out of planetary wells, it seems to me we can have our ion cake and eat it too.

That's very well expressed. Thanks! And just to check the reverse works too? I.e. a spacecraft that has just managed to escape from Mars (but is still in essentially the same orbit around the Sun as Mars) has plenty of time at aphelion to perform an ion "burn" that drops its perihelion down to intersect Earth?
Title: Re: Basic Rocket Science Q & A
Post by: Avron on 01/19/2014 02:36 pm
In a LOX/H2 engine, do the two fluids enter the combustion chamber via the injector plate as gasses or as liquids?
Title: Re: Basic Rocket Science Q & A
Post by: aga on 01/19/2014 03:35 pm
In a LOX/H2 engine, do the two fluids enter the combustion chamber via the injector plate as gasses or as liquids?

in rl-10 they are gases, acc. to this post by ranulfc

http://forum.nasaspaceflight.com/index.php?topic=30547.msg1125002#msg1125002
Quote
...even though at the same time everyone also understood that the RL-10 used "gas/gas" in operation and why....
Title: Re: Basic Rocket Science Q & A
Post by: Avron on 01/19/2014 04:27 pm
In a LOX/H2 engine, do the two fluids enter the combustion chamber via the injector plate as gasses or as liquids?

in rl-10 they are gases, acc. to this post by ranulfc

http://forum.nasaspaceflight.com/index.php?topic=30547.msg1125002#msg1125002
Quote
...even though at the same time everyone also understood that the RL-10 used "gas/gas" in operation and why....

I got this "In main combustion chambers oxygen is injected in its liquid state, whereas
the fuel - used for regeneratively cooling the combustor walls - is injected in the gaseous state.
"

ntrs.nasa.gov/archive/nasa/casi.ntrs.../20020045359_2002079435.pdf‎
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 01/19/2014 06:27 pm
Depends on engine cycle. Normally  pressure fed, gas generator and bleed expander means liquid for both, stage combustion and closed expander means gas for the preburner rich or expanded part (Hydrogen, normally for hydrolox), and both gas.
Now, is is a bit simplified, because usually the temperature and pressure is above the critical point. So the actual state might not be clear cut.
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 01/20/2014 09:35 am
I've read in quite a few places now, that the Isp of different propellant mixes scales with the density of the fuel. So, for example, RP-1 offers more ISP than methane etc. (All other things being equal.. consider some hypothetical engine that burns Metholox as well as it burns Kerolox).

Why is this?

Specific impulse scales with density?  That's a new one on me.  Hydrolox has a notoriously low density yet a very high specific impulse.

Specific impulses are usually quoted for a particular expansion ratio.  In effect, specific impulses given for propellant combinations are actually for particular idealized engines.

More complex molecules (in the exhaust products) will have more non-translational degrees of freedom.  In principle this does make complex exhaust products less attractive, but for a reasonably large expansion ratio, the effect is pretty small.

Dredging this up...

I wonder if this is simply a question of pragmatics.

Certain prop combinations have standout Isp vs density (vs usability).

Imagine a prop combination with the Isp of kerolox, but the density of hydrolox. Given it has the disadvantages of both, you'd expect a designer to choose either kero- or hydro- instead. The commonly used prop combo's are just those that stand out from the crowd somewhere on the Isp/density relationship.

Of course, sometimes usability gets in the way as well, and biases the decision - dangers of hypergolics or  hydrogen/fluorine, hassles of deep cryogens/LH, etc.

cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: mheney on 01/20/2014 03:50 pm
ISp and fuel density impulse are only one of a number of considerations when designing a vehicle.  The Lunar Modules used hypergolics, for example, because of reliability - you want to make as sure as possible that the Ascent Module stage lights when you push the button.  You'd look at different prop combos if/when the infrastructure is there - if you have a base that you can walk back to if you have a launch abort  with in-situ propellant production, yu'd use a different vehicle than you would an "expedition" mission like Apollo.

Title: Re: Basic Rocket Science Q & A
Post by: Hop_David on 02/26/2014 02:51 pm
However earth's orbit about the sun is a much more leisurely degree per day. (360 degrees/365 days). In this case [of a spacecraft in heliocentric orbit] we'd remain in a 60 degree neighborhood of perihelion for two months. 100 micro-gees * 2 months = 5 km/s.

For heliocentric orbits, low thrust, high ISP engines might be able to enjoy some Oberth benefit. Once out of planetary wells, it seems to me we can have our ion cake and eat it too.

That's very well expressed. Thanks! And just to check the reverse works too? I.e. a spacecraft that has just managed to escape from Mars (but is still in essentially the same orbit around the Sun as Mars) has plenty of time at aphelion to perform an ion "burn" that drops its perihelion down to intersect Earth?

Even better for Mars. Angular velocity slows farther from the sun. Mars is about half a degree per day.

Jupiter's about .1 degree per day. Saturn .03 degrees. The more relaxed pace of the outer bodies can give an ion engine a lot more time to do her slow burn in the neighborhood of an peri or aphelion. SEP probably wouldn't work in the outer system though, ion engines would need nukes.

I think the Main Belt might be a sweet spot for SEP. A slow enough pace that SEP can enjoy the Oberth benefit and enough sunlight to power them.
Title: Re: Basic Rocket Science Q & A
Post by: Hop_David on 02/26/2014 03:37 pm
Certain prop combinations have standout Isp vs density (vs usability).

Martin's observation is for chemical engines. I'm wondering if similar considerations can be applied to ion engines.

Here's some of my assumptions on ion engines:

1) Ion engines often use noble gases because the plasma is less corrosive than chemically active elements.

2) Xenon is often used because of higher atomic weight. Increasing molar weight decreases ISP but increases thrust. Ion engines already have plenty of ISP but miniscule thrust. Heavier xenon partially mitigates an ion engine's tiny thrust.

I'm hoping those more knowledgeable than I can tell me if I'm wrong or on the mark.

Mars atmosphere is about 96% carbon dioxide, 2% argon and 2% nitrogen. One of my daydreams is exporting Martian argon to the Main Belt for use as reaction mass.

Title: Re: Basic Rocket Science Q & A
Post by: Adaptation on 03/30/2014 02:55 am
So rocket segments are almost always cylinders and the tanks are cylindrical with domed ends.  Spheres have the optimum volume to surface area ratio.  But just about every design chooses cylindrical rockets so it must have substantial benefits.  I want to understand which reasons are the most important.  I came up with a few please let me know if my order of importance is correct or if I missed any important reasons.

A. Cylinders (pillars) are stronger at holding things up than spherical tanks and the tanks form a major structural component of the rocket.
B. Rockets need to be aerodynamic to reduce stress at max q.
C. Rockets need to be aerodynamic to reduce atmospheric drag.
D. Manufacturing large cylinders is easier than manufacturing large spheres. 
D. Shipping large cylinders around earth is easier to do than large spheres. 
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 03/30/2014 04:23 am
N-1 used spheres for the first 3 stages. Also, spheres are used for things like helium tanks and for some upper stage engines, I think (on Soviet lunar probes, at least).

Cylinders are nice for the reasons you mentioned.
Title: Re: Basic Rocket Science Q & A
Post by: Adaptation on 03/30/2014 04:59 am
Approximately what portion of delta v goes into atmospheric drag?
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 03/31/2014 10:37 pm

Approximately what portion of delta v goes into atmospheric drag?
I think it's around 150m/s.
But ink of the other considerations. You need both fuel and oxidizer. As the N-1 showed, spherical tanks need a lot of support mass, and piping is a mess. And then all they things you said did played a role. Transportation is one of the biggies. But aerodynamic drag is very important for smaller rockets. Drag is proportional to the front surface. But rocket weight is a cubic measure. Since drag losses are drag/thrust, as your rocket grows (and thus thrust), the aspect ratio is less important. Look how thin sounding rockets are vs Shuttle or Saturn V, for example.
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 04/01/2014 07:06 am
Actually, thrust also goes as area, too. That places an upper limit on rocket height given a certain chamber pressure and propellant bulk density.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 04/01/2014 01:18 pm

Actually, thrust also goes as area, too. That places an upper limit on rocket height given a certain chamber pressure and propellant bulk density.
I've calculated at around 1000MN for a sphere at RD-191 MCC pressure / expansion ratio.
Title: Re: Basic Rocket Science Q & A
Post by: pagheca on 04/15/2014 10:29 pm
I know how the specific impulse of a rocket engine is defined.

What I'm asking here is if any of you knows a way to "think" at the specific impulse of a rocket engine. What actually means 282 sec?

For example, one can express the speed of an object in sec, by dividing its length by his speed: m / (m/s) = sec. This number tells how many sec are required for the vehicle to make a single move as long as its own length.

Does anyone knows something like that for specific impulse?
Title: Re: Basic Rocket Science Q & A
Post by: ddunham on 04/15/2014 10:42 pm
From wikipedia...
http://en.wikipedia.org/wiki/Specific_impulse

Quote
This Isp expressed in seconds is somewhat physically meaningful—if an engine's thrust could be adjusted to equal the initial weight of its propellant (measured at one standard gravity), then Isp is the duration the propellant would last.
Title: Re: Basic Rocket Science Q & A
Post by: pagheca on 04/16/2014 01:27 am
Approximately what portion of delta v goes into atmospheric drag?

I do not know if you are referring to something in particular. However, from another post (http://forum.nasaspaceflight.com/index.php?topic=9959.msg189860#msg189860) and, in turn, from this book (http://www.amazon.com/Propulsion-Analysis-Design-Ronald-Humble/dp/0070313202/ref=sr_1_2?ie=UTF8&qid=1397611602&sr=8-2&keywords=Space+Propulsion+Analysis+and+Design):

Ariane A-44L: Gravity Loss: 1576 m/s Drag Loss: 135 m/s
Atlas I: Gravity Loss: 1395 m/s Drag Loss: 110 m/s
Delta 7925: Gravity Loss: 1150 m/s Drag Loss: 136 m/s
Shuttle: Gravity Loss: 1222 m/s Drag Loss: 107 m/s
Saturn V: Gravity Loss: 1534 m/s Drag Loss: 40 m/s (!!)
Titan IV/Centaur: Gravity Loss: 1442 m/s Drag Loss: 156 m/s

Title: Re: Basic Rocket Science Q & A
Post by: aero on 04/16/2014 02:05 am
Approximately what portion of delta v goes into atmospheric drag?

I do not know if you are referring to something in particular. However, from another post (http://forum.nasaspaceflight.com/index.php?topic=9959.msg189860#msg189860) and, in turn, from this book (http://www.amazon.com/Propulsion-Analysis-Design-Ronald-Humble/dp/0070313202/ref=sr_1_2?ie=UTF8&qid=1397611602&sr=8-2&keywords=Space+Propulsion+Analysis+and+Design):

Ariane A-44L: Gravity Loss: 1576 m/s Drag Loss: 135 m/s
Atlas I: Gravity Loss: 1395 m/s Drag Loss: 110 m/s
Delta 7925: Gravity Loss: 1150 m/s Drag Loss: 136 m/s
Shuttle: Gravity Loss: 1222 m/s Drag Loss: 107 m/s
Saturn V: Gravity Loss: 1534 m/s Drag Loss: 40 m/s (!!)
Titan IV/Centaur: Gravity Loss: 1442 m/s Drag Loss: 156 m/s

I wonder if gravity loss correlates with the rocket lift-off T/W ratio, that is, Gross thrust/GLOW.  It seems that it might.
Title: Re: Basic Rocket Science Q & A
Post by: pagheca on 04/16/2014 03:10 am
I got a few questions about gravitational drag.

Frankly speaking, I haven't been able to fully understand how this drag depends on the vertical and horizontal components of the speed. Why GD (gravitational drag) isn't described by an analytic function of the trajectory(x,y), weight(t) and speed(t)?

GD is shown for other rockets in the table above. However, how to extrapolate the approximate value for the 1st stage only of a Falcon 9? Staging happens at a relatively low altitude for the Falcon (~100 km), respect to other rockets. On the other side, I guess DG is a function of rocket mass, so that most of it is "spent" at a low altitude. So, for the 1st stage only, I would say GD is much less than the one relative to a Titan, for example. Something like... 1/2? Is that assumption correct?

Moreover, one should add or not the GD while the rocket is re-entering the atmosphere?

Last but not least, why is the Saturn V value of the Drag Loss so drammatically low??? What made it so efficient - assuming this is not just a typo?
Title: Re: Basic Rocket Science Q & A
Post by: aero on 04/16/2014 06:19 am
I got a few questions about gravitational drag.

snip

Last but not least, why is the Saturn V value of the Drag Loss so drammatically low??? What made it so efficient - assuming this is not just a typo?

A lot of good information about the Saturn V here.

http://www.braeunig.us/apollo/saturnV.htm (http://www.braeunig.us/apollo/saturnV.htm)

Braeunig says that the low aerodynamic drag was because the Saturn V thrust to weight ratio was low resulting in lower velocity within the lower, thicker atmosphere. He also says that the drag coefficient at max q was about 0.51. I note that Cd= 0.51 is quite low for a rocket moving just above Mach 1, which is when max q happens.
Title: Re: Basic Rocket Science Q & A
Post by: aero on 04/16/2014 06:38 am
I got a few questions about gravitational drag.

Frankly speaking, I haven't been able to fully understand how this drag depends on the vertical and horizontal components of the speed. Why GD (gravitational drag) isn't described by an analytic function of the trajectory(x,y), weight(t) and speed(t)?

snip


Oh there is. It's equal to -g + ac where ac is centrifugal acceleration. You just have to integrate it along the trajectory you want to fly. Of course g is a function of altitude and ac is a function of horizontal velocity and altitude so the integration is a little complicated...
Title: Re: Basic Rocket Science Q & A
Post by: fatjohn1408 on 04/16/2014 10:14 am
I got a few questions about gravitational drag.

snip

Last but not least, why is the Saturn V value of the Drag Loss so drammatically low??? What made it so efficient - assuming this is not just a typo?

A lot of good information about the Saturn V here.

http://www.braeunig.us/apollo/saturnV.htm (http://www.braeunig.us/apollo/saturnV.htm)

Braeunig says that the low aerodynamic drag was because the Saturn V thrust to weight ratio was low resulting in lower velocity within the lower, thicker atmosphere. He also says that the drag coefficient at max q was about 0.51. I note that Cd= 0.51 is quite low for a rocket moving just above Mach 1, which is when max q happens.

In addition, drag loss is negatively correlated with vehicle length.
If you have a vehicle with a certain thrust to weight and you double its length, you need to double its thrust to keep the ratio the same and to be able to fly exactly the same launch trajectory.
However, since your diameter hasn't expanded the drag force will not have doubled and the size of your drag relative to your thrust will have been cut in half (or about that). Therefore your drag loss, which is just the total deceleration your rocket undergoes during flight, will also have been cut roughly in half.
Note also that drag loss is totally uncorrelated to vehicle diameter following the same logic.
Drag loss (during gravity turn) is all about length and thrust-to-weight (which dictates your velocity-height profile).
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 04/16/2014 08:05 pm
Drag loss id directly related to diameter, but inversely to thrust. If you scale a rocket linearly on all directions, surface growths quadratically while mass (and thus, thrust) has to grow cubically. Thus, as rockets get bigger they could decrease the drag loss. The fact is that drag is also directly related to aerodynamic stress. And also fatter vehicles are more mass efficient (because of surface to volume reasons) than thinner ones. Thus, you'll notice that bigger vehicles tend to be squatter. Also the reason why the Shuttle had not higher drag losses whiles having such a huge frontal area.
Title: Re: Basic Rocket Science Q & A
Post by: pagheca on 04/16/2014 08:19 pm
Oh there is. It's equal to -g + ac where ac is centrifugal acceleration. You just have to integrate it along the trajectory you want to fly. Of course g is a function of altitude and ac is a function of horizontal velocity and altitude so the integration is a little complicated...

Thanks very much to you and all the others replying on this issue.

Any reference to this equation and how it can be derived?
Title: Re: Basic Rocket Science Q & A
Post by: aero on 04/16/2014 09:17 pm
Oh there is. It's equal to -g + ac where ac is centrifugal acceleration. You just have to integrate it along the trajectory you want to fly. Of course g is a function of altitude and ac is a function of horizontal velocity and altitude so the integration is a little complicated...

Thanks very much to you and all the others replying on this issue.

Any reference to this equation and how it can be derived?

Not really, it's first principles. But you can find an explanation here.

http://www.ux1.eiu.edu/~cfadd/1150/05UCMGrav/Sat.html (http://www.ux1.eiu.edu/~cfadd/1150/05UCMGrav/Sat.html)

You will find knowledge of the acceleration of gravity is needed. It is derived from F = G *M*m/r^2 and is commonly given as g = (G M/ r^2)*r Look for terms here.

http://en.wikipedia.org/wiki/Gravitational_acceleration (http://en.wikipedia.org/wiki/Gravitational_acceleration)

An observation which is convenient to use is that at the surface of the earth, the magnitude of GM/re^2 = go so at altitude h, the magnitude of gh = go*(re/re+h)^2
Title: Re: Basic Rocket Science Q & A
Post by: pagheca on 04/16/2014 10:27 pm
Not really, it's first principles. But you can find an explanation here.


Aero,
I checked that page but found nothing like what I was searching for. I can easily derive the equation including the potential energy, but I read on several sources that the gravitation drag is not just something like mgh and depends on the trajectory: the more vertical you go, the more gravitational drag you pay. This is what I can't really understand.
Title: Re: Basic Rocket Science Q & A
Post by: aero on 04/16/2014 10:55 pm
Not really, it's first principles. But you can find an explanation here.


Aero,
I checked that page but found nothing like what I was searching for. I can easily derive the equation including the potential energy, but I read on several sources that the gravitation drag is not just something like mgh and depends on the trajectory: the more vertical you go, the more gravitational drag you pay. This is what I can't really understand.

It's straight forward. Look at the equation, GD = -g + v^2/r . The more time spent going vertically the larger the integral of the -g term becomes, but when accelerating vertically, the v term remains =0. Remember, the v here is the horizontal velocity. The complete integrand must go to zero before gravity drag stops accumulating and the only way to do that is to accelerate horizontally because you can't orbit if you go so high that the g term goes to zero with no horizontal velocity. In LEO, the magnitude of the g term is about 9.2 m/s^2 so that is what the magnitude of the velocity term must equal. The less time spent achieving that horizontal velocity, the lower the gravity drag.

Of course, the more time spent within the atmosphere accelerating horizontally, the higher the aerodynamic drag becomes so your trajectory must be chosen to trade off those two penalties.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 04/17/2014 12:42 am
Intuitively is the integral of g x cos(b) where b is 0 when going horizontal and Pi/2 when going vertical.
Title: Re: Basic Rocket Science Q & A
Post by: aero on 04/17/2014 04:44 am
Intuitively is the integral of g x cos(b) where b is 0 when going horizontal and Pi/2 when going vertical.

That's just confusing though. Go ahead and derive the full vector formulation, starting with the vectors g and v.

The problem is displaying vector math on this web site.
Title: Re: Basic Rocket Science Q & A
Post by: pagheca on 04/17/2014 09:56 am
Thanks for the help, everyone. I think I got the point. I wonder how I missed it actually, and for so long...  ???

I was doing my math correctly but I was misplaced by something I read everywhere: that the gravitational drag "increases" the more you move vertically. I just misinterpreted this statement.

When people says that by going up vertically grav drag increases is just saying that as it depends on the integral of m(t) x g (scalar product), the less propellant you lift up, the less gravity drag you get. So, at the end of the day, the saving is the potential energy of the amount of propellant you already spent before reaching a certain altitude.

On a no-atmosphere celestial body like the Moon the optimal trajectory - at least from this point of view - would be to accelerate horizontally, while remaining on the ground (I'm not talking about a mass driver, that would be powered by external energy and would not be submitted to the Tsiokolsky equation during the acceleration), and gain altitude only when orbital speed has been achieved. At this point the maximum theoretical amount of propellant would have been used to achieve delta-v, and only a bare minimum of it would be spent in potential energy while gaining altitude at the end of the acceleration period.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 04/17/2014 10:01 am
Look at the equation, GD = -g + v^2/r

Wrong equation.

Intuitively is the integral of g x cos(b) where b is 0 when going horizontal and Pi/2 when going vertical.

Almost right equation. That would give maximum gravity loss while going horizontal and none going vertical. It's the oppostive so change cos into sin and then it is as in Sutton;

g·sin(θ) where θ is angle of flight path with local horizontal and g local gravity. Integrate over flight time.

That's just confusing though. Go ahead and derive the full vector formulation, starting with the vectors g and v.

See image. The vector math reduces to ||g||cos(α) where α is angle between g (vertical) and v, which is just alternate way to write Sutton's equation.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 04/17/2014 03:05 pm
On a no-atmosphere celestial body like the Moon the optimal trajectory - at least from this point of view - would be to accelerate horizontally, while remaining on the ground (no: a mass driver would feature other advantages of course), and gain altitude only when orbital speed has been achieved. At this point the maximum theoretical amount of propellant would have been used to achieve delta-v, and only the bare minimum required would be spent in potential energy while gaining altitude.

That's a very helpful thought experiment.

In the ideal case, the delta-V needed to gain altitude is zero.  Suppose, for example, you want to take off from an airless body into a 200-km circular orbit.  As you mention, you start by thrusting horizontally.  Keep going until you've reached the periapsis speed of a 0 x 200-km orbit.  Then shut the engines off and cost up to 200 km.  Then fire a short (ideally instantaneous burst) to circularize.  The sum of the delta-Vs of the two powered phases is the minimum delta-V to orbit.  It's a little larger than the orbital speed.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 04/17/2014 03:08 pm
Regarding the vectors needed to calculate losses, a neat trick that makes things conceptually simpler is to use the velocity vector and its normal as basis vectors (not my idea -- you'll find it many textbooks covering launch vehicles) throughout powered flight, even though those vectors rotate.  Attached is a little piece about this.  Corrections welcome.

EDIT:  Revised the attachment to correct a mathematical error, fix a typo and streamline the text
Title: Re: Basic Rocket Science Q & A
Post by: pagheca on 04/18/2014 03:26 am
Braeunig says that the low aerodynamic drag was because the Saturn V thrust to weight ratio was low resulting in lower velocity within the lower, thicker atmosphere. He also says that the drag coefficient at max q was about 0.51. I note that Cd= 0.51 is quite low for a rocket moving just above Mach 1, which is when max q happens.

Interesting... I understand that despite the enormous thrust from the 5 F-1 engines, the mass at take-off was overwhelming, so that at the beginning of the flight the acceleration was "only" 1.3 g. As a side effect this allowed to get across the denser atmosphere with a very low atmospheric drag, hence this very low value. Correct?

However, I would like to compare the acceleration profile for a few 1st stages to quantify this. I searched through the internet but found nothing like a real acceleration profile for other rockets. Just a table that compare theoretical - I suppose - T/W ratio to aerodynamic drag values, presumably with the same Cd:

T/W R       Velocity loss
2              336 m/s
3              474 m/s
4              581 m/s

(from here (http://www.princeton.edu/~stengel/MAE342Lecture3.pdf), page 6)

Does anyone knows where to find REAL acceleration profiles for some 1st stages other than the Saturn V one, possibly including the Falcon 9?
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 04/20/2014 12:25 pm
The attachment to my previous post in this thread (http://forum.nasaspaceflight.com/index.php?topic=13543.msg1184133#msg1184133) contained a mathematical error which has now been corrected.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 04/20/2014 12:33 pm
Does anyone knows where to find REAL acceleration profiles for some 1st stages other than the Saturn V one, possibly including the Falcon 9?

The appendix B of the attached paper contains a calculated trajectory for a Saturn IB.
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 05/04/2014 04:20 pm
I've got a couple of questions...

How do you calculate/estimate peak pressure in a solid rocket?

Also, if you were to scale a solid rocket motor, would the throat's surface are be proportional to the volume of the solid rocket fuel? (I basically am wondering if you wanted to scale up a solid rocket motor, what increases linearly, exponentially, etc.)
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 05/04/2014 08:10 pm
I've got a couple of questions...

How do you calculate/estimate peak pressure in a solid rocket?

Dunno.

Quote
Also, if you were to scale a solid rocket motor, would the throat's surface are be proportional to the volume of the solid rocket fuel? (I basically am wondering if you wanted to scale up a solid rocket motor, what increases linearly, exponentially, etc.)

IANARS but my understanding follows.

Rocket thrust is proportional to chamber pressure times throat area times thrust coefficient, the last of which is about 2 and doesn't vary much. So if you want to scale up the thrust of a solid (or liquid) rocket by a factor of X keeping chamber pressure the same you need to increase throat area by X and hence throat diameter by sqrt(X).

When you increase thrust by a factor X you need to increase mass flow rate by the same factor (assuming fixed specific impulse). I believe mass flow rate in a solid, for fixed chamber pressure and propellant composition, is proportional to the area of the propellant/"air" interface in the main combustion chamber. So if you increase the thrust by a factor X increase that surface area by factor X too.

Burn time, for fixed chamber pressure and propellant composition, is proportional to the (initial) thickness of the propellant in the main combustion chamber. If you don't want to change burn time when scaling your rocket you should keep this thickness unchanged - i.e. lengthen the rocket motor without changing its diameter.
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 05/05/2014 12:05 am
I've got a couple of questions...

How do you calculate/estimate peak pressure in a solid rocket?

Dunno.

Quote
Also, if you were to scale a solid rocket motor, would the throat's surface are be proportional to the volume of the solid rocket fuel? (I basically am wondering if you wanted to scale up a solid rocket motor, what increases linearly, exponentially, etc.)

IANARS but my understanding follows.

Rocket thrust is proportional to chamber pressure times throat area times thrust coefficient, the last of which is about 2 and doesn't vary much. So if you want to scale up the thrust of a solid (or liquid) rocket by a factor of X keeping chamber pressure the same you need to increase throat area by X and hence throat diameter by sqrt(X).

When you increase thrust by a factor X you need to increase mass flow rate by the same factor (assuming fixed specific impulse). I believe mass flow rate in a solid, for fixed chamber pressure and propellant composition, is proportional to the area of the propellant/"air" interface in the main combustion chamber. So if you increase the thrust by a factor X increase that surface area by factor X too.

Burn time, for fixed chamber pressure and propellant composition, is proportional to the (initial) thickness of the propellant in the main combustion chamber. If you don't want to change burn time when scaling your rocket you should keep this thickness unchanged - i.e. lengthen the rocket motor without changing its diameter.

Ok. Thanks haha I may need to read that a few time...  ::)
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 05/05/2014 08:16 am
How do you calculate/estimate peak pressure in a solid rocket?

IANARS either but here's a semieducated guess:

You need to know

- the motor grain composition, geometric profile and how it changes during burn
- grain reaction rate / chamber pressure curve (the higher the pressure the faster (millimeters per second or similar unit) it burns)
- throat area

Find the maximum grain surface area. This is not necessarily in the beginning of burn nor in the end.

Take a initial wild (but somewhat educated) guess what the chamber pressure Pc might be.

1. Calculate dot_m_burn (mass burned per second) using Pc, reaction rate curve and max surface area. Calculate the properties of resulting gases at Pc and resulting temperature.

2. Using gas properties and Pc calculate the dot_m_throat, the rate at which the gases can escape through the throat.

3. If difference between dot_m_burn and dot_m_throat is less then preset margin you have solution Pc. If dot_m_burn > dot_m_throat then increase Pc and goto 1, else decrease Pc and goto 1.

So it's an iterative process. The difference to liquid engines is the Pc affecting the burn rate, with proper injector liquid engines can assume static injection (hence reaction) rate.

Warranty void when someone spots obvious error ;)

Title: Re: Basic Rocket Science Q & A
Post by: pagheca on 05/24/2014 06:25 pm
Triggered by the recent Aerojet engine test failure, I would like to ask a couple of questions:

(1) which is the typical failure curve for rocket engines? Bath-tube like (high level of failures at the beginning and at the end of their life cycle)? Proportional to time? Something else?

(2) How much an hot-fire test impacts on the engine lifetime and reliability? Any idea?

(3) It is true that a test may last just a few seconds (as for pre-launch F9 tests), but I guess the start-up procedure is much more demanding than sustained operation time. Is that correct?

Title: Re: Basic Rocket Science Q & A
Post by: ciscosdad on 06/06/2014 04:28 am
Can anyone point to some info on the reentry speed capability of heatshields?

What relationship is there between thickness and weight to the maximum reentry speed?
If reentry speed is raised from 25,000 to (say) 27,000 mph, what effect would this have?
Title: Re: Basic Rocket Science Q & A
Post by: aero on 06/06/2014 05:53 am
Can anyone point to some info on the reentry speed capability of heatshields?

What relationship is there between thickness and weight to the maximum reentry speed?
If reentry speed is raised from 25,000 to (say) 27,000 mph, what effect would this have?

I get a lot of hits when searching for aerodynamic heating equations. Unfortunately many of them are behind a pay wall. Here is one place to look. In simple words, temperature (T-stagnation/T-static) rises as the square of the Mach number. I'll let you convert miles per hour to Mach number.

http://www.dept.aoe.vt.edu/~mason/Mason_f/ConfigAeroHypersonics.pdf (http://www.dept.aoe.vt.edu/~mason/Mason_f/ConfigAeroHypersonics.pdf)

Edit - forgot the link
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 06/06/2014 08:02 am
Can anyone point to some info on the reentry speed capability of heatshields?

What relationship is there between thickness and weight to the maximum reentry speed?
If reentry speed is raised from 25,000 to (say) 27,000 mph, what effect would this have?

I immodestly point to one of my own posts (http://forum.nasaspaceflight.com/index.php?topic=30573.msg1158194#msg1158194), which contains some information about the topic.  The answer is not simple, unfortunately.

Title: Re: Basic Rocket Science Q & A
Post by: ciscosdad on 06/06/2014 08:12 pm
Thanks guys, That's just what I'm looking for.
My curiosity was roused by the upcoming EFT-1. That reentry is from High orbit, so significantly below escape speeds. Is there any way to simulate a higher speed by changing the reentry profile (steeper for example).

The info you supplied shows that its a very different flight regime at escape velocities, and is a complex interplay with radiative and convective and both change with altitude.

Another query: Is a higher entry velocity dealt with only by a thicker (ablative) shield? Can the flight profile be shaped to ease the load? Is that why Apollo used the double "dip" style reentry?
Title: Re: Basic Rocket Science Q & A
Post by: scienceguy on 06/20/2014 05:36 pm
Is it possible for a rocket to go faster than its propellant exhaust velocity?
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 06/20/2014 05:40 pm
Is it possible for a rocket to go faster than its propellant exhaust velocity?

http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

Let's put it this way: if it wasn't possible, no chemical rocket would ever reach LEO.
Title: Re: Basic Rocket Science Q & A
Post by: DarkenedOne on 08/23/2014 10:23 pm
SpaceX seems to do perform a form of full vehicles integration testing when it performs the static fire tests.  Obviously they can do this at little expense due to the fact that their engines are restartable and relatively reusable.   Obviously this type of test cannot be performed with rockets that can only be started once.

It seems like a pretty valuable test to me.  I was wondering how any other launch systems are tested in such a way?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/24/2014 12:29 pm
SpaceX seems to do perform a form of full vehicles integration testing when it performs the static fire tests.  Obviously they can do this at little expense due to the fact that their engines are restartable and relatively reusable.   Obviously this type of test cannot be performed with rockets that can only be started once.

It seems like a pretty valuable test to me.  I was wondering how any other launch systems are tested in such a way?

It was a practice in the 50's that had diminishing returns as the art of rocket science progressed.  Delta and Atlas are doing good with out doing it.  Even WDR's are becoming a thing of the past.  But that is only because the operators are accepting a schedule risk, if there are an problems that would be uncovered during WDR, they still would be found during the actual countdown.
Title: Re: Basic Rocket Science Q & A
Post by: su_liam on 08/28/2014 07:41 am
I'm trying to calculate the Tsiolkovsky delta-V for a parallel stage like the Space Shuttle with greatly different ISP. How would you calculate the "effective" ISP for that kind of arrangement?
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 08/28/2014 07:59 am
I'm trying to calculate the Tsiolkovsky delta-V for a parallel stage like the Space Shuttle with greatly different ISP. How would you calculate the "effective" ISP for that kind of arrangement?
Isp is a ratio of prop consumed vs thrust produced.

You can add the solid + liquid prop consumed, and add the solid + liquid thrust produced. The ratio of these two sums gives you a blended Isp.

Unfortunately, both the solids and SSME were throttled, which makes this more complicated for Shuttle. Any change in the proportion of thrust from the two systems needs a separate calculation of the blended Isp for that portion of the flight.

But, you may be able to get away with just averaging prop use and thrust over duration of solid operation.

Cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: aga on 08/28/2014 08:40 am
Isp is a ratio of prop consumed

just to clarify... it is mass/weight consumed, not volume
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 08/28/2014 09:08 am
Yes, assuming Isp is constant.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 08/28/2014 12:19 pm
Isp is a ratio of prop consumed
just to clarify... it is mass/weight consumed, not volume

Important addition to the original sentence is per unit of time ie. thrust divided by mass flow rate.

For n different engines firing simultaneously with thrusts F1, F2 ... Fn and respective specific impulses of Isp.1, Isp.2 ... Isp.n the math is

Isp.combined = ( F1 + F2 + ... + Fn ) / ( F1 / Isp.1 + F2 / Isp.2 + ... + Fn / Isp.n )
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 08/28/2014 02:47 pm
Isp is a ratio of prop consumed
just to clarify... it is mass/weight consumed, not volume

Important addition to the original sentence is per unit of time ie. thrust divided by mass flow rate.

For n different engines firing simultaneously with thrusts F1, F2 ... Fn and respective specific impulses of Isp.1, Isp.2 ... Isp.n the math is

Isp.combined = ( F1 + F2 + ... + Fn ) / ( F1 / Isp.1 + F2 / Isp.2 + ... + Fn / Isp.n )

Thanks.  And, of course, yes to both.  :-[

cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: liamdoyle on 08/28/2014 03:44 pm
hi new here but have a question, is there any point to having the dragon v2 make a powered landing? seems to me parachutes work just fine and that thrusters will only add weight and complexity
Title: Re: Basic Rocket Science Q & A
Post by: oiorionsbelt on 08/28/2014 03:51 pm
Reusability, accuracy of landing and time to access of of time sensitive cargo are all improved with powered landing.
Title: Re: Basic Rocket Science Q & A
Post by: dglow on 08/28/2014 03:55 pm
Parachutes won't work well on Mars.
Title: Re: Basic Rocket Science Q & A
Post by: Nascent Ascent on 08/28/2014 04:07 pm
The thrusters and fuel have to be there because these serve double-duty.  Used in case of launch abort escape and if not needed for abort, then used for landing.
Title: Re: Basic Rocket Science Q & A
Post by: rpapo on 08/28/2014 04:16 pm
You don't need to send ships out to fish the capsule out of the water, and you have much less chance of seawater leaking in.
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 08/28/2014 04:19 pm
Also, the thrusters don't really add complexity since they're needed for abort. Also, parachutes have a way of failing. Powered landing can be done with parachutes as a backup.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 08/28/2014 04:32 pm
You don't need to send ships out to fish the capsule out of the water, and you have much less chance of seawater leaking in.

You can avoid those also by landing on ground with parachutes like Soyuz and Shenzhou (albeit using last second retro rockets).
Title: Re: Basic Rocket Science Q & A
Post by: SpacexULA on 08/28/2014 04:33 pm
Also, the thrusters don't really add complexity since they're needed for abort. Also, parachutes have a way of failing. Powered landing can be done with parachutes as a backup.

I agree, there are a lot of reasons to do propulsive landing

-increased Landing Accuracy
-decreased impact landing
-Adds no complexity if the LAS is integrated into the capsule
-Adds redundancy if the parachutes fully or partial fail (Soyuz has had partial parachute failures), these lead to high G non fatal impacts at the end of the mission.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 08/28/2014 04:37 pm
Soyuz has had partial parachute failures

Apollo 15 also.

(http://upload.wikimedia.org/wikipedia/commons/thumb/8/85/Endeavour_nears_splashdown_-_GPN-2000-001121.jpg/240px-Endeavour_nears_splashdown_-_GPN-2000-001121.jpg)
Title: Re: Basic Rocket Science Q & A
Post by: Coastal Ron on 08/28/2014 04:38 pm
Strange that NASA is happy to fly their reusable vehicle multiply times, but don't trust a private company's reusable vehicle. NB Orion is also planned to be reused a few times.
Garrett (Reisman, not me; I always find it weird talking about namesakes) said it would complicate the certification process. So the decision not to offer reused Dragons was made by SpaceX, not NASA. Now, whether NASA's certification process deserves criticism is another story ...

Like others have said, I think the decision to not bid reuse at this time is because there are so many unknowns that they still need to discover before they can feel confident that they are ready to offer reusability for Dragon V2.  The first being that they will be flying a brand new design spacecraft, and landing it in a new way (i.e. on land instead of water, or at least that's the goal).

Right now they have a production line set up for building new Dragons, but reusability means they have to set up a facility/workspace and process for refurbishing vehicles that have just flown.  That takes different types of skills, and though you can share some of the same people it makes sense to have dedicated personnel that can become experts on refurbishment and recertification.

I would imagine that they will want to have vehicle reusability ready for service for the follow on Commercial Crew service contract (~2020?).
Title: Re: Basic Rocket Science Q & A
Post by: jtrame on 08/28/2014 04:40 pm
You don't need to send ships out to fish the capsule out of the water, and you have much less chance of seawater leaking in.

You can avoid those also by landing on ground with parachutes like Soyuz and Shenzhou (albeit using last second retro rockets).

In fact, Dragon V2 will initially use the combination of parachutes and retro cushioning at landing.  Eventually, they will make the change to totally propulsive landings.
Title: Re: Basic Rocket Science Q & A
Post by: aero on 08/28/2014 04:43 pm
Don't overlook the possible impact damage for parachute landings, or the added mass to LEO (reduced payload) of the landing thrusters being carried all the way up instead of discarded.
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 08/28/2014 04:52 pm
Don't overlook the possible impact damage for parachute landings, or the added mass to LEO (reduced payload) of the landing thrusters being carried all the way up instead of discarded.
if you don't bring them to LEO, they'd basically have to be expendable.
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 08/29/2014 12:35 am
Soyuz has had partial parachute failures

Apollo has three main chutes while Soyuz has only one. When Apollo 15's chute failed it still had two more to bring it down. The Soyuz chute failure killed the cosmonaut because the spacecraft impacted the ground at terminal velocity.
Title: Re: Basic Rocket Science Q & A
Post by: dglow on 08/29/2014 12:44 am
So Reisman (re-)confirmed terra firma landings for Dv2... but has anyone indicated where such landings might take place?
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 08/29/2014 04:39 am
Soyuz has had partial parachute failures

Apollo has three main chutes while Soyuz has only one. When Apollo 15's chute failed it still had two more to bring it down. The Soyuz chute failure killed the cosmonaut because the spacecraft impacted the ground at terminal velocity.

Soyuz I (like, e.g., Mercury) had a back-up 'chute.  It's just that it failed too, because its compartment was affected by the same manufacturing flaw that prevented the main 'chute from working properly.
Title: Re: Basic Rocket Science Q & A
Post by: mgfitter on 09/04/2014 09:12 pm
Not sure if this is the best thread for this question, so mods, please move it to wherever it needs to be.

After 50+ years of continued developments in human spaceflight technologies, why is it that communications between the ground and crews aboard orbiting spacecraft (and launching ones too) are still so poor in terms of audio quality?

What is the technical bottleneck that causes this?

Surely it must be technically possible to get at least Skype-like audio quality these days?

Are there any plans for Orion/Dragon/CST-100/Dreamchaser to improve things in this area?

-MG.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 09/05/2014 12:47 am
Not sure if this is the best thread for this question, so mods, please move it to wherever it needs to be.

After 50+ years of continued developments in human spaceflight technologies, why is it that communications between the ground and crews aboard orbiting spacecraft (and launching ones too) are still so poor in terms of audio quality?


What makes you think it is poor quality?
Title: Re: Basic Rocket Science Q & A
Post by: SWGlassPit on 09/05/2014 02:47 pm
When ISS does media events, they use live HD video with audio quality equivalent to any other television production.  For day-to-day events, the audio only needs to be comfortably intelligible, not silky-smooth.  Why waste downlink bandwidth on an unnecessarily high audio bitrate when you can use it for more meaningful telemetry?
Title: Re: Basic Rocket Science Q & A
Post by: clongton on 09/05/2014 08:50 pm
Why waste downlink bandwidth on an unnecessarily high audio bitrate when you can use it for more meaningful telemetry?

Who says it's wasted? They have the bandwidth available to use if they wanted to.
Title: Re: Basic Rocket Science Q & A
Post by: SWGlassPit on 09/09/2014 03:37 pm
If the current voice communication is adequate, would you rather use an additional, say, 128 kbps, to make that voice silky smooth, or would you rather use it for systems monitoring?  Keep in mind, also, that ISS isn't the only TDRS customer.
Title: Re: Basic Rocket Science Q & A
Post by: breadfan on 09/20/2014 05:18 am
Realistically, how likely is it that an anomaly will occur due to launching in inclement weather?
Is it realistic that future LV's will be capable of flying all but the most intense weather, just like all other forms of transport? Seems like a necessary development if SF is to become as routine as other regular flight in the future, given how often weather causes delays.
Title: Re: Basic Rocket Science Q & A
Post by: QuantumG on 09/20/2014 05:22 am
Seems kinda like the opposite, the Saturn V and the old reliable Soyuz have happily launched in terrible weather, but these new launch vehicles don't.
Title: Re: Basic Rocket Science Q & A
Post by: breadfan on 09/20/2014 05:28 am
Of course, there was the whole lightning strike and SCE-to-AUX episode.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 09/20/2014 08:29 pm
Realistically, how likely is it that an anomaly will occur due to launching in inclement weather?
Is it realistic that future LV's will be capable of flying all but the most intense weather, just like all other forms of transport? Seems like a necessary development if SF is to become as routine as other regular flight in the future, given how often weather causes delays.

Spacecraft don't like it.
Title: Re: Basic Rocket Science Q & A
Post by: Avron on 09/20/2014 08:56 pm
Seems kinda like the opposite, the Saturn V and the old reliable Soyuz have happily launched in terrible weather, but these new launch vehicles don't.


so much for progress
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 09/20/2014 09:09 pm
Seems kinda like the opposite, the Saturn V and the old reliable Soyuz have happily launched in terrible weather, but these new launch vehicles don't.


so much for progress

Again, look up Apollo 12 for a weather lesson.
Title: Re: Basic Rocket Science Q & A
Post by: Avron on 09/20/2014 09:11 pm
Seems kinda like the opposite, the Saturn V and the old reliable Soyuz have happily launched in terrible weather, but these new launch vehicles don't.


so much for progress

Again, look up Apollo 12 for a weather lesson.

Sometimes things get pushed a little far.. still they all made it..
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 09/20/2014 09:12 pm
Seems kinda like the opposite, the Saturn V and the old reliable Soyuz have happily launched in terrible weather, but these new launch vehicles don't.


so much for progress

Again, look up Apollo 12 for a weather lesson.

Sometimes things get pushed a little far.. still they all made it..

Is that what you tell your customer when his payload, for a change, *doesn't* make it?

"Sorry, we pushed it a little too far."

Seriously...
Title: Re: Basic Rocket Science Q & A
Post by: Avron on 09/20/2014 09:16 pm
Seems kinda like the opposite, the Saturn V and the old reliable Soyuz have happily launched in terrible weather, but these new launch vehicles don't.


so much for progress

Again, look up Apollo 12 for a weather lesson.

Sometimes things get pushed a little far.. still they all made it..

Is that what you tell your customer when his payload, for a change, *doesn't* make it?

Seriously...

seriously - progress is dismal  compared to all other forms of high speed transport..  in some cases it backwards.. Soyuz don't have that problem.. alas problems are elsewhere, it launches in some nasty  stuff 
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 09/20/2014 09:22 pm
seriously - progress is dismal  compared to all other forms of high speed transport.. 

When other forms of "high speed transport" have even a fraction of energy involved here, then we can talk apples-to-apples.
Title: Re: Basic Rocket Science Q & A
Post by: Avron on 09/20/2014 09:35 pm
seriously - progress is dismal  compared to all other forms of high speed transport.. 

When other forms of "high speed transport" have even a fraction of energy involved here, then we can talk apples-to-apples.

  any aircraft in dense atmosphere, say 737 .. they fly is nasty stuff.. or A380 if the engine does not de-construct or a Soyuz, ICBM, f-119.. 
Title: Re: Basic Rocket Science Q & A
Post by: ugordan on 09/20/2014 09:57 pm
This is the kind of thing that happens when Apollo 12 lessons (more generally, reasons for specific LCCs) get "forgotten":
https://www.youtube.com/watch?v=nWFTIHSlo3Y

Sometimes "progress" is not launching when you want, but learning when NOT to launch.
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 09/26/2014 12:21 pm
Also, don't forget AC-67. Lightning strikes have caused LOM's. No reason to risk it. 
Title: Re: Basic Rocket Science Q & A
Post by: GA on 01/19/2015 04:03 pm
Hi, does anyone know if (Tan x = 1) or (sin x/cos x = 1) means anything in astrophysics?
Title: Re: Basic Rocket Science Q & A
Post by: kch on 01/19/2015 04:24 pm
Hi, does anyone know if (Tan x = 1) or (sin x/cos x = 1) means anything in astrophysics?

As far as I know, it means a 45 degree angle.  If there's something more to it than that, I'd be interested to hear about it!  :)
Title: Re: Basic Rocket Science Q & A
Post by: spacenut on 03/30/2015 11:18 pm
How fast would a spacecraft have to go to have 1 G acceleration?  If SEP acceleration can achieve 1 G, or even small chemical or nuclear rockets, is it possible to achieve 1 G acceleration to Mars, then 1 G deceleration getting there?  Also, if this is possible, how long would it take to get to Mars at continuous 1 G acceleration and deceleration?  Would it be easier to get 1/2 G? 

I was just wondering if this would be more practical than a spinning spacecraft on a slower trip.
Title: Re: Basic Rocket Science Q & A
Post by: sandrot on 03/30/2015 11:26 pm
Just dug out this:

http://www.johndcook.com/blog/2012/08/30/flying-to-mars-in-three-days/
Title: Re: Basic Rocket Science Q & A
Post by: QuantumG on 03/30/2015 11:28 pm
I was just wondering if this would be more practical than a spinning spacecraft on a slower trip.

If by practical you mean magical, sure.
Title: Re: Basic Rocket Science Q & A
Post by: spacenut on 03/31/2015 12:30 am
Ok, so we would have to have warp drive. 
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 03/31/2015 12:13 pm
I'm cool with that. Get cracking ;)
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 03/31/2015 03:04 pm
I was just wondering if this would be more practical than a spinning spacecraft on a slower trip.

If by practical you mean magical, sure.
Well, if you could accelerate your exhaust speed to 0.034c, you'd have an isp of 1,000,000s. With that, assuming you can throttle down, you can get a 1g of acceleration during 3 days by using 25% of you craft mass. Thus, while completely unfeasible with known technology, it wouldn't look like it would violate any laws of physics. Fusion engine might enable that in a distant future.
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 04/03/2015 02:57 am
How fast would a spacecraft have to go to have 1 G acceleration?  If SEP acceleration can achieve 1 G, or even small chemical or nuclear rockets, is it possible to achieve 1 G acceleration to Mars, then 1 G deceleration getting there?  Also, if this is possible, how long would it take to get to Mars at continuous 1 G acceleration and deceleration?  Would it be easier to get 1/2 G? 

I was just wondering if this would be more practical than a spinning spacecraft on a slower trip.

1 gee acceleration from Earth surface gets nowhere.
1 gee acceleration from LEO for about 10 minutes, is 5880 meters a second which will reach Mars in about 7 months. And 1 gee for about 20 minutes could reach Mars in less than 4 months. Say 15 mins at LEO and 5 mins at near Mars. Though there different ways of doing this.
And roughly 30 minute of 1 gee starting from LEO could reach Mars in about 3 months.

And from Earth elliptical orbit- say 300 by 200,000 km, with about 10 minutes at 1 gee one reach Mars in 3 to 4 months.

And it seems if got to Mars in about 3 months, one would need less shielding and receive less radiation and one would not need to make artificial gravity- one still have physiological affects from the microgravity
but they would be significantly less debilitating then compared to +6 months.

And to get 2 months- about 40 mins starting from LEO, and 20 mins starting from high orbit [highly elliptical orbit]. Though twice the acceleration for 1/2 the time would be more efficient. 
Title: Re: Basic Rocket Science Q & A
Post by: the_other_Doug on 04/03/2015 06:09 pm
And don't forget that one-G acceleration for five minutes got the Apollo CSM/LM stack from LEO to the Moon in three days...
Title: Re: Basic Rocket Science Q & A
Post by: SkipMorrow on 05/05/2015 06:16 pm
I am following the decay of Progress 27M, and people keep mentioning TLE and space-track.org. I went there and made an account for myself, but there isn't any help on the site as to how to read the data there.

[Edit: deleted quoted TLE info because of possible copyright issues]

If it is too hard to explain in a forum post, that's cool. But perhaps you can point me to a web page or book that I can read that explains it all?
Title: Re: Basic Rocket Science Q & A
Post by: AnalogMan on 05/05/2015 06:22 pm
http://en.wikipedia.org/wiki/Two-line_element_set
Title: Re: Basic Rocket Science Q & A
Post by: satwatcher on 05/05/2015 06:31 pm
T.S. Kelso has a good FAQ on two-line elements: https://celestrak.com/columns/v04n03/ (https://celestrak.com/columns/v04n03/)

Also note that space-track.org typically does not allow you to share elements originating from their database (https://www.space-track.org/documentation#/user_agree (https://www.space-track.org/documentation#/user_agree)). You may want to remove them and use elements from another source.
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 05/05/2015 06:40 pm
Here is a quick and dirty spread sheet I use to calculate crude Apogee/Perigees from TLE's. It's based upon some posts people have made in the past. The numbers are crude and use a circular earth model (Shutter).
Title: Re: Basic Rocket Science Q & A
Post by: Apollo-phill on 05/11/2015 04:19 pm
What was the carrier aircraft for the X 37 test drops at EAFB circa 2005/6 ?

Image capturess from X 37 test flight B roll video
Title: Re: Basic Rocket Science Q & A
Post by: DaveS on 05/11/2015 04:23 pm
Scaled Composites' White Knight One.
Title: Re: Basic Rocket Science Q & A
Post by: e of pi on 05/11/2015 11:49 pm
What was the carrier aircraft for the X 37 test drops at EAFB circa 2005/6 ?
WhiteKnight One (http://www.space.com/1226-rutan-white-knight-carries-37-space-plane-aloft.html), the carrier aircraft Scaled built to win the X-Prize with SpaceShipOne.
Title: Re: Basic Rocket Science Q & A
Post by: CitizenSpace on 05/16/2015 08:48 am
Well, this kind of seems like the right place to ask this...

So, I'm 14 years old and interested in aerospace engineering, been working on rockets for two years or so now, but never really got too far into the proper gritty stuff of it all (Been just building and flying my own rockets and drawing crappy designs). So to further my knowledge on rocket engines, I'm designing a H202/RP-1, 100N rocket engine.
I've used RPA for getting the parameters of it, and I've been following a couple PDF's along the design process. I've so far calculated all the nozzle parameters (half-angles, nozzle throat area, nozzle exit area) and now I'm up to the combustion chamber volume/diameter/length.
So I'm trying to figure out the characteristic length of the chamber, calculated as followed:
L* = Vc/At
Where Vc is chamber volume and At is nozzle throat area.
The problem I am having lies within Vc. You calculate Vc using the following equation:
Vc = Ac * Lc + Cv
Where Ac is nozzle throat area, Lc is chamber length and Cv is convergent volume.

So the problem is this: How do i calculate chamber length? Only information I can find is that L* is largely determined by past experiments, as Lc is very complex to calculate. To calculate it mathematically you need to use:
Vc = Mf * V * Ts
Where Mf is propellant mass flow rate, V is average specific volume, and Ts is propellant stay time.
I know Mf and I can calculate V, but how do I calculate Ts?! I can't find any formula for it! Complexity and length is not a problem for me, I really don't care whether I have to spend 3 or even 4 hours on the one thing.

Unless of course, someone happens to have some past data on the L* of 100-200N class H202/RP-1 engines.

Thanks for the help in advanced.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 05/17/2015 07:38 am
I know Mf and I can calculate V, but how do I calculate Ts?! I can't find any formula for it!

Because there is no neat compact formula for it. Ts is the answer to the question how long does it take for the propellants to properly combust. To mathematically find that out you have to model propellant injection, mixing, atomization and combustion. Things like injector droplet size, droplet vaporization rate and turbulence matter. The choice of propellants is most prominent factor so past experience gets you to the right ballpark.

Quote
Unless of course, someone happens to have some past data on the L* of 100-200N class H202/RP-1 engines.

Huzel'n'Huang (http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19710019929.pdf) to the rescue.

Page 87. HTP/RP-1 L* 60-70 inches including the catalyst bed.
Title: Re: Basic Rocket Science Q & A
Post by: CitizenSpace on 05/17/2015 07:57 am
I know Mf and I can calculate V, but how do I calculate Ts?! I can't find any formula for it!

Because there is no neat compact formula for it. Ts is the answer to the question how long does it take for the propellants to properly combust. To mathematically find that out you have to model propellant injection, mixing, atomization and combustion. Things like injector droplet size, droplet vaporization rate and turbulence matter. The choice of propellants is most prominent factor so past experience gets you to the right ballpark.

Quote
Unless of course, someone happens to have some past data on the L* of 100-200N class H202/RP-1 engines.

Huzel'n'Huang (http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19710019929.pdf) to the rescue.

Page 87. HTP/RP-1 L* 60-70 inches including the catalyst bed.

Aha! You legend, thanks for that! Now I can finally move on with design  :D

Also about the formula, is 'no neat and compact formula for it' mean there can't be one, because of incredibly-hard to model-mathematically variables? Or is it just because it would be very difficult to make one and no can really be bothered when we have past experience at hand. Not saying its bad- just curious.

Also, 1.5m seems a bit long for a 100N engine...
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 05/17/2015 09:02 am
Also about the formula, is 'no neat and compact formula for it' mean there can't be on, because of incredibly-hard to model-mathematically variables? Or is it just because it would be very difficult to make one and no can really be bothered when we have past experience at hand.

According to the revised Huzel'n'Huang (http://www.amazon.com/Engineering-Propellant-Progress-Astronautics-Aeronautics/dp/1563470136/) computer models do exist to estimate the L* but even they have trouble modelling the mixing properly and have to assume things like droplet size. AFAIK the hardness is in the required computational power to get meaningful results out of the model, not that the math is not known. And because working values are known for common propellants there's no pressing need to try to compute it.

Best of luck to your project! Keep us posted how it goes.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 05/17/2015 09:22 am
Also, 1.5m seems a bit long for a 100N engine...

Note that even though it looks like 'length' it is really just another form to express the stay time. The propellants don't know nor care whether they go into 100N engine or 1000,000N engine, the proper mixing and combustion takes about the same time in both cases.

Smaller engines tend to have higher combustion chamber contraction ratio than larger engines in order to achieve required L* without being impracticaly long.
Title: Re: Basic Rocket Science Q & A
Post by: CitizenSpace on 05/17/2015 09:29 am
Also, 1.5m seems a bit long for a 100N engine...

Note that even though it looks like 'length' it is really just another form to express the stay time. The propellants don't know nor care whether they go into 100N engine or 1000,000N engine, the proper mixing and combustion takes about the same time in both cases.

Smaller engines tend to have higher combustion chamber contraction ratio than larger engines in order to achieve required L* without being impracticaly long.

Thanks a bunch, I was getting myself confused with actual length and L* :D Sorry about that. Thanks for all the help, its been great.
Title: Re: Basic Rocket Science Q & A
Post by: msat on 07/10/2015 01:48 am
How are g forces on rockets kept to acceptable levels, at least with humans on-board, when your upper stages have reasonably high T/W and mass ratios but the engines have limited or no throttling? As one example, I'd imagine that the Falcon 9's second stage acceleration near the end of its burn would be uncomfortably high even at 40% throttle (~70k lbf).

I guess this applies to any of the stages involved in getting you to orbit but not necessarily any additional stages once you're already there.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 07/10/2015 12:38 pm
With the stage, trunk, Dragon V2, payload and people you hit maybe 3g, hardly too uncomfortable.
Title: Re: Basic Rocket Science Q & A
Post by: kraisee on 07/10/2015 01:08 pm
R7, unless they do some fairly serious throttling (and lose quite a bit of performance as a result) I think Dragon crews are more likely to see close to 5g at the end of both stage burns.

That is still nothing to worry about for eyes-in flight.   I'll always remember that a few weeks before he passed away aged 84, Skylab 4 astronaut Bill Pogue told me that he could still have easily flown that Saturn rocket up to 4.4g! :)


msat, normally during the design of a new launch system, the stages will be (roughly) sized so that as each stage approaches burnout, the total mass of that nearly-dry stage, the stages above it and the payload together mass the desired amount to produce a maximum peak acceleration.

For example, say a hypothetical first stage had 1,000,000 lbf vacuum thrust.   If the vehicle should not exceed 5g accelerations at any point in the flight (and assuming no throttling), then the first stage burnout mass, upper stages and payload together should mass right around 1/5th of the thrust, in this case no less than 200,000 lb.   That way the vehicle simply can't exceed 5g.

Don't assume this is a hard and fast rule, but it should help when trying to figure out the basics.

In practice, all vehicles play around with this balance in order to get the most out of their specific propellants and propulsion systems, especially if they are capable of throttling their engines.   Atlas-V is a particularly good example of how much it is possible to play around with your optimization.

Ross.
Title: Re: Basic Rocket Science Q & A
Post by: msat on 07/10/2015 04:32 pm
Ah! Thanks, kraisee. That makes sense.

This actually brings up a point I haven't heard the SSTO folks talk about. Using conventional technology, even if one was willing to accept the abysmal payload payload capability, the associated accelerations near the end of it's burn would be extreme unless the propulsion system consisted of numerous identical engines, most of which would get shut down throughout its flight, or incorporate smaller sustainer engines. Having just a few [large] engines would not be possible, assuming a practical lower limit for throttlability. Does that sound right?

Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 07/11/2015 07:36 am
Ah! Thanks, kraisee. That makes sense.

This actually brings up a point I haven't heard the SSTO folks talk about. Using conventional technology, even if one was willing to accept the abysmal payload payload capability, the associated accelerations near the end of it's burn would be extreme unless the propulsion system consisted of numerous identical engines, most of which would get shut down throughout its flight, or incorporate smaller sustainer engines. Having just a few [large] engines would not be possible, assuming a practical lower limit for throttlability. Does that sound right?

Sounds about right to me.
Of course if the returning stage is more massive as it needs wings and/or keeping rocket fuel to de-orbit
it lessens it.
Also if using some sort of assisted boost- like a mothership [though it's not strictly SSTO] one lessen need for as much initial thrust.

And what talking about does not apply for launch vehicles leaving Mars or the Moon- for number reasons- main reason is roughly need 9 rocket fuel to lift 1 payload from Earth. The Moon 1 rocket fuel to 1 payload, and Mars about 2 rocket fuel to 1 payload. Another reason is the low gravity loss for Moon and Mars.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 07/11/2015 08:27 am
The 3g figure was for stated 40% Mvac throttle. But kraisee is right, they probably floor it all the way up.

Having just a few [large] engines would not be possible, assuming a practical lower limit for throttlability. Does that sound right?

You need to define the practical lower limit. LEM descent engine demonstrated 10% throttle using moving pintle. YouKnowWhoX is already using pintle injector, adding a moving sleeve to increase throttleability (is that a word?) wouldn't be a huge step.

Another question is would you need to use the engines for landing too. In that case just a few large ones are problematic.
Title: Re: Basic Rocket Science Q & A
Post by: msat on 07/11/2015 10:14 am


And what talking about does not apply for launch vehicles leaving Mars or the Moon- for number reasons- main reason is roughly need 9 rocket fuel to lift 1 payload from Earth. The Moon 1 rocket fuel to 1 payload, and Mars about 2 rocket fuel to 1 payload. Another reason is the low gravity loss for Moon and Mars.

Yeah, Mars and the Moon are much more conducive to SSTO than the Earth .

The 3g figure was for stated 40% Mvac throttle. But kraisee is right, they probably floor it all the way up.

Having just a few [large] engines would not be possible, assuming a practical lower limit for throttlability. Does that sound right?

You need to define the practical lower limit. LEM descent engine demonstrated 10% throttle using moving pintle. YouKnowWhoX is already using pintle injector, adding a moving sleeve to increase throttleability (is that a word?) wouldn't be a huge step.

Another question is would you need to use the engines for landing too. In that case just a few large ones are problematic.

I guess should have looked it up before I said anything. I didn't expect S2 + Dragon to mass ~20k lbs!

And no, "throttleability" doesn't appear to be a word, but it should be!  ;)

Apparently AJ has managed to get a version of the RL-10 to throttle down to 8%, but it was a bit of a challenge from the sound of it. Compared to the LEM descent engine, I don't know if the difficulty was mainly associated with the injector type, or the use of turbopumps. 

I had sort of expected that a moving pintle/sleeve was necessary to throttle an engine with that kind of injector. Does _____X "simply" reduce the output pressure from the turbopump to achieve their engine throttling?
Title: Re: Basic Rocket Science Q & A
Post by: MP99 on 07/12/2015 10:05 am
Merlin is a pintle engine.

Cheers, Martin
Title: Re: Basic Rocket Science Q & A
Post by: msat on 07/12/2015 01:00 pm
Merlin is a pintle engine.

Cheers, Martin

Yeah, I was aware of that, but R7 made it sound like it's fixed position, so throttling is accomplished by other means.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 07/12/2015 03:11 pm
Merlin is a pintle engine.

Cheers, Martin

Yeah, I was aware of that, but R7 made it sound like it's fixed position, so throttling is accomplished by other means.

Haven't ever heard anything suggesting Merlin pintle would move so assuming it doesn't. No need really for 40% throttle, even conventional injectors can do that. Lowering turbopump pressure is enough.

RL-10 benefits from hydrogen and expander cycle. No GG nor preburner so there's no worry about combustion stability in those things (they have injectors too). Gas-liquid mixing in the main injector provides faster/better mixing than liquid-liquid. Gas-gas is the best, Raptor for the win!
Title: Re: Basic Rocket Science Q & A
Post by: Hoonte on 07/23/2015 10:27 am
My head sometimes askes me the strangest questions :-)

What rocket/booster will have made the overall largest cumulative thrust.
So taken the thrust of 1 type of rocket/booster e.g. Titan II and multiply its thrust by number of launches.

It's a silly question but I guess the proton might be the nr 1 there..
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 07/23/2015 01:16 pm
Wouldn't Soyuz be right at the top of the list, just by sheer number of launches...
Title: Re: Basic Rocket Science Q & A
Post by: kevin-rf on 07/23/2015 01:26 pm
Shuttle would be 135 x 6,780,000 lbf = 9.15x10^8 lbf
Soyuz would be what, ~1700 x (5 x ~183,000 lbf)  = 1.5x10^9 lbf ? *sealevel thrust

Just a thought experiment, you really need to add in time. number of launches x thrust x time at thrust level, but I digress...
Title: Re: Basic Rocket Science Q & A
Post by: msat on 07/27/2015 02:00 am
I thought I had a grasp on gravity losses, but after reading https://en.wikipedia.org/wiki/Gravity_drag#Vector_considerations I'm partially at a loss. I understand the trig fine, but I can't get an intuitive grasp of how this can be true. Basically it's saying that the [vertical] force of gravity acting on the vehicle is affected by the horizontal component of acceleration. How!? If I'm not mistaking, this means that a vehicle capable of 10g of acceleration effectively reduces the force of gravity exerted on the rocket by the planet to .05g, or in other words, 99.5% of the thrust is available for horizontal acceleration, instead of 90% as one would intuitively expect. Can someone shed some light on what's going on here? This is very bizarre to me.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 07/27/2015 08:11 am
@msat, the angle of thrust vector (also points towards direction of flight) relative to local horisontal determines how much of the local gravity works against you ie. is drag.

If your thrust vector points straight up you push against full force of the gravity and the vehicle net acceleration towards the direction of flight is 10g - 1g = 9g

If your thrust vector is completely horisontal then you don't work against gravity at all, gravity loss is zero.

If your thrust vector points straight down then gravity works for you (as long as the direction of flight is intended...  ;) ), drag becomes a bonus and net acceleration is 11g.

Gravity drag is

local_g * sin θ

where θ is the angle between direction thrust and local horisontal.

For 0.05g gravity drag θ must be arcsin (0.05) = ~2.866 degrees flight angle (almost horisontal).

Attaching a scan from Sutton's, hope it helps.
Title: Re: Basic Rocket Science Q & A
Post by: msat on 07/27/2015 10:42 am
Thanks, R7. Your answers are always appreciated  :)

What you said makes sense to me (I think.. heh), but from what I can tell, it doesn't coincide with what's stated in the wikipedia entry (please tell me the wiki is wrong!).

From what I make of your explanation and the diagram from Sutton, the vertical component consists of the sum of mg and D sin θ. So it's only considered "gravity drag" if you're thrusting at some angle less than 90 relative to the gravity vector, but that doesn't mean gravity doesn't have an effect on the vehicle otherwise. In other words, if we're thrusting perpendicular to g, we can say there's no gravity drag, but we're still accelerating perpendicular to the direction of thrust at the rate of g, right?

I'm guessing I still got some stuff wrong. I gotta get ready for work now, but when I get home I'll plug some values into the equations above and see what comes out the other end in an attempt to get a better grasp of what's going on.
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 07/31/2015 04:10 pm
Can someone help me understand the ratio between tank wall/mass and the tensile strength? (I'm not sure if I phrased that right haha)

For example, if I have a 1 meter cylindrical tank with a tank wall of .1 meter thickness (1m OD, .8m ID) that can handle a tensile strength of 1000 psi and I wanted to scale this up, how would it scale? I think this has to do with the square cube law maybe, but I'm pretty confused haha. So I guess I'm asking, could a 1 meter tank with a wall thickness of .1m handle the same internal pressure as a 10m tank with a .1m wall thickness?
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 07/31/2015 07:03 pm
Can someone help me understand the ratio between tank wall/mass and the tensile strength? (I'm not sure if I phrased that right haha)

For example, if I have a 1 meter cylindrical tank with a tank wall of .1 meter thickness (1m OD, .8m ID) that can handle a tensile strength of 1000 psi and I wanted to scale this up, how would it scale? I think this has to do with the square cube law maybe, but I'm pretty confused haha. So I guess I'm asking, could a 1 meter tank with a wall thickness of .1m handle the same internal pressure as a 10m tank with a .1m wall thickness?

For a cylindrical tank with "thin" wall (as compared to radius), looking just at the first order pressure load, the equation of relevance is sigma=P*r/t. Where sigma is the stress in the wall, P is the pressure, r is the tank radius, and t is the wall thickness. For the tank to not burst, sigma needs to be less than the ultimate tensile strength of the material. In practice, a factor of safety is applied as well. Per the ASME Boiler code, that factor is 3.5-4 (depending on edition of code). Space applications tend to use 2.

To answer your question, all else being equal, a tank with a 0.1m wall thickness at 1m diameter would be able to handle 10 times the pressure of a 10m tank with the same wall thickness (though, 0.1m is thick enough for the former case that the thin wall assumption is getting dicey).

As an example, if I have a tank made out of Aluminum 7075-T6, operating at 1000psi, with a diameter of 48 inches, then the minimum wall thickness I need is:

t=P*r/sigma

Sigma=84,000 psi http://www.makeitfrom.com/material-properties/7075-T6-Aluminum (http://www.makeitfrom.com/material-properties/7075-T6-Aluminum)
P=1000 psi
r=48/2=24"

t=1000psi*24"/84000=0.286"

Then, if I want a factor of safety of 2, I would build the vessel with a wall of 0.286*2=0.572"
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 07/31/2015 09:26 pm
Can someone help me understand the ratio between tank wall/mass and the tensile strength? (I'm not sure if I phrased that right haha)

For example, if I have a 1 meter cylindrical tank with a tank wall of .1 meter thickness (1m OD, .8m ID) that can handle a tensile strength of 1000 psi and I wanted to scale this up, how would it scale? I think this has to do with the square cube law maybe, but I'm pretty confused haha. So I guess I'm asking, could a 1 meter tank with a wall thickness of .1m handle the same internal pressure as a 10m tank with a .1m wall thickness?

For a cylindrical tank with "thin" wall (as compared to radius), looking just at the first order pressure load, the equation of relevance is sigma=P*r/t. Where sigma is the stress in the wall, P is the pressure, r is the tank radius, and t is the wall thickness. For the tank to not burst, sigma needs to be less than the ultimate tensile strength of the material. In practice, a factor of safety is applied as well. Per the ASME Boiler code, that factor is 3.5-4 (depending on edition of code). Space applications tend to use 2.

To answer your question, all else being equal, a tank with a 0.1m wall thickness at 1m diameter would be able to handle 10 times the pressure of a 10m tank with the same wall thickness (though, 0.1m is thick enough for the former case that the thin wall assumption is getting dicey).

As an example, if I have a tank made out of Aluminum 7075-T6, operating at 1000psi, with a diameter of 48 inches, then the minimum wall thickness I need is:

t=P*r/sigma

Sigma=84,000 psi http://www.makeitfrom.com/material-properties/7075-T6-Aluminum (http://www.makeitfrom.com/material-properties/7075-T6-Aluminum)
P=1000 psi
r=48/2=24"

t=1000psi*24"/84000=0.286"

Then, if I want a factor of safety of 2, I would build the vessel with a wall of 0.286*2=0.572"
You did a pressure calculation in ACU?  :P
It is important to note that that only takes into consideration the pressure stress and not other kinds of stresses (but the higher the pressure the better it handles anything else). And that the top and bottoms need special calculations later depending on design. BTW, the usual definition of "thin walled" pressure vessel is thickness<1/10 Diameter. So The example that ClaytonBirchenough used was exactly on the limit (which must have some margin in itself).
Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 08/01/2015 04:36 am
Can someone help me understand the ratio between tank wall/mass and the tensile strength? (I'm not sure if I phrased that right haha)

For example, if I have a 1 meter cylindrical tank with a tank wall of .1 meter thickness (1m OD, .8m ID) that can handle a tensile strength of 1000 psi and I wanted to scale this up, how would it scale? I think this has to do with the square cube law maybe, but I'm pretty confused haha. So I guess I'm asking, could a 1 meter tank with a wall thickness of .1m handle the same internal pressure as a 10m tank with a .1m wall thickness?

For a cylindrical tank with "thin" wall (as compared to radius), looking just at the first order pressure load, the equation of relevance is sigma=P*r/t. Where sigma is the stress in the wall, P is the pressure, r is the tank radius, and t is the wall thickness. For the tank to not burst, sigma needs to be less than the ultimate tensile strength of the material. In practice, a factor of safety is applied as well. Per the ASME Boiler code, that factor is 3.5-4 (depending on edition of code). Space applications tend to use 2.

To answer your question, all else being equal, a tank with a 0.1m wall thickness at 1m diameter would be able to handle 10 times the pressure of a 10m tank with the same wall thickness (though, 0.1m is thick enough for the former case that the thin wall assumption is getting dicey).

As an example, if I have a tank made out of Aluminum 7075-T6, operating at 1000psi, with a diameter of 48 inches, then the minimum wall thickness I need is:

t=P*r/sigma

Sigma=84,000 psi http://www.makeitfrom.com/material-properties/7075-T6-Aluminum (http://www.makeitfrom.com/material-properties/7075-T6-Aluminum)
P=1000 psi
r=48/2=24"

t=1000psi*24"/84000=0.286"

Then, if I want a factor of safety of 2, I would build the vessel with a wall of 0.286*2=0.572"
You did a pressure calculation in ACU?  :P
It is important to note that that only takes into consideration the pressure stress and not other kinds of stresses (but the higher the pressure the better it handles anything else). And that the top and bottoms need special calculations later depending on design. BTW, the usual definition of "thin walled" pressure vessel is thickness<1/10 Diameter. So The example that ClaytonBirchenough used was exactly on the limit (which must have some margin in itself).

Thank you both for the replies/comments!

I arbitrarily chose the example above, but I did have "thin" walled tanks in mind. Thanks also for the helpful math strangequark!

With regards to the "other stresses" you mention baldusi, what do you mean? I was starting to imagine that there would be a complex interaction of different stresses like tensile strength and compressive strength. Is there anyway to estimate this interaction and its resulting "felt" stresses? For example, if we use the above 48" diameter aluminum tube/tank used for discussion/calculation purposes, what would the resulting "felt" stress on the tank be if it was accelerating at 4g with a 1000 kg payload on top of it? It's ok if we don't get to the forward closure of the tank, but just assume that the load is dispersed equally around the radius of the cylindrical tank. I also know the math might get a little crazy, but was just wondering if there was some theory/explanation for the "compounding" effects of such forces?

Thank you both again!
Title: Re: Basic Rocket Science Q & A
Post by: gbaikie on 08/01/2015 06:41 am
Can someone help me understand the ratio between tank wall/mass and the tensile strength? (I'm not sure if I phrased that right haha)

For example, if I have a 1 meter cylindrical tank with a tank wall of .1 meter thickness (1m OD, .8m ID) that can handle a tensile strength of 1000 psi and I wanted to scale this up, how would it scale? I think this has to do with the square cube law maybe, but I'm pretty confused haha. So I guess I'm asking, could a 1 meter tank with a wall thickness of .1m handle the same internal pressure as a 10m tank with a .1m wall thickness?

For a cylindrical tank with "thin" wall (as compared to radius), looking just at the first order pressure load, the equation of relevance is sigma=P*r/t. Where sigma is the stress in the wall, P is the pressure, r is the tank radius, and t is the wall thickness. For the tank to not burst, sigma needs to be less than the ultimate tensile strength of the material. In practice, a factor of safety is applied as well. Per the ASME Boiler code, that factor is 3.5-4 (depending on edition of code). Space applications tend to use 2.

To answer your question, all else being equal, a tank with a 0.1m wall thickness at 1m diameter would be able to handle 10 times the pressure of a 10m tank with the same wall thickness (though, 0.1m is thick enough for the former case that the thin wall assumption is getting dicey).

As an example, if I have a tank made out of Aluminum 7075-T6, operating at 1000psi, with a diameter of 48 inches, then the minimum wall thickness I need is:

t=P*r/sigma

Sigma=84,000 psi http://www.makeitfrom.com/material-properties/7075-T6-Aluminum (http://www.makeitfrom.com/material-properties/7075-T6-Aluminum)
P=1000 psi
r=48/2=24"

t=1000psi*24"/84000=0.286"

Then, if I want a factor of safety of 2, I would build the vessel with a wall of 0.286*2=0.572"
You did a pressure calculation in ACU?  :P
It is important to note that that only takes into consideration the pressure stress and not other kinds of stresses (but the higher the pressure the better it handles anything else). And that the top and bottoms need special calculations later depending on design. BTW, the usual definition of "thin walled" pressure vessel is thickness<1/10 Diameter. So The example that ClaytonBirchenough used was exactly on the limit (which must have some margin in itself).

Thank you both for the replies/comments!

I arbitrarily chose the example above, but I did have "thin" walled tanks in mind. Thanks also for the helpful math strangequark!

With regards to the "other stresses" you mention baldusi, what do you mean? I was starting to imagine that there would be a complex interaction of different stresses like tensile strength and compressive strength. Is there anyway to estimate this interaction and its resulting "felt" stresses? For example, if we use the above 48" diameter aluminum tube/tank used for discussion/calculation purposes, what would the resulting "felt" stress on the tank be if it was accelerating at 4g with a 1000 kg payload on top of it? It's ok if we don't get to the forward closure of the tank, but just assume that the load is dispersed equally around the radius of the cylindrical tank. I also know the math might get a little crazy, but was just wondering if there was some theory/explanation for the "compounding" effects of such forces?

Thank you both again!
It seems the safety factor of 2 would allow for 1000 kg at 4 gees- or 4 tons.
A 48" diameter has 1809 square inches, so the ends when it has 1000 psi would have
1000 times 1809 or 1,809, 000 lbs force.
Circumference is 48" times pi 150.7 ". So 150,700 lbs of force pushing outward [radial direction- sideways].

So that seems to indicate your ends are going to need to be made of stronger material [thicker and/or curved- a dome shape]. So 4000 kg is 8818.5 lbs. So had 8000 lb weight evenly distributed on top of the end, it's essential 1,809,000 minus 8000 lbs of strength needed for that end. Of course it starts with only 2000 lbs, and could have vibration gees which could get into negative. But seems minor considering the safety factor of 2.
Now you said load would be on wall only. The walls are having 1,809, 000 lbs force pulling them apart [both ends] it's about the same difference- again with safety factor of two, it should not be problem.
If weakening the end by welding to it- that could be problem.

Other problems are buckling- and 1000 psi should make that less of problem. If you had safety factor of 1.3 which used with rockets, still unlikely a problem. Could be problem with longer lengths.

Anyhow, here's a calculator, for cylinder walls, pipe:
http://www.aerocomfittings.com/barlows.html

Title: Re: Basic Rocket Science Q & A
Post by: ClaytonBirchenough on 08/02/2015 05:25 pm
Thanks!

I'm still thinking about tank pressures and such, so I might fire back another question soon haha. Is there any theories/natural phenomena that, given a certain material and its strength, would allow you to figure a certain ratio between tank side/wall mass and volume for a cylinder?

Also, is there any published information on volumetric efficiency for solid rocket motors that have actual flown and their grain configuration?

Lots of questions, I know haha.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 08/06/2015 09:14 am
What you said makes sense to me (I think.. heh), but from what I can tell, it doesn't coincide with what's stated in the wikipedia entry (please tell me the wiki is wrong!).

Well the wiki page doesn't even mention sine or cosine once  :o A subpar long-winded essay lacking the core math for actual trajectory to orbit.

Quote
From what I make of your explanation and the diagram from Sutton, the vertical component consists of the sum of mg and D sin θ.

Forgot to say but the D is atmospheric drag, the diagram has all the forces. The force of gravity drag is m*g*sin θ.

Quote
In other words, if we're thrusting perpendicular to g, we can say there's no gravity drag, but we're still accelerating perpendicular to the direction of thrust at the rate of g, right?

Correct (as long as direction of thrust is also direction of velocity). When gravity is perpendicular to velocity it causes neither acceleration nor deceleration but only turns the direction of flight. Applies also to thrusting, for example orbital plane change is achieved by perpendicular thrusting.
Title: Re: Basic Rocket Science Q & A
Post by: Hoonte on 08/29/2015 09:33 pm
What was the largest mass launched into low earth orbit in a single launch?
Title: Re: Basic Rocket Science Q & A
Post by: AS-503 on 08/29/2015 09:37 pm
IIRC the Apollo 17 stack comprised of partially spent SIVb with LEM and CSM.
Title: Re: Basic Rocket Science Q & A
Post by: Bob Shaw on 08/29/2015 09:54 pm
The *largest* object launched into orbit was the Skylab workshop, the Saturn V SII stage which accompanied it, and the inadvertently orbited Interstage. Actual mass to orbit is actually unknown, as some of the SII-to-workshop fairing appears to have fallen away after being punctured by debris from the untimely deployment of the workshop meteoroid shield. After attaining orbit one of the semi-deployed workshop solar arrays was blown off the workshop by the SII separation motors - the other was held in place by a strap of debris.

Does anyone have a reasonable estimate of the mass of this enormous object?
Title: Re: Basic Rocket Science Q & A
Post by: Burninate on 08/29/2015 11:12 pm
In terms of LEO *capabilities*:
173T: Energia was capable of launching 88T of payload atop an 85T core stage.  Supposedly.
131.3T: Saturn V was capable of launching 118T of payload atop a 13.3T upper stage.
123T: SSTS was capable of launching 25T of payload inside a 98T upper stage.
105T: Energia-Buran was capable of launching 30T of payload inside a 75T upper stage.
Title: Re: Basic Rocket Science Q & A
Post by: edkyle99 on 08/30/2015 01:38 am
What was the largest mass launched into low earth orbit in a single launch?
The SA-513/Skylab 1 Flight Evaluation Report lists the mass actually inserted into orbit at 147,363 kg, including 88,474 kg for Skylab 1 itself.

The similar report for SA-512/Apollo 17 lists 140,893 kg as the mass inserted into the initial parking orbit.  That included 90,257 kg for the partially used S-IVB stage and 2,027 kg for the Instrument Unit.

 - Ed Kyle
Title: Re: Basic Rocket Science Q & A
Post by: deaville on 08/30/2015 05:56 am
The heaviest spacecraft from the Apollo era is generally reckoned to be Apollo 15 at 52819.5 kilos. This is just for the CSM/LM that might be considered to be simply the 'payload'.
The heaviest in lunar orbit is Apollo 16 at 34523.1 kilos.
At the other end of the scale of manned (personned to be strictly pc  :P) spacecraft is Liberty Bell 7 at 1286.4 kilos.
Title: Re: Basic Rocket Science Q & A
Post by: nadreck on 08/30/2015 08:22 am

At the other end of the scale of manned (personned to be strictly pc  :P) spacecraft is Liberty Bell 7 at 1286.4 kilos.

Crewed is a better term and even more pc
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 08/30/2015 12:09 pm
The heaviest spacecraft from the Apollo era is generally reckoned to be Apollo 15 at 52819.5 kilos.

no, that would be Skylab then.
Title: Re: Basic Rocket Science Q & A
Post by: Bob Shaw on 09/28/2015 07:31 pm
What was the largest mass launched into low earth orbit in a single launch?
The SA-513/Skylab 1 Flight Evaluation Report lists the mass actually inserted into orbit at 147,363 kg, including 88,474 kg for Skylab 1 itself.

The similar report for SA-512/Apollo 17 lists 140,893 kg as the mass inserted into the initial parking orbit.  That included 90,257 kg for the partially used S-IVB stage and 2,027 kg for the Instrument Unit.

 - Ed Kyle

The quoted figures for Skylab may not take account of either the bits which went into orbit that shouldn't, or the bits that should have gone into orbit, but didn't. At least we can make a judgement regarding the *length* of the object which made it into orbit (presumably the length would be taken as prior to the ATM deploying, when it was still above the MDA). Certainly, it was the 'biggest', and from Ed's figures probably the heaviest item to get up there!
Title: Re: Basic Rocket Science Q & A
Post by: ScepticMatt on 11/02/2015 07:24 am
Math question regarding convergent-divergent nozzles:

I understand that  area expansion during subsonic will slow the exhaust gas while it will increase it in the supersonic region.

i.e. -( 1-M^2 )dV/V = dA/A

But how does it break the sound barrier? During sonic chocked flow (M=1), how and why will changing the nozzle area affect exhaust gas velocity?
Title: Re: Basic Rocket Science Q & A
Post by: SWGlassPit on 11/02/2015 04:19 pm
Changing the nozzle area changes the combustion chamber pressure.  If the pressure upstream is too low, a shock wave will form at some point in the nozzle resulting in subsonic exit flow (and probably causing a fair amount of harm to the nozzle itself).  A wider nozzle throat results in lower upstream pressure.
Title: Re: Basic Rocket Science Q & A
Post by: Dante80 on 11/02/2015 05:08 pm
Theoretically speaking, a H2/Be mixture with LOX oxidizer is shown here (https://en.wikipedia.org/wiki/Liquid_rocket_propellant) to give abnormally large Isp figures (something like 540 Isp vac possible).

Has something like this been tried in the past? Does it make sense for LV usage?

Another one shown is H2/Li with Fluorine as the oxidizer. Again, does it make sense?

Title: Re: Basic Rocket Science Q & A
Post by: Jim on 11/02/2015 05:19 pm
Theoretically speaking, a H2/Be mixture with LOX oxidizer is shown here (https://en.wikipedia.org/wiki/Liquid_rocket_propellant) to give abnormally large Isp figures (something like 540 Isp vac possible).

Has something like this been tried in the past? Does it make sense for LV usage?

Another one shown is H2/Li with Fluorine as the oxidizer. Again, does it make sense?


There is a thread on fluorine and it is not a good idea to use it.
Title: Re: Basic Rocket Science Q & A
Post by: ScepticMatt on 11/02/2015 08:21 pm
If the pressure upstream is too low, a shock wave will form at some point in the nozzle resulting in subsonic exit flow.
Interesting. Would you mind going a bit more into the detail as for the conditions needed for this to happen (or to avoid it)?
Title: Re: Basic Rocket Science Q & A
Post by: Dante80 on 11/03/2015 03:36 am
There is a thread on fluorine and it is not a good idea to use it.

Thanks for the hint. It got me to this (http://www.amazon.com/Ignition-informal-history-liquid-propellants/dp/0813507251), and I'm really enjoying reading it. Cheers..C:

a random excerpt..
Quote
(....)  But then Pino, in 1949, made a discovery that can fairly be described as revolting. He discovered that butyl mercaptan was very rapidly hypergolic with mixed acid. This naturally delighted Standard of California, whose crudes contained large quantities of mercaptans and sulfides which had to be removed in order to make their gasoline socially acceptable. So they had drums and drums of mixed butyl mercaptans, and no use for it. If they could only sell it for rocket fuel life would indeed be beautiful.

Well, it had two virtues, or maybe three. It was hypergolic with mixed acid, and it had a rather high density for a fuel. And it wasn't corrosive. But its performance was below that of a straight hydrocarbon, and its odor — ! Well, its odor was something to consider.

Intense, pervasive and penetrating, and resembling the stink of an enraged skunk, but surpassing, by far, the best efforts of the most vigorous specimen of Mephitis mephitis. It also clings to the clothes and the skin. But rocketeers are a hardy breed, and the stuff was duly and successfully fired, although it is rumored that certain rocket mechanics were excluded from their car pools and had to run behind. Ten years after it was fired at the Naval Air Rocket Test Station — NARTS — the odor was still noticeable around the test areas.

California Research had an extremely posh laboratory at Richmond, on San Francisco Bay, and that was where Pino started his investigations. But when he started working on the mercaptans, he and his accomplices were exiled to a wooden shack out in the boondocks at least two hundred yards from the main building.

Undeterred and unrepentant, he continued his noisome endeavors, but it is very much worth noting that their emphasis had changed. His next candidates were not petroleum by-products, nor were they chemicals which were commercially available. They were synthesized by his own crew, specifically for fuels. Here, at the very beginning of the 50's, the chemists started taking over from the engineers, synthesizing nc:w propellants (which were frequently entirely new compounds) to order, instead of being content with items off the shelf.

Anyhow, he came up with the ethyl mercaptal of acetaldehyde and the ethyl mercaptol of acetone, with the skeleton structures:

(http://i.imgur.com/LEAUGHw.png)

respectively. The odor of these was not so much skunk-like as garlicky, the epitome and concentrate of all the back doors of all the bad Greek restaurants in all the world.

And finally he surpassed himself with something that had a dimethylamino group attached to a mercaptan sulfur, and whose odor can't, with all the resources of the English language, even be described. It also drew flies. This was too much, even for Pino and his unregenerate crew, and they banished it to a hole in the ground another two hundred yards farther out into the tule marshes.

Some months later, in the dead of night, they surreptitiously consigned it to the bottom of San Francisco Bay...
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 11/03/2015 10:05 am
If the pressure upstream is too low, a shock wave will form at some point in the nozzle resulting in subsonic exit flow.
Interesting. Would you mind going a bit more into the detail as for the conditions needed for this to happen (or to avoid it)?

In the case of an ideal gas flowing through an ideal nozzle, the pressure drops by a factor of [(k + 1) / 2]k/(k - 1) from the nozzle entrance to the throat, k being the ratio of the specific heat at constant pressure to that at constant volume (see a textbook like Sutton's Rocket Propulsion Elements).  If the ambient pressure is higher than the pressure at the throat (the nozzle exit in a converging-only nozzle), you won't get supersonic flow.  A representative value of k is 1.2, so the chamber pressure needs to be at least 1.8 times ambient.  In a real nozzle with a non-ideal gas, the pressure would need to be a little higher.

EDIT: Clarified pressure condition, prompted by deltaV's comment (https://forum.nasaspaceflight.com/index.php?topic=13543.msg1442488#msg1442488), below]
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 11/05/2015 05:10 am
If the ambient pressure is higher than the nozzle exit pressure, you won't get supersonic flow.

Sshhhhhhh. If SSME hears you it'll stop working at sea level (its nozzle exit pressure (https://en.wikipedia.org/wiki/Space_Shuttle_main_engine#Nozzle) is much below atmospheric) and SLS will have to find a new engine.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 11/05/2015 11:17 am
My reference to nozzle exit pressure was in the context of a converging-only nozzle, which makes sense in the context of ScepticMatt's question.  I absolutely guarantee that the pressure at minimum cross-section of the SSME (ie, the throat) is much greater than atmospheric.
Title: Re: Basic Rocket Science Q & A
Post by: Christian La Fleur on 11/17/2015 03:57 pm
Is it possible to use centrifugal force/energy, to subdue, separate, control and expell atoms in a desired direction/path?
Title: Re: Basic Rocket Science Q & A
Post by: deltaV on 11/19/2015 04:41 am
Is it possible to use centrifugal force/energy, to subdue, separate, control and expell atoms in a desired direction/path?
Your question is a bit vague, but a centrifugal pump (https://en.wikipedia.org/wiki/Centrifugal_pump) connected to a nozzle sort of fits that description.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 11/25/2015 08:36 pm
It's been mentioned that SpaceX subcools first stage propellants on the Falcon 9 and that Antares subcools (or subcooled -- I don't know whether the new version will do the same) lox from 94 K to 78 K (https://forum.nasaspaceflight.com/index.php?topic=33598.msg1170038#msg1170038).

Do we know how much SpaceX subcools either propellant?  Or how much subcooling is used on any other launch vehicles?
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 11/26/2015 07:44 pm
Antares 100 cores (the ones with the NK-33) used subcooled LOX because it was required by the NK-33. Dr. Elias has stated here that he desired to do away with it, but given that the 200 cores are re-engined 100 cores, if they didn't subcool they will loss propellant mass. Given that the 300 cores (the ones designed for the RD-181) have increased propellant load, one would assume that they will keep using the subcooled LOX until the 300 cores.
Title: Re: Basic Rocket Science Q & A
Post by: whitelancer64 on 12/07/2015 09:48 pm
Bumbling about the internet today, I saw someone say that the Antares and the Soyuz 2-1v used LOX at different temperatures (even though they use the same base NK-33 rocket engine), which made me wonder how much of a density difference the engine can handle.

Soyuz 2-1v (-192°C)
Antares (–196°C)

However, I could not find a density chart for liquid oxygen at different temperatures. I had to dig up an online calculator to work this out and I'm not certain of its accuracy. Can anyone help me out with this?

LOX at -192 C = 736.62 kg/m3
LOX at -196 C = 979.81 kg/m3

Is this correct? It seems like an enormous leap in density.
Title: Re: Basic Rocket Science Q & A
Post by: Burninate on 12/08/2015 01:02 am
First reference found:

http://www.engineeringtoolbox.com/cubical-expansion-coefficients-d_1262.html

Does not list oxygen, but the highest value listed for a liquid would offer about a 1% volume increase from a 4K rise in temperature.  The lowest, water, is nearly negligible.

I'm going to bet that your references are based on differing assumptions.
Title: Re: Basic Rocket Science Q & A
Post by: Burninate on 12/08/2015 01:06 am
http://www.apithailand.com/data.html

lists

"Coefficient of thermal Expansion of Liquid @ 25 C   0.00954 1/C"

which would indicate just under 4% expansion from 4K increase.

Liquid oxygen at 25C is under extreme high pressure and may not be representative of cryogenic liquid oxygen: coefficient of thermal expansion in real substances is not perfectly fixed over all temperatures.
Title: Re: Basic Rocket Science Q & A
Post by: Burninate on 12/08/2015 01:15 am
http://encyclopedia.airliquide.com/encyclopedia.asp?LanguageID=11&GasID=48

lists, just to muddy the waters

Liquid density (1.013 bar at boiling point) : 1141.2 kg/m3
Title: Re: Basic Rocket Science Q & A
Post by: Burninate on 12/08/2015 01:37 am
An exhaustive look from probably the best reference you'll find:

http://www.nist.gov/srd/upload/jpcrd423.pdf

gives densities in mol/dm^3 for a variety of pressures.

You'll probably want to pay attention to the 0.2Mpa or 0.3MPa isobar, or 30 to 45 psi, which seems to be the typical range for a launch vehicle tank.  1 atmosphere ~= 0.1MPa.

-192C is 81.15K
-196C is 77.15K
Title: Re: Basic Rocket Science Q & A
Post by: strangequark on 12/08/2015 01:48 am
The NIST Webbook database is the best public resource I've found. Like Burninate indicates, anything NIST is pretty much gold, this is just a little more interactive.

NIST Webbook (http://webbook.nist.gov/chemistry/fluid/)

1204 to 1185 kg/m3 for that shift, or a 1.6% difference. There's some small pressure dependence, but it's negligible at these temperatures and realistic tank pressures (~.1 kg/m3 difference for 30 vs 50 psia).
Title: Re: Basic Rocket Science Q & A
Post by: whitelancer64 on 12/08/2015 03:09 pm
Thank you both! I'm pretty sure now that I was not using the online calculator I had found correctly.

strangequark, that page is exactly what I really needed, thank you!

A 1.6% difference in density makes a lot more sense and that must be well within the tolerances of the NK-33 engine.

For future reference, anyone discussing the merits of LOX densification, etc.

LOX at -184 C = 1141.7 kg/m3 (just below boiling point)
LOX at -192 C = 1185.4 kg/m3
LOX at -196 C = 1204.3 kg/m3
LOX at -210 C = 1268.4 kg/m3
LOX at -217 C = 1299.0 kg/m3 (just above freezing point)
Title: Re: Basic Rocket Science Q & A
Post by: Burninate on 12/08/2015 08:34 pm
Thank you both! I'm pretty sure now that I was not using the online calculator I had found correctly.

strangequark, that page is exactly what I really needed, thank you!

A 1.6% difference in density makes a lot more sense and that must be well within the tolerances of the NK-33 engine.

For future reference, anyone discussing the merits of LOX densification, etc.

LOX at -184 C = 1141.7 kg/m3 (just below boiling point)
LOX at -192 C = 1185.4 kg/m3
LOX at -196 C = 1204.3 kg/m3
LOX at -210 C = 1268.4 kg/m3
LOX at -217 C = 1299.0 kg/m3 (just above freezing point)
Boiling & freezing points at what pressure though?
Title: Re: Basic Rocket Science Q & A
Post by: whitelancer64 on 12/09/2015 08:46 pm
Thank you both! I'm pretty sure now that I was not using the online calculator I had found correctly.

strangequark, that page is exactly what I really needed, thank you!

A 1.6% difference in density makes a lot more sense and that must be well within the tolerances of the NK-33 engine.

For future reference, anyone discussing the merits of LOX densification, etc.

LOX at -184 C = 1141.7 kg/m3 (just below boiling point)
LOX at -192 C = 1185.4 kg/m3
LOX at -196 C = 1204.3 kg/m3
LOX at -210 C = 1268.4 kg/m3
LOX at -217 C = 1299.0 kg/m3 (just above freezing point)
Boiling & freezing points at what pressure though?

Oh, right, the boiling / freezing point also changes due to pressure. I should have mentioned in my previous post that each of the density numbers I listed is at 40 psi.

The calculator on the NIST page doesn't allow you to set temperatures less than -218.789 C, which is the freezing point of Oxygen at sea level air pressure (1 atm or 101.325 kPa). -182.96 C is the the boiling point at sea level air pressure. I just rounded up and down to the nearest degree C to get to just above and just below those points, my intention being to establish upper and lower boundaries for the density information.

It's safe to say that those two data points are not precisely accurate in regards to being just above or just below the freezing / boiling points at 40 psi, and I'm not sure how to figure it out, as I'm quite far outside of my knowledge base here, which is why I asked about it in the first place :D
Title: Re: Basic Rocket Science Q & A
Post by: Dante80 on 01/04/2016 03:51 pm
Here is a question.

Falcon9 has done three GTO missions, where the end orbit was super-synchronous.

12/03/13 Falcon 9 v1.1 F9-7  SES 8                 3.183 CC 40 295x80000x20.8 GTO+
01/06/14 Falcon 9 v1.1 F9-8  Thaicom 6             3.016 CC 40 295x90000x22.5 GTO+
03/02/15 Falcon 9 v1.1 F9-16 Eutelsat 115WB/ABS 3A 4.159 CC 40 400x63300x24.8 GTO+


If I understand correctly, the reason for that is that it makes circularizing to GEO for the payload easier.

I want to know how I can convert the above stats to a - X m/s equivalent. In other words, I want to find out how much dv the payloads had to use to get to their end orbits.

How can I do that?
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 01/04/2016 04:25 pm
Here is a question.

Falcon9 has done three GTO missions, where the end orbit was super-synchronous.

12/03/13 Falcon 9 v1.1 F9-7  SES 8                 3.183 CC 40 295x80000x20.8 GTO+
01/06/14 Falcon 9 v1.1 F9-8  Thaicom 6             3.016 CC 40 295x90000x22.5 GTO+
03/02/15 Falcon 9 v1.1 F9-16 Eutelsat 115WB/ABS 3A 4.159 CC 40 400x63300x24.8 GTO+


If I understand correctly, the reason for that is that it makes circularizing to GEO for the payload easier.

I want to know how I can convert the above stats to a - X m/s equivalent. In other words, I want to find out how much dv the payloads had to use to get to their end orbits.

How can I do that?
If you assume two burns from a super synchronous GTO, you can use my spreadsheet (it is done in LibreOffice). But I guess you could use it for sub synchronous, too. Normal GTO usually have a single burn and thus I'm not sure it will work. It should since it takes the delta-v requirements of two burns which should be the same. But I'm not sure.

BTW:
12/03/13 Falcon 9 v1.1 F9-7  SES 8                 3.183 CC 40 295x80000x20.8 GTO+ -> 1,511m/s
01/06/14 Falcon 9 v1.1 F9-8  Thaicom 6             3.016 CC 40 295x90000x22.5 GTO+ -> 1,504m/s
03/02/15 Falcon 9 v1.1 F9-16 Eutelsat 115WB/ABS 3A 4.159 CC 40 400x63300x24.8 GTO+ -> 1,594m/s
Title: Re: Basic Rocket Science Q & A
Post by: Dante80 on 01/04/2016 04:44 pm
Here is a question.

Falcon9 has done three GTO missions, where the end orbit was super-synchronous.

12/03/13 Falcon 9 v1.1 F9-7  SES 8                 3.183 CC 40 295x80000x20.8 GTO+
01/06/14 Falcon 9 v1.1 F9-8  Thaicom 6             3.016 CC 40 295x90000x22.5 GTO+
03/02/15 Falcon 9 v1.1 F9-16 Eutelsat 115WB/ABS 3A 4.159 CC 40 400x63300x24.8 GTO+


If I understand correctly, the reason for that is that it makes circularizing to GEO for the payload easier.

I want to know how I can convert the above stats to a - X m/s equivalent. In other words, I want to find out how much dv the payloads had to use to get to their end orbits.

How can I do that?
If you assume two burns from a super synchronous GTO, you can use my spreadsheet (it is done in LibreOffice). But I guess you could use it for sub synchronous, too. Normal GTO usually have a single burn and thus I'm not sure it will work. It should since it takes the delta-v requirements of two burns which should be the same. But I'm not sure.

BTW:
12/03/13 Falcon 9 v1.1 F9-7  SES 8                 3.183 CC 40 295x80000x20.8 GTO+ -> 1,511m/s
01/06/14 Falcon 9 v1.1 F9-8  Thaicom 6             3.016 CC 40 295x90000x22.5 GTO+ -> 1,504m/s
03/02/15 Falcon 9 v1.1 F9-16 Eutelsat 115WB/ABS 3A 4.159 CC 40 400x63300x24.8 GTO+ -> 1,594m/s


Many thanks for that. I had excel, tried libre online, didn't work, installed it, saved as xls, uninstalled libre, restarted the PC, opened FF and...saw that you provided the numbers..T_T

Now, I'm thinking of uninstalling office and getting libre..XD

So...judging by the fact that S2 was burned to depletion, the Thaicom 6 mission gives us an estimate of 3,000kg to a GTO-1500 (equivalent) orbit for the v1.1 maxed out.
Lets see how the FT version does in SES9..C:
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 01/05/2016 02:07 pm
Well, if its true that they have 30% more performance, 3.9tonnes should be doable and 4 tonnes may be doable. But this is just from the first launches. They stated they had some performance padding. In fact, NLS II states that the F9 v1.1 could do 5.6 tonnes to a 28.5deg GTO (1,804m/s).. And a 20.2Deg x 88,000km (1,500m/s) NLS II says 3,855kg.
If those numbers are true, then FT could do 5 tonnes to a 1,500m/s GTO!
Title: Re: Basic Rocket Science Q & A
Post by: Dejv on 01/07/2016 05:55 pm
Hi guys,

I am preparing for spaceflight mechanics exam and I am a little bit stuck with these questions. (see attach)

In the first question if we assume the perturbation is -1.7x10^-3 deg/day , and the burn is every 1000s (i.e. 86.4 times a day), then the maximal deviation from non-perturbed motion should be -1.97x10^-5 deg. I don't get it why there is the delta v then... is the slot calculated some other way (should it be in seconds..)?

In the second question, the perturbance is easily calculated, but i dont really know how to connect it with the Tsiolkovsky equation.

Thanks for any help!
(I am not aerospace/spaceflight student, I took this course only from pure interest)

Title: Re: Basic Rocket Science Q & A
Post by: Stan-1967 on 01/07/2016 06:23 pm
1.  Can you drop the class

2.  The second part seems to be describing the pertubation from the ideal orbit with the given parameters.  So it seems like you need to set the rocket equation for deltaV equal to the amount of deltaV caused by the pertubation.

I may help you to know that in the rocket equation, exhaust velocity can be replaced by ISP x Gravity ( 4500 x 9.807 m/s^2) ln(mo/m1) and then solve for m1

Title: Re: Basic Rocket Science Q & A
Post by: Dejv on 01/07/2016 07:12 pm
1.  Can you drop the class

2.  The second part seems to be describing the pertubation from the ideal orbit with the given parameters.  So it seems like you need to set the rocket equation for deltaV equal to the amount of deltaV caused by the pertubation.

I may help you to know that in the rocket equation, exhaust velocity can be replaced by ISP x Gravity ( 4500 x 9.807 m/s^2) ln(mo/m1) and then solve for m1

1. no, i cant

2. well the perturbation is in rad/s, as i know the semimajor axis i can convert that to m/s. so i just pop that into rocket equation and thats it? shouldnt be the time somehow included there as well?

Yeah I know exh. velocity = g x Isp
Title: Re: Basic Rocket Science Q & A
Post by: meetsitaram on 01/25/2016 02:47 pm
Hi Guys, can you suggest a good book that explains basics of rockets and space flight? I was searching on google and found:
Quote
Astronautics: The Physics of Space Flight - Ulrich Walter
Rocket and Spacecraft Propulsion - Martin J. L. Turner
Title: Re: Basic Rocket Science Q & A
Post by: Zaum on 01/25/2016 07:58 pm
Rocket Propulsion Elements - George Sutton, Oscar Biblarz

Covers the basics and a bit more. Very comprehensive.
You can probably see a preview on google books to check if it has what you're looking for.
Title: Re: Basic Rocket Science Q & A
Post by: nicp on 02/21/2016 01:37 pm
I have a really boring question.  ;)
Obviously liquid propellant rockets have valves. Valves for the pre-burner and for the main combustion chamber and a load of others.
I don't think I've ever seen a picture of a main oxidizer or fuel propellant - or perhaps I didn't recognize it.
I mean, take the F-1 - that's got to have some seriously meaty propellant valves. With bloody enormous actuators.

Or perhaps more interestingly, take the original Atlas. Those booster engines are gonna get dropped off, so you have a separation plane too.

So what kind of valves are they - obviously they need to be lightweight, but have to take a very high flow rate, etc..

Just my weird curiosity.. If this is covered elsewhere, apologies, I haven't noticed anything appropriate..
Title: Re: Basic Rocket Science Q & A
Post by: AnalogMan on 02/21/2016 02:16 pm
I have a really boring question.  ;)
Obviously liquid propellant rockets have valves. Valves for the pre-burner and for the main combustion chamber and a load of others.
I don't think I've ever seen a picture of a main oxidizer or fuel propellant - or perhaps I didn't recognize it.
I mean, take the F-1 - that's got to have some seriously meaty propellant valves. With bloody enormous actuators.

Or perhaps more interestingly, take the original Atlas. Those booster engines are gonna get dropped off, so you have a separation plane too.

So what kind of valves are they - obviously they need to be lightweight, but have to take a very high flow rate, etc..

Just my weird curiosity.. If this is covered elsewhere, apologies, I haven't noticed anything appropriate..

Just to give you some idea - here are a couple of slides showing the Main Fuel Valve for the shuttle SSME engine.
Title: Re: Basic Rocket Science Q & A
Post by: mfck on 02/21/2016 09:59 pm
Hi,

I have a Mechanical Engineering (probably) question

It is my understanding, that in most LVs and spacecraft pressure vessels are also a main structural element, due to mass-efficiency considerations. My question is whether there's a physical limit to scaling of such design before mass-efficiency dictates main structural element be external to the pressure vessel, or in other words, how big would your pressure vessel need to be, assuming habitable temperature, pressure and max load (5g?), before you have to unload to external structure to stay mass-efficient?

I am not sure it's even a valid question. Maybe a constraint should be no exotic materials as well?

Thanks
Title: Re: Basic Rocket Science Q & A
Post by: sdsds on 02/22/2016 04:32 am
how big would your pressure vessel need to be [...] before you have to unload to external structure to stay mass-efficient?

I have no answer, but like your question! Is it possible the honeycomb isogrid or orthogrid integrated into many pressure vessel walls performs a function like what you describe? Does it matter if that structure is inside or outside the pressure vessel wall?

A completely unrelated question of my own: does the "residence time" requirement (in the sense of the characteristic length) for a combustion chamber change dramatically when the propellants are being injected as gases, e.g. in a full-flow staged combustion design?
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 02/22/2016 04:43 pm
I have a really boring question.  ;)

(valves)

Not a boring question at all. AnalogMan already gave an example of ball valve. Other types used in rocket engines are at least poppet and butterfly valves. Here's a diagram of F-1 main LOX valve (the engine had two of them and also two similar main fuel valves). They are usually actuated using high pressure pneumatics or hydraulics.

(http://heroicrelics.org/info/f-1/f-1-main-lox-valve/oxidizer-valve-sm.jpg)

More here: http://heroicrelics.org/info/f-1/f-1-main-lox-valve.html
Title: Re: Basic Rocket Science Q & A
Post by: Remes on 02/22/2016 11:06 pm
With bloody enormous actuators.

E.g. fuel Isolation valves are relatively simple. They only fully open or fully close. Most often they are driven pneumatically. Pressurized Helium drives a piston, the piston drives a rack/pinion which in turn runs the valve itself. Force equals pressure times area. So a big enough pneumatic cylinder gives plenty of power. A spring closes the valve.

The ball valve shown from the Space Shuttle is hydraulically driven (with a pneumatic emergency shutdown alternative system). Hydraulic systems work on higher pressures and have a much higher power density. So the Actuator housing must be more robust, but is still smaller than for pneumatic actuation. On L2 there is a manual for SSME, showing lots of details.

Modern engines more often get electric actuators. E.g. several motors are driving the ball type or poppet type valve. Most often they have an emergency pneumatic backup for shutdown. The actuators have a lower power density. But they are easier to install, to test, etc. as keeping the hydraulic system clean is not that easy.
Title: Re: Basic Rocket Science Q & A
Post by: Remes on 02/22/2016 11:29 pm
Does it matter if that structure is inside or outside the pressure vessel wall?
My best guess: If the grids go outwards, the distance between the grid corners become bigger. If they go inwards, they are becoming smaller and therefore provide more stiffness.

Here is a nice video about a buckling test.
https://youtu.be/nUjpVBktTAI?t=203

Also inward pointing grids allow a easier attachment of sensors, pressure vessels, etc.
Title: Re: Basic Rocket Science Q & A
Post by: R7 on 02/27/2016 12:16 pm
A completely unrelated question of my own: does the "residence time" requirement (in the sense of the characteristic length) for a combustion chamber change dramatically when the propellants are being injected as gases, e.g. in a full-flow staged combustion design?

For instance RD-170/RD-180 combustion chamber has characteristic length of 1079.6mm. I would not rate that as dramatic reduction. Gasification is just step one on the complex chain of combustion.

http://lpre.de/energomash/RD-170/index.htm
Title: Re: Basic Rocket Science Q & A
Post by: Oli on 03/05/2016 03:25 pm

I find it curious that so many rockets do parallel staging. Apart from ground-starting the engines on the core there don't seem to be any obvious advantages. You save some thrust on the first stage (often little because the core uses hydrolox) but you lose some performance (more dry mass at separation, lower engine ISP).

Anything else?
Title: Re: Basic Rocket Science Q & A
Post by: Remes on 03/11/2016 09:16 pm

I find it curious that so many rockets do parallel staging. Apart from ground-starting the engines on the core there don't seem to be any obvious advantages. You save some thrust on the first stage (often little because the core uses hydrolox) but you lose some performance (more dry mass at separation, lower engine ISP).

Anything else?
Ariane 5 was developed to be manrated, so the reliability and abort capability of ground start plus  an opportunity to test the liquid propulsion comes in handy. Same goes for Angara 5.

Soyuz is heritage. Delta 4 requires a lot of helium in order to spin up the pumps. On Atlas 5 you save some tvc for the booster, but aside from that the rd-180 has a significant amount of thrust.

Ariane 5 has a blow down hydraulic tvc system, the Falcon 9 and heavy would need some hydraulic source (like e.g. the APU from the shuttle). (Upper stage for Ariane 5, Atlas 5 and Delta 4 is different, it's electric. And even the Europa rocket could live with a battery driven electrohydrostatic system).
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 03/12/2016 04:24 pm
I'd have thought that in the cases Angara, Delta 4 and Atlas 5 (as well as Falcon and the old Titan, for that matter), the attraction of parallel staging is simply the modularity it provides.  It allows one basic design to provide a range of capabilities, depending on what's needed for any particular mission (although it didn't work out so well for the Delta IV, since performance shortfalls have meant that it has several substantially different cores).

Title: Re: Basic Rocket Science Q & A
Post by: sugmullun on 03/13/2016 04:11 pm
I have been assuming that the main advantage was less gravity loss early on because
of the T/W, (for a given total "installed thrust?") of this arrangement, as getting through the vertical component quickly is
always preferable.

wrong?
Title: Re: Basic Rocket Science Q & A
Post by: Remes on 03/13/2016 06:51 pm
the main advantage

the attraction of parallel staging is simply [...]

I think in rocketry as in all other complex systems there is not one reason which is used to make a decision. It's rather a trade off between performance, cost, reliability, development risks, producibility, repeatability, operations, transportability, ground support, etc. Then in addition to technical reasons there are sometimes political issues (who gets to build the thing, who has the biggest power and influence). Most often there is not the one optimum solution, it's rather a set of several solutions and finally (and hopefully) one path is chosen. Decisions are often (on engineering level) influenced by the most influential people. And I believe that every engineer is somehow biased (in good and bad ways) by his/her experience. Then there is heritage: do we use what we have or do we make something better from scratch...

less gravity loss early on because of the T/W, (for a given total "installed thrust?") of this arrangement, as getting through the vertical component quickly is always preferable.
Tradeoff between getting quickly through the vertical component against (e.g., numbers guessed) having a 90% full lox/lh2 tank on an Ariane 5 when staging. Same for D4H. I guess the gravity loss is less influential on performance than having an optimum mass fraction for the center core (e.g. tanks full) after staging.

There is a long discussion on cross feed for F9H.

(And Ariane 5 is a bad example, as in the beginning the thrust/weight ratio is anyway small.)
Title: Re: Basic Rocket Science Q & A
Post by: sugmullun on 03/14/2016 11:33 am
After Googling around I've decided that my question was too simplistic for an inherently complicated problem, so I removed it. 
It seems that parallel staged boosters would be somehow closer to a magical ssto rocket, that sheds dry mass along with propellant, than stacked stages are.

Title: Re: Basic Rocket Science Q & A
Post by: Hirox on 05/17/2016 09:54 pm
Are rockets designed with the intent of using solid rocket boosters from the beginning?

My assumption was that solids are expensive and are designed only to be used on rare heavy payloads when the rocket needs an extra boost from the pad. That way the standard rocket don't have excesses capacity i.e. lower cost. How wrong am I?
Title: Re: Basic Rocket Science Q & A
Post by: shooter6947 on 05/17/2016 10:10 pm
Are rockets designed with the intent of using solid rocket boosters from the beginning?

My assumption was that solids are expensive and are designed only to be used on rare heavy payloads when the rocket needs an extra boost from the pad. That way the standard rocket don't have excesses capacity i.e. lower cost. How wrong am I?

Depends on the rocket.  The Space Shuttle and Delta2 both require solids to leave the pad (they're TWR<1 without).  Delta4 and AtlasV, for instance, can optionally use solids if they need extra oomph.
Title: Re: Basic Rocket Science Q & A
Post by: Hirox on 05/17/2016 10:37 pm
Are rockets designed with the intent of using solid rocket boosters from the beginning?

My assumption was that solids are expensive and are designed only to be used on rare heavy payloads when the rocket needs an extra boost from the pad. That way the standard rocket don't have excesses capacity i.e. lower cost. How wrong am I?

Depends on the rocket.  The Space Shuttle and Delta2 both require solids to leave the pad (they're TWR<1 without).  Delta4 and AtlasV, for instance, can optionally use solids if they need extra oomph.


When the Atlas V rocket was designed, why did they decide to use 2 rd180 instead of 3? An extra engine would remove the need for solids which in turn could reduce their cost by removing a supplier and increasing their production of the engine bringing the cost down. Does one extra engine cost significantly more than the occasional solid rocket booster?
I've been thinking of the falcon 9 and their "one size fits all" approach and been wondering if, for example, the falcon 9 rocket never was designed and they went with the Falcon 5 instead, would it make economic sense for them to occasionally use solid rocket boosters?
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 05/17/2016 11:07 pm

When the Atlas V rocket was designed, why did they decide to use 2 rd180 instead of 3? An extra engine would remove the need for solids which in turn could reduce their cost by removing a supplier and increasing their production of the engine bringing the cost down. Does one extra engine cost significantly more than the occasional solid rocket booster?
I've been thinking of the falcon 9 and their "one size fits all" approach and been wondering if, for example, the falcon 9 rocket never was designed and they went with the Falcon 5 instead, would it make economic sense for them to occasionally use solid rocket boosters?

RD-180 is one engine with two nozzles

The EELV program Medium (single core) vehicle basic requirement was to place 10klb into GTO.  Atlas V did this with the RD-180 and latest version of the single engine Centaur.  The EELV program Heavy (3 cores) vehicle basic requirement was to place 10klb into GSO. The Atlas V Heavy (which was designed but not built) did this with 3 RD-180 cores and the same Centaur. 
The need for solids came later as commercial spacecraft grew.  So Atlas V was modified to accept solids after the basic design was finished (that is why it can only carry 5 solids in an unsymmetrical pattern and why the VIF has intermediate levels between floors 1 & 2)
Title: Re: Basic Rocket Science Q & A
Post by: rocx on 05/18/2016 10:55 am
Are rockets designed with the intent of using solid rocket boosters from the beginning?

My assumption was that solids are expensive and are designed only to be used on rare heavy payloads when the rocket needs an extra boost from the pad. That way the standard rocket don't have excesses capacity i.e. lower cost. How wrong am I?

Solid rockets are actually cheaper to produce and operate than liquid stages of similar performance. What also helps is that solids are the best way to launch nuclear bombs at other continents, so most of their development and production investments are already made for the military.
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 05/18/2016 12:28 pm
Are rockets designed with the intent of using solid rocket boosters from the beginning?

My assumption was that solids are expensive and are designed only to be used on rare heavy payloads when the rocket needs an extra boost from the pad. That way the standard rocket don't have excesses capacity i.e. lower cost. How wrong am I?

Solid rockets are actually cheaper to produce and operate than liquid stages of similar performance. What also helps is that solids are the best way to launch nuclear bombs at other continents, so most of their development and production investments are already made for the military.
Not quite. Solids require vertical integration with the LV (Atlas does the whole stack integration vertically, while Delta IV does most horizontally and just the solids are integrated vertically at the pad).
Also, the LV needs to be designed for the extra thrust. It does allows for a "smother" payload/cost curve. Depending on requirements solids might be cheaper or not. In other words, solids were optimal for Atlas V (even if added after the fact) because they had to comply with the EELV requirements. They didn't appear to be optimal for Falcon 9, Antares nor the Russians. For OrbitalATK, since they actually make them, they now want to go all solid. So no straight al generic answers. Each trade results in different optimizations.
Title: Re: Basic Rocket Science Q & A
Post by: jabe on 06/25/2016 01:31 pm
It seems many companies are working on methane engines.  Would a methane 1st stage be ruled out, or is RP1 too good fuel of choice density/performance wise.
Title: Re: Basic Rocket Science Q & A
Post by: rocx on 06/26/2016 08:38 pm
It seems many companies are working on methane engines.  Would a methane 1st stage be ruled out, or is RP1 too good fuel of choice density/performance wise.
http://spacelaunchreport.com/vulcan.html
The successor to the Atlas V and Delta IV rockets is probably going to have a methane first stage.
Title: Re: Basic Rocket Science Q & A
Post by: jabe on 06/26/2016 11:41 pm
It seems many companies are working on methane engines.  Would a methane 1st stage be ruled out, or is RP1 too good fuel of choice density/performance wise.
http://spacelaunchreport.com/vulcan.html
The successor to the Atlas V and Delta IV rockets is probably going to have a methane first stage.
is it lack of maturity of the tech for methane that it isn't used more?
Title: Re: Basic Rocket Science Q & A
Post by: rocx on 06/27/2016 08:01 am
is it lack of maturity of the tech for methane that it isn't used more?
Yes. There is currently no large scale methane engine, upper stage nor first stage. SpaceX is building Raptor for both roles, and Blue Origin is building BE-4 as a first-stage engine. Masten is developing the Cutlass for use in planetary landers. This takes many years to fully develop.
Title: Re: Basic Rocket Science Q & A
Post by: smoliarm on 07/05/2016 06:52 am
Hi everybody,
I have few questions on Earth’s rotation.
Pretty frequently I see statements in press on launch site choice, where low latitude (close to equator) is considered as a simple advantage – because Earth’s rotation contribution is highest at equator. It seems to me, that this is not quite correct:
First, bonus from Earth’s rotation is minor – I estimated it on the order of percent or less.
Second, the most fuel-efficient launch gives an orbit with inclination equal to launch site latitude. In case your target orbit is different, you have to change the orbital plane – and this could cost significantly more fuel, than *Earth rotation savings*.
However, I did not have a proper Orbital Mechanics course, and the one I had, it was 19 years ago. Therefore, I’d like to check my calculations.

So the first question is about the correct numeric estimate for the contribution of Earth’s rotation.
I take as example a launch vehicle Soyuz-like – 305 metric tons at launch, and I assume Earth’s radius = 6371 km. Then I compare launches of this LV from two launch sites – one at 46° N (Baikonur-like) and the other at 5° N (Kourou-like). In both  cases the target orbit is circular at 200 km with the same inclination as launch site latitude.
My result: the launch from Kourou gives ~ 72 kg advantage in payload mass over the launch from Baikonur, of less than one percent.
Is this close enough?

The second question is about inclination change and its hit on payload mass.
Here I take as example Kourou as launch site, Ariane 5ES as LV and two target orbits:
a) ISS orbit (400x400 km at 52°) and
b) a “launch site inclination orbit” – 400x400 km at 5°.
My estimate: in the first case the penalty for orbital plane change will be 15%. Or –
Performance (a) = 0.85 Performance (b)
The same question – is this a close estimate?
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 07/05/2016 08:30 pm
If you measured the instantaneous speed of an orbit projected over the surface of the Earth, you could write it as a vector of speeds on the latitude/longitude axis (save at the poles). The Earth rotation at the Equator gives something like 465m/s. So it adds a (0m/s , 465m/s) vector to your launch.
To reach orbit you need something like 9,500m/s of delta-v. If you launch westward, you are adding exactly that amount. If you angle you launch pitch, you will have the cos of the pitch angle.
The math is that at around 86deg of inclination, that's about what you need on the west direction if you launched from the Equator. Anything more inclined and you start loosing performance from sites like Kourou. That's why sites like Plesetsk are so good for polar launches. But there are many other considerations for launch sites.
Title: Re: Basic Rocket Science Q & A
Post by: Robotbeat on 07/09/2016 02:35 am
You can do significantly less than 9500m/s (i.e. around 9km/s) if you launch due West and have good thrust/weight ratio and a high ballistic coefficient.
Title: Re: Basic Rocket Science Q & A
Post by: mobile1 on 08/07/2016 10:51 pm
How did the Juno space probe reach 165,000 MPH?
Title: Re: Basic Rocket Science Q & A
Post by: joek on 08/07/2016 11:11 pm
How did the Juno space probe reach 165,000 MPH?
Gravity.  Specifically, Jupiter's gravity.
Title: Re: Basic Rocket Science Q & A
Post by: Nomadd on 08/07/2016 11:11 pm
It didn't really. They published it's speed relative to Earth. It was mostly luck that Jupiter and Earth were at a point where they were receding from each other. Add that to Juno's approach speed. If the spacecraft had gotten there at a different time, it's speed relative to Earth could have been much less than 165k. Just figuring the difference between Earth receding or approaching in it's orbit can make a 120,000 mph difference in relative speed.
 I'm not sure what Juno's orbital speed is at closest approach.
Title: Re: Basic Rocket Science Q & A
Post by: Burninate on 08/07/2016 11:21 pm
How did the Juno space probe reach 165,000 MPH?

"The Ferrari hit Arthur at a relative speed of 104mph.  How did Arthur reach 104mph, a speed no human runner has ever attained before?"

Because Arthur was running at 8mph and the Ferrari was speeding at 96mph.  Only Arthur's interaction with the Ferrari propelled him to such a speed.

It's not a perfect analogy because there's no gravity in it, but it's close enough for government work.

When you're talking about the capabilities of a planetary spaceprobe, ask questions like "What was the mission delta V, the sum of the changes in speed from all the burns of its engine?", but that has relatively little to do with current velocity if the mission traverses multiple gravity wells, and current velocity is only even a useful concept relative to some other object or as a time-varying metric at a particular part of an orbit.
Title: Re: Basic Rocket Science Q & A
Post by: joek on 08/08/2016 12:14 am
I'm not sure what Juno's orbital speed is at closest approach.

At perijove, approximately 130K mph Jupiter-relative and approximately 160K mph Earth-relative.  The difference is due to Earth-Jupiter orbital velocity.
Title: Re: Basic Rocket Science Q & A
Post by: eeergo on 08/08/2016 01:44 am
My references for this issue are jcm's authoritative analysis (http://planet4589.org/space/jsr/back/news.728.txt) and the very complete and didactic report Daniel Marin compiled a couple of weeks ago (http://danielmarin.naukas.com/2016/07/09/cual-ha-sido-la-nave-espacial-mas-rapida/ (in Spanish))

Title: Re: Basic Rocket Science Q & A
Post by: Nomadd on 08/08/2016 02:52 am
I'm not sure what Juno's orbital speed is at closest approach.

At perijove, approximately 130K mph Jupiter-relative and approximately 160K mph Earth-relative.  The difference is due to Earth-Jupiter orbital velocity.
So. the Earth relative speed could have come in at something like 65K to 195K mph depending on orbital positions.
 I think the Jupiter relative speed might have been a more honest one to report.
Title: Re: Basic Rocket Science Q & A
Post by: gin455res on 09/30/2016 09:21 pm
I just read that 70% peroxide freezes at -40degC and 100% at about 0.


Has anyone studied the use of subcooled peroxide?
What kind of density increases does 70% peroxide gain from subcooling?


I also read 60% peroxide goes down to -55degC. 


Has any theoretical work been done on finding interesting mixtures of water, peroxide (and perhaps nitrous) as rocket oxidisers?


I imagine dissolving nitrous in 60%(-55degC) peroxide, and seeing what happens to the freezing point. Perhaps, adding nitrous (and further cooling (if possible)), until the mixture doesn't quite have enough energy to decompose.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 10/01/2016 09:36 am
The characteristics of peroxide-water mixtures have been thoroughly studied (see the Hydrogen Peroxide Handbook (https://www.scribd.com/doc/27376589/Rocketdyne-Hydrogen-Peroxide-Handbook) produced by Rocketdyne in June 1967 as AFRPL-TR-67-144.

While it's true that, to some extent, diluting 100% peroxide with water lowers its freezing point, adding water lowers the density too.  The result is that, for example, 70% peroxide has a lower density (1.36 kg/m3) at its freezing point (-40 oC) than does 100% peroxide (1.47  kg/m3) at its freezing point (-0.4 oC).  And the water does not contribute to specific impulse, so the impulse density of the more dilute mixtures is lower still.

I don't know about peroxide-nitrous mixtures, but since nitrous generally has lower performance as an oxidizer, I don't have high hopes.
Title: Re: Basic Rocket Science Q & A
Post by: adrianwyard on 10/09/2016 12:59 am
For long duration missions cryogenic propellants are problematic because they tend to boil off to a gaseous state. To circumvent this we hear of designs that go to the trouble of keeping it cold enough that it stays liquid, or just switch to another propellant. But is there a straightforward reason that it couldn't be allowed to turn to gas, and then recooled and liquified only when needed?

Perhaps the energy and mass of the equipment needed to reliquify is always going to be demonstrably greater than insulation?
Title: Re: Basic Rocket Science Q & A
Post by: TrevorMonty on 10/09/2016 08:16 am
For long duration missions cryogenic propellants are problematic because they tend to boil off to a gaseous state. To circumvent this we hear of designs that go to the trouble of keeping it cold enough that it stays liquid, or just switch to another propellant. But is there a straightforward reason that it couldn't be allowed to turn to gas, and then recooled and liquified only when needed?

Perhaps the energy and mass of the equipment needed to reliquify is always going to be demonstrably greater than insulation?
The lastest Lockheed Martin Mars mission idea uses active cryogenic cooling to keep LH/LOX  in liquid state. The fuel is pumped from storage tanks into propulsion stage's tanks a few hours before it is required for a burn.

ULA think they can keep LH/LOX cool for days to weeks even months by using lots of insulation and sun shade. Any H gas boil off can be used to help keep rest of fuel cool. Even though it is in gas state the H is still extremely cold and can be used to cool outer skin of tank. This surplus gas can be used in gas thrusters or burnt in H/O gas thrusters for station keeping.

Title: Re: Basic Rocket Science Q & A
Post by: sdsds on 10/09/2016 08:39 am
This surplus gas can be used in gas thrusters or burnt in H/O gas thrusters for station keeping.

Or used in an internal combustion engine to produce mechanical energy, which can be used for a variety of purposes. See the Integrated Vehicle Fluids paper: http://www.ulalaunch.com/uploads/docs/Published_Papers/Supporting_Technologies/Space_Access_Society_2012.pdf
Title: Re: Basic Rocket Science Q & A
Post by: nicp on 10/12/2016 05:41 pm
A quick question. Somewhere I got the impression that the oxygen-rich staged combustion cycle is inherently more efficient than a fuel rich staged combustion cycle.
Perhaps that is incorrect - I can't think of a good reason why it would be.

But if it is more efficient, why?

Something is niggling in the back of my brain now about smaller molecules of oxygen vs kerosene, but my last chemistry lesson was in 1984 when I was 18...

Title: Re: Basic Rocket Science Q & A
Post by: rocx on 10/13/2016 07:41 am
A quick question. Somewhere I got the impression that the oxygen-rich staged combustion cycle is inherently more efficient than a fuel rich staged combustion cycle.
Perhaps that is incorrect - I can't think of a good reason why it would be.

But if it is more efficient, why?

Something is niggling in the back of my brain now about smaller molecules of oxygen vs kerosene, but my last chemistry lesson was in 1984 when I was 18...

I'm not quite the engine expert that some other forum members are, but I can give a first response.
Fuel-rich staged combustion for kerosene does not work because kerosene decomposes and cokes (leaves carbon soot on the engine) at high temperature instead of becoming a hot, clean gas.
For hydrogen engines, fuel-rich staged combustion is actually preferred. See the Space Shuttle Main Engines.
There are other reasons why oxidiser-rich staged combustion is preferred for methalox, but I can't explain it satisfactorily to myself, so I'll leave that to others.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 10/13/2016 08:34 am
To a very rough first approximation, the various gases involved (hydrogen, oxygen, methane) have similar (within a factor of 2) specific heats per molecule.  In other words, it takes a similar amount of energy to raise the temperature of a mole of oxygen molecules from 699 K to 700 K as it does to raise the temperature of a mole of hydrogen molecules from 699 K to 700 K.  That means that the propellant present in the larger quantity (by number, not mass) will tend to provide the most pumping power at a given temperature.  Lox-hydrogen engines run at O/F mass ratios of 5-6, corresponding to number ratios of about 0.35.  Hence, fuel-rich staged combustion offers about three times the power of the oxygen-rich alternative.

When it comes to lox-methane, O/F is usually just slightly fuel-rich, meaning that O/F is a bit over 3 by number.  Hence, there is more pumping power available in the oxygen-rich cycle (though the polyatomic nature of methane raises its molar heat capacity, reducing oxygen's advantage somewhat).
Title: Re: Basic Rocket Science Q & A
Post by: baldusi on 10/13/2016 10:19 pm
To a very rough first approximation, the various gases involved (hydrogen, oxygen, methane) have similar (within a factor of 2) specific heats per molecule.  In other words, it takes a similar amount of energy to raise the temperature of a mole of oxygen molecules from 699 K to 700 K as it does to raise the temperature of a mole of hydrogen molecules from 699 K to 700 K.  That means that the propellant present in the larger quantity (by number, not mass) will tend to provide the most pumping power at a given temperature.  Lox-hydrogen engines run at O/F mass ratios of 5-6, corresponding to number ratios of about 0.35.  Hence, fuel-rich staged combustion offers about three times the power of the oxygen-rich alternative.

When it comes to lox-methane, O/F is usually just slightly fuel-rich, meaning that O/F is a bit over 3 by number.  Hence, there is more pumping power available in the oxygen-rich cycle (though the polyatomic nature of methane raises its molar heat capacity, reducing oxygen's advantage somewhat).
If I might add, the original question was ambiguous. Turbine-pumped rocket engines have the turbine power issue and the main combustion chamber issue. And even then, you have gas generator/staged combustion cycles, the expander cycles and the tap-off cycle. And then you have the more complex full flow, dual expander and staged/expander (like the RD-0162).
So, the answer is, it depends. As it was very well explained by Proponent, the turbine power is, when possible, the defining issue on the O/F ratio for turbines. In the RP-1/LOX staged combustion, it is impossible to do fuel rich precombustion. In the LH/LOX the H2 has better performance and on the CH4/LOX the oxidizer-rich case has more power.
Regarding the main combustion, the specific impulse is directly related to average molecular speed, and thus the lighter the molecule, the better the isp. It is obviously also related to combustion temperature and pressure. But that is usually power and material limited, and thus, your only decision is to use oxidizer rich or fuel rich.
In this case, from my limited understanding, given that the molecular weight of the fuel is usually lower than the weight of the oxidizer, using a fuel rich combustion gives improved specific impulse. The oxidizer-rich case would give you better propellant density (again, only if the fuel molecular weight is lighter than the oxidizer's). And in this case specific impulse usually trumps the density considerations.
Please understand that on rocket engines you have to carry both fuel and oxidizer. On combustion engines on Earth, the oxidizer is taken from the atmosphere, and thus, running leaner gives you much better performance, since you are adding a reactant that's "free" in mass terms. But this does not applies to anything that goes to vacuum.
Title: Re: Basic Rocket Science Q & A
Post by: FishInferno on 04/30/2017 02:36 am
Why is an overexpanded nozzle less efficient than an ambient nozzle?  According to the attached diagram, the nozzle is "full" all the way to the end with both an ambient and overexpanded nozzle (not grossly overexpanded, the bottom one, I know that has other issues).  How is the rocket affected by the exhaust "collapsing" after it has left the engine?

I understand that an underexpanded nozzle is less efficient because it fails to capture the maximum possible energy from the expanding exhaust, but just to be clear, it is still producing the same amount of thrust that it does at ambient, right?  Because both nozzles are "full".
(https://upload.wikimedia.org/wikipedia/commons/4/47/Rocket_nozzle_expansion.svg)
Title: Re: Basic Rocket Science Q & A
Post by: Damon Hill on 04/30/2017 02:44 am
One problem with over-expansion is the turbulent flow as it detaches from the nozzle wall; this can tear up an engine and cause guidance problems--the flow will be asymmetrical.
Title: Re: Basic Rocket Science Q & A
Post by: FishInferno on 04/30/2017 03:06 am
One problem with over-expansion is the turbulent flow as it detaches from the nozzle wall; this can tear up an engine and cause guidance problems--the flow will be asymmetrical.

I thought that was only with grossly over-expanded when it becomes detached from the nozzle like in the bottom picture.  Aren't most rockets designed with engines that are slightly over-expanded at sea level in order to maintain efficiency at a higher altitude? It can't become too unstable at that amount of over-expansion if it is usable on rockets.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 04/30/2017 03:12 am
Aren't most rockets designed with engines that are slightly over-expanded at sea level

under-expanded
Title: Re: Basic Rocket Science Q & A
Post by: dimovski on 04/30/2017 10:37 am
Hello everybody,

I've been lurking around in this forum for quite a while and I sincerely hope that I am not repeating an already asked question or posting in the wrong topic.

Intrigued by the LANTR-concept (http://www.nss.org/settlement/moon/LANTR.html), I tried to replicate the different Isps for different mixture ratios by using the formula Ve^2=2k/(k-1)*R*T/M, which I've found here (http://www.thespacerace.com/forum/index.php?topic=1481.0) and simplified a bit by removing the part with different pressures, as that factor would most likely be rather close to 1.

However, I always get results that are way off, e.g., if I go for pure hydrogen & T=2900, according to the 1st link
I should get an Isp of around 940, so a Ve of around 9000. But what I actually get is: Ve=sqrt(2*1.666/0.666*8.314*2900/1)=347.311m/s, which is off by a factor of 30(!).

Could anybody tell me what exactly I'm missing here?

Thanks in advance and sorry for posting a potentially dumb question.
Title: Re: Basic Rocket Science Q & A
Post by: deruch on 05/01/2017 01:34 am
Why is an overexpanded nozzle less efficient than an ambient nozzle?  According to the attached diagram, the nozzle is "full" all the way to the end with both an ambient and overexpanded nozzle (not grossly overexpanded, the bottom one, I know that has other issues).  How is the rocket affected by the exhaust "collapsing" after it has left the engine?

I understand that an underexpanded nozzle is less efficient because it fails to capture the maximum possible energy from the expanding exhaust, but just to be clear, it is still producing the same amount of thrust that it does at ambient, right?  Because both nozzles are "full".
The general thrust equation for rockets is: F=mdot*Ve + (Pe - P0)*Ae where,
     F=thrust,
     mdot=mass flow rate,
     Ve=velocity of the exhaust at the nozzle exit,
     Pe=exhaust pressure at the nozzle exit,
     P0=ambient pressure, and
     Ae=area of the nozzle exit.

So, if the exhaust pressure is below the ambient pressure, you end up with a negative term in your thrust equation.  i.e. (Pe - P0)*Ae < 0.  And ergo, a reduction in thrust. 
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 05/01/2017 03:50 pm
So, if the exhaust pressure is below the ambient pressure, you end up with a negative term in your thrust equation.  i.e. (Pe - P0)*Ae < 0.  And ergo, a reduction in thrust. 

Yes: if the nozzle is over-expanded, then it is actually sucking the rocket backward to some extent.  If this is hard to imagine, it's because our intuition about fluids is based entirely on subsonic flows.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 05/01/2017 03:56 pm
Intrigued by the LANTR-concept (http://www.nss.org/settlement/moon/LANTR.html), I tried to replicate the different Isps for different mixture ratios by using the formula Ve^2=2k/(k-1)*R*T/M, which I've found here (http://www.thespacerace.com/forum/index.php?topic=1481.0) and simplified a bit by removing the part with different pressures, as that factor would most likely be rather close to 1.

However, I always get results that are way off, e.g., if I go for pure hydrogen & T=2900, according to the 1st link
I should get an Isp of around 940, so a Ve of around 9000. But what I actually get is: Ve=sqrt(2*1.666/0.666*8.314*2900/1)=347.311m/s, which is off by a factor of 30(!).

Could anybody tell me what exactly I'm missing here?

If you're doing the calculation in SI units, than the molecular weight of atomic hydrogen is 0.001 kg/mol.  At 2900 K, though, I would think most of the hydrogen would be molecular, so 0.002 kg/mol might be more accurate, and the specific heats would need to be adjusted too.

It's a relatively minor point, but ideal-gas specific heats may not be very accurate at high temperatures.  As the temperature rises, molecules behave less and less like the rigid rotors, because vibrations and bending become significant.
Title: Re: Basic Rocket Science Q & A
Post by: nicp on 06/23/2017 06:16 pm
An old but vague memory suddenly popped up. Which may very well be utterly wrong.

Somewhere amongst my hundred or so books on rocketry and spaceflight I'm sure there was a reference somewhere (probably dating to events between 1945 and 1970) describing cryogenic (oxygen I think) lines making a lot of scary noise during fueling operations. This might even be X-1 or X-15 territory.

If memory serves the noise was sort of continuous, not sudden bangs or anything and genuinely worrying to those not used to it.

But now I come to think about it I don't remember any launch coverage with audio containing anything much, other than the noise (before launch) of venting gases.

Is my memory incorrect? I've done a quick search and come up blank.
It's entirely possible I'm fooling myself and my memory is wrong...
Title: Re: Basic Rocket Science Q & A
Post by: nicp on 01/04/2018 11:51 pm
What are the reasons for using hydrogen in a first stage? Ever? This is a genuine question.

Ok, Shuttle did it, H-II does it, Ariane V does it. But all with (comparitively poor) Isp boosters.

Hydrogen is _great_ for an an upper stage, but surely a royal pain in the rear in the volumes required (and the necessary insulation for that immense volume) in the warm lower atmosphere for a (necessarily large) first stage.
Never mind embrittlement of alloys et al.

Saturn used H2, but _only_ when high and fast in upper stages, Centaur ditto. No hydrogen low and slow.

Looking back, hydrolox in first (boosted) stages looks like a fad.
Methalox _now_ looks like a fad, but perhaps a much better one, though that is to be proven.

I wonder as a mere software engineer if there ever was, or is, a good reason for a rocket engine to ever be burning hydrogen at close to sea level.





Title: Re: Basic Rocket Science Q & A
Post by: RonM on 01/05/2018 12:46 am
What are the reasons for using hydrogen in a first stage? Ever? This is a genuine question.

Ok, Shuttle did it, H-II does it, Ariane V does it. But all with (comparitively poor) Isp boosters.

Hydrogen is _great_ for an an upper stage, but surely a royal pain in the rear in the volumes required (and the necessary insulation for that immense volume) in the warm lower atmosphere for a (necessarily large) first stage.
Never mind embrittlement of alloys et al.

Saturn used H2, but _only_ when high and fast in upper stages, Centaur ditto. No hydrogen low and slow.

Looking back, hydrolox in first (boosted) stages looks like a fad.
Methalox _now_ looks like a fad, but perhaps a much better one, though that is to be proven.

I wonder as a mere software engineer if there ever was, or is, a good reason for a rocket engine to ever be burning hydrogen at close to sea level.

All three of your examples use large SRBs. For example, Ariane 5 gets 90% of its takeoff thrust from the solids. So, one way of looking at it is the SRBs are the first stage while the hydrolox booster is the second stage. No matter what you call it, when the SRBs are very large, a hydrolox booster makes sense.
Title: Re: Basic Rocket Science Q & A
Post by: incoming on 01/05/2018 06:59 pm
What are the reasons for using hydrogen in a first stage? Ever? This is a genuine question.

Ok, Shuttle did it, H-II does it, Ariane V does it. But all with (comparitively poor) Isp boosters.

Hydrogen is _great_ for an an upper stage, but surely a royal pain in the rear in the volumes required (and the necessary insulation for that immense volume) in the warm lower atmosphere for a (necessarily large) first stage.
Never mind embrittlement of alloys et al.

Saturn used H2, but _only_ when high and fast in upper stages, Centaur ditto. No hydrogen low and slow.

Looking back, hydrolox in first (boosted) stages looks like a fad.
Methalox _now_ looks like a fad, but perhaps a much better one, though that is to be proven.

I wonder as a mere software engineer if there ever was, or is, a good reason for a rocket engine to ever be burning hydrogen at close to sea level.

All three of your examples use large SRBs. For example, Ariane 5 gets 90% of its takeoff thrust from the solids. So, one way of looking at it is the SRBs are the first stage while the hydrolox booster is the second stage. No matter what you call it, when the SRBs are very large, a hydrolox booster makes sense.

Another way to look at it is this - pretty much all rocket fuel choices involve a tradeoff between thrust, ISP, and propellant density (and thus stage dry mass). 

With LH2, you get pretty good thrust, really good ISP, and not so good propellant density.  The "opposite" is SRBs, where you get a whole lot of thrust, high propellant density, but comparatively low ISP.  Kerolox is basically right in between the two when it comes to ISP, thrust, and prop density, with Methalox being similar to Kerolox but slightly less dense.  It just so turns out that Kerolox strikes just a pretty optimal balance between thrust, ISP, and propellant density for 1st stages. 

So...the end result is the overwhelming majority of rockets are either powered by kerolox 1st stages OR a combination of LH2 and SRBs, which, not coincidentally, yields an average propellant density and ISP between the two fuels as being about the same as Kerolox.  The lone exception I can think of is the three core all LH2 Delta IV heavy, which has always seemed to me (visually at least) to really struggle off the pad. 





Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 01/05/2018 07:23 pm
What are the reasons for using hydrogen in a first stage? Ever? This is a genuine question.

Ok, Shuttle did it, H-II does it, Ariane V does it. But all with (comparitively poor) Isp boosters.

To this list we can add Energia.  As RonM points out, in all of these cases hydrogen-fueled systems provide only a fraction of the take-off thrust, so were dealing not so much with first stages but with upper stages that happen to be lit on the ground.

As incoming says, Delta IV Medium and Delta IV Heavy do on the other hand have truly hydrogen-fueled first stages.  My guess is that the sole reason is that at the time they went into development, the US had no modern hydrocarbon engines, having foolishly abandoned development of such after the F-1.

Quote
Hydrogen is _great_ for an an upper stage, but surely a royal pain in the rear in the volumes required (and the necessary insulation for that immense volume) in the warm lower atmosphere for a (necessarily large) first stage.
Never mind embrittlement of alloys et al.

I generally agree, though insulating a large hydrogen-fueled stage might in an important sense be easier than insulating a small one, because the large stage has a lower ratio of surface area to volume.

Quote
Saturn used H2, but _only_ when high and fast in upper stages, Centaur ditto. No hydrogen low and slow.

If the parameters of the first stage are fixed, then a hydrogen-fueled upper stage maximizes payload.  If the design of the first stage is a free parameter, however, hydrogen may not look as attractive.  A hydrogen upper stage will give the smallest first stage, but it may not give the smallest, cheapest launch vehicle overall.

Quote
Looking back, hydrolox in first (boosted) stages looks like a fad.

I wasn't there, but I'll bet many people were seduced by lox-hydrogen's high specific impulse and simply assumed it was the way to go, without careful analysis of hydrogen's drawbacks.

Quote
Methalox _now_ looks like a fad, but perhaps a much better one, though that is to be proven.

I don't think methalox is a fad.  Its major proponents -- Musk and Bezos -- are hard-headed businessmen who are probably more tightly focused on economic efficiency and less likely to be distracted by pure performance specs (e.g., specific impulse) than NASA or DoD.

Consider methane's advantages: methalox is quite a bit denser than hydrolox, and, though less dense than kerolox, it offers higher specific impulse.  It's very cheap, and relatively easily made on Mars.  In contrast to kerosene, methane is chemically very stable, making it suitable for staged-combustion cycles.  Its viscosity is very low (https://forum.nasaspaceflight.com/index.php?topic=31040.msg1519267#msg1519267), reducing pressure losses.  It can be stored at lox temperatures, allowing common-bulkhead tankage with insulation.  Unlike kerosene, methane readily evaporates without leaving a residue, simplifying reuse of engines.

Methalox's clear disadvantage with respect to kerolox is its lower density.  How much of a drawback that is depends on assumptions about structural mass.  With one particular set of assumption (hardly unimpeachable), it's not greatly worse (https://forum.nasaspaceflight.com/index.php?topic=31040.msg1518743#msg1518743).
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 01/05/2018 07:26 pm
With LH2, you get pretty good thrust....

You mean engine thrust-to-weight ratio, right?
Title: Re: Basic Rocket Science Q & A
Post by: TomasMraz on 01/07/2018 09:06 pm
Hi,

please, how can I calculate needed amount of oxygen in gaseous state versus liquid. If you need 1 unit of liquid oxygen, how many units of gaseous it will be? Under atmospheric pressure, or under 200 bar?

Thank you
Title: Re: Basic Rocket Science Q & A
Post by: cppetrie on 01/08/2018 01:07 am
Hi,

please, how can I calculate needed amount of oxygen in gaseous state versus liquid. If you need 1 unit of liquid oxygen, how many units of gaseous it will be? Under atmospheric pressure, or under 200 bar?

Thank you
pv=nrt is the equation you need I think. p is pressure, v is volume, n is number of moles of substance (set n1=n2 if trying to determine ratio of volumes at different v, p or t), r is the gas constant, and t is temp (in Kelvin). Gas constant is the same on both sides of the equation so that’ll cancel out so you’re left with p1v1/t1 = p2v2/t2. I think that’s all correct. Someone correct me if I’m off base. It’s been quite some time since I’ve delved into chemistry.
Title: Re: Basic Rocket Science Q & A
Post by: meberbs on 01/08/2018 01:49 am
Hi,

please, how can I calculate needed amount of oxygen in gaseous state versus liquid. If you need 1 unit of liquid oxygen, how many units of gaseous it will be? Under atmospheric pressure, or under 200 bar?

Thank you
Not sure what you mean by "unit."

If you mean mass, then it is easy, 1 kg of liquid oxygen is 1 kg of gaseous oxygen.

More likely you mean volume, in which case the answer is to determine the density and use that to determine how much mass you have in the one state (liquid or gas) and then use that mass and the density of the other state to determine the final volume.

For determining the volume of a gas, it depends on both temperature and pressure, so you need to use the ideal gas law that cppetrie mentioned. (density is n/ (V*M) where M is the molar mass of the substance)

For liquids, density is also a function of temperature, but at this point you are best off using a lookup table to determine the density at the conditions you care about. (Whether this temperature dependence is significant or not depends on the situation, it is generally a smaller effect than in a gas so it might not make much difference, but SpaceX cools their liquid oxygen extra cold to make it a bit denser and increase how much fuel fits in the rocket.)
Title: Re: Basic Rocket Science Q & A
Post by: deruch on 01/08/2018 05:55 am
Hi,

please, how can I calculate needed amount of oxygen in gaseous state versus liquid. If you need 1 unit of liquid oxygen, how many units of gaseous it will be? Under atmospheric pressure, or under 200 bar?

Thank you

You have X liters of pure LOX. 

1) Multiply by the density to find the mass of O2 present.   
2) Divide by molecular mass of oxygen: 31.998g/mol 
3) Multiply by molar volume of an ideal gas: 22.4L/mol (@1atm & 273K) 
4) Adjust volume for pressure and temperature differences from STP.
     a) Set P1*V1/T1 = P2*V2/T2, and solve for V2.
Title: Re: Basic Rocket Science Q & A
Post by: deruch on 04/01/2018 10:36 am
The general thrust equation for rockets is: F=mdot*Ve + (Pe - Pa)*Ae where,
     F=thrust,
     mdot=mass flow rate,
     Ve=velocity of the exhaust at the nozzle exit,
     Pe=exhaust pressure at the nozzle exit,
     Pa=ambient pressure, and
     Ae=area of the nozzle exit.


When considering the Pa term during the low altitude segment of the flight, is that very strictly the general atmospheric pressure at the rocket's altitude?  I believe the forward travel of the rocket and backward travel of the high velocity exhaust gasses creates a localized low pressure at the base of the rocket (this causes the plume recirculation seen on some launches, right?).  Does this area's lowered pressure need to be taken into account for a higher accuracy calculation of the thrust?
Title: Re: Basic Rocket Science Q & A
Post by: mgfitter on 04/11/2018 09:06 pm
Does anyone know how many rocket engine nozzles have been damaged in flight?

I know STS-93 experienced a frightening event just after liftoff that I understand involved the regen nozzle being punctured through, but are there any other known instances of such problems elsewhere?

-MG.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 04/12/2018 01:38 pm
Maybe the Merlin 1C that failed on an early Falcon 9 flight?  (I seem to recall  that involved implosion of the nozzle, but I'm not sure).
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 04/12/2018 03:20 pm
Does anyone know how many rocket engine nozzles have been damaged in flight?

I know STS-93 experienced a frightening event just after liftoff that I understand involved the regen nozzle being punctured through, but are there any other known instances of such problems elsewhere?

-MG.

One of the Delta III launches. 

Many early Atlas and Titans
Title: Re: Basic Rocket Science Q & A
Post by: whitelancer64 on 04/12/2018 03:40 pm
Maybe the Merlin 1C that failed on an early Falcon 9 flight?  (I seem to recall  that involved implosion of the nozzle, but I'm not sure).

On CRS-1, a Merlin 1C engine had a fuel dome (a structure above the thrust chamber) rupture and lose pressure. Nothing to do with the nozzle.
Title: Re: Basic Rocket Science Q & A
Post by: meberbs on 04/12/2018 06:14 pm
...
When considering the Pa term during the low altitude segment of the flight, is that very strictly the general atmospheric pressure at the rocket's altitude?  I believe the forward travel of the rocket and backward travel of the high velocity exhaust gasses creates a localized low pressure at the base of the rocket (this causes the plume recirculation seen on some launches, right?).  Does this area's lowered pressure need to be taken into account for a higher accuracy calculation of the thrust?
If you want to talk about pressure variations along the rocket body, due to forward flight, then you are discussing terms that count as drag as well. The Pa term basically comes from the atmospheric pressure pushing back on the equivalent area at the front of the rocket, which obviously increases with velocity.

For accurate performance calculations, they would need to account for all of the various effects at the same time, including drag and variations in the thrust performance with altitude. I think your question can be rephrased as "does the thrust performance with altitude variation depend on forward velocity as well?" I think the answer to this is most likely yes, though I am not sure how significant it would be. Note that Pe and Ve vary as the exit flow shape changes, so the variation with altitude isn't simply due to the Pa term changing as one might think.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 04/13/2018 08:03 pm
On CRS-1, a Merlin 1C engine had a fuel dome (a structure above the thrust chamber) rupture and lose pressure. Nothing to do with the nozzle.

But wasn't there something about the nozzle collapsing after the engine shut down?
Title: Re: Basic Rocket Science Q & A
Post by: Nomadd on 04/14/2018 11:56 am
 SpaceX F1-3 2nd stage.
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 04/14/2018 10:27 pm
The general thrust equation for rockets is: F=mdot*Ve + (Pe - Pa)*Ae where,
     F=thrust,
     mdot=mass flow rate,
     Ve=velocity of the exhaust at the nozzle exit,
     Pe=exhaust pressure at the nozzle exit,
     Pa=ambient pressure, and
     Ae=area of the nozzle exit.


When considering the Pa term during the low altitude segment of the flight, is that very strictly the general atmospheric pressure at the rocket's altitude?  I believe the forward travel of the rocket and backward travel of the high velocity exhaust gasses creates a localized low pressure at the base of the rocket (this causes the plume recirculation seen on some launches, right?).  Does this area's lowered pressure need to be taken into account for a higher accuracy calculation of the thrust?

I imagine this could come down to a matter of definition.  What really matters is the total force on the rocket, and how that's divided up between thrust and drag may be a little bit arbitrary.  However, I am inclined to use atmospheric pressure in calculating the pressure term in the thrust equation.  That equation is typically derived by noting that if a rocket engine imparts momentum to its exhaust at a rate
The usual way of deriving the thrust equation quoted above is to argue that if the rocket engine imparts momentum at a rate mdot*Ve, then by Newton's third law momentum in the opposite direction is imparted to the engine.  Then the back-pressure term is rather artificially, in my view, tacked on.

However, the equation can also be derived by considering nothing but the pressure forces on the engine.  As one proceeds aft from the forward dome of the combustion chamber, diverging sections (including the forward dome itself) exert forward forward on the engine wherever the internal pressure exceeds ambient, and converging sections exert rearward force.  Where internal pressure is lower than external, diverging sections exert rearward force and converging sections exert forward force.  Add up all of those forces (i.e., integrate from the top of the combustion chamber to the nozzle exit), and you get the usual expression for thrust.

If you look at thrust this way, then it's natural that the general ambient pressure appears in the equation.

EDIT:  "thats" -> "that's"
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 04/15/2018 08:44 pm
For a simple illustration of my point, above, imagine a very simple rocket engine in the shape of a cylinder with the top end closed and the bottom end open.  We inject fuel and oxidizer near the top.  They burn, increasing the pressure there.  The net force on the engine is cylinder's cross-sectional area multiplied by the difference in pressures across the top end.
Title: Re: Basic Rocket Science Q & A
Post by: parsec on 09/13/2018 11:11 am
I'm working my way through the isentropic flow section of Sutton (9th edition). The equations derived for exhaust velocity and a subsequent plot seem to indicate that specific impulse decreases with increasing k (the ratio of specific heats).

This seems counterintuitive to me. The more degrees of freedom a molecule has, the more k tends towards one, since k is (f/2 +1)/(f/2), where f is the amount of degrees of freedom. However, I would expect that less degrees of freedom result in a higher translational exhaust velocity due to the equipartition of energy between those degrees of freedom.

Wikipedia (I know, not the best source) seems to confirm this qualitative intuition;

https://en.wikipedia.org/wiki/Rocket_engine#Propellant_efficiency

Quote
For a rocket engine to be propellant efficient..... using propellants which are, or decompose to, simple molecules with few degrees of freedom to maximise translational velocity

Sutton's equations and text seem to indicate the contrary. In fact, one of Sutton's equations for the maximum theoretical value for exhaust velocity for infinite expansion produces an infinite velocity (divide by zero) for k=1, i.e. molecules with a large amount of degrees of freedom.

Is there something I am missing here? Why do molecules with more degrees of freedom seem to produce higher specific impulse values according to Sutton?
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 09/13/2018 01:50 pm
Since the expansion ratio is infinite, the temperature at the nozzle exit is zero, and all of the energy that was in all of those degrees of freedom back in the combustion chamber has been converted into bulk kinetic energy, regardless of how many degrees of freedom exist.  If you hold the exit temperature constant at a relatively high value while varying the ratio of specific heats, then flows with lower ratios of specific heats should do better over some range.  When examining the behavior of an equation as one variable changes, one must ask oneself which of the relevant quantities are also varying and which are (perhaps only implicitly) being held constant.

When thinking about the exhaust-velocity equation, by the way, I find that replacing k Cp/(k - 1) with CpT, where Cp is the mass-specific heat at constant pressure and T, of course, is the absolute temperature, aids my intuition.  The point is that CpT is the enthalpy of an ideal gas.
Title: Re: Basic Rocket Science Q & A
Post by: Kosmos2001 on 09/20/2018 12:26 pm
Why some booster nozzles are pointing not vertically but with some angle outside?
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 09/20/2018 01:17 pm
If nozzles point directly backward, then variations in thrust of one engine with respect to another will tend to make the rocket turn a bit.  That  is eliminated if each engine's thrust vector points at the rocket's center of mass.
Title: Re: Basic Rocket Science Q & A
Post by: envy887 on 10/03/2018 08:15 pm
If nozzles point directly backward, then variations in thrust of one engine with respect to another will tend to make the rocket turn a bit.  That  is eliminated if each engine's thrust vector points at the rocket's center of mass.

Also, if the engines are not symmetric about the center of mass (as in the SSME on the Shuttle), then they have to be pointed through the center of mass to prevent an excess torque - even if the thrust is precisely controlled.
Title: Re: Basic Rocket Science Q & A
Post by: gin455res on 11/07/2018 09:41 pm
Has anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.


Lets say the first stage is HTP/kerosene then this has a O/F ratio of about 7:1.  Then if all the fuel was carried in the second stage and it was LOX/Kerosene the ratio would be down to about 2.5:1.


If the first stage + 1st stage fuel is 3ish times bigger than the second
and it has O:F of 126:18 (144).


then the upper-stage -1st stage fuel would be about 48 split about 34:14


The upper stage would need to have tanks for  O:F of 34:32 (14+18)


Would the extra weight of the larger fuel tank kill performance or simplify construction having tanks of 126:34:32 (O:O:F), instead of 126:18:34:14?
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 11/10/2018 02:05 pm
Gravity losses accumulate at the rate g sin γ, where g is, as you guess, the same g that appears in the rocket equation, and γ is the angle between the flight path and the local horizontal.  Some time ago, I put together a comprehensive look at velocity losses in general, attached to this post (http://forum.nasaspaceflight.com/index.php?topic=33439.msg1227484#msg1227484) (warning: it uses a fair amount of math).
Title: Re: Basic Rocket Science Q & A
Post by: cppetrie on 11/10/2018 04:33 pm
This post isn’t really meant to answer the gravity losses question but rather summarize my understanding of what gravity losses are. So if I’ve got something wrong please point it out.

Orbit isn’t about altitude per se but rather velocity. You need an angular velocity such that the centrifugal force is equal to the pull of gravity to essentially nullify it and enter orbit. Given that we have an atmosphere on earth, achieving the required velocity is easier/possible if your altitude places you outside or at least in the very very thinnest parts of the atmosphere. In order to achieve that altitude, you have to spend some of your fuel countering gravity to increase altitude rather than devoting all the fuel to increasing tangential velocity, thus you lose something to gravity. If you had a perfectly spherical body with no atmosphere you could orbit just off the surface given you have the necessary angular velocity, and orbit could be achieved with almost no gravity losses. Is that at least a decent conceptualization of gravity losses?

If I’m understanding the concept correctly the launch trajectory would affect the amount of gravity losses that occur. A more lofted trajectory would have greater losses versus one that pitches down range earlier in the ascent. But there is a sweet spot between fighting atmospheric drag versus gravity losses that would define the “ideal” trajectory. Different trajectories from the ideal might be chosen for various reasons (such as?). What other factors in rocket design and launch trajectories affect gravity losses.

I appreciate the feedback and the tremendous resource NSF is for us amateur armchair rocketeers and space nerds.
Title: Re: Basic Rocket Science Q & A
Post by: envy887 on 11/12/2018 02:24 pm
This post isn’t really meant to answer the gravity losses question but rather summarize my understanding of what gravity losses are. So if I’ve got something wrong please point it out.

Orbit isn’t about altitude per se but rather velocity. You need an angular velocity such that the centrifugal force is equal to the pull of gravity to essentially nullify it and enter orbit. Given that we have an atmosphere on earth, achieving the required velocity is easier/possible if your altitude places you outside or at least in the very very thinnest parts of the atmosphere. In order to achieve that altitude, you have to spend some of your fuel countering gravity to increase altitude rather than devoting all the fuel to increasing tangential velocity, thus you lose something to gravity. If you had a perfectly spherical body with no atmosphere you could orbit just off the surface given you have the necessary angular velocity, and orbit could be achieved with almost no gravity losses. Is that at least a decent conceptualization of gravity losses?

If I’m understanding the concept correctly the launch trajectory would affect the amount of gravity losses that occur. A more lofted trajectory would have greater losses versus one that pitches down range earlier in the ascent. But there is a sweet spot between fighting atmospheric drag versus gravity losses that would define the “ideal” trajectory. Different trajectories from the ideal might be chosen for various reasons (such as?). What other factors in rocket design and launch trajectories affect gravity losses.

I appreciate the feedback and the tremendous resource NSF is for us amateur armchair rocketeers and space nerds.

Even without an atmosphere it would still have gravity losses, due to the need to angle some part of the thrust down to avoid hitting the ground while accelerating up to orbital velocity.

The only way to not have gravity loss is to instantly reach orbital velocity, or to be physically supported (gun barrel, maglev, launch loop, etc) while accelerating. Or to launch horizontally from e.g. a tower with a high enough elevation that the rocket can thrust only tangentially and still reach orbital velocity before hitting the ground.
Title: Re: Basic Rocket Science Q & A
Post by: Tuna-Fish on 12/03/2018 05:58 pm
Has anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.

This entire question makes me feel weird. Just what purpose could this possibly have? Reducing the dry mass of the second stage is absolutely crucial for performance. Why would you ever feed fuel down from there, instead of just making the first stage tanks bigger?
Title: Re: Basic Rocket Science Q & A
Post by: gin455res on 12/03/2018 06:19 pm
Has anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.

This entire question makes me feel weird. Just what purpose could this possibly have? Reducing the dry mass of the second stage is absolutely crucial for performance. Why would you ever feed fuel down from there, instead of just making the first stage tanks bigger?

The point was to remove the fuel tank from the first stage. So only 3 tanks need to be manufactured not 4. And also the tanks would be more similar in size. Perhaps, this would mean all three tanks could be made the same thickness. One long cylindrical oxidiser tank on the first stage, and 2 smaller more spherical-ish tanks of oxidiser and fuel on the upper-stage, perhaps with common bulkheads.
Title: Re: Basic Rocket Science Q & A
Post by: John-H on 12/03/2018 10:36 pm
Has anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.

This entire question makes me feel weird. Just what purpose could this possibly have? Reducing the dry mass of the second stage is absolutely crucial for performance. Why would you ever feed fuel down from there, instead of just making the first stage tanks bigger?

Didn't Atlas essentially do just that?  The first stage was just booster engines, the next stage had the tanks. Of course, the tanks were the lightest possible.

John
Title: Re: Basic Rocket Science Q & A
Post by: meberbs on 12/04/2018 12:59 am
Has anyone ever designed a rocket which cross-feeds the fuel from the 2nd stage to the first stage.

This entire question makes me feel weird. Just what purpose could this possibly have? Reducing the dry mass of the second stage is absolutely crucial for performance. Why would you ever feed fuel down from there, instead of just making the first stage tanks bigger?

The point was to remove the fuel tank from the first stage. So only 3 tanks need to be manufactured not 4. And also the tanks would be more similar in size. Perhaps, this would mean all three tanks could be made the same thickness. One long cylindrical oxidiser tank on the first stage, and 2 smaller more spherical-ish tanks of oxidiser and fuel on the upper-stage, perhaps with common bulkheads.
Basic problem is that starting with any typical 2 stage design, there is something like a 10% penalty to payload capacity for extra dry mass on the first stage. For any final stage, the dry mass penalty to payload is 100% by definition. Even ignoring the complexity that a fluid flow across stage separation entails, that means you are guaranteed to lose performance unless the weight added to the upper stage tank is less than 10% of the mass of the tank that was removed from the first stage. This is unlikely to ever work out.

The Atlas example mentioned above is a stage and a half design, so there was no real stage separation, just dropping some of the engines mid flight, but keeping the full tanks all the way to orbit. This required extremely lightweight tanks due to the near SSTO design.
Title: Re: Basic Rocket Science Q & A
Post by: Slarty1080 on 12/22/2018 08:55 am
I have a question about gravity loss. Am I correct to say that for orbital vehicles there is no gravity loss? If so presumably it follows that the number of engines on an orbital stage should be minimised (excepting the need for redundnacy and thrttling issues)?
Title: Re: Basic Rocket Science Q & A
Post by: Proponent on 12/22/2018 10:42 am
Gravity losses accrue at a rate equal to the local acceleration of gravity multiplied by the sine of the angle between the flight path and the local horizontal.  In practice, then, gravity losses are large during the early phases of launch to orbit, when the flight path is nearly vertical, and smaller layer on. It's even possible to have "gravity gains" when a launch vehicle overshoots the target altitude and gains speed as it descends. Gravity losses still exist in orbit, but because they're small, optimal thrust-to-weight ratios are small, as you surmise. E.g., the top stage of an orbital launch vehicle might have a T/W around 0.5.

Suppose you're in a circular orbit and you want to raise the apogee.  An instantaneous delta-V in the direction of motion will raise the apogee without an losses. In reality, though, no delta-V can be instantaneous.  Altitude will tend to increase during the burn, resulting in some gravity loss.  By directing the thrust a little bit inward, you could maintain a constant altitude while accelerating.  That would eliminate gravity loss but introduce steering loss (thrust not aligned with velocity vector).

Very-low-thrust engines, e.g., ion engines, tend to produce substantial losses (though it's a moot point whether the losses are gravity or steering).  The delta-V needed for a very-low-thrust transfer between two circular orbits is just the difference in orbital speeds of the orbits, which is in general quite a bit more than the delta-V needed for a pure Hohmann transfer.
Title: Re: Basic Rocket Science Q & A
Post by: nicp on 01/03/2019 12:12 am
The Titan 1 rocket used an LR-87 engine running kerosene (well RP-1 I guess) and Lox.
Different variants of the engine ran LH2/LOX and (in the Titan 2) Aerozine 50/N2O4.

Ok, these were different engine variants. But _how_ different? And how much work went into redesigning the engine to run on Aerozine 50/N204 vs kerolox?

I can imagine the liquids have different viscosities, densities, etc, so the turbopump will differ - is this the big problem?
Are the thrust chamber or engine bells different? If so I suspect less so...

My question sort of arises from seeing a nice photograph of a Proton and thinking, 'Could you switch this to kerolox?' (EDIT: I'll add - yes I know you now need an ignition source).

Probably a non-starter for good reasons, but Titan (sort of) did it. The other way.

Title: Re: Basic Rocket Science Q & A
Post by: rakaydos on 04/03/2019 01:18 am
Hopefully this is the right section. I've encountered this term on the forum, but I never took chemestry in high school. Anyone care to explain?
Title: Re: Basic Rocket Science Q & A
Post by: ncb1397 on 04/03/2019 01:25 am
It is the point on a pressure vs temperature graph where a material can be liquid, solid or gaseous. Sort of how New Mexico, Utah, Colorado and Arizona all converge at one point. At this point, with small adjustments to pressure or temperature, you can change the phase of a substance between all three just as you can change between several states with minor adjustments to latitude and longitude where the four aforementioned states meet.

Wikipedia has a better explanation than what I can come up with:
Quote
In thermodynamics, the triple point of a substance is the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium.[1] It is that temperature and pressure at which the sublimation curve, fusion curve and the vaporisation curve meet. For example, the triple point of mercury occurs at a temperature of −38.83440 °C and a pressure of 0.2 mPa.
https://en.wikipedia.org/wiki/Triple_point



Title: Re: Basic Rocket Science Q & A
Post by: Wolfram66 on 04/03/2019 01:39 am
This will give you an idea of the weirdness that goes on at triple point of CycloHexane
https://youtu.be/MP6MVLWuNZQ (https://youtu.be/MP6MVLWuNZQ)

Title: Re: Basic Rocket Science Q & A
Post by: IanO on 04/11/2019 03:38 pm
Why is wind shear a problem for modern launches?  Were there historical launch failures due to wind shear?  Are there different rocket designs which allow launching through wind shear and what are the tradeoffs?
Title: Re: Basic Rocket Science Q & A
Post by: mme on 04/11/2019 05:58 pm
Why is wind shear a problem for modern launches?  Were there historical launch failures due to wind shear?  Are there different rocket designs which allow launching through wind shear and what are the tradeoffs?
I don't have specific numbers but ... Wind shear causes a sudden sideways moment on the rocket, starting at the nose and then proceeding down the vehicle. Both payloads and the rocket itself are much stronger in the vertical axis than they are to lateral forces. Part of the agreement when launching is that both the lateral and axial g-loads will be constrained within known limits and exceeding those constraints could damage the payload itself and lead to mission failure.

And the F9 is really long and skinny which may make it more sensitive to flexing and developing harmonic oscillations. But now I'm mostly hand waving...
Title: Re: Basic Rocket Science Q & A
Post by: intelati on 04/11/2019 08:38 pm
Were there historical launch failures due to wind shear?

Not exactly sure how to ascribe the failure of Challenger between the booster O-Rings, Joint design, and the wind shear, but Challenger hit unprecedented wind shear before the break up of the vehicle

e:Columbia Challenger
Title: Re: Basic Rocket Science Q & A
Post by: sunworshipper on 04/11/2019 08:48 pm
Were there historical launch failures due to wind shear?

Not exactly sure how to ascribe the failure of Columbia between the booster O-Rings, Joint design, and the wind shear, but Columbia hit unprecedented wind shear before the break up of the vehicle
Do you mean Challenger?
Title: Re: Basic Rocket Science Q & A
Post by: cebri on 04/23/2019 02:41 pm
Yes, wind shear was the last nail in the coffin for Challenger. The o-ring had already failed, but both the melted o-rings and the fuel from the SRM had solidified at launch and had created a very fragile seal.  The seal it is believed to have been broken when Challenger encounter unusual strong high winds, the SRB tried to compensate to keep Challenger on course, which put a lot of pressure on the seal until it finally broke. It is hard to say whether Challenger could have survived if it hand't encountered those winds, as the shuttle was also going through max-q when the seal broke off.
Title: Re: Basic Rocket Science Q & A
Post by: Jim on 04/23/2019 06:37 pm
Why is wind shear a problem for modern launches?  Were there historical launch failures due to wind shear?  Are there different rocket designs which allow launching through wind shear and what are the tradeoffs?

Always been a problem.  Longer vehicles are more susceptible to them.
Day of launch I-loads is what is used to deal with them.  Not so much the vehicle design themselves.

But generally a more squat or stiffer vehicle is less affected by shear.
Title: Re: Basic Rocket Science Q & A
Post by: Bananas_on_Mars on 05/18/2019 09:02 pm
I've been wondering wether some of the proposed lunar landers (for example Blue Moon) could do part of TLI and possibly LOI with their landing engine and the use of some kind of "drop tanks"?

Is there any launch system that used or uses some kind of "drop tank" (besides Shuttle external tank)?

My reasoning is that once the lander would be in orbit, you don't have to worry about gravity losses any more, so a small engine would be enough for TLI. And such a drop tank should have much lower dry weight than a second stage, could be a simple stainless steel balloon tank. And since it's dropped before landing, it doesn't need to withstand the landing when empty.

Of course this adds complexity, i understand that. It might make some kind of Apollo style transposition and docking maneuver necessary. What else am I overlooking?
Title: Re: Basic Rocket Science Q & A
Post by: A_M_Swallow on 05/19/2019 04:02 pm
I've been wondering wether some of the proposed lunar landers (for example Blue Moon) could do part of TLI and possibly LOI with their landing engine and the use of some kind of "drop tanks"?

Is there any launch system that used or uses some kind of "drop tank" (besides Shuttle external tank)?

My reasoning is that once the lander would be in orbit, you don't have to worry about gravity losses any more, so a small engine would be enough for TLI. And such a drop tank should have much lower dry weight than a second stage, could be a simple stainless steel balloon tank. And since it's dropped before landing, it doesn't need to withstand the landing when empty.

Of course this adds complexity, i understand that. It might make some kind of Apollo style transposition and docking maneuver necessary. What else am I overlooking?


Refuelling the lander at a propellant depot or from a propellant tanker is the practical equivalent of using drop tanks.
Title: Re: Basic Rocket Science Q & A
Post by: turbopumpfeedback2 on 07/22/2021 09:54 pm
The thrust vector control hobby rockets are very popular right now. It takes a lot of time, knowledge and other stuff to make one.

So I am completely shocked to see a liquid nitrogen rocket without fins and with no tvc fly perfectly straight during the boost phase, after flipping the bottle with liquids, e.g. this video at 7:26:

https://youtu.be/ANGO-gwLV5E?t=446

Why? Why does it not tumble?

It gains at least couple of meters per second (probably about few tens of meters per second) and much much smaller angular velocity. The center of mass due to sloshing should be way off.