Author Topic: Basic Rocket Science Q & A  (Read 282553 times)

Offline kevin-rf

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Re: Basic Rocket Science Q & A
« Reply #1000 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...
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Offline msat

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Re: Basic Rocket Science Q & A
« Reply #1001 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.

Offline R7

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Re: Basic Rocket Science Q & A
« Reply #1002 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.
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Offline msat

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Re: Basic Rocket Science Q & A
« Reply #1003 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.

Offline ClaytonBirchenough

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Re: Basic Rocket Science Q & A
« Reply #1004 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?
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Offline strangequark

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Re: Basic Rocket Science Q & A
« Reply #1005 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
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"
« Last Edit: 07/31/2015 07:07 PM by strangequark »

Offline baldusi

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Re: Basic Rocket Science Q & A
« Reply #1006 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
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).

Offline ClaytonBirchenough

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Re: Basic Rocket Science Q & A
« Reply #1007 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
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!
Clayton Birchenough
Astro. Engineer and Computational Mathematics @ ERAU

Offline gbaikie

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Re: Basic Rocket Science Q & A
« Reply #1008 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
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


Offline ClaytonBirchenough

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Re: Basic Rocket Science Q & A
« Reply #1009 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.
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Offline R7

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Re: Basic Rocket Science Q & A
« Reply #1010 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.
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Offline Hoonte

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Re: Basic Rocket Science Q & A
« Reply #1011 on: 08/29/2015 09:33 PM »
What was the largest mass launched into low earth orbit in a single launch?
« Last Edit: 08/29/2015 09:40 PM by Hoonte »

Offline AS-503

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Re: Basic Rocket Science Q & A
« Reply #1012 on: 08/29/2015 09:37 PM »
IIRC the Apollo 17 stack comprised of partially spent SIVb with LEM and CSM.

Offline Bob Shaw

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Re: Basic Rocket Science Q & A
« Reply #1013 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?
« Last Edit: 08/29/2015 09:55 PM by Bob Shaw »

Offline Burninate

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Re: Basic Rocket Science Q & A
« Reply #1014 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.

Offline edkyle99

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Re: Basic Rocket Science Q & A
« Reply #1015 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

Offline deaville

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Re: Basic Rocket Science Q & A
« Reply #1016 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.
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Re: Basic Rocket Science Q & A
« Reply #1017 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
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Offline Jim

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Re: Basic Rocket Science Q & A
« Reply #1018 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.

Offline Bob Shaw

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Re: Basic Rocket Science Q & A
« Reply #1019 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!