Author Topic: Basic Rocket Science Q & A  (Read 281589 times)

Offline aero

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Re: Basic Rocket Science Q & A
« Reply #900 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

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

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
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Offline pagheca

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Re: Basic Rocket Science Q & A
« Reply #901 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.

Offline aero

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Re: Basic Rocket Science Q & A
« Reply #902 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.
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Offline baldusi

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Re: Basic Rocket Science Q & A
« Reply #903 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.

Offline aero

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Re: Basic Rocket Science Q & A
« Reply #904 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.
« Last Edit: 04/17/2014 04:47 AM by aero »
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Offline pagheca

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Re: Basic Rocket Science Q & A
« Reply #905 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.
« Last Edit: 05/05/2014 04:13 PM by pagheca »

Offline R7

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Re: Basic Rocket Science Q & A
« Reply #906 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.
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Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #907 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.
« Last Edit: 04/17/2014 05:01 PM by Proponent »

Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #908 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
« Last Edit: 04/20/2014 12:22 PM by Proponent »

Offline pagheca

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Re: Basic Rocket Science Q & A
« Reply #909 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, 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?

Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #910 on: 04/20/2014 12:25 PM »
The attachment to my previous post in this thread contained a mathematical error which has now been corrected.

Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #911 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.
« Last Edit: 04/20/2014 12:34 PM by Proponent »

Offline ClaytonBirchenough

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Re: Basic Rocket Science Q & A
« Reply #912 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.)
« Last Edit: 05/04/2014 04:35 PM by ClaytonBirchenough »
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Offline deltaV

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Re: Basic Rocket Science Q & A
« Reply #913 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.

Offline ClaytonBirchenough

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Re: Basic Rocket Science Q & A
« Reply #914 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...  ::)
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Offline R7

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Re: Basic Rocket Science Q & A
« Reply #915 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.

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Offline pagheca

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Re: Basic Rocket Science Q & A
« Reply #916 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?

« Last Edit: 05/24/2014 11:26 PM by pagheca »

Offline ciscosdad

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Re: Basic Rocket Science Q & A
« Reply #917 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?

Offline aero

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Re: Basic Rocket Science Q & A
« Reply #918 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

Edit - forgot the link
« Last Edit: 06/06/2014 05:55 AM by aero »
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Offline Proponent

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Re: Basic Rocket Science Q & A
« Reply #919 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, which contains some information about the topic.  The answer is not simple, unfortunately.