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Hello All,

I recently learned that I'm heading back to school to do some postgraduate study and I need an excuse to revise and touch up my MATLAB skills. 

Simultaneously I have been grappling with the SSTO design problem and the fact that I cannot find any tools that will guide me in the design of an SSTO and payload.

My thought is that while fluid dynamics was not my strong suit (I am a mechanical engineer) the simplified modelling of KSP should make this a reasonable endeavour with a bit of research.

My a first step will be to produce an aircraft design tool that I will input a desired payload, ceiling and perhaps a few other parameters and the program will output required thrust, lift surface area and required coefficient of friction. In this way, the 'art' of aircraft design still exists because you have to design something within those parameter (x lift surface, y kN of thrust etc) so but at least the science gives you a starting point.

Would be keen for thoughts, guidance, advice or be pointed at any existing tools.

Good flying :)

Edited by Crispynaut
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@The Moose In Your House,

Thanks for the video, that's one of the better ones I have seen working through the problem. Thank you for taking the time to post. 

I suppose it also embodies the kind of video that is widespread for SSTO, it has a bit more science but I still think it uses too many rules of thumb. I am trying to avoid empirical figures and develop tools to examine design parameters from first principals and allow a greater level of optimisation without so much trial and error. 

What do you think my chances are?

Edited by Crispynaut

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4 minutes ago, Crispynaut said:

@The Moose In Your House,

Thanks for the video, that's one of the better ones I have seen working through the problem. Thank you for taking the time to post. 

I suppose it also embodies the kind of video that is widespread for SSTO, it has a bit more science but I still think it uses too many rules of thumb. I am trying to avoid empirical figures and develop tools to examine design parameters from first principals and allow a greater level of optimisation without so much trial and error. 

What do you think my chances are?

Every time I create a SSTO and ignore the science stuff the SSTO keeps pulling up, and then stalls and smashes into the ground.

Edited by The Moose In Your House

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1 minute ago, The Moose In Your House said:

ignore the science stuff

I'm with you, I guess what I'm saying (poorly) is I want more science stuff. I want to understand how he came to those numbers.

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1 minute ago, Crispynaut said:

I'm with you, I guess what I'm saying (poorly) is I want more science stuff. I want to understand how he came to those numbers.

What do you mean by "numbers"?

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@The Moose In Your House

How he got to the theoretical max speed of the engine, how he got to 14% of total mass being for fuel, how he decided how many engines to use (why is 1-1.2TWR ideal), how did he work out his maximum payload etc. 

I have no doubt it's out there, I'm on the road so I'll probably have to do more digging when I get home.

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1 minute ago, Crispynaut said:

What do you think my chances are?

Slim, to be brutally frank. A regular rocket launch isn't really solvable analytically, even NASA brute forces it numerically. Add in the additional complications of airbreather performance and it's pretty intractable.

Empirical rules of thumb really work well though.

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4 minutes ago, The Moose In Your House said:

Because it says the max power of the engine on the part.

 

Yep i think we are taking about different things. It's all good I'll probably just tinker and post any results.

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Also if you are having trouble with tsiolkovsky rocket equation read this. (Information from Wikipeda) Newton's second law of motion relates external forces (Fi{\displaystyle F_{i}\,}F_{i}\,) to the change in linear momentum of the whole system (including rocket and exhaust) as follows:

∑Fi=limΔt→0P2−P1Δt{\displaystyle \sum F_{i}=\lim _{\Delta t\to 0}{\frac {P_{2}-P_{1}}{\Delta t}}}\sum F_{i}=\lim _{{\Delta t\to 0}}{\frac  {P_{2}-P_{1}}{\Delta t}}

where P1{\displaystyle P_{1}\,}P_{1}\, is the momentum of the rocket at time t=0:

P1=(m+Δm)V{\displaystyle P_{1}=\left({m+\Delta m}\right)V}P_{1}=\left({m+\Delta m}\right)V

and P2{\displaystyle P_{2}\,}P_{2}\, is the momentum of the rocket and exhausted mass at time t=Δt{\displaystyle t=\Delta t\,}t=\Delta t\,:

P2=m(V+ΔV)+ΔmVe{\displaystyle P_{2}=m\left(V+\Delta V\right)+\Delta mV_{e}}P_{2}=m\left(V+\Delta V\right)+\Delta mV_{e}

and where, with respect to the observer:

V{\displaystyle V\,}V\, is the velocity of the rocket at time t=0
V+ΔV{\displaystyle V+\Delta V\,}V+\Delta V\, is the velocity of the rocket at time t=Δt{\displaystyle t=\Delta t\,}t=\Delta t\,
Ve{\displaystyle V_{e}\,}V_{e}\, is the velocity of the mass added to the exhaust (and lost by the rocket) during time Δt{\displaystyle \Delta t\,}\Delta t\,
m+Δm{\displaystyle m+\Delta m\,}m+\Delta m\, is the mass of the rocket at time t=0
m{\displaystyle m\,}m\, is the mass of the rocket at time t=Δt{\displaystyle t=\Delta t\,}t=\Delta t\,

The velocity of the exhaust Ve{\displaystyle V_{e}}V_{e} in the observer frame is related to the velocity of the exhaust in the rocket frame ve{\displaystyle v_{e}}v_{e} by (since exhaust velocity is in the negative direction)

Ve=V−ve{\displaystyle V_{e}=V-v_{e}}V_{e}=V-v_{e}

Solving yields:

P2−P1=mΔV−veΔm{\displaystyle P_{2}-P_{1}=m\Delta V-v_{e}\Delta m\,}P_{2}-P_{1}=m\Delta V-v_{e}\Delta m\,

and, using dm=−Δm{\displaystyle dm=-\Delta m}dm=-\Delta m, since ejecting a positive Δm{\displaystyle \Delta m}\Delta m results in a decrease in mass,

∑Fi=mdVdt+vedmdt{\displaystyle \sum F_{i}=m{\frac {dV}{dt}}+v_{e}{\frac {dm}{dt}}}\sum F_{i}=m{\frac  {dV}{dt}}+v_{e}{\frac  {dm}{dt}}

If there are no external forces then ∑Fi=0{\displaystyle \sum F_{i}=0}\sum F_{i}=0 (conservation of linear momentum) and

mdVdt=−vedmdt{\displaystyle m{\frac {dV}{dt}}=-v_{e}{\frac {dm}{dt}}}m{\frac  {dV}{dt}}=-v_{e}{\frac  {dm}{dt}}

Assuming ve{\displaystyle v_{e}\,}v_{e}\, is constant, this may be integrated to yield:

ΔV =veln⁡m0m1{\displaystyle \Delta V\ =v_{e}\ln {\frac {m_{0}}{m_{1}}}}\Delta V\ =v_{e}\ln {\frac  {m_{0}}{m_{1}}}

or equivalently

m1=m0e−ΔV /ve{\displaystyle m_{1}=m_{0}e^{-\Delta V\ /v_{e}}}m_{1}=m_{0}e^{{-\Delta V\ /v_{e}}}      or      m0=m1eΔV /ve{\displaystyle m_{0}=m_{1}e^{\Delta V\ /v_{e}}}m_{0}=m_{1}e^{{\Delta V\ /v_{e}}}      or      m0−m1=m1(eΔV /ve−1){\displaystyle m_{0}-m_{1}=m_{1}(e^{\Delta V\ /v_{e}}-1)}m_{0}-m_{1}=m_{1}(e^{{\Delta V\ /v_{e}}}-1)

where m0{\displaystyle m_{0}}m_{0} is the initial total mass including propellant, m1{\displaystyle m_{1}}m_{1} the final total mass, and ve{\displaystyle v_{e}}v_{e} the velocity of the rocket exhaust with respect to the rocket (the specific impulse, or, if measured in time, that multiplied by gravity-on-Earth acceleration).

The value m0−m1{\displaystyle m_{0}-m_{1}}m_{0}-m_{1} is the total mass of propellant expended, and hence:

Mf=1−m1m0=1−e−ΔV /ve{\displaystyle M_{f}=1-{\frac {m_{1}}{m_{0}}}=1-e^{-\Delta V\ /v_{\text{e}}}}M_{f}=1-{\frac  {m_{1}}{m_{0}}}=1-e^{{-\Delta V\ /v_{{\text{e}}}}}

where Mf{\displaystyle M_{f}}M_{f} is the propellant mass fraction (the part of the initial total mass that is spent as working mass).

ΔV {\displaystyle \Delta V\ }\Delta V\ (delta v) is the integration over time of the magnitude of the acceleration produced by using the rocket engine (what would be the actual acceleration if external forces were absent). In free space, for the case of acceleration in the direction of the velocity, this is the increase of the speed. In the case of an acceleration in opposite direction (deceleration) it is the decrease of the speed. Of course gravity and drag also accelerate the vehicle, and they can add or subtract to the change in velocity experienced by the vehicle. Hence delta-v is not usually the actual change in speed or velocity of the vehicle.

If special relativity is taken into account, the following equation can be derived for a relativistic rocket,[3] with Δv{\displaystyle \Delta v}\Delta v again standing for the rocket's final velocity (after burning off all its fuel and being reduced to a rest mass of m1{\displaystyle m_{1}}m_{1}) in the inertial frame of reference where the rocket started at rest (with the rest mass including fuel being m0{\displaystyle m_{0}}m_{0} initially), and c{\displaystyle c}c standing for the speed of light in a vacuum:

m0m1=[1+Δvc1−Δvc]c2ve{\displaystyle {\frac {m_{0}}{m_{1}}}=\left[{\frac {1+{\frac {\Delta v}{c}}}{1-{\frac {\Delta v}{c}}}}\right]^{\frac {c}{2v_{e}}}}{\frac  {m_{0}}{m_{1}}}=\left[{\frac  {1+{{\frac  {\Delta v}{c}}}}{1-{{\frac  {\Delta v}{c}}}}}\right]^{{{\frac  {c}{2v_{e}}}}}

Writing m0m1{\displaystyle {\frac {m_{0}}{m_{1}}}}{\frac  {m_{0}}{m_{1}}} as R{\displaystyle R}R, a little algebra allows this equation to be rearranged as

Δvc=R2vec−1R2vec+1{\displaystyle {\frac {\Delta v}{c}}={\frac {R^{\frac {2v_{e}}{c}}-1}{R^{\frac {2v_{e}}{c}}+1}}}{\frac  {\Delta v}{c}}={\frac  {R^{{{\frac  {2v_{e}}{c}}}}-1}{R^{{{\frac  {2v_{e}}{c}}}}+1}}

Then, using the identity R2vec=exp⁡[2vecln⁡R]{\displaystyle R^{\frac {2v_{e}}{c}}=\exp \left[{\frac {2v_{e}}{c}}\ln R\right]}R^{{{\frac  {2v_{e}}{c}}}}=\exp \left[{\frac  {2v_{e}}{c}}\ln R\right] (here "exp" denotes the exponential function; see also Natural logarithm as well as the "power" identity at Logarithmic identities) and the identity tanh⁡x=e2x−1e2x+1{\displaystyle \tanh x={\frac {e^{2x}-1}{e^{2x}+1}}}\tanh x={\frac  {e^{{2x}}-1}{e^{{2x}}+1}} (see Hyperbolic function), this is equivalent to

Δv=c⋅tanh⁡(vecln⁡m0m1){\displaystyle \Delta v=c\cdot \tanh \left({\frac {v_{e}}{c}}\ln {\frac {m_{0}}{m_{1}}}\right)}\Delta v=c\cdot \tanh \left({\frac  {v_{e}}{c}}\ln {\frac  {m_{0}}{m_{1}}}\right)

Other derivations[edit]

Impulse-based[edit]

The equation can also be derived from the basic integral of acceleration in the form of force (thrust) over mass. By representing the delta-v equation as the following:

Δv=∫t0t1|T|m0−Δmt dt{\displaystyle \Delta v=\int _{t0}^{t1}{\frac {|T|}{{m_{0}}-\Delta {m}{t}}}~dt}{\displaystyle \Delta v=\int _{t0}^{t1}{\frac {|T|}{{m_{0}}-\Delta {m}{t}}}~dt}

where T is thrust, m0{\displaystyle m_{0}}m_{0} is the initial (wet) mass and Δm{\displaystyle \Delta m}\Delta m is the initial mass minus the final (dry) mass,

and realising that the integral of a resultant force over time is total impulse, assuming thrust is the only force involved,

∫t0t1F dt=J{\displaystyle \int _{t0}^{t1}F~dt=J}{\displaystyle \int _{t0}^{t1}F~dt=J}

The integral is found to be:

J ln⁡(m0)−ln⁡(m1)Δm{\displaystyle J~{\frac {\ln({m_{0}})-\ln({m_{1}})}{\Delta m}}}{\displaystyle J~{\frac {\ln({m_{0}})-\ln({m_{1}})}{\Delta m}}}

Realising that impulse over the change in mass is equivalent to force over propellant mass flow rate (p), which is itself equivalent to exhaust velocity,

JΔm=Fp=Vexh{\displaystyle {\frac {J}{\Delta m}}={\frac {F}{p}}=V_{exh}}{\displaystyle {\frac {J}{\Delta m}}={\frac {F}{p}}=V_{exh}}

the integral can be equated to

Δv=Vexh ln⁡(m0m1){\displaystyle \Delta v=V_{exh}~\ln \left({\frac {m_{0}}{m_{1}}}\right)}{\displaystyle \Delta v=V_{exh}~\ln \left({\frac {m_{0}}{m_{1}}}\right)}

Acceleration-based[edit]

Imagine a rocket at rest in space with no forces exerted on it (Newton's First Law of Motion). However, as soon as its engine is started (clock set to 0) the rocket is expelling gas mass at a constant mass flow rate M (kg/s) and at exhaust velocity relative to the rocket ve (m/s). This creates a constant force propelling the rocket that is equal to M × ve. The mass of fuel the rocket initially has on board is equal to m0 - mf. It will therefore take a time that is equal to (m0 - mf)/M to burn all this fuel. Now, the rocket is subject to a constant force (M × ve), but at the same time its total weight is decreasing steadily because it's expelling gas. According to Newton's Second Law of Motion, this can have only one consequence; its acceleration is increasing steadily. To obtain the acceleration, the propelling force has to be divided by the rocket's total mass. So, the level of acceleration at any moment (t) after ignition and until the fuel runs out is given by;

 Mvem0−(Mt).{\displaystyle ~{\frac {Mv_{e}}{m_{0}-(Mt)}}.}{\displaystyle ~{\frac {Mv_{e}}{m_{0}-(Mt)}}.}

Since the time it takes to burn the fuel is (m0 - mf)/M the acceleration reaches its maximum of

 Mvemf{\displaystyle ~{\frac {Mv_{e}}{m_{f}}}}{\displaystyle ~{\frac {Mv_{e}}{m_{f}}}}

the moment the last fuel is expelled. Since the exhaust velocity is related to the specific impulse in unit time as

Isp=veg0,{\displaystyle I_{\rm {sp}}={\frac {v_{\text{e}}}{g_{0}}},}I_{\rm {sp}}={\frac {v_{\text{e}}}{g_{0}}},[4]

where g0 is the standard gravity, the corresponding maximum g-force is

 MIspmf.{\displaystyle ~{\frac {MI_{\rm {sp}}}{m_{f}}}.}{\displaystyle ~{\frac {MI_{\rm {sp}}}{m_{f}}}.}

Since speed is the definite integration of acceleration, and the integration has to start at ignition and end the moment the last propellant leaves the rocket, the following definite integral yields the speed at the moment the fuel runs out;

 ∫0m0−mfMMvem0−(Mt) dt= −veln⁡(mf)+veln⁡(m0)= veln⁡(m0mf){\displaystyle ~\int _{0}^{\frac {m_{0}-m_{f}}{M}}{\frac {Mv_{e}}{m_{0}-(Mt)}}~dt=~-v_{e}\ln(m_{f})+v_{e}\ln(m_{0})=~v_{e}\ln \left({\frac {m_{0}}{m_{f}}}\right)}{\displaystyle ~\int _{0}^{\frac {m_{0}-m_{f}}{M}}{\frac {Mv_{e}}{m_{0}-(Mt)}}~dt=~-v_{e}\ln(m_{f})+v_{e}\ln(m_{0})=~v_{e}\ln \left({\frac {m_{0}}{m_{f}}}\right)}

 

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@Red Iron Crown, acknowledged, I have no doubt they work and I will probably end up reverting to them when I admit defeat. I feel like since KSP is a simplified model of a real world system, it has to be analytically solvable because my computer is doing it. I guess I have to decide how much I want to simplify the system in my model and wether that is just one step ahead of empirical numbers.

Thanks a lot for the input, it does set the stage for me a bit better. I'll have a tinker when I get home and post if I get any results.

Thanks again :)

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4 minutes ago, Crispynaut said:

I feel like since KSP is a simplified model of a real world system, it has to be analytically solvable because my computer is doing it.

There's a difference between calculating what happens and the most optimal way to do it.

4 minutes ago, Crispynaut said:

I guess I have to decide how much I want to simplify the system in my model and wether that is just one step ahead of empirical numbers.

Thanks a lot for the input, it does set the stage for me a bit better. I'll have a tinker when I get home and post if I get any results.

Thanks again :)

Don't let me discourage you, exploring it mathematically can be fun too.

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Just now, Red Iron Crown said:

There's a difference between calculating what happens and the most optimal way to do it.

Haha I respectfully disagree, if you calculate what happens for a wide variety of conditions you can estimate an optimised solution (that being said I will be the first to admit im wrong in 4 days so standby hehe). The model is the hard bit.

Thanks for the feedback, I do enjoy the community here. :)

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2 minutes ago, Crispynaut said:

Haha I respectfully disagree, if you calculate what happens for a wide variety of conditions you can estimate an optimised solution (that being said I will be the first to admit im wrong in 4 days so standby hehe). The model is the hard bit.

That's the numerical approach. :) What I think you'll find is there are so many variables that the number of runs required to bracket the solution is prohibitive.

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