arkie87

In your paper, you say phi_min(t) = arctan(1/TWR) (eq. 8) Why are you using arctan... shouldnt it be arcsin? Or do i have an old version of the paper? T sin(phi) = m*g, no? - - - Updated - - - If we use arcsin instead of arctan, then my numerical model and your analytical solution match perfectly, when i disable centripetal lift. For convenience, the solution of the ODE: dv/dt = TVR*FMR/(1-FMR*t) cos(arcsin((1-FMR*t)/TWR)) is (via Wolfram Alpha, but after manual simplification... which Wolfram Alpha couldnt do for some reason?): v(t) = c1 - TVR/TWR*X(t) - TVR*ln(m(t)) + TVR*ln(TWR*(X(t)+TWR)) where: X(t) = sqrt(TWR^2 - m(t)^2) m(t) = 1 - FMR*t And c1 is the constant of integration which must be found by enforcing v(t=0) = 0 m, t, and v are dimensionless I dont think t can be solved for @ V = 1 i.e. V_orbital, given the non-linear nature of the equation.

You are using my model correctly. I noticed that your model predicts less deltaV under those conditions, and got confused as well, since this shouldnt be possible, and it's what led me to post this question on your thead. However, that might be a trivial problem since your model might not be set up to handle the case when TWR < 1 (after all, mine wasnt until Slashy tested it under those conditions which led me to modify the code to handle it). Nevertheless, i think a better test case is this one: T= 3.2560 Mwet = 2 g = 1.628 Isp = 1,000,000 R = 200 i.e. what does your model predict for ISP --> infinity @ TWR = 1? Efficiency should approach zero, since TWR will hover around 1 for a long-ass time until enough mass is burned to get it moving. When I use my model with centripetal lift disabled, I get 3.72% efficiency since angle is almost 90 degrees during the entire burn. When i enable centripetal lift, i get 23.04% efficiency since upon approaching orbital velocity, angle approaches zero. Moreover, it is my understanding that the only difference between our models is my inclusion of the centripetal lift term. However, i have added a mechanism to turn this term off, and our answers still differ. Thus, i think i need to investigate whether our models really are the same except for this one difference. I will get back to you.

I have created a third (and final) version that uses Macros available here: https://www.dropbox.com/s/16lgb6o9ey4g0xs/Horizontal%20Orbital%20Insertion%20Simulator%20v3%20with%20Macros.xlsm?dl=0 It uses a Macro to perform the numerical integration. The macro can be called as a normal Excel function i.e. type in "=Efficiency(TWR,TVR,?CL)" into any cell, and it will give you the efficiency for that TWR, TVR, and centripetal lift assumption. It also includes total deltaV (which is also the same as expended deltaV), as per LD's request.

Ok, i've tried doing this in your model, but wasnt sure if it was equivalent or not. Thanks. just divide V_orbital by efficiency, as per the formula/definition given: dV_expended = V_orbital / efficiency

Okay. I have updated model to always run at DVR = 1/eta such that it is using approximately the same number of integration steps every time I have also added a "physics assumptions" section, where you can enable and disable "centripetal lift" to see how this term effects results. Happy simulating! https://www.dropbox.com/s/9o3vqf9hlzyzpmf/Horizontal%20Orbital%20Insertion%20Simulator%20v2.0.xlsx?dl=0

EDIT: i take this first point back. I think i agree that slope should be zero in theory. The reason its not is because of the numerical integration i.e. if DVR >> 1, then it wont end up using all 1000 integration steps, and so, results wont be as accurate. I am revising my model to always use 1000 steps, and to calculate necessary DVR and/or FMR (sort of what you are doing). This is better for accuracy. Thanks!

LD: I am using your version of spreadsheet without vlookups. Can you explain this result I obtained? It is giving me an efficiency > 0, which shouldnt be possible since TWR < 1 given starting and end mass?

LD: Any chance you can simplify your model to only integrate up to orbital velocity, to make it more easy to compare to my model? Pretty please

Yes, i noticed that too. First off, I never said the slope equals 0, just that slope is approximately 0 (for all intents and purposes). Besides, there is no reason off the bat to assume that if slope doesnt equal zero, there is something wrong... after all, it was this model itself that actually "discovered" that the slope is approximately zero... Second, the small differences can be due to numerical integration. I used ~1000 time steps (maximum) of first order, forward Euler integration. Errors are inherent in any numerical method used, so it should be no surprise if values change slightly.

the 9.82 figure does not come from the exact value of 1g on Kerbin. It is just a constant. In KSP, it is exactly equal to 9.82 m/s2, and does not vary with altitude, planet/body/moon etc.. Thrust by definition is: Thrust = m_dot*v_exhaust For some reason, engineers thought it useful to define and use specific impulse when discussing rocket performance: v_exhaust = g0*Isp This g0 is just a conversion factor, so it is a matter of convenience and can be rounded to exactly 9.82 (or 9.81 in real life).

This is true -- the overall gist -- increasing TWR is good for efficiency-- is still there, but the specific value of TWR needed for, say, 90% efficiency, will vary between our models, and the difference might change for different planets or engines. You have only provided data for one body so far, and i cannot get your excel sheet to work (perhaps due to vlookup?)

Besides, i have shown/am willing to bet that for high TVR (i.e. even smaller moons), the models will diverge even further, since very little fuel mass is burned during ascent, so TWR @ burnout will be approximately equal to TWR at liftoff. If you dont believe me, give me some data points to compare, and I will gladly do it. (though now anyone can since i have released an excel version)

What is your model's result for TWR = 1, 1.1, 1.2, 1.3, 1.4?

I have shown the practical difference... In practice, difference is small... (though it might get a lot larger for TWR ~= 1)

I think the confusion is that it is well know that burning prograde, when above orbital velocity, is most efficient due to Oberth effect. You do not want to burn horizontal, since we established that horizontal and prograde are two different directions, unless you are in a circular orbit or at apoapsis/periapsis in an elliptical one.

I have shown the practical difference between our two models on the first page of LD's thread: As you can see, the difference is only significant for low TWR. LD has said that his model assumes a small distance traveled/angle changed during the burn, which automatically means his model requires a high TWR for accuracy, or else it violates its own assumptions (how high a TWR it requires is unclear). Our models are both accurate at high TWR. At low TWR, only my model is accurate.

In KSP? Damn.... you are right. All of my modeling work is RUINED!!!! Just kidding.

At higher velocities, means V > V_orbital, which is out of the range of my model. So he clearly wasnt talking about my model's theta, but rather, your model's theta/phi, since mine stops at V=V_orbital. If you are going above V_orbital such that you have an elliptical orbit, after you pass periapsis, "horizontal" will be below "prograde".

Speaking of the devil, guys, guess what I have been working on??? hmmm? An excel version of my Model Get it while it's hot! https://www.dropbox.com/s/sddikc3qzv37zyx/Horizontal%20Orbital%20Insertion%20Simulator%20v1.0.xlsx?dl=0

I would be shocked if there was an analytical solution for t_star... hence my numerical approach.

In my model, theta is always w.r.t. to horizontal.... In general though:

My definition of theta is the same....

I think horizontal is relevant reference direction when below orbital velocity; and prograde is relevant reference direction at or exceeding orbital velocity.

I suppose it's overlooked; i will revise problem statement to make it more clear. Thanks for pointing that out. But yes, you could deduce it only integrates up to orbital velocity from v_bar = 1. If you are going above orbital velocity i.e. in a non-circular orbit, at periapsis and apoapsis, prograde and horizontal directions coincide. The moment after periapsis, prograde direction is above horizontal.

It's of course entirely possible I've made an error somewhere What about t_star? Can you show it is FMR independent analytically? I think not... If we ignore t_star, efficiency should be independent of TWR as well, but this clearly isnt the case. But good work anyway