arkie87

I think the way matlab plots it, the beginning of the contour line is the value (you can tell this from the fact that the colors end at 0.9, but clearly, efficiency increases above 0.9 in the far right). Thus, your estimate from the graph is incorrect. The orange contour begins at 0.7 and ends at 0.8 not begins at 0.65 and ends at 0.75. I have also run the numbers for TWR = 1.1 and TVR = 4.98, and arrived at eta = 0.7326. Why not??? V0 is correct... ?

Yes, both of those the model does not account for. Can you confirm that when you launch with TWR < 1, you dont move for a while (until TWR >1)?

Of course it should! All fuel spent while TWR < 1 is completely wasted! You are basically jettisoning mass to get your TWR >=1... Is that not what happened during your tests?

So when you start with TWR of 0.91, what exactly happens for the first few minutes when you burn with TWR < 1? You should stay in place and just waste fuel. Is that what happened? I dont see 69% or 72% anywhere in that graph that you quoted. Please explain where you are getting these numbers from....

The reason I cannot put it in terms of ISP is because ISP, alone, is not what is important. What is important for estimating efficiency is ISP*g0/v0. So having a certain ISP engine on one planet will perform differently than on others. Thus, i would need to provide a graph of efficiency vs. ISP and TWR for every body in KSP. Defining TVR, on the other hand, allows me to show just one graph, which can be used for all planets and moons as long as the player can do some basic math to find out what their TVR is. This is the advantage of non-dimensionalization. I dont think that result is necessarily obvious, given the complicated, non-linear nature of the equations. It is important to know that if you increase ISP, you efficiency drops, and so while you might gain more deltaV, you will also need more deltaV to get into orbit. If a player didnt know this, they might improve their engine ISP, and reduce FMR (amount of fuel) to compensate to maintain the same total deltaV they had before, without realizing they will need more fuel to get into orbit.. I'm not sure what this has to do with anything. My graphs dont suggest you go to TWR = infinity. They suggest quite the opposite, actually, that TWR > 2.5 are quite pointless (though very fun).

+1 But the results show that MOAR BOOSTERS doesnt pay very much... (at least once TWR > 1.6 and for airless bodies) But it is MOAR FUN!

Well then you will have to trust my result

In the science forum, i developed, edited, tested in-game, and refined the paper. Now it is fully-refined, and serves merely as information for game play. Maybe tutorials is the more proper forum.

I fully understand, though you could use another language, like C++ or Octave (which is basically a free version of Matlab) Who knows... maybe after learning C++, you will write an App that makes billions of $$$$

[tl:dr]To obtain efficiency greater than or equal to 90%, a thrust-to-weight ratio (TWR) of at least 1.4 at launch on an atmosphereless, Kerbin-sized body (like Tylo) is required with standard ISP engines; but for planets with smaller orbital velocity or more efficient engines (nuclear or Ion), TWR at launch to obtain efficiency greater than 90% increases. [/tl:dr] I have made a computer model in Matlab (EDIT: Excel version now available! See Below!) to simulate optimal TWR (thrust-to-weight ratio), TVR (thrust-velocity-ratio), and DVR (delta-V-ratio). I provide non-dimensional contour plots to show results which are globally applicable for the same non-dimensional parameters. Link to the Paper Results: This graph shows the independence of efficiency on DVR. The slope of the curve is essentially zero, until efficiency drops to zero instantly when delta-V-ratio no longer provides enough deltaV to get into orbit due to efficiency. Since efficiency is not a function of DVR (i.e. how much extra fuel you carry into orbit), we only need to vary TWR and TVR. The above plot contains all the information a player needs when designing craft. First, increasing TWR increases efficiency while increasing TVR decreases efficiency. Increasing TWR increases efficiency by allowing the craft to aim more horizontal, thereby, using more of its fuel to accelerate into orbit instead of fighting gravity. Increasing TVR decreases efficiency since, for a given DVR, less fuel is burned. This, in turn, results in a more constant TWR during the flight, and therefore, a longer flight as well as a steeper angle above vertical, causing more fuel to be wasted fighting gravity. Second, for low TVR, lower TWR are needed for a given efficiency; similarly, for a higher TVR, a higher TWR is needed to obtain the same efficiency. This is the direct result of the trends described above, and is the most important result of the simulation: thus, for two crafts with the same ISP engine, the craft on Minmus will require a higher TWR for the same efficiency as a craft on Mun, and so on. Finally, this model should be easy to test. Since efficiency is not a function of DVR (i.e. size or scale), two craft with the same TWR can be compared on the same planet with different ISP engines or on different planets with the same engine, and the craft with the larger ISP engine or equivalently on the planet with a smaller orbital velocity should have reduced efficiency. And in case you dont believe that efficiency is independent of DVR for all TWR and TVR: Excel Versions of the Simulator: Version 1 Version 2 Version 3 with Macros

No one said it was a "real force". It is a phantom force resulting from change of coordinates. Its effect is relevant regardless of the coordinate system used, however, since if you neglect it when calculating the thrust angle, you will steadily gain altitude. If you dont believe me, write your own code to simulate this case for a 2-D Cartesian case (as i have done three posts above this one). So LD's model is accurate in the limiting case of large TWR, but only because burn angle at TWR = infinity equals 0 degrees regardless of the way you calculate burn angle... (nothing to do at all with how many degrees around the circle it travels during the burn).

What did we decide about centripetal lift?

I'm glad the way Iskierka explained it to you made sense. I was trying my best :-( The phantom term exists regardless of which frame of reference we use. If you are using Cartesian coordinates, and you calculate phi using asin(m*g/T), you will steadily gain height for finite TWR. If you subtract phantom term (centripetal lift, centrifugal lift , whatever you want to call it, etc...) you will not gain altitude. I have redone my results for the Mun (instead of Kerbin). The result is essentially the same: If Phi is calculated from asin(m*g/T): If Phi is calculate from asin(m*(g-V^2/R)/T): In the end, as others have said, the difference is small since centripetal lift term only becomes significant near orbital velocity. Furthermore, as you and others have noted, the final altitude gained becomes increasingly less significant for higher and higher TWR (since in the limit of TWR = infinity, phi = 0 for both equations):

You cannot plot the first data point on the contour plot because the TWR for this point is 0.91 but the contour plot starts from TWR of 1.... You can however plot your numbers on this contour plot:

Didnt have time to read the rest (i will later tonight), but in my code, I'm using m = 1 for simplicity, so my formula is correct.

Yes, you convert acceleration to force units by multiplying by m, but this is not what we are discussing.... lol sum(F) = m*a_r From the textbook: a_r = r_dot_dot - V_theta^2/r So, substituting this identity: sum(F) =m*(r_dot_dot - V_theta^2/r) Re-arranging: m*r_dot_dot = sum(F) + m*V_theta^2/r The term m*V_theta^2/r is positive when on the "force" side of the equation, therefore, it acts as lift...

See this response: In order to be considered a "force" it has to be on the same side as the sum(F) i.e. force side. It starts off as negative on the inertia (i.e. m*a) side. We have to switch it over to the force side to treat it like a force, and when we do that, it becomes positive. I assume now you see how ridiculous this post was: I wouldnt recommend responding this condescendingly or rudely, especially since if you are wrong, it will make you look like a fool

LethalDose: The simplest way to explain the need for the "centripetal lift" term is the following: Consider a spacecraft at orbital velocity. By definition, this is the velocity where V^2/R = g. If you leave the spacecraft alone, it will orbit in a circle. If you try to calculate the thrust angle needed to balanced gravity from phi = asin(m*g/T)=asin(1/TWR) you will arrive at a non-zero thrust angle from the horizontal/prograde. This is clearly not correct since at orbital velocity, one can burn at horizontal since gravity is canceled by centripetal lift term. I have made a 2-D Cartesian model that integrates the second-order equations of motion (for position and velocity in x- and y-coordinates). The mode assumes infinite ISP engines, such that mass is constant (not a function of t) for simplicity, and is always equal to 1. For these conditions, if we neglect centripetal lift, the angle, phi, above horizontal needed to cancel gravity would be constant since TWR does not change during flight. The governing equations of motion are: a_x = T*cos(pi/2 + theta - phi) - g*cos(theta) a_y = T*sin(pi/2 + theta - phi) - g*sin(theta) The angle, phi, is calculated from: phi = asin(g/T); The trajectory is as shown below: It is clear the spacecraft lifts off the planets surface since the angle is steeper than needed due to centripetal lift. On the other hand, if we calculate the angle from: phi = asin((g-v^2/R)/T) The trajectory is as shown below: It is clear that the spacecraft stays on the surface, as expected.

Sorry, I didnt post the derivation just the result. I have since posted the derivation as well. I hope that helped explain the notation and nomenclature. Yes, r*theta_dot^2 = V_theta^2/R, since V_theta = R*theta_dot This term is negative, but it is on the same side as m*a (i.e. inertia side); when we bring it over to the sum(F) side, it becomes a positive force r_dot and r_dot_dot are zero since radius is not changing. Acceleration in the radial direction (a_r) is not zero since particle is following a curved path. Sorry for my part in this confusion. I should have posted the derivation instead of the result. My bad. I agree my approach isnt rigorous (though i think the equations in my textbook justify my approach) and my use of x- and y-coordinates is confusing since i'm really using polar coordinates. I will change the document to reflect this (though i think it might be obvious to some people). The images I posted can be zoomed in on/they are plenty large on my screen...

If you get deltaV values below LD's prediction, it means that either KSP is wrong, or LD is wrong. Obviously, I will suggest LD's approach is too conservative because it neglects centripetal lift, which becomes significant once you approach orbital velocity. This, in turn, means that you dont have to aim as high above the horizontal, which makes the burn slightly more efficient.

I agree with LethalDose that my approach isnt "rigorous" w.r.t. to a_theta, but i think if d(theta)/dt is small (d(theta)/dt = V_theta/R << 1), then my approach is accurate enough.

Confused. Here is your table: I don't see a Mun TWR = 1 @ ISP = 290... there is only one @ TWR = 0.91 (in bold). EDIT: I see what you are saying-- that in my table TWR = 0.91 @ISP=290 ~= TWR 1.01 @ ISP = 4200, which shouldnt be possible since ISP =4200 should perform worse than ISP 290 (for the reasons we've discussed). However, that is only at constant TWR. In our case, TWR < 1 is really bad, since it means we are burning fuel until TWR = 1 before we can start moving... if TWR started at 1 (instead of 0.91) then the results would be much different. As a side note, if TWR was 0.91 with ISP = 4200 i imagine efficiency would be zero, since due to high ISP, TWR would never go above 1 to allow the craft to actually move...

I have taken pics (no scanner, sorry) of the derivation of time derivatives in polar coordinates. Sorry for the quality but it will have to do. Please review and get back to me.

I wouldnt recommend being so rude on these forums, especially since nothing you've said "corrected" anything I've said. I don't understand why people on these forums are so rude and condescending so quickly. a_r is negative, due to the radial unit vector changing direction (in the Cartesian plane). But even so, this means you accept the form of the equation I've provided, which includes the V^2/R term i.e. centripetal lift... There are only two forces acting on the body -- thrust and gravity-- but from Newton's Law: sum(F) = m*a, and it is the m*a term that changes when switching to polar coordinates i.e. the V^2/R term. I'm not sure how the reference i supplied isnt an "independent source" -- it is a textbook on dynamics... what else could i possibly provide? I'm not confused about polar coordinates. From your own admission, it's been a long time since you've dealt with them, so i would like to humbly recommend refreshing yourself. You can use either polar or cartersian coordinates to solve this problem. Cartesian does not require a V^2/R term, but polar does. And i'm willing to bet that both approaches will yield identical results...

Your table doesn't have mun with twr of 1 at isp 290. It has twr if 0.91....