arkie87

It DOES display (free-fall) terminal velocity, this value is just inaccurate due to incorrect drag coefficient being used. To get up to terminal velocity in FAR, I would try >10g acceleration. vf^2 = vi^2 + 2*a*h; vi=0; a=98.1 m/s/s, vf=1200 m/s h = 1/2 vf^2/a = 7.34 km altitude So you will reach about 7.34 km altitude with 10 g (which should be plenty of buffer to counter act drag force) such that you should hit terminal velocity at 10 km.

Ferram, what is the general form of the correlation between drag coefficient with Mach number? Is it piece-wise defined? If you give me this relationship, I can write a matlab code in 5 seconds to disprove GoSlash.... Also, does Mach number in FAR factor in temperature from c = sqrt(gamma*R*T)? I suppose I can ignore than dependence for the purposes of this debate though and assume V_sound = 343 m/s independent of altitude.

Even if Cd varies with 1/V^n where n > 2, this still would not matter since decreasing drag coefficient results in increased terminal velocity, not decreased, as GoSlash is suggesting. - - - Updated - - - Nice Avatar. Mind if i copy it?

I said i cannot calculate it since I dont have the information. I did not see that relationship. Please show this relationship to me. unless you are referring to this: http://upload.wikimedia.org/wikipedia/commons/0/0e/Qualitive_variation_of_cd_with_mach_number.png as it was i that posted that. That relationship is qualitative only. Each part has its own relationship, which are, essentially, its aerodynamic properties. What do you mean by "display to flip?" Acceleration does not effect terminal velocity, as Ferram explicitly stated.

Ok. This approximately Mach 2. Drag coefficient according to FAR readout is 0.187. Terminal velocity readout is 1270 m/s, which is about Mach 4. Drag coefficient at Mach 4 will be less than 0.187 (how much less I cannot say since i do not have the relationship between drag coefficient and Mach that FAR uses), since drag coefficient decreases with Mach after Mach 1. Terminal velocity and drag coefficient are inversely proportional, that is, if drag coefficient decreases, terminal velocity increases and vice versa. Thus, in this case, drag coefficient at terminal velocity will be smaller than drag coefficient at current velocity, thus terminal velocity is actually higher than predicted, not lower. Below Mach 1, terminal velocity prediction is an overestimate since drag coefficient is increasing. Even so, terminal velocity prediction, even in this case, is still larger than current velocity, so acceleration should still help. I have done so here. I cannot calculate terminal velocity at any other velocity other than the one FAR reports, since i do not have the relationship between drag coefficient and Mach that FAR uses.

My final attempt is as follows. Even with stock aerodynamics, terminal velocity is optimum during ascent. I would like to know how you manage your ascent velocity during the first 7 km climb before the gravity turn (you mentioned somewhere you usually climb straight vertical for 7 km)? Since you clearly prefer experiments to theory, I would like to propose an experiment: measure deltaV required to get to LKO in stock aerodynamics by climbing however fast you like until 7 km and then performing your gravity turn. Then follow this profile as close as you can, which specifies craft velocity vs. altitude. I guarantee following this chart will require less deltaV since you will be flying at terminal velocity in stock.

We have already countered this argument. The terminal velocity display in FAR is not always 100% accurate, since it is not accounting for the effect of velocity on the drag coefficient. However, since it uses drag coefficient at current velocity, it IS useful for assessing whether we are below or above terminal velocity, since when current craft velocity and the terminal velocity displayed in FAR and are equal, the terminal velocity displayed in FAR is 100% accurate. Thus, while the terminal velocity indicator does not tell us quantitatively how far away from terminal velocity we are, it does always qualitatively tell us if we are above or below it. And since knowing whether we are above or below is what is needed for establishing optimum ascent speed and thrust, the terminal velocity function is adequate as is. I dont know where you get this idea from. Ferram himself has countered this argument, and i have posted a video showing that terminal velocity is not a function of spacecraft acceleration already - - - Updated - - - Ferram always does

Yes, Starman, I, and FERRAM himself have already said this. See Ferram's response in page 8:

I have asked ferram in PM if he thinks it is worth using the current *guess* of terminal velocity to approximate the drag coefficient. That way, as time goes on, the prediction will become more accurate (assuming conditions arent changing too fast). The only downside to this, is that because the multiplier is piece-wise and derivative is discontinuous at mach 1, this method might flip flop back and forth and never converge if subsequent guesses flip flop back and forth on either side of Mach 1. Thus, under-relaxation might be necessary,and is probably not worth it. As you mentioned, all that is needed is to know if your velocity is above or below terminal to know if you should throttle up or throttle down. - - - Updated - - - This is not true. See my and Starman's point above.

Yes, this is an approximation by Ferram so that it doesnt use too much cpu power to get an exact value of terminal velocity. However, if craft velocity is close to terminal velocity, terminal velocity value will be correct. Thus, the indicator should read more as: terminal velocity is faster than current speed or slower....and that is the only information needed to assess launch efficiency.

This is still incorrect. It is showing the instantaneous terminal velocity were it in free fall at that altitude, since this is the definition of terminal velocity.... If you think the velocity profile that would be measured had it been dropped from space is the optimal velocity required for launch, then i ask you this: From what height should it be dropped? 70 km i.e. just outside Kerbin's atmosphere or from the edge of the SoI... These two drops will have vastly different velocity profiles through the atmosphere... Regardless, neither is correct. Instantaneous terminal velocity is the "correct" (read: only) one to use to assess ascent profile and efficiency.

The optimal velocity during ascent is the instantaneous terminal velocity of the craft, were it in free fall at that altitude (which is sorta what you describe). The terminal velocity readout in FAR is the instantaneous terminal velocity of the craft at that altitude (and attitude). One does not need to conduct an experiment to drop it from space and somehow, let it come to terminal velocity while simultaneously remaining at constant altitude the whole time. Terminal velocity can be calculated analytically using the equations i provided a few posts ago.

I like this approach, since the only thing that matters is in career mode is the Kerbucks required to get a certain package into orbit-- and not its deltaV, efficiency, etc... I think this is the crux of my argument, so I'm glad you proposed this test. My proposal is as follows: Build the cheapest craft-- in terms of Kerbucks-- to get the upper stage into orbit around the Mun with FAR aerodynamics installed (if you dont use FAR, you will lose). Upper stage consists of: Mk16 parachute, mk1 command pod, TR-18A stack decoupler, FL-T400 fuel tank, and LV-T30. I will try tonight. I look forward to see your results.

I agree. It's passed my bedtime. I'm glad you are interested in this enough to keep discussing past yours. I think i see the problem: terminal velocity doesnt actually decrease with altitude for the first 5 km; this is merely coincidence since i am flying vertically and accelerating. If I were flying horizontally and accelerating, the same trend would be observed. What is happening in the background is due to the way FAR calculates terminal velocity (i.e. a numerical problem related to the solver): FAR always uses *current* velocity to calculate drag coefficient which it then uses to calculate terminal velocity from: 1/2 rho V_t^2 A Cd = m*g OR V_t= sqrt(2 m*g/(rho A Cd)) However, Cd, drag coefficient, varies with velocity/Mach number as shown here:http://upload.wikimedia.org/wikipedia/commons/0/0e/Qualitive_variation_of_cd_with_mach_number.png However, the drag coefficient is calculated using the current velocity, and not the terminal velocity. To get accurate results for terminal velocity, terminal velocity should be used when calculating drag coefficient or current velocity and terminal velocity must be close; otherwise, drag coefficient used to calculate terminal velocity will be wrong, and might be higher or lower depending on which side of the Cd vs Mach number curve you are on. I have confirmed that his is how it works, by flying a spaceplane at relatively constant/level flight and observing terminal velocity change as drag coefficient is updated as expected. I have recorded this and will upload video to youtube tomorrow as i am too tired now. I think this is actually an interesting observation. Props Slashy, and Good night. Edit: reviewed NathanKell's post regarding FAR, and tried reading the code to see if drag coefficient is calculated based on current velocity or predicted terminal velocity; it seems like my suspicion is confirmed. Perhaps NathanKell can elaborate?

This just isnt true. I would recommend you read up on drag coefficient, terminal velocity, and the way drag force is treated in KSP. Who says objects dont free-fall at Mach 1 at 5 km? If its dense and aerodynamic enough, of course it will! Data about the air? The air has terminal velocity? What? (incidentally, terminal velocity of air is zero, since air is neutrally buoyant). Vt readout is worthless in FAR because you will never reach it, not because it isnt optimal.

I think your assumption is that drag force is only a function of altitude i.e. pressure gradient, and not velocity. If drag force was only a function of altitude, then, you would be right: since atmosphere is not changed, then optimal flight path shouldnt be either.

Terminal velocity demonstrates the balance between gravity force and drag force, not gravity and pressure gradient. FAR alters the drag force (drastically). The good news is, i think i understand where you are coming from and what your misunderstanding is. Did this help clear it up? I think your assumption is that drag force is only a function of altitude i.e. pressure gradient, and not velocity. If drag force was only a function of altitude, then, you would be right: since atmosphere is not changed, then optimal flight path shouldnt be either.

The trend i was describing was the result of this trend, which occurs (qualitatively) for every object, whether in thrust or free fall, whether aiming up, horizontal, or down: http://upload.wikimedia.org/wikipedia/commons/0/0e/Qualitive_variation_of_cd_with_mach_number.png As Mach 1 is approached, Cd increases, thus, terminal velocity decreases. Above Mach 1, Cd decreases, and so, terminal velocity increases again. So yes, it will happen to a free-falling object as well. Ferram can confirm this as well.

No one was arguing about that... Admittedly, i wasnt sure if FAR changed atmospheric parameters, but it is entirely irrelevant to this argument.

Terminal velocity is supposed to depend on drag (i.e. velocity), but not thrust. I have just posted a video, above, showing this to be true.

Watch this video if you think FAR reports "terminal velocity under acceleration". I repeatedly floor the engine and then cut it and terminal velocity does not increase by a factor of TWR, as it would if your definition was correct.

What did you "personally" confirm earlier and how/by asking whom?

KER does not have a terminal velocity readout. Terminal velocity readout comes from FAR directly. Pressure, pressure gradient and gravity are the same; but the interaction of pressure on the craft is completely different in FAR. Furthermore, we have already established in another thread that the rockets in my youtube videos are outperforming the nominal FAR values, so increased drag forces are clearly not a problem, even if i was "reading the KER readout wrong". And once again, there is no such thing as terminal velocity under acceleration from thrust. It varies with Mach number because that is what happens physically, and is accounted for in FAR. It goes to infinity after Mach 1 because drag coefficient is decreasing and air density is approaching zero. Terminal velocity in space above Kerbin is infinity, since air density (and therefore drag) is zero. Anyway, let's wait for Ferram...

FYI: Your imgur gallery didnt work. Very interesting work. I need to review in detail. Looks promising/poignant though. Thanks for posting!

I never said FAR changed anything about the atmosphere, only the interaction between the craft and the atmosphere i.e. drag. My Previous analysis makes no assumptions regarding the planet's atmosphere, or the drag model used-- stock or FAR-- only the relationship between drag force and velocity i.e. quadtratic. This quadratic relationship holds for both FAR and stock, and is the reason that optimal ascent velocity equals (falling) terminal velocity. I assume by "acceleration" you mean "thrust", i.e. m*a = 0 = T - W - D; T = W + D, where D is calculated at V_terminal. Anyone who knows anything about terminal velocity will know this suggestion is preposterous. A quick googling of "terminal velocity" results in the wikipedia definition, which refutes your claim: "The terminal velocity of an object is the velocity of the object when the sum of the drag force (Fd) and buoyancy equals the downward force of gravity (FG) acting on the object. Since the net force on the object is zero, the object has zero acceleration." However, since I know now that you will not listen to anything i say, I have asked Ferram himself (the maker of FAR)... Stay tuned.