Ideal atmospheric escape? (Calling the help of advanced engineering mathematics.)

[Yasmy]

I don't know of any situation where drag forces in a fluid are exponential in velocity.

You'll have to ask your roommate for clarification. Both in air and in liquids,

drag forces tend to be proportional to the square of the velocity.

Though for very small particles at low velocity, the drag may be linear in velocity.

In KSP: Drag

Edited March 29, 2014 by Yasmy

[capi3101]

One of my roommates told me that, in fluid dynamics, the drag acting upon a body in motion in a fluid is an exponential function of velocity, in contrast to the textbook example of DE's involving drag where drag is a function of velocity squared. Does anyone know how drag is modeled in KSP? I know the drag model here does not take into account reference areas, but at least it is related to velocity. Anyone?

Here we go: http://wiki.kerbalspaceprogram.com/wiki/Drag#Drag

Basically, the only difference between KSP's default drag model and reality is that a mass proportionality is used in place of the cross-section area; the equation is otherwise 100% the same, but the difference is enough. It's unrealistic in that during an ascent (as I've mentioned before) your mass will be decreasing as long as you burn. It means that in KSP you can have a bunch of tanks, engines, and what-not set out radially and you can still expect a decrease in the drag; in real life, that wouldn't happen until after you've chucked off some stuff. It also explains why most real rockets are of the long, skinny variety, and at least partially explains the problems that some KSP players have with the notion of asparagus staging (when they say it's "unrealistic", that's the part they're talking about; there are some experiments with fuel ducts going on, so that bit's real).

If you want to play the game with a drag model that uses cross-section instead of mass, there's a mod for that: Ferram Aerospace Research (FAR). I don't use it my own self, but a lot of folks do.

[Zephram Kerman]

In this thread, we've mainly discussed velocity. Another useful thread (from almost a year ago!) leans into the turn shape.

http://forum.kerbalspaceprogram.com/threads/33570-Best-ascent-path-for-mechjeb

At the time, we approached the problem by trial-and-error instead of mathematically, but the discussion should help with the strategy.

[Thogapotomus]

I was under the impression that staying near terminal velocity on ascent was just a by product of Kerbin's soupy atmosphere and the way drag is calculated in game. Pretty much the same reason it seems most efficient to start the gravity turn at 10k even though that's not realistic. If you use ferram, for example, you can start the gravity turn at launch, and with a streamlined rocket exceed terminal velocity by a large margin with minimal losses as opposed to stock.

[Woopert]

I suppose I should point out that gravity is only 9.8 at the surface; it is decreasing with altitude...

Does KSP actually model the gravity losses with altitude? I know it's actually quite small in real life (like 90% gravity in low orbit or something?), and it's also a way to disprove those people who believe space is "zero gravity." I also noticed with Kerbal Engineer Redux it lists surface TWR.

[capi3101]

By my observation, KSP does indeed model changes in gravity with respect to altitude. There isn't anywhere in the game where the g-force is ever exactly zero; I imagine a number of things would break if that ever happened.

KER also lists a Current TWR when you're in space; it wouldn't be able to do that if the gravity was zero.

[GoatRider]

KER also lists a Current TWR when you're in space; it wouldn't be able to do that if the gravity was zero.

I think that's in reference to the Kerbol (sun) sphere of influence.

[Frederf]

TWR is a misnomer. All of the mod readouts of TWR (that I know) are really TMR where mass is expressed in m*g0 (aka units of weight at reference gravity). Expressing weight with respect to local gravity doesn't allow an easy understanding of acceleration since the acceleration perpendicular to gravity would depend on local gravity if expressed as a ratio.

As for minimizing drag, gravity losses are a simple function of time. You start a stopwatch at launch and every second that ticks by represents loss so minimizing time is the goal. Aero drag's losses are some power greater than linear with reduced total time. To reduce gravity drag (total time) in a given segment of flight to 9/10th you have to fly 10/9ths as fast. Aero drag increases at greater than linear with speed (the quantitative function really doesn't matter here). The result is a crossover where at some speed the reduction in gravity drag by going faster results in an equal (and increasing) increase in aero drag for that speed increase. This is by definition terminal velocity.

The difference between the falling body and the ascending body is the drags oppose instead of add but the desires are the same. In both cases each body desires to minimize gravity drag, one because nature and the other by design. The limit on falling is ultimately vacuum freefall but further aero drag. The limit on ascension is vacuum acceleration and further limited by aero drag.

The derivation is to find the function of total drag which is the sum of both parts. Since we expect an optimum (minimum) set the derivative (with respect to v) of the total drag to zero. The only way to reconcile the two parts being zero in sum and non-zero themselves is for them to be equal, aka terminal velocity.

[daniu]

TWR is a misnomer. All of the mod readouts of TWR (that I know) are really TMR where mass is expressed in m*g0 (aka units of weight at reference gravity).

That doesn't make it a misnomer, considering TWR means "Thrust to Weight Ratio", and weight = mass * gravitational acceleration. It's not TMR, where Mass would leave gravity out... its point is to determine whether a craft would lift off.

However, for the calculation asked for from OP, you're right of course. You'd use a function TWR(h) = m*g(h).

Similarly, you'd put drag into a function of height d(h) depending on air density and its decline with altitude.

You'd also have to consider that you lose weight over time at a given thrust due to the fuel consumption.

All those forces add up in integrals dh and dt somehow, but to be honest that's one step of math I'm not ready to go for as I'm pretty convinced it will come down to the terminal velocities anyway

Edited March 31, 2014 by daniu

[BlazeFallow]

I have read through about 4 different threads discussing the issue of ideal velocities and gravity turn heights, and I would like to share my findings and assumptions here.

First K^2 offered mathematical proof that terminal velocity is the optimal ascent velocity, something that many kerbonauts have assumed for a while now.

The solution y'(t) = Sqrt(c(y(t))/k(y(t))) is locally optimal whenever c'(y) = g'(y). In other words, vertical ascent is optimal at terminal velocity even with arbitrary variation in gravity and air density with altitude.

I know everyone suspected as much, but it's nice to have a rigorous proof of it.

The added complexity comes from adding in the gravity turn. The complexity comes from the following list of assertions:

- In KSP gravity as an effect on your craft is not a constant, and thus cannot be treated as a constant, but rather a formula based on the distance of your craft from the surface as compared to surface gravity. Thanks to Capi for this one.

- You get a greater benefit from the Oberth effect from starting your gravity turn earlier.

- You have to contend with more atmospheric drag for a longer time, thus eating away some of your dV by fighting atmo, by starting a gravity turn earlier.

- Your engines will also have a (generally) much higher ISP at higher altitude, thus you will get more dV from your rockets the sooner you can get to that higher altitude.

The last item is much more pronounced at 0-10km at which point your engines are already well above low end ISP and are quickly reaching peak ISP. The rate at which the ISP changes is almost parabolic in nature from what I have noticed.

So the problem that I suppose someone who is MUCH better at math than I am (not saying much as I can barely do any calculus) should try to tackle is this:

Given that a craft will be traveling at terminal velocity v for a given altitude h, what is the dV tradeoff between counteracting drag but gaining more from the Oberth effect, when also accounting for an ISP that is dependent on altitude h

Can someone please tell me if any of my assertions are incorrect or if my maths are being way too fuzzy. Also please comment on whether or not my ending pseudo equation makes sense.

[wilt57]

I was under the impression that staying near terminal velocity on ascent was just a by product of Kerbin's soupy atmosphere and the way drag is calculated in game. Pretty much the same reason it seems most efficient to start the gravity turn at 10k even though that's not realistic. If you use ferram, for example, you can start the gravity turn at launch, and with a streamlined rocket exceed terminal velocity by a large margin with minimal losses as opposed to stock.

Just wanted to clarify something here. Ferram does not allow you to fly faster than terminal velocity. You can always fly faster than terminal velocity, if you want to. Ferram does not allow you to save delta-v by flying faster than terminal velocity, it allows your terminal velocity to be faster. You still want to fly at terminal velocity, it's just that Ferram allows you to have a faster terminal velocity. Why?

What is terminal velocity? Very quickly, it's when the force of drag equals the force of gravity. If your rocket is more aerodynamic, it's drag generated is less, so it's terminal velocity is higher as gravity remains constant. We want to fly at terminal velocity because at that speed the sum of our losses to gravity and our losses to drag is the least. Any faster and we generate more losses to drag than we're saving to gravity. Any slower and we're losing more to gravity than we're saving to drag. It's all about balancing the losses we generate. As we know, it basically takes 4,500 delta-v to reach LKO. At 80 km, my ship is only traveling 2,279 m/s. The missing 2,221 m/s went into drag, gravity, and a little bit of steering losses. It's managing these losses that's the goal of an efficient launch.

[Thogapotomus]

-snip-

Yeah you're right. I guess my point was that in stock KSP it seems like terminal velocity is fairly consistent with little regard to the rocket you're launching, but with ferram a streamlined rocket can easily exceed the terminal velocity it would experience with stock KSP. Also, as far as I know, the drag losses you experience for exceeding terminal velocity in ferram are less extreme than in stock.