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How do you calculate the altitude of max-Q?


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Greetings, gentlemen.

This one doesn't really have much to do with a typical KSP mission, but it's been bugging me for a while and I figured there might be enough steely-eyed missile men around here to give me an answer. The question's in the title: given that the dynamic pressure on a vehicle is one-half the local atmospheric density times the square of the vehicle's velocity (i.e. q = 0.5*rho*v^2), how exactly do you calculate the altitude at which that value is maximized? And why is it important in spaceflight?

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Lordie. Not my expertise, really, but if you'd want to do that in KSP you'd want a graph of your velocity and atmospheric pressure during ascent - there's an addon called telemachus that can actually do that, if you stick a barometer on your ship.

As for why Max Q is important in spaceflight.. i'll quote wiki: "The point of max Q is ... the point at which the airframe undergoes maximum mechanical stress." - given the sheer speed a rocket gets up to and it's necessity to be as light as possible, that's an important design consideration. Don't want yer craft falling apart in atmosphere due to sheer stress.

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Use the Graphotron 2000: it comes with a dynamic pressure gauge as standard, it will give you a nice graph showing how the pressure increases up to a point, then drops off.

I've never seen a spacecraft in KSP fail due to an excessive max q, but then again, that could be the reason behind many of my rapid spontaneous disassemblies! :P

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Because of the placeholder drag model in KSP atm, there isn't really the shearing type effects from atmospheric pressures or air flow that would make max-Q an issue. The drag from atmo is applied to all parts at the same time, regardless of what's around them, so the only time something like this might happen was when there's parts with different drag values working against each other. But they'll also do that regardless of their relative orientation to the craft, airstream, or each other.

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I'm actually looking to figure how it's calculated in RL...I'd like the equation. And I'm not necessarily looking to apply it in KSP, either (though that would probably be the easiest way to figure it out in terms of an example). Thanks for the help anyway, y'all.

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From what reasoning out I've been able to do in my head, it seems like you'd first have to calculate what your craft's ascent trajectory looks like, then plug in the v and rho values you get at each point along that trajectory until you'd find the biggest number. A while back a few other forumites and I put together a few tables and charts for the "index altitudes" of various levels of atmospheric pressure on Kerbin, which might be useful for such a thing, and the scale height equation for Kerbin is on the wiki if you want to know the exact pressure at any given altitude.

Of course, I'm no math whiz, so it would probably take me the better part of a day to work out completely how to calculate a rocket's flight path in the first place.

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Well, lemme ask another question real quick; I know KSP uses a simplified drag model. Does anybody know the equation involved? I saw it not too long ago on the forum but I don't remember where. I could use the kinematic equations to try and brute force an answer, but for that to work, I'd need to know moments of acceleration with time, and I know that gravity and drag are the forces countering thrust. Gravity is easy, thrust is easy, but I don't know how to calculate drag.

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Confusing equation, though...Fd = q (essentially)*d*A, and for A I'm just supposed to use mass?

Maybe it's supposed to be one square meter per kg in the conversion factor for A instead of one cubic meter...square meter would give a result in Newtons, cubic meter gives a result in Joules.

Also looks circular. I have to know velocity to calculate drag, but to get my velocity, I have to know my acceleration rate and to know that I have to know the drag. Maybe V is Vo in this case...I'm going to treat it as such.

Still don't know how to figure out where q is maximized mathematically, but it's a start.

Of course, now I know that max-Q is the point where the drag is maximized...

EDIT: Actually, I can't assume that, can I?

Edited by capi3101
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Also looks circular. I have to know velocity to calculate drag, but to get my velocity, I have to know my acceleration rate and to know that I have to know the drag.

Welcome to the wonderful world of calculus, where everything affects everything else at the same time. Granted, I never did quite wrap my head around it myself.

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  • 2 months later...
EDIT: Actually, I can't assume that, can I?

A simple rearrangement says that Q = Fd /(d*A) = 1/2 * rho * v^2. So, max Q isn't max drag, but the max ratio of drag compared to the drag coef times the, in this case, mass. Assuming a single rocket stage for simplicity's sake means that Q is highest later in the stage, when mass of the stage is lower due to propellant loss and velocity is up because of the burn. I suspect, however, that thrust to weight ratio has a relationship in that it is what can push your velocity higher in relation to rho.

As for the calculus, I'm thinking this leads to partial derivatives in order to get around the circular nature of the acceleration vector being determined by velocity but it's been so long that I can't remember exactly where to go without a refresher.

And then you have to remember that if you're doing a gravity turn, your time at a given pressure altitude is longer. Boy this can get complicated quick. I'm gonna keep working on this. I'm interested in what you find out.

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  • 6 years later...
On 8/14/2013 at 2:53 PM, CalculusWarrior said:

Use the Graphotron 2000: it comes with a dynamic pressure gauge as standard, it will give you a nice graph showing how the pressure increases up to a point, then drops off.

I've never seen a spacecraft in KSP fail due to an excessive max q, but then again, that could be the reason behind many of my rapid spontaneous disassemblies! :P

Realism Overhaul

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Considering that the game's atmosphere has changed significantly since this question was asked, and the information in it no longer applies, we're going to lock this as to not cause any confusion.

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