Interpreting Drag Equations

[FLutterFlight]

The page this comes from can be found HERE

I'm having troubles interpreting this set of equations relating to drag. I'm not sure what 'e' stands for in the the first one, and in the second, I can't find what to do with the 'Kg/(m³*atm)'. I'm not sure how to treat the 'kg' or the 'm³'. Any help is appreciated

Edited October 9, 2013 by FLutterFlight

[Joe32320]

e is a constant approximately equal to 2.71828...

The kg/(m3 * atm) is just the unit of the constant I believe, so you can ignore it for working out the maths. The units kg/m3 is the units for density and the atm in the equation is cancelled by the fact pressure, p is measured in atm as well.

[FLutterFlight]

e is a constant approximately equal to 2.71828...

Huh. Well, that's nice. But what IS e (what does it represent)?

And thanks, by the way. My lack of common knowledge and slow comprehension is sometimes cumbersome. I appreciate your assistance!

Edited October 9, 2013 by FLutterFlight

[Rhomphaia]

Huh. Well, that's nice. But what IS e (what does it represent)?

And thanks, by the way. My lack of common knowledge and slow comprehension is sometimes cumbersome. I appreciate your assistance!

http://en.wikipedia.org/wiki/ℯ

[FLutterFlight]

Oh, I see. that's nice. I'll settle for an estimation.

Thank you all!

I do plan on making a very efficient flight pattern using KoS, and using this pattern to create some large orbital vessels with ease (and keeping a mostly-legit feeling to the game).

[FLutterFlight]

One more thing. In what unit of measurement does this equation output, kN, newtons, or what?

[Supernovy]

It should be Newtons, if density is Kgm^-3, v is m/s and A is square metres.

Kerbal Space Program drag is pretty weird, though. 'd', the drag coefficient, is the drag given ingame (usually 0.2) multiplied by the mass of the part. This means that the unit the comes out of that equation should be Newton-kilograms. Unless the drag is actually the specific drag, in which case mass cancels out. The drag value also takes into account surface area, so the equation is more like Fd = rho*v^2*sum(Di*mi).

That means if you put in mass in tonnes, it'll come out kilonewtons. As long as you keep the rest of the units SI.

[numerobis]

The two usual ways to think about this are either:

d = 0.008, and A (in m^2) is the drag value of your part times the mass in kg. So an LV-N, mass 2.25 tonnes, with a drag value of 0.2 has A = 2250 * 0.2 = 450 m^2.

OR

d = the drag value of your part, and A is 0.008 times the mass in kg. So the LV-N has d = 0.2 and A = 2250 * 0.008 = 18 m^2.

Either way, you multiply all the terms together and the number comes out right. If you write the mass in tonnes rather than kg, you get a result in kN rather than N.

[FLutterFlight]

Hmm... But I could just use .2 as an approximation, right? I've already spent a few days poking at this project, so I'm okay with 'pretty darn close'. And the outcome will be in newtons (.001 of kN).

Bringing up another thing...

Just out of curiosity, why wasn't the '(GM)/r²'(found in the 2^ndand 3^rd) divided from both sides? and if '(GM)/r²' were to be represented by a single value, how would it be written as it is in the 3^rd equation?

Edited October 9, 2013 by FLutterFlight

[Leyvin]

It is there in the final variant but it isn't really presented in a very good means.

http://en.wikipedia.org/wiki/Drag_coefficient

Follow that link and you'll understand the equations being used considerably better.

[numerobis]

I'm not sure what 0.2 is supposed to be an approximation for, so it's hard to day whether it is a good approximation.

Your second and third equations check out fine; the third just isolated v from the second equation. What's the issue?

[aeronaut]

Hmm... But I could just use .2 as an approximation, right? I've already spent a few days poking at this project, so I'm okay with 'pretty darn close'. And the outcome will be in newtons (.001 of kN).

Bringing up another thing...

Just out of curiosity, why wasn't the '(GM)/r²'(found in the 2^ndand 3^rd) divided from both sides? and if '(GM)/r²' were to be represented by a single value, how would it be written as it is in the 3^rd equation?

The third equation solves the second equation for v where v is the object's terminal velocity in free fall. GM/r^2 is the acceleration due to gravity and is still in the right hand side of this equation.

[FLutterFlight]

Oh! that makes loads of sense now. Heh, thanks.

I really do appreciate all of your help. I do believe that will be all.