# Calculating air drag for a massive rocket

1. Sep 26, 2006

### Noone1982

Not just a little rocket that goes a 1000 feet in the air, but one that can get into space. How does one take into account that the air density grows less with altitude, etc?

Anyone know of any good resources for a problem like this?

2. Sep 26, 2006

### ken

Usual practrice in aerodynamics is to work in "indicated airspeed" by assuming sea level density (1.225 kg/m^3). If you want actual velocity you need to account for density. General formulae:

L = 1/2 . p . v^2 . S . Cl

D = 1/2 . p . V^2 . S . Cd

Where :

L = Lift Force
D = Drag Force
p = density (1.225 kg/m^3 @ sea level)
V = velocity
S = surface area
Cl = coefficient of lift
Cd = coefficient of drag

Depending how complicated you want to go will have changing density, acceleration, and Cd with Reynolds number due to velocity so you will have either to estimate by calculation in small incriments or use integration. Good luck.

Ken

3. Sep 27, 2006

### Noone1982

So, I suppose I cant get it as an explicit function of time.

4. Sep 27, 2006

### Dense

When one of your variables changes (like air density at various altitudes), and you have a formula for how it changes (like air density at various altitudes), there's this marvelous mathematical tool that lets you do calculations. It's called "calculus." Get an adult to help. :tongue2:

5. Sep 27, 2006

### rcgldr

An old equation I found for up to 300,000 feet (more boundaries above this):

Code (Text):

pressure

P_0 = 14.7 psi
p= P_0*(1-6.8755856*10^-6 h)^5.2558797    h<36,089.24ft
p_Tr= 0.2233609*P_0
p=p_Tr*exp(-4.806346*10^-5(h-36089.24)) h>36,089.24ft

density

rho_0 = 2.06 lb mass / cubic yard
rho=rho_0*(1.- 6.8755856*10^-6 h)^4.2558797 h<36,089.24ft
rho_Tr=0.2970756*rho_0
rho=rho_Tr*exp(-4.806346*10^-5(h-36089.24)) h>36,089.24ft

http://www.centennialofflight.gov/essay/Theories_of_Flight/atmosphere/TH1.htm

Drag versus speed gets complicated once you're near or beyond supersonic. The range above .95 to 1.00 is different than below .95. The range between Mach 1.0 and Mach 1.4 is different than above Mach 1.4.

Regarding sources, obviously NASA and space oriented companies deal with this stuff all the time, but I wasn't able to find any links with all the required formulas.

Last edited: Sep 27, 2006
6. Sep 27, 2006

### rcgldr

Unlike the abstract world of ideal equations, ballistics invovling high altitudes is too complicated to intergrate directly. So after spending a year learning to solve all sorts of differential equations in a class, you find in the real world that many situations are too complicated to solve directly, and you end up using numerical intergration (like Runge-Kutta). Think of this as "advanced spread sheet math". You have a set of formulas that calculate an acceleration vector given position and velocity vector. Numerical integration is then used to "predict" a new position and velocity vector based on the current acceleration vector over a small step in time. The process is repeated in order to calculate a path. Runge Kutta speeds this process up by "remembering" values from mutlple previous steps.

Then there is real world testing of high speed aerodynamics. Rocket sleds can be fun:

http://www.46tg.af.mil/world_record.htm

Last edited: Sep 27, 2006
7. Oct 1, 2006

### Noone1982

Thanks everyone. I am stepping through the equations at 0.1 second intervals and my results are fairly close to conditions of a Saturn V rocket at 80 seconds into flight.

For the air drag, I intend to use what Ken put forth:

$D_{f}\; =\; \frac{1}{2}\mbox{C}_{d}pV^{2}\mbox{S}$

Now, I can presumably just have the density as a function of height. Now Cd is just the reynolds number as a function of velocity?