- #1

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

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- Thread starter Noone1982
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- #1

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

- #2

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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

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

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- #5

rcgldr

Homework Helper

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

The only link I found.

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

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.

Code:

```
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
```

The only link I found.

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

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.

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- #6

rcgldr

Homework Helper

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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.Dense said: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."

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

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- #7

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

[itex]D_{f}\; =\; \frac{1}{2}\mbox{C}_{d}pV^{2}\mbox{S}[/itex]

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?

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