# Air Drag formula?

1. Sep 26, 2008

### v6maik

Hello,

I'm working on a project about solid fuel rockets and since these are moving through the atmosphere, they experience Air-drag. I'm trying to set up a formula to exactly calculate the height a rocket will achieve. So without the use of any model. The problem I ran into is that I can't get the formula for acceleration to include air-drag, since I ran into the following loop:

-the speed of the rocket (and thereby the acceleration) depends on the air-drag.
-the air-drag depends on the speed of the rocket (and thereby the acceleration)

I found this equasion for air-drag but it's of no use in its current form.

Fd(t)= -0,5 * p * A * Cd * v^2

as you can see, drag depends on the speed relative to the air-mass, which is pretty obvious.

But since v=a * t, this formula is the same as:

Fd(t)= -0,5 * p * A * Cd * (a*t)^2

and since a= F/m this formula is the same as:

Fd(t)= -0,5 * p * A * Cd (((F / m ) *t)^2

note that F is the net Force on the rocket. The net Force at a given time is equivalent to the propulsion force minus gravity minus drag:

Fnet(t)= Fp(t) - Fg(t) - Fd(t)

So the drag formula now is:

Fd(t)= -0,5 * p * A * Cd (((Fm(t) - Fg(t) - Fd(t)) / m ) *t)^2

Notice that this formula Fd(t) involves it's own answer, so it is a differential equasion, right?

Now, I can simplify this formula to this, leaving 3 constants: a, b and c:

Fd(t)= a * ( (b-Fd(t) )/c * t)^2

Any suggestions about solving this problem? Or might there be a different equasion to calculate air-drag at a given time during acceleration?

Kind regards,
Maik

2. Sep 26, 2008

### rcgldr

There isn't any one formula for drag that works for high speeds, especially transonic and supersonic speed. The mathematical model for bullets and cannon shells is complex, relies on the usage of tables from actual measured data, and it wasn't until around 1990 (Desert Storm), that tanks (USA tanks) could reliably shoot other tanks with a single shot with advanced fire control systems.

3. Sep 26, 2008

### olgranpappy

no. that is only true if the acceleration is constant (and if the velocity is zero at t=0). In general
$$a=\frac{dv}{dt}$$
and so
$$v(t)=v(0)+\int_0^t a(t) dt$$