## Air drag with acceleration

Hello and thanks to all who read this. Recently I've just been messing around with air drag equations, trying to extend applied maths problems to include air drag. And I've hit a road block, at least with regards to my knowledge anyway.

I've been using the F_drag = 1/2 P (mass density of fluid) v^2 C_d (drag coefficient) A (area). But my problem with this equation is it's dependence on velocity. If I were to use this equation on a projectile which is under acceleration (under gravity, and also the air drag would slow down the velocity) it would change the initial velocity, making the equation useless to me (I think).

I guessed that air drag on an accelerating body would require a differential equation, so I tried to go about making one.

F = c.v^2 (c is just the constant of pressure, area and drag coefficient etc in the drag equation)

So, I got...
dP/dt = c.v^2
m(dv/dt) = c.v^2
dv/v^2 = c/m dt

Then I went about integrating this trying to get some kind of an equation. But to no avail. I don't have a great physics knowledge as I'm only in school; so could someone be so kind as to help me get an equation which could calculate the air drag on a body that is undergoing acceleration.

I don't know if I'm making much sense in this post; but thanks anyway!

Just thinking about it there: would the best method be to calculate the air drag on the projectile at various time intervals? I.e. every second, then recalculate the air drag at the new lower speed, then, a second later recalculate again? If you get what I mean.

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I'm pretty sure you can't integrate the force analytically when you have a ~v2 force term. So this would require a numerical solution to calculate the position vs. time.

 Just thinking about it there: would the best method be to calculate the air drag on the projectile at various time intervals? I.e. every second, then recalculate the air drag at the new lower speed, then, a second later recalculate again?
Yes, that is the basic idea. There are algorithms of varying sophistication for doing this.

The simplest way is known as Euler's Method. Given x1 and v1, calculate x2 and v2 at a time Δt later as follows:

x2 = x1 + v1Δt + (1/2) a Δt2
and
v2 = v1 + a Δt

where "a" is calculated from F/m, given F(t, x1, v1)

Euler's Method is fairly easily entered into Excel (if you're familiar with it). A more involved, but also more accurate, method is 4th Order Runge-Kutta.

Regards,

Mark
 Recognitions: Homework Help Note that your equation for drag is an approximation at low sub-sonic speeds. In the case of ballistics (bullets, cannon shells, high speed aircraft), there are no simple equations and instead tables (of coefficients) are used.

## Air drag with acceleration

A falling object increases velocity until it reaches it's terminal velocity, assuming it is high enough to start with.
http://en.wikipedia.org/wiki/Terminal_velocity

 Quote by Redbelly98 I'm pretty sure you can't integrate the force analytically when you have a ~v2 force term. So this would require a numerical solution to calculate the position vs. time. Yes, that is the basic idea. There are algorithms of varying sophistication for doing this. The simplest way is known as Euler's Method. Given x1 and v1, calculate x2 and v2 at a time Δt later as follows: x2 = x1 + v1Δt + (1/2) a Δt2 and v2 = v1 + a Δt where "a" is calculated from F/m, given F(t, x1, v1) Euler's Method is fairly easily entered into Excel (if you're familiar with it). A more involved, but also more accurate, method is 4th Order Runge-Kutta. Regards, Mark
Dont do this. It wont work
 Recognitions: Gold Member One site with interesting explanations and formulas is http://www.math.cornell.edu/~numb3rs..._of_watch.html You want the formula under VERTICAL MOTION which reflects acceleration due to gravity... but this formula is for uniform (constant) acceleration......I assume that's what you want....you'll note there the drag force is an exponential function of time and drag coefficient. OR ...Try Googling "air resistance as a function of acceleration" ..lots of hits...