Solving the Drag equation

1. Jun 25, 2009

Hobnob

Does anyone have a solution (approximate is fine) for the Drag equation? What I mean by this is that I need to be able to get position and velocity as functions of time for a body falling through air from rest.

I managed this for a viscous liquid where drag is approximately linearly dependent on velocity, but in air it's quadratic and I'm getting a fairly horrible integral for velocity, and position is eluding me completely!

Any help would be very much appreciated

Thanks
Hob

2. Jun 25, 2009

Hobnob

Okay, I've done some more research and found the answer to my original question (it was on wikipedia all the time, I just missed it). The function is:

v = sqrt(g/r) * tanh(t*sqrt(g*r)), where r is a constant that depends on various factors that don't matter at this stage. This is easily integrated to give me a function for position as well.

But unfortunately, I'm still missing one last step: getting this function to work for an arbitrary initial velocity v0. As far as I can see, the general solution is

v = sqrt(g/r) * tanh(t*sqrt(g*r) + c)

where c is equal to arTanh(v0*sqrt(r/g)). But because arTanh is only defined for -1 to +1, if v0 is too big, this doesn't work.

I'm doing some more calculating, but if anyone can see where (and if) I've gone wrong, that would help a lot.

 Just to clarify, the issue is this: The formula for falling bodies assumes the body is *speeding up* to terminal velocity. I'm interested in what happens at the moment you open your parachute and *slow down* to terminal velocity, at which point the system seems to break down.
Hob

Last edited: Jun 25, 2009
3. Jun 25, 2009

Hobnob

This is turning into more of a blog than a thread...

I think I've solved it, with the help of a dodgy cached document on Google - thanks, Gernot Hoffmann! The solution is to replace tanh with coth and cosh with sinh when the initial velocity is greater than terminal velocity.