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Solving the Drag equation

  1. Jun 25, 2009 #1
    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. jcsd
  3. Jun 25, 2009 #2
    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.

    [Edit] 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
  4. Jun 25, 2009 #3
    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.
     
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