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Air drag equasion

  1. Sep 26, 2008 #1

    I'm having some problems trying to solve the equasion for air drag. I'm currently doing a project about solidfuel rockets and air drag is a big deal =)

    I used this equasion for dragforce (Fd)

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

    which is just a lot of parameters so simplified this is:

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

    I'm not familiar with differential equasions at all but my math teacher told me this is one. To bad he couldn't solve it though.

    Does anybody know how to solve this so it can be used to calculate dragforce Fd at time t?

    Thanks ahead!
    Regards from the Netherlands,
  2. jcsd
  3. Sep 26, 2008 #2
    If a, b, c, are just constants, what you wrote is a quadratic equation for Fd, and this is solved with the usual formula.
  4. Sep 26, 2008 #3
    Hello smallphi,

    could you explain that showing the math? Because I can't come up with anything to solve an equasion involving it's own answer:

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

  5. Sep 26, 2008 #4
    You never solved a quadratic equation in your life? I find that hard to believe if you are doing project about solidfuel rockets.

    You have to expand the square, treat everything except Fd as constants, treat even t as constant, treat Fd as the unknown variable x, and use the formulas for the roots here


    The article has examples, those will be most helpful to you if you haven't solved a quadratic equation before.
  6. Sep 26, 2008 #5
    I'm sorry we must have misunderstood. If the link you posted is what you mean by quadratic equasion then the formula above is no quadratic equasion =)

    Please note that Fd(t) is to be filled into the function, not Fd as constant or t as constant.

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

    Fd(t) means Drag at a given time

    So within the function Fd(t) is another function. In this case also Fd(t).

    This is shurely no quadratic equasion to be solved like as in your link.

    For example, if you were to write out the Fd(t) within the function above, you'd get:

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

    which obviously also has the funtion of Fd(t) in it. So we could write that out as:

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

    its a never ending loop.

    I hope this explains better as to why this is a differential equasion rather than just a quadratic equasion.

    Kind regards,
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