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Need help with D.E to account for air drag of my projectile

  1. May 6, 2008 #1
    This past weekend, for fun, I made a pneumatic air cannon, using compressed air to shoot out a small projectile at very high velocity. I'd post my video, but I have not made 15 posts on this forum.

    What I need to do is find as close as I can to the actual initial
    velocity of a projectile, once it leaves the barrel. If I have the air
    cannon pointed at 90 degrees to the ground, than I could shoot the
    projectile vertical and time hang time, factoring in known
    acceleration of gravity, to find an approx initial velocity. The only
    problem is, with a projectile going this fast, with significant
    surface area, there will be significant air drag to account for.

    As air resistance effects the velocity of the projectile, and as
    decreased velocity coincides with a decrease in drag, I believe I will
    be needing to use differential equations.

    Do you have any idea how I could setup a differential equation to
    account for air resistance in my quest to find as close as I can to
    the projectile's initial velocity? I have never done anything with air drag before. I just finished Calculus II, but I can put in the nessesary time and effort if someone can get me started here.

    Thanks,
    John
     
  2. jcsd
  3. May 7, 2008 #2

    HallsofIvy

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    How are you modeling the air resistance? Two commonly used ways are "linear";F= -kv, where F is the friction force due to air resistance, v is the velocity vector, and k is a constant, or "quadratic"; F= -k |v|v, where |v| is the 'length' of the velocity vector or speed. This is "quadratic" because |F|= -k|v|2.

    The first case is fairly simple: the x coordinate of velocity is given by mvx'= -kvx and the y coordinate of velocity is given by mvy'= -kvy- (1/2)mg. Unfortunately that is only accurate for very low speeds.

    The quadratic case is harder: the x coordinate of velocity is given by

    [itex]mv_x'= -k\sqrt{v_x^2+ v_y^2}v_x[/itex], [itex]mv_y'= -k\sqrt{v_x^2+ v_y^2}v_y- (1/2)mg[/itex].
     
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