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Projectile Drag

  • 23-11-2008 9:17pm
    #1
    Posts: 4,630 ✭✭✭


    I'm trying to figure out the air drag on projectiles (For "fun" I'm just extending applied maths problems to include air drag to see how far they really would travel etc.). And a problem has arisen. How do I determine the drag on an object while it is accelerating? The drag equation just gives you an instantanous drag because the equation is dependent on velocity. But as the drag slows down the object, it's velocity would decrease thus it's drag would change. I'm presuming there is no "simple" formula for this, and that it would require a differential equation?

    Thanks in advance for any responses.


Comments

  • Registered Users, Registered Users 2 Posts: 861 ✭✭✭Professor_Fink


    The drag equation just gives you an instantanous drag because the equation is dependent on velocity. But as the drag slows down the object, it's velocity would decrease thus it's drag would change. I'm presuming there is no "simple" formula for this, and that it would require a differential equation?

    Thanks in advance for any responses.

    Well drag usually looks something like F_drag = -c v^2 weher c is some constant usually involving cross-sectional area, and v is velocity.

    The rate of change of momentum of the projectile is then given by dp/dt = F_tot. Where p = mv, and F_tot is the total force. In this case it will be F_drag + F_gravity. Here the quantities will all be vectors, since you are in a 2D or 3D environment. For the sake of simplicity, I'm going to assume the projectile is fired horizontally, and assume the you only care about the horizontal component of velocity, although it is easy to adapt this approach to suit other situations. In this situation we can drop the force due to gravity, since it only affects vertical acceleration and velocity, not horizontal motion. So we have dp_h/dt = -c v^2.

    -> m dv/dt = - c v^2

    -> dv/v^2 = -c/m dt

    Integrate this over the time period 0 to t, and we get 1/v + 1/u= c t/m, where u is the initial horizontal velocity of the projectile. Hope this answers your question.


  • Posts: 4,630 ✭✭✭ [Deleted User]


    That's exactly what I was looking for, thank you Professor_Fink!

    I'll have to play around with this for a few years probably to be able use it properly, but oh well.

    Thanks again!


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