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Muzzle velocity given test range, time of flight and ballistic properties.

  1. May 9, 2012 #1
    This isn't really homework, but I figure this is the most appropriate place to post this... Please do say if you think there is a more likely place it will get answered - I'm new!

    The problem statement, all variables and given/known data
    I am trying to calculate the muzzle velocity of an air rifle. I can find the time of flight over a known horizontal range. I know the mass and quadratic velocity damping constant due to air drag of the pellet. I also know that over a reasonable range (20m), air drag cannot be ignored, so [itex]\large \dot{x}_0 = \frac{r}{T}[/itex] is not true.

    I will assume the trajectory of the pellet is a parabola, because the trajectory is so flat a parabola is a good enough approximation to real life. I will treat the motion decay problem as one dimensional (along the parabola arc length), so gravity can be ignored from here on.

    Relevant equations
    X, parabola arc length is known, [itex]\large X = T \sqrt{(\frac{r}{T})^{2}+(\frac{g T}{2})^{2}}[/itex], where r is horizontal range, and g is gravity.
    T, time of flight is known
    C, quadratic velocity damping constant is known
    M, pellet mass is known
    Acceleration due to quadratic velocity damping is given by [itex]\Large \ddot{x} = \frac{C \dot{x}^{2}}{M}[/itex]
    [itex]\dot{x}_{0}[/itex], the initial velocity, is unknown.

    The attempt at a solution so far
    I have already worked out the arc length, as seen above. I have come up with the following expression for velocity as a function of time, which may, or may not be useful;

    [itex]\LARGE \dot{x}(t) = \frac{\dot{x}_{0}}{1 + \frac{t C \dot{x}_{0}}{M}}[/itex]

    and an expression for distance, as a function of time;

    [itex]\LARGE x(t) = \frac{M}{C} ln(1 + \frac{t C \dot{x}_{0}}{M})[/itex]

    I am not really sure what to do next, to get what I want; an expression for [itex]\dot{x}_{0}[/itex] as a function of X, T, C and M.
     
    Last edited: May 9, 2012
  2. jcsd
  3. May 9, 2012 #2
    Assuming that all of your math is correct (I haven't gone through the steps), could you not just take your last expression and invert it for the initial velocity by taking the exponential of both sides?
     
  4. May 9, 2012 #3
    Wow, thanks, I am obviously in need of sleep!
     
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