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Air resistance in relation to ballistics

  1. Dec 26, 2008 #1
    Having this discussion on another board, basically stemmed from one member saying that a bullet would be in flight for "over 5 seconds"in response to something else someone said about a 0.50 caliber sniper rifle firing at a target 2 miles away.

    I figured that negating air resistance, the time would be 3.857 seconds (using Vf^2 = Vi^2 + 2ad then putting finding average velocity (Vf + Vi)/2 and substituting that into v = d/t) "d" is two miles, or 10,560 feet, "Vi" is 2,800 feet per second, and "a" is -32.15 feet per second.

    So assuming that was right, I found that for the time to be "over 5 seconds" (I used 5.5), the final velocity would have to be 1,040 feet per second (using v = d/t, solving for v as average velocity, multiplying Va by two and subtracting initial velocity to find the new Vf) is that correct?

    Anyway, none of that really matters, just background, my question is, how do you take wind resistance into account in kinematics, and how high would the wind speed have to be to cause a 63% loss of velocity? I'd preferably like to do this without calculus.

    Thanks in advance.
     
  2. jcsd
  3. Dec 26, 2008 #2

    mgb_phys

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    Wind isn't needed to slow the bullet - just it's motion through the air is enough.
    The force on an object moving through air (at height speed)
    F = 1/2 [tex]\rho[/tex] v 2 A C d

    [tex]\rho[/tex] is the density of air
    v is the velocity
    A is the cross section area
    C d depends on the shape, for a bullet it's around 0.3
     
  4. Dec 26, 2008 #3

    russ_watters

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    Mythbusters did a "can a falling bullet kill you" episode where they found the terminal velocity of a bullet to be around 100fps. You can plug that into the drag equation to find the drag coefficient (the bullet tumbled in their demo, but at least that gives you an upper bound). I think you'll find that the drag needed to explain the 5.5 second flight time is absurdly high.
     
  5. Dec 26, 2008 #4

    rcgldr

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    Bullets from most rifles travel at super sonic speeds. At these speeds, the math to calculate the drag is so complex that tables of coefficients are required to do the ballistics calculations.

    http://en.wikipedia.org/wiki/External_ballistics

    In the case of 50 caliber sniper rifles, the muzzle velocity is very high, enough that the bullet will still be well above supersonic at impact to improve the accuracy.

    Must have been a very small caliber weapon. Hunters sometimes get pelted by tiny birdshot falling from above without much harm, but people have died from falling bullets. I'm pretty sure that a 50 cailiber bullet would have a much higher terminal velocity than 100fps.
     
    Last edited: Dec 26, 2008
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