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Calculating the final velocity of a compressed-air launched projectile

  1. Dec 21, 2007 #1
    Ehhhhh.... this really isn't a "homework" problem but something derived out of curiosity(But I guess it could be considered homework-esque).
    1. The problem statement, all variables and given/known data

    Basically, not accounting for drag or friction, what'd be the final velocity of a projectile launched from a compressed air chamber of volume v_0 connected DIRECTLY to a tube of radius R with launch length L with initial pressure P_0, with the projectile launcher set at angle [tex]\theta[/tex] from the horizontal?


    2. Relevant equations

    m is the mass of the projectile
    g is the gravitational acceleration.
    x will be the length from the beginning of the tube to the given position of the projectile inside the tube.
    Area = pi*R^2

    P_x = V_0*P_0/(V_0+x*Area) [The pressure where the projectile is down the tube at x-length]
    F = pressure*Area
    Work = [tex]\int[/tex] F ds
    (With the Y-axis perpendicular to the projectile's motion and X-axis parallel to it.)
    F_x = Area*P_x -mgsin([tex]\theta[/tex])

    3. The attempt at a solution

    The kinetic energy of the initial state of the projectile is 0. The net work done along the X-axis, that being along the length of the tube, will be the final kinetic energy. Since the net work is the integral of the Force applied along the X-axis(As there's no net Y_force with my set-up) in respect to distance, then

    K_f= Work = [tex]\int[/tex]_L,0 F ds =
    [tex]\int[/tex]_L,0 Area*P_x -mgsin([tex]\theta[/tex]) dx =
    Area*[tex]\int[/tex]_L,0 P_x dx -mgsin([tex]\theta[/tex])[tex]\int[/tex]_L,0 dx=
    Area*[tex]\int[/tex]_L,0 V_0*P_0/(V_0+x*Area) dx -L*m*g*sin([tex]\theta[/tex])= (Area is now A)
    V_0*P_0[log(V_0+A*L)-log(V_0)] - L*m*g*sin([tex]\theta[/tex])

    Now that the work ergo final Kinetic energy has been found, relating to the kinetic energy formula 1/2*m*v_f^2= K_f,
    v_f = (2K_f/m)^(1/2) =
    (2/m)^(1/2)*(V_0*P_0[log(V_0+A*L)-log(V_0)] - L*m*g*sin([tex]\theta[/tex]))^(1/2)

    Is that right? I didn't think the solution would be that simple as I imagine a time-based solution would be much more difficult(even though I haven't even envisaged the solving procedures.). Some of the equations were derived from my own thinking, such as the pressure at length-x equation, so I fear some of my premises may be flawed and so the final solution.
     
    Last edited: Dec 21, 2007
  2. jcsd
  3. Dec 21, 2007 #2

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  4. Dec 21, 2007 #3
    Yes, interesting, though most of the content appears to be devoted to addressing oversights that I haven't over-sighted. There are some fairly similar functions, however, that might be useful in relating. Thank you.
     
  5. Dec 23, 2007 #4
    After reading ahead, it appears that my model is a little bit too simplistic. Upon reading about the pressure being directly related to the sum of the force provided by the collision of air-molecules against the sealing membrane, I remembered that an air-launched cannon has a "top speed" since air molecules have a "natural speed limit"(Not an *ultimate one*, but just one under natural physical circumstances) and that's why one team decided to use nitrogen for an ultra-long range projectile. This then shows that the amount of force exerted on the projectile is just not simply due to the "amount of air-molecules inside" and the volume that contains those air molecules(And the area the projectile takes up), but also on the very speed of the projectile(!) as the relative average speed of the air molecules' collisions on the projectile decreases as the projectile's velocity increases, regardless of pressure. So this must be accounted for for an accurate model to be developed(And thus might've been a substantial source of error in others experimentally derived values and expected values that didn't account for this difference.). Any ideas how to mathematically account for it?(Yeah, I know, so does the tube-on-projectile friction and air drag.)
     
  6. Dec 24, 2007 #5

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    The speed of air molecules will have to be taken into effect when the speed of the projectile inside the launching barrel is of the order of the speed of sound in the gas being used under that pressure. I never thought that you wanted such high speeds. I feel in that case many other significant factors would have to be considered. You have to talk to experts.
     
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