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Internal Ballistics

  1. Jun 16, 2009 #1
    Hi guys,

    I am hoping some one can help me out here and after some forumlas and very basic ones at that on internal ballistics. There is a heap of information out there and they tend to jump straight to the hard stuff which tends to be tough on a novice like myself.

    I have done a heap of googling and the formulas are either totally over my head or not what I am after.

    The formulas are based around ww2 tank guns, so caibres of 20mm up to 120mm

    I have some real world figures below, but cannot work out the relationship that can determine the muzzle velocity from the powder mass

    diameter=37mm shellweight=.600kg cartridgesize=145mm velocity=670m/s barrellength=37.7 calibre

    diameter=50mm shellweight=2.06kg cartridgesize=419mm velocity=835m/s barrellength=60 calibre

    diameter=75mm shellweight=6.8kg cartridgesize=495mm velocity=740m/s barrellength=43 calibre


    The formula I have come up with just does not cut it. I was using X as a pseudo factor to get the rough estimates so Force would equal Force. As you can see I am far from a physics genius and that why I need help.

    P = Volume of Powder
    X = Explosive force Factor
    m = shell mass
    v = velocity

    PX = mv^2

    sqr (PX / m) = v

    Now I know there huge amount of factors that add and subtract but I am after even ballpark forumla based on powder being the same quality and power, shells the same aerodynamic velocity.

    What I am trying to achieve is if I know the powder volume and the shell weight and the calibre, what is the muzzle velocity

    Please remember to keep it simple and a example is always great, thanks for any help that can be given.
     
  2. jcsd
  3. Jun 16, 2009 #2
    I am no ballistics expert but I am not sure if your formula works. When the explosion happens then the chemical energy gets converted into heat and into pressure, but this is only half the story. The explosion also produces an expansion speed, which the projectile cannot surpass.
    I don't know how much the explosion shock wave matters when shooting large caliber, but if it does, then maybe the barrel length is adjusted such that we get some target high speed.

    If the shock wave doesn't matter I would assume that the explosive reaches the same temperature in all cases and produces some amount of gas proportional to the mass of the explosive and try to calculate the work done (equaling the kinetic energy) by adiabatic expansion of the pipe volume.
     
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