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How to calculate the velocity of a bullet fired by a rifle?

  1. Jun 25, 2011 #1
    Hi everybody,
    I'm trying to calculate the spontaneous energy of a projectile fired by a rifle, handgun, or a canon.
    The amount of the powder in the cartridge is known and with a reasonable assumption of the following we can calculate the initial pressure produced in the chamber as the rifle fired:

    "For every 10g of powder, there will be 3.2 litters of gas produced in the chamber"

    The next step is to find the Force acting behind the bullet with known caliber. Next, we can calculate the initial acceleration using F=m*a. So, now that we have the acceleration, how do we calculate the velocity of the bullet right at the start of firing. I have forgotten the formula. Could somebody please answer this question. I'm grateful if you would be kind enough to write the formula.
    P.S. All dimensions are known: Calibre, mass of bullet, length of the barrel, etc.
    Last edited: Jun 25, 2011
  2. jcsd
  3. Jun 25, 2011 #2
    This is my uneducated guess, since no one seems to have an expert answer at the moment:

    You need more information on the total free volume of the chamber behind the projectile, but even with that there is no simple solution. Assuming instantaneous combustion and production of your 3.2 liters, and that the powder has no volume, you will think how many times that volume right behind the bullet in the shell casing will go into 3.2 liters, and you can multiply that result by atmospheric pressure/square measure and calculate the force using the back end area of the projectile.

    But THEN it gets messy, as the pressure will half with each doubling of the volume as it goes down the barrel, even discounting that the heat/pressure relationship may not be linear.
  4. Jun 25, 2011 #3
    You could use v^2=2*a*s
    where s is length of barrel
    But if you calculate acceleration just from initial force on bullet and exclude air friction you'll probably get some silly speed. I would try this FPr-FFr=m*a and then
    a= FPr-FFr/m
    FPr is force caused by pressure of expanding gas
    FFr is friction(air friction and friction of barrel surface)
    Or you could try to calculate speed from potential energy of powder that's transferred to
    kinetic energy, but here you also have loses because some of that energy is transferred into heat.
  5. Jun 25, 2011 #4

    Andrew Mason

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    That does not tell you much. You have to know the temperature and pressure of the gas.

  6. Jun 25, 2011 #5
    first of all using gas laws find the pressure of the gas inside tube. now we know from the definition of pressure that it is equal to force*area so from here you got net force on the bullet(we can assume pressure as constant). now it depends on the length of pistol,rifle that how long this force exerted on bullet. since barrel may not too long so we can consider pressure constant. now you have acceleration distance so
    using v2=2as
    find final velocity.
    here friction are not considered.
    so you are done.
  7. Jun 25, 2011 #6
    He wants the velocity at the start of firing. I assume by 'spontaneous' energy, kinetic energy is meant. There seems to be 2-3 questions.

    My answer to the start of firing, of course, would be 0. So maybe that's not what was meant. Perhaps it was acceleration at start.
  8. Jun 25, 2011 #7

    Andrew Mason

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    Pressure is not constant. This would be an adiabatic expansion so:

    [tex]PV^\gamma = K = \text{Constant} = P_0V_0^\gamma[/tex]

    So if you know the initial pressure and volume:

    [tex]Work = \int_{V_0}^V PdV = \int_{V_0}^V \frac{P_0V_0^\gamma}{V^\gamma}dV[/tex]

    That will give you the bullet energy at the muzzle.

    [Addendum: That integral will give you the work done by the expanding gases. This work is done on the bullet and on the atmosphere, so you will have to subtract the latter: [itex]P_{atm}\Delta V[/itex] to get the work done on the bullet. There are losses due to friction and rifling (which adds rotational kinetic energy to the bullet) which would have to be estimated in order to determine the muzzle energy of the bullet.]

    Last edited: Jun 26, 2011
  9. Jun 26, 2011 #8
    we ignore for now the friction and the final speed of the bullet. I know the final speed formula
    my question is you have initial acceleration and the mass of the bullet. there is no distance or time involved yet. we have only the initial acceleration and the mass. since there is acceleration, then, there should be a Velocity. How do we calculate this very first velocity?

    The way I tried before, was that I calculate the velocity at the time the bullet left the muzzle then, using excel program, calculated the velocity and increasing acceleration (decreasing acc. forward) for every mm that the bullet is distant from the MUZZLE.

    THANK YOU ALL FOR YOUR RECOMMENDATIONS BUT FOR THE TIME BEING WE IGNORE EVERY DETAIL ABOUT FRICTION, HEAT, DELTA VS AND DELTA Ps. I just would like to know is there a formula that one can calculate the speed of a bullet just at the START OF FIRING. say 1mm from the chamber.
    Last edited: Jun 26, 2011
  10. Jun 26, 2011 #9
    That's why I said at the start of firing it would be 0 acceleration, but I wasn't sure what you meant by start of firing. If there is a formula for the 1st mm, it's probably some very complicated and arcane empirical thing in the hands of weapons engineers, and is probably very hard to derive from first principles; it would be an engineering question, more than a physics one. However, since we are talking milliseconds here, we can assume we can eliminate any adiabatic heat/pressure trade-offs and calculate the pressure from the liters of gas which you've given above, compressed into the volume of of the shell casing plus 1 axial mm • the area of the casing. Get that pressure and the area of the back of the bullet and you have your force:


  11. Jun 26, 2011 #10

    Andrew Mason

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    The initial velocity = 0. The change in momentum of the bullet after time [itex]\Delta t[/itex] is [itex]\Delta p = m\Delta v = ma\Delta t [/itex] assuming that [itex]\Delta t[/itex] is small enough so that a is more or less constant over that time period. There can be enormous acceleration (force) but very little change in speed if the time is small.

    That is irrelevant. The speed starts at 0 and by the time it reaches the end of the barrel it reaches its maximum speed. The speed change is logarithmic. The acceleration is greatest at the beginning where its speed is slowest. That bullet energy is determined by the work done by the expanding gas. That work is key to determining the energy of the bullet at the muzzle.

    What you really need to know is the initial volume and number of moles of the gas immediately after firing.

    Addendum: The physics of firing a bullet is similar to the physics of an internal combustion engine. Expanding gas does work on a piston sending it down a cylinder. (This transmits energy to the wheels so the piston speed does not increase like a bullet's speed). The compression ratio is key to determining the work done by the down stroke: that is Vf/V0. In the case of the rifle, V0 is the initial volume of the combustion chamber (this would be the volume of the casing behind the bullet, assuming complete combustion) . Vf is the volume of the entire gun barrel. [itex]V_f = V_0 + L\pi d^2/4[/itex] where L is the barrel length and d is the bore diameter.

    Last edited: Jun 26, 2011
  12. Jun 26, 2011 #11
    He wants us to ignore all that (our meat is his poison) and focus on the calculation of the value of the acceleration 'a' at a travel distance of 1mm from the start of firing. He's given us the volume of (room pressure) gas, and we only have to cram that into the shell casing volume (plus the added volume the extra 1mm would give us).

    That would give us the immediate pressure, from which we can derive the force f on the back area of the projectile. a=f/m. He only needs the mass of the slug.

    I would ask why this particular question. Just curiosity.
  13. Jun 26, 2011 #12

    Andrew Mason

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    Why acceleration at 1 mm? That tells you nothing. You need to find the acceleration (ie. you need to know pressure) as a function of the distance that the bullet has travelled down the barrel. And to determine that you need to know the initial volume of the casing (the space behind the bullet that contains the gun powder).

    That does not give us the pressure. You need the temperature as well as the number of moles of gas and the initial volume to determine the pressure.

    Me too.

  14. Jun 26, 2011 #13
    I forgot about the title question, but he seemed to change the to just initial a. I will treat the whole thing as a freakin' air-gun, and stick with the 1 mm, though, since he's let me off the hook. The higher physics is beyond me.
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