Q: In earlier issue you answered a question about free recoil
energy between a light, fast bullet and a heavy slow one where
both had the same muzzle energy. Could you run the calculation
again using the IPSC Power Factor Formula of bullet weight x
velocity/1,000?Johnno JohnsonA: I think it would be a lot easier to understand if I showed
you the calculation for free recoil energy. Since the power
factor formula is just a variation of momentum (mass x velocity)
which is also a component of recoil calculations, there is an
indirect relationship. If the power factor is higher, free
recoil will also be higher, but I'm not aware that any other
valid comparison can be drawn. The calculation of recoil energy actually has three parts. This
is: I=(WBxVB +4,000xWC) 225,400 "WB" is the weight of the bullet
in grains; "VB" is the velocity of the bullet in FPS; "WC" is the
weight of the powder charge in grains; 4,000 is the assumed
velocity of the powder gas: 225,4000 is a constant to convert
units to allow the use of grains. The second calculation is for the velocity of the recoiling gun
VG): VG=32.2xI/WG. "WG" is the weight of the gun in pounds; 32.2
is the acceleration of gravity. The final calculation gives you the free recoil energy of the gun
(EG) in foot/lbs. These equations can be found in Understanding
Firearm Ballistics by Robert A. Rinker which offers a more complete
explanation than I can go into here.
I am not sure of what you want to accomplish by comparing power
factor with free recoil energy. There is a lot longer discussion
of recoil in Hatcher's Notebook which is also valuable reference.
 
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