View Single Post
Old December 18, 2009, 10:44 AM   #44
brickeyee
Senior Member
 
Join Date: December 29, 2004
Posts: 3,342
The recoil energy (what the shooter feels) is easily computed.
You conserve momentum between the ejecta (bullet and powder weight) and the gun itself.

This produces a velocity for the recoiling gun.
This velocity is then used with 1/2*m*v^2 to compute the kinetic energy of the recoiling gun.

This is the energy the shooter must absorb.

The bullet has a much higher energy than this, since while its mass is a fraction of the guns, its energy is based on v^2.

You need a lot of assumptions and simplifications to determine the force delivered by the bullet.

If a bullet had 2000 ft-lbf of energy, and you brought it uniformly to a stop in 1 foot, the force delivered over that foot would be 2000 pounds.
If you know the bullet diameter you can obtain a pressure (force / area).

The problem is that we are far from uniform, and bullets in mixed tissue do not "uniformly" come to a halt.

The loss of energy is not 'smooth' since the materiel is not uniform.

Conservation of momentum works all the time, but does not indicate energy.

Conservation of energy is a much larger problem, since ALL sources of energy (including the energy released by the gunpowder) must be considered, and ALL loss mechanisms (friction, heating, the work done on the bullet to accelerate it, even the work done in creating the muzzle blast, work done deforming bodies).

Conservation of energy is often not a useful calculation to bother with since there are so many loss mechanisms that must be accounted for.
brickeyee is offline  
 
Page generated in 0.04864 seconds with 7 queries