Originally posted by 1911Tuner:
Nope. You didn't read anything except the one sentence...and ignored the rest.
Velocity at the muzzle is the result of the acceleration AFTER the punch that gets the bullet up to the speed that allows the small percentage of added accelleration that occurs...and provides final velocity at bullet exit.
That initial punch is where 90% of the velocity occurs...for pistols, it's within the first half-inch of bullet travel...and probably more than 90% of the total recoil impulse.
Go back and re-read the whole thing.
Don't believe it? Try this. Warning:
It's illegal...but it will demonstrate the reality in neat fashion.
Go buy a cheap single-shot 12 gauge shotgun. Cheap is good, because you'll be destroying the barrel shortly after you do this.
Lop off the barrel flush with the end of a fired shell, so that no projectile acceleration will be gained after the initial punch. Add weight to the stock so that the gun's mass will be equal to the original.
Go fire the gun.
I don't recommend placing your face on the stock's comb unless you relish the thought of a black eye and/or a bloody cheek. Ask me how I know...
Cheers!
I read your entire post prior to making my first post (#15) above.
The experiment that you propose (and no I won't try it 'cause I an awfully fond of my freedom
)
simply changes the location of the muzzle so that it is at the same point that the shotgun shell ends. In other words, it is still the "muzzle" (a.k.a. the end of the barrel) and it is the point at which the highest velocity of the projectile will be achieved since the shot/wad is no longer being accelerated by the high pressure propellant gasses remianing trapped behind it. The only change effected will be that the bullet (or shot charge) will not exit the "muzzle" as fast as it would as if it had the entire length of the barrel in which to continue to accelerate from the continued force (F) of the expanding gasses produced by the burning propellant. While the shot and wad will most certainly leave the 'new muzzle' (created by cutting the barrel off so that it is now the same length as the shot shell) at high velocity that veocity will not be as high as it would have been had the shot and wad been allowed to accelerate the full length of the barrel. I could not have come up with a better example with which to illustrate the point that I have been trying to make.
Arguably, there will be some recoil with the arrangement that you suggest because you
still have a mass (m) namely the shot and wad leaving the 'new muzzle' that you have created at a lower velocity (v) that will
still result in momentum as well as K.E. being generated in both directions.
The fact is, that regardless of how short the barrel is made, there will always be velocity (v) since the mass (m) is moving out of the barrel regardless of its length which means there will always be momentum (ρ) and K.E.
I even find you (quoted again for the sake of clarity) to be in agreement with what I just explained above:
"Velocity at the muzzle is the result of the acceleration AFTER the punch that gets the bullet up to the speed that allows the small percentage of added accelleration that occurs...and provides final velocity at bullet exit."
The "bullet exit" is the muzzle.
The acceleration of a bullet (or shot and wad) is a single impulse of exceedly short duration and momentum (not to mention K.E.) is a result of the final velocity (v) obtained of the mass (m) as it exits the muzzle even if the "muzzle" coincides with the end of the cartridge casing or shotshell mouth. It also a fine example of Newton's Second Law of Motion:
"The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object."
Expressed mathematically, Newton's Second Law of Motion is:
F = ma
where "F" equals force (this is the "PUNCH" to which you refer above in your quoted statement), "m" equals mass and "a" equals acceleration.
Additionally, if you substitute the value 0 (fps) for the velocity variable (v) in any of the following equations you will see that since any number multiplied by zero equals zero that you cannot possibly have momentum or kinetic energy without having velocity.
∫ mv δv = m
∫ v δv = m ½v² = ½mv² = (KE)
ρ = mv
which proves the following statement false:
Muzzle velocity has little to do with recoil energy.
because it would violate the Newton's Third Law of Motion:
"For every action, there is an equal and opposite reaction."
Mathematically expressed Newton's Third Law of Motion is:
mv = mv
which proves again that muzzle velocity has
everything to do with momentum (ρ) as well as the kinetic energy (KE) of both the bullet (or shot/wad) and the recoil of the gun.
Regards,
