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To: KayEyeDoubleDee

I understand what you’re saying, I just don’t see it though. If this were the case, then every time I shot my .45 ACP I would get bowled over by the force “-X”, because the force from the projectile will certainly knock the recipient over regardless of penetration (even if shot in a BPV that could stop the bullet).

The heat dissipated can’t amount to that much negative energy, because I would experience numerous FTFs and FTEs because the gun would get so hot it wouldn’t function.

Like I said, I’m just not buying that it works how you’re explaining it.

It would be interesting to see something like this explored on mythbusters, or at least explained by a gun manufacturer.

Perhaps howstuffworks.com drills this down in some detail.


182 posted on 03/23/2009 1:55:19 PM PDT by jurroppi1 (We need to reward the people that carry the water instead of the people that drink the water!)
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To: jurroppi1
If this were the case, then every time I shot my .45 ACP I would get bowled over by the force “-X”, because the force from the projectile will certainly knock the recipient over regardless of penetration (even if shot in a BPV that could stop the bullet).

What I am almost certain is happening here is that all of this recoil stuff is actually absorbing the energy of the blast which took place in the firing chamber, and that, as a result, the round is leaving the muzzle with LESS energy [i.e. at a lower velocity] than would would be the case in an old-fashioned gun without the recoil stuff.

Remember, conservation of energy is a zero sum game - if you are feeling less recoil, then that [almost certainly] means that the round must be departing the muzzle with less "oomph" than would be the case otherwise.

188 posted on 03/23/2009 2:29:54 PM PDT by KayEyeDoubleDee
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