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Old 01-25-2014, 05:22 PM   #91
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I would say to get the term "knock down power" out of your head. It does not exist. All handguns aren't all that great for stopping people. If you want to look at the ballistics of all three, the differences are marginal at best. I honestly don't feel that it matters which you carry from that stand point of stopping an attacker.

My advice would be to shoot all three and see which you're most comfortable with. I went with 9mm because I hate .40 S&W and .45 ACP is too damn expensive. For me the choice was easy.

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Old 01-25-2014, 07:00 PM   #92
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Just seems like the same old caliber debate. The Glockophiles will say any caliber as long as it comes from one of Gaston's Creations,..the Staunch 1911 guys(And I was One for a long time!) will say "1911 In 45" the folks who swear by 40's will say the same. But It's where you put the shot......Not that I'm shooting a 500 Lb animal Rushing at me with a 22LR....

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Old 01-25-2014, 08:21 PM   #93
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I would agree with the article with one exception.

Quote:
If it had the energy to do so, then equal energy would be applied against the shooter and he too would be knocked down.
This is not true - and is a misconception that is often stated. You can't examine only one part of the entire energy equation and conflate Newton's second and third laws of motion with the kinetic energy of the bullet.

The problem with the statement is that he is using Newton's second and third laws of motion to describe the acceleration of the bullet and the equal and opposite force applied to the gun - but ignoring the kinetic energy of the bullet after it has been accelerated to its top speed, and the energy contained in the bullet at its impact speed.

Kinetic energy is (simply) calculated as KE= 1/2 x m x v^2. That's why the amount of energy of a bullet is described in foot pounds at impact - this is NOT the same as the amount of energy expended to accelerate the bullet.

Newton's second and third laws applied to the bullet and the gun do not cancel the KE = 1/2 x m x v^2 equation. Or, the more complex formula often used for bullets of: KE = 1/2 x weight (mass) x v^2 / 7000 / 32.175.

As an example, a .357 magnum, 158 grain bullet accelerated to 1250 feet-per-second, the above equation gives a calculated 548 ft/pounds of energy at the muzzle - but that is NOT the force applied to the gun - that is the amount of energy contained in the bullet due to the speed generated over the time it was in the barrel by the force applied to it.

Whether you are knocked over by the guns' recoil generated during acceleration of the bullet has no bearing on the amount of energy generated by the bullet when it impacts another object.

The bullet weight is very small (relatively) and is at rest when the gun is fired. The lightweight bullet is accelerated over time (the time it is in the barrel) and an equal force is generated in the opposite direction against an object (the gun) over the same time. Because of the gun's mass, it is accelerated much less than the bullet.

In comparison, a 158 grain bullet weighs 0.0226 pounds. Given a .357 magnum pistol that weighs 2.75 pounds means the gun weighs 121.7 times more than the bullet. Or, the bullet is 0.0082% the weight of the gun - however you want to describe the difference.

Looking at it in a different way - with a .357 magnum a 158 grain bullet is accelerated to 1250 feet- per second by the forces applied to it. The gun it is fired out of weighs 2.75 lbs and is accelerated to 14.3 feet-per-second
generating 8.7 ft/lbs of recoil. That's why you are not knocked over by the recoil - but that has NOTHING to do with the energy contained in the bullet at impact.
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Old 01-25-2014, 10:41 PM   #94
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Quote:
Originally Posted by buckhorn_cortez View Post
I would agree with the article with one exception.



This is not true - and is a misconception that is often stated. You can't examine only one part of the entire energy equation and conflate Newton's second and third laws of motion with the kinetic energy of the bullet.

The problem with the statement is that he is using Newton's second and third laws of motion to describe the acceleration of the bullet and the equal and opposite force applied to the gun - but ignoring the kinetic energy of the bullet after it has been accelerated to its top speed, and the energy contained in the bullet at its impact speed.

Kinetic energy is (simply) calculated as KE= 1/2 x m x v^2. That's why the amount of energy of a bullet is described in foot pounds at impact - this is NOT the same as the amount of energy expended to accelerate the bullet.

Newton's second and third laws applied to the bullet and the gun do not cancel the KE = 1/2 x m x v^2 equation. Or, the more complex formula often used for bullets of: KE = 1/2 x weight (mass) x v^2 / 7000 / 32.175.

As an example, a .357 magnum, 158 grain bullet accelerated to 1250 feet-per-second, the above equation gives a calculated 548 ft/pounds of energy at the muzzle - but that is NOT the force applied to the gun - that is the amount of energy contained in the bullet due to the speed generated over the time it was in the barrel by the force applied to it.

Whether you are knocked over by the guns' recoil generated during acceleration of the bullet has no bearing on the amount of energy generated by the bullet when it impacts another object.

The bullet weight is very small (relatively) and is at rest when the gun is fired. The lightweight bullet is accelerated over time (the time it is in the barrel) and an equal force is generated in the opposite direction against an object (the gun) over the same time. Because of the gun's mass, it is accelerated much less than the bullet.

In comparison, a 158 grain bullet weighs 0.0226 pounds. Given a .357 magnum pistol that weighs 2.75 pounds means the gun weighs 121.7 times more than the bullet. Or, the bullet is 0.0082% the weight of the gun - however you want to describe the difference.

Looking at it in a different way - with a .357 magnum a 158 grain bullet is accelerated to 1250 feet- per second by the forces applied to it. The gun it is fired out of weighs 2.75 lbs and is accelerated to 14.3 feet-per-second
generating 8.7 ft/lbs of recoil. That's why you are not knocked over by the recoil - but that has NOTHING to do with the energy contained in the bullet at impact.
I understand what you are saying....sorta. But doesn't your same rule work in reverse when the slowing 158 gr. bullet hits the maybe 200 lb. target and isn't that the reason bullets don't "knock" much of anything down? I seem to remember that Rich Davis of Second Chance body armor used to shoot himself with a .44 mag to demonstrate his vest.

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