Recoil Equation?

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How does one factor in the "felt recoil" differences between different types of grips and such, if there really is any?
 
^^ Too much depends on the grips, the meat in the shooter's hand, the height of the bore above... let's say, the trigger. As someone above quite wisely said, "Just shoot the freakin' thing."

"Felt" recoil is an individual thing.

And what happens to the momentum of the moving slide? It gets transfered to the frame by the recoil spring and by hitting whatever part of the frame stops its movement. Generally, a self-loader will stretch out the time recoil forces act in the system and this affects felt recoil.

Don't forget that the relatively light slide will pick up much more energy than if the recoil just went to the whole gun.

Let's face it, if the slide weighed as much as the bullet, it would come back just as fast as the bullet went forward, picking up exactly the same amount of energy as the bullet --a hairy situation, no doubt.

(This, neglecting of course, the other recoiling parts, like the barrel and spring and the energy that goes into tipping the barrel down... if one really is inclined to pick at nits.)
 
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hammerbite, thanx for the save, one would think i would have commited that to memory by now but it was getting late. i took the equation staight out of the lymans #48 reloading manual pg.398.
 
if the slide weighed as much as the bullet, it would come back just as fast as the bullet went forward, picking up exactly the same amount of energy as the bullet

Indeed. The momentum of the slide and the bullet are equal at the onset of action/reaction. Momentum must be conserved, and will be conserved...and objects in motion will remain in motion unless and until influenced by an outside force. The spring represents an outside force. Friction is another one. Even a fly sitting on top of the slide when the gun fires will constitute an outside force. Anything outside the system that acts to delay...reduce the speed and momentum...or block the movement will do just that.

If you fire a gun in the weightless void of outer space, the bullet will continue on at the exit velocity forever unless and until it either strikes something or enters the gravitational field of another object.
 
230RN...Too late! It came in on the E-mail notification before ya whacked it!

:D

The response is...not unless the gun is fired by magic from afar. If your hand is on it...or even your finger and thumb in a pinch-fire...if the opposite velocity and momentum is effected for even a split nanosecond...the gun's rearward momentum is reduced, and it will conserve that momentum, and continue in motion AFTER the outside force influenced it.

If ya wanna get all technical...

:)
 
I whacked it 'cause I started to get too detailed in my Notepad composing for the post and found myself going into "This presumes the recoil forces are directed directly through the center of mass of the gun/shooter system... otherwise part of the energy would be converted to rotational energy and one would spin forever.... et cetera, et cetera, et cetera..."

But I decided the heck with that kind of stuff.

Wisely, if I may add.

I edit interminably, so one of my posts may change from moment to moment.
 
I guess I'm too old fashioned to use speadsheets for my more complex calculations. I still write my own BASIC and QBASIC programs, just like my ancestors did.

:D
 
McCall911,

Shhh... Top Seekrit:

I hate to mention this, but people are constantly amazed at how fast I can turn out results. Y' know why?

'Cause I frequently use good old Gee-Whiz Basic to analyze data.

One of the best and fastest ways to manipulate strings, 'cept for the fact that it doesn't handle Unicode or ASCII beyond character string 255 without a lot of fussin'.

Some complex algorithms using powers sometimes get screwed up though. Best to do a reality check on your results sometimes.
 
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