From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Newsgroups: rec.guns
Subject: Re: Trajectory question
Date: 6 Nov 92 20:00:09 GMT
Article-I.D.: dog.27270
Organization: Lawrence Berkeley Laboratory, California

In article <9211051831.aa15759@claudius.pica.army.mil> 
mfolenta@pica.army.mil writes:

#     John Bercovitz wrote recently (yesterday, if the header is right):
##In a sense, a semi-auto can actually have more recoil.  Let's take 
##that 45 auto and fire it normally.  Let's say we're firing a 200 gn
##bullet at 900 fps.  So the momentum of the bullet is (200/7000)*900
##or 25.7 lbm-ft-sec^-1.  Now let's idealize the gun for calculation
##purposes and say the slide weighs 1.2 pounds and so does the frame.
##So the slide gets a velocity of 25.7/1.2 = 21.4 fps from the bullet.
##The slide weighs the same as the grip frame so it hits the frame and
##comes to a dead stop and the frame acquires the 21.4 fps.  Now the
##frame is the part you've got to stop and it's carrying a kinetic
##energy of 1/2 mv^2 or 0.5*(1.2/32.174)*21.4^2 = 8.55 ft-lbs.  OK, so
##this is really idealized like the frame was being loaded through its
##center of gravity or something.  Anyhow, this is just an illustration.

##Now let's somehow lock the slide to the frame so it can't move relative
##to the frame.  Now the mass that the gasses are pushing against weighs
##2.4 pounds instead of 1.2 pounds.  So the velocity the gun acquires 
##in the momentum transaction is 25.7/2.4 = 10.7 fps.  Now the kinetic
##energy your hand has to absorb is 0.5*(2.4/32.174)*10.7^2 = 4.28 ft-lbs,
##only half as much as in the semi-auto version.
#
##QED  No legerdemain or nothin'.  8-)


#     Well, legerdemain or not, you are off by a factor of 2 in the
#first case (slide moves first and then whacks into frame)!

Oh, not really.  It's merely a consequence of our different assumptions.
Your calculations and mine are both correct if their initial assumptions are 
followed.

#     Again, using your idealized gun and ammo (let's assume
#[stuff deleted] a slide and a frame that EACH
#weigh 1.2 lbs.

#     So, by conservation of momentum, mV=mV, 25.7 lbm-ft-sec^-1 means
#the slide moves at 21.4 fps initially.  Now, that moving slide hits the
#slide stops and whacks the frame.  Now we have BOTH the slide and frame
#moving with a mometum transfer of mV=25.7 blahs.  Which gives us a gun
#velocity of 25.7/2.4=10.7 fps, which then gives us .5mv^2=4.27 ft-lbs.
#Which assumes that the slide and frame both move back together (perfectly
#inelastic collision) as opposed to where the slide STOPS in inertial space
#after hitting the frame and transfers all of its momentum to the frame
#(perfectly elastic collision).  I think the real world is in between but
#closer to the inelastic condition (the slide and frame both move back 
#together).

I can't disagree that the real world would be closer to an inelastic collision
than an elastic collision because I don't have good data one way or the other.
Certainly in my own 45 auto, I have made an attempt to cause the collision
to be as inelastic as possible so that the result is closer to your 
calculation than mine.  That's why I put in a high hystersis shock buffer.
But that wasn't the point of my calculation.  I was trying to show that
recoil, if you count your recoil in energy, could be higher for a semi-auto
in the idealized case.  Several kind souls have pointed out that Hatcher 
has made this argument, which I have tried to make, much more eloquently
than I.  Could one of you kind souls point out a reference?

#     And all of this ASSUMES that we have a perfect model and look at just
#momentum transfer.  In the real world, impulse is the item which must
#conserved to start of this analysis, then we jump to momentum, and then
#to energy.  A gun with a recoiling mass (like most semi-autos) lowers
#the peak forces that must be opposed to stop the rearward motion by
#spreading out the application of those forces over both distance (the
#recoiling distance) and time.  Impulse, momentum, and energy ALLWAYS
#must be accounted for and conserved properly.

Yup, yup.

#     Tsk, tsk.  I'm surprised at you John. Usually your math and logic
#are impeccable.  Oh well, we all slip up once in a while (I still remember
#arguing once that 2+2 = 4 was wrong. Don't ask.)

I'm sorry to have disappointed you!  8-)   I disagree that my math and
logic are off.  My conclusions follow from my asumptions.  I conceded
that my assumptions are not realistic in my previous post when I said
it was an ideal gun.  You used different assumptions.  Absolutely nothing
wrong with your math or logic either.  

Because of the ideal gun assumption, I get to say that the slide hitting
the frame is like two billiard balls hitting each other (on a line of
action connecting their centers), one billiard stops and the other
proceeds with the first one's velocity.

So I think your disagreement is with my assumptions, not my math or
logic.  I picked my assumptions to show a worst case, you picked yours
to show the exact opposite.

The next step would be for us to pick over the assumptions if we want to
change this into a realistic scenario.  However, absolute realism wasn't 
my purpose; demonstration was.

John Bercovitz     (JHBercovitz@lbl.gov)