From: mfolenta@pica.army.mil
Newsgroups: rec.guns
Subject: Re: Trajectory question
Date: 6 Nov 92 01:58:07 GMT


     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)!

     Again, using your idealized gun and ammo (first let's assume
a perfectly spherical, homogenious cow - OK, non-sequiteur bad joke),
200 gr bullet at 900 fps (ignoring the charge weight or just lumping
it in with that 200 gr mass for now) and 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).

     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.

     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.)