From: weed@sweetpea.jsc.nasa.gov (daniel weed 283-4162)
Newsgroups: rec.guns
Subject: Re: Trajectory Question
Date: 6 Nov 92 01:57:22 GMT


--
responding to article by John Bercovitz     (JHBercovitz@lbl.gov)

John, I'm sorry but you've completely missed it:

 >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.

Well, your assumptions of a linear gun (no rotational dynamics), and no
recoil spring, and no friction, invalidate the model - its just not
realistic enough.  Nevertheless, the assumptions make the math easy,
so lets look at it further:

 >So the slide gets a velocity of 25.7/1.2 = 21.4 fps from the bullet.

Correct, but we are also assuming the frictionless slide does not hit the
frame before the bullet exits - OK with this model.

 >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.  

Lots of legerdemain here John: this is not correct.  Again, momentum is
conserved,  The momentum of the system after the slide contacts the
frame is the same as the momentum of the slide alone before it hits
the same.  The total mass of the system after contact is 2.4 lbm, the
momentum is 3.555127 N*s (those SI units again, same as 25.7 lbm-ft-sec^-1).
the final velocity of the system then is 10.714 ft/sec, not 21.4 ft/sec.
I assume the slide and the frame remain in contact after the collision.

(also assuming no impact energy losses due to heat generated, internal 
elastic waves, and the generation of sound energy).

 >Now the frame is the part you've got to stop 

Wrong, you must stop both the slide and frame.  It would be a very poor 
assumption that the slide vanishes into thin air, or bounces off the frame 
like a billard ball. ( take that gun to the smith! ;-) ).

 >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.

And a poor one at that.  Again, wrong, since the real velocity of the
slide/frame is 10.714, if you re-calculate the kinetic energy, you
get 4.28145

 >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,

Correct.

 >only half as much as in the semi-auto version.

Tis exactly the same in both cases.


 >QED  No legerdemain or nothin'.  8-)

Houdini would be proud. ;-)

QED

--- Dan



------------------------------------------------------------------
Daniel Weed                             weed@sweetpea.jsc.nasa.gov
McDonnell Douglas Space Systems Co. - Houston
          "My comments are my own, not my employer's"



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