From: weed@sweetpea.jsc.nasa.gov (daniel weed 283-4162)
Newsgroups: rec.guns
Subject: Re: Trajectory Question
Date: 9 Nov 92 18:52:49 GMT
Article-I.D.: sweetpea.9211091828.AA04413



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

 >Sounds like your gun is pretty loose.  This is 7.5 inches at the 25 yard
 >standard.  One can't readily evaluate a system which has the extra degrees 
 >of freedom which this gun is evidencing.  I don't think we can say that its 
 >results are experimental evidence of anything other than what that particular
 >gun will do under these particular circumstances.

I was amused by your comment that my gun must be "loose", because all of
your analysis is an attempt to "prove" that a .45 auto with a VERY loose slide
and a VERY loose grip, will still not show any variation in impact point
with different bullet energies.

A "loose" .45 will shoot inaccurately and inconsistently.  Mine, while not
blueprinted, gives me results consistent enough to notice different loads
vary in mean impact point.

I agree that my evidence shows only what my gun will do, when I shoot it,
from one stance, on Saturdays.  However, I remind you that I am not the 
first and only persion to notice that higher energy loads hit the targets
lower.  This whole thread started because someone noticed just that, and
several others have also commented on same.

In an article in the December 1992 issue of the "Shooting Times", in an 
article on recoil on page 14, Rick Jamison writes:  
"While recoil can affect accuracy adversly by inducing a flinch for a shooter 
with any firearm, it affects the handgunner's point of impact in yet another 
way.  You may already have noticed how bullet weight and recoil affect the 
point of impact.  Heavy-bulleted, hard-kicking handgun loads generally hit 
higher on target than milder loads."

 >I'm not sure what relevance the sights not being changed has to the argument,

Only that the experimental conditions were not changed, other than the load.

 >I don't think you understand what I wrote.  The only resistance to slide 
 >motion other than its inertia is the recoil spring and a little bit of slide-
 >to-frame friction.  Do you see any other resistance to its motion?

Most certainly.  I agree slide friction is the smallest force.  Slide
momentum is significant, the recoil spring is very significant, and the
most significant force is the one you are missing - that of the mainspring.
Remember the gun is not cocked after firing, and the slide must cock the
hammer.  This about doubles the energy required to move the slide over the 
recoil spring alone.

 >Have you ever seen high speed motion pictures or even appropriate stills of a
 >45 going off?  You will see the gun stays nearly level until the slide
 >strikes the frame.  Certainly it stays level during the short slide
 >motion which occurs during the period the bullet is in the barrel. 

Keep this in perspective.  You are looking for the barest fraction of an inch 
upward motion of the barrel over around 0.0006 second time.  That had better
be a damn high speed camara - and stills won't catch it.  You are correct
in that the gun "stays NEARLY level".

In the Shooting Times article, again on page 14, Rick Jamison writes to 
explain why heavier bullets hit higher:
"The reason is that the bullet is still in the bore as the muzzle begins to 
elevate under recoil.  This can be calculated mathematically.  But what left 
no question in my mind was an X-ray photograph of the firing of a .45 auto. 
The bullet was clearly seen still in the barrel as the muzzle was elevating."

Unfortunately for us, he left out the mathematics.

 >If you assume the slide's motion is negligible, than you may be assuming
 >it is fixed in the gun which is an error.

Lets contrast my analysis with John's. I assumed that the slide can be
considered fixed to the frame while the bullet is in the barrel.  While
this may raise eyebrows for a semi-auto, it is a good assumption for
revolvers.  I chose this assumption because I know that the barrel/slide moves
less than 1/8 of an inch while the bullet is in the barrel (for a 1911 style 
.45 auto), and because I believe the countering effects of slide mass, recoil 
spring and mainspring are significant.  By the way, the barrel is completely
unlocked from the slide after the slide travels 1/4 to 3/8 of an inch - in
a Colt 1911 .45 auto.

John takes the opposite approach and assumes the slide is completely free to
move backward while the bullet is in the barrel.  Totally free of any
counter forces or friction.

The reality of course, is somewhere in between (between NO forces being
applied to the frame, and ALL the forces being applied to the frame).
To look at what this implys, lets create some number "e", between zero and 
one, which indicates how much impulse is applied to the frame by the bullet 
momentum, divided by the total impulse.  Then lets give "e" an important 
sounding name, like, say, the "coefficient of restitution". ;-)

If "e" is zero, (John's assumption) then the slide movement is completely independent of the frame, and the bullets will all strike in the same place 
(varying only by the external ballistic trajectory).  If "e" is one, then the 
slide and frame are locked together - or its a revolver.  

I have shown that the impulse applied to the frame ("e" = 1) is higher
for the heavier bullet, causing the change in impact point.  This must
also be true for ANY value of "e" OTHER than ZERO.  Therefore, even though
I don't know what "e" is, my basic analysis is still correct, and is 
supported by the preponderance of physical evidence.

Since "e" is at its maximum (one) for revolvers, I think both John and I can
agree revolvers are likely to be more sensitive to bullet impulse than semi-
autos, when it comes to changes in impact point.  Since "e" can never be
zero, I maintain semi-autos are also affected, to a somewhat lesser degree,
but still noticable to the shooter.

To conclude any more than this will require scrapping both assumptions
and digging in the the slide/frame equations of motion in MUCH greater
detail.  I may get around to it in time.  

Anyone else who can shed some light on this, show equations, data, or 
references, please join in!

Regards,

--  Dan


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