From: bercov@csg.lbl.gov (John Bercovitz)
Subject: Re: recoil springs (was: 10mm report)
Organization: Lawrence Berkeley Laboratory, California

#I've noticed that my (recent, Portuguese) HP has a _very_ stiff recoil
#spring. It's stiffer than any of the older HPs I've shot, and stiffer than
#just about _any_ other 9mm that I've ever shot. I don't like it--racking
#the action is hard work; my wife can't rack it at all.
#
#So...would it be advisable to replace this spring with a lighter one,
#assuming I don't shoot any horrendous +P heavy bullet loads through it? If
#so, does anyone know of a source for lighter HP springs?

This a WAG (guess) but I would imagine they put a stiffer spring in the
HP to allow the gun to more properly handle the cartridge du jour, the
147 gn JHP.  If this is the case, I see no problem with replacing the
current spring with the earlier version if you stick to the lighter
bullets.  Your action cycle time will go up, however.

The HP has a slotted underlug so it is very easy to determine how far
back the slide moves before it starts to unlock.  If you then weigh the
slide and determine the total bullet travel (the distance from the base
of the bullet as it sits in the chambered cartridge to the muzzle) then
you can calculate whether or not the slide starts to open before the
147 gn bullet has exited the muzzle.  For most autoloaders this is a near
thing.

John Bercovitz     (JHBercovitz@lbl.gov)

From: bercov@csg.lbl.gov (John Bercovitz)
Subject: Re: recoil springs (was: 10mm report)
Organization: Lawrence Berkeley Laboratory, California

Peter Cash writes:

#> I've noticed that my (recent, Portuguese) HP has a _very_ stiff recoil
#> spring. It's stiffer than any of the older HPs I've shot, and stiffer than
#> just about _any_ other 9mm that I've ever shot. I don't like it--racking
#> the action is hard work; my wife can't rack it at all.
#>
#> So...would it be advisable to replace this spring with a lighter one,
#> assuming I don't shoot any horrendous +P heavy bullet loads through it? If
#> so, does anyone know of a source for lighter HP springs?

John Bercovitz writes:

# This is a WAG (guess) but I would imagine they put a stiffer spring in
# the HP to allow the gun to more properly handle the cartridge du jour,
# the 147 gn JHP.  If this is the case, I see no problem with replacing
# the current spring with the earlier version if you stick to the lighter
# bullets.  Your action cycle time will go up, however.
#
# The HP has a slotted underlug so it is very easy to determine how far
# back the slide moves before it starts to unlock.  If you then weigh the
# slide and determine the total bullet travel (the distance from the base
# of the bullet as it sits in the chambered cartridge to the muzzle) then
# you can calculate whether or not the slide starts to open before the
# 147 gn bullet has exited the muzzle.  For most autoloaders this is a near
# thing.

I got hold of an older HP ('73?) and by golly it doesn't have a slotted
underlug; it has ramps.  Oops!  8*)  Musta got it mixed up with something
else.  Anyhow, the slide weighs 6500 grains and the total slide travel
until unlocking starts is .140".  The bullet travel is 4.06" to exit for
a 125 gn bullet seated long.  So the slide moves back
(125/6500)*4.06 = .078"
before unlocking starts.  That means there is still .140"-.078" = .062"
of surface holding the barrel up into the locking lugs.  If a 147 gn
bullet is substituted, there is only .048" of surface left to hold the
barrel up into the locking lugs.  That doesn't seem like enough of a
reduction to cause the manufacturer to increase the spring force to
compensate.  So I withdraw my WAG.  Maybe they wanted to decrease cycle
time or maybe they were afraid someone would try to fire 162 gn, is it?,
bullets in it to make major.

John Bercovitz     (JHBercovitz@lbl.gov)

From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Recoil? When?
Organization: Lawrence Berkeley Laboratory, California

In article <9211021550.AA02295@hpfcmgw.fc.hp.com> bartb@hpfcmgw.fc.hp.com
(Bart Bobbitt) writes:

#The recent responses to the question of when a firearm recoils has been
#interesting to read.  Especially the ones that mention semiauto functions.
#Comments stating that semiautos start to cycle before the bullet leaves
#the barrel are most interesting.
#For example, the Colt M1911 and Browning P35 (high power).  These pistols
[stuff deleted]
#As soon as the slide starts back, this link turns and pulls the back of
#the barrel down, unlocking the lugs.  Seems to me that if the cycle starts
#before the bullet is clear of the barrel, several thousand pounds of
#pressure could escape as the breech is unlocked.
#So, if the slide starts back while the bullet is still in the barrel, I
#don't think it would be in exactly the same place for each shot.  This
#means accuracy would not be too good.  It seems that good accuracy could
#only be obtained while the slide is in battery, holding the barrel in the
#same place from shot to shot, before the bullet exits the muzzle.

You're close to the right answer here.  The Colt M1911 has a link as you
say, but it's a toggle and it doesn't cause much vertical movement of the
barrel until the barrel and slide have traveled a good ways backward in
recoil.  The Browning has a slot for the pin to ride in rather than having
a link.  The effect is the same because the slot is straight for a while
and then kicks up at the end.  This allows the barrel/slide unit to recoil
for a distance before any unlocking takes place.  In another post which
you probably haven't seen yet, I mentioned that the gun and bullet maintain
a relationship with each other such that their center of mass does not
move, neglecting outside influences, and so the barrel/mass unit moves
a distance equal to the distance the bullet moves times the value, bullet
mass divided by the slide/barrel mass.

Let's take an example.  The barrel/slide unit of the Colt M1911 weighs
around 18 ounces (working from memory here) (what memory?).  The bullet
weighs 230 gns or (230/7000)*16 = 0.526 ounces.  All right you SI users,
I can hear you snickering; please snicker more quietly.  The bullet
travels some 4.4" from rest position until the base of it clears the
muzzle.  The ratio of bullet mass to slide mass is .526/18 = .029.  So
the slide/barrel unit moves .029*4.4 = .129" while the bullet is still
in the barrel. Of course the gas pressure is quite low when the bullet
is near the muzzle.  The slide/barrel unit has to move some 0.2" in
the Colt M1911 before it unlocks. But I won't cheat you here: if the
link holes are tight on the pins, theoretically the back end of the
barrel does lower a little before the bullet leaves.  So you're right,
it could have a small effect on accuracy.  In actual practice, I think
there's slop in the pin holes and so the barrel and slide stay locked
together a bit longer due to friction of case head against breech face.
But I'm only speculating here.  I know there's enough pin hole slop in
my very tight Gold Cup for this to happen, and I don't see any evidence
of case heads scraping against my breechface.  Of course in the Browning
with its slotted cam on the bottom of the barrel, there is no theoretical
reason that the barrel has to move downward while it and the slide  move
backward.  One should also consider that the 9 mm Para has very light
bullets in a typical load so the barrel/slide unit doesn't move far before
the bullet has left.

#This same thing seems to be the case with semiauto rifles, such as M1s
#and M14s.  I don't think the operating rod moves at all until the bullet
#is some distance from the muzzle.  Although the gas system has pressure
#in it, the operating rod doesn't move until after the bullet exits the
#muzzle.  The time from when the bullet goes past the gas port until it
#leaves the muzzle is about 30 microseconds in an M1; about 150 microseconds
#in an M14.

I haven't worked on an M-14 personally though I've worked on a lot of
analogues.  As I recall, the operating rod has to move some distance
backwards before it actually contacts the "bolt handle" and starts to
unlock the gun.  The guns which have early gas ports have bolt carriers
which have to travel some distance before they start to unlock the action,
I think.  Some semi-auto rifle 'smith ought to verify this for us but
if you have one of these guns at hand, it would be a simple matter to
prove it to yourself.

John Bercovitz     (JHBercovitz@lbl.gov)

From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Trajectory question
Organization: Lawrence Berkeley Laboratory, California

In article <8572@lee.SEAS.UCLA.EDU> osmith@abbott.SEAS.UCLA.EDU
(Owen I. Smith) writes:

#Actually neither one of these explanations is correct. Certainly the gun
#must react to the acceleration of the bullet in the barrel, so as the
#second poster points out, recoil must begin as soon as the bullet begins
#to move.  However, it's certainly not true that once the bullet leaves
#the barrel there is no force acting on the gun.  In fact, the expansion
#of the highly compressed propellant gas out the muzzle leads to a very
#significant force.  At the point the bullet leaves the barrel, you're basically
#holding onto a rocket.  You might expect that the mass of propellant gas
#is so much lower than that of the bullet that the force caused by its
#acceleration out the muzzle would always be negligible.  A little reflection
#makes it clear that this is not always the case.  Take a typical .308 rifle,
#for example.  You might have a 120 gr bullet and 40 gr of propellant.
#Assuming all the propellant is vaporized, the ratio of masses is only 3:1.
#The bullet is accelerated to perhaps 2600 fps over a period of a few
#milliseconds.  However, once the bullet leaves the muzzle, the extremely high
#pressure ratio will cause the gas to accelerate to far, far higher velocities.

I've read in many places that the average exit velocity for the gasses
may be considered to be around 4 or 5 thousand feet per second for the
purposes of calculating momentum transfer/recoil.  There's got to be
sonic flow because you're a long way from the criticial pressure ratio.
The speed of sound even in a polyatomic molecule like that should be pretty
high at combustion temperatures.  Of course you'd get quite a bit of cooling
or quench when the gas expands so its pressure is only a few thousand
psi when the bullet approaches the muzzle.  Sorry about the stream-of-
consciousness style; hope it's understandabale.

Anyway let's take your 308 with 120 bullet and put 55 gns of ball powder
behind it.  I would hope we could get 3000 fps out of this load.  The
momentum of the bullet would be 120*3000 = 360,000 gn-fps what ever the
heck that is and the momentum of the gasses would be 55*5000=275,000 gn-fps.
So the gas momentum is a _very_ significant fraction of the total recoil.
Sounds like a muzzle brake is very much in order if you're already deaf.
In the present case you could almost make a recoilless rifle with the proper
brake.

John Bercovitz     (JHBercovitz@lbl.gov)

From: bercov@csg.lbl.gov (John Bercovitz)
Subject: Re: Trajectory Question
Organization: Lawrence Berkeley Laboratory, California

In article <9211022109.AA21263@sweetpea.jsc.nasa.gov> weed@sweetpea.jsc.nasa.gov
(daniel weed 283-4162) writes:

#I typically load two .45 acp rounds for practice with a 1911:  A 200 gr LSWC
#and a 230 gr LRN.  I load both up to Major Power (weight in grains times
#velocity in fps = 175,000).  Because of the different bullet weights,
#I must load the 200 gr to about 850 fps, and the 230 gr to about 760 fps.

#At 10 yards, the faster 200 gr bullet consistently hits the target about
 ^^^^^^^^^^^             ^^^^^^                     ^^^^
#one inch lower than the 230 gr.  A bullet drop of one inch in 360 inches is
 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
#NOT due to the external ballistics alone, but mostly due to the movement
#of the gun before the bullet leaves the barrel.

#Momentum (mass times Velocity) is conserved.  So the instant the bullet
#begins to move down the barrel, the gun also begins to move backwards and
#up because of the torque and the fact that the pivot point is below
#the line of the barrel.  Recoil begins with the motion of the bullet,
#not when it exits the barrel.

I have trouble understanding how this could be happening.  Perhaps you
could explain in a little more detail how you performed this test and
then maybe someone can figure out why you're getting this anomalous
behavior.  As I said in an earlier post, semi-autos constructed like
the 45 government model shoot all bullets to the same point of impact
because the mass of the slide is more or less coaxial with the bore
and the slide delivers force to the grip frame only thru a light spring.
Hence there really aren't any forces to create a torque on the gun while
the bullet is still in the bore.  (After the bullet leaves, the slide
slams into the frame and _really_ torques the gun upward.)

There are a number of reasons I believe your case is anomalous.  I'll list
a few:

1) I've never seen this myself.  (How's that for anecdotal evidence?)
I shoot offhand at 25 yards a lot and I shoot 230 hard ball and a 200 H&G
SWC.  And yes, my Gold Cup is plenty accurate enough to see the effect
you mention.

2) Unlike the revolver where the front sight is much higher above the
bore centerline than the rear sight, the government model's front and
rear sights are nearly exactly the same height above the bore line.
To put some numbers on it, the revolver's sight line is tipped away
from its bore line about 15 parts in a thousand while the government
model's sight line is parallel to its bore line (and 1/2 to 5/8 inch
above it) to about 1 part in a thousand, typically.

3) When I fire all sorts of different loads out of the government model
with the government model mounted in a machine rest, all of the loads
hit the same point at 25 yards.

4) I've just now done a little back-of-the-envelope calculation which
tends to confirm what I've said.  To give us a worst case, I'll assume
the shooter has a very loose grip.  So any forces tending to tip the
gun are unopposed.

Let me try a little intuition first, though, to see how far I can get.
I think the first thing to notice is that revolvers' and single-shot
pistols' barrels are directly connected to the frame.  So if you have
a 357 magnum (cross sectional area 1/10 square inch) the force from
the combustion is all delivered directly to the frame.  Let's say a
hot 357 averages 15,000 psi while the bullet is in the barrel.  That
means the gas is pushing back on the gun with around 3/4 ton of force.
Sort of irresistible.  Now in the 45 government model, when the maybe
1/2 ton of force average is applied to the breechblock, most of it
goes into accelerating the slide straight back; the grip frame can
only push back with 8 or 10 pounds of force at the initial part of
slide movement (when the bullet is still in the barrel).  So there
is very little force trying to torque the gun and a large force trying
to accelerate the slide straight back.

If I've been insufficiently persuasive, let me apply some numbers.
Let's look at the momentum as you've suggested.  By rearranging the
equation for kinetic energy, we get impulse = momentum or:
integral F(t) dt = integral M dV.  To convert this to the angular
case, you would change the M as in mass to I as in polar moment of
inertia and the F would have to change to angular impulse which is
merely F(t) times the radius it's acting at.  Also, V changes into
omega, which is angular velocity.

The polar moment of inertia about the center of gravity of my Gold
Cup half-loaded (one up and three down) is .047 lbf-sec^2-inch.
The inertia doesn't change appreciably when the slide is pushed back
the 1/10 of an inch or so that occurs while the bullet is still in
the barrel.

Now to get the impulse.  F(t) is easy because it's nearly constant at
8 pounds in my Gold Cup during the time the bullet is in the barrel.
The time isn't too tough either, it's just the time the bullet is in
the barrel.  From the Le Duc equation, we know that the velocity curve
is sort of an ersatz parabola so the average velocity of the bullet
while it's in the barrel is 2/3 of the peak velocity.  Bullet travel is
4.4 inches.  So the barrel times for your 760 and 850 fps loads are
.00049 seconds and .00043 seconds.  Since the bore line is 1" above the
center of gravity of the gun, the impulses of your two loads in my gun are
8lb * .00049sec * 1" = .0039 lb-sec-in and for the other, .0034 lb-sec-in.

From the above we can find omega.  Omega is angular impulse divided
by polar moment of inertia.  So the omegas for the two loads are
.0039/.047 = .083 radians/second and .0034/.047 = .072 rps.

To find out how far the gun tipped back, we have to multiply the average
omega by the time it was in effect.  The average omega is again 2/3 of
the final omega so the average omegas are .055 and .048 rps.  Multiplying
these figures by their barrel times, .055*.00049 = .000027 radians and
.048*.00043 = .000021 radians.

Now all we have to do is to find the tangents of these angles to find
out how much higher the bullet lands on the target than it would if
the gun didn't recoil a little while the bullet is still in the barrel.
Let's find out how much higher they are a 25 yards (900 inches).
For the 760 fps load: 900 inches * tangent .000027 rad = .024 inches
For the 850 fps load: 900 inches * tangent .000021 rad = .019 inches

Well, I've made some simplifying assumptions as you can readily see.
I've assumed that the cg of the slide lies _exactly_ on the bore line
which makes things appear slightly more favorable to my argument and
I've assumed that the shooter has a very loose grip which is unfavorable
to my argument.  Any and all are certainly welcome to refine the math.
But I think you'll find that I'm not terribly far off.

John Bercovitz     (JHBercovitz@lbl.gov)

From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Trajectory question
Organization: Lawrence Berkeley Laboratory, California

In article <199210310402.AA08205@ucsu.Colorado.EDU> fcrary@ucsu.Colorado.EDU
(Frank Crary) writes:

#...... (By the way, semi-autos
#have some parts which are accelerated while the bullet is in the
#barrel. They soak up some of the force of the recoil and some of
#the momentum, at least for a short time. That's why they have less
#recoil...)

I guess it depends on what one means by "less recoil".  Certainly
the recoil of a semi-auto is more spread out in time.  That's why
my 45 auto takes so long to settle down after I fire it.  My 357
revolver just does it and gets it over with; no klinkety-klankety
hippity-hop dance.

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-)

John Bercovitz     (JHBercovitz@lbl.gov)

From: bercov@csg.lbl.gov (John Bercovitz)
Subject: Re: Trajectory question
Organization: Lawrence Berkeley Laboratory, California

In article <9211042257.AA22000@sweetpea.jsc.nasa.gov> weed@sweetpea.jsc.nasa.gov (daniel weed 283-4162) writes:
#replying to an article by John Bercovitz  (JHBercovitz@lbl.gov)

# >I have trouble understanding how this could be happening.  Perhaps you
# >could explain in a little more detail how you performed this test and...

#Well, I never performed any "test", this was simply an observation made
#over several months of shooting these loads.  I'm shooting from a
#Weaver stance, with a target distance of about 10 yards.  Since I re-load,
[stuff deleted]
#at old targets.  I usually can shoot about a 13 shot (para-ordinance frame),
#3-inch group, and the center of the groups are typically around 1 inch

Hey, an observation is a test.  I was just curious about the conditions.
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.

# >...there really aren't any forces to create a torque on the gun while
# >the bullet is still in the bore.  (After the bullet leaves, the slide
# >slams into the frame and _really_ torques the gun upward.)

#Sure there are.  The slide is not moving freely while the bullet is
#in the barrel.  It had better not be moving much at all.  Also, if
#the muzzle jump were only due to the motion of the slide, then revolvers
#wouldn't have any.  Now I don't own one (yet), but I'm pretty sure
#they do.

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?  After the
slide has moved to the rear fully, it strikes the frame and tilts the gun back.
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.
Revolvers, which are solid, start tipping back as soon as the bullet starts
to move because the line of action is not through the c.g.  (And I agree,
if muzzle jump were only due to the slide, revolvers wouldn't have muzzle
jump because they don't have slides.  No sir, can't argue with that one!
8-))

# >2) Unlike the revolver where the front sight is much higher above the
# >bore centerline than the rear sight, the government model's front and
# >rear sights are nearly exactly the same height above the bore line...

#All very interesting, but since the sights didn't change between loads,
#I don't see how its relevent.  Anyway, I have adjustable sights, sighted
#in for the 230 gr load.

I'm not sure what relevance the sights not being changed has to the argument,
but allow me try to explain further what I meant in 2) above.  If the gov't
model's sight tops form a plane parallel to the bore centerline, then the bore
is pointing at the nominal impact point on the target very nearly when the gun
is fired.  Now if the bullet actually does hit that nominal impact point, I
take it as evidence that the bore was pointed at the impact point when the
bullet left the barrel.  Again, this is an approximation.  [[For starters,
the sight plane is a half inch above the bore centerline so the gun must
tilt enough to compensate for that.  Secondly, the bullet is under the
influence of gravity after it has left the barrel and so it drops according
to s = (1/2) g t^2.  If the bullet travels at 900 fps, then it is in free
fall for 1/12 th second while on the way to a target 900 inches away.  So
the drop is s = (1/2)*386*(1/12)^2 = 1.3 inches.  If you add the two
together, you get the fact that the bullet must have started out at an angle
of arctan 1.8/900 = 0.1 degrees or 2 parts in 1000.  I had really hoped to
leave this out so I wouldn't add more marginally significant stuff to the
post, but I can see I'll never get it by you! ]]  Now let's compare that to
a revolver.  Its sights tilt the gun's bore downward 15 parts in 1000 as a
sort of nominal figure.  15/1000*900 = 13.5  This means that a revolver
sighted in for 25 yards is actually pointed over a foot below its eventual
impact point at the time of firing.  Does any of this argument make any
sense to you?  What I hope to have shown is that the revolver tips back a lot
while the bullet is in the bore and the gov't model does not.  The fact
that the pistol in reality tips back 1 or 2 parts in 1000 does not negate
my arguments of yesterday because what I was showing was the absolute change
due to different bullet weights, not the absolute amount of tip.  You can
check the construction of yesterday's argument to verify this.

# >3) When I fire all sorts of different loads out of the government model
# >with the government model mounted in a machine rest, all of the loads
# >hit the same point at 25 yards.

#Ah, but using a machine rest is cheating :-)
#The gun cannot be considered independently when examining its motion.
#The gun is attached to a flexible hand, attached to a flexible wrist,
#attached to a flexible arm, and a flexible body.  The gun/hand/wrist/arm/body
#is not a rigid system, and has multiple pivot points.  I wager that
#the machine rest is MUCH stiffer than any human.  So, even though it is
#much easier to mathematically model the gun motion when mounted to a
#machine rest, that model won't match the real world.  I would not be
#suprised if the motion of a gun in a machine rest is nearly identical
#for all loads.

So you're saying my Ransom rest is a rigid system and the human is a non-
rigid system.  Ok, I'll buy that.  However, while gov't models throw
different weight bullets to the same point of impact in the machine rest,
revolvers do not.  In fact, both types of gun's sights very nearly agree with
point of impact in the machine rest.  The 45 in particular can be sighted
in on the Ransom rest (25 yards) and then fired by hand with no change.  The
revolver is slightly different but the dispersion of heavy vs. light bullets on
the target is the same as if you were firing by hand.  A long time ago Charlie
Ransom (may he RIP) told me he had spent some effort to get this similitude.

#Now, let's see what I would have to do mathematically to prove the gun
#barrel moves differently with different loads.  Lets take two identical
#scenarios, differing only in the bullet weight and velocity.  Call (A) the
#230 gr, 760 fps round, and (B) the 200 gr 850 fps round.  Assume everything
#else is equal and consistent with my test case (weaver stance, 10 yards,etc).

I think I can agree to all of this except incorporating your test results.
I think your test is flawed so I wouldn't want to start with it and work
backwards.

#Lets further assume the motion of the slide (if any), is negligable, and
#that the conservation of momentum rule applies.  Assume friction is not a
#major factor.  External ballistics are not a factor since I'm only
#interested in the motion of the gun up to the time the bullet leaves the
#barrel.

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

#Since the gun is attached to a flexible body, I don't even want to think
#of modelling the actual motion of the barrel.  However, I can look at the
#total impulse applied to the bullet and frame, and compare the impulse
#for both cases.  Then, knowing that the rotation point (NOT the CG) is below
#the centerline of the barrel (an unknown distance), I know the resultant
#impulse will result in a torque that will tend to rotate the barrel UP.

You're forgetting the slide is not locked to the grip frame.  It is only
loosely connected thru a spring delivering 8 pounds force.

You don't want to model the actual motion of the barrel but that's what
we're discussing.  I think you need to model it.  Make simplifying
assumptions where you can and clearly state them so all can judge their
validity.

#Since the observed data indicates (A) hits higher than (B), then the total
#impulse due to firing (A), causes more barrel rotation than firing (B), meaning
#that the total impulse of (A) is greater than (B), since the mass properties
#and rotation points are the same (whatever they are).

The observed data was not rigorously obtained.  Based on my more rigorously
obtained data, I believe your data is flawed.

#So, lets see if this is true.
#
#The total impulse, from ignition to the bullet leaving the barrel, is equal
#to the total change in linear momentum.  If the gun is considered a closed
#system, the mass does not change, and the total momentum is zero, then I
#can determine the total impulse applied to the gun is equal and opposite
#to that applied to the bullet.  Impulse = rate-of-change of momentum,
#equals mass times change in velocity.

The gun is not a closed system.  It is a slide and a spring loosely
connected during the time the bullet is in the barrel.  You may treat
it as a closed system to determine the resulting configuration after
it has done its thing and settled down; that's what I did as evidenced
by my saying it tilts back when the slide hits the frame which occurs
at the end of the cycle.

#For case (A), the total impulse is 3.3452 N*s  (I prefer SI units)
#For case (B), the total impulse is 3.3576 N*s
#
#So the impulse for (B) is about 3% lower than for (A).  This demonstrates
#that the motion of the barrel in the two cases is different, significant,
#and in the right direction consistent with the emperical evidence.  But I
#do not attempt to quantify the actual barel motion.  That would require
#A LOT MORE WORK, and I'm not getting paid for this.

It does demomonstrate they are different but it does _not_ demonstrate
that this is a significant difference in the present case.  I would
contend that what you have done is say something akin to: "John has sent
10 cents to the IRS for the purpose of paying off the national debt while
Dan has sent 25 cents.  Dan has made a 150% larger contribution than John
has.  This is significant."

All you have done is quantify the impulse.  You have not shown its effect.
You have not completed your argument.  But, hey, wouldn't it be neat if
we _did_ get paid for this?

#Impulse is Force times Time, or the "area under the curve".  It really
#doesn't matter what the curve looks like since we can get the Impulse
#directly from the final velocity of the bullet.

Yes, the bullet momentum is the impulse that is delived to the closed
system.  But that's not what we were talking about.  We were talking
about the impulse which tilts the gun during the time the bullet is
in the barrel.  The force of that impulse is contributed by the recoil
spring.  That's why I calculated impulse rather than taking the bullet's
momentum.  The impulse due to the bullet (actually the gas) is merely
driving the slide backwards during the period of interst.

Things just ain't always quite that simple is the problem.  Sometimes
you get caught and you gotta analyse a little deeper.  If you pursue
your present path, you will calculate the closed system response, the net
effect when all is said and done.  Sure as hell the gun will pointing
upward at that time.  But we're not talking about that; we're talking
about how much it's tipped up at the instant the bullet leaves.

# >The polar moment of inertia about the center of gravity of my Gold
# >Cup ...
# >Since the bore line is 1" above the center of gravity of the gun,
# >the impulses...
#
#Not relevent since the gun does not pivot about the CG.

As you said above, the human attached to the gun is a non-rigid system.
In fact, considering the low spring constant of the hand's grip, the gun
can initially be considered to be floating in space to a good approximation.
So in fact it does very nearly pivot around the c.g. during the period the
bullet is still in the bore.

For revolvers this is also a good approximation.  Since their borelines
are above their c.g.s, they tip back.  A 45's boreline is on the c.g. of
the only thing significant it sees.  Both of these are approximate
statements because we all know that a _very_ loose grip will have _some_
effect on impact point.  But this effect is small compared to the dominant
effect.

# >From the Le Duc equation, we know that the velocity curve
# >is sort of an ersatz parabola so the average velocity of the bullet
# >while it's in the barrel is 2/3 of the peak velocity...

#I've not heard of Le Duc, could you site a reference? Sounds interesting.

# >From the above we can find omega.  Omega is angular impulse divided
# >by polar moment of inertia.

#You can't even begin to estimate the moment of inertia, nor the moment arm.
#The gun cannot be taken to be floating in space when determining its motion.

I didn't estimate the moment of inertia; I measured it.  The gun can be
considered floating in space because the slide is loosely connected to the
frame and the frame is loosely connected to the hand at least during the
time the bullet is in the bore.  The frame does not compress the flesh of
the hand significantly until the slide is much further back in its travel.
The hand is soft for the first little bit of compression.  It takes little
force to move the gun even in a hard hold if you don't have to move it far
because the flesh of the hand is an initially-soft nonlinear spring.

#Well, this is quick back of the envelope stuff.  I welcome anyone to
#jump in and correct me.

Critique: I think it was a little too loose.  I've outlined the weak areas
above.  Please get a new envelope and give it another try.

#Suggested experiment:  Load (B) to 874 fps and try again.  This time, load
#(B) will have the same impulse as (A).  Would It shoot lower?

Go for it, Dude!  Let us hear from you.

#Net Question:  Which is more important to felt recoil and muzzle jump,
#Momentum (Impulse) or Energy?  True also for a compensated gun? (that
#is: ligher bullet = more powder = more gasses = better comp even though
#higher energy?)

The British historically tend to judge by momentum and in the US it's usually
energy.  It's a great controversy and I fear it will never be resolved.  My
personal feeling is that the momentum is sort of the shock of the thing, in
a way, while the energy is what you have to absorb or contend with before
you can get off that second shot.  This is just a trend, a feeling and certainly
far from rigorous or universally applicable; I've never analyzed this but I
sure ought to sometime.

#Can anyone cite a good reference on internal ballistics?

Yuh gotta ask Henry.  He's the net bibliographer.  (And semi-omniscient.)

Sorry if the response sounded a little like a harangue.  I was trying to
seize every opportunity to show you the way I see it, the way I believe
I've proven it is.

John Bercovitz     (JHBercovitz@lbl.gov)

From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Trajectory question
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)

From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Trajectory Question
Organization: Lawrence Berkeley Laboratory, California

In article <9211051549.AA01004@sweetpea.jsc.nasa.gov> weed@sweetpea.jsc.nasa.gov (daniel weed 283-4162) writes:

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

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

[stuff deletd]

# >center of gravity or something.  Anyhow, this is just an illustration.

#And a poor one at that.  Again, wrong, since

[stuff deleted]

# >QED  No legerdemain or nothin'.  8-)

#Houdini would be proud. ;-)

#--- Dan

Hey Dan, this is the kinder and gentler newsgroup.  By all means take
exception to everything I've said but let's hold down the insult crap,
huh?  I'm guilty of getting frustrated at the handicap of having to
communicate well in short sound bytes on this infernal machine same as
you or anybody else, but let's try to be reasonable here and use this
frustration to increase our efforts to honestly communicate.
'Nuff said.

As to your objections to my simplification of the problem:  Absolutely.
I agree completely that my assumptions were idealistic.  I was only
trying to demonstrate that it is possible to increase recoil energy by
having the momentum initially taken up by a lighter mass.  So I picked
the worst case.  Let me try it a different way:

If you have a gun that has the weight of a car, the momentum transferred
to the gun when it is fired causes very little increase in the gun's kinetic
energy.  When you have a gun that has the weight of a derringer, a large
amount of kinetic energy is transferred to it.  If that momentum is
transferred to yet another mass of equal weight by elastic collision,
then this second mass now has the velocity and energy which the first
mass had.  All very ideal, I agree, but the point was demonstration of
a possibility, not absolute realism.  And in fact, semi-autos do have
more recoil energy.  No where near as much as I've shown, but a little more.

Now who was it that said all this was better put in Hatcher's writings?
Which writings?  What page of what book or magazine?  Is it the same
argument differently approached or a different argument?

John Bercovitz     (JHBercovitz@lbl.gov)

From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Finding the polar moment of inertia
Organization: Lawrence Berkeley Laboratory, California

Ever the glutton for punishment 8-), I'm responding to an off-net
query as to how I measured the polar moment of inertia of my Gold
Gold Cup about its center of gravity.  I've started with standard
formulae which should be in your physics book, rearranged them and
combined them.  So if you will, take a look in your physics book
for that which precedes the formulae below.

I found the polar moment of inertia by measuring the frequency of
the gun's oscillation about a point which is a known distance from
the c.g.  (I stuck a knife edge inside the trigger guard.)  I found
the location of the c.g. in the usual way: I suspended the gun
from two points and drew vertical lines through those points onto
the gun.  The intersection of the two lines locates the c.g.
(Suspending a gun on a knife edge, drawing lines on a gun;  is
nothing sacred?!)

Inputs:

"acceleration" of gravity = 386.1 inch/sec^2
mass of half-full gun: 40.62 ounces = .00658 lb-sec^2/inch
distance from pivot point to c.g: 1.50 inch
frequency of oscillation: 60 cycles/48.0 seconds = 7.85 rad/sec


Symbols:

g = "acceleration" of gravity
m = mass of gun
i = polar moment of inertia about pivot point
p = polar moment of inertia about c.g.
r = distance from pivot point to c.g.
q = distance from pivot point to center of percussion
w = frequency of oscillation  (using w for the standard omega)


Standard formulae:

w = (g/q)^0.5            Note that q is not only the distance from the pivot
from which:              point to the center of percussion; it is also the
q = g/w^2        (1)     length of an equivalent simple pendulum.

q = i/(mr)
from which:
i = mrq          (2)

i = p + mr^2
from which:
p = i - mr^2     (3)


Calculations:

To reduce to one formula, plug (1) into (2) and (2) into (3):

p = mrg/(w^2) - m(r^2) = mr((g/(w^2))-r)
  = (.00658)(1.5)((386.1/(7.85^2))-1.5) = .047 lb-inch-sec^2


Extremely rough check:

p = (m/12)(a^2+b^2)   (also in your physics text)

where a and b are the length and width of some rectangle which
approximates the firearm in some way.  I guessed a 3x8 inch
rectangle.

p = (.00658/12)(3^2+8^2) = .04 lb-inch-sec^2

So we're in the ball park.

John Bercovitz     (JHBercovitz@lbl.gov)

From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Trajectory Question
Organization: Lawrence Berkeley Laboratory, California

 Hey Dan, when you or I send in an opus and Jim moderates and sends it out
 from magnum@cs.umd.edu, does that make it a "magnum opus"?  Hyuk, hyuk,
 hyuk, hyuk, I slay myself.

 Sorry.

 [MODERATOR:  AAAAAAAAUGYHHHHHHH!}

In article <9211091828.AA04413@sweetpea.jsc.nasa.gov> weed@sweetpea.jsc.nasa.gov
(daniel weed 283-4162) writes:

#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

#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.

No, I was assuming a low-friction slide and a loose grip.  Low-friction and
loose are not necessarily partners.  The spindles on your car have low-friction
but are not loose while farm equipment is often loose but has high friction.
I assumed the low friction to favor my side of the argument and the loose
grip to favor your contention.  Couldn't be fairer.  8-)

#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.

The guy who started the thread noticed the phenomenon in a rifle.  Actually,
what he noticed is that one of his light bullet loads struck much lower while
one did not.  This is pretty standard since barrel whip is a major factor
in point of impact for rifles.  In ordinary handguns, the barrels are too
short to whip.  Many other people have noticed the difference in point of
impact in handguns but not for the gov't model, I think.  I sure haven't in
the 38 super, 10 mm and 45 gov't models I've owned.

# >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.

Good point.  The hammer does take quite a bit of force and it happens just
at the period in the cycle when the bullet is still in the barrel.  However,
this force is not too much larger than the recoil spring force as I recall.
Since I got zilch barrel rise from the recoil spring, I expect that a few
times zilch is still small.  However, I think there is another factor missing
which may be modestly significant, but more about that below.

#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."

I think I'd better grab that article and take a look.  Sounds interesting.
How much tilt was he observing?  As I showed in an earlier post, the
barrel must tilt enough to cover the bullet's drop and the height of the
sights above the slide.  That amounted to some 2 parts in a thousand
while the revolver rose some 15 parts in a thousand.  All I've been
trying to say since the beginning (so who can remember the beginning?)
is that revolvers have wild dispersions due to different bullet weights
while 45 autos have virtually none.

# >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.

Well, not totally free; I said the recoil spring was acting on the slide.
But as I've said many times, I was making an ideal case to demonstrate
the tendency.  There was no attempt to come up with the exact numbers.
I just wanted to show why a revolver tips a lot while an auto of this
type tips very little.  I think it's fairly common to idealize a case to
show the direction of the argument.  Of course we can always agree to
disagree on how much was lost when I idealized the problem.

#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.

You have shown that the heavier bullet tends to strike higher but you have
not shown that it causes it to strike significantly higher.  The physical
evidence you cite is that of your gun and Rick's article.  I think the former
is somewhat suspect as generally-applicable evidence for the reasons I've
mentioned before.  The latter must be quantified to count.  I'll get the
article and take a look.  I have a lot of respect for Jamison's gun knowledge,
but although I haven't talked to him in a coon's age, and people change, I've
always thought of him as more conversant in rifles than handguns.

#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.

If you do get a round tuit:
I think perhaps a significant thing I left out is that there really may
be quite a bit of friction between slide and rails.  It may be enough
to account for the small barrel rise which I showed earlier must occur.
I think the fact that the boreline does not go directly through the
c.g. of the slide means the slide wants to rotate as it's pushed back.
This will cause increased friction.  I don't think you would see very
much of this in a long-railed gun like a CZ or one of those gorgeous
Neuhausen's.  However, a short-railed gun such as a gov't model should
tend to get a little self-actuation like a properly designed brake.

I have to say you've done a fair job of contrasting our analyses.
I think we now agree on everything except the magnitude of the effect.
And that we can determine by test.

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

You bet!  Also, anyone else feel like shooting a few groups at a reasonable
distance with 200 grain and 230 grain bullets and comparing the points
of impact?  Dan and I have done our tests and they differ.  Obviously
we need more tests.  What we need is group sizes and displacements of
their centers at given distances.

John Bercovitz     (JHBercovitz@lbl.gov)

From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Trajectory Question
Organization: Lawrence Berkeley Laboratory, California

#In article <9211091828.AA04413@sweetpea.jsc.nasa.gov> weed@sweetpea.jsc.nasa.go
#(daniel weed 283-4162) writes:

##Most certainly.  I agree slide friction is the smallest force.  Slide
[stuff deleted]

In article <27334@dog.ee.lbl.gov> bercov@bevsun.bev.lbl.gov
(John Bercovitz) writes:

#I think perhaps a significant thing I left out is that there really may
#be quite a bit of friction between slide and rails.  It may be enough
[stuff deleted]

Thinking about it again, I think you're right.  If the slide/frame friction
were significant, then the state of lubrication of the interface would make
a difference in the impact point.  Though I can't recall if there was a
difference when I tried high pressure lube, I don't think I've seen a differ-
ence using ordinary gun oils or Break Free.  So I doubt this friction is really
significant.  It could and should be tested in a tight gun.  I'll do that
next time I drag out the ol' Ransom rest.

John

From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: Re: Trajectory Question
Organization: Lawrence Berkeley Laboratory, California

I've had a couple of killer experiments suggested which could very possibly
settle some of the questions which have been raised in this thread.  This
is a call for critique of these suggestions and also a call for volunteers
if anyone feels especially generous.

From Doug:
If in fact the gov't model's frame is loosely connected to the hand during
the bullet's time in the barrel so that the hand doesn't greatly affect
the point of impact, then if said gov't model is suspended by strings and
soft springs, its point of impact should likewise not be greatly affected.

From Alan:
If the gov't model's slide moves essentially straight back because it is
disconnected from the frame while the revolver tips back because its barrel
is solidly connected to the frame, then if you somehow lock the slide of the
gov't model to its frame, the gov't model should now have a higher point of
impact.

Great suggestions, huh?  Any comments?  Any holes to be poked in these
suggested procedures?  Any volunteer experimentalists?  (Hope, hope)

I have one comment on Alan's suggestion and that is that since the c.g. of
the gov't model is so much closer to the boreline than the c.g. of the
revolver, I don't think you will see a 1 foot rise in impact at 25 yards as
you do with the revolver.  Maybe only half that or a little less because
as I established earlier, the gov't model already rises almost 2" at 25 yards
from the aim point.  This be no means invalidates the experiment, just
quantifies it a little.

John Bercovitz     (JHBercovitz@lbl.gov)