From: bercov@csg.lbl.gov (John Bercovitz)
Newsgroups: rec.guns
Subject: Re: Trajectory question
Date: 6 Nov 92 01:57:56 GMT
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)