From: bercov@bevb.bev.lbl.gov (John Bercovitz)
Newsgroups: rec.guns
Subject: Colt GM dynamics
Keywords: physics, Colt
Date: 15 Aug 90 01:04:50 GMT

			       GM Dynamics

As a result of recent discussion on the net, several people have responded
privately with the request that I post an analysis of the dynamics of the
government model on the net to serve as a basis for further discussion and
development (a target to shoot at).  I'll do the best I can but it is hoped
that folks who really "know" will eventually respond and then correct and
expand upon this attempt.  Are folks at Aberdeen, Quantico, etc. on the
net?

In an effort to not exclude any potential participants, I will try to write
thorough explanations.  I'll assume all have had high school physics and
math.  These explanations may necessarily run to some length so that the
article will best be broken up into several parts.  Do feel free to jump in
at any time.  Need I have mentioned that?

Part one:

The first thing to get squared away is a review of impulse-momentum.
Starting with the familiar equation, F = ma:

F = ma = m dv/dt.

Rearranging: F dt = m dv.  [More exactly, if F is varying: F(t) dt = m dv]

The integral of the term on the left is impulse and the integral of the term
on the right is momentum.  These equations are for linear action but
equivalent equations exist for the angular case.

If you've ever seen stop-action photos of government models in the
process of firing, you know that during the initial stages of the slide's
opening, the firearm appears to be still pointed at the target: no angular
deflection is obvious though some small amount must exist.  This is
evidence that the first part of the cycle can be described by the linear
impulse-momentum equation given above.

For this first article, let's ignore the force of the recoil spring and the
mass of the propellant.  The force at any given time on the back of the
bullet is the chamber pressure at that time multiplied by the cross
sectional area of the bullet.  The force at any given time on the slide is
exactly equal to the force on the bullet even in the case of a bottle-necked
cartridge as long as the action is locked.*  So integral F dt, the impulse, is
the same in magnitude but opposite in sense for the bullet and the slide.

Since at any given time the magnitude of the impulse is the same for
bullet and slide, the magnitudes of the momenta of the bullet and slide are
also equal.  Lacking subscripts, let's call M and V the mass and velocity of
the slide and call m and v the mass and velocity of the bullet.  It's just
been shown that:

M dV = m dv

or: M dS/dt = m ds/dt

Integrating this equation with respect to time:

M S = m s for any period of time you care to name.  Let's take the period of
time that starts when the bullet starts to move and ends when the bullet
clears the barrel:

The mass of the slide and other moving parts (when barrel and slide are
locked together) is about 18 ounces or 510 grams.  Let's say the mass of
the bullet is 200 grains or 13 grams.  In a government model the travel of
the bullet until it clears the barrel is around 4.4 inches or 110
millimeters.  So how far does the slide/barrel assembly travel to the rear
while the bullet travels to the end of the barrel?  It would travel
13/510*110 = 2.8 mm or 0.11 inches.  (To be more exact, one should take
into account that the barrel backs off the bullet a little, decreasing the
distance the bullet travels to get to the end of the barrel.)  Simplistically
speaking, after the slide/barrel assembly in the government model travels
roughly 0.09 inch (2.3 mm) it starts to unlock.  Unlocking is completed
after roughly 0.2 inch (5.1mm) of travel.  So the bullet has left before
unlocking is complete.

For this simple model, it has been shown that the distance the slide
travels until the bullet exits depends only on the bullet's mass and is
totally independent of the bullet's exit velocity.

It is amusing to extrapolate the consequences of the equation, MS = ms,
further.  If we have a gun in some void in intergalactic space and this gun
has its barrel's centerline going through the center of gravity of the gun,
after we fire the gun, M*S will be equal to m*s for eons before anything
disturbs either component.  Another way of saying the same thing is that
at any time after firing, the center of gravity of the two components
considered as a system remains at the same point in space it occupied
prior to firing.

* Why the force on the slide is equal to the force on the bullet even when a
bottle-necked cartridge is used: Visualize the area of the bore projected
axially onto the interior of the cartridge case head.  The remaining annular
area inside the cartridge case head is compensated by a similar annular
area (more or less the shoulder of the case) around the neck end of the
cartridge case.  Since these two annular areas are equal, the pressure-
induced forces on them oppose and cancel out.


JHBercovitz@lbl.gov

From: bercov@bevb.bev.lbl.gov (John Bercovitz)
Newsgroups: rec.guns
Subject: Re: Colt GM dynamics
Keywords: physics, Colt Government Model
Date: 24 Aug 90 17:45:30 GMT

     A quick review:  In part one it was noted that the equations of linear
motion apply to recoil-operated pistols.  Evidence cited for this was that
stop action photos which show pistols in the initial stages of recoil do
not show the pistol tipping up.  Other evidence is that, unlike revolvers,
borelines and sightlines of recoil-operated pistols are nearly parallel.
From the equation, F = m a, the equation MS = ms was derived.  This latter
shows that the distance the barrel-slide assembly travel times its mass
is equal to the distance the bullet travels times the bullet's mass
regardless of the velocities involved.

     In part one the effect of the powder's mass and the effect of the recoil
spring's force were ignored.  So what is the effect of the powder's mass?
For a specific example, let's consider 7 grains (0.45 grams) of powder and
a 200 grain (13 gram) bullet.  As noted before, the center of gravity of the
200 grain bullet moves approximately 4.4" (110mm) from start of motion
to exit from muzzle of the 45 ACP Government Model.  The center of
gravity of the powder mass doesn't move nearly as far.  It starts its
motion in the center of the chamber volume and has moved to the center of
the barrel-plus-chamber volume at the time the bullet exits.  If the
cartridge case has approximately the same inside diameter as the barrel,
the distance the center of gravity of the powder moves is approximately
half the distance the bullet moves.  Proof of this is left as an exercise for
the curious.  (That's those who ARE curious, not those who are considered
curious.)  So the MS for the powder mass is 15 grain-inches or 25
gram-millimeters.  This figure is less than 2% of the figure for the 200
grain bullet traveling 4.4".  While the MS of the powder can't be ignored in
instances of cartridges with a high ratio of powder mass to bullet mass,
it seems safe to ignore it in this instance.

From: bercov@bevb.bev.lbl.gov (John Bercovitz)
Newsgroups: rec.guns
Subject: Re: Colt GM dynamics
Keywords: physics, Colt Government Model, spring
Date: 24 Aug 90 17:55:42 GMT

     One prerequisite for looking at the effect of the recoil spring is to
review the mechanics of springs.  There are a couple of types of wire coil
compression springs.  Those with large wire diameter compared to outside
diameter, such as a rifle's striker spring, are a bit more complicated to
describe mathematically than lighter weight springs such as recoil
springs; only lighter weight springs will be considered here.
     Three spring properties should be reviewed: force/displacement,
energy/displacement, and spring constant/physical dimensions.  Springs
are linear devices: the force a spring responds with is directly
proportional to the amount (distance) it is compressed from its free
length.  The distance and force are related by the spring constant which is
a property of a given spring.  In the case of the recoil spring of the GM,
there are three dimensions which are of prime importance other than the
spring constant.  They are the free length, the installed length, and the
fully-compressed length.  The difference between installed length and
free length can be called X1 and the distance between fully compressed
length and free length can be called X2.  When the action is closed, the
force tending to keep it closed will be the spring constant times X1.
When the slide is drawn fully back, the force trying to return it will be
the spring constant times X2.
     Work is force times distance or: integral F(s)ds.  If you plot the
function of force versus distance on Cartesian coordinates, the area under
the curve will be this integral.  Since the spring is a linear device, the
area under the curve is a triangle; the area of a triangle is one half the
ordinate times the abscissa.  Since kx is the force of a spring, the
potential energy of a spring (the work absorbed by it) is 1/2 k X^2.
Therefore the work absorbed by a recoil spring when the slide moves from
fully closed to fully open is = 1/2 k ((X2^2)- (X1^2)).  This is equivalent to
subtracting the area of the triangle associated with the installed
compression of the spring from the area of the triangle associated with
full compression of the spring.
     This article is running much too long so I'll get into numerical
examples of the above in part 4 of this series.
     Since recoil spring manufacturers don't supply conventional spring
data, it is useful to know how to calculate a spring constant.  The obvious
way is to put a weight on the spring and measure the spring's deflection.
Without a good set up this can be difficult to do.  Another approach is
calculation.  The formula for light weight springs is k= (G*d^4)/(64 R^3*n).
G is the modulus of rigidity; its value is around 11.6*10^6 pounds per
square inch for steel spring wire.  It is different for other materials, such
as beryllium copper or stainless steel.  d is the diameter of the spring
wire.  R is the radius from the axis of the spring to the center of the
spring wire; numerically, that would be half the quantity: (outside
diameter of the spring minus the diameter of the spring wire).  n is the
number of active coils (don't count coils  which are touching each other at
the end of the spring).  Try working the following example: Steel spring,
outside diameter = 0.435 inch, wire diameter = 0.043 inch, number of
active coils = 31.5.  Ans: 2.6 lb/in.  Note that the length of the spring does
NOT, repeat NOT, affect the spring constant.  Permanently stretching out a
spring to make its free length longer only changes X1; this causes the
force at the installed length to be higher, but it does not change the spring
constant.
     From the above formulae you can see that if you cut one active coil off
the return spring in the rebound slide of your S&W, you will lighten the
trigger pull since X1 and X2 will be smaller but you will make the trigger
force curve peakier because the spring constant will be greater.


               JHBercovitz@lbl.gov

From: bercov@bevb.bev.lbl.gov (John Bercovitz)
Newsgroups: rec.guns
Subject: Colt GM dynamics - conclusion
Keywords: physics, Colt Government Model
Date: 30 Aug 90 17:03:13 GMT

     This series is running much too long so here's the concluding article:
25 lbs in a 5 lb bag.  To make up for the data compression, I'll be happy to
answer wudja-mean-by's at the address at the end of this article.
     Part 1 of this series showed that MV = mv and that MS = ms.  Part 2
showed that the effect of the mass of the powder gas is negligible during
the time the bullet is in the barrel.  Part 3 showed that the force exerted
by a spring is kX and that the energy absorbed by the spring is
1/2 k ((X2^2)- (X1^2)).
     Since knowledge that kinetic energy (KE) is equal to 1/2 m V^2 is so
well disseminated, I don't think it needs to be covered here.  This is the
same formula that gives bullet energy (bullet energy = mass of bullet
times bullet velocity squared times 2.22 divided by 10^6 where bullet
mass is in grains, bullet velocity is in feet per second and bullet energy is
in foot-pounds).
     Now for a specific example:  The moving mass (slide, barrel, etc)
of my Gold Cup weighs 17 ounces (1.06 lb.) of which 3.25 ounces is barrel.
The recoil spring weighs 1/4 ounce so I ignore it, but technically one
should add about half of its mass to that of the rest of the moving mass
depending on the dynamics of the situation.
     The recoil spring currently in my GC has the following characteristics:
  free length = 6.480"   X1 = 2.833"   X2 = 4.873"    k = 2.6 lb/in
So its total capacity for absorbing energy is, by 1/2 k ((X2^2)- (X1^2)),
equal to 20.4 in-lbs.
     If my 200 grain (0.457 oz) bullet leaves the barrel at 950 feet/second,
then the velocity of the moving mass in the opposite direction, by MV = mv,
is 25.5 fps (306 inches per second).  (950fps*0.457oz/17oz = 25.5fps)
The kinetic energy of the moving mass, by KE = 1/2 m V^2, is 10.7 ft-lb or
129 in-lb.   (1/2 * (1.06/32.174) * 25.5^2 = 10.7ft-lb)
     But during the time the bullet travels down the barrel the slide moves
back 0.11" (from part 1).  By the potential energy formula for springs we
see that the energy absorbed by the spring during this period is
1/2*2.6*(((2.833+0.11)^2)- (2.833^2)) or 0.83 in-lb.  Comparing this with the
129 in-lb the slide carried, I think we can safely ignore this effect.
     The force pushing the slide back varies quite a bit since the chamber
pressure rises, peaks and falls.  The peak pressure is around 14,000 psi in
a typical load.  The peak force, then, is the area of the bullet's base times
the peak pressure: 0.16 square inches times 14000 psi = 2240 pounds.
(Same magnitude of force acts on bullet and slide, refer to part 1.)
     So with how much force is the recoil spring resisting this peak of
2240 pounds?  (Average F) = k*(average x) = 2.6lb/in * 2.888" = 7.5 pounds.
Again, I think we can safely ignore this effect.
     Gettin' glassy-eyed yet?  Me too.
     After the slide travels a short distance, the barrel drops out of it,
impacting the frame.  Though the coefficient of restitution of steel is
high, eventually this energy is burned up in multiple collisions with frame
and slide (equivalent of totally inelastic collision).  I haven't had the
pleasure of following the paper trail on the impulses, but by the principle
of conservation of momentum, the total momentum of the system (frame,
slide, and barrel) must be the same before and after the barrel's multiple
collisions with frame and slide.  Put another way, the momentum the
barrel carries as a component will be found in the system after all is said
and done.  One question is how much time this momentum transfer occupies
in the firing cycle.  Anyone care to pursue this?
     The remaining mass (17 - 3.25 = 13.75 ounces) continues traveling at
25.5 feet per second or 306 inches per second.  By KE = 1/2 m v^2, this
mass is carrying 104 in-lbs of kinetic energy.  At the time the remaining
mass impacts the frame, the kinetic energy is 104 - 20.4  = 84 in-lbs.
(Remember spring absorbs 20.4 in-lbs.)  Running the equation for KE in
reverse, we find that the remaining velocity at impact is 275 inches per
second.  All this assumes that the frame doesn't move during the interim
under the influence of the recoil spring.  Not true and I haven't proven that
it can "safely" be ignored.  Anyone care to tackle this problem?
     I think almost all of the tilting back of the pistol occurs at the time
the slide runs into the frame.  This is when a recoil buffer helps.  It helps
dissipate some of the 84 in-lb but mostly cuts the peakiness of the
impulse loading.
     I didn't account for the jet effect of the propellant gases leaving the
barrel.  Jets are easy enough to calculate if you have inputs of pressures,
times, and mass flow rates.  These inputs come from chemical and
thermodynamic calculations and are hairy to calculate; better to just
measure the momentum of the recoiling firearm and ascribe that part of
the momentum which is not accounted for by the momentum of the bullet
to the propellant gases.  Some authorities just assign a flat 4000 fps as
the velocity of the gases.  In this case the impulse on the moving mass due
to gas escape would be 7 grains of powder times 4000 fps (as compared to
200 grains of bullet times 950 fps for the impulse from the bullet).  This
is about 15% of the impulse due to the bullet.  So we need to go back and add
15% to the moving mass' momentum if we accept this approach.
     The purpose of this series was twofold: respond to the requests for an
analysis, and provoke a discussion of gun dynamics so I can learn more on the
subject.  If you're interested and can add to the calculations outlined above,
please consider doing so.


      Thanks.
                  Wudja-mean-bys to: JHBercovitz@lbl.gov

From: bercov@bevb.bev.lbl.gov (John Bercovitz)
Newsgroups: rec.guns
Subject: Re: Light trigger pull: the down side
Keywords: spring
Date: 19 Sep 90 19:33:06 GMT

In article <45206@ism780c.isc.com> bercov@bevsun.bev.lbl.gov (John
Bercovitz) writes:

> 3) What do "16" and "20" lbs, as examples, refer to in a recoil spring?

A friend lent me a W. C. Wolff Company spring catalog.  I looked through it to
find the answer to my question (above).  Although Wolff makes wonderful springs,
their catalog uses words a little loosely; notably, it seems to confuse the
meanings of the words "force" and "energy".

So I called up Wolff and asked them how they make the measurement to determine
the rating of a particular spring.  It turns out that the spring rating is
merely the force the spring exerts on the slide when the slide is fully drawn
to the rear.  That is, the force required to compress a spring to approximately
1.6 inches in the case of a government model.

As discussed before, the simplest model of a spring has two degrees of freedom
so two figures are required to describe it.  The present rating system of
spring manufacturers does not, therefore, completely describe a spring.  But
one could always ask for the free length of the spring also; then you'd have it.

As an aside, I found Wolff personnel extremely courteous and helpful.


            JHBercovitz@lbl.gov
"The knowing presentation of conjecture as fact is the act of a poseur." 8-) JB

From: bercov@bevb.bev.lbl.gov (John Bercovitz)
Subject: Re: GM trigger problems
Organization: Lawrence Berkeley Laboratory, Berkeley

In article <33094@mimsy.umd.edu> megatek!avf@UCSD.EDU (Andy Funk) writes:
#What difference is there between dropping the slide on an empty chamber
#(purported to damage custom trigger jobs), and dropping the slide when
#chambering a round (as when firing the pistol) with regard to trigger
#damage/stress?

Of course this is my question and that of several others, too.  Dropping the
slide like this is harder on other parts of the pistol, so I don't do it,
but how it's harder on the trigger, I don't know.  

Just for reference, though, some of the recoil spring's stored energy is
used to strip the top round from the magazine thereby leaving less energy
to accelerate the slide.  The result, then, is that the slide doesn't slam
home as hard as it would if it _didn't_ have to strip and feed a round.  The
impact of the slide slamming home is taken up by the shaft of the slide stop.
So I imagine that the slide stop's frame hole will be elongated in time by the
slamming of the slide.  Another area which takes up the impact is the barrel-
hood/breech-face contact.  I think this area would also be beaten up by
habitual slide-dropping.  I also don't like to do this on "mechanical aesthetic"
grounds: that sort of clanking is against my religion.  8-)
                          
                                      John Bercovitz

From: bercov@bevb.bev.lbl.gov (John Bercovitz)
Subject: Re: GM trigger problems 
Organization: Lawrence Berkeley Laboratory, Berkeley

In article <33070@mimsy.umd.edu> charlesg@tybalt.caltech.edu (Charles Grosjean) writes:
#.............  it seems that if you drop the slide without holding the
#trigger down, then the disconnector is in the way of the slide, that is, it
#is pushing against the rear of the slide harder than it is after you've pulled
#the trigger. ..........

I'm not sure if I'm exactly on point here, but if your slide has more trouble
going forward when the trigger is not pulled than when it is, you've got a
small fit-up problem in the disconnector area which can lead to a little more 
slop in the trigger pull.  When you aren't pulling the trigger, the disconnector
sticks up out of the frame higher than when you are pulling the trigger.  So
if there isn't enough ramp on the back side of the disconnector, or if the
disconnector sticks up farther than it should, or if that teeny-tiny ramp
at the bottom of the breech face is missing, then the disconnector gets
banged hard before it pops down into its hole.  If this keeps up, over a long
period of time the disconnector hole will be beaten into an oval shape - it
will be stretched forward.  Since the upper end of the disconnector moves
forward when you pull the trigger, there will be a slightly longer distance
to pull before the upper end of the disconnector runs into the front of the
now-elongated hole.  For a fix, I would caution you against removing material
from the upper very top of the disconnector unless you're really sure of 
yourself; this is a really good way to go full-auto in the most unreliable
way.  And remember: the third one goes up yer nose.  At least it will if 
you're not used to occasional doubling.  Oh yeah - if you want to check for
this problem most easily, turn your muzzle bushing, let the recoil spring plug
out, and cycle your slide.  It's a lot easier to feel the problem when you're
not simultaneously trying to overcome the force of the recoil spring.
       
                    John Bercovitz

From: bercov@bevsun.bev.lbl.gov (John Bercovitz)
Subject: gm recoil springs

This weekend, faced with an enormous amount of reloading, I 
opted to see if I could characterize differences between 
various recoil springs used in Colt Government models.  The 
springs I looked at were: the stock spring used in the 45 ACP 
Gov't Model, the 18.5# "HD" spring supplied with the Wilson-
Rogers one-piece full-length recoil spring guide, and the 
dual-spring set up that comes with the Colt Delta Elite.  I 
had a little success, but not a lot: I can infer some of the 
effects of the various springs, but I don't know whether 
these are the essential effects or merely incidental.  I'd 
appreciate others' speculations.

Gov't model background: As I've shown before in various 
articles, the slide under recoil initially carries 70 to 150 
in-lb of kinetic energy according to how hot the load you're 
firing is (slow 45 ACP target or smoking 10 mm).  Of this, 10 
to 30 in-lb is given up to the recoil spring according to its 
type, some is lost in various forms of friction, and the rest 
is still there when the slide bangs into the stop (which may 
or may not have a shock buffer for frame protection).

Spring background: Fundamentally, a spring has two 
characteristic numbers: its free length and its spring rate.  
Its force is proportional to its deflection (its deflection 
is the difference between its free length and its compressed 
length); the proportionality constant is the spring rate.

Here are the data; symbols are defined below.

       Stock Colt      Wilson-Rogers      Stock Delta
       45 ACP Gov't    18 1/2# "HD"       Elite dual
       Model           with buffer        springs

X0      6.53              5.92               NA
X1      2.75              2.33               NA
S       2.18              2.05              2.12 
X2      4.93              4.38               NA
K       2.65              4.30              7.15
F1      7.29             10.0               7.65
F2     13.1              18.8              22.8
TSE    22.2              29.6              32.3
KEC     7.60              8.72             11.7
VC     83.1              89.1             103

NA  is "not applicable" because the Delta has 2 springs; it 
    actually has one value for each spring
X0  is the free, unrestrained, length of the spring in inches
X1  is the initial compression of the spring (free length   
    less installed length)
X2  is the final compression (free length less fully-
    compressed length)
S   is the stroke of the slide (X2-X1)
K   is the spring constant in pounds per inch
F1  is the force in pounds on the spring when it is compressed X1
F2  is the force on the spring when it is compressed X2
TSE is the total energy in inch-pounds required to compress the
    spring to compress it from X1 to X2
KEC is the kinetic energy the slide has on its return at the 
    point in time that the bullet contacts the feed ramp
VC  is the velocity the slide has in inches per second when the
    bullet contacts the feed ramp

Observations:  The Delta's dual-spring set up is the 
equivalent of a very stiff spring with little initial 
compression.  This results in a "peaky" force curve: it 
starts relatively low and rises relatively high.  The other 
two spring types are effectively longer with consequent flatter 
force curves.  The Wilson-Rogers spring absorbs approximately 
the same amount of energy as does the Delta dual spring.

What effects does using the Delta's peaky spring assembly 
have compared to using the Wilson-Rogers spring?  Well for 
one thing, the slide using the Delta spring will have a 
little higher velocity at the time the case runs into the 
ejector.  This will result in smarter ejection.  The slide 
has higher velocity because less energy has been removed from 
it because spring forces on it are lower during the early part 
of its travel.

Another effect is that on its return, the slide will have 
more velocity at the time it is pushing the cartridge's 
bullet up the feed ramp.  This is because the slide will have 
been operating in the high-force range of the spring's force 
curve: the peaky area.

What about cycle time?  Qualitatively speaking, as the slide 
goes rearward in recoil, the Delta's peaky spring set allows 
it to move at higher velocity during the early part of the 
travel; it doesn't "pull the velocity out of the slide" until 
later.  So the slide's average velocity to the rear will be 
higher.  On the slide's return forward, the Delta's peaky 
spring set gets the slide up to a given speed sooner 
resulting in higher average slide velocity forward.  Because 
of these two similar effects, total cycle time should be lower.

Are these minor effects the whole point of Colt's putting the 
dual spring set in the Delta Elite?  Have I missed a subtler 
(to me) effect?  I could use some help here.  Thanks 


Notes:

The reason for the different values of S is that the recoil
spring guides (some with buffers) are different thicknesses.
The Wilson-Rogers, with the thickest buffer & guide, has somewhat 
lower VC than it would have if the slide had more of a running 
start at the job.

The Wilson-Rogers spring started out with 6.2 inch free 
length.  It now seems to be stabilizing at 5.92.  So the 
spring was overloaded and is now operating at yield stress.  
This may or may not be a good situation; I don't know.

The Delta Elite has a recoil buffer built into its recoil 
spring guide.  The buffer's effect is to eat up a lot of the 
slide's energy when it is rammed rather than allowing the
slide to bounce off of the frame stop and thereby regain some
of its velocity (but with the velocity's sense reversed).

The Delta Elite's spring set is composed of a stock gov't 
model spring plus a small-diameter stiff inner spring which 
is compressed very little in the initial, installed position.

The W.C. Wolff company sells variable rate springs.  These 
should have an effect similar to the effect of the Delta's 
spring set only more so.  This is because the force curve is 
no longer straight-line; it is now "swamped" in the middle so 
forces will stay even lower, relatively speaking, at the 
beginning of the slide's travel in recoil, and will be 
relatively even higher at the end of the slide's travel.

  JHBercovitz@lbl.gov    (John Bercovitz)