Index Home About Blog
From: John Bercovitz <>


This project was the natural outgrowth of the series of articles on 
government model dynamics found in rec.guns last August.  The purpose of 
the present project was to find the pressure vs. bullet-travel curve for 
typical 45 auto loads.  With this curve, one can can take the first steps 
towards estimating the locking force required of the 45 auto at various 
relative positions of barrel/slide assembly to the frame.  An unsuccessful 
literature search had been made for this data or velocity vs. barrel length 

A series of velocity measurements was made on a government model 45 auto 
barrel cut to increasingly shorter lengths.  Seven loads were tested.  Using 
the Le Duc equation and its differential, pressures were found with respect 
to bullet positions.  

Two surprises were found: 1) Regardless of powder type, velocities of loads 
with similar bullets and the same muzzle velocities in a full-length barrel 
tracked together as the barrel was shortened.  2) The curve of standard 
deviation vs. bullet travel reached a minimum at a short barrel length.  

From the tests it appears that the pressure in the 45 auto cartridge peaks 
after the bullet has moved 0.3 inches.  This causes an insignificant 
movement of the slide and hence no degradation of the state of the 
barrel/slide lock up.  

A donated 45 auto barrel in good condition (extremely fine pitting - more 
gray than pitted) was turned round over the chamber and glued into an ad hoc 
adapter block for a T/C Contender.  At this time the chamber end of the 
barrel was shortened to achieve a minimum chamber length.  Also, without 
removing any bore, the barrel was re-crowned flat.  After installation, the 
headspace was .899 inches.  Later, sized WW brass was selected for length.  
The range of acceptable brass lengths was .002 inches (from .892 to .894).  
The T/C was mounted in a Ransom Rest and fired over an Oehler model 12 
chronograph with 6 foot screen spacing and located so that the center of the 
chronograph was 15 feet from the average position of the muzzle.  The 
chronograph had been modified for ease of use by the addition of an LED 
digital readout.

The barrel has an average groove to groove diameter of .4515 inch.  Its bore 
is .4450.  Since the grooves are twice as wide as the lands, the bore area is 
.159 in^2.

The barrel overall length after shortening the chamber and removing the 
toroidal-section crown was 5.013 inches, down from 5.022 inches stock 

The #4515 Hornady bullets had a diameter of .4510 inch.  The #68 H&G 
bullet had a sized diameter of .4520 inch.  The driving bands on these two 
bullets were the same overall length within .005 inch.  So by seating their 
bases to the same depth, I was able to make them engrave and exit the 
muzzle together.  The average diameter of the Federal 230 gr. ball bullet 
was .4492 inch; its maximum diameter minus its minimum diameter was 
.0005 inch.  The 4515 and 68 both weighed within 0.2 gr. of 200 grains.  I did 
not weigh the 230 grain bullet.

From the loading manuals, loads bracketing 900 fps at 15 feet were made up 
using the #68 H&G bullet and its jacketed look-alike, the Hornady #4515.  
Three powders were used: Bullseye, Unique, and Blue Dot.  After the 
bracketing loads were fired, 75 each of new loads determined by linear 
interpolation of the bracketing data were made up.  Of these new loads, 5 
successfully met the 900 fps criterion.  It is not known why one of the loads 
failed to fly at 900 fps (probably operator error).  That one load is 
undoubtedly an overload.  In addition to the six loads noted above, Federal 
American Eagle 230 grain ball ammo was fired (Lot #38A-0411).

The barrel was fired at full length and then shortened twice by 0.8 inches 
and four times by 0.5 inches for bullet travels (distance from base of bullet 
to muzzle) of approximately 4.4, 3.6, 2.8, 2.3, 1.8, 1.3, and 0.8 inches.  The 
muzzle crown used was a smooth, straight cut (planar and perpendicular to 
the axis) and since several jacketed bullets were fired as warm-ups after 
each cut, it is doubtful that any burrs existed at the time of any test.  
Certainly no burrs were visible even under magnification.

It was intended that 10 rounds of each load be fired at each length.  
However, data for a few rounds were lost due to equipment problems.  The 
minimum number of rounds fired for a given load at a given barrel length 
was 8.  The vast majority of data points represent 10 or more rounds. 

    Potential sources of error were legion.  However, none were seen as 
-The Le Duc equation seems to be a curve-fit-up job.  I don't see a 
theoretical basis for it.  This is probably not important since the curve 
seems to fit the data quite well.
-If we don't know the actual friction between bullet and bore, we can't know 
the actual pressure in the gun from the type of data gathered.
-As the bullet's body exits the muzzle,  there is a decrease in the amount of 
friction of bullet in bore.  This is because at any given area along its bearing
surface, the bullet exerts a radial force per unit area (pressure) on the bore. 
So as area decreases, friction decreases.  This should result in a slight 
increase in bullet acceleration as the bullet exits the muzzle.
-Eight to ten firings per data point is not overly generous.
-The sigmas (standard deviations) of the firings were higher than desirable 
in some instances.
-The chrono screens failed two thirds of the way through the series - was 
the failure catastrophic or gradual?  If gradual, is it even possible for this 
to have an effect?
-The weather varied a little - does the chrono have a significant 
temperature coefficient?  How about the ammo?
-Does the chrono have any significant built-in systematic errors?

I made a fixture to hold the barrel vertically on top of a bathroom scale and 
then drove samples of the test bullets through the barrel with a 7/16 inch 
diameter rod held in a drill press chuck.  The barrel had been "shot in" some 
during load development prior to this test.  Ignoring the engraving forces, 
the running force required to push the bullets through the dry bore were as 
follows (this is the dynamic friction):  

#4515: 175 lb;  #68: 115 lb;  230 ball:  260 lb

One certainly expects that the force required to overcome bullet-bore 
friction will be higher under actual firing conditions due to the effects of 
acceleration on the bullet.


bullet.	           muzzle velocity in feet/second          
4.394    906	   903	    905	   907    891    1053
3.595	   867	   855	    882	   875	   862	    991
2.795	   840	   823	    856	   836	   814	    958
2.295	   797	   781	    795	   789	   772     891
1.795	   749	   743	    752	   742	   741	    854
1.295	   689	   684	    694	   683	   678	    772
0.795	   597	   575	    591	   589	   591	    671
0	         0      0       0      0      0       0

	    standard deviation (sigma) feet/second                 
4.394	  13.8	  20.8	   19.2    7.3	   26.9	  47.9
3.595	  19.2	  16.7	   24.2	  12.5	   19.5	  17.3
2.795    8.8   27.0    28.4   12.9    14.7   19.3
2.295   10.3   14.0    12.3    5.1    18.9   16.9
1.795    6.5   11.5    14.5    4.8    12.0   16.2
1.295    8.1   21.6     9.0   11.2     9.6   15.8
0.795    9.0   11.2    18.5    8.9    33.3    9.2
0        0      0       0      0       0      0

JKT - jacketed Hornady #4515         CST - cast #68 H&G
BE - bullseye          UNQ - unique            BD - blue dot


    bullet trav.   muzzle veloc.,fps   std. dev.,fps
       4.420              789             12.1
       3.621              755             10.3
       2.821              731              8.9
       2.321              701              8.7
       1.821              655              5.3
       1.321              607              6.5
       0.821              518              9.8
       0                    0              0

Note: The above velocities were corrected from instrumental 
velocities (at 15 feet from the muzzle) to muzzle velocities using the 
Sierra ballistics program.

The sigmas are gotten with the usual formula for the small-population 
estimator of sigma, the formula that has n-1 in it.  Caveat: There are 
formulae which may be better estimators in some cases.  This may be one of 
those cases.

Another caveat: The data looks a little lumpy, especially in the region of the 
second and third tests..  Either there was a chronograph problem or the 
barrel had an uneven finish in the 2 1/2 inch region.

I bought a book from Wolfe Publishing while the testing was being done.  In 
reading the book, "Firearms Pressure Factors" by Brownell, I found an 
equation relating the velocity of the bullet to its position in the barrel. 
It is called the "Le Duc equation":  

>>>        V = (ax)/(b+x).  

Dramatis Personae:
V = muzzle velocity
x = bullet travel (distance from bullet base to muzzle in this case)
a, b = constants cut to fit the data.  

The book said without substantiation that "b" is twice as large as the 
distance the bullet travels from its rest position to the position it occupies 
when the pressure in the barrel is maximum (see proof below).  By 
inspection of the equation we see that "a" is the velocity which "V" 
approaches as "x" gets large.

It struck me that this was just the equation I'd been trying to devise from 
the preliminary data.  I could use it to find pressure in the barrel vs. the 
position of the bullet in the barrel.  All I had to do was differentiate the 
equation with respect to time to get acceleration.  This is because the net 
force on a bullet is related to its acceleration through its mass by the 
equation F = mA.

Finding acceleration as a function of bullet travel:

V = (ax)/(b+x)             the Le Duc equa.	       		(1)

d (u/v) = (v du - u dv) / v^2                  	  		(2)
where in this case:
           u = ax      and      v = b+x
          du = a dx            dv = dx

Therefore: dV = { (b+x)(a dx) - (ax)(dx) }/(b+x)^2		(3)
              = (ab dx) / (b+x)^2 = [ab/(b+x)^2] dx		(4)

Since:      A = dV/dt 
              = { ab/(b+x)^2} dx/dt = { ab/(b+x)^2} V	
But:     V = (ax)/(b+x)
Therefore:  A = { ab/(b+x)^2} { (ax)/(b+x) } 

>>> Acceleration, A = a^2*b*x / (b+x)^3				(5)

In order to prove A is maximum at x=b/2, the thing to do is differentiate the 
acceleration equation.  This result, set to 0, will give the value of x at 
which A is maximum since the slope of tangents to a curve must be zero at 
all maxima and minima.

A = a^2*b*x / (b+x)^3
A = kx/(b+x)^3  where k =a^2*b 					(6)

Again: d (u/v) = (v du - u dv) / v^2				(7)
where u = kx      and      v = (b+x)^3
     du = k dx            dv = 3 (b+x)^2 dx 

dA = [(b+x)^3 * k dx - kx * {3 (b+x)^2 dx}] / (b+x)^6

As I'm only trying to find out if dA = 0 when x = b/2, I can ignore the divisor,
(b+x)^6, since 0 divided by anything is 0.  And since the right half of the 
equation is no longer dA, I'll just call it "SUM".  So ignoring the divisor:

SUM = (b+x)^3 * k dx  -  kx * {3 (b+x)^2 dx}			(8)

By similar reasoning, I can delete factors common to both terms:

SUM =  (b+x)^3  -  x * 3 (b+x)^2				(9)

By inspection we can now see that if we substitute b/2 for x, SUM = 0.


The first thing I did was plug the data into an Excel spreadsheet so I could 
operate on it.  Since the data for the three jacketed bullet loads were the 
same within the margin of error, I averaged them to increase the 
significance of the data.

I did not use a least squares fit or anything else quite so nifty.  I merely 
adjusted the constants of the Le Duc equation until it ran through two of the 
data points.  I got the best results when I ran the curve through the data for 
the highest and lowest velocities.  This approach appeared to give the best 
fit at the lower end of the curve where pressures are highest.  To determine 
how well the curve was fitting, I used Cricket Graph to plot the Le Duc curve 
and the Stineman-interpolated data curve on the same graph (barrel travel 
on the abscissa, velocity on the ordinate).  I used my Mark I eyeball to 
determine the goodness of fit.

The Algorithm:
Looking at the Le Duc equation, V = ax/(b+x), you can see that if you know a 
"V" at an "x" and you make a guess at "b", you can calculate "a".  What I ended 
up doing was: using V at x equal to the maximum bullet travel (uncut barrel), 
guess at "b", calculate "a", then compute the velocity for the shortest bullet 
travel in the series.  I then iterated on the guess for "b" until I got the 
correct velocity for the shortest bullet travel.  I converted velocities to 
inches per second to keep their units consistent with the unit of bullet 
travel; this was important later for calculating peak acceleration.

I did not try setting "a" equal to the maximum velocity one might expect 
from a long 45 auto barrel.

From above:
Bore area = .159 in^2
f = 175 lb for #4515 Hornady
f = 115 lb for #68 H&G
f = 260 lb for 230 ball

From the curve-fitting calculations for the average of the jacketed bullet 
data: b came out to be 0.594 inches which gave a = 12324 inches/second in 
order for the muzzle velocity to come out to 10856 ips at bullet travel of 
4.394 inches and approximately 7052 ips at 0.795 inches of bullet travel.  
Using these numbers as input:

Apeak = A(x =.594/2) = a^2*bx/(b+x)^3
Apeak=(12324^2 *.594*.594/2)/(.594+.594/2)^3=3.7878*10^7 in/s^2
	                                (approximately = 100,000 G)

Since the bullet mass is:
m = 200 gr/[(7000 gr./lb)(386.1 in/sec^2)] = 7.4 * 10^-5 lb-sec^2/inch

The maximum net force on the bullet is:
Fnet = mApeak=(7.4 *10^-5 lb-sec^2/inch)*(3.7878 *10^7 in/sec^2) = 2803 lb

Adding in friction:
Fpeak = 2803 + 175 = 2978 lb

And the pressure:
Ppeak = F/bore area = 2978/.159 = 18,730 psi

A similar calculation follows for the 200 gr. #68 H&G/blue dot overload:
a = 14453 inches/sec         b = .632 inches
Apeak = A(x =.632/2) = a^2*bx/(b+x)^3 
Apeak = (14453^2 *.632 *.632/2) / (.632+.632/2)^3 = 4.8966 * 10^7 in/sec^2
m = 200gr/[(7000 gr./lb)(386.1 in/sec^2)] = 7.4 * 10^-5 lb-sec^2/inch
Fnet= m A = (7.4*10^-5 lb-sec^2/inch)*(4.8966 *10^7 in/sec^2) = 3623 lb
Ppeak = (Fnet +175)/area = (3623+115)/.159 = 23,500 psi

A similar calculation for the 230 gr. factory load:
a = 10753 inches/sec         b = .600 inches
Apeak = A(x =.6/2) = a^2*bx/(b+x)^3 
Apeak = (10753^2 *.6 *.6/2) / (.6+.6/2)^3 = 2.8550 * 10^7 in/sec^2
m = 230 gr/[(7000 gr./lb)(386.1 in/sec^2)] = 8.51*10^-5 lb-sec^2/inch
Fnet= m A = (8.51*10^-5 lb-sec^2/inch)*(2.855 *10^7 in/sec^2) = 2430 lb
Fpeak = Fnet + Ffriction = 2430 + 260 = 2690 lb
Ppeak = (Ffriction)/area = (2690)/.159 = 16,920 psi

The peak pressures appear reasonable for their respective loads.  Remember 
these pressures are in psi, not CUP, so on that basis they should be slightly 
higher than what one usually sees.

The original purpose of all this work was to find the relative position of the 
slide/barrel assembly to the frame during the peak of the pressure curve.  If 
we believe all of the foregoing calculations, the peak pressure of the 230 
grain ball load is about 16,900 psi and it occurs when the bullet has 
traveled 0.3 inches from its rest position.  The slide/barrel unit of the 
government model weighs about 18 ounces (7875 grains) so if the 230 grain 
bullet has moved 0.3 inches, the slide has moved: 0.3*(230/7875) = .0088 
inches when the pressure peaks.  If you inspect the locking mechanism of a 
45 auto, you'll see that this amount of movement has no significant effect 
on the state and strength of the lock-up.

In the case of the 230 grain ball load, the net force was shown to be 2430 
pounds.  This is the force which accelerates the slide/barrel assembly 
backward and the bullet forward.  But to find the force on the parts which 
lock the barrel to the slide, it's easiest to look at the forces on the barrel. 
The 2690 pound peak force isn't acting on the barrel; it's acting on the 
bullet.  Of the 2690 pounds of force, 260 pounds of force, the friction, is 
acting on the barrel trying to push it in the same direction the bullet is 
going.  Remember, this 260 pound force was measured under unrealistic 
circumstances: the acceleration at the time of measurement was near zero.  
The other significant force on the barrel is that of the slide accelerating 
the barrel backward with it.  If the net force on the slide/barrel assembly is 
2430 pounds and the mass of the assembly is = (18 oz/16)/386.1 = .0029 lb-
sec^2/in, then the peak acceleration of the slide/barrel, As/b = F/m = 
2430/.0029 = 834,000 in/sec^2 (2160 G).  Since the barrel is going along for 
the ride, this is also the peak acceleration of the barrel caused by the force 
of the slide yanking it backwards.  The barrel weighs 3.25 oz so it has a 
mass of .000526 lb-sec^2/in.  Ergo, the force required to cause this peak 
acceleration of the barrel backward is F = mA = .000526*834,000=440 
pounds.  The force acting on the locking parts is, then, the sum of the 
frictional force of the bullet on the barrel and this accelerating force:  
summation forces = 260 + 440 = 700 pounds.  Again, please keep in mind 
that the 260 figure is low by an unknown amount so the 700 pound figure is 
also low by that same amount.  There is yet another force which will affect
this total.  This is the force with which the case, gripping the walls of the
chamber, can pull the barrel to the rear.  I say "can" because the amount of
grip depends on a lot of variables, some of which were explored in an earlier
article.  So the 700 pound figure might be considered the "oiled-cartridge"
figure for the forces on the locking parts.

A good title for this article might have been "Internal Ballistics for the Ill-
Equipped".  With the advent of cheap chronographs, I think this indirect 
approach to characterizing internal ballistics may be a good one for the 
home experimenter.  But the validity of this approach has not been 
established and can't be until some direct confirming experimental data is 
found.  The confirming data should preferably originate in a ballistics lab 
which has at least a piezo gauge and a high speed X-ray motion picture 
camera.  Does any of you out there have this sort of data?  Does anyone have 
any personal experience working with the internal ballistics of the 45 auto?    (John Bercovitz)

From: (John Bercovitz)
Subject: Re: Summary: Ideal Concealed Carry Weapon (long)

In article <> sixhub!
(Wm E. Davidsen Jr) writes:
#In article <>
(John Bercovitz) writes:
#> If two rounds using the same bullet are loaded for the same velocity in the
#> same given barrel but are loaded one with a fast powder and one with a slow
#> powder, these two rounds will always have the same velocity if they are fired
#> through any given barrel regardless of that barrel's length. (Found by
#> personal experiment; see below.)

#Not having data to refute this, I can only say that my understanding
#of interior ballistics gives me a gut feeling that this is not correct.
   [explanation of cause of gut feeling deleted]

I agree with you completely; my problem is this data I've very carefully
experimentally determined.  8-)  The only way I can think of to reconcile
your explanation with my results would be to say that what you've described
is somehow a minor effect and that my experimental results are somehow not
sensitive enough to reveal what you've described.  One minor point is that
for my experiment I loaded all rounds to have the same muzzle velocity with
a 5" barrel regardless of powder type or bullet used.  (Well, at least I
tried to.)  If someone else would be so kind as to repeat the experiment
and report the results here, I for one would be everlastingly grateful.
It may well be that there is something about a radically "underbore"
cartridge like the 45 ACP which tends to keep these fine effects from
evidencing themselves.  A good comparison experiment would be testing a
handgun cartridge with a relatively large boiler room such as the 357 Mag
or Max.

John Bercovitz     (

Index Home About Blog