SEMI-EXPERIMENTALLY-DETERMINED INTERNAL BALLISTICS OF THE 45 AUTO 91/03/02 PURPOSE
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 data.
SYNOPSIS
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.
APPARATUS AND DIMENSIONS
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 length.
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.
METHOD
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.
RECOGNIZED AND POTENTIAL LIMITATIONS OF THE METHOD
Potential sources of error were legion. However, none were seen as calamitous.
-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?
DATA
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.
FOR THE 200 GRAIN BULLETS:
bullet. muzzle velocity in feet/second
trav.
inches JKT/BE JKT/UNQ JKT/BD CST/BE CST/UNQ CST/BD
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
ABBREVIATIONS:
JKT - jacketed Hornady #4515 CST - cast #68 H&G
BE - bullseye UNQ - unique BD - blue dot
FOR THE FEDERAL "AMERICAN EAGLE" 230 HARDBALL:
bullet trav. muzzle veloc.,fps std. dev.,fps
inches
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.
EQUATIONS
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
or:
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.
QED
COMPUTATIONS
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 ballFrom 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.
CONCLUSION
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?
JHBercovitz@lbl.gov (John Bercovitz)
In article <45052@mimsy.umd.edu> sixhub!davidsen@crdgw1.ge.com
(Wm E. Davidsen Jr) writes:
#In article <44538@mimsy.umd.edu> bercov@bevsun.bev.lbl.gov
(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 (JHBercovitz@lbl.gov)