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Independent of the sight radius, will a longer barrel on a handgun increase accuracy?

Discussion in 'Handguns: Autoloaders' started by JLStorm, Aug 24, 2007.

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  1. JLStorm

    JLStorm Member

    Jun 19, 2004
    I am considering two guns, both exactly the same except one has a .66" longer threaded barrel and o-ring. I initially figured that in the least the longer barrel would give me better velocity and more accuracy, but I have heard mixed opinions. I am certainly considering a suppressor in the future, but I havent decided on caliber yet as the dry suppression of 45 ACP isnt as good as I would like. Given that, I dont want the only reason to buy this gun to be the option for suppression. The different barrel lengths are 3.80" vs 4.46" caliber is 45ACP. The sight radius of each gun is exactly the same.
  2. Shear_stress

    Shear_stress Member

    Apr 27, 2005
    Unless the barrel is so short that it reduces muzzle velocity so much that the bullet isn't spinning fast enough to stabilize (not likely to happen), the barrel length won't make a difference in mechanical accuracy. I wouldn't worry about it.
  3. W.E.G.

    W.E.G. Member

    Sep 26, 2006
    all over Virginia

    From: John Bercovitz <JHBercovitz@lbl.gov>


    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?

    JHBercovitz@lbl.gov (John Bercovitz)
  4. W.E.G.

    W.E.G. Member

    Sep 26, 2006
    all over Virginia
  5. JLStorm

    JLStorm Member

    Jun 19, 2004
    Ok I started going cross eyed after trying to read that entire thing, so I started to skim (Thank you very much for the info though). From the figures he listed it looks like there its a VERY minute amount. Did I get that right?
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