CAST BULLET BLOW UP OR BULGE
Do cast bullets "blow up" because they are spinning/rotating due to the rifling? Do cast bullets change shape or "bulge" because of the spinning?
Calculations and explanations can be found in two excel workbooks called "CAST BULLET BLOW UP OR BULGE" and "CAST BARREL HEAT-BLOW UP OR BULGE” in the excel workbooks.
Cast bullets can and rarely do blow up at very high RPMs/velocity.
Cast bullets change shape, bulge, at high RPMs/velocity.
Cast bullets are hot as they exit the muzzle, and may/do get hotter going through the air, for some distance-this varies with the load.
Blow up or bulging of cast bullets occurs in a certain RPM area, and is independent of velocity or twist rate and varies slightly with caliber. Plotting the curves of "Centripetal Force = Tensile Strength", and working with the arithmetic shows this to be true. For instance, if a certain alloy bullet will blow up at 400 degrees F, then:
.224 bullets blow up at 126,400 RPM
.308 bullets blow up at 114,320 RPM
.375 bullets blow up at 106,400 RPM
.457 bullets blow up at 98,360 RPM
This range, from 98,360 to 126,400, is relatively slight. See the chart "C.F. = T.S. (RPM)" on the " CAST BULLET BLOW UP" workbook.
This fact, that cast bullet blow up or bulge is primarily related to RPM, was far from obvious or expected in the beginning-but is obvious now.
I don't know if this has anything to do with Larry Gibson's "RPM Threshold" theory, but suspect that it does.
I start with a cylinder of a certain alloy, diameter = caliber and length also = 1 caliber.
(For example, for a .308" caliber the cylinder diameter would be .308" and cylinder length would also be .308".)
This cylinder may be considered to consist of a tube with .025" wall, diameter = caliber, length = 1 caliber, and a small cylinder with diameter of caliber-.050" and length of 1 caliber.
(Graphics by Glenn Stewart)
(For the .308" example, the tube O.D. = .308", I.D. = .258", length = .308"; the small cylinder O.D. of .258", length 308".)
We're interested in figuring out when the tube would "blow up" and leave the small cylinder.
Centripetal force = CF = mv^2/r where CF is in pounds, m = mass in slugs = pounds weight/g with g = 32.15, v = fps, and r = the radius of the small cylinder.
(In the .308" example, r = .258"/2 = .129".)
CF is easily calculated for any caliber/cylinder diameter, tube dimensions, tube length velocity and twist.
Lead has a tensile strength of ~ 1920 pounds per square inch. (BHN 4 * 480 = 1920 pounds per square inch tensile strength)
I think that when the CF = the tensile strength of the inside area of the tube, that the bullet will blow up