Originally Posted by
MTWeatherman
Joe:
Damn, should have let common sense be my guide and kept my mouth shut. However, an honest question deserves an honest answer and will try to provide one.
Understanding the math and physics prepared by someone else requires far less operating expertise than developing those equations from scratch. Although my background is in math and physics, it will be 39 years this June since I last saw the inside of university classroom (other than an alumni tour). Although physics was required in my profession, computers made that knowledge superfluous rather quickly as I was in the realm of applied atmospheric physics, not research. Suffice to say math and physics is very much the language of science, and like any language, when not used it becomes rusty and in my case is rapidly approaching little more than a red stain. There are much better physicists than I on this board...I’ve seen their work.
With that disclaimer aside, lets consider what is involved from the standpoint of the pure physics involved and any perceived “blowup” associated with a rotating bullet. That rotating bullet contains rotational energy as well as linear energy associated with its forward motion. The amount of that rotational energy is related to it’s mass (weight for our purposes) and rotational speed (directly related to RPM and diameter). For any bullet fired from a firearm with a given twist, once we know the forward velocity of the bullet we can calculate the RPM. From the RPM it’s pretty easy to calculate the velocity of any point on or in the bullet by knowing its distance from its rotational axis. You’ve already done that so well understand the principle.
I believe the implication of rotational velocity needs to be taken one step further. The important consideration here is rotational energy...that’s what we are demanding that the bullet contain if it is not to fly apart. One of the implications of Newton’s First Law is that any object in motion at a certain speed and direction will remain at that same speed and direction unless acted on by an outside force. Any particle of metal on the surface of the bullet or within it wants to continue moving in exactly the same direction it was the millisecond before. That is the force commonly viewed as “centrifugal force”. The force that keeps that bullet from flying apart is the bond that holds the lead together...once exceeded it will in effect “blow up” as each particle takes the tangential path it is currently on...not the continual turning path the rotating bullet is trying to induce on it. Run that rotational energy high enough and any bullet will fly apart...even a solid copper one...just that in the case of the latter is way beyond what is capable of being accomplished in a rifle. As you’ve already indicated, the velocity on the bullets surface, is dependent on caliber. Take a .22 caliber rifle and a .44 caliber rifle (true calibers here, not actual firearm calibers) with the same twist and firing a bullet at the same velocity. Upon firing, both bullets are rotating at the same RPM. From the formula for calculating the circle circumference, we know any particle on the surface of the .44 has to travel twice the distance of the .22. Its speed has doubled...but not its energy. We know from the formula for kinectic energy that if mass remains constant, a doubling of velocity will quadruple the energy since it is determined by the square of the velocity. Every particle on the surface of that .44 bullet has four times as much contained energy as the .22 ... attempting to move it off in a tangential direction to the actual bullet. In effect to attempt to “blow it up”. Also, on the .44, any particle of lead between the radius of the .22 bullet (0.11) and the .44 will have a like increase in energy induced because of that increased velocity. This is the whole principle of the flywheel...to maximize stored energy for any given mass, you want to maximize diameter...most of the energy contained in a rotating flywheel is stored in the outer sections where mass velocity is the highest. The good news is that increase in “gyroscopic” effect with diameter means that you don’t need to rotate a larger bullet as fast to stabilize it. That’s why large caliber rifles can get away with slower twists even while firing relatively long bullets.
I believe that the real issue for on an accuracy threshold may be found to be more directly related to rotational energy of the bullet ...which is dependent on bullet diameter as well as RPM. Run that energy high enough and the bullet will fly apart...at some point below that, depending on how perfect the bullet is, inaccuracy will begin to appear as imbalanced rotational energy adversely affects the bullets flight to the extent it becomes unacceptable. An unbalanced tire remains a good analogy to an imperfect or unbalanced bullet. The less uniform the tire, the bigger the wheel, the faster the rotation and the more adverse the affects on smooth rotation become. An imperfect tire or bullet prefers to rotate around the center of mass and that center is not the dimensional center that the car axle or firearm barrel is forcing upon it. Wheel weights (which we all love)...simply adjust the center of mass toward the center of rotation. We don't balance our bullets...just load and shoot them...weight sorting is as close as we can come to indirectly achieve balance. I suspect that there is a bit of an "average" quality that is produced in a cast bullet...and that's why we have an apparent threshold. This threshold is not linear with twist rate because the associated energies are not linear. In theory, an absolutely perfect bullet should have no upper velocity limit for accuracy...just the disintegration limit when it flies apart. The true energy contained in a rotating bullet would require mathematic integration of the mass and velocities involved. However, for our purposes, I believe a simple calculation of a surface particle energy would suffice. If mass remains the same in each case, then one would simply square the rotational velocity at the surface of the bullet to get it...and use then use this "rotational energy index" to perhaps establish a better upper threshold than pure RPM will provide. Not a hard and fast rule due to differing bullet alloys, degrees of bullet perfection, etc. but likely a useful one nevertheless. Also, other variables become introduced such as a tendency for long thin bullets to bend, etc.
No, I do not have the resources to test that theory myself. I have zero data to try to back it up. As it stands, Larry still has the best substantiated theory out there. It’s why I am extremely grateful to people like Larry who are willing to share their data with us.
So, now we’re back to your actual question, Joe, and the complex world I alluded to in my initial post. As I understand your question, in order to apply science to determine the value at which that cylinder would leave the core, or in effect disintegrate as a lightly constructed .22 condom might do when driven at high speed from a fast twist rifle, one would first need to determine the energy value at which the rotational energy of the lead core is high enough that the lead, without the jacket, would wish to deform (yield strength) and then fail...then one would need to determine the energy or force the yielding and failing lead core would transfer to the jacket that is now holding it all in place, then calculate how much internal force the jacket is able to contain as it itself is under increasing strain from its own increasing rotational energy. However, in order to do that one would have to have an intimate knowledge of metallurgy. Does the tensile strength value of a metal adequately measure its total failure point when the internal rotational energy or force is constantly changing and increasing as one moves outward...and is not a constant force applied externally to the material itself? Do the tensile strength values apply at all in this case? The physics required are complex and beyond me. However, I’m sure it is a problem that metallurgists and engineers have tackled before...and done empirical studies to determine the values. I’ll leave those experts, hopefully someone on this board, to answer it.
No I don’t believe RPM itself is the direct culprit here...I believe it to be an indirect means of measuring the real culprit which is bullet imperfection and our limits in controlling it. However, that may be better measured by rotational energy. Rotational energy better measures our limits by more accurately measuring the stress and accompanying distortion inflicted on the bullet while it is in the barrel and serves to magnify any bullet imperfections, both initially present and barrel induced, once that bullet leaves the muzzle.
However, RPM does provide a good indication of a caliber dependent threshold that the “average caster” can accept as an upper limit for accuracy. It’s a damn good indicator if not the direct cause. Kind of like the health experts telling you that if you’re male you have a reasonable chance of living to be 75 but a poor one of reaching 90. You won’t die of your age...it will be cancer, heart disease, stroke, you name it. Age is a good indicator...not the cause. Same here.
Time to really shut up this time. I’ve said enough to fill 20 average posts.