joeb33050
05-23-2008, 05:44 AM
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 on the book site, http://sports.groups.yahoo.com/group/CB-BOOK/
in "FILES", in "ERRATA". There are two EXCEL workbooks called "CAST BULLET BLOW UP" and "BARREL HEAT" dealing with this project..
Conclusions
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. VS. 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.
The Model
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.
(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
Bullet Heat
Cast bullet blow up has to do with the temperature of the bullet. As the temperature increases the tensile strength decreases, and the velocity/rpm at which centripetal force equals tensile strength decreases-making blow up more possible/likely.
I found one site talking about measuring the temperature of a bullet as it left the muzzle;
http://www.rangerats.org/bullet.html
where they found the temperature of a 5.56mm NATO bullet leaving the muzzle was 267 degrees C = 513 degrees F.
The author mentions that "The bullets cooled as they traveled away from the gun."
"Garandsrus" on CAST BOOLITS was kind enough to send me a copy of the article "Comet Tails" by Gardner Johnson, from the April 1998 PRECISION SHOOTING. The article includes a table of Tensile Strength of lead in temperature.
I used this table to construct a graph of Tensile Strength(lb/in^2) vs. Temperature (Degrees F) for lead, WW and Lino.
A cast bullet is heated up by:
the powder gas as it travels through the barrel (I don't know how to estimate this.)
the friction of the bullet rubbing on the barrel
the air friction after the bullet leaves the muzzle
Barrel/bullet friction
How much does friction heat up the bullet and barrel?
I tapped a 31141 bullet into the bore of my 30/30 bench gun, then tapped and pushed it back and forth in the barrel. It never loosened up; it always had a lot of stiction = static friction. I had planned to use a scale on the rod to get a notion of the force required to move the bullet, but that didn't work. A better setup is required for good readings.
So I did the arithmetic, and made a spreadsheet that would answer the question of how much the barrel and bullet might heat up at various amounts of force.
If it takes 20 pounds of force to move a 170 grain bullet through the 2' barrel, there's 40 foot pounds of work done. (20 pounds is in the ball park, it ain't 2 pounds and it ain't 100 pounds.) This work turns into heat that goes into the bullet and the barrel. Let's say the barrel weighs 6 pounds.
If all the heat goes into the bullet, (and it doesn't), then this work raises the bullet temperature 1221 degrees C.
If all the heat goes into the barrel, (and it doesn't), then this work raises the barrel temperature 1.28 degrees C.
Somewhere between these two is the point of equal increased temperature of both the barrel and the bullet. This point is where the barrel and the bullet both increase temperature by 1.275 degrees C.
It seems to me that the tendency would be toward equal temperature increases; toward the 1.275 degree C increase for both bullet and barrel.
All if I did the arithmetic correctly.
The workbook is up on the book site; I'm hoping that somebody who knows about this stuff will review my arithmetic and point out the mistakes.
Bullet/air friction
Energy loss
As a bullet loses velocity and energy as it goes through the air; the energy has to go to heating up the bullet and moving (and heating) the air. We know how much energy the bullet has, and loses. How much energy does it take to heat up a bullet?
The Lyman Cast Bullet Handbook, 3rd edition, tells us that the 311291 at 2500 fps loses 377 foot-pounds of energy in the first 50 yards of travel. The same bullet starting at 1400 fps loses 107 foot-pounds of energy in the first 50 yards.
If the calculations are correct, it takes 12.8 foot-pounds of energy to raise the temperature of a 170 grain lead bullet 500 degrees Fahrenheit.
The bullet loses 377(mv = 2500 fps) or 107(mv = 1400 fps) foot pounds of energy in the first 50 yards of travel, and it would take 12.8 foot pounds of energy to raise the temperature 500 degrees Fahrenheit.
Doesn't mean that the temperture does go up, just that the energy loss is there, sufficient and available.
A skin temperature estimating formula
The article "Comet Tails" by Gardner Johnson is from the April 1998 PRECISION SHOOTING.
While I have some problems with the article, Johnson explains that there is a formula for predicting skin temperature of high speed aircraft.
Fiddling with the table of temperatures and velocities in the article, I find the formula to be: Delta T =:Temperature rise above ambient (Degrees F) = V^2(fps)/13310. Means that skin temperature equals velocity squared, divided by 13,310.
Increases to T are:
V(fps) Delta T
500 19
1000 75
1500 169
2000 300
2500 469
In the example of lead/1.5% antimony with a melting point of 600 degrees F, and an ambient temperature of 75 degrees F, the skin of a lead alloy bullet would become liquid at 2646 fps.
The calculated skin temperature falls as bullet speed falls; and that's pretty fast. For example, a .202 BC bullet with 2700 fps mv has skin temperature of 548 degrees F above ambient at 50 yards; and 179 degrees F above ambient at 300 yards when v has fallen to 1545 fps.
It seems unlikely that the center of the bullet reaches anything like the skin temperature.
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 on the book site, http://sports.groups.yahoo.com/group/CB-BOOK/
in "FILES", in "ERRATA". There are two EXCEL workbooks called "CAST BULLET BLOW UP" and "BARREL HEAT" dealing with this project..
Conclusions
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. VS. 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.
The Model
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.
(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
Bullet Heat
Cast bullet blow up has to do with the temperature of the bullet. As the temperature increases the tensile strength decreases, and the velocity/rpm at which centripetal force equals tensile strength decreases-making blow up more possible/likely.
I found one site talking about measuring the temperature of a bullet as it left the muzzle;
http://www.rangerats.org/bullet.html
where they found the temperature of a 5.56mm NATO bullet leaving the muzzle was 267 degrees C = 513 degrees F.
The author mentions that "The bullets cooled as they traveled away from the gun."
"Garandsrus" on CAST BOOLITS was kind enough to send me a copy of the article "Comet Tails" by Gardner Johnson, from the April 1998 PRECISION SHOOTING. The article includes a table of Tensile Strength of lead in temperature.
I used this table to construct a graph of Tensile Strength(lb/in^2) vs. Temperature (Degrees F) for lead, WW and Lino.
A cast bullet is heated up by:
the powder gas as it travels through the barrel (I don't know how to estimate this.)
the friction of the bullet rubbing on the barrel
the air friction after the bullet leaves the muzzle
Barrel/bullet friction
How much does friction heat up the bullet and barrel?
I tapped a 31141 bullet into the bore of my 30/30 bench gun, then tapped and pushed it back and forth in the barrel. It never loosened up; it always had a lot of stiction = static friction. I had planned to use a scale on the rod to get a notion of the force required to move the bullet, but that didn't work. A better setup is required for good readings.
So I did the arithmetic, and made a spreadsheet that would answer the question of how much the barrel and bullet might heat up at various amounts of force.
If it takes 20 pounds of force to move a 170 grain bullet through the 2' barrel, there's 40 foot pounds of work done. (20 pounds is in the ball park, it ain't 2 pounds and it ain't 100 pounds.) This work turns into heat that goes into the bullet and the barrel. Let's say the barrel weighs 6 pounds.
If all the heat goes into the bullet, (and it doesn't), then this work raises the bullet temperature 1221 degrees C.
If all the heat goes into the barrel, (and it doesn't), then this work raises the barrel temperature 1.28 degrees C.
Somewhere between these two is the point of equal increased temperature of both the barrel and the bullet. This point is where the barrel and the bullet both increase temperature by 1.275 degrees C.
It seems to me that the tendency would be toward equal temperature increases; toward the 1.275 degree C increase for both bullet and barrel.
All if I did the arithmetic correctly.
The workbook is up on the book site; I'm hoping that somebody who knows about this stuff will review my arithmetic and point out the mistakes.
Bullet/air friction
Energy loss
As a bullet loses velocity and energy as it goes through the air; the energy has to go to heating up the bullet and moving (and heating) the air. We know how much energy the bullet has, and loses. How much energy does it take to heat up a bullet?
The Lyman Cast Bullet Handbook, 3rd edition, tells us that the 311291 at 2500 fps loses 377 foot-pounds of energy in the first 50 yards of travel. The same bullet starting at 1400 fps loses 107 foot-pounds of energy in the first 50 yards.
If the calculations are correct, it takes 12.8 foot-pounds of energy to raise the temperature of a 170 grain lead bullet 500 degrees Fahrenheit.
The bullet loses 377(mv = 2500 fps) or 107(mv = 1400 fps) foot pounds of energy in the first 50 yards of travel, and it would take 12.8 foot pounds of energy to raise the temperature 500 degrees Fahrenheit.
Doesn't mean that the temperture does go up, just that the energy loss is there, sufficient and available.
A skin temperature estimating formula
The article "Comet Tails" by Gardner Johnson is from the April 1998 PRECISION SHOOTING.
While I have some problems with the article, Johnson explains that there is a formula for predicting skin temperature of high speed aircraft.
Fiddling with the table of temperatures and velocities in the article, I find the formula to be: Delta T =:Temperature rise above ambient (Degrees F) = V^2(fps)/13310. Means that skin temperature equals velocity squared, divided by 13,310.
Increases to T are:
V(fps) Delta T
500 19
1000 75
1500 169
2000 300
2500 469
In the example of lead/1.5% antimony with a melting point of 600 degrees F, and an ambient temperature of 75 degrees F, the skin of a lead alloy bullet would become liquid at 2646 fps.
The calculated skin temperature falls as bullet speed falls; and that's pretty fast. For example, a .202 BC bullet with 2700 fps mv has skin temperature of 548 degrees F above ambient at 50 yards; and 179 degrees F above ambient at 300 yards when v has fallen to 1545 fps.
It seems unlikely that the center of the bullet reaches anything like the skin temperature.