hwingo

01-05-2009, 02:51 AM

I believe I am mixing apples and oranges and I need the help of a math genius to get me on the “straight and narrow”. Here’s my problem:

I want to figure the amount of backward force (pressure??) on the bolt face of a rifle when a round is discharged and here is where my confusion begins. Since the bolt’s locking lugs are in direct contact with the locking lugs on the receiver, would it not stand to reason that the backward force applied to the bolt would be the same force applied to the locking lugs on the bolt? Let’s take this from the sublime to the ridiculous; if the locking surfaces were enormously large but thickness of the locking lugs (bolt and receiver) were far less than customary (make the locking lugs 5 feet tall and 4 feet wide if you so desire) then would it not stand to reason that backward force is being distributed over a huge surface area and in theory detonation would be safe?

Let me state this differently because I have “target fixation” and I can’t seem to see this distribution of force any other way. Let’s say I walk out on a frozen pond weighing 250 pounds and my foot size is 11. Admittedly the ice is thin but I don’t break through the ice. Would it be safe to send my wife of 140 pounds onto the ice wearing ice skates? Remember, all her weight is being applied to very thin blades where as my mass was distributed over a larger area. And if the ice were very, very thin would I not stand a far better chance of staying on top of the ice if I stood on a full sheet of ply wood distributing my weight across the surface of the ply wood?

Two significantly different size cartridges generate identical pressures, e.g. 52,000 PSI. One cartridge is very small with a base diameter of .350” and a case length of 1.4” while the other is very large having a base diameter of .750” and case length of 3 7/8”. Obviously the larger case holds more powder thus shouldering this rifle will be punishing when discharged as compared to the small cartridge. Similarly, more backward force is applied to the bolt face and locking lugs thus it stands to reason that locking lugs will need to be significantly stronger than the locking lugs on a significantly smaller cartridge although both generate the same pressure when discharged. Pressure is pressure right? One would think so but it is certain that the larger case requires very heavy duty locking lugs less the gun blows to pieces. There’s something more in play than pressure and I am missing the boat.

I was told today by an electrical engineer that the amount of backward force is equal to PSI x surface area or (pi R square). Using his formula (provided he is correct) I figured the larger case to generate 22,961 pounds of backward force on the bolt face. The smaller case was calculated to be only 5000 pounds of backward force. Assuming he is correct, one would think that if the surface area of the locking lugs were greatly increased, hence distributing force over a larger area, then the larger the locking lugs (greater surface area) the more force one could safely apply. But mathematically, in accordance with his formula, this is simply not the case. Taking the same large cartridge, with the same powder charge but changing the cartridge base to 1” dia, nearly doubles backward force (40,820 lbs).

Something is simply not right with this picture. Some one please explain where I am going wrong with either my thinking or the math formula.:o

Thanks,

Harold

I want to figure the amount of backward force (pressure??) on the bolt face of a rifle when a round is discharged and here is where my confusion begins. Since the bolt’s locking lugs are in direct contact with the locking lugs on the receiver, would it not stand to reason that the backward force applied to the bolt would be the same force applied to the locking lugs on the bolt? Let’s take this from the sublime to the ridiculous; if the locking surfaces were enormously large but thickness of the locking lugs (bolt and receiver) were far less than customary (make the locking lugs 5 feet tall and 4 feet wide if you so desire) then would it not stand to reason that backward force is being distributed over a huge surface area and in theory detonation would be safe?

Let me state this differently because I have “target fixation” and I can’t seem to see this distribution of force any other way. Let’s say I walk out on a frozen pond weighing 250 pounds and my foot size is 11. Admittedly the ice is thin but I don’t break through the ice. Would it be safe to send my wife of 140 pounds onto the ice wearing ice skates? Remember, all her weight is being applied to very thin blades where as my mass was distributed over a larger area. And if the ice were very, very thin would I not stand a far better chance of staying on top of the ice if I stood on a full sheet of ply wood distributing my weight across the surface of the ply wood?

Two significantly different size cartridges generate identical pressures, e.g. 52,000 PSI. One cartridge is very small with a base diameter of .350” and a case length of 1.4” while the other is very large having a base diameter of .750” and case length of 3 7/8”. Obviously the larger case holds more powder thus shouldering this rifle will be punishing when discharged as compared to the small cartridge. Similarly, more backward force is applied to the bolt face and locking lugs thus it stands to reason that locking lugs will need to be significantly stronger than the locking lugs on a significantly smaller cartridge although both generate the same pressure when discharged. Pressure is pressure right? One would think so but it is certain that the larger case requires very heavy duty locking lugs less the gun blows to pieces. There’s something more in play than pressure and I am missing the boat.

I was told today by an electrical engineer that the amount of backward force is equal to PSI x surface area or (pi R square). Using his formula (provided he is correct) I figured the larger case to generate 22,961 pounds of backward force on the bolt face. The smaller case was calculated to be only 5000 pounds of backward force. Assuming he is correct, one would think that if the surface area of the locking lugs were greatly increased, hence distributing force over a larger area, then the larger the locking lugs (greater surface area) the more force one could safely apply. But mathematically, in accordance with his formula, this is simply not the case. Taking the same large cartridge, with the same powder charge but changing the cartridge base to 1” dia, nearly doubles backward force (40,820 lbs).

Something is simply not right with this picture. Some one please explain where I am going wrong with either my thinking or the math formula.:o

Thanks,

Harold