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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

Fasttrack
01-05-2009, 03:34 AM
I don't know anything about guns, but your EE told you right. The smaller cartridge will produce less force on the bolt head than will the larger. Why do you say

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).

Let me preface this by saying I don't know what a locking lug is and my mental picture of a bolt head is the mechanical fastener, not a bolt on a gun.

Anyway, we don't have a constant force, we have a constant pressure. (I'm assuming that the barrel is closed at the other end and that we've turned this into a pressure vessel at 52 kpsi). The greater the cross-sectional area, the greater the force felt. Thus you would want a very tiny cross-section. Now in a real world situation, the pressure generated by a shell is not constant. The bullet exits the barrel.

As far as strength goes, you can make a tiny "plug", which will feel a certain amount of force. This force could then be distributed to a frame over a large surface area. By distributing the 22,000 lbs over a large area, you've decreased the pressure that the material has to deal with. When you look at strength of materials, they are given in psi. This lets you know how much pressure it can stand and then, if you know the amount of force it is subject too, you can determine what size fastener/material you need.

This is just like hydraulic systems. A small plunger with tiny cross-sectional area can produce a HUGE force if cylinder has a large diameter. That is how car brakes work. You take a tiny plunger (maybe .5" in diameter) and pump some brake fluid to four cylinders that have a diameter of maybe 2". Although the volume displaced by the plunger is small and hence the movement of those larger cylinders will be small, they will provide tremendous force compared to the force applied to the plunger.

Frank K
01-05-2009, 07:07 AM
Something that also needs to be considered. When the cartrige is fired, the case momentarily expands and grips the walls of the chamber transferring a considerable amount of recoil force to the receiver. After firing, the brass case relaxes a bit allowing easy extraction of the fired round - hence the need to resize cases before reloading. Not all of the recoil force is transferred via the bolt face.

Bguns
01-05-2009, 07:17 AM
Another Major consideration.

The OD of Case does not matter. Only the ID that the Powder can push against counts for these Calcs...

There is the Case Friction as listed above, Inertia, and a host of other Calcs.

huntinguy
01-05-2009, 07:25 AM
I think the confusion is you have two different things going on here, beside the case friction issue.

On a 50k pressure, you have 50k of pressure but as that is distributed over the face of the bolt you have less pressure in the surface area. Same issue as high heels on airplane floors, your young 125lb bride, in her high heels, puts thousands of psi on the floor (bolt face) where as my over weight frame spread across my size ll feet puts very little. That being said the support frames of the floor (the bolt lugs) will have less pressure on them with your young bride that my fat carcass.

So to answer the question: Bolt force is an issue of available inches to distribute the pressure; the lugs must be strong enough to support the overall force. As you can see, you have two different math problems.

Now that being said. You must figure out the shear force of the cartridge and calculate the bolt dimension against that, do not forget to calculate in moment of elasticity, yield and strength into you formula.

As for me, I let Ruger figure it all out.

Bguns
01-05-2009, 07:29 AM
Not distributed evenly over face of Bolt...

The Gas can only PUSH THROUGH BRASS. THE EXTRACTOR GROOVE EXERTS NO REARWARD PRESSURE. Only the ID if rimmed, or Brass under Extractor groove can transmit Rearward Force..

Not a Hunting Guy, a Gunsmith, Army Ordanance School Grad...

Biggest Build, 30mm... biggest Live Cannon I have worked on, 8 inch Bore... 203mm M110 SP

hitnmiss
01-05-2009, 09:10 AM
Check out this book. Extremely well done.

http://www.amazon.com/Rifle-Accuracy-Facts-Harold-Vaughn/dp/1931220077/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1231160928&sr=8-1

An engineers approach to rifle physics and dealing with them. One of the best technical books I've ever read.

BigBoy1
01-05-2009, 03:03 PM
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

Harold,

The confusion here is between pressure and force. Pressure is defined as force per unit area. So the formula given to you was correct. The force on the base of the case is the chamber pressure times the inside area of the base of the case. The other topics about the reduction of the forces are correct but for our case I'll neglect them. For your example of the chamber pressure of 52,000 psi and the .35" and .75" base diameters, the forces calculated 5,000 and 23,000 lbs are correct. You have to remember that FORCE and not pressure is being applied to the locking lugs. Increasing the size of the locking lugs permits them to carry more of the FORCE load. The maximum stress (pressure) carried by the metal is a fixed value. Therefore, if greater force is applied, the greater the area of metal is required to safely carry the applied force.

I hope you understand what I'm trying to say.

Bill

Bob Ford
01-05-2009, 07:55 PM
In Stuart Otteson’s The Bolt Action Volume I pages 275 - 277. He has explained the problem well.

Think in terms of a hydraulic jack if you are trying to lift 5 tons and you have only mud to set the jack on you use boards or something to spread out the force, but you are still applying 5 tons to the mud. If you made a action with larger locking lugs and thicker receiver walls it would do the job. If you make the case head larger to spread the load you would still be applying the same force to the locking lugs. Like a hydraulic cylinder it is the inside diameter of the cylinder that does the work.

Bob

hwingo
01-05-2009, 10:31 PM
SUPER!!!!

Ok guys, I'm finally seeing the light. Trying to make sense of this, my mind had more clouds than a thunder storm. I can assure you I will have more questions as time continues.

At least you have validated the formula and calculations. This make me feel a little better.

Thanks again guys.

Harold