Home Forums Hunting & Shooting Ask The Gunwriters What difference is there with breach pressure and chamber pressure?

Is one more important than the other in terms of saftey?

no difference

When you mention breach pressure, were you thinking about bolt thrust? That thrust is caused by the pressure in the case
and as Ken says the pressure inside the case is the same against back as well as sides of case...But bolt thrust is a
factor of the pressure times the area of the base of the case,
against the bolt in simple terms.Ed.

Hubel, when calculating bolt thrust, do you use the surface area inside the case, or do you just use the diameter of the case base?



Al

AJ--I use a simple formula that is within 5% of complicated
ones.Base outside diameter times pressure times 2/3.
Example- 30-06 with .470 base has area of .1735 sq inches,
times a 60,000 psi load, times 2/3 = 6950 lbs of thrust.Ed.

Since it's the powder gas that's exerting the force, the flat portion of the inside surface of the web (exclusive of wall thickness, etc) is the active piston surface. Using the diameter of the rim produces an area figure that is too large.



In lieu of direct measurement, I methodologically estimate the area of this net surface by using 90%-95% of the outside diameter. This allows for the fact that the case wall and the fillet (curved transition from web to wall) are not subject to the straight-back force.



Peak chamber pressure (lb/sq in.) times this net web area (sq in.) equals the force (lb) pressing the case rearward against the resistance of the locked lugs.



Obviously, actually measuring the diameter of the flat area of the web produces a figure that is more accurate than my hasty estimate.

Thanks guys. I figured it was the surface area inside the case, just wondering how you went about calculating it. I'm anxiously awaiting release of Ken's book so I can get edjumacated beyond my intelligence <img src="/ubbthreads/images/graemlins/grin.gif" alt="" />



Another queston, which would have a higher force, say the inside measurement was .375" diameter, flat surface, or if the surface was shaped in a semi-hemispherical, .375" diameter, with the height of the crown being .050". My line of thinking is that you would have to concider the force vector on the curved surface, total force would be lower than on a flat surface. It's been too long since my math courses.



Al