Computing Breech Pressures
in Conventional Guns

(Created May 2010)
(Finally Uploaded October 2017)

Calculating Pressure at the Breech if Projectile Base Pressure is known

PB = (1 + ε (MC/MP) ) PP

Where:

PB = Pressure at the breech in bar

PP = Pressure at the base of the projectile in bar

ε = Fraction of the propellant accelerated. Usually 0.5

MC = Mass of the Propellant Charge in kilograms.

MP= Mass of the Projectile in kilograms.

EXAMPLE: A M735 APFSDS projectile is fired from a 105mm M68 gun. The weight of the projectile is 5.797 kg, the mass of the propellant is 5.67 kg and the projectile base pressure was 2,800 bar. What is the Breech pressure?

(1 + 0.5(5.67/5.797))2800 = 4169.33 bar

Reference:
Technology of Tanks by R.M. Orgorkiewitz

CONTEXT: To help put all this into context, here are some breech pressures for well-known weapons:

90mm F1 Gun (French Low-Pressure Gun): 1,200 bar
76mm L23 Gun (UK Scorpion): 1,600 bar
75mm M2 Gun (M4 Sherman): 2,620 bar
75mm L70 Gun (Panther): 3,200 bar
105mm M68 Gun (Patton): 5,100 bar
120mm M256 Gun (Abrams): 6,300 bar (maximum design fail pressure is 7,100 bar)

Breech pressures can't get much higher than current technical levels – 6,300 bar is about 91,373~ PSI and the best steel suitable for gun barrels (at least in the early 2000s) had a compressive strength of around 122,000 PSI (8,411~ bar).

It's possible that if safety rules are relaxed somewhat, higher breech pressures could be achieved; but there's a limit to how much relaxation can occur; due to the need for the vehicle crew to have total confidence in their weapon since it is (currently) in the fighting compartment with them and a breech failure would be catastrophic for them.

If unmanned turrets become more of a going thing, it's highly likely that these safety rules would be relaxed even further; since a catastrophic breech failure would no longer kill or severely wound the crew.