Max pounds force of revolver cylinder against frame?

[Maillemaker]

Hi all,

With a 140 grain bullet, a 30 grain charge, and a muzzle velocity of 900 fps, a 2.8 pound revolver has a free recoil of about 8.6 ft*lb of energy.

I would like to figure out how this translates into a force pushing against a circle .44 inches in diameter (the back of the chamber).

Can anyone help out with this bit of engineering magic?

Steve

[Charlie Hahn]

I am making aa bunch of assumptions.

The recoil force/rate is about 17.9 lbs per second. Assuming just the cylinder loaded at about 8 oz's, that would be about 143.2 pounds of force on the frame. Now there are a bunch of other factors all but impossible to quantify with out a very expensive test but that is my thumb nail guesstimate

[Maillemaker]

I worked it another way. I figured an internal max pressure of 20,000 PSI, and applied that to the area of a .44" diameter circle to get pounds of force at the back of the chamber wall. It was around 3,000 pounds as I recall.

Steve

[Charlie Hahn]

I do not think internal pressure and recoil force can be factored this way. I would think if we had that amount of force we might have broken bones to show for it.

[Maillemaker]

Yeah I guess my answer is not logical.

If there is an internal pressure of 20,000 PSI in the chamber, and the chamber is .45" in diameter, then the rear of the chamber has .159 square inches of surface area.

When we apply 20,000 PSI to the .159 square inches, that comes out to 3,180 pounds of force on the back wall of the cylinder. While true, it's also there for only an instant. So impulse is the key here. I don't know anything about calculating impulse.

It's like those Estes rockets as a kid. They develop high thrust but only for an instant.

https://en.wikipedia.org/wiki/Specific_impulse

I never really understood it.

Steve

[Muley Gil]

Yeah I guess my answer is not logical.

If there is an internal pressure of 20,000 PSI in the chamber, and the chamber is .45" in diameter, then the rear of the chamber has .159 square inches of surface area.

When we apply 20,000 PSI to the .159 square inches, that comes out to 3,180 pounds of force on the back wall of the cylinder. While true, it's also there for only an instant. So impulse is the key here. I don't know anything about calculating impulse.

It's like those Estes rockets as a kid. They develop high thrust but only for an instant.

https://en.wikipedia.org/wiki/Specific_impulse

I never really understood it.

Steve

​You'll shoot your eye out!! :D

[Chris Sweeney]

I think 20k psi is pretty high for a black powder revolver. I am many years separated from any math or physics classes , but I believe you have to consider more than just the area at the rear of the cylinder; the pressure is also being applied to the chamber walls and the surface area of the back side of the ball

[Maillemaker]

I think 20k psi is pretty high for a black powder revolver.

I just did a quick Google search for black powder chamber pressures. Did not turn up anything really directly useful but found that value and went with it.


I am many years separated from any math or physics classes , but I believe you have to consider more than just the area at the rear of the cylinder; the pressure is also being applied to the chamber walls and the surface area of the back side of the ball

The pressure is indeed applied uniformly on all walls of any pressure vessel, but the pounds in any one direction can be easily found by applying the pounds per square inch over the square inches in the direction of interest. This is how hydraulic cylinders apply pressure. But, they are essentially constant-force devices.

Steve

[Hal]

I'm not sure what the pressure is, but for the one you picked, the force is correct. But I think impulse is more where you need to look. That force acts on the ball for a VERY short amount of time compared to the amount of time recoil is working on the gun and your hand. It's spreading it out over a longer amount of time and that is why you're not getting broken bones. There was a time when I could probably have figured it out, but it's been a long time since Physics class for me too.

[Kevin Tinny]

Hello:

In the mid-late '70's, Ken Ramage edited two LYMAN blackpowder "manuals" that contained articles written by bp shooting experts. In the back of one were ballistics tables that included pressures measured in LUP's (Lead Units of Pressure), in a Modern Bond Pressure rig. Photo's of that rig and a description of its operation were included. LUP's were used, as they are for shotguns, because of the relatively lower pressures that would not register well in "CUP's, copper units of pressure.

I gave my copy away and have not seen one lately. Pity.
It had velocities, pressures, trajectories and foot pounds of impact force.

Maybe someone has a copy and can share the handgun portion of those tables.
Cannot find it used or via Google, either.

Kevin Tinny
13667, 42 NY

[Chris Sweeney]

Just looked through my Lyman Black Powder handbook:Load data for 44 revolver (which starts at 19 gr. and goes up to 37!) pressure is listed in the 5000 LUP range. It lists the Italian National PROOF load as giving 10,677 psi. Of course that's with Curtiss and Harvey powder, so YMMV.

Just as a reference, looking through some of my reloading manuals, a 200 grain 44/40 @ 1100 fps with smokeless powder, generates 12500 psi. Factory 230 grain 45 ACP loads range from 17000 to 19500 psi.

I think we are trying to correlate pressure, force, energy, work, and time into some "ouch" factor. If the weather ever clears up so we can shoot, we'd stop thinking all these dangerous thoughts

[Hal]

Say it was 10,000 psi. That would be 1590 lbf that the cylinder put on the frame, but that's only if the frame were mounted rigidly to something big hard and heavy and did not move backward due to recoil. Conservation of Momentum would come into play. Mass x Velocity of the ball would equal Mass x Velocity of the gun, but only if the gun were floating freely in space. With it held in your hand, there is also the added mass of your hand and arm and whatever else is moving backward and none of that is inelastic so the computations get complicated.