doublegunshop.com - home Forums General Discussion DoubleGun BBS @ doublegunshop.com Pressure vs recoil in old guns

R'Man; now just think about this one a minute, in this scenario you would obtain closed cell pressures. Certainly this would give higher pressure than an ordinary firing of the load. OK, differing assumptions. I was assuming firing a shell loaded to the same end pressure in the closed cell as the peak pressure of the shell in an open barrel.


I can see no place for the recoil to start other than from the energy imparted to the standing breech from the back thrust of the shell head. The friction of the charge being pushed down the bore would in fact impart a forward motion to the bbls, but a lesser force than is being imparted to the breech. Many years ago in my National Guard days I fird a 3.5 Rocket Launcher & there was in fact a distinct forward pull as the rocket went down the tube, which of course was totally open to the rear. This same condition should exist in the shotgun bbl, but of course is un-feelable due to the offsetting effect upon the gun's breech. Thus when the charge is fired there are actually opposing forces trying to seperate the breech from the bbls. Excellent example, thank you. If we accept this line of reasoning, and it does seem logical, the recoil reaction starts at the fences. Would there be any logic in the recoil forces transmitting into the action bar as opposed to straight back into the stock? Seems more logical to go to the stock and imagine the action bar and locking joints keeping the action shut against the opening moment (opening torque) of back-thrust above the hook/pin center line.

Now I do believe we are mostly in agreement, with the exception of the part time plays in the mix. In looking at the peak pressure in available pressure curve drawings with the faster powders this peak is a virtual point, but with slower powders more of a rounded hump. Now I think we all know this peak of pressure will not compress a lead piston to the same extent it would be by a static load of the same pressure, thus LUP is lower than PE pressures from an identical load. Now the only real difference I can see here is the steel of the action is being placed under tension rather than compression. I see a bit of a problem in the analogy. First, force can transmit within the PE crystal fast enough to easily detect fast powder peak pressures. Steel is crystaline in nature and would transmit forces at speeds similar to the PE crystal sensor. BTW, for anyone not clear on force transmission rate, the issue is how long would it take for a sensor on the end of a mile long steel rod to know the other end got whacked by a hammer? Air transmits force at the speed of sound, Mach 1, about 750 mph (1125 fps), about 5 seconds per mile. Steel, being much less compressible than air, transmits force (sound is a pressure wave) much faster. The lead crusher system has the delay of a large movement, relatively speaking, of the crusher piston plus the inertia of that piston.

If a 1200 fps load was given absolute uniform accelration & allowing for 29" of travel to clear the bbl then total bbl time would be .004 sec. Since acceleration occurs quicker near the breech actual time is more on the order of .003 sec. Now the top of that pressure peak is just a "Dot" on that 29" curve. Yet, the forces transmit and compress the PE sensor fast enough to easily detect said "dot." Lets say that the "dot" occured within the first one ten thousandeth of a second after firing ignition and that there was three inches of steel between the face and the stock head. A transmission rate of 2500 fps would tell the stock head about the "dot" within another 0.0001 seconds. 2500 fps is just a tad more than double sonic velocity of air and way slow compared to steel. "Knowledge" of firing forces within the action steel will happen way faster than the pressure curve changes as I see it.

Yes I believe time does play a very important role in this situation. If it did not I do believe you would find a very large pile of destroyed guns, but you don't. If you calculate aformulas for strength of steels vs wall thicknesses, hinge pin shear, frame deflections etc, etc you will find many running on the Ragged edge, or over, yet they just keep on keeping on with loads far heavier than they were designed for.
Could you cite an example of this, please?

While just a machinist (don't you start that "just a machinist" stuff, you are one of the smartest, best informed, and most logicial guys I know) & not an engineer (that line of education does offer a jump-start in some areas of knowledge and reasoning, but is not the end-all/trump card in discussions such as this; it counts only if understandable points are made for all to consider) I have studied everything I could come up with on this & similar subjects for about the last 40 years. I have come up with no other satisfactory explanation. Please comment on the above points. Good discussion!