2-p has my unwavering respect as a well learnd, experienced and resourceful person. He and I have not, over the years agreed completely on all subjects. However, we are always willing to discuss and learn.

Originally Posted By: 2-piper
Yes, I totally agree if we pluged the bore & made the chamber strong enough not to burst but the breech was sealed by the cartridge & it did not rupture then it would be correct to say that the pressure & area it acted upon would be the total consideration, recoil would not enter as was correctly there would be none. We agree to here. This however is not the manner in which our shotguns function. This would in efect be dealing with a static pressure load, but in firing we don't have a constant static pressure. "IF" the full pressure a normal powder charge is capable of generating was applied to the gun there is likely not a gun in existance which could stand the strain, either the chambers or the breech, it would just be a matter of which let go first. However as soon as the charge starts down the bbl pressure begins to fall away & we generate the "Curve". To go back to basics as you say R'Man I think you would find if you placed a hydraulic piston hooked to the hinge pin with its ram pushing against the standing breech, centered on the firing oin hole & having same bore as the internals of the case & pumped up the pressure equal to that developed by firing a standard shell with proper gauges hooked up to measure deflection of breech, hinge pin or any other critical, you would find the deflection greater than firing the shell which produced identical "Peak" pressure. Don't agree with this position. The force will transmit through the steel at sonic velocity for steel and that is very fast; will have to look up the actual value. Now this will be because the peak falls away before its full effect can act upon the breech. Don't think so. Force transmits very fast; small movement like strain happens fast, displacement size movement is slow. Now I am not sure I can word this in a technically correct manner, but I am throughly convinced that;
#1 recoil is of course brought about by the presure acting inside the gun. As noted it requires movement in opposite directions of different parts to occur. Agree
#2 the same force which is creating this recoil is also acting upon the parts of the gun under axial load. The firing pressure is accelerating the payload via force on the wad base and is straining the action via axial force on the fences. Sounds like we agree.
#3 I am in total agrement that under your scenario of plugging the bbl while recoil would be stopped the thrust against the gun would not only not be stopped, but rather increased. Agree, but I doubt increased.
#4 I am still convinced that as long as the gun is operating as intended with the charge rushing down the bore the amount of strain put upon the parts of the gun for all the above mentioned reasons willbe more in proportion to the recoil than the "Peak Pressure". Disagree. I think the strain is proportional to the pressure as the strain movement is small and joint closure movements are small. Small equals quick.
#5 I am totally convinced that a 1oz 1200 fps load @ 10k psi will not load the gun as much as a 1¼oz 1330fps load @ 9k psi. This one is tricky. Where does the recoil force actually start? With the barrels? If so, then the rearward force of the barrels will unload the hook and hinge pin joint. In the example I gave of the "sled" being swatted, now swat the barrels and you can see that this will drive the hook away from the pin. Prima facia, that doesn't seem reasonable. If we say recoil force emminates from the action, then there is no force on the hook/pin bolts/bites except pressure force resistance. Sooooo, what really happens? Seems to me that we have the action of the firing pressure with very little off-setting recoil force. As the pressure curves down, the recoil force is building up; at some point, the hook/pin would unload if recoil force exceeded pressure back-thrust force. However, i don't see recoil force doing much to the metal. It will sure beat on the wood, though.
#6 "IF" someone has the recourses & equipment & knowledge to set up a test & can show that the higher pressure load will indeed load the gun more than the lower pressure load producing heavier ballistics I will offer my most humble appologies. Until then we will just have to agree to disagree. We may have to disagree if the above did not convince you.


PS; I might well add that under the conditions of the plugged bbl scenario the 1oz load with the fast powder which uinder ordinary conditions produced the higher pressure would in this case produce the lower. Agree - very astute observation! Reason; under this condition the heavier load will contain a considerably heavier charge (more chemical energy) of a slower burning powder will have suffcient time to fully burn & deliver its total potential. Of course it should be understood that even under normal conditions it is doing more work, thus a higher total "Average Pressure" just a lower peak. I do believe R'Man you are delving just a little too deep in total theory & ignoring a few actual occurances. With all due respect, 2-p, can you cite an example. Surely, I do want to understand.

PPS; I have to the best of my knowledge been talking about the "WHYS" for this entire thread. Agree that you have.

Last edited by Rocketman; 08/10/09 11:45 PM.