doublegunshop.com - home Forums General Discussion DoubleGun BBS @ doublegunshop.com Smokeless loads for 2 & 1/2" 12 Bore

KirkD, welcome.

Many of us load low pressure 2 3/4" hulls for our 2 1/2" chambered guns with good satisfaction. The issue with 2 3/4" hulls in a 2 1/2" chamber gun is when higher pressure loads are used. Higher pressure loads tend to have a higher pressure differential than do low pressure loads.

The 7000 psi recommendation is a sort of rule of thumb based on being conservative with black powder loads and service proof as a guideline. BTW, many BP loads have high chamber pressure; they are not universally low pressure as is sometimes said. There are many low psi 2 3/4" loads in the various loading manuals and web sites. Pick one below 7000 psi (more or less) that you have exact components for and have fun. I shot five-stand last Sunday with a Parker "D" with damascus barrels and can still count to ten on my fingers!!

DDA

Woodchucks?

Last Dollar: Most fellows around here use high speed rifles that fire a bullet at nigh unto 4,000 fps and shoot them from 400 yards away. That makes it way too easy. I like to stalk up to about 20 yards away and then let them have it. I've done this for a number of years with a 20 Bore, but now that I have this old hammer gun, I want to use it for the Woodchucks. They are making big holes in my alfalfa field.

Men, I wrote to Sherman Bell to see if he had tested black powder cartridges to see what sort of pressures they generate. He mentioned that he had compared smokeless and Black powder pressures in two articles in the Winter 1999 and Summer 2002 issues of Double Gun Journal. He also mentioned that full power black powder loads produce from 6,000 to 8,000 psi breech pressure. By 'breech' pressure, I think he is referring to the hoop stress in the barrel near the breech (from how he uses that term in another article he wrote).

So that fits with the advice that has been given in this thread, but it is nice to have an idea where the upper limit is so one knows whether they have 7,000 plus or minus 5 psi to play with or 7,000 plus or minus 800 psi.

For thin walled cylinders, hoop stress is calculated by multiplying the radius by the pressure in psi divided by the wall thickness. If pressure and wall thickness remain constant, a larger radius causes a larger numerator which would be divided by a constant demominator; hence, a larger hoop stress. Based on this, if you have a smaller radius (12g. down to 20g for example)you can have higher pressure with the same wall thickness and still maintain the same hoop stress. Well, lets just work the numbers. We do need to use mean radius (mean radius is half ID + half wall thickness, say 0.730/2 + 0.100/2. For 12 bore = 0.365" +0.050" = 0.415", assuming wall thickness at 0.100", Assume pressure at 5,000 psi. Then 12 bore hoop stress equal (5,000 X 0.415)/0.100 = 20,750 psi stress (pressure and stress have the same measurement units). For 20 bore, r = 0.615"/2 + 0.100/2 = 0.3575"; hoop stress = (5,000 X 0.3575)/ 0.100 = 17875 psi or, r for = (5,000 X 0.3505)/ 20,377 = 0.086". Yes, you are right!!
This from Wikipedia.
Also from Wikipedia: Fracture is governed by the hoop stress in the absence of other external loads since it is the largest principal stress. Note that since the hoop stress is largest when r is smallest (nope, when r is largest), cracks in pipes should theoretically start from inside the pipe (nope, the outside should fail first).

Seems contadictory. Am I misunderstanding this? Yes, I think so - as above. You were right to start with and then went astray. OK?.