Although it would seem to be a strange thing for a grown man to do, this topic got me to wondering how much pressure would be needed to press a lead ball into the chamber of a Cap & Ball pistol.
I couldn't find a good formula for shearing off the outside of an object so I settled on the formula for punching an object out of a plate. (I'm sure just shearing off some material would take far less pressure than punching out an object like a round disk so take my answers to be on the high side. High side or not, the percentages in the amount of force needed to do the job will be very close to the real world.)
The formula for punching something like a washer out of a steel plate is Diameter X Thickness X 85.
That gives the number of tons to punch a hole in a mild steel plate.
If the plate was brass, the formula is Diameter X Thickness X 65 = Tons of force needed.
These two factors, the 80 and 65 have a relationship that corresponds with the materials tensile strength.
Based on that correlation I find the factor for lead to be 3.25. (Remember, shearing off the outside of an object needs less pressure than the punching out formula predicts).
OK. So how much "thickness" is there when you press a larger round ball into the cylinder?
The "thickness" is the length of the cylindrical surface after the ball is pressed in place.
That answer is pretty easy to calculate if you know the diameter of the ball and the diameter of the chamber.
It equals √ ball²-chamber². (the square root of the diameter of the ball squared, minus the diameter of the chamber squared).
If the chamber for your .44 measures .446" and the ball is .451" the length of the flat area on the sheared ball will be .067" long.
For a .454" ball in a .446" chamber the length is .084"
For a .457" ball in a .446" chamber the length is .100.
So, to figure the amount of pressure needed to press the ball into the hole, if I use the washer formula I have, Dia X Length X 3.25/2000 to get pounds of force.
The .451 ball would need 194 pounds.
The .454 ball would need 243 pounds.
The .457 ball would need 289 pounds.
If the projectile was a bullet with a land length of .250" (1/4") the force would be 724 pounds.
Using the 194 pound figure for the .451 ball as a baseline, the .454 diameter ball requires a force 125% greater and the .457 diameter ball requires a force 149% greater.
The bullet with its .250 long cylindrical diameter would require a force that is 373% greater.
(That's why I recommend using a loading stand if you want to load bullets into your C&B chamber.)
OK. I think I'll go take an aspirin. For some reason I have a headache. :grin: