I was thinking about my "Woodcraft" book by Nessmuk (nom de plume for George W. Sears) in the middle of the night last night. He went on a trek across a wilderness area of Michigan, carrying very little except his rifle, hatchet, some food and trail items in a knapsack. The rifle was what got me up to dig my copy of his book out. This is his only description:
"My rifle was a neat, hair-triggered Billinghurst, carrying sixty round balls to the pound, a muzzle-loader of course, and a nail-driver."
So what caliber was this underhammer muzzle-loader? This morning I dug out my Hoag's "Engineering and Technical Handbook" and found the unit weight for lead is 708 lbs./cubic foot and (for a refresher, I learned this in high school) the volume of a sphere is 4/3 x pi x radius3.
To convert the unit weight to one pound I divided 1728 cubic inches in a cubic foot (12 inches cubed) by 708 and got 2.440 in3 make one pound of lead. I then divided that by 60, since there were 60 round ball to the pound for Nessmuk's rifle, and got 0.0407 in3 per ball.
I plugged that number in for volume in the formula and solved for the radius:
0.0407 in3 = 4/3 x pi x r3 r3 = .0407 in3 x 3/(4 x pi)
r3 = 0.00971 in3, taking the cube root gives us a radius of 0.21335 inch.
Since bore is the diameter, we multiply the radius by two and find that it is 0.427 caliber.
Therefor Nessmuk's rifle was a tad larger than Rob's .40, and a tad smaller than Hanshi's (and my) .45.
~WH~
P.s. - Any engineering types on the site please let me know if I screwed something up!