What is the standard bore, twist, and number of grooves for a production 1803 Rifle? I have seen .54 caliber, 7 groove, with a 1:48 twist sighted as the specification, but the citation is Flayderman's Guide to Antique American Firearms. I don't have that book, so I don't know their source or if it is accurate.
If Dearborn actually specified the rifle should “carry a ball of one thirtieth of a pound.” (as quoted in Notes on United States Ordnance, Vol. 1: Small Arms, 1776–1946) then that I take that to mean that the ball "carried" (meaning fired??) would be .538 in diameter. Which means the bore would need to be larger. A 25 (.571) or 26 (.564) bore would be the size of the bore if you follow the normal windage practices of the time. If you just use the other rule of thumb and go up two bore sizes, the bore would be 28 (.550) bore. If it was a very tight ball to bore fit, a 29 bore it would have been .544 in diameter, which would be impractical if one actually shot a .538 ball.
The ball size that is commonly cited as the standard US Army size for that period is a 1/32 lb (.526) ball. That sounds like it would work in a 30 bore barrel. So did the Army not follow the specification to use a rifle ball a 1/30 lb (.538) ball or am I interpreting the specification incorrectly, and what Dearborn actually meant was that the hole should be exactly the same size as a 1/30 lb (.538) ball?
And then what is the twist and the number of grooves? And how do we know? Did Dearborn specify those as well?
Mystified Mike