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SECT. XVI. To know what quantity of Powder should be allowed to a Piece of Ordnance not truly bored.
ADmit the diameter of the Metal of the Piece at the Touch-hole be 16 inches, and the diameter at the bore is 5 ¼ inches, the weight of the Piece 4850, as you may see by Chap. 7, such a Piece you may find in the ninth Chap. requires 11 l. for her due Charge, being near two Diameters of her bore in Powder; But by my Instru∣ment in the general Figure, with the two Rammers heads at the two ends LM (at the Rammer end L that was in the Gun at the Touch-hole, I find by the prick at S on the Rammer, the soule or bore to be 1 inch out of his place, or 1 inch from the middle of the Metal; then I conclude, that the thinnest part of the Metal is 4 inches 3/2 parts, and the thickest side 6 and ⅜ parts; by which it appears, that one side is just 2 inches thicker than the other side, as you may see plainly by this Figure; the Line AB divided is the diameter or greatest thickness at the Touch-hole, every Division signifies an inch from the inward Circle to the outward Circle, is the thickness of the Metal; the inward Circle signifies the bore of the Piece, which you may see is just an inch from the true bore or Centre of the outmost Circle; therefore you must work as if the Piece were fortified no more than only so much as the thinnest part of the Metal is, which here doth appear to be 4 inches ¾ parts, the ½ of the diameter of the bore is 2 ⅝ inches added, makes 7 from A to D, the Centre of the bore being the thinnest part of the Metal, the whole diameter being 14, which is the true diameter, by reason the thinnest side of the Metal is but 4 ⅜ inches thick.
And by this you must proportion your Charge by the former being 16 inches, if the bore had been placed at C in the true Centre, then evermore by these Rules.
The Logarithm of greatest diameter 16 is | 220412 |
The Logarithm of the less diameter 14: 0 l. | 214612 |
The difference decreasing, | 5800 |
3 | |
The triple of the difference Substracted | 17400, |
from the Logarithm of 11 l. Powder 0 ounces, | 204139 |
Leaves the Logarithm of the Powder 7 4/1•• pound | 186739 |
So that 7 pound 4/19 or 6 ounces, is a sufficient Charge for such a false bored Piece; or extend the Compasses from 16 to 14, the same distance 3 times repeated from 11, will reach 7 4/19 pound, as before.
C 14 is •• 2744 × 11/C 16 is / 4096 = 7 pound 6 ounces, as before.