20 × 38·2 8 × 8 = 11 15 16 inches.
To find the content, and weight of a piece of ordnance.
Divide the length of the gun into as many sections as may be found necessary. Find the content of each (by preceding rules) and from their sum subtract the content of a cylinder, whose length is equal to that of the bore, and its diameter equal to that of the calibre of the piece; multiply the difference (if it be a brass gun) by 5·0833, (if an iron gun) by 4·2968, and the product will be the weight in ounces.
| Note.—A cubic inch | of gun metal | weighs 5·0833 ounces. |
| Ditto | of cast iron | 4·2968 ounces. |
To find the content of a cask.
Multiply half the sum of the areas of the two interior circles, viz. at the head, and bung, by the interior length, for the content.
Or, to the area of the head add twice the area at the bung, multiply that sum by the length, and take one-third of the product.
Example.—Required the content of a cask, its greatest interior diameter being 24 inches, its least interior diameter 20 inches, and the interior length 30 inches.
24 × 24 × ·7854 = 452·3904, area of large circle.
20 × 20 × ·7854 = 314·1600, area of small circle.
452·3904 + 314·1600 2 = 383·2752, half sum.