A treatise on mensuration, both in theory and practice

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1788 Excerpt: ...9: 15 = the semi-transverse axe of the less end. Then(2Tc--tc)X na--(2 X 50X30 + 30'X 18) X-2618 X 20 r 118 X 30 X 20 X-2618 = 70800 X-2618 =: 18535-44 cubic inches. RULE II. From 3 times the square of the semi-axe perpendicular to the ends of the frustum, subtract the square of the height of the frustum; then multiply the difference by of the height, and the product by 3-14159 &c; and call the last product p. Then 1. If the ends be parallel to the fixed axe, As the revolving axe is to the fixed axe, so will F be to the content of the frustum. 2. IVhen the ends are 'perpendicular to the fixed axe, As the square of the fixed axe is to the square of the revolving axe, so is p" to the content of the frustum. That is, x phrr will be the frustum 3// whose ends are perpendicular to the fixed axe. And 3rr hh X phf the frustum whose ends are 3r parallel to the fixed axe: / being the fixed, and r the revolving semi-axe, b the height, and p--S'14159 EXAMPLE I. If the axes of an oblong spheroid be 50 and o, required the content of a frustum whose ends are perpendicular to the fixed axe, and one of them passing through the center, the height of the frustum being 20 inches. Here Here A-X pbrr--3-2-----X 3-141 iff r 3 X 25 3 t 3V X20X 15 = 1475 X 314159 X T = 1112123799 = the content required. EXAMPLE II. If from the fame spheroid be cut a frustum whose height is 9, the ends being parallel to the fixed axe, and one of them passing through the center, what will be its content? Herex/= x 9X25X3-14159 = 594 x 5 X 3-14159 = 93353l8 = the content required. EXAMPLE III. If from an oblate spheroid, whose axes are 50 and 30, a frustum, whose height is 9, and its ends perpendicular to the fixed axe, be cut; what will be its solidity? Here pbrr = 3 "/x 9 X 25 x 3ff 3...