multiply their sum by one third of the altitude or length of the frustum, and the product will be the solidity of the frustum of a cone, or pyramid. Note. — These are the exact rules for measuring round timber, and should be adopted. See page 330. 41. What are the contents of a stick of timber, whose length is 40 feet, the diameter of the larger end 24 inches, and the smaller end 12 inches ? Ans. 731 feet, nearly. PROBLEM XVIII. 42. What is the solidity of a sphere, whose-axis or diameter is 12 inches ? Ans. 904.78+ inches. 43. Required the contents of the earth, supposing its circumference to be 25,000 miles. Ans. 263858149120.06886875 miles. PROBLEM XIX. 44. Required the convex surface of a globe, whose diameter or axis is 24 inches. Ans. 1809.55+ inches. 45. Required the surface of the earth, its diameter being 7957 miles, and its circumference 25,000 miles. Ans. 198943750 square miles. PROBLEM xx. To find how large a cube may be cut from any given sphere, or be inscribed in it. RULE. — Square the diameter of the sphere, divide that product by 3, and extract the square root of the quotient for the answer. Demonstration. - It is evident, that, if a cube be inscribed in a sphere, its corners or angles will be in contact with the sur. face of the sphere, and that a line passing from the lower corner of the cube to its opposite upper corner will be the diameter of the sphere ; and that the square of this oblique line is equal to the sum of the squares of three sides of the inscribed cube is evident, from the fact that the square of any two sides of the cube (suppose two sides at the base) is equal to the square of the diagonal across the base; and that the square of this diagonal (which we have just proved to be equal to the square of two sides at the base) and the square of the height of the cube are equal to the square of the diagonal line, which passes from the lower corner of the square to the opposite upper corner, which line is the diameter of the sphere. Therefore, the square of the diameter of any sphere is equal to the sum of the squares of any three sides of an inscribed cube; or of the square of the diameter of any sphere is equal to the square of one of the sides of an inscribed cube, Q. E. D. 46. How large a cube may be inscribed in a globe 12 inches in diameter? Ans, 6.928+in. in the side of the cube. 12 x 12 =48; 748 = 6.928+ Answer. 47. How large a cube may be inscribed in a sphere 40 inches in diameter? Ans. 23.09+ inches. 48. How many cubic inches are contained in a cube, that may be inscribed in a sphere 20 inches in diameter ? Ans. 1539.6+ inches. SECTION LXXVI. GAUGING is the art of finding the contents of any regular véssel, in gallons, bushels, &c. PROBLEM I. To find the number of gallons, &c., in a square vessel. RULE. – Take the dimensions in inches; then multiply the length, breadth, and height together; divide the product by 282 for ale gallons, 231 for wine gallons, and 2150.42 for bushels. 1. How many wine gallons will a cubical box contain, that is 10 feet long, 5 feet wide, and 4 feet high? Ans. 1496-8-gal. 2. How many ale gallons will a trough contain, that is 12 feet long, 6 feet wide, and 2 feet high? Ans. 8821.gal. 3. How many bushels of grain will a box contain, that is 15 feet long, 5 feet wide, and 7 feet high? Ans. 421.8bu. PROBLEM II. To find the contents of a cask. Rule. - Take the dimensions of the cask in inches ; namely, the dr ameter of the bung and head, and the length of the cask. Note the difference between the bung diameter and the head diameter. If the staves of the cask be much curved between the bung and the head, multiply the difference by .7; if not quite so much curved, by .65; if they bulge yet less, by .6 ; and if they are almost straight, by .55; add the product to the head diameter; the sum will be a mean diameter by which the cask is reduced to a cylinder. Square the mean diameter thus found, then multiply it by the length; divide the product by 359 for ale or beer gallons, and by 294 for wine gallons. 4. Required the contents in wine gallons of a cask, whose bung diameter is 35 inches, head diameter 27 inches, and length 45 inches. 35 – 27 X.7 = 5.6 32.6 x 32.6 x 45 = 47824.20 27 +5.6 = 32.6 47824.20 201 = 162.66 wine gallons. 5. What are the contents of a cask in ale gallons, whose bung diameter is 40 inches, head diameter 30 inches, and length 50 inches? Ans. 185.55+ ale gallons. PROBLEM III. To find the contents of a round vessel, wider at one end than the other. Rule. — Multiply the greater diameter by the less; to this product add } of the square of their difference, then multiply by the height, and divide as in the last rule. 6. What are the contents in wine measure of a tub, 40 inches in diameter at the top, 30 inches at the bottom, and whose height is 50 inches ? Ans. 209.75 wine gallons. SECTION LXXVII. TONNAGE OF VESSELS. CARPENTER'S Rule. – For single-decked vessels, multiply the length, breadth at the main beam, and depth in the hold together, and divide the product by 95, and the quotient is the tons But for a double-decked vessel, take half of the breadth of the main beam for the depth of the hold, and proceed as before. 1. What is the tonnage of a single-decked vessel, whose length is 65ft., breadth 20ft., and depth 10ft. Ans. 13619 tons. 2. What is the tonnage of a double-decked vessel, whose length is 70 feet, and breadth 24 feet ? Ans. 2124 tons. GOVERNMENT Rule. If the vessel be double-decked, take the length thereof from the fore part of the main stem to the after part of the stern-post above the upper deck ; the breadth thereof at the broadest part above the main wales, half of which breadth shall be accounted the depth of such vessel, and then deduct from the length of the breadth; multiply the remainder by the breadth, and the product by the depth, and divide this last product by 95, the quotient whereof shall be deemed the true contents or tonnage of such ship or vessel ; and if such ship or vessel be single-decked, take the length and breadth, as above directed, deduct from the said length of the breadth, and take the depth from the under side of the deck-plank to the ceiling in the hold, and then multiply and divide as aforesaid, and the quotient shall be deemed the tonnage. 3. What is the government tonnage of a single-decked ves. sel, whose length is 70 feet, breadth 30 feet, and depth in the hold 9 feet? Ans. 14715 tons. . 4. What is the government tonnage of a single-decked vessel, whose length is 75 feet, breadth 22 feet, and depth in the hold 12 feet? Ans. 17115} tons. 5. What is the government tonnage of a double-decked vessel, which has the following dimensions : length 98 feet, breadth 35 feet? Ans. 49647 tons. 6. Required the government tonnage of a double-decked vessel, whose length is 180 feet, and breadth 40 feet. Ans. 131313 tons. 7. Required the government tonnage of a single-decked vessel, whose length is 78 feet, width 21 feet, and depth 9 feet. Ans. 130233tons. 8. What is the government tonnage of a double-decked vessel, whose length is 159ft., and width 30ft.? Ans. 66717 tons. 9. What is the government tonnage of Noah's ark, admitting its length to have been 479 feet, its breadth 80 feet, and its depth 48 feet. Ans. 1742119 tons. 10. What is the government tonnage of a vessel, whose length is 200 feet, and breadth 35 feet? Ans. 1154;tons. 11. The new ship Montezuma is 280 feet in length, and 40 feet in breadth. Required the government tonnage. Ans. 215518 tons. Section LXXVIII. MENSURATION OF LUMBER. PROBLEM I. To find the contents of a board. RULE. - Multiply the length of the board, taken in feet, by its breadth taken in inches, divide this product by 12, and the quotient is the contents in square feet. 1. What are the contents of a board 24 feet long, and 8 inches wide ? Ans. 16 feet. 2. What are the contents of a board 30 feet long, and 16 inches wide ? Ans. 40 feet. PROBLEM II. To find the contents of joists. RULE. — Multiply the depth and width together, taken in inches, and their product by the length in feet; divide the last product by 12, and the quotient is the contents in feet. 3. How many feet are there in 3 joists, which are 15 feet long, 5 inches wide, and 3 inches thick ? Ans. 564 feet. 4. How many feet in 20 joists, 10 feet long, 6 inches wide, and 2 inches thick ? Ans. 200 feet. PROBLEM III. To measure round timber. We have inserted below the rule usually adopted by surveyors of lumber ; but it is a very unjust rule, if it is intended to give only 40 cubic feet of timber for a ton. For, if a stick of round timber be 40 feet long, and its circumference be 48 inches, it is considered by surveyors to contain one ton, or 40 feet; whereas, it in reality contains, according to the following correct process, 50-14% cubic feet. OPERATION. 48 X.31831=15.27888; 15.27888; 2= 7.63944; 7.63944 x 24 = 183.34656 ; 183.34656 x 40 = 7333.8624; 7333.8624 • 144 = 50, that is, it contains as many cubic feet as a stick 5018 feet long and 1 foot square. RULE. — Multiply the length of the stick, taken in feet, by the square |