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. Ans. 23.09 47. How large a cube may be inscribed in a sphere 40 inches in diameter ? 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. GAUGING is the art of finding the contents of any regular vessel, 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? 2. How many ale gallons will a trough feet long, 6 feet wide, and 2 feet high? 3. How many bushels of grain will a 15 feet long, 5 feet wide, and 7 feet high? Ans. 1496 gal. contain, that is 12 Ans. 8821 gal. box contain, that is Ans. 421.8bu. PROBLEM II. To find the contents of a cask. RULE. Take the dimensions of the cask in inches; namely, the diameter 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. 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. 13618 tons. 2. What is the tonnage of a double-decked vessel, whose length is 70 feet, and breadth 24 feet? Ans. 212 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 sel, whose length is 70 feet, breadth 30 hold 9 feet? 4. What is the government tonnage of sel, whose length is 75 feet, breadth 22 hold 12 feet? a single-decked vesfeet, and depth in the Ans. 14715 tons. a single-decked vesfeet, and depth in the Ans. 171 tons. a double-decked ves 5. What is the government tonnage of sel, which has the following dimensions: length 98 feet, breadt 35 feet? Ans. 4961 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. 130 tons. 8. What is the government tonnage of a double-decked vessel, whose length is 159ft., and width 30ft.? Ans. 6671 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. 17421 tons. 10. What is the government tonnage of a vessel, whose length is 200 feet, and breadth 35 feet? Ans. 11543 tons. 11. The new ship Montezuma is 280 feet in length, and 40 feet in breadth. Required the government tonnage. Ans. 215515 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. 561 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 cubic feet. OPERATION. 48 x .31831 15.27888; 15.27888÷27.63944; 7.63944 × 24=183.34656; 183.34656 × 40=7333.8624; 7333.8624 ÷ 144 509, that is, it contains as many cubic feet as a stick 50 feet long and 1 foot square. RULE. = - Multiply the length of the stick, taken in feet, by the square of the girth, taken in inches; divide this product by 144, and the quotient is the contents in cubic feet. NOTE. — The girth is usually taken about § of the distance from the larger to the smaller end. 5. How many cubic feet in a stick of timber, which is 30 feet long, and whose girth is 40 inches? Ans. 20 feet. 6. If a stick of timber is 50 feet long, and its girth is 56 inches, what number of cubic feet does it contain? Ans. 68 feet. 7. What are the contents of a log 90 feet long, and whose circumference is 120 inches? Ans. 562 feet. SECTION LXXIX. PHILOSOPHICAL PROBLEMS.* PROBLEM I. To find the time in which pendulums of different lengths would vibrate; that which vibrates seconds being 39.2 inches. The times of the vibrations of pendulums are to each other as the square roots of their lengths; or their lengths are as the squares of their times of vibrations. RULE. - As the square of one second is to the square of the time in seconds in which a pendulum would vibrate, so is 39.2 inches to the length of the required pendulum. EXAMPLES. 1. Required the length of a pendulum that vibrates once in 8 seconds. 12: 82: : 39.2in. : 2508.8in. 2. Required the length of a pendulum times a second. 209 feet, Ans. that shall vibrate 4 Ans. 2 inches. 3. Required the length of a pendulum that shall vibrate once a minute. Ans. 3920 yards. 4. How often will a pendulum vibrate whose length is 100 feet? Ans. once in 5.53+ seconds. PROBLEM II. To find the weight of any body, at any assignable distance above the earth's surface. * For demonstrations of the following problems, the student is referred to Enfield's Philosophy, or to the Cambridge Mathematics. |