2 ft. under feet. Having added the two products, we have 16 feet, 11 primes, 9 seconds, 4 thirds, for the total product. 342. We have seen (110) that the content of any superficies is estimated in squares, and is the product of the length and breadth, each taken in the same measure. Artificers' work, such as glazing, painting, plastering, paving, roofing, &c., is usually paid by the square yard or square foot; and, as the dimensions are taken in feet and inches, being generally measured with a carpenter's rule, such work is conveniently calculated by duodecimals. Parts less than seconds are commonly omitted in practice. Suppose the roof of a green-house to be 100 ft. 9 in. long and 21 ft. 6 in. wide, what is the amount of glazing, and what will it cost at 9 cts. per square foot? 144 18 The scholar must take particular notice, that in the product, the prime 1' is of a square foot, or of 144 square inches, and is consequently 12 square inches; that the 6", being a number of hundred and forty-fourths, is 6 square inches, and hence, that 1' 6"}+ƒ==;or, recollecting that 11 d. of a shilling, he will see that I' 6" or 11⁄2 twelfths of a s. f. = of a s. f. 1' 6" Therefore, 2166° 1′ 6′′ 2166 s. f.; which, as dollars and cents are decimals, we may express thus: 2166,125 s. f. Then, multiplying by 9, we have $194,95125 or $194,951 for the required cost. 8 Examples. 1. Suppose a room to be 47 ft. 2 in. long, and 13 ft. 6 in. wide; what will the ceiling of it cost, at 25 cts. per square yard? Ans. $17,683 2. How many yards of carpeting, that is yard-wide, will carpet a room that is 36 ft. 8 in. long, and 25 ft. 6 in. wide, and what will it cost, at $1,12 cts per yard? Ans. 1038 yds., and the cost $116,871. 3. How many square yards of paving are there in a courtyard, which is 77 ft. 3 in. long, and 66 ft. 8 in. wide; how many bricks will pave it, allowing each brick, when laid, to cover a space of 8 in. long and 4 in. wide; and how much will the paving of it cost at 561 cts. per square yard? Ans. 5723 s. yds.; 23175 bricks, and the cost $321,871. 343. Duodecimals may also be applied to quantities of three dimensions, the content or capacity of which, as we have seen (170,) is estimated in cubes, and is found by multiplying together their length, breadth, and thickness. Required, the content, in bushels, of a granary bin, the length being 53 ft. 9 in.; the breadth 6 ft. 3 in. and the depth 4 ft. 5 in. 2 ΤΣ The scholar must here observe with attention, that in the product 1483° 8' 8" 3"", the number 1483 is a number of cubic feet (170); also, that each unit in the 8', being of a cubic foot, or of 1728 cubic inches, is 144 cubic inches; each unit in the 8", being of 1', or of 144 cubic inches, is 12 cubic inches; and lastly, that each unit in the 3"", being 2 of 1', or of 12 cubic inches, is 1 cubic inch. We can, therefore, reduce these parts to cubic inches, thus: 8' +8" + (8. X 144) + (8 × 12) +3=1152 +96 +3: cubic inches, or 1251-132 of a cubic foot. 3"" Therefore, 1483° 8′ 8′′ 3′′′′ 1483132 cubic feet. = 1251 Having reduced the cubic feet to bushels, by the method given in art. 288, we have 1192° 6', or 1192 bush. for the required content. Proof by Fractions. = 53° 9'53321.5; 6° 3'6125; 4° 5' 4,5 53; then, subtracting from one of the factors, namely the factor 25, which makes it 20 5, and adding to one of the factors, namely to 215, which makes it 21654; we have, 53 × 54 × 53X2X5238511921 for the numX X 5 ber of bushels. = Parts less than thirds are not noticed in common practice. Examples. 1. What is the content of a cave, the length of which is 12 ft. 9 in.; the breadth 7 ft. 6 in., and the height 8 ft. 3 in.? Ans. 788 c. f. 10' 10" 6"", or 78832 c. f. 2. Suppose that the body of a cart is 7 ft. 2 in. long, inside, 3 ft. 6 in. wide, and 1 ft. 1 in. deep; how many times can the cart be filled from a pit that is to be dug in the earth, 45 ft. 3 in. long, 10 ft. 2 in. wide, and 6 ft. 1 in. deep; allowing the earth in the pit and in the cart to be equally compact; and how many cubic yards of earth does said pit contain? 65 Ans. The cart can be filled 1027741, or nearly 103 times, and the pit contains 1035998, or not quite 104 cubic yards. 344. The transportation of merchandise in ships is called freight, and the freight of all irregular packages, such as bales, boxes, &c. is charged by the cubic foot; the parts less than half a cubic foot being neglected, and those amounting to half à cubic foot or more accounted a cubic foot each; that is to say, the content of each package is calculated to the nearest cubic foot. When the packages are uniform, the freight is charged by the barrel, bag, &c. Examples. Required the content of the following bales, boxes, or packages, the length, breadth, and depth being given in feet and inches : 6. Suppose the bushel to contain exactly 2150 cubic inches, and that the inside of a box is the exact cube of 5 feet; how 5 ft. 7 in. 334 much must the edge of the box be planed down, that it may contain just 100 bushels? Ans., or not quite of an inch. To Measure Cord-wood. 345. The cord by which wood is measured is 128 c. f., or the content of a pile of wood 8 feet long, 4 ft. wide, and 4 ft. high. The content, therefore, of any quantity of cord-wood, is calculated by comparing it with a pile of the above dimensions. Hence, the rule, multiply the length, breadth, and height together, all in feet, and divide by 128, which will give the content in cords. Or rather, set down, in continued multiplication, the length, breadth, and height, as a numerator; and 8 X4 X 4, the dimensions of a cord, as a denominator, and find the result as in multiplication of fractions. Required the content, in cords, of a pile of wood that is 60 ft. long, 16 ft. wide and 10 ft. high. 16 × 60 × 10 4 X 4 X 8 15 X 575 cords. Answer. A cord is sometimes said to contain 8 feet of wood, each foot being consequently, 16 cubic feet. Now, on a base 8 ft. long and 4 ft. wide, it is plain that every 6 inches in height will give 1 foot of wood. For, 8 X4 X 1 = 16 cubic feet. 346. When the dimensions are feet and inches, we may find the content by duodecimals, and divide by 128. What is the content of a pile of wood, the length of which is 20 ft. 9 in., the breadth 3 ft. 7 in., and the height 9 ft. 8 in. ? 20° 9' We here divide by 128, for cords, and the remainder by 16, for feet of wood. As we only calculate to the nearest foot of wood, in dividing by 128, we neglect the parts inferior to cubic feet, which, as they never can equal of a foot of wood, cannot affect the result. In dividing by 16, as the fraction exceeds, we add 1 to the feet, which gives for the content 5 cords, 5 feet, or 5ğ cords, nearly. Examples. 16 1. What is the content of a pile of wood, 24 ft. long, 20 ft. broad, and 18 ft. high? Ans. 671 cords. 2. What is the content of a pile, 17 ft. 8 in. long, 9 ft. 4 in. broad, and 10 ft. 7 in high? Ans. 13 C. 5 f., or 13g cords. 3. What is the content of a pile, the length being 101 ft., the breadth 17 ft., and the height 8 ft. ? Ans. 107 C. 23 ft., or 107,5 cords. 4. What is the content of a pile, the length being 203 ft., the breadth 14 ft. 9 in., and the height 7 ft. 6 in. ? Ans. 175 C. 31 ft., or 175 cords, nearly. Board Measure, &c. In this is included the measurement of planks, scantling, rafters, and hewn timber. To measure boards, the usual thickness being one inch : 347. The length, in inches, multiplied by the breadth, in inches, will evidently give the number of square inches in the surface of the board. This divided by 144, the number of square inches in a square foot, will give the content of the surface of the board in square feet. Put the length of the board, in feet; then, X 12 = the length, in inches; now put w = the width, in inches, then Xw X 12 the content of surface, in square inches, and 1 X w X 12 1 X w or, 12 × 12 1 X w X 12 144 = 12 content in square feet. Now, if 7=12, that is, if the board is 12 feet long, then = 12 =w; that is, the number which shows the width in inches, is the number of square feet contained in the surface of a board, when the board is twelve feet long. Again, if w12, then 12 =7, or the length. |