The width and depth given, to find how much in length to equal a solid or cubical foot. RULE. Divide a solid foot, or the number of inches that a solid foot contains, by the product of the width and depth in inches, and the quotient will be the required length of the same denomination as the dividend. EXAMPLES. 1. If a stick of scantling be 6 inches in depth, and 6 inches in width, how many feet in length will it require to equal 1 cubic foot? 2. Of scantling that is 8 inches in depth and 6 inches in width, how many feet in length will it require to equal 10 solid feet? Ans. 360in. or 30 feet. OBLONG SQUARE. The quantity and one side given, to find the other side. RULE. Divide the quantity reduced to the same denomination of the side given, by the given side, and the quotient will be the length of the side required. EXAMPLES. 1. A sells half an acre of ground to B, and B is to have 8 rods in front-required the length, (or the other side.) an acre 80÷8=10 rods length. Ans. 2. Suppose a person to buy 10 square rods of ground on Market-street, in Bridgeton, to have 30 feet front-required the other side, or length back? Ans. 90.75, or 90ft. 9in. PAVING AND PLASTERING. Paving and Plastering are measured by the yard. Multiply the length in feet by the breadth in feet, and divide by 9 for the answer in yards. To ascertain the number of bricks or stone, to any number of square feet given, Reduce the length and breadth of the square feet given, H2 to inches separately, multiply one by the other; divide by the product of the length and breadth of the brick. Or, divide by the length of the brick, and that quotient by the width of the brick, for the number of bricks. EXAMPLES. 1. Suppose a gentleman's yard, 30 feet square, to be paved with brick 8 inches long, and 4 inches in width-I require the number o square yards, the cost of paving, at 2 cents per square yard, ard the number of bricks it will require. Ans. 100sq.yds.; cost, $2; and 4050 bricks. 2. A certain yard, 15 feet square, is to be paved with stone that are 6 inches square-I require the number of stone, the number of square yards, and the price of the paving at 4 cents per yard. Ans. 900 stone; 25sq.yds.; and the paving, 1 dollar. 3. What will the plastering of a room come to, at 10cts. per square yard, measuring as follows, viz. two sides 14.5 feet by 8.5 feet each, and two ends 12 feet 6 inches by 8.5 feet each, with a deduction for a door that is 6 feet long by 34 wide, ond 4 windows 3 feet by 4 each. Ans. $4.244+ The number of brick and size given, to find the admeasurement of the pavement in square inches, feet, or yards. Multiply the number of brick by the product of the length and breadth of one brick; the quotient will be the square inches; divide the inches by 144 for feet, or 1296 for yards. EXAMPLE. 1. Suppose a certain pavement contain 4050 bricks, 8 inches long, and 4 inches wide-I require the number of square yards? Ans. 100 square yards. SHINGLE OR ROOF MEASURE. To calculate the number of shingles for a roof. 1. Reduce the length and breadth of the space to be roofed to inches, separately. 2. Divide the breadth by the average width of the shingles, and the quotient will be the number of shingles in one course. 3. Divide the length by the number of inches you intend laying the shingles, or courses to the weather, and the quotient will be the number of courses. 4. Multiply the number of courses and the number of shingles in one course, and you will have the number of hingles required. EXAMPLES. 1. Required the number of shingles for a roof 16ft. square, the shingles to average 5in. in width each, and the courses to run 8in. to the weather. Ans. 921.6 shingles. 2. How many shingles will it require to make a roof 26 feet in length and 22 feet in breadth, with shingles 5 or 5.5 inches wide, and of such a length as to admit them to be laid 8 inches to the weather? Ans. 1872. 3. Required the number of shingles to roof a shed 28ft. in length and 20ft. in breadth, the shingles 5.5in. wide, the courses to run 8in. to the weather. Ans. 1832.744=18324+ CIRCLE MEASURE. The diameter of a circle being given, to find the circumference. RULE 1. As 7 is to 22, so is the diameter to the circumference. That is, multiply the diameter by 22, and divide by 7; this gives the circumference, nearly. RULE 2. As 113 is to 355, so is the diameter to the circumference. That is, multiply the diameter by 355, and divide by 113. RULE 3. Multiply the diameter by 3.1416, and the product will be the circumference, nearly. EXAMPLE. If the diameter (A B) of a circle be 154, what is the circumference? By Rule 1st, 484. Rule 2d, 483.8+ Rule 3d, 483.8+ Circumference being given, to find the diameter. RULE 1. As 22 is to 7, so is the circumference to the diameter. That is, multiply the circumference by 7 and divide by 22. RULE 2. As 355 is to 113, so is the circumference to the diameter. RULE 3. Divide the circumference by 3.1416, and the quotient will be the diameter. EXAMPLES. 1. If the diameter of a circle is 154, what is the circumference, by Rule 1st. Ans. 484. 2. If the circumference is 484, what is the diameter, by Rule 1st. Ans. 154. NOTE. It is found by calculation, that in a circle of which its diameter is 1, the circumference is 3.14159, and its area .7854. If the area .7854 be divided by the square of (3.14159) its circumference, it gives .07957. One-sixth of the diameter is .16667, which multiplied by 3.1416 (circumference) gives .5236. The solid content of all bodies, which are of a uniform bigness throughout, whatever may be the form of the ends, is found by multiplying the area of one end into its height, or length. To find the area of a globe or ball, multiply the whole circumference by the whole diameter.-The area of a globe multiplied by one-sixth of its diameter gives its solid contents. To find the area of a circle. RULE 1. Multiply half the circumference by half the diameter, and the product will be the area. RULE 2. Take of the product of the whole circumference and diameter. EXAMPLES. 1. What is the area of a circle whose diameter is 42 and circumference 131.946? Ans. 1385.433. ROUND TIMBER, &c. The cubical or square content of round timber, grindstones, or grain in a circular bin, is found by multiplying the superficial area by the thickness or length. EXAMPLES. 1. Required the cubical and square feet in a round log, 30 feet in length, and its circumference or girth 3 feet 3 inches or 3.25 feet. Operation. 22: 7:: 3.25: 1.034 diameter. 1.625.517 × 30=25.20375=251c.ft. and 25.20375× 2. Suppose a round stick of timber of equal size from one end to the other, whose diameter is 21 inches, and length 20 feet-required its solid content and square feet of an inch thick. Ans. 48.125c.ft. and 577.5sq.ft. 7:22:21: 66in.=5.5ft.2-2.75ft. and 21in.÷2= .875ft. &c. 3. Required the solid content of a roller 2 feet in diameter, and 6 feet in length. Ans. 20.4+c.ft. Measurement of a sphere or globe, or any round solid body. RULE 1. Cube the axis or diameter, and multiply its cube by .5236 for the solid content. RULE 2. Multiply the circumference by the diameter for the superficial content; then multiply the surface by one-sixth of the diameter, and it will give the solid content. RULE 3. Take 1 of the cube of its axis for the solid content. EXAMPLE. What are the solid content of a sphere or globe whose diameter is 20? 20 × 20×20=8000×.5236-4188.8 solid content. |