Example. Required the tonnage of a ship of war, length 100 feet, breadth 32 feet, and depth 16 feet. Ans. 512 tons. Note 2.-To find the length of the mast of a ship, add the breadth of the beam to two-thirds the length of the keel, and the sum will be the length of the main mast. Example. 1. What is the length of the main mast for a ship of 105 feet keel, and the breadth of the beam 38 feet? Ans. 108 ft. STONE MEASURE. Stone and stone work, or mason's work, is measured by the standard perch of 24.75 cubical or solid feet, which is 16 feet long, 1 feet wide, and 1 foot high. RULE. 1. Divide the continued product of the length, width and height in feet by 24.75, and the quotient will be the number of perches. Ör, divide the continued product of the length and height, in feet, and width in inches, by 297, and the quotient will be the number of perches required. If the wall be no more than the standard thickness, multiply only the length and height together, and divide the product by 16.5. 2. Place the length and height of the wall in feet, and the width in inches, for dividends, and the numbers 3, 9 and 11, for divisors; divide each of the dividends by one of the divisors, and the continued product of the quotients will be the number of perches required. It makes no difference in the result which divisor and dividend be used together; the work will, however, be abridged, by using such together as will leave no remainder, (if there be any such.) Examples. 1. Required the number of perches in a pile of stone 27 ft. long, 9 feet high, and 55 inches wide. 3279 9 91 quotients. 11 555 9x1x5=45 perches. Ans. 2. Required the number of perches in a wall 94.5 ft. long, 22 feet high and 21 inches wide. 10.5×2×7=147 perches. Answer. 3. What quantity of stone is there in a wall 64 feet long, 24 feet high and 20 inches wide? 3 | 647.11 =64÷ 9 9 248 =248 11 20 1.818=20÷11 7.11x8x1.818=103.4 perches. Answer. Note 1.-In mason's work, the dimensions of a building are taken from corner to corner, that is, the girth of a building is reckoned the length of the wall. The dimensions of chimnies are taken separately, the girth and half girth of which is reckoned the length of the wall. Note 2.-When the wall is less than 1 feet in width, no deduction is to be made, but reckoned at the standard thickness; nor is any deduction to be made for doors, windows and corners, except in reckoning the quantity of stone actually made use of, when all doors, windows, &c. are deducted. 4. A certain building measures as follows, viz: side walls each 90 feet 6 inches, ends each 74 feet 3 inches; thickness of each 20 inches; a partition wall 70 feet 11 inches, and 1 foot thick, and all 18 ft. 9 in. high; in the outside walls are four doors, 7 ft. 3 in. by 3 ft. 3 in. each, and twelve windows 5 ft. 4 in. by 3 ft. each, and in the partition wall are 2 doors, 6 feet 6 in. by 3 feet 2 in. each-Required the number of perches of mason's work, also the number of perches of stone. Answer. S Mason's work, 496.622 perches. Stone, 439.242 do. To measure stone in a well. Add the thickness of the wall to what it measures in the clear; then say, As 7 is to this sum, so is 22 to the circumference or length of the wall. Examples. 1. Required the number of perches of stone in a well. whose diameter in the clear is 3 feet 6 inches, thickness of the wall 18 inches, and depth 44 feet 11 inches. Ans. 423 perches. 2. A certain well measures 4 feet 9 inches in the clear, the wall is 15 inches thick and 48 feet deep-Required the number of perches of stone. Ans. 45.7 perches + 3. Required the number of perches of stone in a well 30 feet deep, diameter in the clear 2 feet 4 inches, and thickness of the wall 1 foot 2 inches. Ans. 15.555 perches + PAVING AND PLASTERING. Paving and plastering are measured by the square yard. RULE. Reduce the area of the pavement to square inches, and divide this by the product of the length and breadth of the brick or stone. Or, divide by the length of the brick, and that quotient by the breadth, for the number of bricks. Or, multiply the length of the pavement by the breadth, and divide by 9, for the area in square yards. Examples. 1. A certain yard 16 feet square, is to be paved with brick 9 inches long and 4 broad; required the number of bricks, the number of square yards, and price of paving at 8 cents a square yard. 16×1612724, and 2721÷9=304 sq. yards. = area of a brick. 39204-401968=number of brick. 30 ×.08=82.42 = price of paving. 2. What will the plastering of a room come to at 9 cents a square yard, and measuring as follows, viz. two sides 13.25 feet by 8.75 feet each, and two ends 10.5 by 8.75 feet each, with a deduction for a door that is 6 feet long by 3.25 feet wide, and 3 windows 3.25 by 4 feet each. Ans. $3.571. 3. Suppose a certain pavement contains 3240 stones, 10 inches long and 8 inches wide, I require the number of square yards. Ans. 200 sq. yards. SHINGLE OR ROOF MEASURE. To compute the number of shingles for a Roof. RULE. 1. Divide the breadth of the space to be roofed, in inches, by the average width of the shingles, and the quotient will be the number of shingles in a course. 2. Divide the length of the space to be roofed, in inches, by the number of inches the shingles are to be laid to the weather; this quotient will be the number of courses. 3. Multiply the number of shingles in a course by the number of courses, and the product will be the number of shingles required. Examples. 1. How many shingles will be required for a roof 25 feet square, the courses to be laid 7 inches to the weather, and the shingles to average 6 inches in width? 25 feet = 300 inches. 3006=50 number of shingles in a course. and 50×40=2000 Answer. 2. Required the number of shingles it will take to make a roof 24 feet long and 20 feet broad, the shingles to be laid 8 inches to the weather, and to average 71⁄2 inches in width. Ans. 1152. 3. How many shingles will be required to roof a house 40 feet in length and 36 in breadth, the average width of the shingles 4 inches, and to be laid 10 inches to the weather? Ans. 4608. To find the side of a square piece of timber, that may be hewn or sawed from a round piece. RULE. the Extract the square root of half the square of the diameter. Or, multiply the girth by 9 and divide by 40, for the side of the square. Or, multiply the girth by 2 and divide by 9, quotient will be the side of the square, near enough for common purposes. Examples. 1. The girth of a tree is 8 feet 9 inches-I wish to know the side of a square piece of timber that may be hewn from it. 8 ft. 9 in. x9=78 ft. 9 in. and 78 ft. 9 in.÷401 ft. 11 in. Answer. Or, 8 ft. 9 in. x2=17 ft. 6 in. and 17 ft. 6 in. 9-1 ft. 11 in. 2. I wish to procure a piece of timber inches-What girth will be required? 20 in. x9=180; Answer, nearly. that will square 20 and 1802-90 in. 7 ft. 6 in. Answer, nearly. 3. The diameter of a tree is 30 inches-How much will it square? 30×30=900; and 9002-450, 450-21.2 in. Answer. MECHANICS. OF THE LEVER. There are three varieties of the lever, whereby the prop, moving power or weight may be applied differently to the inflexible bar or vectis, so as to effect mechanical operations in a convenient manner. Of the Lever of the first order. A lever of the first order has the power applied at one end, the weight to be raised at the other, and the prop or fulcrum at some point between them, as the common handspike or steelyards; therefore, the power applied at one end of this order of levers will be reciprocally proportional to the distance of the fulcrum or prop from those ends: Or, as the distance from the point of suspension is from the weight. Examples. 1. What weight will a person be able to raise who presses with the force of 120 pounds on the end of an equipoised handspike 12 feet long, which is to meet with a convenient prop, just 9 inches above the end of the handspike? |