70. A cylinder is a body of uniform diameter throughout its entire length, whose ends are equal parallel circles. (Fig. 30.) FIG. 30. 71. The altitude of a prism, or of a cylinder, is the perpendicular distance between its bases. 72. A right prism is one whose sides are perpendicular to the bases. 73. A right cylinder is one in which the line joining the centers of the two circular bases is perpendicular to those bases. 74. In the case of plane figures, we have had to do with perimeters and areas. In the case of solids, we have to do with the areas of their outside surfaces, and with their contents or volumes. 75. The entire surface of any solid is the area of the whole outside of the solid. 76. The convex surface of a solid is the same as the entire surface, except that in the case of prisms and cylinders the areas of the ends are not included. 77. Rule.-To find the convex surface of a prism or cylinder, multiply the perimeter of the base by the altitude. EXAMPLE.-A block of marble is 24 inches long and its ends are 9 inches square; what is the area of its convex surface? = SOLUTION.- 9 X 4 = 36 in. = the perimeter of the base; 36 × 24 864 sq. in., the convex area. Ans. 78. To find the entire area of the outside surface, add the areas of the two ends to the convex area. Thus, in the last example, the area of the two ends 9 × 9×2 = 162 square inches; 864 + 162 1,026 square inches. = = EXAMPLE. (a) What is the convex surface of a cylindrical tank with flat ends 23 feet long and 4 feet 6 inches in diameter? (b) What is the entire surface? SOLUTION.-Perimeter of end = 43.1416 = 14.137 ft. (a) Convex surface = 14.137 × 23 = 325.151 sq. ft. Ans. Area of one end = .7854 X (4})* = 15.904 sq. ft. (b) Entire surface = 325.151 +2 X 15.904 = 356.959 sq. ft. Ans. 79. The volume of a solid is the quantity of space it occupies. As shown in Arithmetic, Part 4, the measuring unit is a cube whose edges are equal in length to a linear unit; it may be a cubic inch, cubic foot, cubic yard, or cubic meter. Fig. 31 represents a rectangular prism 4 feet long, 3 feet wide, and 2 feet thick. Dividing the prism by lines, as shown, it is seen that there are four equal slices, each of which is made up of 2×3 6 cubes. In all, there are 4 X 2 X 3 = 24 cubes, each containing 1 cubic foot; that is, the volume of the prism is 24 cubic feet. It is seen that the number of cubes in each horizontal layer is just equal to the number of square feet in the base; and the number of layers is equal to the number of feet in the altitude. The same reasoning holds true for a prism with triangular base, or for a cylinder. FIG. 31 = 80. Rule. To find the volume of a prism or cylinder, multiply the area of the base by the altitude. In applying this rule, all dimensions must have the same unit. EXAMPLE 1.-A packing box is 4 feet long, 4 feet wide, and 34 feet deep; what is its volume? = SOLUTION-Area of base 4X4 18 sq. ft. Altitude = Volume, or cubical contents = 18 × 3: 58 cu. ft. Ans. = 34 ft. EXAMPLE 2.-(a) How many cubic feet of water will a circular cistern contain that is 8 feet in diameter and 10 feet deep? (b) How many gallons will the cistern hold? SOLUTION. (a) The problem is to find the volume of a cylinder whose altitude is 10 ft., and whose bases are 8 ft. in diameter. (6) According to Arithmetic, Part 4, 1 gal. contains 231 cu. in. 502.65 X 1,728 Hence, the cistern can hold 3,760 gal., very 231 nearly. Ans. = 81. The dimensions of a rectangular solid are spoken of as length, breadth, and thickness. According to Art. 79, the volume of the solid is the product of these three dimensions. EXAMPLE. A brick is 8 inches long, 4 inches wide, and 2 inches thick; what is its volume? = SOLUTION.-Volume = length X breadth X thickness 64 cu. in. Ans. = 8 X 4 X 2 MASONRY. 82. In estimating the cubical contents of stone walls, the perch of 24 cubic feet is used. As stated in Arithmetic, Part 4, the perch is often assumed to be 25 cubic feet. 83. Rule.-To find the number of perches of masonry in a wall, divide the volume of the wall in cubic feet by 243. EXAMPLE. How many perches in a wall 8 rods long, 4 feet high, and 2 feet thick? 84. = = 8 × 16 1,188 cu. ft. In estimating the contents of stone foundations for buildings, the length of the wall is measured on the outside, ~-20 FIG. 32. thus counting each corner twice. This is illustrated in Fig. 32. If a wall 2 feet thick measures 12 feet by 20 feet on the outside, and we assume that the corners are parts of the longer sides, we have two walls each 20 feet long, and two walls each 8 feet long. = The actual length is, therefore, 2 x 20 + 2 x 8 56 feet. = The length estimated on the outside is 2 X 20+ 2 x 12 64 feet. To find the actual length of such a wall, subtract four times the thickness of the wall from the length measured on the outside. Thus, in the above case, actual length = 64 - 4 X 2 56 feet. = Usually, masons make no allowance for windows or doors in estimating their work. In estimating the quantity of stone required for the wall such allowance should be made. EXAMPLE.— (a) How many perches of stone are required to build the walls of a church 60 feet long by 32 feet wide, the walls being 24 feet high and 24 feet thick? There are 8 windows, each 5 feet wide and 11 feet high, and 2 doors, each 6 feet wide and 9 feet high. (6) What is the cost of laying the walls at $3.50 per perch? SOLUTION. = = = 2 × 60+2 × 32 = 175 ft. = 184. Length of wall (outside) 175 X 24 X 24 = 9,450 cu. ft. = = 332. Ans. (b) Since in estimating the cost of the work, no allowance is made for corners, doors, and windows. 85. In brickwork, the unit of measurement is one thousand (M) bricks. The dimensions of an ordinary brick are 8 in. X 4 in. x 2 in. In some localities, they are made smaller; in others, larger. To allow for mortar,inch is added to the length and to the thickness in making calculations. On this assumption, the ordinary brick with its mortar has a volume of 8 X 4 X 2 74 cubic inches. Since a cubic foot contains 1,728 cubic inches, it takes 1,728 ÷ 744 231 bricks to make a cubic foot of wall. = 86. Rule. To find the number of ordinary bricks in a wall, multiply its volume in cubic feet by 23. EXAMPLE. How many bricks are required in a wall 80 feet long, 16 feet high, and 4 feet thick? 87. In estimating the cost of brickwork, it is customary in most localities to use the outside, or gross, length of the wall, and to allow for doors and windows. The practice, however, is not uniform, and in some cases no allowance is made for corners or openings. EXAMPLE.-What will be the cost of erecting the walls of a building 64 feet long and 40 feet wide, the wall being 36 feet high and 3 bricks (= 1 ft.) thick? Allowance is to be made for 40 windows, each 6 ft. X 2 ft. 9 in., and 8 doors, each 8 ft. × 3 ft. 6 in. The bricks cost $5.75 per M, and the laying costs $1.40 per M, based on outside length of walls. SOLUTION.-Outside length of wall = 64 × 2 + 40 × 2 = 208 ft. Net length of wall = 2084 X 1 = 204 ft. = EXAMPLES FOR PRACTICE. 88. Solve the following examples: 1. Find the cost of building a stone wall around a rectangular yard 160 feet long and 108 feet wide. The wall is 9 feet high and 2 feet 6 inches thick, and the price of laying is $2.25 per perch. Ans. $1,096.36. 2. How many thousand bricks are required for a house 18 feet wide, 38 feet long, and 32 feet high, walls 3 bricks thick, making allowance for 3 doors, each 3 ft. 4 in. x 7 ft. 6 in., and 16 windows, each 3 ft. by 6 ft.? Ans. 71.983 M. 3. Philadelphia bricks are 8 in. × 41 in. × 2 in. Allowing inch on length and thickness for mortar, how many of these bricks are required to make a cubic foot? Ans. 18, nearly |