The Axis of a cylinder is a straight line passing through the cylinder and terminating at the centers of the two bases. (In Fig. 44 the line a b is the axis of the cylinder, and in Fig. 45 the line eƒ is the axis.) The Altitude of a cylinder is the perpendicular distance between the two bases. (In Fig. 44 a b is the altitude, and in Fig. 45 g h is the altitude.) The Diameter of a cylinder is a straight line perpendicular to and passing through the axis, and terminating at both ends at the convex surface. (In Fig. 44 c d and c1 d1 are diameters of the cylinder, and in Fig. 45 ik is a diameter.) The Circumference of a cylinder is the circumference of any plane that is perpendicular to its axis. (In Fig. 46 a b c d, a1 b1 c1 d1 and a2 b2 c2 d2 are circumferences of the cylinder. In Fig. 45 i j k l is the circumference of the cylinder.) A Lune or Crescent is a plane bounded by two arcs of a circle. (In Fig. 51 is shown a portion of a cylinder whose cross-section a cbd is a lune.) 144.-Propositions Relating to Cylinders. Prop. XXXVI.-The area of the convex surface of a cylinder is equal to the circumference, multiplied by the length of the axis. Prop. XXXVIII.—The volume of a segment of a cylinder is equal to the area of the segment, of the circular plane perpendicular to its axis, multiplied by the length of the axis. Prop. XXXIX.-The area of a lune is equal to the difference between the areas of the segments of the two circles whose arcs form the lune, and which arcs have a common chord. In Fig. 49 suppose the lune abcd and its component circles, etc., have the following dimensions: Radius gc and ga of larger circle 77 in.; radius fc and fa of smaller circle 52 in.; chord a c 102 in.; perpendicular ej = 10.1 in.; perpendicular e g = 57.7 in.; angle a gc=83'; angle a fc = 162'. d FIG. 49 In the same manner we find the area of the segment a bce: Radius of smaller circle fc (52 in.) X 2 diameter of smaller circle (104 in.). The area of smaller circle = square of diameter (1042 = 10,816) X .7854 = = 8,494.8864 sq. in. The sector a b c f is measured by the angle a fc (1622), therefore, the sector a b c f = of the area of the circle (8,494.8864 sq. in.) or 3.822.6988 sq. in. The area of the triangle a c f = the base a c (102 in.) X half the altitude e ƒ (5.05 in.) = 515.1 sq. in. The area of the segment abce the area of the sector a b c f (3,822.6988 sq. in.) the area of the triangle a cf (515.1 sq. in.) 3,307.5988 sq. in. 162 360 We have now found the areas of the segments a dce and a bce. The area of the lune a b c d the area of the segment a b c e (3,307.5988 sq. in.) — the area of the segment a dce (1,351.7537 sq. in.) 1,955.8451 sq. in. 145.-Problems Involving Areas and Volumes of Cylinders. Prob. 39. Suppose Fig. 44 to be a cylinder of the following dimensions: The axis a b 240 in.; the circumference c1e d1ƒ: area of the convex surface of the cylinder? = = 500 in.; what is the Ans. 120,000 sq. in. From Prop XXXVI. The circumference (500 in.) X the axis (240 in.): sq in., which is the area of the convex surface of the cylinder. 120,000 Prob. 40. Suppose a cylindrical steel water tank has the following dimensions: Diameter 20 ft.; depth 30 ft.; number of transverse lap joints 3; number of longitudinal lap joints 1; amount of lap in each joint 3 in.; weight of sheet steel per square foot 10 lb.; weight of rivets used in construction of tank 84 I.; weight of interior brace rods 162 lb.; what is the total weight of the steel water tank? Ans. 22,966.978648 lb. From Prop. XXXVI. The axis of the cylinder, or depth of the tank (30 ft.) + the two lap joints between the convex sheets (3 in. X 2 = 6 in.) 30 ft. 6 in., or 366 in., which is the vertical extent of the convex sheets. The diameter of the tank (20 ft.) X 3.141662.832 ft., or 753.984 in., which is the circumference of the tank. The longitudinal lap joint adds 3 in. to this circumference, which makes 756.984 in. The longitudinal extent of the convex sheets has been shown to be 366 in., which multiplied by the circumference the longitudinal lap joint of 3 in. (756.984 in.) = 277,056.144 sq. in., which is the area of the steel in the convex sheets. The diameter of the bottom of the tank (20 ft.) the turned-up transverse lap joint (3 in. X 2 = 6 in.) = 20 ft. 6 in., or 246 in., which is the diameter of the bottom sheet. The square of the diameter of the bottom sheet (2462 = 60.516) X .7854 = 47.529.2664 sq. in, which is the area of the steel in the bottom sheet of the water tank. The weight of the sheet steel per square foot (10 lb.) the number of square inches in a square foot (144) = .07 ib. (nearly), which is the weight of 1 sq. in. of the sheet steel = Prob. 41. if the water tank described in Prob. 40 was half full of water, considering that 231 cu. in. = 1 gal., how many gallons of water would be in the tank, making no allowance for the water displaced by the interior brace rods? Ans. 35,251.2 gal. From Prop. XXXVII. The depth of the water, or axis of the one-half the total depth of the tank (30 ft. ÷ 2 = 15 ft.), or 180 in. cylinder of water, is The diameter of the cylinder of water is 20 ft., or 240 in. The square of the diameter (24057.600) X .7854: 45.239.04 sq. in., which is the area of the surface of the water. The area of the surface of the water, or the area of the cross-section of the cylinder of water (45.239.04 sq. in.) X the depth of the water, or axis of the cylinder of water (180 in.)8, 143,027.2 cu. in., which is the volume of the cylinder of water. The volume of water in cubic inches (8,143,027.2) the number of cubic inches in 1 gal. (231) = 35.251.2 gal., which is the number of gallons of water in the tank. Prob. 42.-If a cylindrical oil tank has the following interior dimensions: Length 38 ft.; diameter = 6 ft.; how many gallons of oil will it hold? Ans. 8,037.2736 gal. From Prop. XXXVII. The area of a cross-section, or of a circumference, of the oil tank is the square of the diameter (62: 36) X .7854 28.2744 sq. ft. The area of the cross-section (28.2744 sq. ft.) X the length of the tank 38 ft.) 1,074.4272 cu. ft., which is the capacity of the oil tank, or volume of the cylinder in cubic feet, which multiplied by the number of cubic inches in a cubic foot (1,728) = 1,856,610.2016 cu. in. This divided by the number of cubic inches in 1 gal. (231) = 8,037.2736 gal., which is the number of gallons the tank will hold. Prob. 43. If in the oil tank described in Prob. 42 it was found that the depth of the oil at the center, or deepest part, was only 28 in., and the tank was perfectly level so that this depth was the same the entire length of the tank, how many gallons of oil would there be in the tank? cr 1,719.08352 sq. in. The base a c of the triangle a gc (72 in.) X one-half the altitude gh (4 in.) == 288 sq. in., which is the area of the triangle a &c. The area of the sector @gcf (1,719.08352 sq. in.) - the area of the triangle a gc (288 sq. in.): 1.431.08352 sq. in., which is the area of the segment of the circle, or cross-section of the body of oil a cf. The area of the cross-section of the body of oil (1,431.08352) X the length of the oil tank (456 in.) 652,574.08512 cu. in.), which is the volume of oil in the tank. The volume of oil (652.574.08512 cu. in.) the number of cubic inches in 1 gal. (231) = 2,825 gal. (nearly), which is the number of gallons of oil in the tank. NOTE. If the tank was more than half full of oil it would be necessary to find the volume of the upper, or empty space, and then subtract that volume from the volume or capacity of the entire tank. When "Down Aroun' the Depo' When the Keers Cum In" Down aroun' the depo' when the keers cum in. What a hustle an' a bustle an' a clatter an' a din, Engine kinder puffin', an' a blowin' off its steam, Drayman sorter fussin' an' a cussin' at his team. Boy a sellin' papers an' a shoutin' out the 'Nother one a wantin' fer to blacken up cum in. the depo' when the keers Down aroun' the depo' when the keers cum in, Folks 'at never crack a smile an' sum 'at allus grin, Settin' there a-waitin' fer to hear the whistle blow, Some a-wishin' they could stay, an' some 'at they could go; A A woman dressed in mournin'; 'nother as a bride, banker an' a beggar a-settin' side by side, Some 'at never loses an' a lot 'at never win, Down aroun' the depo' when the keers cum in. Down aroun' the depo' when the keers cum in, Ever stir a lot of ants an' see 'em all begin, A-runnin' here an' ever'wher' 'sif they didn't know Which way they thought they orter or hadn't orter go? Well, that's the way with people, fer purt nigh every day, I go down there an' see 'em a-doin' that away; Ain't like any other place 'at I have ever .bin, Down aroun' the depo' when the keers cum in. -Anon. |