XXXI. IRREGULAR SOLIDS. 304. We will now explain methods by which we may, in some cases, determine the volumes of solids that are not included in any of the Rules which have been given. 305. Suppose the solid is one which will sink in water, but will not be injured by water. Put the solid inside a vessel of convenient shape, such as a rectangular parallelepiped or a cylinder. Pour water into the vessel until the solid is quite covered; and note the level at which the water stands. Remove the solid and note the level at which the water then stands. The volume of the solid is of course equal to the volume of the water which would be contained in the vessel between the two levels; and this can be easily calculated. Or we might state the process thus: fill the vessel full of water; put the solid into it gently and measure the volume of the water which runs over. 306. If the solid is composed entirely of the same substance we may estimate its volume by means of its weight thus: weigh the solid, also weigh a cubic inch of the same substance as the solid; divide the weight of the solid by the weight of the cubic inch, and the quotient will be the number of cubic inches in the solid. If instead of ascertaining the weight of a cubic inch of the substance we ascertain the weight of any known volume of the substance, we can determine by a proportion the volume of the proposed solid. Some examples of the principle that volumes of solids of the same substance are in the same proportion as their weights, have been given at the end of Chap. xxix. 307. A Rule resembling that given in Chapter XVII. may be used for finding approximately the volumes of certain solids: Divide the length of the solid into any even number of equal parts; and ascertain the areas of sections of the solid through the points of division perpendicular to the length of the solid. Add together the first area, the last area, twice the sum of all the other odd areas, and four times the sum of all the even areas; multiply the sum by one-third of the common distance between two adjacent sections. 308. The preceding Rule will in general be more accurate the greater the number of sections that are made; and the Rule ought not to be trusted if the solid be very irregular in form. The difficulty of ascertaining the areas of the sections, except in a few simple cases, renders the Rule of little practical value. 1. The radius of the base of a cylindrical vessel is 10 inches; a block of stone is placed in the vessel and is covered with water; on removing the block the level of the water sinks 6 inches: find the volume of the block. 2. If a cubic foot of marble weighs 2716 ounces: find the volume of a block of marble which weighs 4 tons 8 cwt. 3. A cask full of water weighs 3 cwt.; the cask when empty weighs 40 lbs.: find to the nearest gallon the capacity of the cask. 4. Five equidistant sections of a solid are taken, the common distance being 3 feet; the areas of these sections in square feet are 3.72, 528, 6'96, 877, and 10-72: find the volume of the solid between the extreme sections. 5. Five equidistant sections of a solid are taken, the common distance being 6 inches; these sections are all circles, and their circumferences are respectively 57 inches, 63 inches, 69 inches, 76 inches, 83 inches: find the volume of the solid between the extreme sections. XXXII. SIMILAR SOLIDS. 309. Similar solids are such as are alike in form though they may differ in size. In common language the fact that one solid is similar to another is often expressed by saying that one is a model of the other. 310. All cubes are similar solids. All spheres are similar solids. 311. It is easy in various cases to give tests for determining whether two solids, which are called by the same name, are similar. For example: If the three edges of one rectangular parallelepiped which meet at a point are respectively double, or treble, or any number of times, the three edges of another which meet at a point the two rectangular parallelepipeds are similar. If the height and the diameter of the base of one right circular cone are respectively double, or treble, or any number of times, the height and the diameter of the base of another, the two right circular cones are similar. The same test will serve for two right circular cylinders. We may express these statements thus: Two rectangular parallelepipeds are similar when their edges are proportionals. Two right circular cones, or two right circular cylinders are similar when their heights and the diameters of their bases are proportionals. 312. The following most important proposition holds with respect to similar solids: The volumes of similar solids are as the cubes of corresponding lengths: For example, suppose the diameter of one sphere to be 5 inches, and the diameter of another sphere to be 4 inches; the volume of the first sphere is to the volume of the second as the cube of 5 is to the cube of 4, that is, as 125 is to 64: so that the larger cube is almost double the other. Persons who have not had their attention drawn to such a fact, often find a difficulty in realising the rapid rate at which the volumes of solids increase, as some assigned dimension of them is increased. 313. We will now solve some exercises. (1) The edge of a cube is 1 foot: find the number of feet in the edge of another cube of double the volume. The cube of the required number is to the cube of 1 as 2 is to 1; so that the required number is the cube root of 2: this will be found to be 1.2599210. Thus we see that a cube with its edge 1.26 feet is rather more than double a cube with its edge 1 foot. (2) The height of a pyramid is 12 feet: it is required to cut off a frustum which shall be a fourth of the pyramid. Since the frustum is to be a fourth of the pyramid, the remaining pyramid will be three-fourths of the original pyramid; and these two pyramids are similar. Therefore the cube of the height of the remaining pyramid must be three-fourths of the cube of the height of the original pyramid, that is, of 1728, that is, 1296; thus the height in feet of the remaining pyramid is the cube root of 1296: it will be found that this is 10.9027. Hence the height of the frustum in feet is 12-10-9027, that is, 1.0973. (3) The diameters of the ends of a frustum of a cone are 6 feet and 10 feet, and the height of the frustum is 3 feet: it is required to divide the frustum into two equal parts by a plane parallel to the base. In the diagram of Art. 210, let AB and CD be the diameters of the ends. As in that Article we find OK=45, and OM=7·5. Let OL denote the perpendicular distance of the required plane from O. Thus we shall find that the cube of OL must be equal to half the sum of the cubes of OK and OM. The cube of 45=91.125; the cube of 7.5=421.875. Thus the cube of OL of 513=256'5; and therefore the number of feet in OL the cube root of 2565; it will be found that this is 6.3537. Hence the distance of the required plane from the smaller end of the frustum in feet 6.3537-4.5=1.8537. (4) A frustum of a circular cone is trimmed just enough to reduce it to a frustum of a pyramid with square ends: find how much of the volume is removed. This Exercise may be conveniently placed here although not strictly connected with the subject of similar solids. Suppose that the radius of one end is 2 feet, and the radius of the other end 3 feet; and that the height is 12 feet. By Art. 268 the volume of the frustum of the cone in cubic feet= 1 3 When the frustum of the cone is trimmed the ends become squares, the diagonals of which are 4 feet and 6 feet respectively: by Art. 268 the volume of the frustum of a pyramid in cubic feet x 38 x 12 = 152. The volume removed is = 1 3 therefore 86.7616 cubic feet; and the fractional part of the Now on examining this process it will be immediately seen that the result will be the same whatever be the height of the frustum; and by trial it will be found that the result will also be the same whatever be the radii of the ends of the frustum of a cone. Thus the result is true for any frustum of a right circular cone. |