To find the volume of a prismoid. 122. A PRISMOID is a frustum of a wedge. Let L and B denote the length and breadth of the lower base, 7 and b the length and breadth of the upper base, M and m the length and breadth of the section equidistant from the bases, and h the altitude of the prismoid. m M L Through the edges L and l', let a plane be passed, and it will divide the prismoid into two wedges, having for bases the bases of the prismoid, and for edges the lines L and l'. The volume of the prismoid, denoted by V, will be equal to the sum of the volumes of the two wedges; hence, or, V = ¿Bh (l + 2L) + fbh (L + 21); V = h (2BL + 2b1 + Bl + 6L); Because the auxiliary section is midway between the bases, we have 2M = L + 1, and 2m = B+b; hence, 4Mm = (L + 1) (B + b) Substituting in (A), we have V = h (BL + bl + 4Mm). But BL is the area of the lower base, or lower section, bl is the area of the upper base, or upper section, and Mm is the area of the middle section; hence, the following. RULE.-To find the volume of a prismoid, find the sum of the areas of the extreme sections and four times the middle section; multiply the result by one sixth of the distance between the extreme sections; the result will be the volume required. This rule is used in computing volumes of earth-work in railroad cutting and embankment, and is of very extensive application. It may be shown that the same rule holds for every one of the volumes heretofore discussed in this work. Thus, in a pyramid, we may regard the base as one extreme section, and the vertex (whose area is 0), as the other extreme; their sum is equal to the area of the base. The area of a section midway between them is equal to one fourth of the base: hence, four times the middle section is equal to the base. Multiplying the sum of these by one sixth of the altitude, gives the same result as that already found. The application of the rule to the case of cylinders, frustums of cones, spheres, &c., is left as an exercise for the student. Examples. 1. One of the bases of a rectangular prismoid is 25 feet by 20, the other 15 feet by 10, and the altitude 12 feet required the volume. Ans. 3700 cu. ft. 2. What is the volume of a stick of hewn timber, whose ends are 30 inches by 27, and 24 inches by 18, its length being 24 feet? Ans. 102 cu. ft. MENSURATION OF REGULAR POLYEDRONS. 123. A REGULAR POLYEDRON is a polyedron bounded by equal regular polygons. The polyedral angles of any regular polyedron are all equal. 124. There are five regular polyedrons (Book VII., page 219). To find the diedral angle contained between two consecutive faces of a regular polyedron. 125. As in the figure, let the vertex, O, of a polyedral angle of a F A B tetraedron be taken as the centre of a sphere whose radius is 1: then will the three faces of this polyedral angle, by their intersections with the surface of the sphere, determine the spherical triangle FAB. The plane angles FOA, FOB, and AOB, being equal to each other, the arcs FA, FB, and AB, which measure these angles, are also equal to each other, and the spherical triangle FAB is equilateral. The angle FAB of the triangle is equal to the diedral angle of the planes FOA and AOB, that is, to the diedral angle between the faces of the tetraedron. In like manner, if the vertex of a polyedral angle of any one of the regular polyedrons be taken as the centre of a sphere whose radius is 1, the faces of this polyedral angle will, by their intersections with the surface of the sphere, determine a regular spherical polygon; the number of sides of this spherical polygon will be equal to the number of faces of the polyedral angle; each side of the polygon will be the measure of one of the plane angles formed by the edges of the polyedral angle; and each angle of the polygon will be equal to the diedral angle contained between two consecutive faces of the regular polyedron. To find the required diedral angle, therefore, it only remains to deduce a formula for finding one angle of a regular spherical polygon when the sides are given. E D Let ABCDE represent a regular spherical polygon, and let P be the pole of a small circle passing through its vertices. Suppose P to be connected with each of the vertices by arcs of great circles; there will thus be formed as many equal isosceles triangles as the polygon has sides, the vertical angle in each being equal to 360° divided by the number of sides. Through P draw the arc of a great circle, PQ, perpendicular to AB: then will AQ be equal to BQ, and the angle APQ to the angle QPB (B. IX., P. XI., C.). If we denote the number of sides of the spherical polygon by n', the angle APQ will be equal to 360° 2n'' or 180° In the right-angled spherical triangle AQP, we know the base AQ, and the vertical angle APQ; hence, by Napier's rules for circular parts, we have denoting the side AB of the polygon by s', and the angle PAQ, which is half the angle EAB of the polygon, by A, we have 180° = n' = 36°, and ts' = 30°; To find the volume of a regular polyedron. 126. If planes be passed through the centre of the polyedron and each of the edges, they will divide the polyedron into as many equal right pyramids as the polyedron has faces. The common vertex of these pyramids will be at the centre of the polyedron, their bases will be the faces of the polyedron, and their lateral faces will bisect the diedral angles of the polyedron. The volume of each pyramid will be equal to the product of its base and one third of its altitude, and this product multiplied |