ner as angles determined by planes are measured by the triangles or polygons, they mark out upon the same, or an equal sphere. In all cases, the maximum limit of solid angles, will be the plane towards which the various planes determining such angles approach, as they diverge further from each other about the same summit; just as a right line is the maximum limit of plane angles, being formed by the two bounding lines when they make an angle of 180°. The maximum limit of solid angles is measured by the surface of a hemisphere, in like manmer as the maximum limit of plane angles is measured by the arc of a semicircle. The solid right angle (either angle, for example, of a cube) is 4 (= #4) of the maximum solid angle: while the plane right angle is half the maximum plane angle. The analogy between plane and solid angles being thus traced, we may proceed to exemplify this theory by a few instances; assuming 1000 as the numeral measure of the maximum solid angle = 4 times 90° solid = 360° solid. 1. The solid angles of right prisms are compared with great facility. For, of the three angles made by the three planes which, by their meeting, constitute every such solid angle, two are right angles; and the third is the same as the corresponding plane angle of the polygonal base; on which, therefore, the measure of the solid angle depends. Thus, with respect to the right prism with an equilateral triangular base, each solid angle is formed by planes which respectively make angles of 90°, 90°, and 60°. Consequently 90° + 90° + 60° – 180° = 60°, is the measure of such angle, compared with 360° the maximum angle. It is, therefore, one-sixth of the maximum angle. A right prism with a square base has, in like manner, each solid angle measured by 90° + 90° 4- 90° – 180° = 90°, which is 3 of the maximum angle. And thus it may be found, that each solid angle of a right prism, with an equilateral triangular base is + max. angle = + , 1000. square base 1S + . . . . . . .... = + , 1000. Hence it may be deduced, that each solid angle of a regular prism, with triangular base, is half each solid angle of a prism with a regular hexagonal base. Each with regular square base = 3 of each, with regular octagonal base, pentagonal = # ................ decagonal, hexagonal = + . . . . . . . . . . . . . . . . duodecagonal, m — 4 im gonal =::=3,............. m gonal base. Hence again we may infer, that the sum of all the solid angles of any prism of triangular base, whether that base be regular or irregular, is half the sum of the solid angles of a prism of quadrangular base, regular or irregular. And, the sum of the solid angles of any prism of tetragonal baseis = $sum of ang.in prism of pentag.base, pentagonal . . . . =# . . . . . . . . . . . . . . . . hexagonal, hexagonal .... = . . . . . . . . . . . . . . . . . heptagonal, m gonal . . . . . . ===} - - - - - - - - - - - - - - (m+1)gonal. 2. Let us compare the solid angles of the five regular bodies. In these bodies, if m be the number of sides of each face; n the number of planes which meet at each solid angle; #O = half the circumference or 180°; and a the plane angle made by two adjacent faces; 2n ... I Sin 2m O for the plane angle formed by every two contiguous faces of the tetraédron, 70°31'42"; of the hexaedron, S0°; of the octaëdron, 109°28' 18"; of the dodecaedron, 116° 33' 54"; of the icosaédron, 138° 11’ 23”. But, in these polyedrae, the number of faces meeting about each solid angle, are 3, 3, 4, 3, 5, respectively. Consequently the several solid angles will be determined by the subjoined proportions: then we have sin #A = • This theorem gives, Solid Angle. 360°:3.70°31'42” – 180°::1000: 87-73611 tetratdron, 360°:3.90° - 180°::1000:250. hexaedron. 360°:4.109°28'18"–360°::1000:216-35185octaedron. 360°:3.116°33'54"–180°::1000:471.395 dodecat dron. 360°:5.138°11'28"—540°::1000:419.30169 icosačdron. 3. The solid angles at the vertiges of cones, will be determined by means of the spheric segments cut off at the bases of those cones; that is, if right cones, instead of having plane bases, had bases formed of the segments of equal spheres, whose centres were the vertices of the cones, the surfaces of those segments would be measures of the solid angles at the respective vertices. Now, the surfaces of spheric segments, are to the surface of the hemisphere, as their altitudes, to the radius of the sphere; and, therefore, the solid angles at the vertices of right cones will be to the maximum solid angle, as the excess of the slant side above the axis of the cone, to the slant side of the come. Thus, if we wish to ascertain the solid angles at the vertices of the equilateral and the right angled comes; the axis of the former is # v3, of the latter, 3 v2, the slant side of each being unity. Hence, Angle at vertex. 1 : 1 – 3 v3:: 1000: 133.97464, equilateral cone, 1 : 1 – 3 v2::1000: 292.89322, right angled cone. 4. From what has been said, the mode of determining the solid angles at the vertices of pyramids will be sufficiently obvious. If the pyramids be regular ones, if N be the number of faces meeting about the vertical angle in one, and A the angle of inclination of each two of its plane faces; if n be the number of planes meeting about the vertex of the other, and a the angle of inclination of each two of its faces: then will the vertical angle of the former, be to the vertical angle of the latter pyramid, as NA — (N – 2) 180°, to na — (n − 2) 180°. If a cube be cut by diagonal planes, into six equal pyramids with square bases, their vertices all meeting at the centre of the circumscribing sphere; then each of the solid angles, made by the four planes meeting at each vertex, will be # of the maximum solid angle; and each of the solid angles at the bases of the pyramids, will be no of the maximum solid angle. Therefore, each solid angle at the base of such pyramid, is one..fourth of the solid angle at its vertex: and, if the angle at the vertex be bisected, as described below, either of the solid angles arising from the bisection, will be double of either solid angle at the base. Hence also, and from the first subdivision of this inquiry, each solid angle of a prism, with equilateral triangular base, will be half each vertical angle of these pyramids, and double each solid angle at their bases. The angles made by one plane with another, must be ascertained, either by measurement or by computation, according to circumstances. But, the general theory being thus explained and illustrated, the further application of it is left to the skill and ingenuity of geometers; the following simple examples, merely, being added here. Ea. Let the solid angle at the vertex of a square pyramid be bisected. 1st. Let a plane be drawn through the vertex and any two opposite angles of the base, that plane will bisect the solid angke at the vertex; forming two trila teral angles, each equal to half the original quadrilateral angle. 2dly. Bisect either diagonal of the base, and draw any plane to pass through the point of bisection and the vertex of the pyramid; such plane, if it do not coincide with the former, will divide the quadrilateral solid angle into two equal quadrilateral solid angles. For this plane, produced, will bisect the great circle diagonal of the spherical parallelogram cut off by the base of the pyramid; and any great circle bisecting such diagonal is known to bisect the spherical parallelogram, or square; the plane, therefore, bisects the solid angle. Cor. Hence an indefinite number of planes may be drawn, each to bisect a given quadrilateral solid angle. Ex. 2. Determine the solid angles of a regular pyramid with hexagonal base, the altitude of the pyramid being to each side of the base, as 2 to 1. Ans. Plane angle between each two la Solid angle at the vertex 89.60648 h The max. angle Each ditto at the base .. 218-19367 being 1000. CHAPTER VII. Logarithmic Computation of Spherical Triangles. 1. For the purposes of exemplifying the rules and formulae in the preceding chapter, and of assisting the student in deducing the logarithmic computations from the analytical expressions, a few problems are here added. |