In like manner, since P3 and λ3; and P3 and 90-λ, are the polar distances and longitudes of the faces 1 m n and m 1 n referred to C, as north pole, and C,C,C, as equator, 2 cos &=cos p, cos p3+sin p, sin p, cos (90-2 λg), which gives as above Referring to (fig. 40*, Plate IV.*), and remembering from $117, that P1=f C3 P2=f C2 P3 fC1 From the spherical triangle ƒg C3, we have sin f C3 sin fg sin P1 . sin (90-P3)_cos P3 sin fg C sin f C39 or sin 90 Therefore sin p1 sin λ=cos P. = sin A, Also from spherical triangle Cif Cз, we have sin 1 sin f Ca sin f C1 = sin λ sin (90-λ1) cos λ1 1 P3 sin λ3. sin f CC sinf C ̧С1 Therefore sin P1 cos λ=sin Hence √ sin=sin p1 cos sin p1 sin λ=sin p ̧ sin λ-cos P3 P1 (-06) sin λ=√2 sin+sin and 0, find P3 and A. (Fig. 42*, Plate IV*.)—Let a be the pole of 1mn, a2 that of 1 nm, and a, that of m 1 n. Join a, a, by arc of great circle cutting oC, in ƒ, and a, a, by arc of great circle cutting od in e; also C1, a1 by C1 a1 cutting dC, in g, and Oa, cutting dC, in h. Then Ca1=P3, С39=λ3, C10=54° 44′, and C2od=60°; and let oa1=P, Coa=L. sin L sin L sin sin (60°-L) sin 60° cos L-cos 60° sin L 1 Therefore 3 cot L=1+tan and tan L=√3 cos2 a=tan 60° cos2 a. cos2 a Also from spherical triangle Coa1 we have cos Ca=cos C1o cos oa, + sin C1o sin oa, cos C1oa1; or cos p=cos 54° 44′ cos P+ sin 54° 44' sin P cos (120°+L). cos 54° 44' cos P-sin 54° 44′ sin P cos (60°— L). =cos P [cos 54° 44′-sin 54° 44' tan P cos (60°-L)]. Let tan ẞ=tan P cos (60°-L). Therefore cos p3=cos P {cos 54° 44'-sin 54° 44' tan ẞ} and sin (45°-λ3)= sin P3 for determining P, and X, in terms of 4 and 0; all the formulæ being adapted for logarithmic computation. P3 and A, being determined from the values of p, 0, and 4, m and n can be expressed in terms of p, and A. 127. By the formulæ given in § 124, § 125, and § 126, any two of the angles of inclination such as 4, 0, and 4, over the edges of a six-faced octahedron, having been observed by the goniometer, and λ, can be determined. Again, by formulæ in § 118, 11 and 1, P2 and λ, can be obtained from the values of p, and A.. P3 1'3 and A, being determined, m and n can be obtained. Now all the forms of the cubical system are derived from those of the six-faced octahedron. Hence by determining 0, 4, and 4 for any form of the cubical system, we can obtain the values both of p, and λ ̧, and also of the indices 1, m, and n. As we advance in this treatise we shall show good reasons for preferring the polar circular co-ordinates p, and λ, to the linear ratios or fractions m and n, 128. The problems of crystallography being resolved for the most part into those of spherical trigonometry, may be solved by means of lines drawn on the surface of a solid sphere. This being inconvenient in practice, it is usual to project the points or poles on the surface of the sphere upon those of a plane, just as geographical and astronomical maps are projections from the surface of the sphere upon the plane of the paper on which the map is drawn. There are three principal projections of the sphere, the steregraphic, orthographic, and gnomic. The steregraphic when the eye is supposed to be placed on the surface of the sphere and the points in the hemisphere furthest from the eye are projected on the plane of the equator; considering the point of sight or projection, the pole of the great circle on which the projection is made. In this projection the projections of circles on the sphere are either straight lines or circles. The orthographic where the eye is supposed to be placed at an infinite distance from the sphere. In this projection points on the surface of the sphere are projected on the plane of the equator by perpendiculars from those points to that plane. In this case all great circles inclined to the equator are projected into ellipses on the plane of projection. The gnomic where the eye is placed in the centre of the sphere, and the plane of projection is a plane touching the surface of the sphere. In this projection all great circles are projected into a straight line. From the difficulty of describing arcs of ellipses the orthographic projection is not suited to crystallographical problems. The steregraphic is that mostly used by Professor Miller and other distinguished crystallographers, but there is some trouble in finding the centres of the arcs of great circles on the sphere of projection. The most simple projection for most purposes is the gnomic. By either the steregraphic or gnomic projection, many problems may be very expeditiously solved by simple geometrical con structions. 129. Comparing (fig. 14, Plate II.) with (fig. 27, Plate IV.), we see that if we take 4, the centre of the cube, for the centre of the sphere of projection, and Ao1, 40,, &c., Ao as equal radii of that sphere, the eight faces, C1, C2, C3, &c., of the octahedron will each be tangent planes, touching the sphere in the eight points 01, 0, &c., og. Because each of these plane faces are respectively perpendicular to Ao1, Ao2, &c., at the points 0, 0, &c. The projections on the faces of the octahedron will be the same as in the former case if we regard the sphere of projection as the sphere inscribed in the cube touching the cube in the points C1, C2, &c., C' ̧. All the poles, therefore, of all the forms of the cubical system can therefore be projected on to the planes of the octahedron inscribed in the cube,-one octant of the sphere upon each face. In (fig. 14, Plate II.), as shown in perspective, and (fig. 33, Plate IV.), on the plane of the paper,-the equilateral triangle CCC, represents the gnomic projection of an octant of the sphere of projection. 3 3 CCC, being the projections of three poles of the cube, |