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AN0ET1IIC SYSTEM. 47 208. a,b,c are the poles of 100,010, 001; m, n any two poles of different forms not in the zone-circle 001,100. Having given mb, nb, mu, and the symbols of m, n, to find the angular elements of the crystal. Let mb, nb, mn meet ca in p, q, b, the symbols of m, n being known, those of p, q, it are known, mb, nb, Mn being known, pq may be found, FIG. 65. tan| (pB — qb) _ sin (nb — mb) tan|(pB + qr) ~~’ sin(NB + mb) ’ ''Yhioh gives pe, qe. Whence ap, cp may be found by (22) or (24). From qa, nb, qo, or pa, mb, pc, the angular elements may be found by (208), 209. To change the axes, Let u, v, w bo the indices of any face, it', v, w' its indices when referred to the axes of the zones porfiio, 001,100, coy,010 as crystallographii e axes ; then Pi‘5 *»*++ u’ = pw — m, v' = v, w’ — (ju — etc. 210. The form 010 has two parallel faces. 211. The form hoi hns two faces parallel to each other, 212. In a combination of the forms hoi, 010, the faces of hoi are perpendicular to the faces of 010. . 213. The form hkl has four faces, in two parallel pairs, form- ln g a zone. 214. In a combination of the forms hkl, 010, normals to any tace of hkl and the adjacent face of 010 make with each other an angle, which is measured by the arc pb, where pb has the v alue assigned to it in (200). The faces of the form 010 are in the same zone with those of hkl. 215. The hemihedral form a hkl has the two faces of the form hkl, adjacent to one of the faces of the form 010, ANOETMC SYSTEM. _ _2lG. In the anorthic system, the form hkl has the two faces hkl, 217. To find the position of any pole. n, E, g, 11 are any four poles, of which no three are in one zone-eircle; f the intersection of the zone-circles ed, gh.