280 Pevfield — Solution of Prohlems m Crystallography 



tbns determining the base-line Oy, and from y^ (the distance 

 Oy transferred to the horizontal diameter) lay off the p angle 

 of ,9, 29° 36', which intercepts the lower radius {h axis) at one- 

 half ; hence the axial ratio of s is ooa : ^h : c, the indices being 

 021. For a pyramidal face, for example X, through unity on 

 the a axis, a line at right angles to the radius through A was 

 drawn, which in figure 28 was found to intercept the vertical 

 axis at 0'587, or exactly %c. Thus the intercept on the vertical 

 axis and the base-line Oz are determined. From b\ on the 

 horizontal diameter, a second line was drawn at an inclination 

 of 12° 43^, the p value of A, which intersects the lower radius 

 at O'lll, or -^b. The axial ratio of ?. is therefore —ai-^h: %e, 

 the indices being 2*1 8*3. Thus it makes no difference how 

 complex the symbol may be, it is only necessary to keep 

 in mind a few simple principles and the result is quickly 

 obtained. 



Finally, having made a projection such as shown in figure 

 27, the small circle protractor may be applied to it and the 

 distances between any desired poles determined. Some results 

 actually obtained in this way are as follows : 



Meas. Cal. Meas. Cal. Meas. Cal. 



C/sm, 88° 10' 88' uy Cy^k,n°25' 71° 25' Uy^p, 50° 50' 51° 00' 



C/^p, 52 30 52 28 C/sO, 17 24 77 25 c^'/^A;, 45 05 44 55 



C/sA, 68 10 68 12 7n^p,35 45 35 42 a' y^o, Ql 05 60 59 



Triclinic System. 



For illustrating the application of graphical methods to this 

 system a somewhat complex crystal of anorthite has been 

 chosen, shown in orthograpliic and clinographic projection in 

 figure 29. The crystal was studied and figured by Mr. J. C. 

 Blake. The following were selected as fundamental measure- 

 ments for making the stereographic projection, figure 30, and 

 solving the problem : 



b/^77i, 010^110 = 58° 04' m/yc, 110/x001 = 65° 53' 



w^/^Jf, 110/s 110 = 59 29 5/^6,010/^021 = 43 12 



5/vC, 010/^001 = 85 50 



Having located 5, m and iM on the divided circle, the posi- 

 tion of 100 was first determined by the relation of four faces 

 in a ^one, explained on page 260, h, m and 31 being, respect- 

 ively, the first, second and fourth faces, and 100 the unknown 

 third face. This problem, as solved by itself on a separate 

 sheet, is shown in figure 31. The lines of reference Ox and 

 Oy are at right angles respectively to bV and MM' . A line 

 drawn from a?, at right angles to the radius through m, inter- 

 sects Oy "at a point marked \. By solving the equation given 

 on page 262 the ratio of the cotangents is found to be \\ 1, 

 hence a line from x to the point 1 on the line Oy determines 

 the direction of the pinacoid 100. A line from the center at 



