376 A. C. LANE — THE ANGLES OF CRYSTALS. 



Hence — 



rt -f 6 = 0.7210 log. a -\- b-^ 1 = 0.2358 



2 



log. (a-\-b + ly = 0.4716. 



a + h + r = 2.9621 ; 

 Subtract 4b 1.9744 



0.9877. 



Log. 0.9877 = 9.9946, and dividing by 2 we get the log. i/(a + b -{- 1)- — 4 b 

 = 9.9973, of which the antilog. is .9938. Therefore— 



cos2<p= 0.7210 ± 0.9938 = — 0.2728 ; 2 c? = 105.8 <p --= 52°.9 ; while 

 cos 2 ^ = 1 — "^^4931^ = 0.4732 ; 20= 118°.2, and = 59°.l. 



Moreover, the direction of the 4- extinction in the one half (I) is 28°. 6 ; 

 in the other (II) it is 310°.5. The polarization color in I corresponds to 

 a retardation of about mm. 0.000,48 ; in II, of about mm. 0.000,95. The 

 hyperbolas close into a cross at 315°. This direction, therefore, is parallel 

 or perpendicular to (010). 



Figure 2 is the stereographic projection drawn with these data and illus- 

 trating the case figured in figure 2a and solved above. S is constructed 

 with the above found {(f, 0). SC^, SC.^, SC.^, etc., are the radii of the zone- 

 circles SP^, SP^, SP.,, and are the directions of the traces of the corresponding 

 faces and cleavages. SX^ and SX., are the directions of the extinctions. 

 Figure 2 and figure 2a harmonize well. 



We may also check our solution by solving one of the spherical triangles, 

 of which we have four parts given, as, for example, SP^P.. Here P^P^ = 

 46° 27', ^P, = 59°.2, and < P,SP, = 26°.2. SP,P, = 142°.9 by our previous 

 solution, and by spherical trigonometry we have — 



log. tan 26°.2 = 9.6920, log. cot 46° 27' = 9.9780 

 log. cos 59°.2 = 9.7106, log. taii 59°.l = 0.2247 



9.4026 

 = 75°.9 log. cos = 9.3891 



0^ = 67°.l log. cot 0^ = 9.5901. 



SP,P,=^OJrO, = U2°.d. 



