94 
POSITION OF ANY POLE IN EACH OF THE SIx SYSTEMS OF 
CRYSTALLIZATION, 
Position of any pole in the cubic system. 
28. In this system the ares joining every two of the poles 
100, 010, 001 are quadrants, and the ares joining the pole 
111 and each of the poles 100, 001, 001 are all equal. 
Let A, B, C, O, P be the poles 100, 010, 001, 11 Lo wke 
respectively. Then BC, CA, AB are quadrants, 0A =OB=O0, 
and the right angles 4, B, C are bisected by OA, OB, OC. 
Hence the equations in (20) become 
tan BAP = : , tan CBP= : , tan ACP = i ‘ 
But tan AP = tan ACP sec BAP, 
tan BP = tan BCP sec ABP, 
tan CP = tan CBP sec ACP. 
Whence 
2 
aIP\ se 
(cos a = FiPSE: 
Position of any pole in the pyramidal system. 
29. In this system the ares joining the poles 100, 010, 
001 are quadrants, and the ares joining the pole 111 and each 
of the poles 100, 010 are equal. Let A, B, C, G, P be the 
poles 100, 010, 001,111, Akl respectively. Then BC, CA 
ABare quadrants, and 4G = BG. Consequently BAG =ABG, 
and ACG =BCG. The are BG intersects CA in the pole 101. 
Putting the are 001, 101=Z, and observing that the angles 
A, B, Care right angles, the equations in (20) become 
k 
tan BAP = : cot #, tan ABP =; cot FE, tan ACP = i: 
