96 
In like manner 
(1-2) V3 
tan BOP = Sy eae 
_ (h—k) /3 
and tan COP = T= hak 
Hence, 
he bet 4 
cos AOP = V {2 (k-1?7 +2 (l—h)? +2 (h—h)*’ 
2k—l—h 
os BOF Ra Gaal 2 Cees 
2l1—-h—k 
Cee V{2(b—0? +2 (l— A+ 2 (hw) 
Let the zone-circle OP meet the zone-circle CA in H and 
the zone-circle ABin J. The symbol of J will be h—1, k—1,0. 
The symbol of OB is 101, and that of C4 will be 0 10. There- 
fore (27), 
cot OP —cot OH _ noe 
coh OF— cat OH SET 
Let OA =D. Then 
tan OF = cos 60° tan OA = } tan D; 
cot OH = cot OL cos EOP 
hak Hairs lL+th—2k 
V{2(k—-1)? +2 (L—h)* + 2 (h—k)*}’ 
cot OF = cot OF cos FOP 
h+k—2l 
+2 (h—k)"} 
ae ie eae (Al)? +2(l—h)? 
Hence 
_ {2 (b-1? +2 (1— hy? +2 (h—- k)?} 
tan OP Tay rae) tan D. 
The arc D may be taken for the element of the crystal 
Position of any pole in the prismatic system 
51. In this system the ares joining any two of the poles 
100, 010, 001 are quadrants. Let A, B, C, G, P be the 
poles 100,010, 001, 111, hkl respectively. Then BC, CA 
AB are quadrants. 
