Prof. Miller's Crystallographic Notices. 



515 



right-angled spherical triangles having QR, RP, PQ for their 

 hypothenuses, and that P, Q, R are the poles of KX, KY, KZ, 

 we obtain 



cos RX = s 



— cosRY=s 

 cos QX = s 



— cos QZ = s: 

 cos PY = s 



— cos PZ = s 



n RP sin P = sin ZKX sin KX, 

 n RQ sin Q = sin YKZ sin K Y, 

 n PQ sin P= sin XKY sin KX, 

 n RQ sin R = sin YKZ sin KZ, 

 n PQ sin Q= sin XKY sin KY, 

 n RP sin R = sin ZKX sin KZ. 



P, Q, R are the intersections of the zone-circle uvw with the 

 zone-circles BC, CA, AB. Hence the symbols of P, Q, R will 

 be Owv, wOu, viiO. Therefore 



-cosPY=--cosPZ, 



W V 



— cos QX = cos QZ, 



w u 



- COS RX = cos RY. 



V u 



Hence 



sinKX , sinKY sin KZ 



aw -— ,,^^„ = OV -; — „,^^^ = cw- 



sinYKZ 



sin ZKX 



sin XKY" 



Construct a parallelopiped UVW 

 having OK, the axis of the zone, 

 *or a diagonal, and three of its edges 

 OU, OV, OW coincident with OX, 

 OY, OZ, the axes of the crystal. Let 

 KL, KM, KN be the edges respect- 

 ively parallel to OU, OV, OW. 



an bv cw 

 Hence, the axis of the zone uvw is the diagonal of a parallelopiped, 

 the edges of which coincide with the axes of the crystal, and are 

 respectively proportional to au, bv, cw. 



