102 Prof. Miller on the Anharmonic Ratio of Radii 



Their symbols, when referred to the new axes, are 

 fw — gv kw — Iv qw — rv, 

 e\v — gu hw — lu pw— ru, 

 ev— fu hv — ku pv — qu. 

 The symbol of the zone containing the second and third of 

 these becomes 



(hv — ku) (pw — ru) — (pv — qu) (hw — lu), 



(pv- qu)(ew — gu) — (ev— fu) (pw — ru), 

 (ev — fu) (hw — lu) — (hv — ku) (ew — gu) . 



Hence, multiplying out, and rejecting the common factor u, 

 u' = (kr-lq)u+ (Ip— hr)v+ (hg — kp)w, 

 v'=(qg— rf)u+(re— pg)v+(pf— qe)w, • 

 w'=(fl— gk)u + (gh-el)v+(ek — fh)w. 



Or 



u' = eu +_/V + ^ w, ^ 



v'=Au + *v + /w, I (12) 



w'=jou + gv-f rw. J 



To find u, V, w in terms of u', v', w'. 



Let the faces 100, 010, 001 be referred to the axes of the 

 zones efg, hkl, pqr as crystallographic axes. Then (10) their sym- 

 bols become ehp, fkq, glr. Therefore (12), 

 u = en' + hv' + pw', ^ 



v = fu'+kv'+qw', I (13) 



w = gu' + Iv' + rw'. J 



Having given the symbols of five poles, and the positions of 

 four of them, to find the position of the fifth. 



Let efg, hkl, pqr, mno, uvw be the symbols of the poles 

 D, E, F, G, P (fig. 4). Let D, E, F, G be given in position. 

 Therefore the segments into which any of the angles EDF, FED, 

 DFE are divided by the arcs GD, GE, GF, are known. Let 

 X, Y, Z be the poles of EF, FD, DE; and let jw'n'o', u'v'w' be the 

 symbols of G, P when referred to the axes of the zone-circles 

 EF, FD, DE as crystallographic axes. Then 



a' b' c' 



—7 cos GX = -7 cos G Y = -, cos GZ, 



7n' n' 



a' b' c' 



— r cos PX = -r cos PY = -; cos PZ. 



Hence 



w 

 m' cos PX n' cos PY o' cos PZ 



m' cosGX v' cos GY iv' cos GZ' 



