SOLUTION FOR THREE TRACED FACES. ^ 369 



figure 3, for example, the angles between the traces of the three faces rep- 

 resented by C, 0^ and 0,^, in the section S, are the angles between the zone- 

 circles CS, 0,S and 0,S. These are equal to CT,S and CT,S. In plane 

 trigonometry the solution is simple, since the locus of a point at which two 

 given points subtend a given angle is a circle. Hence S would be found as 

 the intersection of two circles. In spherical trigonometry, or in the stereo- 

 graphic projection, the locus of the face-point of the section in which a given 

 solid angle (a) traces a given plane angle is a curve of the third or fourth 

 degree. 



Figure 6 shows a set of these curves for a= 45° and several different 

 values of/?. In other words, if P^ and P^ are two faces of a crystal, making 

 a normal angle of 45° with each other, the traces of those faces in a section 

 whose face-point lies on a curve marked, say 60, will be at a corresponding 

 angle, i. e., 60°. This diagram applies nearly to augite or diallage, letting 

 Pj be (100) and P^ (HO). Their general equation is (if the plane of projec- 

 tion is perpendicular to the faces, r = CS = tan 2 <; CS, (f = angle from 

 the bisectrix of P,CP, to CS, «= < P,CP, and /3 = < P,SP,) : 



CSC a cot a — cot 3 



r - 2 r' cos 2 ^ --f-^,„~r^ + ,,t a + cot ,3 = 0. (1) 



This equation is obtained as follows: Let S be {x,y),P^ = (?7i, n) and 

 P2 = (w, — n), and let the center of the zone-circle SP^ be at (a^, b^ of 

 the zone-circle SP^ (a^, h^). The equation of the circle SP^ is — 



^x-a,y-^y-h,y = r^; (2) 



whence, substituting {in, n), since P^ is on the circle — 



(m - a,r -f (n - b,f = r\ (3) 



Subtract (3) from (2) and we have — 



x' — m' — 2 a, {X — m) + tf — n' — 2 b, {y — n) = 0; (4) 



but it is easy to prove that — 



TT = = — cot i a. (5) 



Substitute (5) in (4) and find a^ : 



x' -\-y'- (m'-hn') 



(" T.y) 



(6) 



