CKYSTALLOCIKAI'HY 



45 



Fio. 69. 



projection of a point on the sphere at the opposite end of a diame- 

 ter from P; therefore draw a circle through BSP, and cSd will be 

 the projection of the zonal circle re- 

 quired. 



Problem IV. Given the projec- 

 tion of any two poles, to find the 

 aiiiile between them. In Fig. 69 

 let a, b be the given poles; by 

 Problem III draw the zonal circle 

 dabc, and find its pole P by Problem 

 I ; draw Pa and Pb, extending them 

 to meet the primitive at a'b', then 

 the arc a'b' will measure the angle 

 between the normals ab. 



Problem V. Given the zone circle and the projection of one 

 pole in the zone, to find the projection of a face in the same zone 

 at a given angle from it. 



This is the reverse of III. In Fig. 69, let dac be the zone circle, 

 a the given pole, to find the pole b, 80 from it. Find the pole p 

 of dac by I ; draw Paa' and make a'b' = 80 and draw b'P ; 



where it crosses the zonal circle dac 

 as at b will be the projection of the 

 pole 80 from a as required. 



Problem VI. To locate the pro- 

 jection of any pole, the axial ratio 

 and the indices of the face being 

 given. 



The axial ratio of barite is & : b : c 

 = .815 : i : 1.313 ; locate the pole of 

 y = (122) = 2 a: b : c. 



In Fig. 70 draw the primitive and 

 the two diameters aa', bb' at 90. 

 Lay off ob" = b, the intercept on 

 the axis a is oa" = 2 (b X .815) ; draw b"a" and oc at 90 to a"b", 

 then oc will be the projection of a zonal circle at 90 to the primi- 

 tive with its pole at P, on which the pole of 122 will lie. Lay 

 off oc' = b X 1.313, or the unit on the vertical axis, and od = oc, 

 connect d and c' ; the angle c'do = the angle between the normal 

 of y, (122) and the normal to the base ooi ; draw Poc", and make 

 c"oy' = c'do, draw y'P ; where it crosses oc as at y will be the 

 projection of the pole of 122 as required. 



Fia. 70. 



