S. L. Penjield — Crystal!) rawing. 



51 



19 



twinning plane should be vertical and have the position corre- 

 sponding to that of the side pinacoid h (010) of an orthorhom- 

 bic crystal. The solution depends upon the angle of base on 

 twinning plane, c ^f = 63° 7', from which the inclination of 

 the vertical axes, 53° 46' from one another, or 26° 53 / from 

 the twinuing plane placed in vertical posi- 

 tion, as shown in figure 19, is derived. 

 As indicated by figure 20, the inclinations 

 of the vertical axes, c and O, 26° 53', 

 from the perpendicular, are determined 

 by the graduation of the vertical ellipse 

 through B. Also the intersections of 

 the planes of the horizontal axes with 

 the same ellipse are located at Xand X' ', 

 and Z" and Y\ 26° 53' from B and -B. 

 In order to have the twinning plane cor- 

 respond with the side pinacoid 010 of 



the orthorhombic system, it is necessary to make one of the 

 horizontal axes — « 3 , a s of the hexagonal system correspond with 

 the front and back or a axis of the orthorhombic system. The 



