264 Penfield — Solution of Problems in Crystallography 



fore, the problem may be plotted to scale and the length of 

 the vertical axis Oc determined. The foregoing simple trigo- 

 nometric relations may now be applied to figure 15, where the 

 divided circle is employed for the double purpose of a stereo- 

 graphic projection and for solving problems in plane trigo- 

 nometry. Let it be supposed that the poles of the three pina- 

 coids 100, 010 and 001, and of a pyramid jp have been accu- 

 rately located in stereographic projection. Through 001 and J9 

 a line is drawn which is the projection of a great circle (a meri- 

 dian) determining the point ^tn on tlie divided circle. Since 

 the axial inclinations are 90°, the value of r (compare figures 

 6 and 8) is measured by the arc 100 /\m^ which in figure 15 is 

 39° 11^ Considering the radius of the divided circle as unity 



loo 



and as equal to the h axis, the tangent of r (100 A ^^^) gives the 

 intercept on the a axis ; accordingly this may be found by 

 drawing a chord from 010 to a point on the divided circle 

 equal to twice 100 Aw, measured from 010. Applying scale 

 No. 4 of the printed sheets along the direction of the a axis, 

 the length of the axis is found to be 0-815. The chord ha, 

 figure 15, determines also the point x on the radius passing 

 through jj>, and the distance from x to the center is the base- 

 line corresponding to Ox of figures 13 and 14. The numeri- 

 cal value of the length of the base-line^ which is equal to the 

 sine of r, need not be determined, but the distance may be 

 transposed to x' on the horizontal line by means of dividers. 

 Applying the stereographic scale to the radius passing through 

 ^j*, the angle of 001 A jp (the slope of the pyramid) may be 

 measured,' equal to 46° 6' in figure 15, and a line draw^n at this 



