208 jS. L. Penfield — Drawing of Crystals from 



in figure 1 would appear as in figure 4, and the parallel-per- 

 spective drawing, figure 5, could then be derived from figure 

 4 in exactly the same manner as figure 2 was derived from 

 figure 1. This is, for example, because the great circle through 

 m, s and /', figure 4, intersects the graduated circle at x, where 

 the pole of a vertical plane in the same zone would fall, pro- 

 vided one (artificially constructed or otherwise) were present ; 

 hence the intersection of such a surface with the horizontal 

 plane, and, consequently, the direction of the edges of the 

 zone, would be parallel to a tangent at x : In other words, fig- 

 ure 5 is a plan of a crystal in the position represented by the 

 stereographic projection, figure 4. Although not a difficult 

 matter to transpose the poles of a stereographic projection so 

 as to derive figure 4 from figure 1, it takes both time and skill 

 to do the work with accuracy, and it is not at all necessary to 

 go through the operation. To find the direction of the edges 

 of any zone in figure 3, for example m s r, note first in figure 

 1 the point x, where the great circles m s r and SES' cross. 

 During the supposed revolution of 80° about the axis SS\ the 

 pole x follows the arc of a small circle and falls finally at x' (the 

 same position as x of figure 4) and a line at right angles to a 

 diameter through x', as shown by the construction, is the 

 desired direction for figure 3. Similarly for the zones pr, 

 MrM' and MspM: ', their intersections with SES' at w, y and 

 z are transposed by the revolution of 80° to to', y' and z' . 

 The transposition of the poles w, x, y and z, figure 1, to w', x', 

 y' and z' may easily be accomplished in the following ways: — 

 (1) By means of the transparent, small-circle protractor 

 described by the writer,* the distances of w, x, y and z from 

 either jS or S' may be determined and the corresponding num- 

 ber of degrees counted off on the graduated circle. (2) Find 

 first the pole P of the great circle SES ' , where P is 90° from 

 E or 80° from C, and is located by means of a stereographic 

 scale or protractor : A straight line drawn through P and x 

 will so intersect the graduated circle at x', that 8 f x and S'x' 

 are equal in degrees. The reason for this is not easily com- 

 prehended from figure 1, but if it is imagined that the projec- 

 tion is revolved 90° about an axis AA,' so as to bring S' at 

 the center, the important poles and great circles to be con- 

 sidered will appear as in figure 6, where P and C" are the 

 poles, respectively, of the great circles ES'E' and AS' A', and 

 x is 41J° from S' as in figure 1. It is evident from the 

 symmetry of figure 6 that a plane surface touching at C, P 

 and x will so intersect the great circle ASA' that the dis- 

 tances S'x and S'x' are equal. Now a plane passing through 

 C, P, x and x', if extended, would intersect the sphere as a 

 * This Journal (4), xi, p. 17, plate I, 1901. 



