70 S. L. Penfield — Crystal Drawing. 



surface, a plane surface tangent to the sphere at p would be 

 parallel to the crystal face under consideration, and, if ex- 

 tended, it would intersect the plane of the equator on a line at 

 right angles to a radius drawn through the intersection of the 

 meridian of^> and the equator. In the lower right-hand corner 

 of figure 55 the arc OE is supposed to represent a portion of 

 the meridian through p ; O is the north pole of the sphere, 

 OE the trace of the plane of the equator and tt the trace of 

 the tangent at p : If now the tangent plane is shifted, without 

 change of direction, until it intersects O {unity on the vertical 

 axis) it will intersect the radius OE in the plane of the equator 

 &t p'. The linear projection of p is therefore found by deter- 

 mining the point p', where a plane parallel to the tangent at^> 

 and intersecting the vertical axis at cuts the radius drawn 

 through j?, and then drawing the line of the linear projection, 

 U, at right angles to the radius. Knowing the distance to p 

 in degrees, the point p' where the line 11 crosses the radius 

 through j? may be readily found in three ways : (1) Graphi- 

 cally, as shown in the lower right-hand corner of figure 55 ; 

 (2) From the same figure it is evident that Op' is the cotan- 

 gent of the angle Op' or of the arc Up ; the value of the 

 cotangent may be found from a table of natural tangents and 

 cotangents and laid off on the radius through p by means of a 

 scale of decimal parts ; (3) A cotangent scale may be prepared, 

 based on the radius of the circle as unity, and the distance Op r 

 laid off directly from the graduation. The latter method is 

 probably the best, and a scale for laying off cotangents may be 

 easily had by a simple modification of the stereographic scale, 

 No. 3, of the engraved sheets described by the writer.* The 

 basis of the stereographic scale is that the distance from the 

 center to any pole, for example, to p, figure 55, is equal to 

 the tangent of half the arc O^p ; hence in order to prepare a 

 scale for laying off tangents and cotangents it is only necessary 

 to take a stereographic scale and renumber it, making 20° of 

 the one equal to 10° of the other. On applying such a scale to 

 a radius of the graduated circle for laying off cotangents, 90° 

 is located at the center (the cotangent of 90° =0), and 0° falls 

 at infinity. The reason for using a cotangent instead of a tan- 

 gent scale (when the numbering would run in the opposite 

 direction) is that cotangents are better adapted to the 4> and p 

 angles of the two-circle goniometer. Having a second pole a, 

 37° 50' from O, figure 55, its linear projection is the line VV. 

 The two lines of the linear projection 11 and VV intersect at i, 

 and the direction of the edge made by the intersection of^> and 

 q will be parallel to the line joining C and i. 



*Loc. cit. 



