PROPERTIES OF THE STEREOGR APHIC PROJECTION. 367 



(6) The angle two zone-circles make with each other is equal to the exterior 



angle between the two sides of the crystal face (represented by the 

 point of intersection of the zone-circles), common to the two zones and 

 corresponding to the two zone axes. This is important for us, since, 

 if we can find the exact position of a section of a cr^^stal referred to 

 its axes, we can locate its projection (i. e., section-point) S. Then, if 

 we join by means of zone-circles (SP^, SP.^, SP.^, etc., figure 2) this 

 section-point and the face-points (Pp P.,, P.^) of the planes we think 

 we recognize, the directions of these circles at S will correspond to 

 the directions of the traces of those planes in the section of the crystal, 

 and their radii from S (viz., SC^, SC^, SC^, etc.) may be taken as the 

 traces. 



(7) The centers of all zone-circles meeting at a given point S lie on a straight 



line perpendicular to the radius through ^S at a distance from C equal 

 to cot <C C : S. If S is within the primitive, C lies between it and 

 this center-line. This line is a common chord of a circle about C 



with any radius, x, and a circle about S with radius j/x'^ -f- 1. 

 (8) The angle between two face-points {0,^0.^, figure 3) is measured by the 

 arc of the primitive (B^B.^) lying between straight lines (^O^B^, O.^B^) 

 drawn from the face-point J. of a face perpendicular to the given 

 faces. 



By means of these propositions we can solve graphically every ordinary 

 problem in spherical trigonometry, and therefore in crystallography. We 

 can construct a zone-circle through two given points {0^0^, since its center 

 is the intersection of their center lines. We can construct a zone-circle 

 (<SXi, figure 2) through a given point (aS), making a given angle with a 

 given zone circle {PSm), since their radii (0^*5 and C^^S) at the point S 

 will make the same angle, and the center of the circle will be the intersec- 

 tion of the radius C^^S, with the center line of S. We can find the locus 

 of any point at a given angular distance from a given point, thus : First, 

 find two such points {L and 31, figure 4) on the same ground-circle radius 

 (l3y) with it (y), which may be done most quickly by computing their 

 distances from /?. Then LM is a diameter of the locus sought. We 

 can thus readily construct the projection of a spherical triangle from the 

 requisite data. 



One problem only needs especial mention, viz., that in which one side (y/3, 

 figure 4), the opposite angle (ySi3), and the length of another side (-yS) are 

 known. In this case we take the projection on the face opposite the side of 

 known length, i. e., on /?; thus 3/ is radial. We can construct first the circle 

 on LM, which is the locus of aS', taking Ly and 3fy as equal to the known 

 value of jS. Let be the center of that circle; then < (3 80= <^ f3 S y 



