in the Study of Crystallography. 167 



hedron, which consists of six rhombs with plane angles of 60° 

 and 120°. 



When this diagram is looked on as a radial projection, it 

 is that of an octahedron in rhombohedral position, the two 

 basal faces being equilateral triangles, and the six prismatic 

 faces being isosceles triangles in which each of the equal sides 

 is double of the base. 



Let us now consider the pentagonal dodecahedron as it 

 occurs in Nature as pyrites, and is designated by Miller 

 7r{012]-, and by Kopp ^(2a : a : cea). We shall consider 

 one face, which shall be represented by circle No. 0, and the 

 five contiguous faces, Nos. 1, 2, 3, 4, and 5, which by their 

 intersections give No. its pentagonal shape. From the 

 specification or measurement of the crystal we have the 

 inclination of the normal of to the normal of four of the 

 other faces 66° 25 / 'o or 66% and to the fith 53° 7'*8 or 53°. 

 Let No. 1 be inclined to No. at 53°, then Nos. 2, 3, 4, and 

 5 are all equally inclined at an angle of 6C^°. Place the 

 poles of these faces on the globe. Draw the great circles of 

 which they are the poles, then we have great circles Nos. 0, 1, 

 2, 3, 4, and 5 respectively parallel to and representing the 

 faces of the same number. We may also proceed directly to 

 draw the great circles. Place the metrosphere on the globe : 

 describe great circle No. coincident with its equator, and 

 mark the nodes (0) and (0, 1/), also the pole of No. 0. 

 Clamp the quadrant at 90° of azimuth on the equator and 

 incline the metrosphere round the common axis of the equator 

 and meridian until the angle between the meridian and circle 

 No. is 53°. The meridian is now in the position of great 

 circle No. 1. Let it be drawn, and let its pole be marked. 

 The two circles cut one another at an angle of 53°, and the 

 diameter (0, 1) represents the singular edge of pentagons 

 Nos. and 1. 



By specification the inclination of faces 1 and 2 is 66^°, 

 and equal to the inclination of 2 and ; we have now to 

 place a third great circle on the globe, which shall cut 

 both the others at an angle of 66h°. Clamp the movable 

 quadrant at an azimuth of 66^° on the equator. Bring the 

 quadrant to coincide with circle No. and let it slide along 

 it. Then, if it is carried over the whole semicircle of No. 0, 

 the meridian must somewhere coincide with the position of 

 circle No. 2. While the quadrant is being slid along No. 0, 

 the pole of the meridian is describing a small circle parallel 

 to No. 0, which may be drawn on the globe. Now bring the 

 quadrant to coincide with circle No. 1 and let it slip along it, 

 marking the small circle parallel to No. 1 which the pole of 



