BERKELEY 
 
 LIBRARY 
 
 UNIVERSITY OF 
 CALIFORNIA 
 
 ITY OP I 
 RNIA J 
 
 EARTH 
 
 SCIENCES 
 
 LIBRARY 
 
TREATISE 
 
 ON 
 
 CRYSTALLOGRAPHY, 
 
 BY 
 
 W. H.jMILLER, M.A., F.R.S., F.G.S., F.C.P.S., 
 
 FELLOW AND TUTOR OF ST JOHN'S COLLEGE, 
 
 AND 
 PROFESSOR OF MINERALOGY IN THE UNIVERSITY OF CAMBRIDGE. 
 
 CAMBRIDGE: 
 
 PRINTED AT THE PITT PRESS: FOR 
 
 J. & J. J. DEIGHTON, TRINITY STREET 
 
 LONDON: 
 JOHN W. PARKER, WEST STRAND. 
 
 M.DCCC.XXXIX. 
 
 I 
 
EOLOGICAL SCIENCES 
 
M 
 
 SOENCES 
 LIBRARY 
 
 ADVERTISEMENT. 
 
 THE Crystallographic Notation adopted in the fol- 
 lowing Treatise is taken, with a few unimportant 
 alterations, from Professor WhewelPs Memoir "On a 
 general method of calculating the angles of Crystals," 
 printed in the Transactions of the Royal Society for 
 1825. 
 
 The method of indicating the positions of the faces 
 of a Crystal hy the points in which radii drawn per- 
 pendicular to the faces meet the surface of a sphere, 
 was invented by Professor Neumann of Konigsberg 
 (Beitrage zur Krystallonomie) and afterwards, together 
 with the notation, re-invented independently by Grass- 
 mann (Zur Krystallonomie und geometrischen Combi- 
 nationslehre). The use of this method led to the 
 substitution of spherical trigonometry for the processes 
 of solid and analytical geometry in deducing expressions 
 for determining the positions of the faces of crystals 
 and the angles they make with each other. The ex- 
 pressions which in this Treatise have thus been obtained, 
 
 477 
 
IV ADVERTISEMENT. 
 
 are remarkable for their symmetry and simplicity, and 
 are all adapted to logarithmic computation. They are, 
 it is believed, for the most part new. For the conveni- 
 ence of calculation the position of one face with respect 
 to another is represented by the angle between normals 
 to the faces, or by the supplement of the angle be- 
 tween the faces, according to the commonly received 
 definition of the angle between two of the planes that 
 bound a solid. 
 
 Arts. 22 24, 28 31, may be omitted by those 
 readers who are satisfied with such a knowledge of the 
 subject as is sufficient for the purpose of finding the 
 elements and the symbols of the faces of a given 
 crystal, or of determining the form and angles of a 
 crystal, having given its elements and the symbols of 
 its faces. 
 
CONTENTS. 
 
 CHAPTER I. 
 
 IMG! 
 
 GENERAL GEOMETRICAL PROPERTIES OF CRYSTALS 1 
 
 Arts. 1. CRYSTAL. Faces. Cleavage Planes. 2. Law to which the mutual 
 inclinations of the faces and cleavage planes are subject. 3 Cleavage planes 
 are parallel to possible faces. 4. Axes. Parameters. Indices. Symbol of 
 a face. 5. Elements of a Crystal. 6. Different systems of Axes and Parame- 
 ters. 7- Position of a face determined by points in the axes. 8. Angles 
 between a normal to any face and the axes. 9 Sphere of projection. Poles. 
 10, 11. Signs of the Indices. 12 14. Condition that a point may be in a 
 great circle passing through two poles. 15. Symbol of the^ point in which two 
 such great circles intersect. 16, 17- Indices of a zone in terms of the indices 
 of two of its faces, and of a face common to two zones in terms of the indices 
 of the zones, have the same form. 18 20. Indices of a face common to two 
 zones in terms of the indices of two faces of each of the two zones. 21. Con- 
 dition that a face may belong to a zone. 22. To find the poles in a given 
 zone-circle, and the zone-circles passing through a given pole. 23, 24. Any 
 pole having indices greater than unity is the intersection of zone-circles 
 through poles having lower indices. 25 27. Relations between the distances 
 and indices of four poles in the same zone-circle. 28 To change the axes. 
 29. To change the parameters. 30, 31. Having given the symbols of four 
 poles referred to each of two systems of axes, and the symbol of a fifth pole 
 referred to one of the system of axes; to find its symbol when referred to 
 the other. 32. Systems of Crystallization. 33. Distinguished by their sym- 
 metry. 34 Forms. Holohedral forms. HemiKedral forms. Combinations. 
 35 Elements of a crystal. 36 Data requisite for the determination of the 
 elements of a crystal. 
 
 CHAPTER II. 
 
 OCTAHEDRAL SYSTESI 22 
 
 Arts. 37. Axes and Parameters. 38. Holohedral forms. 39 41. Hemi- 
 hedral forms. 42 46. Position of any pole, and distance between any two 
 poles. 47 51. Arrangement of the poles of holohedral and hemihedral forms. 
 52, 53. To find the indices of a form. 5467- To find the figures and angles 
 of different forms. 68. Table of the distances of the poles of different forms 
 from the nearest poles of the cube, dodecahedron and octahedron. 69 79. 
 Examples. 
 
VI CONTENTS. 
 
 CHAPTER III. 
 
 PYRAMIDAL SYSTEM 41 
 
 Arts. 80. Axes and parameters. 81. Holohedral forms. 8284. Hemi- 
 hedral forms. 85. To find the position of any pole. 8690. Arrangement 
 of the poles of holohedral and hemihedral forms. 91 94. To find the distance 
 between two poles of the same form. 95, 96. To find the indices of a form. 
 97 99. To find the distance between any two poles. 100. To find the para- 
 meters. 101. To change the axes. 102 117- To determine the figures and 
 angles of different forms. 118121. Examples. 
 
 CHAPTER IV. 
 
 RHOMBOHEDRAL SYSTEM 55 
 
 Arts. 122. Axes and parameters. 123. Holohedral forms. 124126. Hemi- 
 hedral forms. 127129. To determine the position of any pole. 130132. Ar- 
 rangement of the poles of holohedral and hemihedral forms. 133135. Dirhom- 
 bohedral forms. 136 137- Direct and inverse zones and forms. 138 140. To 
 find the distances between poles of the same form. 141 143. To find the 
 indices of a form. 144. To find the distance between any two poles. 145. Hav- 
 ing given the distance between any two poles ; to find the distance of a pole of 
 {100} from the nearest pole of {111}. 146. To find the position of any pole, hav- 
 ing given its distances from two of three equidistant poles of a form. 147- To 
 change the axes. 148 164. To determine the figures and angles of different 
 forms. 165 171. Examples. 
 
 CHAPTER V. 
 
 PRISMATIC SYSTEM 77 
 
 Arts. 172. Axes. 173. Holohedral forms. 174, 175. Hemihedral forms. 
 176. To determine the position of any pole. 177 179. Arrangement of the 
 poles of holohedral and hemihedral forms. 180 183. To find the distance 
 between two poles of the same form. 184, 185. To find the distance between 
 any two poles. 186, 187- To find the indices. 188, 189. To find the para- 
 meters. 190 198. To determine the figures and angles of different forms. 
 199201. Examples. 
 
 CHAPTER VI. 
 
 OBLIQUE PRISMATIC SYSTEM 87 
 
 Arts. 202. Axes. 203. Holohedral forms. 204. Hemihedral forms. 205. To 
 determine the position of any pole. 206. Arrangement of the poles. 207, 208. 
 To find the distance between any two poles. 209 212. To find the inclination 
 
 of the axes and the parameters. 213. To change the axes. 214 217. To 
 
 determine the figures and angles of different forms. 218 220. Examples. 
 
 CHAPTER VII. 
 
 DOUBLY-OBLIQUE PRISMATIC SYSTEM 95 
 
 Arts. 221. Forms. 222 224. To determine the position of any pole. 225. 
 To find the distance between any two poles. 226, 227. To find the inclinations 
 of the axes and the parameters. 228, 229. Examples. 
 
CONTENTS. Vll 
 
 CHAPTER VIII. 
 
 PAGE 
 
 TWIN CRYSTALS 103 
 
 Arts. 230. Law according to which the two crystals constituting a twin are 
 united. Twin axis. 231. Arrangement of the poles of a twin. 232. To find 
 the twin axis. 233. Distance between the poles of a twin crystal. 234 263. 
 Examples. 
 
 CHAPTER IX. 
 
 GONIOMETERS, &c 113 
 
 Arts. 264. Carangeau's Goniometer. 265 Wollaston's Goniometer. 266. 
 To measure the angle between two faces of a crystal. 267. To determine the 
 faces in a given zone. 268. Adjustments of Wollaston's Goniometer. 269. 
 Signals. 270. Error arising from excentricity. 271. To eliminate the error. 
 272. Expansion of -crystals when heated unequal in different directions. 273. 
 Plane angles of the faces of crystals. 274. Optical properties of crystals 
 belonging to different systems. 275. Elements of a crystal in terms of the 
 distances between certain poles. 276. Table shewing how poles which have 
 no index greater than 7 may be determined by the intersection of zone-circles 
 through poles having lower indices. 277. Comparison of different systems of 
 crystallographic notation . 
 
 CHAPTER X. 
 
 DRAWING CRYSTALS AND PROJECTIONS 129 
 
 Arts. 278282. Drawing crystals. 283. Projection of the sphere on which 
 the poles of a crystal are laid down. 284295. Stereographic projection. 
 296 302. Gnomonic projection. 
 
ERRATA. 
 
 PAGE LINK 
 
 36 8 after Let insert the. 
 
 39 8 from bottom, insert a comma after Octahedron. 
 
 39 6 from bottom, for p'e' read p',e'. 
 
 50 2 for { 4 1 } read {430}. 
 
 81 19 for (182) read (183). 
 
 86 6 after o, a?, />, / insert belong. 
 
 86 8 for wanting read want. 
 
 87 5 from bottom, for 87 read 88. 
 
 88 7 from bottom, /or K reac? <r. 
 92 11 for g read 7. 
 
 107 10 from bottom, for Scheelale read Scheelate. 
 
 114 At the end of Art. 266, add The distance of the crystal from the eye of 
 the observer should be from one inch to two inches; the image of A formed by 
 reflexion at either of the faces p, g will then be seen clearly, while the crystal 
 itself remains unseen. It is essential that the signal B and the image of A should 
 be seen distinctly. When the distance of distinct vision of the observer's eye differs 
 from AC or BC, the eye must be armed with a lens, or a small telescope having 
 a power of from one to three, capable of shewing the signals distinctly. 
 
CRYSTALLOGRAPHY. 
 
 CHAPTER I. 
 
 GENERAL GEOMETRICAL PROPERTIES OF CRYSTALS. 
 
 1. MANY natural substances and many of the results 
 of chemical operations occur in the form of solids which 
 are bounded by plane surfaces, and which commonly exhibit 
 a tendency to separate, when broken, in the directions of 
 planes, passing through any point within them, either parallel 
 to some of their bounding planes, or else making given 
 angles with them. 
 
 Solids of this description are called crystals. The planes 
 by which they are bounded are called their faces, and the 
 planes in which they have a tendency to separate, their 
 cleavage planes. 
 
 2. The mutual inclinations of the faces and cleavage 
 planes of a crystal are subject to a law which we now proceed 
 to enunciate. 
 
 Through any point O (fig. 1) within the crystal, let 
 planes be drawn parallel to each of its faces and cleavage 
 planes. Let OX, OF, OZ be any three intersections of those 
 planes, not all in one plane ; and let any face or cleavage 
 plane meet OX, OF, OZ in A, B, C. Then, if any other 
 face or cleavage plane meet OX, OF, OZ in H, K, L, and 
 if we consider HO, KO, LO to be positive or negative 
 according as they are measured in the same directions as 
 AO, BO, CO, or in different directions, it will be found that 
 1 AO 1 BO l CO 
 
 where h, k, I, may be any positive or negative whole numbers, 
 one or two of which may be zero 
 1 
 
2 
 
 When one of the numbers h, k, I becomes zero, one of the 
 distances HO, KO, LO corresponding, becomes indefinitely 
 great, and, therefore, is parallel to the line along which that 
 distance is measured. 
 
 3. Since the positions of the faces and cleavage planes 
 are subject to the same law, it follows, that a cleavage plane 
 is always parallel to a plane which either is, or may be, a face 
 of the crystal. This being the case, in speaking of the faces 
 of a crystal, we shall always suppose the cleavage planes to 
 be included. 
 
 4. OX, OF, OZ are called the axes of the crystal, 
 their origin ; AO, BO, CO, or any three lines in the 
 same ratio, the parameters of the crystal ; h, k, I the indices 
 of the face HKL. This face will be denoted by the symbol 
 (h k I), a negative index being distinguished by a minus sign 
 placed over it. 
 
 5. The indices h, k, I, which by taking different integral 
 values determine the positions of the different faces of the 
 same crystal, are seldom large. When the axes and par- 
 ameters are properly chosen the highest index does not com- 
 monly exceed six. 
 
 The inclinations of the axes and the ratios of the par- 
 ameters are the same, at a given temperature, for all crystals 
 of the same species. The symbols of the faces may be dif- 
 ferent. Hence, the angles YOZ, ZOX, XOY, which the 
 axes make with each other, and the ratios of two of the 
 parameters AO, BO, CO to the third, are five elements by 
 which each crystalline species is characterized. 
 
 6. It will be sufficient for the present if we suppose 
 the law stated in (2) to hold when the axes are three given 
 lines in which the planes drawn through a point within the 
 crystal, parallel to its faces, intersect each other, and the 
 parameters are the portions of the axes cut off by a given 
 face. Hereafter we shall prove that, if through a point 
 within the crystal, planes be drawn parallel to all the 
 
possible faces of the crystal, and that the law stated in (2) 
 hold when the axes are three given intersections of those 
 planes, and the parameters are the portions of the axes cut 
 off by a given face ; it will also hold, when any three inter- 
 sections are taken for axes, and the portions of them cut off 
 by any plane are taken for parameters. 
 
 7- The law enunciated in (2) may be put, as follows, 
 under a different form, which, though not quite so simple, 
 presents a clearer view of the relative positions of the faces 
 of a crystal. 
 
 Let OX, OF, OZ (fig. 2) be the axes of a crystal, 
 , 6, c its parameters. 
 
 In OX take OA = a, OA 2 =^a y OA 3 ^ a, ... towards X^ 
 OA_ l = a, OA_2=ija, OA_^ ^a, ... in the opposite direction, 
 and OA Q = -J- a - co in either direction. And let the points 
 
 >, ->_!, -O 2 ? -O-25 ->33 -O-3?5 -O 5 ^5 i",!' 25 ^-25 
 
 C 3 , C_ 3 ,..., C , be determined in the same manner from 
 b and c respectively. Then, any face of the crystal will be 
 parallel to a plane drawn through three points thus deter- 
 mined, one being taken in each of the three axes. 
 
 For if the face (h k /) meet the axes OX, OF, OZ in 
 H, K, L. Then (2) 
 
 l OA _ 1 OB _ 1 OC 
 ~hOH~~k~OK~~l~OL' 
 
 But, according to the notation we have adopted, 
 
 , 
 
 these distances being measured towards X, F, Z, or in the 
 opposite directions, according as the corresponding indices 
 /, k, /, are positive or negative; 
 
 OA h OB k OCt 
 
.-. the face (h k I) is parallel to the plane passing through 
 the three points A h , B k , C t . 
 
 8. To find the ratios of the cosines of the angles which 
 a perpendicular to the face (h k I) makes with the axes of 
 the crystal, in terms of the indices of the face and the para- 
 meters of the crystal. 
 
 Let the axes OX, OF, OZ, (fig. 3) meet the surface 
 of a sphere described round O as a center in X, Y 9 Z, 
 and let OP, drawn perpendicular to the face HKL, the symbol 
 of which is (hkl), meet HKL in p, and the surface of the 
 sphere in P. Then, 
 
 Therefore, substituting these values of HO, KO, LO 
 in (2), if AO = a, BO = 6, CO = c, we have 
 
 7 cos PX = - cos PY = C - cos PZ. 
 h k I 
 
 In all problems of Crystallography which may hereafter 
 present themselves, we shall refer the faces of crystals to 
 the surface of a sphere by means of radii drawn perpendicular 
 to the faces, and all our calculations will be performed by 
 spherical trigonometry applied to expressions deduced from 
 the above equations. 
 
 9. The sphere to the surface of which the faces of a 
 crystal are referred, will be called the sphere of projection. 
 The extremity of a radius of the sphere drawn perpendicular 
 to any face, will be called the pole of that face. A face 
 and its pole will be usually denoted by the same letter and 
 by the same symbol. The points in which the axes of the 
 crystal meet the surface of the sphere of projection will be 
 invariably denoted by X, F, Z. 
 
10. Let the axes of any crystal meet the surface of the 
 sphere of projection in X, Y, Z, (fig. 4). Let a, 6, c be the 
 parameters of the crystal, ABC the polar triangle of XYZ. 
 Then, AY, AZ are quadrants, .-. cos AY=0, cos AZ = ; 
 
 .-. - cos AX = - cos AY - - cos AZ; 
 1 
 
 .-. (8) A is the pole of (l 0). Similarly, B is the pole of 
 (0 1 0), and C is the pole of (0 1). 
 
 11. Let P be the pole of the face (h k 1). Then, (8) 
 
 - cos PX = - cos PF= - cos PZ. 
 
 h k I 
 
 When P and A are on the same side of the great circle 
 EC, PX is less than a quadrant, therefore cos PX is posi- 
 tive. When P and A are on opposite sides of BC, PX is 
 greater than a quadrant, therefore cos PX is negative. Hence, 
 if we assume h to be positive in the former case, it will be 
 negative in the latter. In like manner Jc will be positive or 
 negative according as P and B are on the same or opposite 
 sides of CA. And I will be positive or negative according 
 as P and C are on the same or opposite sides of AB. 
 
 When P is in the great circle BC, PX is a quadrant, 
 therefore cos PX = ; therefore h = 0. When P is in CA, 
 cos PY = 0, therefore k = 0. When P is in AB, cos PZ = 0, 
 therefore I = 0. 
 
 Hence, if diameters AA', BB f , CC', PP' be drawn, the 
 symbols of the points A, B, C, P being A (l 0), B (0 1 0), 
 C (0 1), P (h k I), those of A', B', C', P' will be A'(l 0), 
 B' (oTo), C'(ooi), p'(hk~l). 
 
 12. X, Y, Z (fig. 5.) are any three points on the surface 
 of a sphere; P, Q, R any three points in a great circle; 
 to find the relation between the distances of P, Q, R from 
 each of the points X, F, Z. 
 
6 
 
 From the spherical triangles PQAT, J?QA", we have 
 cos PX = cos QX cos PQ + sin QX sin PQ cos PQA", 
 cos RX= cos QX cos QR + sin QX sin QR cos RQX. 
 
 Multiply the first equation by sin Qj?, the second by 
 sin PQ, and add, observing that 
 
 cos PQA" + cos R QX = 0, and that 
 sin QR cos PQ + cos QR sin PQ = sin PR. 
 
 The equation thus obtained, and two others deducible 
 from it by writing Y and Z successively in the place of X, 
 
 cos PA" sin QR + cos R X sin PQ = cos QA" sin PR, 
 cos PF sin QR + cos R Y sin PQ = cos QF sin PR, 
 cos PZ sin QR + cos RZ sin PQ = cos QZ sin PR. 
 
 Whence, eliminating sin PQ, sin PR, sin QR, successively, 
 [cos PA" cos Q F - cos PF cos 
 
 are 
 
 [cos PA" cos ^ F - cos PF cos 
 
 s 
 
 sin 
 
 - cos 
 
 sin/> [cos PZ cos Q^T - cos PX cos QZ] 
 [cos PZ cos .ff JT - cos PA" cos 7? Zl 
 cos ^^ - cos Q^T cos 
 
 . 
 
 sin 
 
 [cos PF cos QZ - cos PZ cos QF] 
 
 t cos PycosRZ- cos PZ cos R F] 
 
 1 
 sinQ7g [cos QF cos ,RZ - cos QZ cos RY] 
 
Eliminating two of the quantities sin PQ, sin PR, sin QR, 
 between two of the preceding equations, we obtain 
 
 (cos PY cos RZ - cos PZ cos R Y) cos Q X 
 + (cos PZ cos RX - cos PXcos R Z) cos QY 
 + (cos PX cosRY- cos PF cos RX) cos QZ = 0. 
 
 13. Let X, F, Z be the points in which the axes of 
 a crystal meet the surface of the sphere of projection ; 
 P, R the poles (h k /), (p q r) ; a, 6, c the parameters of 
 the crystal. Then (8) 
 
 - cos PX= - cos PF = - cos PZ, 
 
 h k I 
 
 - cos R X= - cos R Y = C - cos RZ. 
 p q r 
 
 Therefore eliminating cos PX, cos PF, cos PZ, cos RX, 
 cos R F, cosRZ between the above equations and the equation 
 at the end of (12), we have 
 
 ua cos QX + vb cos QY+ we cos QZ = 0, 
 where 
 
 u = kr Iq, v = Ip hr, w = hq - kp. 
 
 14. The great circle passing through the poles (hk I), 
 (p q r), may be denoted by the symbol [u v w], where u, v, w 
 have the values assigned to them in (13). Since the poles 
 P, R may be denoted by the indices h, k, I; p, q, r, or 
 by any numbers proportional to A, k, /; p, q, r respectively, 
 it follows, that the great circle PR may be denoted by any 
 three numbers proportional to u, v, w. When u, v, w have 
 a common measure, it will be found convenient to employ as 
 indices, the lowest whole numbers in the required ratio. 
 
 15. Let [hkl], [p q r] be the symbols of two great 
 circles, each of which passes through the poles of any two 
 faces not parallel to each other ; and let the great circle 
 
[h k 1] meet the great circle [p q r] in the point Q. Then, 
 since Q is a point in each of the great circles, (13) 
 
 ha cosQX + kb cosQF+lc cos QZ = 0, 
 pa cos QX + q&cos QF + rc cos QZ = 0. 
 
 Whence, eliminating cos QJf, cos QF, cos QZ succes- 
 sively, we obtain 
 
 - cos QX = - cos QF = - cos QZ, 
 u v w 
 
 where 
 
 The indices h, k, 1, p, q, r are integers; .-. u, v, w are 
 integers, and therefore (8) Q is the pole of the face (uvw). 
 Hence, it appears that a face may always exist having its 
 pole in the intersection of any two great circles, each of 
 which passes through the poles of any two faces not parallel 
 to each other. 
 
 16. When three or more faces of a crystal have their 
 poles in the same great circle, they are said to form a zone. 
 The great circle passing through the poles of any two faces 
 not parallel to each other, and which, therefore, passes through 
 the pole of any other face in the same zone with them, will 
 be called a zone-circle. The diameter which joins its poles 
 will be called the axis of the zone. A zone and its zone-circle 
 will be denoted by the same symbol. 
 
 17. It appears from (13) that if [u v w] be the symbol 
 of the zone containing the faces (hkT), (pqr), 
 
 u = kr-lq, v = lp-hr, w = hq - kp, 
 
 and from (13) that if (u v w) be the symbol of the face com- 
 mon to the zones [h k 1], [p q r], 
 
 7* = kr-lq, w = lp-hr, ^ = hq-kp, 
 
5) 
 
 or, that the expressions for u, v, w, in terms of h, k, I* 
 p, 7, r, are precisely similar to the expressions for u y v, w, 
 in terms of h, k, 1, p, q, r. 
 
 We shall sometimes find it convenient to denote the 
 zone containing the faces (h k I), (p q r), by the symbol 
 [Iik I, pqr~\, and the face common to the zones [hkl], 
 [p q r], by the symbol (hkl, p q r). 
 
 18. The intersections of the faces of a zone, or of the 
 faces produced, are parallel to the axis of the zone, and 
 therefore, to one another. In many cases the parallelisms of 
 the edges resulting from the intersections of a series of faces 
 belonging to the same zone can be ascertained by simple 
 inspection. The method of determining by observation 
 whether a face does or does not belong to a zone contain- 
 ing two given faces, when it does not meet them, or when 
 the edges it makes with them are so short that their 
 parallelism is doubtful, will be described when we come to 
 explain the use of Wollastori's Goniometer. 
 
 When, by observing the parallelism of the edges or other- 
 wise, it has been ascertained that a face is in the same zone 
 with two given faces, and that it is also in the same zone with 
 two other given faces, the symbols of the two zones, and, 
 from these, the symbol of the face which is common to 
 them, may be found by the methods of (14) and (15). 
 
 19. The points in which any two zone-circles intersect 
 are the opposite extremities of a diameter of the sphere of 
 projection. Hence, (ll) the symbol of one point of intersection 
 is obtained from that of the other by merely changing the 
 signs of all the indices. 
 
 20. If [u v w] be the symbol of the zone-circle through 
 the poles (h k I), (p q r), it is easily seen that [u v wj will 
 be the symbol of the zone-circle through the poles (hkl), 
 (pqr)\ and also that if the zone-circles [h k 1], [p q r] in- 
 tersect (u v w)> the zone-circles [h k 1], [p q r] will intersect 
 
 2 
 
10 
 
 in (uvw). Hence, if the zone-circles [h k I, pqr], [h 1 k f 1' 9 
 p q' T~\, intersect in (uvw), the zone-circles [h k I, pqr\, 
 [h'k'l', p'q'r] will intersect in (uvw). 
 
 If the zone-circles [h k I, pqr], [h'k'l', p 1 q' r'} intersect 
 in (uvw), it is manifest that [Ihk, rpq], \l r tik', r p q'] 
 intersect in (wuv), and that [h Ik, p r q], [h'l'k', p r q] 
 intersect in (u w v). 
 
 21. Let Q be the pole of a face (uvw), in the zone 
 [u v w]. Then, (IS), (8) 
 
 ua cos QX + vb cos QF-f- we cos QZ 0, 
 - cos QX = - cos QY = - cos QZ ; 
 
 U' V W 
 
 This equation expresses the condition that the face 
 (u v w) may be in the zone [u v w]. Any whole numbers 
 which, when substituted for u, v, w, satisfy the above equation, 
 are the indices of a face in the zone [uvw]; and any three 
 whole numbers which, when substituted for u, v, w, satisfy 
 the same equation, are the indices of a zone containing the 
 face (uvw). 
 
 22. When the zone-circle [u v w] passes through the 
 pole (uvw), (21) 
 
 UM + \v + vfw = 0. 
 
 Hence, in order to find the poles which lie in a given 
 zone-circle, or the zone-circles passing through a given pole, 
 we must find the integral values (one or two of which may 
 be zero), of at, y, %, which satisfy the equation 
 ax + by + c% 0; 
 
 where a, b, c are the indices of the given zone-circle in the 
 former case, and of the given pole in the latter. 
 
 Let the coefficients c, b be prime to each other. Trans- 
 form c-r-6 into a continued fraction, and let c'b' be the 
 
11 
 
 last but one of the resulting converging fractions. Then, 
 by the rule for solving indeterminate equations of the first 
 degree, y =t (cax me), % = =t (mb b'ax) ; where the 
 upper or lower sign is to be taken, according as cb' is greater 
 or less than be. The value of x being assumed, the corre- 
 sponding values of y and % may be obtained by substituting 
 different positive or negative whole numbers for m. 
 
 23. If the zone-circle [h k 1] passing through the poles 
 (h k /), (h r k 1 1'), intersect the zone-circle [p q r] passing 
 through the poles (pqr), (pV r ') ^ n ^ e P^ e (www); values 
 of A, fc 9 /, h', &', I', p, q, r, p', q', r', can always be found, 
 such that the three indices of each pole shall be severally 
 numerically less than the indices u, v, w, or not greater than 
 unity. 
 
 The values of h, k> I, h' 9 k', 1' 9 p, q, r, p\ q', r must 
 satisfy the equations 
 
 uh + vk + wl = 0> up + wq + wr = 0, 
 hh + kk + H = 0, ipp + q</ -f-rr = 0, 
 hti+kk'+ II' = 0, p/+ q^-f r/= 0. 
 
 We have, therefore, to shew that if a, b, c be any whole 
 numbers, the equation ax + by + ess = 0, may be satisfied by 
 two sets of integral values of a?, y 9 #, such that, if either of 
 them be a, /3, 7, the equation ax H- fly + 7* = 0, may be 
 satisfied by two sets of integral values of #, y, % which are 
 severally numerically less than , 6, c, or not greater than 
 unity. The most unfavourable case is that in which the 
 largest of the three numbers a, 6, c is prime to each of the 
 other two, and no two are equal. Let c be greater than -6, 
 and b greater than a. Since a is less than 6, x may have a 
 value which is either unity or some number less than a, and 
 which makes a x numerically less than by. y = (c'ax me), 
 therefore y may be made less than c, and the signs of aw, 
 by different. In which case, since -cz = by + ax, % will be less 
 than 6, and the sign of ex different from that of by, and there- 
 fore the same as that of a.v. Also, since i * - (mb - b'ax), 
 
12 
 
 with either the same value of a?, or a different value, which is 
 either unity or a number less than a, a value of z may be 
 found less than b, and making the signs of a#, cz different. 
 And therefore the corresponding value of y will be less than c. 
 Hence, if either of the above values of a?, y, % be a, /3, y re- 
 spectively, a will be less than a, or unity, j3 less than c, and y 
 less than b. In like manner, the equation ax+fiy+y%=0 9 may 
 be satisfied by two sets of values of a?, v, * less than a, /3, 7, 
 or not greater than unity, and therefore less than , 6, c, or 
 not greater than unity. Hence values of A, &, /, A', A;', ', 
 p, q, r, p', q, /, can always be found capable of satisfying 
 the equations by which they are connected with w, v 9 w, 
 such that the three indices of each pole shall be lower 
 numbers than u, v, w, or not greater than unity. 
 
 24. It appears from the preceding investigation, that the 
 pole of any face (u v w) is the intersection of two zone-circles, 
 each of which passes through the poles of two faces having 
 indices more simple than those of (u v w). In the same 
 manner, the poles of each of these faces is the intersection of 
 two zone-circles passing through the poles of faces having 
 indices still more simple, and so on, till at last we arrive at 
 the poles of four faces the indices of which are either unity 
 or zero. 
 
 25. Having given the distances between four poles in 
 the same zone-circle, and the symbols of three of them ; to 
 find the symbol of the fourth. 
 
 Let P, Q, R, S, (fig. 6), be the four poles in one 
 zone-circle, and let their symbols be P (efg), Q (&&/), 
 R(jpq r), S (u v w) ; X, F, Z the extremities of radii drawn 
 parallel to the axes of the crystal ; a, b, c the parameters of 
 the crystal. Then, 
 
 - cos PX = - cos PY = - cos PZ, 
 
 e f g 
 
 ~ cos QX= - cos QF- - cos 
 n if I 
 
13 
 
 - cos RX = - cos RY = - cos RZ, 
 p q r 
 
 - cos SX = - cos SY = - cos SZ. 
 
 U V W 
 
 P, Q, R are in the same great circle, and PQ is less than 
 PR, therefore, (12), 
 
 - ["cos P^cos QY - cos PFcos QX] 
 smPQ L 
 
 = -7-^ [cos Q^Tcos RY - cos QFcos RX], 
 sm QR 
 
 [cos PZ cos Q X - cos PX cos QZ] 
 L 
 
 sin PQ 
 
 -; ^r^ fcos QZ cos ^^T - cos Q^Tcos JffZ], 
 sin QR L 
 
 . * [cos PFcos QZ - cos PZ cos QY] 
 sm PQ L 
 
 [cos QFcos RZ - cos QZ cos RY]. 
 L 
 
 sm 
 
 P, *S f , ^ are in the same zone-circle, and PS is less than 
 PR, therefore, (12), 
 
 [cos P^"cos SF- cos PFcos SX] 
 
 sin 
 
 [cos ^^cos RY- cos STcos RX], 
 
 sm 
 1 
 
 [cos PZ cos SX - cos PX cos SZ] 
 
 sm 
 
 . net 
 
 sin PS 
 
 [cos Z cos JF2 JT - cos SX cos J?Z1, 
 [cos PFcos ^Z - cos PZ cos 
 
 Whence, dividing each of .the equations between sin PS\ 
 >in SR* by the corresponding equation between sin PQ, 
 
14 
 
 sin QR, and substituting for the ratios of the cosines the 
 values given above, 
 
 [P, Q] [S^ff ] = [Q,fl] [P,S] 
 sin PQ sin SR sin QR sin P# ' 
 where 
 
 [P, Q] _fl-gk gh-el _ek-fh 
 [Q, jR] "" kr-lq ~ Ip-hr ~ hq-kp ' 
 
 *, j vr wg' ^jt> ur uq up 
 
 Sin PQ sin SR, sin QR sin P^ must be to each other 
 as some whole numbers. Let 
 
 sin PQ sin SR _ m [S, R] _ m [Q, R] 
 
 sinQR sin PS ~ n" [P, S] ~ ~n [P, Q] " 
 
 Whence, 
 
 , Q], 
 , Q], 
 
 When P*^ is greater than Pjff, we have 
 
 [P, Q] [j, ^] = [Q, Jg] [P, ^ 
 sin PQ sin RS sin Q# sin PS ' 
 where 
 
 [P, S] _fw -gv ^gu - ew _ ev -* fu 
 [-ff, S] qw - rv ru-pw pv - qu 
 Whence 
 
 , R'] -np[P, Q], 
 
 ] -nr[P, Q]. 
 
 26. Having given the symbols of four poles in the 
 same zone-circle, and the distances between three of them ; 
 to find the fourth. 
 
15 
 
 Let 
 
 [P,Qm*]s _ 
 
 ~' - 
 
 tan (PS - %PR) = tan 1P# tan (- - 0\ . 
 
 27. Sin Q# = sin PR sin PQ [cot PQ - cot PR], 
 
 sin # = sin PR sin P#[cot PS - cot PR]. 
 
 Therefore 
 
 cot PS - cot P# _ [P, Q] [tf, #] 
 cotPQ-cotPtf " [Q, fl] [P, S] ' 
 
 28. The axes of any three possible zones, not being 
 in one plane, may be employed as crystallographic axes. 
 
 Let X> Y 9 Z (fig. 7) be the points in which the axes 
 of the crystal meet the surface of the sphere of projection ; 
 a, ft, c the parameters of the crystal; X'^ Y', 7! the poles 
 of any three zone-circles intersecting in A, B, (7; P the 
 pole of any face of the crystal ; M the intersection of 
 CA, BP-, N the intersection "of AB, CP. Let the symbols 
 of the poles be A(efg), B(hkl), C(pqr), P(uvw), 
 v), Nfrpv). Then (14), (15), 
 
 X = (re - pg) (hv - ku) - (pf - qe) (lu - hw), 
 u - hw ) ~ ( - Pg)(kw - Iv), 
 
 TT = (gh - el)(pv - qu) - (ek -fh)(ru - pw), . 
 p = (ek-fh)(qw-rv) -(fl-gk)(pv - qu). 
 
 A N B 9 AM C are great circles ; therefore, writing 
 X'i K, Z' for X) F, Z in (12), and dividing the equations 
 containing JT, Y, Z' by the corresponding equations con- 
 taining X) F, Z, we have 
 
16 
 
 cos AX' cos NY' - cos AY' cos NX* 
 cos AX cos NY cos A Y cos NX 
 
 cos BY' cos jyjT - cos EX' cos JVT' 
 
 cos AX' cos J/Z' - cos AT! 
 cos AX cos Jf Z cos AZi cos JWJf 
 
 cos CZ' cos J/Jf - cos CX' cosMZ' 
 cosCZcosMX-cosCXcosMZ 
 
 But 
 
 - cos AX = - cos AY = cos 
 e f g 
 
 - cos EX = - cos #F 
 
 A A; ^ 
 
 - cos CX = - cos CF = - cos CZ, 
 p q r 
 
 - cos J/JT a - cos Jlf F = - cos J/Z, 
 
 A yU V 
 
 - cos J^^ = - cos NY - cos NZ, 
 TT p a- 
 
 = 0, cos AZ' =0, cos EX' = 0, 
 
 Also 
 
 cos MX' sin MB cosMZ' cos NX' sin JVC __ cosJVF' 
 cos PX' = sin P5 * cos PZ' ' cos P^T' ~ sin PC = cos PT ' 
 
 ^ (ep - fir) cos 
 
 cos AX cos P.<r cos EX cos PF' ' 
 
 e(r\ - pv) cos AX' p(ev-g\)cosCZ' 
 cos AX cos PX' " ^ c 
 
17 
 
 kir - hp *= (/A - ek) [ew -I- fv + 
 ep -fir = (fh - ek) \\iu + kv + 
 r\pv- (pg re) [ew + f v + gw>], 
 ev - g\ = (pg re) [pw + qw + r0]. 
 Where 
 
 e = kr-lq, f=lp-hr, g = hq - kp, 
 h = qg-rf, k = re-pg, l=pf-qe, 
 
 Hence 
 
 /i/ r 
 
 a -, cos PX' = , cos PY' = , cos 
 w r w ' 
 
 where a', 6', c' depend only upon the angles between the 
 old and new axes, and 
 
 u = eu + fv + gw, 
 
 v = hu + kv + !?, 
 z^' = pw + qv + r^. 
 
 w', u', w' are whole numbers, and therefore OX ', OF', OZ' 
 may be taken for crystallographic axes. 
 
 The coefficients of w, v, w> in the expressions for u ', t>', w', 
 are the indices of the zone-circles BC, CA, JB, their sym- 
 bols being EC [efg], CA [h k 1], J5 [pqr]. 
 
 29. To change the parameters of a crystal. 
 
 Let (h k I) be the symbol of a face P, with parameters 
 , 6, c; (ti k' I') the symbol of P, when referred to the 
 same axes, with parameters a', 6', c'* 
 
 Then 
 
 and 
 
 cos P^ = cos PF = - cos PZ, 
 h k I 
 
 a b c 
 
 -,cosPX=- cosPF=-7 cosPZ; 
 
 ' ~L r 1 f 
 
 a a o b c c 
 
 ' ' i/ t" ' 7^ ** 7" ' T^ =I T * 
 
 h h k k I I 
 
18 
 
 30. P, Q, .ft, S are four poles in the same zone-circle. 
 When they are referred to a system of axes meeting the sur- 
 face of the sphere of projection in X, F, Z, their symbols 
 are P(efg), Q(hkl) R(pqr), S(uvw). When they are 
 referred to a system of axes meeting the surface of the 
 sphere of projection in X\ F', Z', their symbols are P(ef'g), 
 
 'A?'0, R(pq'r') 9 S(u'vw). If 
 
 [P, Q] fl-gk gh -el _ ek - fh 
 -Ar = hq-kp' 
 
 [Q, R] kr-lq 
 [P, ] /w ^" 
 
 - ew ev - fu 
 
 j?] vrwq wp ur uq vp 
 
 we have (25) 
 
 [P, Q] 
 
 [P, 
 
 sin PQ sin 
 
 sn 
 
 sin PS 
 
 And if 
 
 [P ; , Q'] 
 
 -g'k' qti - el' e'k' -fh' 
 
 [P, S'] f'w'-g'v g'u'-e'w e'v'-g'u 
 [S r , R'] = v'r' - w'q' = w'p' - u'i> = u'q' - v'p' 
 
 we have 
 
 Let 
 
 [p 
 
 [Q, g] [P,S] [F, Q'] m 
 [P, Q] [5, 5] [Q', R'] n ' 
 
 whence 
 
 [*". 
 
 n 
 
19 
 
 31. Having given the symbols of four poles P, Q, R, S, 
 (fig. 8.) when referred to each of two systems of axes 
 OX, OF, OZ; OX', OF', OZ', and the symbol of any 
 other pole referred to the axes OX, OF, OZ ; to find its 
 symbol when referred to OX', OF', OZ'. 
 
 Let P, Q, R, S be the poles ; ABC the polar triangle 
 of X' F' Z'. Therefore the symbols of A, B, C, when 
 referred to OX', OF', OZ', will be A (l 0), B(0 1 0), C(0 l). 
 Let PQ, RS meet each other in T, and let them meet EC 
 in U, V. The symbol of T may be found when referred to 
 OX, OF, OZ and also when referred to OX', OF', OZ'; 
 and the symbols of 7, V may be found when referred to 
 OX', OV, OZ'. The symbols of P, Q, T when referred to 
 OX, OF, OZ, and to OX', OF', OZ', and the symbol of U 
 when referred to OX', OF', OZ', being known, the symbol 
 of U may be found when referred to OX, OF, OZ. In like 
 manner may be found the symbol of V when referred to 
 OX, OF, OZ. Hence may be found the symbol of the zone- 
 circle BC when referred to OX, OF, OZ. In like manner 
 may be found the symbols of CA, AB referred to OX, OF, 
 OZ. Hence, the symbols of the zone-circles being known, 
 the symbol of any pole when referred to OX', OF', OZ' 
 may be found by (28). 
 
 32. In many crystals axes may be discovered which 
 make right angles with each other ; in others, axes of which 
 one is perpendicular to the other two, and in others axes 
 making equal angles with each other. In the crystals with 
 equiangular axes, and in some of the crystals with rectangular 
 axes, equal parameters may be found, and, among the remain- 
 ing crystals with rectangular axes, some which have two of 
 the parameters equal. Upon the differences in the positions 
 of the axes with respect to each other, and in the relation 
 between the parameters, above enumerated, is founded the 
 arrangement of crystals in systems. 
 
 1. In the Octahedral system the axes are rectangular 
 and the parameters equal. 
 
20 
 
 2. In the Pyramidal system the axes are rectangular 
 and two of the parameters equal. In this system we shall 
 always suppose a and b equal. 
 
 3. In the Rhombohedral system the axes make equal 
 angles with each other, and the parameters are equal. 
 
 4. In the Prismatic system the axes are rectangular. 
 
 5. In the Oblique-Prismatic system one axis is perpen- 
 dicular to each of the other two. We shall always suppose 
 the axis OF perpendicular to each of the axes OZ and OX. 
 
 6. The Doubly-Oblique-Prismatic system includes all 
 crystals which cannot be referred to either of the preceding 
 systems. 
 
 33. The different systems of crystallization are further 
 distinguished by the various kinds of symmetry observable 
 in the distribution of the faces of the crystals belonging to 
 them. For, if a face occur having the symbol (h k I), it 
 will generally be accompanied by the faces having for their 
 symbols certain arrangements of A, i k, /, determined 
 by laws peculiar to each system, and which will be fully 
 explained when we come to describe each system separately. 
 
 34. " A form " in crystallography is the figure bounded 
 by a given face and the faces which, by the laws of symmetry 
 of the system of crystallization, are required to coexist with 
 it. A form will be denoted by the symbol of any one of its 
 faces enclosed in braces. Thus, the symbol JA k l\ will be 
 used to express the form bounded by the face (h k I) and its 
 coexistent faces. 
 
 The " holohedral forms " of any system are those which 
 possess the highest degree of symmetry of which the system 
 admits. " Hemihedral forms" are those which may be derived 
 from a holohedral form by supposing half of the faces of 
 the latter omitted according to a certain law. 
 
 The figure bounded by the faces of any number of forms, 
 is called a " combination" of those forms. 
 
21 
 
 35. The elements of a crystal are the inclinations of the 
 axes FZ, ZX, XY, and the ratios of two of the parameters 
 a, 9 by c to the third. In the octahedral system, where the 
 axes Are rectangular and parameters equal, all the elements 
 are determined. In the pyramidal system, where the axes 
 are rectangular and two of the parameters are equal, the 
 ratio of either of them to the third, is the only variable 
 element. In the rhombohedral system, where the axes make 
 equal angles and the parameters are equal, the angle between 
 any two of the axes is the only variable element. In the 
 prismatic system, where the axes are rectangular, the ratios 
 of two of the parameters to the third, are two variable 
 elements. In the oblique-prismatic system, where one of the 
 axes is at right angles to the other two, the inclination of 
 the two axes which are perpendicular to the third, and the 
 ratios of two of the parameters to the third, are three 
 variable elements. In the doubly-oblique-prismatic system, 
 the angles between the axes, and the ratios of two of the 
 parameters to the third, are all variable. 
 
 36. The angle between two faces, the symbols of which 
 are known, may be expressed in terms of the indices of the 
 faces, the inclinations of the axes, and the ratios of the 
 parameters. Therefore, one observed angle between two 
 known faces, in the pyramidal and rhombohedral systems; 
 two observed angles in the prismatic; three in the oblique 
 prismatic, and five in the doubly-oblique-prismatic system, 
 are sufficient to determine the variable elements in the re- 
 spective systems. In the three latter systems, however, 
 except in particular cases, the above direct method of find- 
 ing the elements of a crystal is impracticable on account of 
 the high dimensions of the resulting equations. Methods of 
 deducing the elements of a crystal from the requisite number 
 of angles between faces properly selected, adapted to each 
 particular system, will be given in the chapter devoted to 
 that system. 
 
CHAPTER II. 
 
 OCTAHEDRAL SYSTEM. 
 
 37. IN the octahedral system the crystallographic axes 
 are at right angles to each other, and the parameters , 6, c 
 are all equal. 
 
 38. The holohedral form \hkl\ is bounded by all the 
 faces having for their symbols the different arrangements 
 of =t A, J= &, /, taken three at a time. When A, A;, / are 
 all different, they afford the forty-eight arrangements con- 
 tained in the annexed table. When the values of any two 
 of the indices are equal, or when one of them is zero, the 
 number of arrangements will reduce itself to twenty-four. 
 When two of the indices are equal, and the third is zero, 
 the number will be twelve. When the three indices are 
 equal, it will be eight, and when two indices are zero, it will 
 be six. 
 
 h k I 
 
 k I h 
 
 Ihk 
 
 Ikh 
 
 k h I 
 
 h Ik 
 
 hll 
 
 klh 
 
 Ihk 
 
 Ikh 
 
 khl 
 
 hll 
 
 hkl 
 
 h I h 
 
 Ihk 
 
 Ikh 
 
 Ihl 
 
 h I h 
 
 hi I 
 
 klh 
 
 Ihk 
 
 Ikh 
 
 kh I 
 
 hlk 
 
 111 
 
 klh 
 
 Ihk 
 
 Ikh 
 
 khl 
 
 hll 
 
 hk I 
 
 k Hi 
 
 Ihk 
 
 Ikh 
 
 kh I 
 
 h Ik 
 
 hi I 
 
 klh 
 
 Ihk 
 
 I kh 
 
 kh I 
 
 hlk 
 
 hkl 
 
 k 1 7i 
 
 I hi 
 
 I k h 
 
 khl 
 
 h I I 
 
-23 
 
 If we suppose // to be the greatest, and / the least of the 
 three unequal indices /*, &, /, fig. 9, will represent the dis- 
 tribution of the poles of the form {hkl\ on the surface of 
 the sphere of projection. Fig. 10. exhibits the poles of the 
 forms obtained by substituting zero for one of the indices, 
 or by making two of them equal. Both figures shew the 
 poles of the forms {lOOJ, {ill} and {011^. 
 
 39. The form bounded either by all the faces of \hkl\ 
 which have an odd number of positive indices, or by all the 
 faces of {h k 1} which have an odd number of negative in- 
 dices, is said to be hemihedral with inclined faces, and will 
 be denoted by the symbol ic{hkl} 9 where (h k I) is the 
 symbol of any one of its faces. The hemihedral form bounded 
 by the faces which have an odd number of positive indices, 
 is said to be direct. The form bounded by the faces which 
 have an odd number of negative indices, is said to be in- 
 verse. The upper and lower halves of the table in (38), 
 contain the symbols of the faces of the direct and inverse 
 forms respectively. 
 
 If the surface of the sphere of projection be divided 
 into eight triangles by zone-circles through every two of 
 the poles of the form {l 00}, the poles of the direct hemi- 
 hedral form will be found in four alternate triangles, one 
 of which contains the poles of (ill); and the poles of the 
 inverse hemihedral form will be found in the remaining four 
 alternate triangles 
 
 40. The form bounded either by all the faces of {h k l\ 
 the indices of which stand in the order h k I h k, or by all the 
 faces of \h k /} the indices of which stand in the order Ik h I k, 
 is said to be hemihedral with parallel faces, and will be 
 denoted by the symbol ir\hkl\^ where (h k I) is the sym- 
 bol of any one of its faces. The form is said to be direct 
 or inverse according as the numerical values of the indices 
 of one of its faces are in ascending or descending order. 
 The symbols of the faces of the direct and inverse forms are 
 
24 
 
 contained respectively in the right and left halves of the 
 table in (38). 
 
 If the surface of the sphere of projection be divided 
 into twenty-four triangles by zone-circles passing through 
 every two of the poles of the form {l 1 1^, the poles of the 
 direct hemihedral form will be found in twelve alternate 
 triangles one of which is (l 1 1) (0 1 0) (i i 7), and the poles 
 of the inverse form will be found in the remaining twelve 
 alternate triangles. 
 
 41. Any number of holohedral forms may occur in 
 combination with each other, and with any hemihedral forms 
 with inclined faces, or with any hemihedral forms with 
 parallel faces. It is said that hemihedral forms with in- 
 clined faces have never been observed in combination with 
 hemihedral forms with parallel faces. 
 
 42. To find the position of the pole of any face. 
 
 Let the axes of the crystal meet the surface of the sphere 
 of projection in X^ F, Z, (fig. 11); and let P be the pole of 
 the face (h k 1). The axes are rectangular ; therefore FZ, 
 Z X> XY are quadrants, therefore cos FZ = 0, cos ZX = 0, 
 cos XY = 0, therefore X, F, Z are the poles of the faces 
 (100), (010), (0 1) respectively. The quadrantal tri- 
 angles PYX, PYZ give 
 
 (cos PY) 2 = (sin PX) 2 (cos PXY)\ 
 
 (cos pzy = (sin p^) 2 (cos pxzy. 
 
 Add, observing that (cos PXY)* + (cosP^Z) 2 = 1, and that 
 (cos PX)* + (sin PX) Z =1, and we have 
 
 (cos PX)~ + (cos PY) 2 + (cos PZ) 2 = 1. 
 The parameters are equal, therefore (8), 
 
 - cos PX = - cos PY = - cos PZ, 
 
 h K I 
 
25 
 
 h* 
 
 
 43. To find the distance between the poles of any two 
 faces. 
 
 Let P (fig. 11) be the pole of (h k I), Q the pole of (p q r). 
 cos PXQ = cos PXY cos QXY + sin PXY sin QXY, 
 = cos P^Fcos QXY + cos PXZ cos QXZ. 
 
 Substituting this value of cos PXQ in the equation 
 
 cos PQ = cos PX cos QX + sin PX sin Q X cos PXQ, 
 and observing that 
 
 sin P^cos PXY = cos PF, sin Q^cos Q^F= cos QF, 
 sin P^Tcos PXZ = cos PZ, sin Q^T cos QJTZ = cos QZ, 
 we get 
 cos PQ = cos P^cos QX + cos PFcos QF + cos PZ cos QZ. 
 
 But (42) 
 
 .-. cos PQ 
 
 4 
 
44. The three quadrantal triangles FPZ, ZPX, XPY, 
 give 
 
 cos PX = sin PY cos PYX = sin PZ cos PZX, 
 
 cos PF = sin PZ cos PZF = sin PXcos PJfF, 
 cos PZ = sin PXcos PXZ = sin PFcos PFZ. 
 Whence 
 
 tanP^F=- 5 tan PFZ =7, tanPZ^=^. 
 k I h 
 
 45. Let be the pole of the face (l 1 l), /. (42), 
 
 = , (cos OF) 2 = 1, (cosOZ) 2 = l; whence (43), 
 
 OJT=OF=OZ, and FZ, Z^T, -^F are quadrants. 
 Hence, the angles FOZ, ZOX, XOY are each equal to 
 120, and OX, OF, OZ bisect the right angles F^Z, 
 ZYX 9 XZY. From the triangle POX we have 
 
 cot P-^sin XO = cos XO cos OJTP + sin O^TPcot POX. 
 
 If we write F, Z successively in the place of X, and sub- 
 titute for the trigonometrical ratios their values in terms 
 of h, k> I, we obtain 
 
 46. It appears from the form of the expressions in 
 (43), that the distance between the poles of the faces 
 (hkl)> (pq r) is equal to the distance between the pole of 
 any face of the form [hkl], and the pole of any face 
 
27 
 
 of the form {pqr}, in the symbols of which the order 
 and signs of h, k, I are the same as the order and signs 
 of p, q, r. 
 
 47. It appears from the expressions in (42), that if 
 the symbols of two poles of the form \h kl\ differ only 
 in the sign of h, the two poles will be equidistant from 
 (0 1 0) and also from (0 l), therefore the arc joining the two 
 poles will be bisected at right angles by the zone-circle 
 [0 1 0, 00 l]. Hence the poles of \h k l\ are symmetri- 
 cally arranged with respect to the zone-circle [0 1 0, l]. 
 In like manner it may be shewn, that the poles of \hkl\ 
 are symmetrically situated with respect to any one of the 
 three zone-circles that can be drawn through every two of 
 the poles of {l 00^. 
 
 48. It appears from (43), that if the symbols of two 
 poles of the form {h k l^ differ only in the arrangement 
 of the 2nd and 3rd indices, the poles will be equidistant 
 from (I i i) and also from (l 1 l), therefore the arc joining 
 the two poles will be bisected at right angles by the zone- 
 circle [111, ill]. Hence the poles of \h k 1} are sym- 
 metrically arranged with respect to the zone-circle [I l l, l l l]* 
 In like manner it may be shewn, that the poles of {A k 1} 
 are symmetrically arranged with respect to any one of the 
 six zone-circles that can be drawn through every two of the 
 poles of {l 1 l}. 
 
 49. If zone-circles be drawn through every two of the 
 poles of |lOO|, and through every two of the poles of 
 { 1 1 1 J , they will divide the surface of the sphere of pro- 
 jection into forty-eight right-angled triangles. The poles of 
 
 ^h k /| are symmetrically arranged with respect to any side 
 of any one of the triangles. Hence the arrangement of the 
 poles will be symmetrical in any two adjacent triangles, and 
 similar in any two alternate triangles. 
 
 50. If zone-circles be drawn through every two of the 
 poles of {100} , and through every two of the poles of 
 
28 
 
 ^ 1 1 1 } , the zone-circles of one set bisect symmetrically the 
 triangles formed by the zone-circles of the second set. Hence, 
 the poles of K \hkfy are symmetrically arranged with respect 
 to the zone-circles drawn through every two of the poles of 
 {l 1 ij; and the poles of TT \hkl} are symmetrically arranged 
 with respect to the zone-circles drawn through every two 
 of the poles of {lOO}. 
 
 51. If we examine the situations of the poles of two 
 hemihedral forms, either with inclined or parallel faces, de- 
 rived from the same holohedral form, one of them being 
 direct and the other inverse, we shall find that the two forms 
 are identical in all respects, position excepted, and that one 
 of them may be brought into the position of the other by 
 making it revolve through a right angle round any one of 
 its crystallographic axes. In like manner, the combinations 
 of a holohedral form with a direct and inverse hemihedral 
 form differ only in position. But in the combinations of any 
 two hemihedral forms with inclined faces, or of any two 
 hemihedral forms with parallel faces, with each other, the 
 poles of the two forms lie in the same or in different tri- 
 angles according as they are of the same or different deno- 
 minations. Hence a combination of two direct, or of two 
 inverse hemihedral forms, is essentially different from the 
 combination of the same hemihedral forms, when one of 
 them is direct and the other inverse. 
 
 52. If the distance between the poles of any two faces 
 of either of the forms {hkO\, {hkk} be given, and we ex- 
 press the cosine of the given distance in terms of the indices 
 of the faces, we obtain an equation from which the ratio of 
 the indices may be deduced. 
 
 53. If the distances between the pole of any face of 
 the form \hkl\, and the poles of each of two other faces 
 of the same form be given, and we express the cosines of 
 the given distances in terms of the indices of the faces, we 
 shall obtain two equations from which the ratios of the 
 indices may be readily found. 
 
54. To determine the figure and angles of the form 
 \hkl\, when h, k, I take particular values. 
 
 The angle between normals to any two faces, which is 
 measured by the angular distance between their poles, is 
 found by substituting the indices of the faces for A, A;, /, 
 p, q, r in the expressions of (43). The letter placed upon 
 the edge resulting from the intersection of any two faces, 
 in the figures which accompany the description of each par- 
 ticular form, will be used to denote the angle between nor- 
 mals to the two faces. The same letter will be placed upon 
 all the edges at which equal angles are formed by the in- 
 tersecting faces. The relative positions of the poles of the 
 different forms are shewn in figs. 9, 10, and in fig. 37, which 
 is the gnomonic projection of one of the octants into which 
 the surface of the sphere of projection is divided by the 
 zone-circles through every two of the poles of the form 
 {l 00}. The number of faces of each holohedral form has 
 been already determined in (38). 
 
 55. The form {lOO} (fig. 12) has six faces, and is 
 called a cube. 
 
 Hence, the faces of the form {l 00} are parallel to those of 
 a cube. 
 
 56. The form {ill} (fig. 13) has eight faces, and is 
 called an octahedron. 
 
 cosZ> = l, .-. Z> = 70.3l',7. 
 
 Hence, the faces of the form {ill} are parallel to those of 
 a regular octahedron. 
 
 The cosine of the angle between normals to any face of 
 the octahedron and either of the adjacent faces of the cube 
 is \^%. Hence, a normal to any face of the octahedron 
 makes an angle of 54. 44', 15 with the normals to each of the 
 adjacent faces of the cube, and an angle of 125.15',85 with 
 the normals to each of the three opposite faces of the cube. 
 
30 
 
 57- In the hemioctahedron with inclined faces /c{l 1 l} 
 (fig. 14), 
 
 cos T = - ^, .'. T = 109.28',3. 
 
 Hence, the hemihedral form K \ 1 1 1 } is a regular tetrahedron. 
 
 58. The form {01 l (fig. 15) has twelve faces, and is 
 called a dodecahedron. 
 
 cos 
 
 If normals to any two alternate faces, meeting at their 
 acute angles, make an angle D with each other, cos D = 0, 
 
 The cosine of the angle between normals to any face of 
 the dodecahedron and either of the adjacent faces of the 
 cube is i> -y^' The cosine of the angle between normals 
 to any face of the dodecahedron and either of the faces of 
 the cube, not parallel to the two adjacent faces, is 0. Hence, 
 a normal to any face of the dodecahedron makes an angle 
 of 45 with the normals to each of the two adjacent faces 
 of the cube; an angle of 135 with the normals to each of 
 the two opposite faces, and an angle of 90 with the normals 
 to each of the two remaining faces. 
 
 The cosine of the angle between normals to any face of 
 the dodecahedron and either of the adjacent faces of the 
 octahedron is -^ ^/6. The cosine of the angle between 
 normals to any face of the dodecahedron and either of the 
 faces of the octahedron, not parallel to the adjacent faces, 
 is 0. Hence, a normal to any face of the dodecahedron makes 
 an angle of 35.15',85 with the normals to each of the two 
 adjacent faces of the octahedron; an angle of 144. 44', 15 
 with the normals to each of the two opposite faces, and an 
 angle of 90 with the normals to each of the four remain- 
 ing faces. 
 
 59. The form \h k 0} (fig. 16) has twenty-four faces 
 and is called a tetrakishexahedron. 
 
 2 hk h 2 
 
 cos F = , cos G = . 
 
31 
 
 In 1 2 1 oj, cos F = - , cos G = - , 
 5 5 
 
 .\ F=36.52',2, G = 36.52',2. 
 In {310}, cos.F = , cosG= , 
 
 .'. / 1 =53.7',8, G = 25.50',5. 
 
 12 
 
 In {3 20}, cosF = , cos (3 = , 
 
 1 > 13 
 
 20 25 
 
 In {520|, cosF= , cosG = , 
 29 29 
 
 .'. F = 46.23 ; ,8, G = 30.27'. 
 
 The tangent of the angle between normals to any face 
 of {hko} and the nearest face of {100} is k-r-h. 
 
 60. In the hemitetrakishexahedron with parallel faces 
 (fig. 17), h being greater than A;, 
 
 A 9 
 
 = -, cos{7=-, 
 
 5 5 
 
 In7r{023|, cosD= , cos (7= , 
 13 13 
 
 7 12 
 
 In 7r{034{, cos D = , cos U = , 
 
 25 25 ' 
 
 .-. D=x 73. 44,4, f7 = 6l 
 
32 
 
 61. The form {h k k} (fig. 18), h being greater than A?, 
 has twenty-four faces, and is called an icositessarahedron. 
 
 cos D = - - , cos F = 
 
 4< 5 
 
 In |21l}, cosZ> = -, cosF=-, 
 
 .'. D = 48. 11,5, .F=33 .33',4. 
 
 Q 7 
 
 In {311}, cosZ> = , cos/" = , 
 
 .-./> = 35. 5', 8, F= 50.28',7. 
 
 62. In the hemiicositessarahedron with inclined faces 
 K [hkk} (fig. 19), 
 
 In:{21l}, cosF=-, cosT 7 ^:-, 
 
 .-. F=33.33',4, 7 7 =70.3l',7. 
 
 7 7 
 
 In/c{31l}, cosF= , 0087" = , 
 
 .*. jP = 50. 28',7, T = 50. 28',7. 
 
 63. The form {h h k} (fig. 20), h being greater than A:, 
 has twenty-four faces, and is called a triakisoctahedron. 
 
 cos D = -T T cos 
 
 8 7 
 
 In {22lJ, cosG=- ? cosZ)=- 5 
 
 i/ y 
 
 .-. G = 27.l6', Z)=38 .56,5. 
 
33 
 
 15 17 
 
 In J331}, cosG= , cosD= , 
 
 ^ i/ J-J/ 
 
 .-. G = 37. 51' ,8, D = 26. 3l',5. 
 
 64. In the hemitriakisoctahedron with inclined faces 
 K {hhk} (%. 21), 
 
 o 
 
 In /e{22l, cosG = -, cosjP=0, 
 y 
 
 .-. G = 27. 16', 7 7 =90. 
 
 65. The form {h kl\ (fig. 22) has forty-eight faces, and 
 is called a hexakisoctahedron. 
 
 h 2 +k 2 +l 2 
 
 12 13 13 
 , cosD= , cosF= , 008^ = , 
 14 14 14* 
 
 = 31.0',2, J F=21.47',2, 6? = 21.47',2. 
 
 24 25 22 
 
 InH3l}, cosZ> = , cosF = , cosG= , 
 2o 26 26 
 
 .-. D = 22. 37',2, F = 15. 56 r ,5, G = 32. 12',2. 
 
 19 17 20 
 
 In {4 2l}, cosZ>= , cosF= , cosG= , 
 
 A /) = 25. 12',5, F = 35. 57', G = 17. 45,1. 
 
 57 43 55 
 
 In{731J, cos/> = , cosjF= , cos G = , 
 
 .'. D = 14. 57',7, F = 43. 12 ; ,8, G = 21. 13',2. 
 5 
 
34 
 
 Injll53|, cosZ)= , cosF= , cos G = , 
 155 155 155 
 
 .-. D = 27. 53',2, JF = 39. 50',9, G=13.2',7. 
 
 66. In the hemihexakisoctahedron with inclined faces 
 *Jj (fig. 23), 
 
 13 13 5 
 In;{32l}, cosj^ , cosG J = , cos T= , 
 
 14 14 14 
 
 .-. jP = 21. 4/,2, 6r = 21.47',2, 7 7 =69. 4',5. 
 
 31 SI 1 Q 
 
 In/c{53l}, cos^, cosG= , 087"= , 
 o5 35 35 
 
 .-. F = 27. 39' 5 7, G = 27. S9',7, T=577 / ,3. 
 
 67. In the hemihexakisoctahedron with parallel faces, 
 Tr{lkh} (fig. 24), 
 
 h*+k*-l* h*-k*+P kl + lh 
 
 ' cosw = 
 
 In7r{l23|, cos W= , cosZ> = , cosC7= 
 .-. J7=64.37',3, Z) = 30.0' ? 3, C7=38. 12',8. 
 
 In TT \ 1 2 4 } , cos W= , cos D = , cos U = , 
 21 21 21 
 
 /. PT= 51.45',3, D = 25. 12',7, U= 48. 11 ',3. 
 
 17 ^? ^? Q Q 
 
 In7r{l35}, cos TF= , cosZ>= , 008^7 = , 
 35 35 35 
 
 .-. PT=29.3,5, Z>=19.27',8, U=*8.5!>'. 
 
35 
 
 68. Of the preceding forms, those which have the simplest 
 indices, the cube {lOO}, octahedron {ill} and dodeca- 
 hedron {oil} occur much more frequently than the others. 
 It does not appear that any cleavages have been observed 
 except parallel to the faces of one or more of the three forms 
 {100}, .{Hi}, {OH}. 
 
 TABLE of the distances between the poles of the common- 
 est forms, and the nearest poles of the forms JlOOj, 
 {Oil}, {111}. 
 
 
 100 
 
 010 
 
 001 
 
 Oil 
 
 101 
 
 110 
 
 111 
 
 210 
 
 ' 
 
 26.34 
 
 o / 
 63.26 
 
 ' 
 
 90. 
 
 i 
 
 71.34 
 
 / 
 
 50.46 
 
 f 
 
 18.26 
 
 t 
 
 39.44 
 
 310 
 
 18.26 
 
 61.34 
 
 90. 
 
 77. 5 
 
 47.52 
 
 26.34 
 
 43. 5 
 
 320 
 
 33.41 
 
 56.19 
 
 90. 
 
 66.54 
 
 53.58 
 
 11.19 
 
 36.49 
 
 410 
 
 14. 2 
 
 75.58 
 
 90. o 
 
 80. 7 
 
 46.41 
 
 30.58 
 
 45.34 
 
 430 
 
 36.52 
 
 53. 8 
 
 90. 
 
 64.54 
 
 55.33 
 
 8. 8 
 
 36. 4 
 
 520 
 
 21.48 
 
 68.12 
 
 90. 
 
 74.47 
 
 48.58 
 
 23.12 
 
 41.22 
 
 540 
 
 38.40 
 
 51.20 
 
 90. 
 
 63.47 
 
 56.29 
 
 6.20 
 
 35.45 
 
 211 
 
 35.16 
 
 65.54 
 
 65.54 
 
 54.44 
 
 30. 
 
 30. 
 
 19.28 
 
 311 
 
 25.14 
 
 72.27 
 
 72.27 
 
 64.46 
 
 31.29 
 
 31.29 29.30 
 
 122 
 
 70.31 
 
 48.11 
 
 48.11 
 
 19.28 
 
 45. 
 
 45. 
 
 15.48 
 
 133 
 
 76.44 
 
 46.30 
 
 46.30 
 
 13.16 
 
 49.33 
 
 49.33 
 
 22. 
 
 321 
 
 36.42 
 
 57.41 
 
 74.30 
 
 55.28 
 
 40.54 
 
 19. 6 
 
 22.13 
 
 421 
 
 29.12 
 
 64. 7 
 
 77.24 
 
 62.25 
 
 39-31 
 
 22.13 
 
 28. 8 
 
 431 
 
 38.20 
 
 53.58 
 
 78.41 
 
 56.19 
 
 46. 6 
 
 13.54 
 
 25. 4 
 
 531 
 
 32.19 
 
 59-32 
 
 80.16 
 
 61.26 
 
 44.11 
 
 17. 1 
 
 28.35 
 
 731 
 
 24.18 
 
 67. 1 
 
 82.31 
 
 68.24 
 
 42.34 
 
 22.59 
 
 34.14 
 
36 
 
 EXAMPLES. 
 
 69. In a crystal of Tin-white Cobalt Pyrites, (fig. 25.) 
 the normals to the faces a, a', &c. make right angles 
 with each other, and a normal to any of the faces d, d', &c. 
 makes an angle of 45 with a normal to either of the 
 adjacent faces a, a', &c. Hence, (55), (58), a, a', &c. are 
 the faces of the cube {lOOJ, and d, d', &c. are the faces 
 of the dodecahedron \ 1 1 } . Let symbols of the faces 
 be a (100), a' (010), a" (001). Therefore d (Oil), 
 d' (l 1), d" (110). o is in the zone ad, and also in 
 the zone ad'. The symbols of these zones are ad [o 1 T] 
 ad' [Tlo](l4). Therefore (15) o is (l 1 1), therefore o is a 
 face of the octahedron Jill}. Hence the crystal is a com- 
 bination of the forms {l 0}, {011}, {in}. 
 
 70. In a crystal of Fluor (fig. 26), normals to any two 
 adjacent faces o make with each other an angle of 70. 3l'l, 
 therefore (56) o, o', &c. are the faces of the octahedron {ill}. 
 A normal to any face / makes an angle of 22 with a 
 normal to the adjacent face o. The edges in which any 
 two adjacent faces f meet each other and the adjacent faces 
 o are parallel, therefore the faces f are in the same zone 
 with two adjacent faces o. Let o be (l 1 1), o (I l 1), there- 
 fore the zone oo will be [o 1 I], therefore (21) the symbol of 
 / will be (khh), and that of / (khh). If P 9 Q be the 
 poles of /, /, PQ = 70. 31' 1-2.22= 26. Si' 1 , and 
 
 2 A 8 -A? 8 17 h 2 h 
 
 ^-^ = cosPQ = cos260.31 i = -, ,.-= 9 , - - 3. 
 
 Therefore / is a face of the triakisoctahedron { 3 3 1 } . 
 Hence the crystal is a combination of the forms |lll}, 
 {33l}. It is cleavable parallel to the faces of {ill}. 
 
 71. In a crystal of Fluor (fig. 27), the angle between 
 normals to any two adjacent faces n are alternately 35. 57' 
 and 17. 45', the angle between the normals to the faces 
 
37 
 
 that meet in a long edge being the greater of the two. 
 Let X, O be the poles of (l 0), (ill); {hkl\ the 
 symbol of the form bounded by the faces n; P the pole 
 of (hkl), Q the pole of (k h I) and R the pole of (hlk). 
 Then PQ = 35.57', PR = 17. 45'. 
 
 cosPQ 
 
 > = 1Z 
 ' 21' 
 
 20 
 - 
 
 cosPJZ- 
 
 (A -&) 
 
 (A;-/) 
 
 Whence 
 
 (ft-0 2 
 
 h 2 +k 2 +l* 
 
 9_ 
 21 
 
 k*+2hl 12 (h+k+l) 
 
 21 
 
 Whence & = 2/, A = 4/, .-./=!, A; = 2, A 
 w, w', &c. are faces of the form {42l|. 
 
 49 
 21 
 
 4. Hence 
 
 72. In a crystal of Boracite (fig. 28), normals to any 
 two adjacent faces a makes right angles with each other, 
 they are therefore the faces of a cube. Let their symbols 
 be a (l 0), a (0 1 0), a" (0 l). A normal to d'' makes an 
 angle of 45 with a normal to either of the faces a, a', 
 therefore (58), d" is (l 1 0). Similarly d' is (l l), and d 
 is (0 1 1). o is common to the zones ad, ad', therefore (56) 
 o is (l 1 1). The number of faces o is four. But the ho- 
 lohedral form {ill} has eight faces. Therefore o, o, &c. 
 are the faces of the hemioctahedron AC { 1 1 1 } . Hence the 
 crystal is a combination of the forms ^100}, Jl!0},/c{lll}. 
 
 73. In a crystal of Magnetic Iron Oxide, (fig. 29), 
 normals to any two adjacent faces o make with each other 
 an angle of 70. 31 7 ^. Therefore o, o', SEC. are the faces of 
 the octahedron { 1 1 1 \ . Let the symbols of the faces be 
 o (1 1 1), o' (i i l), o" (l 1 1), o" (III). A normal to any 
 face d makes angles of 35. 16' with normals to any two 
 adjacent faces o, therefore d, d' 9 &c. are the faces of the 
 
38 
 
 dodecahedron JO 1 lj, e is common to the zones oo'", do", 
 the symbols of which are oo"' [Toi], do" [21 l], therefore 
 (15), e is (131); therefore e, e, &c. are faces of the ico- 
 sitetrahedron ^3 1 l\. Hence the crystal is a combination of 
 of the forms {ill}, {oil}, {31]}. 
 
 74. In a crystal of Garnet (fig. 30), the normals to 
 any two adjacent faces d make with each other an angle 
 of 60, therefore d, d' ', &c. are the faces of the dodecahedron 
 {Oil}. Let their symbols be d (01 l), d' (l 1), d" (1 1 0). 
 e is in the zone dd', and makes equal angles with d, d'. 
 The zone-circle [ill, ill] bisects the arc dd f (48), there- 
 fore it contains the pole of e. Therefore e is (l 2 1). 
 Similarly e t is (12!). s is in the zone e'e t \ it is also in 
 the zone dd', therefore (15), s is (12 3). Hence the crystal 
 is a combination of the forms {oil}, {21l}, {l23}. It 
 is cleavable parallel to the faces of {0 1 l}. 
 
 75. In a crystal of Silver-white Cobalt from Tunaberg 
 (fig. 31), normals to any two adjacent faces a make an 
 angle of 90 with each other, they are therefore the faces of 
 the cube. Let their symbols be a (l 00), a (0 1 0), a" (001). 
 The angle between a normal to any face o and a normal to any 
 of the adjacent faces a is 54. 44, therefore (56) o is (l 1 1). 
 The edges which d" make with a and a are parallel, therefore 
 d" is in the same zone with , a'; therefore the symbol of d" 
 will be (hkO). It is found that normals to #', d" make an 
 angle of 26. 34'. Therefore (42) 
 
 k 2 h 
 
 - = (cos 26. 34 X ) 2 , .-. - = tan 26. 34' = -L .'. d" is (1 2 0). 
 A 2 + k* k 
 
 The number of faces d is twelve, the number in the holohc- 
 dral form {012} being twenty-four. Hence the crystal is 
 a combination of the forms {lOO}, {ill}, ?r{oi2}. The 
 crystal is cleavable parallel to the faces of {l 0}. 
 
 76. In a crystal of Silver- white Cobalt (fig. 32), the 
 symbols of p, a, r, k are p (l 0), a (l 1 l), c (l 2 0), k (l 40). 
 
39 
 
 i is in the zone ac, and normals to a, i make, according to 
 Phillips, an angle of l6.33'. Let (h k I) be the symbol of i 
 The symbol of the zone ac is [2 1 l], therefore (21) 2A = k + /. 
 
 Als () * 2 + + = < cos 16 - 33 ') 2 - whence 7 - T' ver y 
 
 /*+/+ I i 
 
 nearly, therefore / = 7, & = 15, ^ = 11. The number of faces 
 e, adjacent to* , is three, therefore e is a face of the form 
 7r{7 11 15}. 
 
 77- In a crystal of Yellow Iron Pyrites (fig. 33), a is 
 a face of the cube, o a face of the octahedron. Normals 
 to s, a make with each other an angle of 57. 41, and normals 
 to s, o make with each other an angle of 22. 13'. Let the 
 symbols of the faces be a (l 0), o (l 1 l), s (h k I) ; and let 
 J, O, S be the poles of a, o, s. Then SA = 57. 4l', SO = 
 
 Whence 2& = 3h, 2l = h, .'. I = 1, ^ = 2, & = 3, .-. sis (231). 
 In the holohedral form { 1 2 3 ^ , and in the hemihedral form 
 with inclined faces, there are six faces on one angle of the 
 cube. In the present case there are only three. Hence 
 s, s', s" are faces of the direct hemihexakisoctahedron with 
 parallel faces ?r{l23}. The crystal is cleavable parallel to 
 the faces of the forms }l 00}, {ill}- 
 
 78. In a crystal of Yellow Iron Pyrites (fig. 34), p, p\ p" 
 are faces of the cube, d a face of the octahedron py" e" p', 
 p" sfds'e", p'f'y", pod, dfe are zones; and the normals 
 to p f e' make an angle of 26. 34'. Let the symbols of p 9 p', p" 
 be (1 0), (0 I 0), (0 1), then d will be (l 1 1). e" is in the 
 zone pp', therefore its symbol will be of the form (hkO); 
 
 therefore if F, P be the poles of p' 9 e", (cos PF) 2 = ** , 
 
 rt -f- rC 
 
 .'. - =tanPF=tan26.34' = i, .-. e" is (120). In like 
 
 K 
 
 manner e' is (201), and e is (012). o is common to the 
 
40 
 
 zones pd, p'e, therefore o is (2 1 1). s is common to the zones 
 pe, p" e\ therefore s is (1 2 4). /is common to the zones de, 
 p"e") therefore f is (l 2 3). y is common to the zones pf, p'p", 
 therefore y is (0 2 S). There is only one face e, and one face y, 
 between every two adjacent faces of the cube ; and only three 
 faces/, and three faces s, round each of the faces d; therefore 
 e, q, /, s belong to hemihedral forms with parallel faces. 
 Hence, the crystal is a combination of the forms Jill}, 
 {100}, {211}, 7r{012}, 7r{023}, 7r{l2S}, 7r{l24}. 
 
 79. In a crystal of Fahlerz (fig. 35), /, /, /' are the faces 
 of the cube. Each of the faces d, d', &c. is in the same zone 
 with the two adjacent faces a, and makes equal angles with 
 them, therefore d, d', Sec. are faces of the dodecahedron. Let 
 /, /, /" be (1 0), (0 10), (003) respectively, therefore d, d', 
 &c. will be d (0 1 1), d f (l l), d" (l 1 0), d t (l o I), p is com- 
 mon to the zones f f d^f'd r > therefore p is (i i 1). r" is com- 
 mon to the zones dd f , f ft d", therefore r" is (112). In like 
 manner we have / (l 2 1), r t (2 1 T) ? r n (l 2 I), s is common to 
 the zones//', rV, therefore s is (l 3 0). n is common to the 
 zones f ff d" 9 r f r ff9 therefore n is (3 3 1). The edges and solid 
 angles formed by r, /, T" are not truncated by any faces 
 corresponding to p, n, therefore p 9 n belong to hemihedral 
 forms with inclined faces. Hence the crystal is a combination 
 of the forms {lOO}, {no}, /c{lll} ? {310}, {21]}, 
 ~ 
 
CHAPTER III. 
 
 PYRAMIDAL SYSTEM. 
 
 80. IN the pyramidal system the crystallographic axes 
 make right angles with each other, and two of the parameters, 
 a, b are equal. 
 
 81. The holohedral form {hkl} is bounded by all the 
 faces which have for their symbols the different arrangements 
 of =fc h 9 A;, /, in which / holds the last place. When 
 h, k 9 I are all different they afford the sixteen arrangements 
 contained in the annexed table. When one of the indices is 
 zero, or when h, k are equal, the number will be eight. When 
 I is zero and h = k, or when one of the indices h 9 k is zero, the 
 number will be four. When h and k are zero it will be two. 
 
 Jik I 
 
 kh I 
 
 hkl 
 
 khl 
 
 hk I 
 
 khl 
 
 hkl 
 
 khl 
 
 khl 
 
 hkl 
 
 kh I 
 
 hkl 
 
 khl 
 
 hkl 
 
 kh I 
 
 hk I 
 
 n> iv v lit n i> n, it/ v 10 n / 
 
 If we suppose h to be greater than k 9 fig. 38 will represent 
 the arrangement of the poles of the form {h k l\ on the surface 
 of the sphere of projection. 
 
 82. The form bounded either by all the faces of {hk 1} 
 which have an odd number of positive indices, or by all the 
 faces of \h k l\ which have an odd number of negative indices, 
 is said to be hemihedral with inclined faces, and will be denoted 
 by the symbol K\h k /}, where (h k I) is the symbol of any one 
 6 
 
42 
 
 of its faces. The hemihedral form bounded by faces which 
 have an odd number of positive indices is said to be direct. 
 The form bounded by faces having an odd number of negative 
 indices is said to be inverse. The symbols of the direct form 
 are contained in the first and second columns, those of the in- 
 verse form in the third and fourth columns of the table in (81). 
 
 If the surface of the sphere of projection be divided into 
 eight triangles by zone-circles through the poles of { 1 } and 
 {lOO}, the poles of the direct form will be found in four 
 alternate triangles one of which contains the pole of (ill). 
 The pole of the inverse form will be contained in the remain- 
 ing four alternate triangles. 
 
 83. The pyramidal system admits of a second hemihedral 
 form with inclined faces, which is bounded by all the faces of 
 {hkl\, in which the order of h, k changes with the sign of 
 I, and which will be denoted by the symbol K '{h kl}, where 
 (h k I) is the symbol of any one of its faces. The form is said to 
 be direct or inverse, according as the first index is greater or 
 less than the second, when the three indices have the same sign. 
 The symbols of the faces of the direct form are contained in 
 the first and fourth columns, those of the inverse form in 
 the second and third columns of the table in (81). 
 
 If the surface of the sphere of projection be divided into 
 eight triangles by great circles through the poles of {001} 
 and \1 1 0}, the poles of the direct form will be found in four 
 alternate triangles, one of which contains the pole of (l 1). 
 The poles of the inverse form will be found in the remaining 
 four alternate triangles. 
 
 84. The form bounded by all the faces of {h kl}, in 
 which the order of h, k is the same or different according as 
 A, k have the same or different signs, is said to be hemihe- 
 dral with parallel faces, and will be denoted by the symbol 
 7r{h k l^j where (h k 1) is the symbol of any one of its faces. 
 The form is called direct or inverse according as the first index 
 is greater or less than the second, when the two indices have 
 
43 
 
 the same sign. The symbols of the faces of the direct form 
 are contained in the first and third columns, those of the in- 
 verse form in the second and fourth columns of the table in 
 (81). 
 
 If the surface of the sphere of projection be divided into 
 eight lunes by zfcne-circles through the poles of (0 1) and the 
 poles of Jl Oof, {ll Q}J the poles of the direct hemihedral 
 form will be found in four alternate lunes, one of which is 
 (l 0), (110); and those of the inverse form in the remaining 
 four alternate lunes. 
 
 85. To find the position of any pole. 
 
 Let the axes of the crystal meet the surface of the sphere 
 of projection in X, F, Z (fig. 39). Let a, a, c be the parame- 
 ters of the crystal ; P the pole of (hkl). Since the axes 
 of the crystal are rectangular, FZ, ZJT, XY are quadrants, 
 therefore cos FZ = 0, cosZ^T=0, cosJ5TF=0, therefore JT, 
 F, Z are the poles of (l 0), (0 1 0), (0 1), and the angles 
 at X^ F, Z right angles. 
 
 cos PX = sin PY cos PYX = sin PZ cos PZX, 
 cos PY = sin PZ cos PZF = sin PJTcos PXY, 
 cos PZ = sinPJTcosP^Z = sin PFcos PFZ. 
 
 cot PX = tan PZF cos P^F - tan PFZ cos PXZ, 
 cot PF = tan PXZ cos PFZ = tan PZXcos PF^, 
 
 cot PZ = tan PYX cos PZ^T = tan P^Fcos PZF. 
 But 
 
 - cos PX = ? cos PF = - cos PZ. 
 h k I 
 
 Whence 
 
 tan P^F= \ - , tan PF^= [ - , tan PZX = \ . 
 kc he h 
 
44 
 
 cot PX= - cos PXY = h - cos 
 k la 
 
 cot PF = - cos PF^T = - - cos PFZ, 
 A e a 
 
 cot PZ = - - cos PZJT = - - cos PZF, 
 
 ^ c k c 
 
 86. The poles of { 1 1 0} bisect the arcs joining any two 
 adjacent poles of {l 00^. For if N be any pole of {l 1 OJ; 
 X> Y the adjacent poles of {lOO}, it will be found that 
 cot NX =1, cotJVF=l, therefore NX, NY are each 45, 
 therefore N bisects XY. 
 
 87- It appears from the form of the expressions in (85), 
 that the distances of the poles of \hkl} from the nearest of 
 the two poles (001), (ool) are all equal; and that the 
 angles subtended at (0 l) or (o 1) by the arcs joining any 
 pole of {h kl} and the nearest pole of {lOO} are all equal. 
 Hence, it may be easily shewn, that the poles of the form 
 | A k /| are symmetrically arranged with respect to each of the 
 five zone-circles that can be drawn through every two of the 
 poles of the three forms {00 l}, {l 00}, {l 1 0}. 
 
 88. If the surface of the sphere of projection be di- 
 vided into sixteen triangles, by the five zone-circles drawn 
 through the poles of every two of the forms {O0lj, {lOOj, 
 {l 1 o}, the poles of {h k /} will be symmetrically arranged 
 with respect to any side of any one of the triangles. Hence 
 the arrangement of the poles of ^hkl\ will be symmetrical, 
 in any two adjacent triangles, and similar in any two alternate 
 triangles. 
 
 89- The poles of x{hkl} are symmetrically arranged 
 with respect to the two zone-circles, drawn through the poles 
 
45 
 
 of {001} and the poles of {l 1 0}. The poles of K '\hkl\ 
 are symmetrically arranged with respect to the two zone-circles 
 through the poles of {001} and the poles of {l 00}. The 
 poles of Tr\hkl\ are symmetrically arranged with respect to 
 the zone-circle passing through the poles of {lOO}. The 
 arrangement of the poles of ir\hkl\ will be similar in any 
 two triangles on the same side of the zone-circle through the 
 poles of [l oj, and symmetrical in any two triangles on oppo- 
 site sides of it. 
 
 90. The direct and inverse forms differ only in position. 
 For, if the sphere of projection be made to revolve through 
 two right angles round any two opposite poles of {l OOj, the 
 poles of the direct form will change places with the poles of the 
 inverse form in the hemihedral form with parallel faces and in 
 the first hemihedral form with inclined faces. And, in the 
 second hemihedral form with inclined faces, if the sphere of 
 projection be made to revolve through two right angles round 
 any two opposite poles of {l 1 0}, the poles of the direct form 
 will change places with the poles of the inverse form. 
 
 91. In the form {h k 0} , if the distance between two poles 
 be JT, F or Jf, according as their symbols differ only in the 
 sign of k, the order of the indices h, k, or in the order of the 
 indices h, k, and sign of one of them, then (85) 
 
 92. In the form |AO/|, if L be the distance between 
 two poles differing only in the sign of /, F the distance 
 between two poles differing only in the arrangement of A, 0, 
 
 tan \L = - - , cos F = (sin A) c . 
 he 
 
 93. In the form \h h l\ , if JT, L be the distances between 
 two poles differing only in the signs of h, I respectively, 
 
 tan !/,= -- cos 45, cos K = (sin I/,) 2 . 
 
46 
 
 94. Let H, K, L be the distances between any two poles 
 of the form \h kl\, differing only in the signs of h, fc, I re- 
 spectively. Let F be the distance between any two poles 
 having the indices h, k in different order, and the signs of the 
 first, second and third indices in one, the same as the signs of 
 the first, second and third indices in the other. Let G be the 
 distance between any two poles having the indices A, k in 
 different order, the signs of the first and second indices in one, 
 different from the signs of the first and second indices in the 
 other, and the sign of the third index the same in both ; and 
 let M be the distance between any two poles differing only in 
 the order of the indices A, &, and in the sign of one of them. 
 Then 90-^, 90 -Iff, \L will be the distances of any 
 pole of {h k 1} from the nearest two poles of {l 0} and the 
 nearest pole of {O0l}. F, G subtend at (0 1) angles of 
 90- 20, 90+ 20 respectively, where is the angle subtended 
 at (001) by the distance of any pole of {hkl} from the 
 nearest pole of Jl 0^. M subtends an angle of 90 at (001). 
 Hence (85) 
 
 if tan = - , tan \L = cos d>, 
 h he 
 
 sin \K = cos \L sin 0, sin ^H = cos ^L cos 0, 
 sin ^G cos ^L sin (45 + 0), 
 sin i>F = cos ^L cos (45 -f 0). 
 cos M = (sin 
 
 95. If the distance between two poles of either of the 
 forms \hko}, {hQl}, {h hi} be given, the distance between 
 the two poles, or its supplement, will be one of the arcs 
 -F, K, L, whence, from the expressions in (91) (93), the 
 ratio of the indices may be found. 
 
 96. If the distance between any pole of the form {hkl} 
 and each of two other poles of the same form be given, the 
 
47 
 
 three poles not being in a great circle, the given distances, or 
 their supplements, will be two of the arcs H 9 K> L, F, G, M 9 
 which being known, (f> and Z/, and thence, from the expres- 
 sions in (94), the ratios of the indices may be found. 
 
 97. Let P be the pole of (h k /), Q the pole of (p q r). 
 Then (85) 
 
 tan PX = - cos PXY = - - cos PXZ, 
 K la 
 
 tan QX = ? cos QXY = - - cos QXZ. 
 q r a 
 
 Let Q be in the zone-circle PX. Then QXY=PXY, 
 QXZ = PXZ, therefore 
 
 h tan PX _ k _ I 
 p tan Q X q r ' 
 
 Similarly, when Q is in the zone-circle PF, 
 
 k tan PY I _ h 
 q tan QY ~ r ~ p ' 
 
 And when Q is in the zone-circle PZ, 
 
 / tanPZ _ h _ k 
 r tan QZ p q ' 
 
 98. Let P be the pole of (h k /), Q the pole of (p q r), 
 Z the pole of (0 1). Then (85) 
 
 Z 2 (tanPZ) 2 
 
 99. To find the distance between any two poles. 
 
 Let the axes of the crystal meet the surface of the 
 sphere of projection in X, Y, Z, therefore (85) X is the 
 
48 
 
 pole of (1 0), F the pole of (0 1 0), X the pole of (0 1). 
 Let P, Q (fig. 39) be the poles of (h kl), (pqr); and let 
 PQ meet the zone-circle XY in M. Then, the symbol of M, 
 and the tangents of MZX, PZX, QZX, PZM, QZM may be 
 found in terms of h, k, I, p, q, r. PZ may be found in 
 terms of a, c, h, k, L ZJf=90, /. cosPJf = cos PZM sinPZ, 
 
 tan QM : tan PM = tan QZM : tan PZM. 
 
 Whence PQ, which is either the sum or difference of PM, 
 QM, is known. 
 
 100. Having given the distance between any two poles, 
 not both in the zone-circle l 0, 1 o], to find the ratio 
 of the parameters a, c. 
 
 Let (hkl), (pqr) be the symbols of P, Q (fig. 39), and 
 let tan PZJf, tan QZM be expressed in terms of A, k, /, 
 p, q, r. Then 
 
 tan QM : tan PM = tan QZM : tan PZM , therefore 
 
 sin(QJ/ + PJ/) _ tan QZM + tan PZM 
 sin(QJf-Pilf) ~ tan QZM -tan PZM' 
 
 The given distance PQ is either QM + PM or QM - PM, 
 .-. PM, QM are both known. Cos PM = cos PZM sin PZ. 
 PZ being known, the ratio of c to is given by the equa- 
 tion 
 
 101. To find the indices of any face when referred to 
 the axes of the zones [T l 0, l], [l 1 0, l], [l 0, l] 
 as crystallographic axes. 
 
 The symbols of the three zones are [l 1 o], [l I o]> 
 [0 l] respectively, 
 
 -'. (28) e=l, f=l, g=0, h=l, k = -l, 1=0, p=0, q=0, r=l. 
 
Hence, if w, v, w be the indices of any face when referred to 
 the original axes, u ', ', w' its indices when referred to the new 
 axes, 
 
 u = u + v, v'=u v, w'=w. 
 
 102. To determine the figure and angles of the form 
 \hkl\ 9 when h, k, I take particular values. 
 
 The angle between normals to any two faces is obtained 
 from the expressions in (91)... (94), and will be denoted by the 
 letter which, in the accompanying figure, is placed upon the 
 edge formed by the intersection of the given faces. The 
 arrangement of the poles of the different forms is shewn in 
 (fig, 38). The number of faces is given in (81). 
 
 103. The form { 1 } has two faces (0 1), (o 1), which 
 are parallel to each other. 
 
 104. The form {l 0} (fig. 40) has four faces. Normals 
 to any two adjacent faces of Jl 00}, and to either face of 
 {00 l}, make right angles with each other, therefore F = 90. 
 
 105. The form {l 1 0} (fig. 41) has four faces. tanlJT=l, 
 therefore K'= 90. 
 
 A normal to any face of {l 1 o| makes an angle of 45 
 with a normal to an adjacent face of {l 0^, and an angle of 
 90 with a normal to either face of {o l}. For (85) the co- 
 tangents of the angles are 1 and respectively. 
 
 106. The form {h k 0} (fig. 42) has eight faces. 
 
 In {2 1 0} , tan K = - , .\ K = 53. 7',8, F = 36. 52',2. 
 3 
 
 In {$ 10}, tan^T=-, .-. AT.= 36.52',2, F==53.7>8. 
 7 
 
50 
 
 lu {320}, tanJT=- , .-. K = 67.22',8, F = 22.37',2. 
 5 
 
 In {410}, tanJST= , /. JT = 73.44',4, F = l6.15',6. 
 In {510}, tanJT= , .-. JT= 22.37',2, F = 67. 22',8. 
 
 In {530}, ttmK= , .-. ^T= 6l.55',7, / T =28.4',3. 
 8 
 
 In {710}, tan JT = , .-. JSr= l6.15',6, F = 73. 4d',4. 
 
 A normal to any face of {^A;0} makes with normals to 
 either face of {0 1 } , and the nearest faces of{lOO}, {lio}, 
 angles of 90, 1JT, \F respectively. 
 
 107. The hemihedral form with parallel faces ir\h k 0} is 
 bounded by the alternate faces of \hkQ}, the normals to 
 which make right angles with each other, 
 
 108. The form {h 1} (fig. 43) has eight faces. 
 
 I a 
 
 tan ^ L - - , cos F = (sin ^L) 2 . 
 h/ c 
 
 109. In the hemihedral form with inclined faces K {h /} 
 (fig. 44), 
 
 Z, F= 180- F. 
 
 110. The form {h h l\ (fig. 45) has eight faces. 
 
 tan 1L = - - cos 45, cos K = (sin i) ? . 
 h, c 
 
 111. In the hemihedral form with inclined faces x{h h l\ 
 (fig. 46), 
 
 112. Let P 9 Q be two adjacent poles of either of the 
 forms {hhl} 9 {pOr} 9 equidistant from Z 9 the pole of (0 l) ; 
 
51 
 
 and let the arc of the zone-circle joining P, Q, contain S, a 
 pole of the other form. Then SZ bisects the right angle 
 PZQ, and the angle PSZ is a right angle, 
 
 113. The form {h k 1} (fig. 4-7) has sixteen faces. 
 
 * k Id 
 
 If tan <b = - , tan \L = T - cos 0, 
 h he 
 
 sin K = cos \L sin 0, sin \F = cos \L cos (45 + < 
 
 114. In the first hemihedral form with inclined faces 
 K {hkl] (fig. 48), 7 7 =180-J7, 
 
 sin ^G = cos ^Z sin (45 + 0), cos ^ T 7 = cos ^L cos 0. 
 
 115. In the second hemihedral form with inclined faces 
 K f {hkl] (fig. 49), r=180-G, 
 
 sin ^H = cos \L cos 0, cos \V = cos ^L sin (45 + 0). 
 
 116. In the hemihedral form with parallel faces ir\h k 1} 
 (fig. 50), the distance M subtends an angle of 90 at the pole 
 of (0 l), therefore 
 
 117. The cleavages, in crystals belonging to the pyrami- 
 dal system, are parallel to the faces of one or more of the forms 
 {001}, {lOO}, {l 10}, {hOl}, 
 
EXAMPLES. 
 
 118. IN a crystal of Idocrase (fig. 51), the faces p, m, m 
 make right angles with each other, d is in the zone m m, 
 and makes equal angles with m, m; therefore if m be (10 0), 
 m' (0 1 0), ^(00 1), d will be (l 1 0). In like manner d' will 
 be (l 1 0) ; c is in the zone pd, therefore its symbol will 
 be the form (hhl). Let the symbol of c be (l 1 1). Simi- 
 larly c will be (i i i). v is common to the zones pm', cm, 
 therefore (15) v' is (0 1 1). Similarly v is (l 1). s is com- 
 mon to dc', me, therefore s is (31 1). w is common to 
 dv, me, therefore w is (2 1 1). b is common to wm, pc, 
 therefore b is (22 1). e is common to mb, dc', therefore 
 e is (42 l). r is common to em, pc, therefore r is (44 1). 
 n is common to em', me, therefore n is (4 1 l). a is common 
 to dv, cd', therefore a is (31 2). h is common to ps y mm', 
 therefore h is (31 0). / is common to pw, mm, therefore 
 / is (2 1 0). Hence the crystal is a combination of the forms 
 
 J001J, 
 {311}, 
 
 100}, {ill}, {HO}, {210}, {310}, {211}, 
 
 411}, {221}, {441}, {321}. It is cleavable 
 
 parallel to the faces of {00 l}, {l 00}, {0 1 l}. 
 
 Let m, p, c, &c. (fig. 52), be the poles of the faces 
 m, p, c, &c. pc = 37.7'. The symbols of the poles are 
 p (00 1), m (10 0), c (1 1 1), d (1 1 0), / (2 1 0), h (3 1 0), 
 b (2 2 1), r (4 4 1), w (2 1 1), s (3 1 l), n (4 1 l), a (3 1 2), 
 e (4 2 1). Therefore (85), tan dm = 1, tan/m = -|, tan & w = 1, 
 therefore dm = 45, /m = 26.34', Awz=18.26'. From (97) 
 we have tanc/> = itan&p= ^tanrp ; .'. bp=56.33 f , rp=71.43'. 
 
 2 1 
 
 From (98) we have tan cp = -> tan wp = tan ep = 
 
 A/2 2 1 2 
 
 - tan vp = -- tan op = -r- tan sp = j tan cp. 
 2 /Y/5 -y/5 \/17 
 
 .'. wp = 50. 6', ep = 67.19', v/> = 28.9' 5 6^ = 40. 14', 
 sp = 59. 25'. From (85) we have cos cm = sin cp cos 45, cos bm 
 
53 
 
 sin bp cos 45, cos r m = sin rp cos 45, therefore cm = 65. 44^, 
 bm = 53. 5l', rm = 47. 49', tan cm = 2 tan wm = 3 tan 5m 
 = 4tanwm(97). IK w?m = 46.40', sm=35.14 / l, ram = 27 .55'. 
 tan fem = 2 tan em, therefore em = 34. 23'^. tan wm= 2 tan 6m', 
 therefore wm 69. 56'. tan rm' = ^tan em'=^tar\nm', .'.em' 
 = 65.37'-|, wm'= 77. 14'. , s are in the zone-circle hp, km 
 = 18. 26', .-. cosam=sina^cosl8.26', cosam'= sinjocos71<34', 
 cossw'= sin sjo cos 71. 34', .-. am - 52 .13', am'=78.13', sm 
 = 74. 12'. The distance between the poles of (001) and 
 (l 1 1) is 37. ?'j therefore (100), a, a, c being the parameters 
 of the crystal, if a = 1, c = 0,53511. 
 
 119. In a crystal of Anatase (fig. 53), the face c makes 
 equal angles with each of the four faces p, p, p" 9 p'" 9 and 
 of these any two adjacent faces make the same angle with 
 each other. We may therefore assume c to be (0 1), p (l l 1), 
 p (l 1 l). v is in the zone cp, and s in the zone vp'. Let 
 c i P) P'-> ^5 * be the poles of the faces c, p, p, v, s. Then, 
 by measuring, it is found, that pc = 68.18' ? vc = 19.45', 
 sc = 25.30'. Let the symbol of v be (p q r), v is in the 
 zone-circle pc. 
 
 , r 2 (tansc) 2 (tanpc) 2 
 
 Let s be (;? ? r), .-. (98) ^ /- = ^ i-i- , 
 
 p + q 2 
 
 26 r 2 = 361 (p 2 + q 2 ). s is in the zone-circle vp, the symbol 
 of which is [3 4 T] 5 therefore (21) 3p + 4>q - r = 0. Whence 
 p = 5, g = 1, r = 19. Hence the crystal is a combination of 
 the forms JOO l}, {l 1 l}, {l 1 7}, {5 1 19}. The crystal is 
 cleavable parallel to the faces df j 1 1 1 J. The parameters of 
 the crystal being a, a, c, if a = 1, c = 1,777- 
 
 120. In a crystal of Copper Pyrites, (fig. 54), c, c', c", c'" 
 are the faces of a square pyramid. Let their symbols be 
 c (1 1 1), c (1 1 1), c" (IT i), c'" (ill), p is in the zone cc'", 
 and makes equal angles with c, c"', therefore (87), jo is also in 
 
54 
 
 the zone [001, 1 o] ; therefore p is (1 1). Similarly p is 
 (o 1 1). b t is common to the zones c' V", pp; therefore b t is 
 (1 1 2). There is no face of jl 1 j, between c and c', in the 
 zone c e', therefore p belongs to the hemihedral form with 
 inclined faces K'\I ij. Hence the crystal is a combination 
 of the forms {l 1 l}, {l 1 2}, K ' {l l}. It is cleavable pa- 
 rallel to the faces of the forms {ill}, {o 1 J . 
 
 If we change the axes by the rule in (101), the symbols of 
 the faces will become c (021), b / (ToT) 5 p (111). By 
 changing the axes, the hemihedral form, of which p is a 
 face, becomes /c{l 1 l}. 
 
 121. In a crystal of Scheelate of Lime, (fig. 55), pis (l 1 1), 
 n is (0 2 l). p, g, n, a are in one zone. If p, g, n, a be the 
 poles of p, g, n, a, it is found by measuring that pg= 22. 8l' 9 
 pn = 39.40', pa = 68.6 f . 
 
 Let X, K, Z be the poles of (l 0), (0 1 0), (0 1). Let 
 pn meet the zone-circle ^Fin m. Then m will be (l 1 0), 
 and pm = 90. If (u v w) be the symbol of any pole S in the 
 zone-circle pm, substituting p, n^ m and their indices, for 
 P, Q, R and their indices, in (27), we have 
 
 tan pS v w 
 tan pn w 
 
 Hence, if (u v w) be the symbol of g, we have = - . The 
 
 symbol of the zone-circle pm is [l l 2]? " (21) u 
 Whence w= l,w = 3, = 2; therefore g is (I 3 2). In like 
 manner a is (2 4 l). Of the faces g, a y those only occur which 
 have their poles in alternate lunes of the sphere of projection, 
 therefore g, a belong to the hemihedral forms with parallel 
 faces 7r{3 I 2^, 7r{24l}. 
 
CHAPTER IV. 
 
 RHOMBOHEDRAL SYSTEM. 
 
 122. IN the rhombohedral system the axes make equal 
 angles with each other, and the parameters are equal. 
 
 123. The holohedral form {h kl\ is bounded by all the 
 faces which have for their symbols the different arrangements 
 of + h, + k, + /, together with those of h, k, /. When 
 h, k, I are all different, the number of arrangements will be 
 twelve, as shewn in the annexed table, except when the indices 
 are 0, 1, 1. In this case, and also when two of the indices 
 are equal, the number will be six, and when all three are equal, 
 the number will be two. 
 
 hkl 
 
 Ikh 
 
 hk I 
 
 Ikh 
 
 klh 
 
 kh I 
 
 klh 
 
 khl 
 
 Ihk 
 
 h Ik 
 
 ~lhk 
 
 hlk 
 
 If h be algebraically the greatest, and I the least of three 
 unequal indices A, &, /, fig. 56 will represent the arrangement 
 of the poles of the form [hkl} on the surface of the sphere 
 of projection. 
 
 124. The form bounded either by all the faces which have 
 for their symbols the different arrangements of 4- h, + k, + 1 9 
 or by all the faces which have for their symbols the different 
 arrangements of h, k, - /, is said to be hemihedral with 
 inclined symmetric faces, and will be denoted by the symbol 
 K\hkl}, where (h k I) is the symbol of any one of its faces. 
 A form is said to be direct or inverse according as the alge- 
 
56 
 
 braic sum of its indices is positive or negative. When the 
 sum of the indices is zero, the form will be called direct or 
 inverse according as the largest index is positive or negative. 
 The symbols of the faces of the direct form are contained in 
 the first and second columns, those of the inverse form in the 
 third and fourth columns of the table in (123). 
 
 If the surface of the sphere of projection be divided into 
 two hemispheres by the zone-circle through the poles of 
 {oil}, the poles of the direct form, when the sum of its 
 indices is finite, will be found in the hemisphere which con- 
 tains the pole of (l 1 l), and the poles of the inverse form in 
 the other hemisphere. When the sum of the indices is zero, 
 if the surface of the sphere of projection be divided into six 
 lunes by great circles through the poles of { 1 1 1 } and those of 
 {oil}, the poles of the direct form will be found in three 
 alternate lunes, one of which contains the pole of (l 0), and 
 the poles of the inverse form will be found in the remaining 
 three alternate lunes. 
 
 125. The form bounded either by all the faces of {h k 1} , 
 the indices of which stand in the order hklhk, or by all the 
 faces of [h k l\ , the indices of which stand in the order Ikhlk, 
 is said to be hemihedral with parallel faces, and will be de- 
 noted by the symbol ir\hkl\, where (h k I) is the symbol of 
 any one of its faces. The symbols of the faces are contained 
 either in the first and third, or in the second and fourth 
 columns of the table in (123). 
 
 If the surface of the sphere of projection be divided into 
 twelve lunes by zone-circles through the poles of j l 1 l } , and 
 those of each of the forms {2 1 1}, {oil}, the poles of 
 ir\hkl} will be found in six alternate lunes, except when the 
 algebraic sum of two of the indices is equal to twice the third. 
 And if the sphere of projection be divided into twelve triangles 
 by zone-circles through every two of the poles of j 1 1 1 1 , 
 {sTT}, the poles of Tr\hkl\ will be found in six alternate 
 triangles, except when the sum of the indices is zero. 
 
57 
 
 126. The form bounded either by all the faces of {h k l\ 
 which have for their symbols the arrangements of +A, + k 9 4- /, 
 which stand in the order hklhk, and those of - A, &, I, 
 which stand in the order Ikhlk, or by all the faces which 
 have for their symbols the arrangements of + h, + &, + I 
 which stand in the order Ik hi k, and those of - A, A;, - / 
 which stand in the order hklhk, is said to be hemihedral 
 with inclined asymmetric faces, and will be denoted by the 
 symbol a{hkty, where (hkl) is the symbol of any one of 
 its faces. The symbols of the faces are contained either in 
 the first and fourth, or in the second and third columns of 
 the table in (123). 
 
 If the surface of the sphere of projection be divided into 
 six lunes by zone-circles through the poles of {ill} and 
 I those of ^2 1 7}, the poles of a\hkl\ will be found in three 
 alternate lunes. 
 
 127. To determine the position of any pole. 
 
 Let the axes of the crystal meet the surface of the sphere 
 of projection in X, F, Z, (fig. 57), let O be the pole of (1 1 1), 
 P the pole of (hkl). Since O is the pole of (ill), and 
 a = 6 = f, cos O^T=cosOF=cosOZ, .-. OJT=OF=OZ. 
 The axes make equal angles with each other, therefore FZ = 
 ZX=XY. Hence FOZ, ZOX, XOY are each equal to 
 I 120; 
 
 .-. cos PO Y - cos POZ = ,y/3 sin POX, 
 cos POY + cos POZ = - cos POX. 
 
 cos PX = cos PO cos XO + sin PO sin XO cos POX, 
 cos PY = cos PO cos YO + sin PO sin YO cos POY, 
 
 cos PZ = cos POcos ZO + sin PO sin ZO cos POZ. 
 
 8 
 
58 
 
 Whence 
 
 cos PF - cos PZ = i/S sin PO sin XO sin POX, 
 cos PF + cos PZ = 2 cos PO cos XO - sin PO sin XO cos PO^T, 
 
 cos PX + cos PF + cos PZ = 3 cos PO cos XO, 
 3 sin PO sin XO cos POX = 2 cos PX - cos PF - cos PZ. 
 
 P is the pole of (h k /), therefore 
 
 - cos PX = - cos PF = - cos PZ. 
 ^ k I 
 
 Whence 
 
 tan PO tan ^TO cos PO X = 
 
 h + k + I 
 Similarly 
 
 tan PO tan FO cos POF = - ~ , 
 
 h + k + / 
 
 and 
 
 tan PO tan ZO cos POZ = , 
 
 (tan PO) 2 (tan .TO) 2 =4 
 
 I) 
 
 128. Let A, B, C be the poles of (l 0), (0 l 0), (0 1). 
 Then (127) tan AOX = 0, tan BOY = 0, tan COZ = 0, 
 tan AO tan XO = 2, tan BO tan FO = 2, tan CO tan ZO = 2. 
 Therefore A, B, C lie in the great circles OX, OF, OZ, 
 OA = O# = OC, and the expressions in (127) become 
 
tan POA 
 
 59 
 
 A;-/ 
 
 2 tan PO cot AO cos POA = -, - - - - 
 
 2k- l-h 
 
 2 tan PO cot BO cos FOB = - - - , 
 
 h + k + / 
 
 2 tan PO cot CO cos POC = - - - - r , 
 
 h + k + I 
 
 /tanPO\ 2 _ 
 VtanJOJ = 
 
 129. If Jf , JV be any poles of the forms {2 T 1} , {o 1 1 } ; 
 A any pole of {l 0}, O a pole of {l 1 l}, the expressions in 
 (128) shew that M O, NO are quadrants ; that MO A is a mul- 
 tiple of 60, and that NO A is an odd multiple of 30. Hence 
 the poles of [2 IT} are six equidistant points in which a zone- 
 circle having (l 1 l) for one of its poles, is intersected by the 
 zone-circles through the poles of \l 1 l} and those of {l OOj; 
 and the poles of {oil} bisect the arcs joining every two 
 adjacent poles of {all}. 
 
 130. From the form of the expression for tan PO, it 
 appears that the distances of the poles of {h kl\, which have 
 the indices + h, + k, + I, from (l 1 l), and of those which 
 have the indices - h, - k 9 -I, from (TIT), are all equal. 
 By interchanging the indices A, A;, /, and changing their signs 
 in the expressions for tan POA, tan PO#, tan POC, it appears 
 that the angles subtended at (l 1 l), by the arcs joining any 
 pole of |/i kl\ and the nearest pole of |l OOj, are all equal. 
 Hence, the poles of ]hkl^ are symmetrically arranged with 
 respect to each of the three zone-circles passing through the 
 poles of {l 1 1 \ and those of \2 TIj . 
 
60 
 
 131. If the surface of the sphere of projection be divided 
 into twelve triangles, by zone-circles through every two of the 
 poles of |l 1 1 j and J2 i i}, the arrangement of the poles of 
 {A k / j will be symmetrical in any two adjacent triangles on 
 the same side of the zone-circle [2 1 1, T I 2], or in any two 
 alternate triangles on different sides of it ; and similar in any 
 two adjacent triangles on different sides of the zone-circle 
 [211,11 2]? or in any two alternate triangles on the same 
 side of it. 
 
 132. The arrangement of the poles of /c{& k l\ will be 
 symmetrical in any two adjacent lunes, composed of two adja- 
 cent triangles on different sides of [2 IT? IT 2]? and similar in 
 any two alternate lunes. The arrangement of the poles of 
 7r\h k /} in any two triangles is similar or symmetrical, accord- 
 ing as the triangles are on the same side of the zone-circle 
 [2 IT, IT 2] or on different sides of it. The poles of a {h k l^ 
 are similarly arranged in each of the triangles in which they 
 occur. 
 
 133. If P be the pole of any face, and if in PO produced 
 OQ be taken equal to OP, a face may always exist of which Q 
 is the pole. 
 
 We have from (127) 
 
 2 cos QX - cos QY - cos QZ = 3 sin QO sin OX cos QOX, 
 cos Q X + cos QY + cos QZ = 3 cos QO cos OX, 
 
 cos QZ - cos QY = ^/s sin QO sin OXsinQ OX, 
 sinQO=sinPO, cosQO^= - cosPO^T, sin QOX= - sin POX. 
 
 Hence, if (h k I) be the symbol of P, 
 
 2 cos QX - cos QF - cos QZ Zh-k-l 
 cos QX + cos QF+ cos QZ " ~h~+~k + I ' 
 
 cos QZ - cos QF l-k 
 
 cos QX + cos QF-f cos QZ = " 7T+T+'/ ' 
 
61 
 
 Whence 
 
 3 cos QA^ 
 
 cos QX+ cos QY+ cos QZ h + k + I 
 
 ScosQY Qh-k + 2l 
 
 cos QXTcos QF + cos QZ " h + k + l 
 
 ScosQZ _ 2 A + 2Ar-/ 
 
 cos QJ5T + cos QF + cos QZ h + fc + I 
 
 .-. - cos Q^T - - cos QY = - cos QZ, 
 
 p 7 r 
 
 where 
 
 j9 = - h + Zk + 2/, g = 2A - k + 2/, r = 2A + 2A; - I. 
 
 p, q, r are whole numbers, therefore a face may exist of 
 which Q is the pole. 
 
 134. When A, k, /, /?, </, r are connected by the equations 
 p = - h + 2& + 2/, g = 2^ - Ar + 2/ 5 r = 2h + 2k- I, 
 
 and, therefore, the arc joining the poles (A A; /), (p </ r), is 
 bisected by (11 1), the forms \hkl}, \p q r| are said to be 
 transverse with respect to each other. In certain crystals be- 
 longing to the rhombohedral system, combinations of pairs of 
 transverse forms occur frequently. Such a combination is 
 termed dirhombohedral. The arrangements of its poles is 
 shewn in fig. 58. 
 
 135. If the surface of the sphere of projection be di- 
 vided into twenty-four triangles by zone-circles through every 
 two of the poles of {l 1 1 j, {o 1 I}, {2 IT}? the poles of the 
 dirhombohedral combination \hkl}, {P9 r }y r K^hkl}, 
 K \PV r }> w ^l ^ e symmetrically arranged in any two adjacent 
 triangles, and similarly arranged in any two alternate triangles. 
 The poles of the dirhombohedral combination 7r{hk l},7r\p qr\ 
 will be similarly or symmetrically arranged in any two tri- 
 angles, according as the triangles are on the same side of the 
 
62 
 
 zone-circle [jj i i, i i 2]? or on different sides of it. The pole* 
 of the dirhombohedral combination a\h k Zj, a\p q r\ are simi- 
 larly arranged in all the triangles in which they are found. 
 
 136. In the hemihedral forms with parallel faces, and in 
 the hemihedral forms with inclined asymmetric faces, the faces 
 of different forms of the same kind occasionally occur in zones. 
 If MQM' (fig. 59) be the zone-circle of one of these zones ; 
 MLM' the zone-circle through the poles of {2 IT}, and QML 
 an acute angle, the zone is called direct or inverse, according as 
 the poles of its faces lie nearer to M or M' . The zones in a 
 combination of hemihedral forms with parallel faces are direct 
 on one side of the zone-circle through the poles of ^2 ~\\\ , and 
 inverse on the other side of it. In a combination of hemihe- 
 dral forms with inclined asymmetric faces, the zones are either 
 all direct or all inverse, and the combination is named accord- 
 ingly. 
 
 137- The direct and inverse hemihedral forms with in- 
 clined symmetric faces, and the two hemihedral forms with 
 parallel faces differ only in position. For if the sphere 
 of projection be made to revolve through two right angles 
 round any two opposite poles of {o 1 l} ? the poles of one half 
 form change places with those of the other half form. The 
 direct and inverse hemihedral forms with inclined asymmetric 
 faces are essentially different. 
 
 138. Let P, A be any two adjacent poles of \h k &}, 
 {l 00], O the nearest pole of {l 1 l}. Then (128) tanP0J=0, 
 therefore P, A, O are in one zone-circle. Let PO= T, AO = D. 
 Therefore, making / = k in (128), we have 
 
 h-k 
 tan T = - - tan D. 
 
 h + 2k 
 
 The signs of tan T, tan />, will be the same or different ac- 
 cording as the directions in which PO, AO are measured from 
 O are the same or different. 
 
63 
 
 If V be the distance between any two of three poles of 
 | h k k^ which have the indices h, k, k, V subtends an angle 
 of 120 at (ill); and the two sides that include the angle of 
 120 are each equal to T 7 , therefore 
 
 sin -| V = sin 60 sin T. 
 
 The distance between any two adjacent poles one of which 
 has the indices 4- A, + k, + Z, and the other the indices 
 _ h, -k, -I, will be 180 - V. 
 
 139. If P be any pole of \fikl}, where h + k + I = 0, 
 A the nearest pole of {l 00}, PO = 90 (128). Therefore, if 
 H be the distance between any two adjacent poles of {hkl} 
 having the indices A, k, 1 9 ^H = POA, therefore 
 
 k I 
 tan = 
 
 Zh - k - I 
 
 140. Let the arc joining any two poles of [h k /}, having 
 the indices + h, + k, + 1 9 be H, 7T, L or F, according as 
 h, k, I or neither of the indices holds the same place in the 
 symbols of the two poles ; T the distance of either of the poles 
 from (ill); D the distance of any pole of \ 1 0} from the 
 nearest pole of {ill}; 20, 20, 2>|/ the angles subtended at 
 (1 1 1) by H, K, L. The angle subtended by F will be 120. 
 Then (128) 
 
 k-l 
 
 l-k 
 
64 
 
 The triangles having the vertex (ill), and bases //, 
 L 9 V are isosceles, therefore 
 
 sin Ijy = sin 9 sin Y 1 , sin K = sin <p sin T 7 , 
 sin \L = sin \// sin T 7 , sin ^ F = sin 60 sin T. 
 
 141. If F, the distance between any two of three equi- 
 distant poles of [h k &}, be given, we have 
 
 sin 1 V = sin 60 sin T, tanT = m tan D, 
 
 (h - k) = m (h + 2k) (138), 
 
 the upper or lower sign being taken according as T and D are 
 measured from (l 1 1) in the same, or in different directions. 
 Whence, m being known, the ratio of h to k may be found. 
 
 142. In the form \h k l^ 9 where h + k + I = 0, the dis- 
 tance between any two poles, not a multiple of 6*0, being- 
 known, we can find the distance of one of them from the 
 nearest pole of {2 1 I}. If this distance be $, the ratios of 
 h 9 k, I may be found from the equations 
 
 k I 
 tan 9 = </ 3 - - r 9 h + k + I = 0, 
 
 143. If the distances between any pole of the form 
 \h k /} and each of two other poles of the same form be 
 given, the three poles not being in one zone-circle, the given 
 distances, or their supplements, will be two of the arcs If, JT, 
 Z, V. By eliminating T between the equations for determin- 
 ing H, K, L, V in (140), observing that cf>-0 = 60, ^ + 0=60, 
 we have 
 
 tan 9 _ tan 1 (K - L) sin 9 _ sin 
 tan 6'0 ~ tan 1 (JT + Z) ' sin 60 ~ ^ir 
 
 tan <p tan ^ (Z, + H) sin sin 
 tan 60^ = tan (Z, - H) ' sin 6() = sir 
 
 tan \|/ tan ^(K - H) sin \// sin 
 tan 60 ~ tan~l (K + #) ' sin 60 ~ sin 
 
65 
 
 Two of the four distances H, K, L, V being known, T 
 and one of the angles 0, (jf>, \js may be found ; and then the 
 ratios of the indices may be found from the equations 
 
 k I %h k I 
 
 tan = A/ 3 -i - ; - , > 2 tan T cos = - tan Z>, 
 2h-k-l h+k + l 
 
 , l-h zk-l-h 
 
 tan = A/3 - - - , 2 tan T 7 cos = - - tan D, 
 2k -l-h h + k + l 
 
 h-k Ql-h-k 
 
 tan \(, = A/3 - - - , 2 tan T 1 cos \k = - tan D. 
 21 -h-k h + k + l 
 
 144. To find the distance between any two poles. 
 
 Let P, Q (fig. 60) be the poles of (h k /), (p q r) ; O, A 
 poles of {ill}, {lOO}. Let PQ meet the zone-circle 
 [o 1 1, T l] in M. Then, the symbol of M, and the tangents 
 of MOA, POA, QOA 9 POM, QOM may be found in terms 
 of h, k 9 /, jo, q, r ; and PO may be found in terms of 
 h, k, I. MO is a quadrant, therefore 
 
 cos PM = cos POM sin PO, 
 
 tan QM _ tan 
 tanPJW ~ tan POM ' 
 
 Whence PQ, which is either the sum or difference of 
 PJ/, QJ/, is known. 
 
 145. Having given the distance between any two poles, 
 not both in the zone-circle [o 1 T, To l], to find D, the dis- 
 tance of any pole of { 1 J from the nearest pole of 1 1 1 1 } . 
 
 Let P, Q (fig. 60) be the given poles, (h k /), (p q r) 
 their symbols, then, retaining the construction in (144), let 
 tan POM, tan QOM be expressed in terms of h, k, I, p 9 g, r. 
 
 tan QM tan QOM 
 tan PM = tan POM ' 
 
66 
 
 therefore 
 
 sin (QM + PM ) _ tan QOM + tan POM 
 sin (QM - PM) ~ tan QOM - tan POM ' 
 
 One of the arcs QM - PM, QM + PM is the given distance 
 PQ, therefore PM, QM are both known. 
 
 cos PM = cos POM sin PO. 
 
 PO being known, D is given by the equation 
 h* + k 2 + P-kl-lh- hk 
 
 ^ p y = - pnnrfoi - (tan *> ' 
 
 146. To determine the position of any pole, having 
 given its distances from two of three equidistant poles of any 
 form. 
 
 Let P (fig. 61) be the given pole; A, B, C three equidis- 
 tant poles of any known form ; O the pole of (l 1 1). Let 
 AE meet the zone-circles [o 1 1, I l], CO in M, E, and let 
 PM meet CO in N. AE is bisected in E ; and ME, MN are 
 perpendicular to EO. 
 
 sin AE = sin 60 sin AO, tan EO = cos 60 tan AO, 
 cos PA = cos JE cos PE + sin JE sin PE cos PE J, 
 cos P5 = cos BE cos PJ + sin BE sin PE cos FEB. 
 
 If we express cos PA - cos PI?, cos PA + cos PJ? in terms 
 of products of trigonometrical ratios, observing that BE = AE, 
 - cos PEB = cos PEA = sin PEN, and that sin PJV = sin PE 
 sin PEN, we obtain 
 
 sin JE sin PN = sin 1(PJ5 + PA) sin 
 cos AE cos PE = cos (PB + PJ) cos 
 
 Whence, PE, PN being known, JVE and therefore NO 
 may be found. 
 
 cos PO = cos PJVcos NO, cot POJV = sin NO cot PJV. 
 
67 
 
 147. To find the indices of any face when referred to the 
 axes of the zones containing the faces (hkk), (k h A?), (k k it), 
 as crystallographic axes. 
 
 The indices of the three zones reduced to their simplest 
 terms are 
 
 h + k, k, - k ; - k, h + k, - k ; A;, - A?, h + k, 
 
 therefore (28), e = h + k, f = - k, g = - k, h = - k, k = 
 h + k, 1 = - &, p= - k, q=-&, r =h+ k. Therefore, if 
 u, v, w be the indices of any face when referred to the 
 original axes, u, v', w' 9 its indices when referred to the new 
 axes, 
 
 u (h + k) u kv - kw, 
 
 v'= - ku + (h 4- k) v - kw, 
 w'= - ku - kv + (h + k)w. 
 
 148. To determine the figure and angles of the form 
 {h k fy, when A, k, I take particular values. 
 
 The angle between normals to any two faces of the same 
 form may be computed by the formulae in (138), (139), (140), 
 and will be denoted by the letter placed upon the correspond- 
 ing edge of the accompanying figure. The arrangement of 
 the poles, when the three indices are unequal, is shewn in 
 fig. 56. The poles of {A k k} lie in zone-circles through the 
 poles of {ill} and those of {2 1 1}. The number of faces 
 is given in (123). 
 
 149. The form { 1 1 1] has two parallel faces. A normal 
 to the faces makes equal angles with the three axes. 
 
 150. The hemihedral forms K{I I l}, K {TIT} consist of 
 the faces (l 1 1), (TIT), respectively. 
 
 151. The form [hkk\ has six faces, and is called a 
 rhombohedron. The three poles which have the indices 
 + A, + k, + k are equidistant from each other, and are dia- 
 
68 
 
 metrically opposite to the three poles having the indices 
 h, k, k. Hence, a rhombohedron is bounded by three 
 pairs of parallel faces making equal angles with each other. 
 Let T be the distance of any pole of {h k k} from the nearest 
 pole of { 1 1 1 } ; ' V the distance between two adjacent poles 
 equidistant from (ill); W the distance between two adjacent 
 poles not equally distant from (l 1 l), and D the distance of a 
 pole of {100} from the nearest pole of {l 1 l}. Then (138) 
 
 h-k 
 tan T = - tan D, 
 
 h + 2k 
 
 sin 1 V = sin 60 sin T, W = 180 - V. 
 
 The position of the rhombohedron [h k k} is said to be 
 parallel or transverse according as tan T, tan D have the 
 same or different signs, that is, according as T, D are mea- 
 sured from (l 1 l) in the same direction or in different direc- 
 tions. 
 
 If we take fig. 62 to represent { 1 0} , then ^01 1 \ , which 
 is in transverse position, will resemble fig. 63. In {o 1 ij, 
 tan T = - fL tanD. The poles of {o 1 1 } bisect the arcs join- 
 ing the adjacent poles of {l oj. 
 
 In {211} which is in parallel position, tan T = J tan D. 
 The poles of \Z 1 ij bisect the arcs joining the adjacent poles 
 of {Oil}. 
 
 In {31l}, which is in parallel position, tan T = ^j- tan D. 
 
 In {I 2 2\, which is in transverse position, tan 7 7 = - tanD. 
 Hence, position excepted, the form of {I 2 2] is the same as 
 that of {l 00^. 
 
 In {111} (fig- 64), which is in transverse position, 
 tan T = - 2 tan D. The poles of { l o] bisect the arcs join- 
 ing the adjacent poles of {I i i J . 
 
 In {3 IT}, which is in parallel position, tan T= 4 tan IX 
 The poles of Jl l ij bisect the arcs joining the adjacent poles 
 of {3 1 l|. 
 
69 
 
 152. The hemihedral form with inclined symmetric faces 
 ic{h k k\, has the three faces having the indices + h, + A;, + A;, 
 and which make equal angles with each other. The other 
 half form, has the three faces with the indices -A, A?, - k. 
 
 153. Let P, Q, R be three poles of a rhombohedron, 
 equidistant from 0, the pole of (ill); and let the zone- 
 circle through P, Q, contain S, a pole of another rhombohe- 
 dron. S is in the zone-circle RO which bisects the angle 
 POQ, and the arc PQ. POQ = 120, therefore POS = 60 ; 
 PSO = 90. Whence tan PO = 2 tan #0. 
 
 154. The form {2 1 l} has six faces, the poles of which 
 (129) are the six equidistant points in which a great circle, 
 having (ill), (TIT) for its poles, is intersected by the zone- 
 circles through the poles of Jill}, and those of |lOOJ. 
 Therefore the distance between any two adjacent poles of 
 {21 l} is 60. 
 
 155. The hemihedral form with inclined symmetric faces 
 /c{2 1 l} is bounded by the alternate faces of |<2 i i}. The 
 distance between any two of its poles will be 120. 
 
 156. The form {o 1 l} has six faces the poles of which 
 (129) bisect the arcs joining every two adjacent poles of 
 {21 l}- The distance between any two adjacent poles of 
 {oil} is 60. 
 
 157. The form {h kl}, where h + k + I = 0, (fig. 65) 
 has twelve faces, the poles of which lie in the zone-circle 
 through the poles of {21 l}. 
 
 If H be the distance between any two poles adjacent to 
 a pole of {2 1 l} ? h the largest index, 
 
 - k- I 
 
 The distances of any pole from the adjacent poles of {2 1 
 {oil} are I//, ^G respectively. 
 
70 
 
 In {2 13}, tan ^H = ^3, .: H = 21. 4?',2. 
 
 In {si 4}, tan^JET^v/3, .-. H= 32. 12',3. 
 
 In {4 15}, tan |- # = j .y/3, .'. #= 38. 12',8. 
 
 In {325}, tan| J H' = ^ v / 3, . #= 13. 10',4. 
 
 In {5 16}, tan^ J H r =| v /3 5 .-. #= 42. 6',4. 
 
 In { 5 2 ?} , tan \ H = \ ^/3, .'. H = 27. 4/,7. 
 
 In {7 18}, tan H = </3, .-. # = 46. 49',6. 
 
 158. In the hemihedral form with inclined symmetric 
 faces K\h k 1}, where h + k + / = 0, the distances between the 
 poles of two adjacent faces, are alternately H and 120- H. 
 
 159. The hemihedral form with parallel faces 7r{h k /}, is 
 bounded by the alternate faces of ^hkl^. The distance be- 
 tween any two adjacent poles is 60. 
 
 160. The form {h k I] (fig. 66) has twelve faces. Let 
 h, I be algebraically the greatest and least of the three indices. 
 Then (140), if T, D be the distances of any poles of {h Jcl], 
 {lOO} respectively from the nearest poles of jl 1 l}, 
 
 A + W + V - kl - Ih - hk , 
 
 < tan ^ = - - (tan D) ' 
 
 k-l 
 
 h-k 
 
 = sin^sin T, sin \K= sin sin T, sin \L = sin v// sin T, 
 
 G, F will be respectively equal to the greatest and least of the 
 two angles H, L 9 and TF= 18() n - K. 
 
71 
 
 When the algebraic sum of two of the indices is equal to 
 twice the third, = ^, therefore G = F. 
 
 161. The hemihedral form with inclined symmetric faces 
 K{hkl^, has the faces of one of the two pyramids which, 
 joined base to base, constitute the holohedral form \h k l\. 
 
 162. The hemihedral form with parallel faces ir\ 
 
 is bounded by the alternate faces of \h k /} which occur in 
 three parallel pairs making equal angles with each other. If 
 the distances between two adjacent poles equally and unequally 
 distant from (l 1 l) be V, W, 
 
 V= sin 60 sin T, W = 180 - F. 
 
 163. In the hemihedral form with inclined asymmetric 
 faces a\h kl], if the distance between adjacent poles equally 
 distant from (l 1 1), be F; and if the distances between ad- 
 jacent poles enequally distant from (l 1 l) be U, W, 
 
 sin IF = sin 60 sin T, U = 180 - H, W = 180 - K. 
 
 164. The cleavages, in crystals belonging to the rhombo- 
 hedral system, are parallel to the faces of forms which have 
 two or all of their indices equal. 
 
EXAMPLES. 
 
 165. In a crystal of Calc Spar (fig. 67), the faces p, p, 
 p" are parallel to three cleavage planes, the poles of which are 
 distant 74. 55' from each other. Let p be (l 0), p f (0 1 0), 
 p" (0 1). Let p, p', &c. (fig. 70), be the poles of p, p', &c. 
 g is in the zone-circle pp", and bisects the arc pp"', there- 
 fore g is in the zone-circle p'o, o being the pole of (l 1 1), 
 therefore g is (l 1). c f is one of six faces in a~ zone, the 
 axis of which makes equal angles with normals to p, p\ p". 
 c is also in the zone p'g, therefore c is (l 2 l). In like man- 
 ner c is (2 IT). Let / be the pole of (h k k), then /' will be 
 the pole of (khk), and the zone-circle//" will intersect the 
 zone-circle cc in e, the pole of (l T o). p" is in the zone- 
 circle ff, therefore f is the intersection of ep", p r g, there- 
 fore/' is (i i i). cp, pp", intersect in r, therefore r is (2 T). 
 S c <i> PP intersect in #, therefore t is (31 0), t' (3 l). tt f , 
 p"c" intersect in 0", therefore <p" is (332). Hence the crystal 
 is a combination of the forms |lOO},{oil^, {ill}, {2 T l}, 
 {332}, {310}, {20!}. 
 
 166. In a crystal of Calc Spar (fig. 68) the faces p, p, p" 
 are parallel to the three cleavage planes. Let p be (l 0), 
 p (0 1 0), p" (001); and let p, p', &c. (fig. 70) be the poles 
 of jp, p, &c. o is equidistant from p, p',p", therefore (128), 
 o is the pole of (ill). 0^ = 90, and c ti is in the zone- 
 circle op", therefore c y/ is (i i 2)* c is (2 1 l). cp', pp" 
 intersect in r, therefore r is (2 T). / is (2 T 0) and r" is 
 (02!)- rr' 9 po intersect in m, therefore m is (3 IT). 
 wzc //? pp" intersect in or, therefore cr is (403). 
 
 167. In a crystal of Calc Spar (fig. 69), p is (100), 
 p f is (010), p" is (001), c is (2! T), and c fl is (l 1 2). 
 p, p' &c. (fig. 70) being the poles of p, p', &c. cp, pp" in- 
 tersect in r, therefore r is (2 T). r" is (o 2 1), rr', cp 
 intersect in m, therefore m is (31 i). Let y be (h k /), then 
 y' will be {h I k). yy', cc ti intersect in e /t , the pole of (o 1 l) 
 
73 
 
 m is in the zone-circle yy\ therefore y is in the zone-circle 
 vne^ y is also in the zone-circle pr, therefore y is (302). 
 sr, % are in the zone-circle yy'. %%' - 37. 8', therefore 
 m% - 18. 34'. sin mo = tan %m cotzom. Whence tan 
 
 - v3 , .*. = - . But ss is in the zone-circle my, the 
 
 5 2A K I 5 
 
 symbol of which is [233], therefore (21) 2h + 3k + 3l 0. 
 
 Whence h = 15, k = l, / = - 9, therefore % is (15 1 9). 
 
 168. To determine the positions of the following poles 
 of a crystal of Calc Spar. 
 
 n (2 1 1), g (0 I 1), / (T l l), m(3l l), I (I 3 3), (I 3 3), 
 
 d (5 3 3), A (4 5 5), (3 2 0), * (3 1 0), w (4 1_0), X (4 l), 
 
 r (2 l), ^ (3 2), o- (4 3), (6 5), * (l5 1 9)> (2 1 2), 
 
 5 (3 1 3), b (7 3 5), q (5 3 5). 
 
 Let p, p', p" be the poles of (l 0), (0 1 0), (0 l) ; 
 o the pole of (ill), and let the poles in op' and the 
 sector eoc /5 op" and the sector c'oc y , be distinguished by one 
 and two accents respectively. pp"= 74. 55', pop" 120. 
 og bisects pp" and pop", therefore, since po=p"o, sinlpp" 
 = sin 60 sinpo. Whence po = 44. 36',6. Tan go = ^ tan po 
 (138), therefore go =26. 15'. Sin lg-# = sm60sin#o, there- 
 fore gg'= 45. 3'. In like manner no = 13. 52', nn'= 23. 56'; 
 fo = 63. 7', //= 101. 9' ; mo = 75. 47', m m'= 114. 10'; lo = 
 38. 17, ^'=64. 53'; 0o = 50. 58', 00'= 84. 33'; do =82. 47', 
 dd'= 118. 27'; Ao = 55. 57', hti= 91. 42'. eo = 90, ec'= 90, 
 therefore e^'= 90, p#'= 37. 27',5, pe = 52. 32',5. Let (u v w) 
 be the symbol of any pole S in the zone-circle pp", between 
 e and g. Therefore, substituting e, p, g and their indices 
 for p, q, r and their indices in (27), we have tan Se 
 
 tan pe. Whence, since 0, cr, y, r, X, w', ', v f are in 
 
 the zone-circle pp", 0e = 6. 46', ere =10. 34', ye = 14. 38', 
 re =23. 3l', \e = 33. 8', <t/e= 65. 19', t'e = 69. 2', v'e = 81. 17', 
 pe is known, therefore the distances of 0, cr, &c. from p may 
 be found by subtraction. 
 10 
 
74 
 
 112 
 (Tan bof = (tan PO) 2 (128), therefore 60 = 64. 24',5. 
 
 tan 6op = -| -v/ 3 ( 128 )> therefore bop = 40. 53',6, sin 
 sin 6op sin bo, therefore bb f = 72. 22',5, |6o&" = 19. 6', 4, 
 whence 66" = 34. 20'. In like manner, zo = 76. 32', %op = 
 19. 6',4, ##'=37. 8'. q, a?, S, /" are in the zone-circle 
 ep', ep = 90 Q , ef'= 34. 25', 5. If (wvw) be the symbol 
 of any pole S in the zone-circle ep', then (27) tan Se 
 
 - - tan ef". Whence ge = 22. 2l' 5 #e = 18. 55', Se =12. 52'. 
 
 169. In a crystal of Tourmaline, the two ends of which 
 are represented in figs. 71, 72, c, p, p', p" are (l 1 1), 
 (100), (01 0), (0 1) respectively. n" is common to the 
 zones pp, cp", therefore n" is (1 1 0). Similarly n is (l 1), 
 and n is (0 1 1). s is common to the zones pp", nn", there- 
 fore is (i o I)- Similarly s" is (o 1 T), and s' is (l i o). 
 I is common to the zones ss', cp, therefore I is (211). In 
 fig. 72. p, Pp p^ are respectively parallel to p, p', p", there- 
 fore p is (To 0), p / (0 1 0), p lt (0 T). g is common to the 
 zones sp' 9 sp", therefore g is (ill). The faces parallel to 
 c, /, g are wanting, therefore (124) c, /, g belong to hemihe- 
 dral forms with inclined symmetric faces. Hence the crystal is 
 a combination of the forms {l 0^, {o 1 l}, /cjsll}, /c|oi 1 J ? 
 
 170. p, m os, Szc. (fig. 74), are the poles of the faces 
 p, m, a?, &c. of a crystal of Apatite (fig. 73). Let #, #', at" 
 be the poles of (100), (0 1 0), (001) respectively; and let 
 p be the pole of (ill). ODX', px" intersect in r 3 , therefore 
 r 3 is (l 1 0). In like manner r 2 is (1 l), and r t is (0 1 1). 
 m is in pa?, and pm = 90, therefore (129) m is (2 IT), m is 
 (1 2 I), m" is (IT 2)? and m 3 is (l 1 2). mx, px" intersect 
 in #3, therefore ,v 3 is (2 2 I). wr 3 , px intersect in r, therefore 
 r is (l 1 4). woo", mm' intersect in e, therefore e is (i o T). 
 ex, px" intersect in % 3 therefore %. A is (i 1 1). m% 3 , px in- 
 tersect in %, therefore % is (5 IT). mx f , xm z intersect in 
 s, therefore s is (41 2). ww' 9 mx" intersect in u ', therefore 
 
75 
 
 **' is (2 1 o). im>Zy> xm 3 intersect in w, therefore u is (524). 
 pu 9 mm' intersect in c, therefore c is (5 4, i). The arcs rr , 
 *',, a?* /5 w^ are bisected in /?, therefore (134) the forms 
 to which the faces r, <#, &c. belong are dirhombohedral. 
 If twelve lunes be formed by zone-circles through p and 
 each of the poles ra, e, the poles c, c y , u 9 u^ in the alter- 
 nate lunes are wanting, therefore (125) c, c f &c. belong to 
 hemihedral forms with parallel faces. Hence the crystal (fig. 
 73) is a combination of the forms {ill}, {2llj 5 {oil}, 
 {100}, {T22},_{011},_{411}, {Til}, {5lT}, {412J, 
 7r{2 1 o}, 7r\5 2 4} 5 7TJ5 4 1 \. It is cleavable parallel to the 
 faces of the forms {ill}, {2!!}. The faces u 9 s, at form 
 an inverse zone. 
 
 The symbols of the other poles shewn in (fig. 74) are 
 a (5 2 I), d (7 i 5), / (3 I 2~), b (2 1 2~), V (8 4~I). 
 
 The expressions in (128) give tan epm = ^\/3 9 there- 
 fore the zone-circle pasde makes an angle of 30 with pm. 
 tan cpm = \\/3-> therefore the zone-circle cu t puc makes an 
 angle of 40. 53 r ,6 with pm. tan bpm = |\/3, therefore the 
 zone-circle b^pb makes an angle of 46. 6' with pm. If xp=D, 
 
 2 
 we have from (128), tan D = 2 tan rp = ^ tan zp = -j- tan ap 
 
 1111 
 
 tan sp = - tan dp = tan up = - tan op. 
 
 mpm 3 60, .-. e^m 3 = 30, up m 3 19. 6',4, bpm 3 13. 4'. 
 
 Sin ap = cos a f m 3 e f - tan 60 cot xm z = tan 30 cot sm 3 
 
 = tan 19.6',4 cot um 3 = tan 13. 4' cot &ra 3 = tan 40. 53 ; ,6 cot urn. 
 
 Sin sp = cos s'm 3 e'= tan 60 cot #w 3 = tan 30 cot dm 3 . 
 
 Cos rm 3 = cos 60 sin r^>, cos am 3 = cos 30 sin ap. 
 
 171. f? P? *> &c. (fig. 76) are the poles of the faces 
 r, p, *, &c. of a crystal of Quartz (fig. 75). The distance 
 between any two adjacent poles r, r 2 , r 3 , &c. is 60. The 
 arcs pr, p'r, p"r" are each 38. 13', and are perpendicular 
 to rr a , therefore they pass through o, the pole of rr 2 . There- 
 
76 
 
 fore, if p be (10 0), p (0 1_0), p" (0 l), o will be (l 1 1), 
 r (2 T T), r' (I 2 !),_ r" (I I 2). rp', r> intersect in #", 
 therefore #" is (2 2 l)- f*p 5 r"p intersect in *, therefore s 
 is (4 2 l). ar = 18. ll', therefore tanao = 4 tan po, therefore 
 (138) a is (3 1 1). ar l = ar, therefore ao = ao, therefore 
 (133) a t is (755). n, x are in the zone-circle pr 2 . xr 2 = 
 18. 29', nr 2 = 12. If (uvw) be the symbol of any pole S 
 in ^r 2 , substituting r 2 , ^, ^ /r and their indices for P, Q, 7? 5 
 and their indices in (27), and observing that cot#'V 2 =-cotpr 2 , 
 we have (Zu 5v) tan Sr 2 = (2u + v) tan pr 2 . The symbol 
 of the zone-circle pr 2 is [0 1 2], .-. v+2w=0, tan pr 2 
 = 7tan<2?r 2 , .*. at is (22l), tan^r 2 = 13 tanwr 2 , .*. n is (s 10 5). 
 pz, aa i are bisected in o, therefore (134) p, a belong to 
 dirhombohedral combinations. There are no poles s, #, w 
 in the alternate lunes formed by zone-circles through o and 
 each of the poles r, therefore (126) s 9 a, n belong to hemi- 
 hedral forms with inclined asymmetric faces. Hence the crys- 
 tal (fig. 75) is a combination of the forms {21!}, {lOO}, 
 {122}, {3 IT}, {755}, {42l},a{22l} 5 a{8lo5}. The 
 zone formed by the faces p, s, #, n, r 2 is direct. 
 
 The symbols of the other poles shewn in (fig. 76) are 
 6 (i32_2), &, (78 8), m (722), m t (54J^), e (i65_5), e t (433), 
 c (5 2 2), c t (13 8 8), w (5 4 2), y (7 8 4), # (3 4 2). 
 
CHAPTER V. 
 
 PRISMATIC SYSTEM. 
 
 172. IN the prismatic system the axes make right angles 
 with each other. 
 
 173. The holohedral form \h k 1} is bounded by all the 
 faces having for their symbols the different combinations of 
 
 A, &, =M, each index having always the same place, 
 When h 9 A;, I are all finite, the form {h k /J will have the 
 eight faces 
 
 hkl Jill hkl hll 
 
 ll~l hkl Jill hlcl 
 
 When one of the indices is zero, the number of faces will be 
 four. When two of the indices are zero, the number of faces 
 will be two. 
 
 The arrangement of the poles of \hkl\ on the surface 
 of the sphere of projection is shewn in fig. 77. 
 
 The form bounded by all the faces of {hkl} which 
 have either an odd number of positive indices, or an odd 
 number of negative indices, is said to be hemihedral with in- 
 clined faces, and will be denoted by *\hkl\, where h k Us 
 the symbol of any one of its faces. The forms are said to 
 be direct or inverse according as they have an odd number 
 of positive or of negative indices, and their symbols are con- 
 tained respectively in the upper and lower lines of the table 
 above. 
 
 If the surface of the sphere of projection be divided into 
 eight triangles by zone-circles through every two of the poles 
 
78 
 
 of (1 0), (0 1 0), (0 1), the poles of the direct form will 
 be found in four alternate triangles, one of which contains 
 (1 1 1), and those of the inverse form in the four remaining 
 triangles. 
 
 175. The form bounded by all the faces of {/*&/}, in 
 the symbols of which the sign of one of the indices re- 
 mains unchanged, is said to be hemihedral with symmetric 
 faces, and may be denoted by prefixing to {A&/J, where 
 (h k 1} is the symbol of one of its faces, o^, a~ 2 or o- 3 , accord- 
 ing as the first, second or third index preserves its sign 
 unchanged, and may be called direct or inverse according as 
 that index is positive or negative. 
 
 The poles of the hemihedral form with symmetric faces 
 will be found in one of the hemispheres into which the sur- 
 face of the sphere is divided by a zone-circle through two 
 of the three poles (100), (0 1 0), (001). 
 
 176. To determine the position of any pole. 
 
 Let the axes of the crystal meet the surface of the sphere 
 of projection in X, F, Z (fig. 78). Let a, 6, c be the 
 parameters of the crystal. P the pole of (hkl). 
 
 Since the axes of the crystal are rectangular, FZ, ZX, 
 XY are quadrants, .*. cos YZ = 0, cos ZX = 0, cos^TF=0, 
 therefore X, F, Z are the poles of (l 0), (0 1 0), (0 l), 
 and the angles at Jf, F, Z right angles. 
 
 cos PX = sin PY cos PYX = sin PZ cos PZX, 
 cos PY = sin PZ cos PZF = sin PXcos PXY, 
 cos PZ = sin PX cos PXZ = sin PY cos PFZ. 
 
 cot PX = tan PZFcos PXY = tan PFZ cos PXZ, 
 cot PF = tan PXZ cos PFZ = tan PZ^cos PYX, 
 cot PZ = tan PYXcos PZX = tan PXYcos PZF 
 
79 
 
 But 
 
 - cos PX = - cos PY = - cos PZ. 
 h k I 
 
 Whence 
 
 tanP^F=--, tanPFZ=--, tanPZ^=--. 
 A; c /a h b 
 
 cot PX=-- cos P^TF = - - cos PJTZ, 
 A; a /a 
 
 cot PF = - - cos PFZ = 7 - cos PYX, 
 I b h b 
 
 cot PZ = - - cos PZ^ = - - cos PZF. 
 he k c 
 
 177- It appears from (l?6), that the distance of any pole 
 of \hkl\ from the three nearest poles of {lOO}, {pioj, 
 {O0lj are respectively equal to the distances of any other 
 pole of \h k 1} from the three nearest poles of {l 0}, {0 1 0}, 
 |001}. Hence the poles of {hkl} are symmetrically ar- 
 ranged with respect to each of the three zone-circles through 
 every two of the poles of the forms {l 0), {Oio}, {O0l. 
 
 178. The arrangement of the poles of {&/!, and of 
 a \hfcl\y will be symmetrical in any two adjacent triangles 
 formed by zone-circles through every two of the poles of 
 {100^, {OIO}, {O0l}, and similar in any two alternate 
 triangles. The arrangement of the poles of K\hkl\ will be 
 similar in each of the triangles in which they occur. 
 
 179. In either of the hemihedral forms, the poles of the 
 direct and inverse forms may be made to change places with 
 each other, by causing the sphere of projection to revolve 
 through two right angles round the poles of one of the forms 
 {l 00}, {01 0}, {O0l}. 
 
80 
 
 180. In the form {okl}, if L be the distance between 
 two poles differing only in the sign of Z, 
 
 . 
 k c 
 
 181. In the form {hoty, if L be the distance between 
 two poles differing only in the sign of I, 
 
 i lo, 
 tan \L = --- 
 he 
 
 182. In the form {hkO}, if H be the distance between 
 two poles differing only in the sign of /&, 
 
 hb 
 
 tan IH = - - . 
 k a 
 
 183. In the form \hkl\, if H, K, L be the distances 
 between any two poles, the symbols of which differ only in the 
 signs of h, k, I respectively, 
 
 if tan = , tan \L - - cos 0, 
 ho he 
 
 sin 1/T = cos ^L sin <, sin H = cos L cos <f>. 
 
 184. Let P be the pole of [h k 1} , Q the pole of {p q r } . 
 Then as in (97), when Q is in the zone-circle 
 
 h tan PX _ k_l_ 
 p tan Q X ~ q~ r* 
 
 When Q is in the zone-circle PF, 
 
 k tan PY I _h 
 q tan QY r q 
 
 When Q is in the zone-circle PZ, 
 
 I tan PZ _ h _ k 
 r tan QZ p q ' 
 
81 
 
 185. To find the distance between any two poles. 
 
 Let P 9 Q (fig. 78) be the poles of (h k /), (p q r) ; X, F, Z 
 poles of the forms {O0l}, {o 1 0}, {lOO}. Let PQ meet 
 the zone-circle of which Z is a pole in M. Then the tangents 
 of MZX, PZX, QZXcau be found in terms of h, k, I, p, q, r 
 and of two of the parameters a, 6, c ; therefore PZM, QZM 
 are known. PZ can be found in terms of h> k, I, and of the 
 parameters a, 6, c. MZ is a quadrant, therefore 
 
 cos PM = sin PZ cos PZM , 
 
 tan QM _ tan QZM 
 tan PM ~ tan PZM ' 
 
 Therefore, PM, QM being known, PQ which is their sum or 
 difference is known. 
 
 186. If the distance between any two poles of either of 
 the forms {okl}, \7iOl}, {hko} be given, the ratio of the 
 indices may be obtained from the expressions in (l 80)... (182). 
 
 187- In the form {h k I], the distances between any pole 
 and each of two others, or their supplements, will be two of the 
 arcs H, K, L ; therefore two of the arcs H, JT, L being known, 
 and thence the ratios of h, k, I may be found from (182). 
 
 188. The ratios of the parameters may be found from 
 the expressions in (180)... (182), having given the distances 
 between the poles of two of the forms {o k I}, {h l\, {h k 0}; 
 or from the expressions in (183), having given the distance 
 between any pole of \h k I], and each of two others not all in 
 the same zone-circle. 
 
 189. The ratios of the parameters may also be found 
 from the distances between three given poles in one zone- 
 circle. 
 
 Let P, Q, R (fig. 79) be the three poles. Let PR meet 
 FZ, ZX, XY in L, M, N respectively. Then, the symbols 
 of P, Q, R being known, the symbols of the poles L, M, N 
 11 
 
82 
 
 may be found by (17). Therefore, PL, PM, PN may be 
 found by (27). Therefore the distances between Z, M 9 N 
 are known. 
 
 tan LY _ tan NL tan M Z _ tan LM tan NX _ tanMN 
 tan LZ ~ tan L M ' tan M. X ~ tan M N ' tan NY ~ tan NL ' 
 
 Hence, the places and symbols of Z, M> N being known, the 
 ratios of a, b, c may be found by (180)... (182). 
 
 190. To determine the figure and angles of the form 
 {hkty, when h, k, I take particular values. 
 
 The angle between normals to any two faces of the same 
 form is obtained from the expressions in (180)... (183), and 
 will be denoted by the letter which, in the accompanying figure, 
 is placed upon the edge formed by their intersection. The 
 arrangement of the poles is shewn in fig. 77- The number 
 of faces is given in (173). 
 
 191. The three forms {l 00}, {o 1 o}, {00 l}, have each 
 two parallel faces, the faces of any one form being perpen- 
 dicular to those of each of the other two. 
 
 Either of these forms may become hemihedral according to 
 the second law. 
 
 192. The form \0kl} (fig. 80), has four faces perpen- 
 dicular to the faces of Jl 0}. 
 
 I b 
 
 kc ' 
 
 193. The form [h ty (fig. 81), has four faces perpen- 
 dicular to the faces of {o 1 o}. 
 
 I a 
 he' 
 
 194. The form {h k o} (fig. 82) has four faces perpen- 
 dicular to the faces of {o 1 ] . 
 
83 
 
 195. Either of the preceding forms may become hemi- 
 hedral with symmetric faces. The half-form will consist of 
 any two adjacent faces. 
 
 196. The form {h k 1} (%. 83) has eight faces. 
 
 If tan = - - , tan \L = - - cos 0, 
 ho he 
 
 sin ^K = cos ^L sin 0, sin \H = cos JZ, cos 0. 
 
 197 The hemihedral form with inclined faces K\hkl\ 
 is an irregular tetrahedron, the edges of which are parallel to 
 the faces (l 0), (0 1 0), (0 l). If normals to the faces 
 that meet in these edges make with each other the angles 
 T, V, W respectively, 
 
 198. The hemihedral form with symmetric faces consists 
 of four faces which make one of the solid angles of fig. 83. 
 
84 
 
 EXAMPLES. 
 
 199. LET m, k, p, &c. (fig. 85), be the poles of the faces 
 m, k, p, &c. of a crystal of Aragonite (fig. 84). It is found 
 that the zone-circles mm ', k k r intersect at right angles in h, 
 and that the poles of the faces are symmetrically arranged 
 with respect to each of the circles mm ', kk' and a circle YZ 
 which intersects mm', kk' at right angles in F, Z. Let h be 
 (1 0), k (10 1), m (1 1 0). Then F, Z will be (0 1 0), (0 l) 
 respectively. It appears from the symmetrical arrangement 
 of the poles with respect to the great circles mm', kk', YZ, 
 that pp"i pp" pass through F, Z respectively, k, m are in 
 pp", pp' respectively. Hence p is the intersection of Yk, 
 Zm, therefore p is (l 1 l). s is the intersection of hp, mk, 
 therefore s is (211). s" is (2ll), s m is (2 T l). i is the 
 intersection of hk, ss" f , therefore i is (201). n is the inter- 
 section of pk, ss", therefore n is (21 2). x is the intersection 
 of mn, hk, therefore a? is (10 2). Hence the crystal is a com- 
 bination of the forms {l 00}, {l l}, {20l}, {l 02}, {l 1 o}, 
 {ill}, {21l}, {212}. The crystal is cleavable parallel to 
 the faces of {lOO}. {l Ol}, {l 10}. 
 
 It is found that mm'= 63. 5 0', kk'= 71. 34/ 9 very nearly. 
 Therefore mh=58. 5', M=54. 13'. tan kh=2 tan ih = -J tan<r^ 
 (183). Hence ih=34P. 45', a}h=70. ll'. kZ=35. 4?', cospZk= 
 ta,n A; Zcot j9 Z, sin kZ=cotpZk tan p &. Whence j9 Z = 53. 44',5, 
 jpA? = 43. ll x ,5, therefore ^y= 107. 29', p/"= 86. 23'. cos^A = 
 cos wA cos j9m, therefore j!)7i = 64. 46', pp'= 50. 28'. tan w F= 
 2 tanp F, tan j9^ = 2 tan sh (183), therefore ^ F=64. 51 ; , nk = 
 25.9 r , A = 46. 42'. tan^ZA = 2tansZA ( 175), therefore 5 Zh 
 = 38. 45',5. sin i Z = tan si cot 5 Zh, cos s Zh = tan iZcots Z, 
 therefore si = 33. 24',5, sZ=6l.35'. tansZ=2tanrcZ (183), 
 therefore nZ = 42. 45', cosnh = cosnk cos A; A, therefore wA = 
 58. 2',5. 
 
 Let , 6, c be the parameters of the crystal. Then, since 
 p is the pole of (l 1 1), and h the pole of (l 0), a cosph = 
 bcospY= ccospZ. Therefore a, b, c are proportional to 
 
85 
 
 secj9/*, secpYy secpZ, or to the numbers 2,3457, 1,4610, 
 1,6908 respectively. 
 
 If we change the parameters by the rule in (29) so that s 
 may be (l 1 1), the symbols of the other faces will be h (l 0), 
 m (1 2 0), i (1 1), & (1 2), oc (l 4), p (l 2 2) n (l 1 2). 
 The new parameters will be proportional to the numbers 
 1,1728, 1,4610, 1,6908. 
 
 200. In a crystal of Sulphate of Magnesia with seven 
 proportionals of water (fig. 86), the zones which serve to deter- 
 mine the symbols of the faces are nlte, vlsp, mil") nsm, vtm. 
 eis (10 0), p (0 1 0), n (0 1 1), v (1 l). Therefore (17) the 
 symbols of the remaining faces are / (l 1 1), l" (III), m (1 1 0), 
 t (2 1 1), s (1 2 1), q (0 2 1), r (2 l). The forms to which s, t 
 belong want the faces which have their poles in the alternate 
 octants formed by the zone-circles through every two of the 
 poles (1 0), (0 1 0), (0 1), therefore (173) they are hemihe- 
 dral with inclined faces. Hence the crystal is a combination 
 of the forms {lOO}, {oio}, {0 1 l}, {lOl}, {lio}, {ill}, 
 {021}, {201}, /c{21 l}, K {l2l}. It is cleavable parallel to 
 the faces of {lOO}. 
 
 The form to which the faces / belong is frequently hemi- 
 hedral with inclined faces. 
 
 If e, in, I be the poles of the faces e, m 9 I, em = 45. 15', 
 ml =51. 
 
 201. In a crystal of Topaz (fig. 87), the zones which 
 serve to determine the symbols of the faces are ulmm, 
 ynpriy', mosps"o"m'\ mdsps"d"m", mnx'"d"m", 
 m'oxnm", uoriu", uo'y', lxpx"l", xssx'. p is perpendicular 
 to the faces of the zone mm ', therefore if p be (0 l), the 
 faces (100), (0 1 0) will belong to the zone mm. Let o be 
 (1 1 1), o' (1 1 1> Then o" will be (111), o"' (l 1 1). m, m f 
 will be (l 1 0), (i l o) respectively, n will be (2 1), u (3 1 0), 
 y (40l). If m> m ', /, I' be the poles of the faces m, m', I, t, 
 it is found that tan ^11' = 2 tan^ww', therefore I is (210). 
 Hence x is (423), of (423), s (223). Hence the crystal 
 
86 
 
 is a combination of the forms {O0l}, {lio}, J2 1 o}, 
 {SIO}, {201}, {401}, {ill}? {223}, {423}. The 
 
 crystal is cleavable parallel to the faces of {O0l|, {20l}, 
 {021}. 
 
 The forms of Topaz are sometimes hemihedral with sym- 
 metric faces (174). Thus the forms to which o, <#, p, t (t is 
 common to the zones oo'", nn', its symbol is (101)) occa- 
 sionally wanting the faces on one side of the zone mm'; and 
 the form of which i is a face (i is common to the zones 
 mo', m'o, its symbol is (021)), has been observed to want the 
 faces on one side of the zone nn '. 
 
 If m, u, p, Sec. be the poles of the faces m, u, p, &c., 
 mm'= 55. 4l', ll'= 93. 8', uu'= 115. 29', pn = 43. 30',5, 
 py = 62. 13', po = 45. 27',5, ps = 34. f, psc = 41. 4'. 
 
CHAPTER VI. 
 
 OBLIQUE PRISMATIC SYSTEM. 
 
 202. IN the oblique prismatic system one axis OF, is 
 perpendicular to each of the other two axes. 
 
 203. The holohedral form {h k 1} is bounded by all the 
 faces which have for their symbols the different combinations 
 of h, k y I, in which each of the three indices has always 
 the same place, and the first and second indices are taken with 
 the same signs. When k is finite the form has the four faces 
 
 hie I hkl hi I hll 
 
 When k is zero, or when each of the other two indices are zero 
 the number of faces will be two. 
 
 204. The hemihedral form which we shall denote by 
 \hkl\, where (hkl) is the symbol of one of its faces, is 
 
 bounded by the faces of \hkl\ in the symbols of which k 
 has the same sign. 
 
 The poles of the two half-forms lie on different sides of the 
 zone-circle [100, 001]. 
 
 205. To determine the position of any pole. 
 
 Let the axes of the crystal meet the surface of the sphere 
 of projection in X, F, Z (fig. 87). Let C be the pole (001), 
 A the pole of (l 0), P the pole of (h k I). 
 
 The axis OF is perpendicular to each of the other two 
 axes, therefore X F, YZ are quadrants ; therefore cos XY 0, 
 cos ZZ = 0. Therefore F is the pole of (0 l 0). cos CX = 0, 
 
88 
 
 cosCF=0, cosJZ = 0, cos AY = 0. Therefore CX, CF, 
 AZ, AY are quadrants. Therefore C, A are in the great 
 circle ZX, and CA + ZX = 180. 
 
 cosPX= sin PFcos PFJT= sin PFsin PFC, 
 
 cos PZ = sin PFcos PFZ = sin PFsin PYA. 
 But 
 
 ?cosP^=-cosPF=-cosPZ; 
 h k I 
 
 .-. - sin PFC = - sin PYA, .-. if tan = - - , 
 hi la 
 
 tan 1 (PFC - PYA) = tan 1C^ tan (45- 9). 
 
 cot PF= \- sin PFC = 7 7 sin 
 h b to 
 
 cos PJ = sin PFcos PFJ, 
 cos PC = sin PFcos PFC. 
 
 206. The arc joining two poles of \hkl\> the symbols of 
 which differ only in the sign of &, is manifestly bisected at 
 right angles by the zone-circle [001, 100]. Hence, if the 
 surface of the sphere of projection be divided into two hemi- 
 spheres by the great circle [00 1, 1 00], the arrangement of 
 the poles of \h k 1} in the two hemispheres will be symmetrical. 
 
 The arc joining the two poles of K{!I kl^ will be bisected 
 in a pole of the form {o 1 oj. 
 
 207. Let P be the pole of (h k /), Q the pole of (p q r), 
 and let Q be in the zone-circle PF. Then PFC = QFC, 
 PYA = QYA, therefore (205). 
 
 &tanPF_ I _ h 
 q tan QF r ~ p ' 
 
 208. To find the distance between any two poles. 
 
89 
 
 Let P, Q (fig. 89) be the poles of (h k /), (p q r), A, Y, C 
 the poles of (l 0), (0 1 0), (0 l). Let PQ meet CA in M. 
 Then the symbol of M may be found, and MY A, PYA, QYA, 
 PY may be found in terms of A, k, /, p, q, r, , c and the 
 angle between the axes. MY is a quadrant, therefore 
 
 cos PM = cos PFM sin PF, 
 tanQJf tanQFJf 
 
 tan PM tan 
 whence, PM, QM being known, PQ is known. 
 
 209. Having given the distances of the pole of (h k I) 
 from the poles of (l 0), (0 1 0), (0 1) ; to find ZJT, and 
 the ratios of a, 6, c. 
 
 Let P, A, F, C (fig. 88) be the poles of (h k I), (l 0), 
 (0 1 0), (0 1). Then (205) 
 
 cos PA = sin PFcos PYA, 
 cos PC = sin PFcos PFC. 
 
 Whence PYA, PFC, and, therefore, AC and ZX become 
 known. The ratio of a to c is given by the equation 
 
 - sin PFC = 7 sin PF^ ; 
 
 h I 
 
 and the ratio of b to a or c by the equations 
 
 cot PF = - - sin PFC = 7 - sin PF^. 
 h b I b 
 
 210. P, Q, R (fig. 90) are three poles in the zone-circle 
 CA, C, A being the poles of (001), (100); T, T r are two 
 poles of the same form. Having given PQ, Q/?, TT', and 
 the symbols of P, Q, R', T, to find the inclination of the axes 
 and the ratios of the parameters. 
 
 Let TT f meet PQR in S. P, Q, R, S, A, C are in the 
 same zone-circle, and their symbols are known, therefore (26), 
 
 12 
 
90 
 
 the distances of S 9 C 9 A from P and R may be found. 
 Whence, CA being known, the inclination of the axes is 
 known. Y bisects TT', therefore, TV, CS 9 AS being known, 
 TC, TA may be found, and the ratios of the parameters 
 determined as in (209). 
 
 211. M, M' (fig. 91) are the poles of two faces of any 
 form equidistant from Y"; N 9 N r are the poles of two faces 
 of any other form, also equidistant from F. Having given 
 MM', NN', MN; to find the inclination of the axes, and 
 the ratios of the parameters. 
 
 Let MM', NN f , MN meet CA in P, Q, R, C, A being 
 the poles of (00 l), (l 00). The symbols of M, N being 
 known, those of P, Q, R may be found. MN, YM, YN be- 
 ing known, PQ, which measures the angle MYN, may be 
 found. 
 
 sin PR = cot R cot YM, sin QR = cot R cot YN, 
 
 tan 1 (PR - QR) sin (NY -MY) 
 tan 1 (PR + QR) " sin (NY + MY) ' 
 
 Whence, knowing PQ, PR, QR are found. PQ, Q# be- 
 ing known, the places of C, A 9 and the ratios of the parameters 
 may be found by the methods of (209) and (210). 
 
 212. P, Q, R (fig. 92) are three poles in the same zone- 
 circlec T, T' two poles of the same form equidistant from Y. 
 Having given PQ, QR, TT f , and the symbols of P, Q, R, T ', 
 to find the elements of the crystal. 
 
 Let PR meet ZX in M and TF in S\ and let TF, RY, 
 PY meet ZJT in s 9 r, p. Then the symbols of M, S, p, r, s 
 can be found. PQ, PR and the symbols of M, P, Q, R, S, 
 are known, therefore NP, MR, SP, SR may be found by (26). 
 R Y, RM determine RM Y. RMY, PM, RM, SM determine 
 pM, rM, sM. Whence, knowing the symbols of p, r, s, 
 the places of C, A can be found by (210), the elements of the 
 crystal may then be found by the methods already given. 
 
91 
 
 213. To find the indices of any face when referred to 
 the axes of the zones [e Og, 01 o], [0 1, 10 o], [pOr, 1 o] 
 as crystallographic axes. 
 
 The symbols of the three zones are [r p\, [0 1 0], [gO #], 
 therefore (28) 
 
 e = - r, f = 0, g = p, h = 0, k = 1, 1 = 0, p = g, q 0, r = - e, 
 
 Hence, if u, v, w be the indices of any face when referred to 
 the old axes, u f , v', w its indices when referred to the new 
 axes, 
 
 ru, v' = v 9 w'gu ew. 
 
 214. The form {o 1 0} has two parallel faces. 
 
 215. The form \hQl\ has two faces parallel to each 
 other, and perpendicular to the faces of {010|. 
 
 216. The form [h kl\ has four faces. Normals to two 
 faces adjacent to (0 1 0) make with each other an angle 180 JT, 
 where 90 - \ K = PY. 
 
 217. The hemihedral form a^h k 1} has two faces, normals 
 to which make with each other an angle 180 - K. 
 
92 
 
 EXAMPLES. 
 
 218. IN a crystal of Epidote (fig. 93), the zones which 
 determine the symbols of the faces are metlrm, mkoo'k'm', 
 tuxz'u't', lyqqyl'i ln%'l\ rnn'r', mdzqnocm, muym', 
 ryxox'r', tynod't', edd'e. Let m be (100), I (001), 
 n (ill), n (ii i). Therefore (17) the symbols of the faces 
 will be r (1 0_l), q (0 1 l), % (l 1 1), t (l l), o (2 1 0), y (0 1 2), 
 tf (3 1 l), i (l 3), M (2 J 2), A; (4 1 0), d (3 1 l), e (30 l). In 
 some crystals a face / has been observed common to the zones 
 mt, un\ a face s common to the zones mt, ky, and a face b 
 common to the zones mo, lg. Therefore / is (103), s is 
 (201), and 6 is (0 1 0). The crystal is cleavable parallel to 
 the faces m, t. 
 
 Let m, I, r, &c. (fig. 94), be the poles of the faces 
 m, /, r, &c. Having given rt = 5l. 4l', tm = 64>. 36', nn 
 70. 33'; to find the positions of the remaining poles. 
 
 Let (u v w) be the symbol of any pole S in the zone-circle 
 rtm\ then, substituting r, t, m and their indices for P, Q, R 
 and their indices in (27), 
 
 tan rS tanrm 2w 
 tan r t tan rm u + w 
 
 .'. (u + w) tanr*S'= 2w tanr+ (u - w) tanrwz. 
 
 Whence Zr=25. 44',5 ; /r=34. 55'5 ; er=Sl. 34 ; ir=29. 2l',5 ; 
 sr=18. 6'. If nml (j), tan wr = tan ^>sinmr, tanzv = 
 tan sin m^. Whence t> = 41. S9',5, 0t/= 96. 4l'. In like 
 manner qq = 64. 46', **'= 70. 9', dd'= 96. 10'. It ntr=^, 
 tan^r = tan>|/sin #r, tanom = tan\|/sin #m. Whence ow = 
 58. 26', oo'= 63. 8'. Tan % = 2tan bq (207), therefore by = 
 51. 45', yy'= 103. 30'. Tan b u = 2tan bz, tan 6A; = 2 tan&o. 
 Whence 6^=54. 33 X , w^'= 109. 6', bk=50. 5l',5, ^^=101. 43'. 
 
 If the zones which intersect mt in/, 5, had not been found, 
 the distances of /, s from the poles of some known face in mt 
 would have been requisite in order to determine their symbols, 
 
93 
 
 Suppose /, ts had been measured, and that we had found 
 tf= 19. 37', ts = 69. 47'. We have tr = 51. 4l', rm'= 63. 43'. 
 Therefore, substituting #, r, m' and their indices for P, Q, 7? 
 and their indices, and /for S in (27), if (u v w) be the symbol 
 of /, u = 1, v = 0, w = 3. Therefore / is (103). In like 
 manner 5 will be (201). 
 
 To find the inclination of the axes OZ, OX, and the 
 parameters. 
 
 Let OZ, OX meet the surface of the sphere of projection 
 in Z, X. Then, since I is (00 1) and m is (l 00), mZ = 90, 
 IX = 90. ml = 90. 32',5, therefore ZX *= 89. 27',5. The axis 
 OF meets the surface of the sphere of projection in b. If the 
 parameters of the crystal be a, 6, c, since u is (212), ^acos uX 
 = b cos ub = ^c cos ^ Z. Sec w X sec ^ cosec tl, secuZ = 
 sec ut cosec tm. Therefore a, 6, c are proportional to Zsecut 
 cosec /, secwo, sec ^ cosec ?w. 
 
 To find the symbols of the faces referred to the axes of 
 the zones %%\ mt, oo' as crystallographic axes. The symbols 
 of the zones ##', mt, oo', when referred to the old axes, are 
 [1 l], [0 1 0], [001] respectively. Therefore (28), if (u v w) 
 be the symbol of any face referred to the old axes, (u 1 v w 1 ) 
 its symbol when referred to the new axes, v! u w, v'= v 9 
 w w. If the new axes OZ', OX' meet the surface of the 
 sphere of projection in Z', X ', Z'X'= 180- mt = 115. 24'. 
 
 219. In a crystal of Felspar (fig. 95), the zones which 
 determine the symbols of the faces are tzmz't', pqxyp, 
 pnmn'p, xomox, qoznq, potp, yonty^ m is perpendicular 
 to the faces p, q, #, y, therefore m is (010). Let t be 
 (1 10), t' (Tio), o (111). Then (17) the symbols of the 
 remaining faces will be p (001), w (0 2 l), y (20 1), as (l 1), 
 x (130), g (203). Hence the crystal is a combination of 
 the forms Joio}, {OOl}, {lio}, {l30}, {02l}, {20l}, 
 {lOl}, {203}, {l 1 l}. It is cleavable parallel to the faces 
 of the forms J00l}, {l 1 0}, {o 1 0^. 
 
94 
 
 Having given tt', pt, px, t, p, or, &c. (fig. 96), being the 
 poles of the faces t, p, x, &c., to determine the positions of 
 the poles q, y, n, o, ss. 
 
 Let tt f , px meet in a. Then a will be (100). mt = 
 \tt\ cospt = sin mt cos pa. The right-angled triangles 
 xpo, apt, having a common angle at p, give sinpacotom = 
 sin pa; cottm. 2 cotpq 3 cot poo cot pa, 2 cot py = cot px 
 + cot pa (27). Tan mt = 3 tan m% (207). The right-angled 
 triangles tya, n'yp', having a common angle at y, give 
 sin aycotmn'= sin p'y cot mt. If tt'= 118. 49', p = 67. 44', 
 p<# = 60.20', we shall have pa = 63.5S / , OTW = 63. 7', ^ = 
 34. 13', py = 80 Q .23 f , m% = 29. 25', raw' =45. 3'. 
 
 220. In a crystal of Oxalic acid (fig. 97), the zones 
 pacp', peep' have their axes at right angles to each other. 
 a em a', cemc are zones, p, a, c, &c. being the poles of the 
 faces p, a, c, &c. pa = 50. 40', |?c = 76. 45, raw'= 63. 5'. 
 If the zone-circles ^a, mm' meet in d, it appears from ap- 
 proximate measurements that 2 cot^d = cot pa cot pc (27). 
 Whence pd=73.4<3'. The right-angled triangles ecp, 
 mcd, having a common angle at c, give sinpccotjpe = 
 sin dc cot md. Whence pe = 72. 44', ee- 34. 32'. It fre- 
 quently happens that there are no faces parallel to the faces 
 , e'. In this case (204) the form to which the faces e be- 
 long is hemihedral. 
 
CHAPTER VII. 
 
 DOUBLY OBLIQUE PRISMATIC SYSTEM. 
 
 221. In the doubly oblique prismatic system the form 
 {h k 1} has the two faces (h k I) (hk 7). 
 
 222. To determine the position of any pole. 
 
 Let the axes of the crystal meet the surface of the sphere 
 of projection in X, F, Z (fig. 98). Let A, B, C be the poles 
 of (1 0), (1 1), (0 1), P the pole of (h k I). XYZ, ABC 
 are supplemental triangles. 
 
 cos PX = sin PEC sin PB = sin PCS sin PC, 
 cos PY - sin PC A sin PC = sin PAC sin PA, 
 
 cos PZ = sin PAB sin PA = sin PBA sin PB. 
 But 
 
 
 - = - cos PY= - cos PZ. 
 
 h k I 
 
 Whence 
 
 - sin PAC = j sin PAB, 
 
 K v 
 
 f* CL 
 
 - sin PB A = - sin PBC, 
 I h 
 
 a b 
 
 - sin PCB = - sin PC A. 
 h k 
 
 .-. tan 1 (PAB - PAC) = tan \BAC tan (45 - 0), 
 
 k c 
 
 where tan - - - ; 
 / o 
 
96 
 
 tan 1 (PBC - PEA) = tan \CEA tan (45 - 0), 
 
 where tan d> = ; 
 A c 
 
 tan 1 (PCJ - PCB) = tan 1JC5 tan (45 - >/,), 
 
 h b 
 
 where tan v//- = . 
 k a 
 
 Whence, knowing the angles A, B, C, the segments into 
 which they are divided by PA, PB, PC become known. 
 
 223. Let PX, PF, PZ meet the sides of ABC opposite 
 to Ay B, C, in #, y, %. The angles at <a?, y, % are right 
 angles, therefore 
 
 sin 5i? sinCtf sin Cy smAy sin Ax sinBz 
 cot PBC ~ cot PCS' cot PC A = cotPAC* cotPAB = cotPBA ' 
 
 sn 
 
 Whence the segments, into which the sides of ABC are 
 divided by perpendiculars P#, Py, Pz, become known. 
 
 cot PA = cos PAB cotAz = cos PAC cot Ay, 
 cot PB = cos PBC cot Ex = cos PEA cot Bz, 
 cot PC = cos PC4 cot Cy = cos PC# cot Cat. 
 
 224. Let PJ, PB, PC meet the sides of ABC in #, JT, L. 
 The symbols of the points thus determined will be (Ok I), 
 (ho I), (hko). 
 
97 
 
 whence 
 
 Similarly 
 
 and 
 
 cos HY _ sin HC sin C __ sin HC sin AB 
 cos HZ ~ sin HE sin 5 ~ sin HE sin C J ' 
 
 k sin ## / sin HC 
 
 b sin 
 
 sn 
 
 c sin CL4 
 h sin 
 
 a sn 
 
 h sin Z, A k sin 
 
 Therefore 
 
 tan 
 
 tan 
 
 a sin CA b sin 
 
 - HC) = tan \BC tan (45 - a), 
 
 k c sin CA 
 
 where tan a = - -- , 
 I b sin AB 
 
 tan 1 CJ tan (45 - /3), 
 
 _ / a sn 
 where tan /3 = --- 
 h c sin 
 
 tan \(LA - L5) = tan \AB tan (45 - 7), 
 
 h b sin 
 where tan y = --- : 
 
 A; a sin C J 
 
 Having determined the segments into which H, K, L 
 divide the sides of ABC, by means of the preceding equations, 
 we can find AH, BK, CL. 
 
 sin AP sin APC = sin AC sin PC A, 
 sin HPsiuHPC = sin HC sin PCB, 
 sin JP sin APB = sin J# sin PBA, 
 sin #P sin HPB = sin # sin PBC. 
 
 sin PC# = - sin PC A, 
 
 h k 
 
 13 
 
 
 sin PBA = 7 sin 
 I It 
 
98 
 Whence 
 
 
 sinPH 
 
 h b sin HC h c sin HB 
 
 
 sin PA 
 
 k a sin CA 1 a sin AB 
 
 Similarly 
 
 
 
 
 sinPK 
 
 k c sin KA k a sin KC 
 
 
 sinPB 
 
 I b sin AB h b sin BC * 
 
 and 
 
 
 
 
 sin PL 
 
 1 a sin LB I b sin L A 
 
 
 sin PC 
 
 h c sin BC k c sin CA 
 
 Therefore 
 
 
 
 tan 
 
 (PA J 
 
 H) = tan i#J tan (45 - 
 
 where tan 
 
 h b sin #C h c sin 
 
 - - - - - - - 
 k a sm CA I a sin AB 
 
 tan (PB - PK) = tan 1 KB tan (45 - p), 
 
 k c sin JT^C k a sin 
 where tan a = -- - - = --- : - ; 
 I bsm AB I b sm BC 
 
 tan -KPC - PL) = tan 1JLC tan (45 - o-), 
 
 / a sin LB I b sin LA 
 where tan cr = - - - ^ = - - - 777 . 
 ^ c sm BC k c sm CJ 
 
 Whence the segments into which AH, BK, CL are divided 
 by P become known. 
 
 In the preceding investigations we have supposed the 
 indices of P, and of the nearest poles of the forms {lOO|, 
 ^Oioj, {001} to be all positive. Therefore, when any of 
 the indices of P are negative, we must change their signs and 
 the signs of the corresponding indices in the symbols of the 
 other poles of the crystal, 
 
 225. To find the distance between any two poles. 
 
 Let P, Q be the two poles. Then, having found the 
 distances of P, Q from one of the angles of the triangle ABC, 
 and the angles these distances make with one of the adjacent 
 
99 
 
 sides of ABC, and, therefore, the angle they make with each 
 other, we have two sides of a spherical triangle and the in- 
 cluded angle, from which the third side PQ may be found. 
 
 226. To determine the inclinations of the axes and the 
 ratios of the parameters. 
 
 When the distances of A, B, C, the poles of (l 0), 
 (010), (0 1), from each other, and the distance of any two 
 of them from P, the pole of (h k I), are known, the distance 
 of the third from P, as well as the angles which PA, PB, 
 PC make with the sides of the triangle ABC may be found. 
 Cos PX, cos PF, cos PZ may then be found in terms of 
 PA, PB, PC, and of the angles into which A, B, C are divided 
 by PA, PB, PC. The ratios of the parameters are then 
 given by the equations 
 
 - cos PX - cos PY = - cos PZ. 
 h k I 
 
 The triangles ABC, XYZ are supplemental to each other ; 
 
 227. Having given the symbols of four poles D, E, F, G 
 (%. 99), and five of the arcs *DF, FE, EG, GD, DE, FG; 
 to find the inclinations of the axes and the ratios of the 
 parameters. 
 
 Let A, B, C be the poles of (l 0), (0 1 0), (0 1). Let 
 DE, FG intersect in H, and let them meet the sides of 
 ABC respectively opposite to J, B, C in L, M, P, Q, R, S. 
 Five of the arcs DF, FE, EG, GD, DE, FG being known, 
 we are enabled to calculate the sixth, and also the arcs DH, 
 HE, FH, HG, and the angles they make with each other. 
 The symbols of D, E, F, G, A, B, C being known, the sym- 
 bols of H, L, M, P, Q, R, S may be found by (17). Hence 
 DH, HE, GH, HF being known, we can find HL, HM, 
 HP, HQ, HR, HS, and thence the sides of the triangle 
 ABC, and the distances of H from its angles. This being 
 done, FZ, ZX, XY, and the ratios of a, b, c may be found 
 by (226). 
 
100 
 
 EXAMPLES. 
 
 228. IN a crystal of Axinite (fig. 100), the following 
 zones have been observed; mpdfemf, mltvwm', mrsxycm, 
 mgom, focwnf, fgyvf, fxtf, fslf, ecqve, pyqwp', 
 pxvnp, pstp, prlp. 
 
 Let m be (100), / (0 1 0), v (001), x (ill). Then 
 (17), y will be (0 1 1), t_(l l), p Jl 1 0), e (i i o), * (21 l), 
 / (2^)1), r (311), q (l 1 2), c (ill), w (l l), n (ill), 
 o (121)5 g* (021). If the distances between m, p, d, f, the 
 poles of the faces m 9 p, d,f, be measured, it will be found 
 that cot md - cot mf= 2 (cot mp - cot mf). Therefore (27), 
 if (uvw) be the symbol of d, u = 2, v = 1. d is in the zone- 
 circle mf, therefore w = ; therefore d is (l 2 0). 
 
 Let m, , ^?, &c. (fig. 101) be the poles of the faces 
 m, t, p, &c. Having given moo = 49. 32', o?y = 29. 52', w = 
 44. 35', ta = 32. 55', yv = 40. 5l'; to determine the places of 
 the remaining poles. 
 
 In the triangle ymv, the sides of which are known, we find 
 ymv = 44. 4l',5, yv m = 88. 10',5. xm, tm, xmt give xtm = 
 84. 14',5. ^, /W, /^ give fv = 97. 36'. /v, vm, fvm 
 give v/ra = 97. 58',5, /m = 89. 55'. VTW, w.27, vw# give 
 o?vm = 44. 44'. vm, xvm, vmp give w^p = 45. 12'. The 
 formula expressing the relation between the distances of four 
 poles in the same zone-circle (27) gives 
 
 cotfe = 2 cotfm cotfp ; cot m w = 2 cot mv cotmt ; 
 cot me = 2 cot my cot mx. Whence fe'- 135. 12', m w = 
 119. 50', me = 115. 35'. mw, mp, wmp give pw = 115. 50',5. 
 my, mp, ymp give py = 58. 53'. 2 cot^g = cotp^ + cotpy; 
 
 2 cot md = cot mf + cot W> ; cot me = 2 cot mf cot ra^ ; 
 cotfg = 2 cotfy cotfv ; cot ms = 2 cot m<a? cot m?/ ; 
 
 cot mr Scot m x 2 cot my, cot ml = 2 cotmt cotmv ; 
 4 cot m A; = 5 cot w# cotmv. Whence pq = 86.35 f , md- 
 63. 34', me = 134, 43', fg = 34. 53',5, ms = 33. 20',5, mr 
 25. 2/, w^ = 28. 57', mk = 39. 30 X . 
 
101 
 
 If the axes meet the surface of the sphere of projection 
 in X, F, Z, mfv will be the polar triangle of XYZ, and 
 FZ, ZJT, XY will be the supplements of the angles fmv, 
 vfm, /?;m(226); fv, vm, mf give vfm = 78. 16'. Hence 
 
 
 
 FZ = 82. l',5, ZX= 101. 44', ^F= 91 
 
 Let , 6, c be the parameters of the crystal. Then, since 
 x is (1 1 l), a cos no X = 6 cos x Y = c cos # Z. cos<#J5T= sin xv 
 sin a? v/, cos x Y - sin #u sin ,wvm, cos <# Z = sin a?w sin xmf. 
 Whence 
 
 2,023 _ 1,976 __ 1,580 
 a b c 
 
 To find the symbols of the faces when referred to the 
 zones mp, pt, tm as crystallographic axes. 
 
 The symbols of the three zones when referred to the old 
 axes are [00 l], [I l l], [0 1 0]. Therefore (28), if u, v, w be 
 the indices of any face when referred to the old axes, u' 9 v, w 
 its indices when referred to the new axes, u = w, v'=u+v+w, 
 w'= v. Hence the symbols of the faces will be / (01 l), 
 ra(010), t (100), v (110), w (120), I (llo), p (001), 
 e (021), o (142_), y (121), c (l 3 1), x (ill), * (101), 
 T (ill), n (ill), q (241), g (132), d (012). 
 
 229. In a crystal of Blue Vitriol (fig. 103) the zones are 
 mntrm^ rvkoqwr, rxpr', mpvm', npkn' 9 tpot', txvt'. 
 Let k be (001), n (010), v (101), m (T 1 o). Then (l?) 
 r will be (l 00), and p will be (0 1 1). If the distances be- 
 tween the poles of r, v, &, o, q, w be measured, it will be 
 found that, r, v, &, &c. being the poles of the faces r, v, k, &c, 
 cot rv cot r k = (cot ro cot rk) = - ^(cot rq cot rk) = 
 - ij(cot rw cot rk). Therefore (27) o is (I o l), q is (2 l), 
 and w is (30 l). Hence (17) t is (11 0), x (2 1 1). h, s, % 
 (fig. 104) are the poles of faces which do not occur in the crystal 
 (fig. 103). Their symbols are h (2 1 0), s (I 1 l), % (3 1 l). 
 
102 
 
 Having given nt, tr, nk, pt, pr to find the positions of 
 the remaining poles. 
 
 rt, rn give rh (26) or (27). pt, tr, rp give ptr and prt. 
 pi, tn, ptr give pn and pnt. kn, rn, pnt give kr and fern. 
 km, ptr, tr give ro. kr, r/i, krn give khr. khr, prt, rh 
 give roc. rx, rp give rs, rz (26) or (27). In like manner, 
 rk, ro give rv, rq, rw, The values of the five distances 
 which we have supposed to be used in calculating the places of 
 the poles are nt=30 Q .5l', tr = 69.50 f , nk=109.38 f , pt= 
 52. 20', pr = 76. 33'. 
 
 The three zone-circles rt, rp, rk pass through all the 
 poles of the crystal. The angles which the zone-circles make 
 with each other, and the distance of each pole from r are 
 known. Therefore the distance between any two poles, being 
 one side of a spherical triangle of which two sides and the in- 
 cluded angle are known, may be easily calculated. 
 
 If the axes meet the surface of the sphere of projection 
 in X, Y, Z, 
 
 YZ=l80-nrk, 
 
 Let a, b, c be the parameters. Then since t, p are (l 1 0), 
 (Oil) respectively, a cos tX = b cos tY, bcospY = c cospZ. 
 cost X= sin tnsintnp, costY= sintr sin trk, co$pY=smpr 
 sinprk, cospZ = sinprsinprt. Whence 
 
 1 sin tr sin trk 1 IsinprA; 
 a sintnsintnp b c sin prt 
 
CHAPTER VIII. 
 
 TWIN CRYSTALS. 
 
 230. A TWIN crystal is composed of two crystals joined 
 together in such a manner that one would come into the posi- 
 tion of the other by revolving through two right angles round 
 an axis which is perpendicular to a plane which either is, or 
 may be, a face of either crystal. The axis will be called the 
 twin axis, and the plane to which it is perpendicular the twin 
 plane. 
 
 231. Let the poles of the two crystals be projected upon 
 the same sphere (fig. 104). Let T, T' be the extremities of 
 a diameter perpendicular to the twin plane. Let P, p be the 
 poles of any corresponding faces of the two crystals, p' the pole 
 opposite to p. Since p may be made to coincide with P by 
 turning the crystal through 180 round TT', PTp must be 
 an arc of a great circle bisected in T. In like manner if 
 Q, q be the poles of any other corresponding faces of the two 
 crystals, Qq will be bisected in T. Hence the arcs joining the 
 poles of any corresponding faces of two crystals forming a twin 
 crystal, are bisected by the poles of the twin face. If p\ q f be 
 the poles of the faces opposite to p, q respectively, it is mani- 
 fest that Pp\ Qq f are bisected at right angles by the great 
 circle MN having T, T' for its poles. Hence the poles of the 
 opposite faces of the two crystals are symmetrically arranged 
 with respect to a great circle the plane of which is parallel to 
 the twin face. 
 
 232. In order to find the twin axis in any given twin 
 crystal, when it cannot be done by simple inspection, we must 
 determine by measurement or by the observation of zones, the 
 
104 
 
 intersection of two great circles each of which passes through 
 the poles of corresponding or opposite faces of the two crystals. 
 If the intersections of the circles be the poles of corresponding 
 faces of the two crystals, they will be the poles of a twin plane. 
 Let P, Q (fig. 104) be the poles of any two faces of one crystal ; 
 p 9 q the poles of the corresponding faces of the other, p f , q the 
 poles of the faces opposite to p, q ; T 7 , T' the intersections of 
 the great circles pPp, qQq. Then, if T be the pole of 
 corresponding faces of the two crystals, the triangles PTQ, 
 p Tq are equal and similar, and therefore p, q may be made to 
 coincide with P, Q by turning the crystal to which p, q be- 
 long through 180 round TT '. Hence T 7 , T are the poles of 
 the twin plane. 
 
 233. When the twin plane, and the angles between the 
 faces of one of the crystals are given, the angle between any 
 faces of each of the two crystals can be readily determined. 
 
 Let P, P' be the poles of opposite faces of the two crystals. 
 Then PT = PT 7 ', therefore PP'= 180- 2P T. When PT is 
 greater than 90, the faces P, P' will form a re-entering angle. 
 Let P, Q' be any faces of each of the two crystals, PT 7 , QT 7 , 
 PQ being known, PTQ may be found. TQ' = 180- TQ. 
 PT, Q'T, PTQ being known, PQ' may be found. 
 
 Examples of Twin Crystals belonging to the Octahedral System. 
 
 234. In twin crystals belonging to the octahedral system, 
 the twin axis is either perpendicular to a face of { 1 1 1 } or to 
 a face of {01 l}. 
 
 235. In a twin crystal of Magnetic Iron Oxide (fig. 105), 
 the two octahedrons are joined in such a manner that the face 
 o of one crystal is parallel to the face o t of the other, and the 
 faces o, o', o y ', o / belong to the same zone. One crystal will 
 evidently come into the position of the other by revolving 
 through 180 round an axis perpendicular to the faces o, o, 
 and which is, therefore, the twin axis, o, o', </, o { being the 
 
105 
 
 poles of the faces denoted by the same letters, oo = 70. 3l',7, 
 oV=180 -2oo' = 38 .56',5, o"o=109. 28',3, o"o"=l8QP-2oo"= 
 - 38. 56',5. The faces o", o" form a re-entering angle. 
 
 236. In a twin crystal of Zinc Blende (fig. 106), the indi- 
 viduals are dodecahedrons. Six faces of one crystal coincide 
 with six faces of the other. One crystal will come into the 
 position of the other by revolving through 180 round a line 
 parallel to the intersection of d, d', and which is perpendicular 
 to a face of the form Jl 1 1 j. Hence this line is the twin axis. 
 
 237. In a twin crystal of Fluor (fig. 107) the individuals 
 are cubes. If , ', a", /5 aj, a," be the poles of the faces 
 denoted by the same letters, it is found that aa' = a t a\ 
 a" a" - 109. 28',3. The arcs aa^ aaj joining the poles of cor- 
 responding faces of the two crystals intersect in o, the bisec- 
 tion of the arc a" a" . oa"=54P. 44', 8, and oa" evidently bisects 
 the angle a a" a'. Therefore o is the pole of a face of the 
 octahedron. Therefore the twin axis is a normal to that face 
 of { l 1 1 } which truncates the solid angle formed by the faces 
 
 , a', 
 
 238. In a twin crystal of Gold (fig. 108), the individuals 
 are the icositessarahedrons {3 1 l}. The zones pr, sq of one 
 crystal coincide with the zones jpV, sq of the other. But the 
 face of the octahedron adjacent to the faces m, r, s is common 
 to the zones pr, qs. Therefore the great circle through the 
 poles of p, p intersects the great circle through the poles of 
 q, q' 9 in o, the pole of that octahedron, a normal to which is, 
 therefore, the twin axis. Then, p 9 q, p, q being the poles of 
 the faces p, q, p', q', we have pp f = 180 - 2po = 20. 4', 
 qq' = 20. 4> r . The faces qq form a re-entering angle. 
 
 239. A twin crystal of Diamond (fig. 109) is composed of 
 two hemioctahedrons, the faces of which are parallel to the 
 alternate faces of the same octahedron. One half form comes 
 into the position of the other by revolving through 180 round 
 a normal to a face of the form {o 1 l}, which is therefore the 
 twin axis. 
 
 14 
 
106 
 
 240. A twin crystal of Yellow Iron Pyrites (fig. 110) is 
 composed of two hemitetrakishexahedrons, the faces of which are 
 parallel to the faces of the same holohedral form. It is easily 
 seen by inspecting fig. 10, that the poles of one form will coin- 
 cide with those of the other after revolving through 180 round 
 any two opposite poles of {oil}. Therefore the twin axis 
 is a normal to any face of {o 1 l}. 
 
 241. In the first four of the preceding examples we have 
 found the twin axis to be perpendicular to a face of the octa- 
 hedron. This is perhaps the only way in which the individuals 
 of this system composing twin crystals are united. Twins 
 analagous to the last two examples, in which the twin axis is 
 perpendicular to a face of {o 1 l}, can only be produced by 
 the union of hemihedral forms. In such crystals, (Mohs 
 Naturgeschichte des Mineralreichs 154 158), the crystallogra- 
 phic axes of one of the presumed individuals are parallel to 
 those of the other, and the cleavages of one are parallel to those 
 of the other, or continued into it without interruption. We 
 cannot therefore determine with certainty whether such crystals 
 are to be considered as twins or as single crystals the faces of 
 which are repeated with a certain degree of regularity. 
 
 Examples of Twin Crystals belonging to the Pyramidal System. 
 
 242. In twin crystals belonging to the pyramidal sys- 
 tem, the twin axis is perpendicular to a face of one of the 
 forms {l 00}, {l 1 o}, {h O/}, [hhl\. 
 
 243. In a twin crystal of Oxide of Tin (fig. Ill), the 
 two pyramidal faces s, s' of one crystal are respectively pa- 
 rallel to the corresponding faces s', s of the other. One 
 crystal comes into the position of the other by making a 
 half revolution round an axis perpendicular to the face which 
 belongs to the zone ss' 9 and makes equal angles with the 
 faces s, s. This line, therefore, is the twin axis. Let c be 
 the pole of (0 1), t the pole of the face which truncates 
 the edge formed by s, s'. cs 43. 38'; whence (112) ct = 
 
107 
 
 .33. 59', s = 29. 12', ts"= 71. 23', tg = 63. 35', *#"= 119. 12'. 
 Whence ^= 52. 50', g"g"= - 58. 12', 5^= 121. 36', s'V'= 
 37. 14'. 
 
 If we assume the symbols of the faces s, s f to be (l 1 l), 
 (l 1 1), the symbol of t will be (l 1). If we assume s, s 
 to be (101), (Oil), t will be (112). 
 
 244. In a twin crystal of Copper Pyrites (fig. 112), the 
 zone pp coincides with the zone p,p/j and the face p is 
 parallel to the face p^ In this case the twin axis is mani- 
 festly perpendicular to p, a face of the form {ill}. 
 
 If p, p, p", &c. be the poles of the faces p, p, p", &c., 
 pa = 54. 20', pp"= 108. 40', pp'= 70. ?', pp"' = 109. 53', pb = 
 29. 45', pc = 39. 5',5. Hence ,= 71. 20', p"p'/= - 37. 20', 
 p'pf= 39. 46', p'"p/"= - 39. 46', &X= 20. 30', cc,= 109. ll'. 
 
 245. The individuals composing a twin crystal of Copper 
 Pyrites (fig. 113) are square pyramids, having a hemihedral 
 character in consequence of the very unequal size of their 
 alternate faces. The large faces of one crystal are parallel to 
 the small faces of the other. One crystal will evidently come 
 into the place of the other by revolving through 180, round an 
 axis perpendicular to any one of the faces of a square prism 
 truncating the edges pp 9 p'p'. The twin axis will, therefore, 
 be a normal to a face of|lOO}or|llo}, according as we 
 consider p to be a face of 1 1 1 } or of {ill}. 
 
 246. In a twin crystal of Scheelale of Lime (fig. 114), the 
 individuals are similar to the crystal represented in fig. 55, 
 and are joined so that the faces p of the two crystals coincide 
 with each other. One crystal will come into the place of the 
 other by revolving through 180 round an axis perpendicular 
 to any one of the faces of a square prism truncating the edges 
 PP> P'P'J or nn - H e ce the twin axis is a normal to a face of 
 either of the forms { 1 0}, { 1 1 0} . 
 
 247. In the last two examples the axes and cleavage 
 planes of the two crystals are parallel. The propriety of con- 
 
108 
 
 sidering them twin crystals is, therefore, doubtful. On the 
 other hand, in the last example, the hemihedral forms are dis- 
 similar, and striae, which run parallel to the intersection of 
 a with p 9 meet in a line which divides the faces pp^ p'pf into 
 two parts each of which appears to belong to a different crystal. 
 
 Examples of Twin Crystals belonging to the Rhorabohedral System. 
 
 248. In twin crystals belonging to the rhombohedral 
 system, the twin axis is either perpendicular to a face of 
 { 1 1 1 } , or to a face of one of the rhombohedrons. 
 
 249. In a twin crystal of Calc Spar (fig. 115), the in- 
 dividuals of which are combinations of the forms {o 1 1^ (#), 
 {l!2^ (c), the faces c of the two crystals coincide with each 
 other. One crystal comes into the position of the other by 
 revolving through two right angles round a line parallel to 
 the intersection of c, c. But this line is perpendicular to 
 the face (l 1 l), and is, therefore, the twin axis. 
 
 250. The individuals composing the twin crystal of 
 Calc Spar (fig. 116), are rhombohedrons, the faces of which are 
 parallel to the cleavage planes. The two obtuse angles which 
 the faces of one crystal make with those of the other are 
 equal ; and the angle between normals to the two faces p, p / 
 is equal to twice the angle between a normal to p 9 and a nor- 
 mal to the face (l l 1). Hence one crystal will come into the 
 place of the other by revolving through two right angles 
 round a normal to (l 1 l), which, therefore, is the twin axis. 
 
 251. The individuals composing the twin crystal of Calc 
 Spar (fig. 117), are combinations of the forms |l 1 l| (o), 
 {112} (c). The zone co coincides with the zone cp t . cc = 
 52. 30',5. Let the arcs cc /5 cV, through the poles of opposite 
 faces of the two crystals meet in , t'. t'c t - 90 - \cc = 63. 45'. 
 ot = 90- t'c = 26. 15'. Therefore t is the pole of a face of the 
 form ^01 l}. Hence the twin axis is perpendicular to a face 
 of the form $01 l}. 
 
109 
 
 252. In the twin crystal of Calc Spar (fig. 118), the indi- 
 viduals of which are combinations of {112} (c), {20l| (r), 
 {110} (#), {ill} (/)> the zone#c in one crystal coincides 
 with the zone gp t in the other, and the distance between the 
 poles of g, g t =38. l6',4. Let t be the intersection of the 
 circles gg^ rr drawn through the poles of opposite faces of 
 the two crystals. Then tg=90- \gg = 70. 5l',8, therefore t is 
 the pole of a face of the cleavage rhombohedron { 1 0} . Hence 
 the twin axis is perpendicular to a cleavage plane. 
 
 C 3 > r > f being the poles of the faces c, g, r, /, tc = 
 45. 23',4, therefore cc y = 89. 13',2. tc = 68. 5?', therefore cV= 
 42. 6'. tg'= 37. 27',5, therefore gg f '= 105. 5'. tf= 107. 44', 
 therefore ff= - 35. 28'. */'=50. 34',5, therefore ///= 78. 5l'. 
 
 253. The individuals composing a twin crystal of Calc 
 Spar (fig. 119) have the form ^20l}- The faces r, r of one 
 crystal are respectively parallel to the faces r /5 r' t of the other. 
 The places of r, r are interchanged, and one crystal comes into 
 the position of the other by revolving through two right angles 
 round a normal to a face of { i i i } , the pole of which bisects 
 the arc joining the poles of r, r. Hence the twin axis is per- 
 pendicular to that face of {ill} which truncates the edge 
 formed by r, r. 
 
 254. In a twin crystal of Ruby Silver (fig. 120), the 
 faces ss 9 z of one crystal coincide with the faces # /5 xj of 
 the other. One crystal will therefore come into the place of 
 the other by revolving through two right angles round a per- 
 pendicular to a face, the pole of which bisects the arc joining the 
 poles of % 9 %'. The poles of bisect the arcs joining the ad- 
 jacent poles of the cleavage planes. %%' 42. 2l'. If we make 
 the cleavage rhombohedron {l 00}, #, % will be poles of the 
 form J 1 1 1 , and the twin axis will be perpendicular to a face 
 of {21 l}. 
 
 Examples of Twin Crystals belonging to the Prismatic System. 
 
 255. In a twin crystal of Aragonite (fig. 121), the zone 
 mm' of one crystal coincides with the zone mm\ of the other. 
 
110 
 
 and the faces m, m t of the two crystals are parallel. Hence 
 one crystal will come into the place of the other by revolving 
 through two right angles round a normal to m, which is, 
 therefore, the twin axis. 
 
 m, h, k being the poles of the faces m, h, k, we have 
 wm'=63. 50', mh=58.5 f , hk=54?.13 f . cosmk 
 therefore mk = 71. 59,5, mk'= 108. o',5. Whence 
 52. 20', hh = 62. 50', ,= 36. l', k'k'= - 36. l'. 
 
 256. In a twin crystal of Staurolite (fig. 122), the zone 
 or of one crystal coincides with the zone or t of the other, and 
 the distance between the poles of 00= 60. 36'. mm'= 50. 40', 
 2>r = 55. 22'. From these data it appears that, if we make 
 o (100), p (001), m (110), r (Oil), a point in the great 
 circle oo /? 90 from the bisection of oo /? and therefore 120. 18' 
 from o, in the zone-circle or, will be the pole of (3 2 2). 
 Hence the twin axis is perpendicular to the face (322). 
 
 Let t be the pole of the twin face, m, p, r the poles of 
 the faces m, p, r, tfm = 64. 46', tm'= 115. 14', tp = 6<P.3f, 
 tr = 30. 18'. Whence mm = 50. 28', m'm'= - 50. 28', pp t = 
 58. 46', rr r 11 9. 24'. 
 
 257. In a twin crystal of Staurolite (fig. 123), the zone 
 po of one crystal coincides with the zone p t o t of the other, and 
 the distance between the poles of o, o t - 88. 24'. Hence 
 a point bisecting the arc joining the poles of o, o /5 and there- 
 fore 45. 48' from o, in the zone-circle po, is the pole of (302). 
 Hence the twin axis is perpendicular to the face (3 2). 
 
 Examples of Twin Crystals belonging to the Oblique Prismatic System. 
 
 258. In a twin crystal of Felspar (fig. 124), the zones 
 mt, py of one crystal coincide with the zones mt, p i y l of 
 the other. Hence one crystal will come into the place of 
 the other by revolving through two right angles round . an 
 axis perpendicular to a face common to the zones mt, py, 
 which, therefore, is the twin axis. 
 
Ill 
 
 Let a be the pole of the face to which the twin axis is 
 perpendicular, p, y the poles of the faces p, y. Then ay = 
 35. 44',5, aj5=ll6. l' . Therefore yy = 108. 3l', ^=-52. 14'. 
 
 259- In a twin crystal of Felspar (fig. 125), the zone mt 
 of one crystal coincides with the zone mt^ of the other. The 
 distance between the poles of m, m /5 opposite faces of the 
 two crystals, is 121. 10'. Therefore a point in the great 
 circle tt^ 90 distant from the bisection of tt^ is distant 
 29. 25' from the pole of w, and therefore is the pole of the 
 face % (fig. 95). Hence the twin axis is perpendicular to 
 the face %. 
 
 %t = 29. 59', %t'= 91. 11', **"= 88. 49', *p = 102. 29', 
 xy = 66. 31. Therefore tt = 130. 2', t't'= - 2. 22', "#,"= 
 2. 22', pp^ - 24. 58', yy= 46. 58'. 
 
 260. In a twin crystal of Felspar (fig. 126), the zones 
 pm, oy of one crystal coincide respectively with the zones 
 pm^ o i y l of the other. Therefore one crystal will come into 
 the position of the other, by revolving through two right angles 
 round an axis perpendicular to the face n (fig. 95), common to 
 the zones pm, oy, which is therefore the twin axis. 
 
 p, t, m, n, being the poles of the faces p, t, m, n, we have 
 pn = 44. 56',5, tn = 84. 46', mn = 45. 3',5. Therefore pp,= 
 90. /, tt = ltf. 28', 7^=89. 53'. pp t is found to be very 
 nearly 90, and therefore mn very nearly 45. The difference 
 between this and the value of mn given above, is probably due 
 to a small error in the determination of some of the angles 
 from which mn was computed. 
 
 261. In a twin crystal of Felspar (fig. 127), the faces m, p 
 of one crystal are respectively parallel to the faces m l9 p, of 
 the other. One crystal will come into the position of the other 
 by revolving through two right angles round a normal to />, 
 which, therefore, is the twin axis. 
 
 p, , %, x being the poles of the faces p, t, #, #, we 
 have px = 50. 19', pt = 112. 16', p% = 102. 29'. Therefore 
 xx = 79. 22', tt= - 44. 32', xx= - 24. 58'. 
 
112 
 
 Example of a Twin Crystal belonging to the Doubly Oblique System. 
 
 262. In a twin crystal of Cleavlandite (fig. 128), the zone 
 Imt of one crystal coincides with the zone l^J, of the other, 
 and the faces m, m i are parallel. Hence one crystal will come 
 into the position of the other by revolving through two right 
 angles round a normal to w, which, therefore, is the twin axis. 
 
 Let p, m, , / be the poles of the faces p, m, t, /, the faces 
 p 9 m being parallel to the two most perfect cleavages. It is 
 found that mt = 62. ?', ml = 60. 8', mp = 93. 36'. Therefore 
 tt = 54. 56', //,= 59. 44', pp = - 7. 12'. 
 
 263. Occasionally several crystals are joined together in 
 such a manner that every two adjacent crystals are united 
 according to the law stated in (230). In many twin crystals, 
 as the preceding examples shew, faces of the two individuals 
 make with each other re-entering angles. The occurrence, 
 however, of such angles is not always to be taken for a 
 character of a twin crystal ; for two or more faces of a simple 
 crystal may be repeated, and thus form a re-entering angle 
 In some instances the faces are repeated several times, forming 
 a corresponding number of parallel grooves, which, when the 
 faces are very narrow, produce the striae observed upon the 
 faces of certain forms of some crystals. 
 

 CHAPTER IX. 
 
 GONIOMETERS, &c. 
 
 264. CARANGEAU'S Goniometer consists of two small metal 
 rulers (fig. 129), fastened together by a pin so as to move 
 stiffly on each other. In order to measure the mutual inclina- 
 tion of two faces of a crystal, the rulers are placed with their 
 edges touching the faces of the crystal in lines perpendicu- 
 lar to the intersection of the faces. The rulers are then 
 applied to a graduated semi-circle (fig. 130), without altering 
 their position with respect to each other, so that the intersec- 
 tion of their edges may coincide with the center of the gradu- 
 ation. The portion of the graduated arc, contained between 
 the edges of the rulers, will then measure the inclination of 
 the planes, or the supplement of the distance between their 
 poles. This instrument is obviously incapable of affording 
 accurate results. 
 
 265. The Reflective Goniometer of Wollaston is repre- 
 sented in fig. 131. A graduated circle L, the divisions of 
 which may be read off to minutes by means of a vernier N 9 
 is fixed on a hollow axle which may be turned round by 
 the milled head M. An axle CS passing through the hol- 
 low axle of L 9 and which either turns with the circle L 9 or 
 may be turned independently by means of the milled head 
 at S, carries a crooked arm CF. The part FG is connected 
 with 'CF by a joint which permits it to turn round an axis 
 perpendicular to CS, passing through CS produced, and has 
 a collar at G in which the pin HK turns and slides with 
 its axis perpendicular to the axis of FG, and passing through 
 the point in which SC produced intersects the axis of FG. 
 The crystal is fastened by means of a soft cement to a thin 
 plate of metal, fixed in a slit at K. 
 
 15 
 
114 
 
 266. To measure the angle between two faces of a 
 crystal. 
 
 Let p, q be two faces of a crystal, (p,q) the angle be- 
 tween lines drawn from any point within the crystal perpen- 
 dicular to p, q respectively, or the distance between the poles 
 of p, q. Make the intersection of p, q parallel to the axis 
 of the circle, and as nearly coincident with it as possible, by 
 means of the angular motion of FG, and the angular and 
 sliding motion of HK. Place the instrument upon a firm 
 stand ; and let A, B (fig. 132) be two signals in a plane pass- 
 ing through a point C in the intersection of the faces p, q, 
 and perpendicular to the axis of the circle. Turn the circle 
 till the image of one of the signals A, seen by reflexion in 
 the face p, coincides with B viewed directly, and read off 
 the arc at which zero of the vernier stands. Now turn the 
 circle till the image of A, seen by reflexion in <?, coincides 
 with B seen by direct vision, and read off the arc at which 
 zero of the vernier stands. The difference of the two read- 
 ings will measure the angle (p,q), and will, therefore, be the 
 supplement of the angle between the faces p, </, according to 
 the usual definition of the angle between two planes that 
 bound a solid. 
 
 For, if CE be drawn bisecting ACS, and CD perpen- 
 dicular to CE in the plane ACB, CD will be perpendicular 
 to p at the first observation, and to q at the second. Hence 
 the crystal, and, consequently, the circle to which it is at- 
 tached, must have been turned in the interval between the 
 observations through the angle (p,q) contained between per- 
 pendiculars to the faces p, q. 
 
 267- If a face r belong to the zone of p, q, it will be 
 parallel to the axis of the circle, and, therefore, in some one 
 position of the circle, the image of A seen by reflexion in 
 f will coincide with B seen directly. Hence in order to find 
 the faces which belong to the zone containing two given faces, 
 we must adjust the crystal as for the purpose of measuring 
 the angle between those faces, and then, while the circle makes 
 
115 
 
 one revolution, observe the faces that afford by reflexion im- 
 ages of A passing through B. 
 
 268. In order to make ABC, the plane through the two 
 signals and the crystal, perpendicular to the axis of the circle,, 
 turn the stand of the instrument, or move A, till the image 
 of A seen by reflexion in the plane surface of the circle, or 
 in any bright plane surface fixed parallel to it, coincides with, 
 a point A seen directly, the distance of which from A, in 
 a line parallel to the axis, is equal to twice the distance of 
 the crystal from the plane of the circle, or from the reflect- 
 ing surface. Let a solid having two bright parallel surfaces be 
 fixed in the place of the crystal, and adjusted till the image 
 of A seen by reflexion in either surface, describes the same 
 path while the circle revolves, and place the lower signal 
 B in the path traced out by the image of A. Having 
 thus made the plane ABC perpendicular to the axis of the 
 instrument, the intersection of the faces p, q is known to be 
 parallel to the axis, when, on turning the milled head S, 
 the images of A, seen by reflexion in p, q, are observed to 
 pass through B. The adjustment of the edge p, q is most 
 easily made by cementing the crystal to the plate K with 
 one of the two faces, p for example, nearly parallel to the 
 plate, which is to be fixed in the slit at the end of HK, 
 so that intersection of p, q may be nearly perpendicular to 
 HK, and, therefore, HK nearly perpendicular to CS. By 
 turning HK round its axis, the path of the image of A seen 
 in p, may be made to pass through B ; and then by turning 
 FG on its axle at F, the path of the image of A seen in q 
 may be made to pass through B. Should the latter adjust- 
 ment disturb the former, the process must be repeated. 
 
 269- The distances of the two signals from the crystal 
 should be nearly equal, and not less than six or eight feet. 
 The upper signal, when the observations are made in the 
 day time, may be a narrow black bar, placed in the upper 
 part of a window, parallel to the axis of the circle ; and the 
 lower signal, a white line on a black ground also parallel 
 
116 
 
 to the axis of the circle. In some cases an image of the 
 sun formed by a lens of short focal length, or a small round 
 hole in a screen through which the light of the sky is seen, 
 may be used with advantage for the upper signal. In ob- 
 serving at night, two narrow slits in screens, through one of 
 which is seen the flame of a candle, and through the other 
 a sheet of paper illuminated by a candle placed behind it, 
 answer extremely well for the upper and lower signals re- 
 spectively. 
 
 270. When the points of p, q at which the reflexions 
 take place are not equally distant from the axis of the circle, 
 the angle through which the circle revolves between the two 
 observations will differ slightly from the angle between nor- 
 mals to p 9 q. 
 
 Let A, B (fig. 133) be the two signals ; PE, QE the lines 
 in which p, q meet a plane through A, B perpendicular to 
 the axis of the circle at the first and second coincidences of 
 B with the reflected image of A ; P, Q the points at which 
 the reflexions take place; (p,q) the angle between normals 
 to the faces; V the angle, not greater than 180, through 
 which the circle revolves between the two observations. Then, 
 if PEQ E, we have (p,q) V =p E, where the upper or lower 
 sign is to be taken according as the circle is turned in the 
 direction PQE or QPE. 
 
 Let AP, BQ meet in D, then 2 PEQ = 2 PED + 2 QED = 
 2EPA - 2EDA + 2 EQB - 2 EDB = APB - ADB + AQB - 
 ADB = PBQ + PAQ ; .% PEQ = \ sum of PAQ, PBQ. 
 
 When one of the points P, Q is within the angle which 
 AB subtends at the other, we have PEQ = 1 difference be- 
 tween PAQ, PBQ. 
 
 271. To eliminate the error arising from the excentricity 
 of the points P, Q at which the reflexions take place. 
 
 Let A) B be equidistant from C 9 the point in which the 
 axis of the circle meets the plane through A, B perpendicular 
 
117 
 
 to it. When the crystal is on the left of the circle, let V be 
 the angle through which the circle has revolved between the 
 two observations ; E the error ; P, Q the points at which the 
 reflexions take place. Turn the instrument half round in 
 azimuth so that the crystal may be on the right of the circle, 
 and repeat the observations. Let V' be the angle through 
 which the circle revolves ; E* the error ; P', Q' the points at 
 which the reflexions take place. We shall now have PQ, P'Q' 
 very nearly equal, and equally inclined to EC, but in contrary 
 directions. Hence PAQ, = P'BQ', PBQ = P'JQ', and, there- 
 fore, E, E f are very nearly equal. Now the directions in which 
 the circle is turned in its first and second positions are opposed 
 to each other, therefore E, E f will have different signs, there- 
 fore F, V are one greater and the other less than (p,q). 
 Hence if (p,q)=V=?E, (p,q) = F E' . Therefore, since 
 E, E' are very nearly equal, (p,q) = 1(F + V f ) very nearly. 
 
 When either of the faces is large, it should be blackened 
 over, except at the point where it is intended the reflexion should 
 take place. If this precaution were neglected, we should not 
 be at liberty to assume that PQ = P'Q', and that PQ, P'Q' are 
 equally inclined to CE. Any error that may arise from im- 
 perfect centering of the circle, will be eliminated, if the obser- 
 vations at a given face, in the two positions of the instrument, 
 be made with zero of the vernier at points distant nearly 180 
 from each other. 
 
 When AC, BC are equal, AC=ECcosACB nearly; 
 therefore, if PQ make an angle with EC, AC sin PEQ = 
 PQcosACBsm0. 
 
 272. In many crystals, not belonging to the octahedral 
 system, Mitscherlich discovered that the angles between certain 
 faces vary slightly with the temperature of the crystal ; thus the 
 directions of the cleavages of Calc Spar, which, at the ordinary 
 temperature of the atmosphere, make angles of 105. 5 f with 
 each other, become more nearly right angled by 8',5, when the 
 temperature of the crystal is increased 100 C. For the same 
 
118 
 
 increase of temperature a line perpendicular to the face (ill) 
 expands 0,00286, and a line parallel to the face (ill) contracts 
 0,00056 of its length. The distance between the poles of the 
 faces m, m of a crystal of Aragonite (fig. 84), is diminished 
 2',8, and the distance between the poles of the faces &, k' is in- 
 creased 5', 5 by increasing the temperature of the crystal 100 C. 
 In a crystal of Gypsum, which belongs to the oblique pris- 
 matic system, the angle between the two axes to which the 
 third axis is perpendicular, and the ratios of the parameters, 
 vary with a change of temperature. 
 
 273. To find the plane angles of a face of a crystal. 
 
 Let the face p meet the faces ^, r ; and let P, Q, R be the 
 poles of the faces. The edges in which p meets #, r are per- 
 pendicular to the planes of the great circles PQ, PR. There- 
 fore the plane angle formed by the edges in which p intersects 
 q, r is the supplement of the angle QPR. The lengths of the 
 edges are not subject to any known law. 
 
 274. The system to which any given crystal belongs, is 
 best determined by observing the kind of symmetry that pre- 
 vails in the distribution of the poles of its faces. When the 
 crystal is transparent, its optical properties frequently help to 
 determine its system. Hauy ascertained that crystals belong- 
 ing to the octahedral system have no double refraction. 
 Sir David Brewster discovered that crystals belonging to the 
 pyramidal system have one optic axis parallel to the crystallo- 
 graphic axis 0Z, which is perpendicular to the axes along 
 which the two equal parameters are measured : that crys- 
 tals belonging to the rhombohedral system have one optic 
 axis which makes equal angles with the crystallographic axes ; 
 and that crystals belonging to the three remaining systems 
 have two optic axes. In crystals belonging to the prismatic 
 system, the optic axes lie in a plane through two of the crys- 
 tallographic axes, and the angle they make with each other is 
 bisected by one of the crystallographic axes. In crystals be- 
 longing to the oblique prismatic system, the optic axes are 
 either in the plane of the two crvstallographic axes OZ, OX 
 
119 
 
 which are perpendicular to the third OF, or in a plane passing 
 through OF, and make equal angles with OF. 
 
 Thus in Idocrase the optic axis is perpendicular to the 
 face p (fig. 51). In Calc Spar the optic axis is perpendicular 
 to the face o (fig. 68). In Apatite it is perpendicular to the 
 face p (fig. 73). In Quartz it is parallel to the intersec- 
 tion of the faces r. A crystal of Quartz which has faces of 
 a hemihedral form in the zones containing two adjacent faces 
 /?, #, is found to be optically right-handed or left-handed 
 according as the zones are direct, as in fig. 75, or inverse. 
 (Transactions of the Cambridge Philosophical Society, Vol. i. 
 p. 43. Vol. iv. p. 79.) Aragonite has two optic axes perpendi- 
 cular to the axis of the zone kk' (fig. 84), making angles of 
 9. 7' with the axis of the zone mm'. In Sulphate of Magnesia 
 the optic axes are perpendicular to the axis of the zone mm 
 (fig. 86), and make angles of 25. 50' with a normal to the face 
 e. In Saxon Topaz the optic axes make angles of 31. 9' with 
 a normal to p, in a plane through a normal to p and the axis 
 of the zone nri '. In Epidote the optic axes are perpendicular 
 to the axis of the zone mt (fig. 93) ; one makes an angle of 
 5. ll' with a normal to r towards a normal to I; the other makes 
 an angle of 18. 5' with a normal to m towards a normal to t. 
 In Felspar the optic axes are parallel to the face p (fig. 95), and 
 make angles of about 58 with a normal to the face m. In 
 Oxalic Acid the optic axes are perpendicular to the axis of the 
 zone pe (fig. 96), and make angles of 56 with a normal to p. 
 Axinite and Blue Vitriol have each two optic axes, the posi- 
 tions of which with respect to the faces of the crystals have not 
 been well ascertained. 
 
 275. The geometrical description of a crystal may be 
 considered complete when we have given The angles its axes 
 make with each other : the ratios of its parameters : the 
 symbols of the simple forms of which it is a combination, and 
 the symbols of the cleavage forms. We may substitute for 
 the inclinations of the axes and ratios of the parameters certain 
 angles from which the position and mutual inclination of the 
 
120 
 
 faces of a crystal may be more readily calculated. In the 
 pyramidal system, the distance of the pole of (0 l) from the 
 pole (l 1 l) or that of (l 1), may be given instead of the ratio 
 of the parameters. In the rhombohedral system, the distance 
 between the poles of (l 0) and (1 1 l) may be given instead of 
 the inclination of the axes. In the prismatic system, the dis- 
 tance of the pole of (l 1 l) from the poles of two of the three 
 faces (100), (010), (001), may be given instead of the ratios 
 of the parameters. In the oblique prismatic system, the dis- 
 tance between the poles of (l 1 1) and (0 1 0), and the angles 
 which this distance makes with the zone-circles through the 
 pole of (01 0) and the poles of (001) and (l 0), may be 
 given instead of the inclination of the axes OZ, OX, and the 
 ratios of the parameters. In the doubly oblique system, the 
 distances between the poles of every two of the three faces 
 (100), (010), (001), and of any two of them from the pole 
 of (l 1 l), may be given instead of the inclinations of the axes 
 and ratios of the parameters. 
 
 276. It has been proved in (23) that any given pole, 
 which has an index greater than unity, is the intersection of 
 two zone-circles passing through poles that have lower indices 
 than the given pole. The following table shews how the posi- 
 tion of any pole, having no index greater than 7? may be 
 determined by the successive intersections of zone-circles drawn 
 through poles having lower indices, commencing with the poles 
 (ill), (T 1 1), (l T l), (l 1 1). If we suppose the pole T to 
 be the intersection of the zone-circles PQ, RS, the first column 
 will contain the symbol of T, the second and third columns 
 will contain the symbols of P, Q, and the fourth and fifth 
 those of R, S. When the three indices of T are numerically 
 the same as the three indices in any line of the table, but their 
 order and signs different, the symbols of P, Q, R, S may be 
 found from the symbols given in the table by the application 
 of the rules in (20). 
 
121 
 
 00 
 
 1, 
 
 1 1 
 
 1 1 1 
 
 1 1 1 
 
 1 1 1 
 
 554 
 
 1 1 1 
 
 001 
 
 1 
 
 21 
 
 21 1 
 
 1 
 
 1 
 
 00 
 
 010 
 
 1 1 1 
 
 001 
 
 610 
 
 100 
 
 010 
 
 i 
 
 01 
 
 212 
 
 1 
 
 1 
 
 1 1 
 
 loT 
 
 100 
 
 01 
 
 611 
 
 1 1 1 
 
 100 
 
 3 
 
 10 
 
 01 1 
 
 1 1 
 
 1 
 
 01 
 
 1 1 
 
 1 1 1 
 
 1 00 
 
 621 
 
 100 
 
 02 1 
 
 
 
 Ti 
 
 201 
 
 2 1 
 
 
 
 01 
 
 1 1 1 
 
 100 
 
 021 
 
 631 
 
 001 
 
 2 1 
 
 1 
 
 1 
 
 2Tl 
 
 10 
 
 
 
 01 
 
 010 
 
 ol 
 
 21 1 
 
 632 
 
 001 
 
 2 10 
 
 1 
 
 1 
 
 2T2 
 
 1 1 
 
 1 
 
 1 1 
 
 100 
 
 1 1 
 
 1 1 
 
 641 
 
 ill 
 
 1 10 
 
 
 
 1 1 
 
 210 
 
 20 
 
 | 
 
 00 
 
 01 
 
 il 
 
 21 1 
 
 643 
 
 010 
 
 201 
 
 1 
 
 1 1 
 
 20T 
 
 21 
 
 I 
 
 01 
 
 1 10 
 
 1 1 
 
 Toi 
 
 650 
 
 100 
 
 01 
 
 2 
 
 1 1 
 
 023 
 
 22 
 
 1 
 
 1 1 
 
 1 00 
 
 20 
 
 101 
 
 651 
 
 1 10 
 
 101 
 
 2 
 
 1 1 
 
 021 
 
 31 
 
 1 
 
 1 1 
 
 1 00 
 
 20 
 
 ill 
 
 652 
 
 1 10 
 
 1 02 
 
 2 
 
 1 1 
 
 021 
 
 32 
 
 1 
 
 1 1 
 
 001 
 
 21 
 
 ilo 
 
 653 
 
 1 1 1 
 
 102 
 
 
 
 10 
 
 201 
 
 1 
 
 1 
 
 00 
 
 010 
 
 101 
 
 212 
 
 654 
 
 1 1 1 
 
 10! 
 
 2 
 
 1 1 
 
 021 
 
 1 1 
 
 1 
 
 1 1 
 
 100 
 
 oil 
 
 210 
 
 655 
 
 1 1 1 
 
 100 
 
 
 
 T2 
 
 221 
 
 2 I 
 
 1 
 
 Ti 
 
 1 10 
 
 001 
 
 21 1 
 
 661 
 
 1 1 1 
 
 001 
 
 1 
 
 20 
 
 22 1 
 
 30 
 
 1 
 
 00 
 
 01 
 
 1 1 1 
 
 2 12 
 
 665 
 
 1 1 1 
 
 001 
 
 2 
 
 1 1 
 
 221 
 
 31 
 
 1 
 
 01 
 
 1 10 
 
 1 1 1 
 
 102 
 
 710 
 
 I 00 
 
 010 
 
 2 
 
 01 
 
 1 13 
 
 32 
 
 1 
 
 1 1 
 
 Toi 
 
 010 
 
 21 1 
 
 7 1 1 
 
 1 1 1 
 
 1 00 
 
 2 
 
 10 
 
 121 
 
 33 
 
 /i 
 
 1 1 
 
 100 
 
 oil 
 
 22 1 
 
 720 
 
 1 00 
 
 01 
 
 1 
 
 02 
 
 3 1 T 
 
 41 
 
 i 
 
 1 1 
 
 001 
 
 201 
 
 121 
 
 721 
 
 100 
 
 121 
 
 1 
 
 11 
 
 1 2T 
 
 43 
 
 i 
 
 1 1 
 
 001 
 
 021 
 
 21 1 
 
 722 
 
 1 1 1 
 
 100 
 
 1 
 
 12 
 
 210 
 
 1 
 
 :i 
 
 00 
 
 1 
 
 1 12 
 
 201 
 
 730 
 
 100 
 
 010 
 
 2 
 
 iT 
 
 103 
 
 1 1 
 
 i 
 
 ] 1 
 
 100 
 
 ill 
 
 21 
 
 731 
 
 101 
 
 210 
 
 1 
 
 00 
 
 1 31 
 
 20 
 
 i 
 
 00 
 
 010 
 
 2 ll 
 
 102 
 
 732 
 
 ill 
 
 1 10 
 
 1 
 
 oT 
 
 1 12 
 
 2 1 
 
 i 
 
 00 
 
 121 
 
 101 
 
 1 ll 
 
 733 
 
 1 1 1 
 
 100 
 
 1 
 
 Ti 
 
 230 
 
 22 
 
 -A 
 
 1 1 
 
 100 
 
 1 20 
 
 201 
 
 740 
 
 100 
 
 010 
 
 1 
 
 02 
 
 32T 
 
 30 
 
 i 
 
 00 
 
 010 
 
 211 
 
 1 12 
 
 741 
 
 1 1 1 
 
 i ol 
 
 1 
 
 21 
 
 21 1 
 
 31 
 
 i 
 
 1 1 
 
 10! 
 
 il 
 
 1 10 
 
 742 
 
 100 
 
 12 1 
 
 1 
 
 iT 
 
 21 1 
 
 32 
 
 i 
 
 01 
 
 1 10 
 
 1 1 
 
 20! 
 
 743 
 
 I 10 
 
 101 
 
 1 
 
 21 
 
 eTo 
 
 33 
 
 i 
 
 1 1 
 
 100 
 
 il 
 
 221 
 
 744 
 
 1 11 
 
 100 
 
 1 
 
 02 
 
 123 
 
 40 
 
 i 
 
 00 
 
 010 
 
 22 
 
 102 
 
 750 
 
 1 00 
 
 010 
 
 1 
 
 T3 
 
 22T 
 
 >41 
 
 i 
 
 01 
 
 1 10 
 
 Ti 
 
 2 1 
 
 751 
 
 111 
 
 1 10 
 
 1 
 
 1 1 
 
 121 
 
 42 
 
 i 
 
 00 
 
 021 
 
 1 10 
 
 102 
 
 752 
 
 1 10 
 
 101 
 
 1 
 
 1 1 
 
 310 
 
 i43 
 
 i 
 
 1 1 
 
 10! 
 
 101 
 
 120 
 
 753 
 
 1 1 1 
 
 10! 
 
 1 
 
 1 1 
 
 120 
 
 144 
 
 i 
 
 1 1 
 
 1 00 
 
 1 20 
 
 112 
 
 754 
 
 1 1 1 
 
 Tl 2 
 
 1 
 
 21 
 
 iTo 
 
 51 
 
 i 
 
 1 1 
 
 001 
 
 2ll 
 
 120 
 
 755 
 
 1 1 1 
 
 1 00 
 
 1 
 
 2T 
 
 lT3 
 
 152 
 
 i 
 
 1 1 
 
 001 
 
 321 
 
 ilo 
 
 760 
 
 100 
 
 010 
 
 1 
 
 03 
 
 22 1 
 
 >53 
 
 i 
 
 1 1 
 
 00 1 
 
 121 
 
 2 To 
 
 761 
 
 101 
 
 1 10 
 
 I 
 
 2 1 
 
 31 1 
 
 
 
 16 
 
 
 
 
 
 
 
 
 
 
122 
 
 762 
 
 I 1 
 
 102 
 
 1 2 
 
 1 
 
 201 
 
 772 
 
 l 
 
 1 00 1 
 
 1 30 
 
 
 
 1 1 
 
 763 
 
 00 
 
 121 
 
 1 1 
 
 1 
 
 1 03 
 
 773 
 
 1 
 
 1 001 
 
 2lo 
 
 1 
 
 31 
 
 764 
 
 00 
 
 032 
 
 1 1 
 
 1 
 
 2 ll 
 
 774 
 
 L 
 
 1 001 
 
 121 
 
 3 
 
 To 
 
 765 
 
 1 1 
 
 10! 
 
 1 2 
 
 1 
 
 20 1 
 
 775 
 
 i 
 
 1 001 
 
 1 2 1 
 
 3 
 
 Ti 
 
 766 
 
 1 1 
 
 100 
 
 22 
 
 1 
 
 103 
 
 776 
 
 i 
 
 1 001 
 
 1 2 1 
 
 C 2 
 
 3 1 
 
 771 
 
 1 1 
 
 00 1 
 
 13 
 
 1 
 
 32 1 
 
 
 
 
 
 
 
 If we commence with (l 1 1), (l 00), (0 1 0), (00 1), from 
 which the positions of the poles of crystals belonging to the 
 rhombohedral system are most obviously deducible, we must 
 substitute for the first two lines of the table 
 
 110 111001 100010 
 llo 011101 1 00 01 
 
 Tii 001110 100011 
 
 277- Tables for converting the symbol of any form in 
 different systems of crystallographic notation into the equiva- 
 lent symbol in the notation used in the preceding pages. If 
 the indices, when expressed in numbers, appear as fractions, 
 they must be multiplied by the least common multiple of 
 their denominators. 
 
 1. Modified notation of Hauy employed by Mr Brooke, 
 (Enc. Metropolitana, Art. Crystallography). The same table 
 serves also to translate the notation of M. Levy, (Description 
 d'une Collection de Mineraux formee par M. H. Heuland), 
 the indices of which have the same ratios as in the preceding 
 notation, but often differ in magnitude. 
 
 Octahedral System. 
 
 P \\ 00} 
 A {vll} 
 
 B {vio} 
 
 B,B 9 'B r " {qr,rp,pq 
 
123 
 
 Pyramidal System. 
 
 M {ioo\ B [iov\ 
 
 P {001} G {vio} 
 
 A {llv} B p B q 'G r {qr,rp,pq\. 
 
 Rhombohedral System. 
 i 100 * E (Levy) {- 
 
 t) 
 
 E 
 
 |.. 
 
 } 
 
 V 
 
 O 
 
 f-.l 
 
 1} 
 
 V 
 
 
 
 B 
 
 \v 1 0} 
 
 
 (Prisme hexaedre ; Levy) 
 P {111} 
 
 i" + 3 '"'- 
 
 Prismatic System. 
 
 J/ {110} A, {-!, + !,!} 
 
 P I 001 ! ^ {-I, + l,o} 
 
 ^ {0, 2t), 1} ^ {o + l, -1,0} 
 
 B {lie} B p B,'H r {qr-rp,qr+rp,pq\ 
 
 E {8., 0,1} B,S 5 "G 
 
 + !, 0-1, lj 
 
124 
 
 Oblique Prismatic System. 
 
 Jf {110} A n {0 + 1, 0-1, -if 
 
 V 
 
 O ^2 00 1 ' i ^ ~~ ^? ^ ~i~ ^5 -^ ( 
 
 O, {0 + 1,0-1, 1} JJ {0 + 1, -.1,0} 
 
 4 {20, 0, ij G {0- 1, + 1. 0} 
 
 V f 
 
 B p B q 'H r ' \rp + qr,rp-qr,-pq\ 
 B p D q G, (rp-qr,rp+qr,pq\ 
 
 jj x 
 
 E {0,20,1} 
 
 Doubly-oblique Prismatic System. 
 
 T 7 jlOOl 
 
 ^ {I 
 
 J/ {010} 
 
 P {001^ ^ {i 
 
 O 
 
 H 
 
 D [oiv] p q 
 
 B p D q G r 
 
 F 
 
 The first, second and third indices become respectively 
 negative when E, I and A are substituted for O. 
 
 2. Notation of Mohs. 
 
125 
 
 Octahedral System. 
 
 H 
 
 {100} 
 
 B> 
 
 {332} 
 
 O 
 
 {111} 
 
 c, 
 
 {211} 
 
 D 
 
 {011} 
 
 C 2 
 
 {311} 
 
 4 
 
 \320\ 
 
 T, 
 
 {231} 
 
 A., 
 
 \210\ 
 
 T 
 
 {531} 
 
 A, 
 
 \310} 
 
 T 3 
 
 {421} 
 
 Pyramidal System. 
 
 [P+co] { 10 } 
 
 P-CC 5001} 
 
 P+OC 5110} 
 
 (P+oo)* {mio} 
 
 [(P+ co)'"] \m + l, 
 
 P ,:-- {111} 
 
 jr2 n ,0,l} 
 |r2 R + 1 ,0,l} 
 - 1 {r2",r2 M , l} 
 
 { 
 (rP + 2n- l) m \(m 
 
126 
 
 Rhombohedral System. 
 
 {in} R {100} 
 
 {21l} P m \m + l,0,l-w| 
 
 {oil} (P+co) w {3m + l,-2,l -3m] 
 
 rf + rais {h k k}, where (A - k) -~- (A + 2 &) = r2". If Q be a 
 pole of any rhombohedron, (Q) its symbol in the notation of 
 Mohs or Naumann ; D an adjacent pole of |o 7 1 } ; S a pole of 
 {& + /, - A;, - /J adjacent to Q and Z>; O a pole of {l 1 1 }; 
 T the intersection of QZ>, OA^, the symbol of T, in the 
 notation of Mohs or Naumann, will be (Q)" 7 ? where 
 m = (k + 2l)-r-3Jc. 
 
 Prismatic System. 
 P + n {2",2%l} 
 
 {0, (m 
 
 {(m+ l)2 n ,0,2} 
 
 (Pr-f ri) m \(m- l)2",(m + 1)2%2} 
 
 (Pr + n) m {(m + l)2 n 5 (m - l)2",2}. 
 
 Oblique Prismatic System. 
 
 Let (P) denote the symbol, according to Mohs, of the 
 
 (P) (P) 
 form \h kl} in the prismatic system. Then --- , - 
 
 will be [hkl], [hJcl] respectively, in the oblique prismatic 
 system. When the form \h k 1} has the same number of faces 
 in either system the 2 in the denominator is omitted. 
 
127 
 
 Doubly-oblique Prismatic System. 
 Let (P) be the symbol, according to Mohs, of the form 
 
 \hkl\ in the prismatic system. Then -r will be [h k /}; 
 
 -I- - \hll}', r^ * - {ft&7{; l{hJcl\,iu the doubly- 
 
 oblique system. When \hk 1} has twice as many faces in one 
 system as in the other the denominator is changed to 2, and 
 when it has the same number of faces in either system it is 
 omitted. 
 
 3. Notation of Naumann. 
 
 Octahedral System. 
 
 coOco {100} mOm {mil} 
 
 O \ lll \ O {mml} 
 
 co O { 011 } mOn {m,mn<,n]. 
 
 co On \n I 0} 
 
 Pyramidal System. 
 
 co Pco I 100 } >P \mml\ 
 
 OP { 01 } mPco {mOl} 
 
 co P I 110 } co Pw \n 1 0} 
 
 P I 111 } mPra |m,m^,n|. 
 
 Pco {101} 
 
 Rhombohedral System. 
 
 OR {ill} R" 1 {m + 1,0, 1-m} 
 
 co R {sll} 2R m {m, 1, -m} 
 
 coP2 {oil? coP'" {-m, m-1, 1} 
 
128 
 
 Prismatic System. 
 
 P {HI? mPn \r 
 
 OP { l \ ^ n [lnn\ 
 
 co P { 110 } mPn \mn,m,n\ 
 
 mP \mm\\ p n \nln\. 
 
 Oblique Prismatic System. 
 
 OP \ 001 \ -mPn [-mn,m,n} 
 
 P [ lll \ Pn {nln} 
 
 mP \mrnl} (mPn) \m,mn,n] 
 
 mP \-mm\\ (P^) \lnn\. 
 
 C * ' C 3 
 
 mPn \mn,m,n\ 
 
 Doubly-oblique Prismatic System. 
 
 P' {ill}; ; P {Til}; P t {ill?; f {ill}. The 
 symbols of the other faces are derived from P', 'P, P , P as 
 in the prismatic system. 
 
 4. Notation of Weiss. 
 
 Octahedral System. 
 
 1 
 
 1 
 
 l 
 
 
 
 - a : 
 
 Q 
 
 ', 
 
 ', \h 
 
 k l\ . 
 
 h 
 
 k 
 
 I 
 
 
 
 Pyramidal 
 
 System. 
 
 
 i 
 
 l 
 
 1 
 
 
 
 JL 
 
 
 
 ' !* 
 
 kl}. 
 
 Rhombohedral System. 
 c 
 
 h + k 
 a 
 
 Prismatic System. 
 
 h a: 
 
CHAPTER X. 
 
 DRAWING CRYSTALS AND PROJECTIONS. 
 
 278. THE figures of crystals are usually projections, upon 
 a plane, of the edges in which their faces intersect, by lines pa- 
 rallel to a given straight line. Since the lengths of the edges 
 of a crystal are not subject to any known law, the edges in the 
 figure need not be either equal or proportional to the projec- 
 tions of the edges of any given crystal. In the rhombohedral 
 system the plane of projection is generally parallel to one axis, 
 and makes equal angles with each of the other two axes. In 
 the remaining systems the plane of projection is generally 
 parallel to two of the axes. The following rules for drawing 
 crystals will not require any demonstration. 
 
 279. To draw the axes of a crystal. 
 
 When a plane of projection is parallel to the axes OZ, OX-> 
 draw ZOZ', XOX' (fig. 135), making with each other an angle 
 equal to the angle between the corresponding axes of the crys- 
 tal, and any line YOY' making finite angles with ZOZ', XOX'. 
 Then, the three lines XOX', YOY', ZOZ' will represent the 
 three axes of the crystal. 
 
 If in OZ, OX, we take OC, OA proportional to the 
 parameters c, a respectively, and OB of any magnitude, OA, 
 OB, OC will represent the parameters of the crystal. 
 
 In the rhombohedral system, make the angle ROC (fig. 136) 
 equal to the angle between a normal to (l 1 l) and either of 
 the axes of the crystal. Draw CRS perpendicular to OR, 
 making RS - ^ CR. Through S draw ASB making any 
 angle with CS, and in ASB take A, B at any equal distances 
 
 17 
 
130 
 
 from 5. Then, the three straight lines OA, OB, OC will 
 represent the axes, and the distances OA, OB, OC will re- 
 present the parameters. 
 
 When the position of the plane of projection with respect 
 to the axes is arbitrary, the axes may be represented by any 
 three lines meeting in one point. 
 
 280. To draw lines parallel to the intersections of the 
 plane of the face (h k I) with the planes through the axes. 
 
 Let OA, OB, OC (fig. 135.) represent the parameters of 
 the crystal. Take OH, OK, OL respectively proportional 
 
 to - OA, - OB, - OC. Then KL, LH, HK will be parallel 
 
 ft r ' 
 
 to the lines in which the plane of the face (h k I) intersects the 
 planes YOZ, ZOX, XOY. 
 
 When one of the indices h, k, I is zero, the corresponding 
 axis will be parallel to two of the intersections. 
 
 281. Let the sides of the triangles HKL, PQR (fig. 137.) 
 be parallel to the intersections of the planes of two given faces 
 with the planes YOZ, ZOX, XOY. Let the intersections 
 with YOZ, ZOX, ^OFmeet in U, V, ^respectively. Then 
 a line through any two of the three points U, V, W, will pass 
 through the third, and will be parallel to the projection of the 
 edge formed by the intersection of the two given faces. 
 
 If we draw lines, by the above method, parallel to the 
 projections of all the edges of a crystal, meeting each other 
 in the order in which the edges meet, the figure thus obtained 
 will be the figure of the crystal. 
 
 282. To draw the axes of a twin crystal. 
 
 Construct fig. 138. so that OU, 0V, OW may be propor- 
 tional to the parts of the axes of either crystal cut off by the 
 twin plane, and VOW, WOU, UOV equal to the angles YOZ, 
 ZOX, XOY respectively. From the points O draw lines per- 
 pendicular to the adjacent sides of the triangle UVW, meeting 
 
131 
 
 in T, and draw the lines UTu, VTv, WTw. Let OX, OF, OZ 
 
 (fig. 139.) represent the axes of one of the crystals; U 9 V 9 W 
 the points in which they are intersected by the twin face. 
 Divide two of the sides of UVW into segments proportional 
 to the corresponding sides of UVW (fig. 138.), and let the lines 
 through the points of division and the opposite angles intersect 
 in T. Then OT represents a perpendicular on the plane 
 UVW. Take OO'=<2OT. Then O'U, O'V, O'W will re- 
 present the axes of the second crystal. Let (u v w) be the 
 symbol of the twin face. Take O'A'= u . OU, O'B'= v.O'V, 
 O'C'=w.O'W-, then 'A ', O'J9', O'C' will represent the pa- 
 rameters of the second crystal. The axes and parameters 
 having been projected, the faces of the two crystals may be 
 drawn by the rules already given. 
 
 283. No representation of a crystal is capable of exhi- 
 biting the relative positions of its faces, the zones which they 
 form, and the kind of symmetry with which they are arranged, 
 so clearly as the figure of a sphere having the poles of the faces 
 of the crystal laid down upon its surface. This method of 
 representing a crystal, which was invented by Professor Neu- 
 mann of Konigsberg, possesses the additional advantage of 
 enabling us to investigate all the geometrical properties of 
 crystals by spherical trigonometry alone, without the aid of 
 either solid or analytical geometry. 
 
 When the figure of the sphere is either its stereographic or 
 gnomonic projection, many problems of crystallography may 
 be very expeditiously solved by simple geometrical construc- 
 tions. On this account the following investigation of the 
 principal properties of the stereographic and gnomonic pro- 
 jections has been given. 
 
 284. In the stereographic projection, points and circles 
 upon the surface of a sphere are referred to the plane of a 
 great circle of the sphere, called the primitive, by lines drawn 
 to one of the poles of the great circle. To an eye placed in 
 this pole of the primitive, the projection of any point will 
 
 be the picture of that point on the plane of the primitive. 
 
132 
 
 285. Let O (fig. 140) be the center of a sphere which is 
 to be projected stereographically ; E, C the poles of the pri- 
 mitive, the eye being at E ; P 7 , Q' any two points on the 
 surface of the sphere. The straight line EC meets the plane 
 of the primitive in 0; .-.0 is the projection of C. Draw 
 the straight lines EP, EQ' meeting the plane of the primitive 
 in P, Q; .'. P, Q are the projections of P', Q'. Let r be the 
 radius of the sphere. Then OP = r tan OEP = r tan 1 CP'. 
 In like manner OQ = r tan 1 CQ'. 
 
 The angles QOP, Q'CP' are manifestly equal. 
 
 A straight line drawn from E to any point in the great 
 circle CP' will meet the plane of the primitive in OP. Hence 
 a great circle passing through the poles of the primitive is 
 projected into a straight line passing through the center of 
 the primitive. 
 
 286. Let Q' be any point in a circle of which P' is the 
 farther pole; P, Q the projections of P', Q'. Then 
 
 cos P'Q' = cos P'C cos Q'C + sin P'C sin Q'C cos Q'CP' 
 
 r QO 
 
 whence, 
 
 (cos P'Q' + cos P'C) QO 2 - 2r sin P'C cos QOP. QO 
 
 + r 2 (cos P'Q 7 - cos P'C) = 0. 
 Therefore Q is a point in a circle having its center in OP. 
 
 287- Let the given circle meet CP' in M', N' ; and let 
 M, N be the projections of M', N'. Then M'N will be a 
 diameter of the projection of the circle. 
 
 Let K be the center of the circle M QN. Then 
 2^Q = r tan 1 (P'Q' + CP') + r tan 1 (P'Q' - CP'), 
 2 .fiTO = r tan \ (P'Q' + CP') - r tan 1 (P'Q' - CP'). 
 
 When Q' is a point in a great circle, P'Q' is a quadrant. 
 Therefore KQ, = r sec CP', ^TO = r tan CP'. 
 
133 
 
 When Q r is a point in a small circle the pole of which 
 is in the primitive, CP' is a quadrant. Therefore 
 
 KQ, = r tan P'Q', KO = r sec P'Q'. 
 
 A circle passing through E, the place of the eye, will 
 manifestly be projected into a straight line. 
 
 288. To draw the projection of a great circle through 
 two given points. 
 
 Let Q be the most distant of the two points P, Q (fig. 141); 
 O the center of the projection. Draw OE perpendicular to 
 OQ meeting the primitive in E ; EQ meeting the primitive in q; 
 q O meeting the primitive in s ; Es meeting QO in S. A circle 
 through QRS will be the projection of a great circle. For QS 
 is the projection of an arc equal to qs ; .'. Q, S are the projec- 
 tions of opposite points of the two extremities of a diameter of 
 the sphere. Therefore the circle projected into QRS is a 
 great circle. 
 
 289. Having given the projection of a great circle; to 
 find the projection of its pole. 
 
 Let GMH (fig. 142) be the projection of a great circle, 
 meeting the primitive in G 9 H, and, therefore, GH a diameter 
 of the primitive. Through O, the center of the primitive, 
 draw MO perpendicular to GH. Draw GM meeting the 
 primitive in m. Make mp a quadrant ; and draw Gp meet- 
 ing M O in P. Then P will be the point required. For MP 
 is the projection of a quadrant mp, and G, H are the poles 
 of the circle projected into MP. Therefore GMH, P are 
 the projections of a great circle and its pole. 
 
 290. If a great circle and its pole be projected into GQH 
 and P (fig 143) ; and if straight lines PQ, PR be drawn 
 meeting the primitive in q, r, the arc qr will be equal to the 
 arc projected into RQ. For PQq, PRr are the projections 
 of small circles passing through the pole of the circle projected 
 
134 
 
 into GQH and through the place of the eye, which is one 
 pole of GqH. But two small circles drawn through the poles 
 of two equal circles manifestly intercept equal arcs of the 
 equal circles. Therefore rq is equal to the arc projected 
 into RQ. 
 
 291. To find the angle between two great circles; 
 having given their projections. 
 
 Let GR, LR (fig. 144) be the projections of two great 
 circles intersecting in R. Let the poles of the circles be 
 projected into P, Q. Draw the straight lines RP, RQ meet- 
 ing the primitive in p, q. The angle between the circles 
 projected into GR, LR is measured by pq. For R is the 
 projection of the pole of the great circle projected into PQ. 
 Therefore pq measures the distance between P, Q, the poles of 
 the circles projected into GR, LR, and therefore it measures 
 the angle between the circles which are projected into GR, 
 LR. 
 
 292. Let O be the center of the primitive MN (fig. 145) ; 
 MQ the projection of a great circle MQ' ; K its center; C 
 the farther pole of the primitive; CQ' a great circle meeting 
 MQ' in Q' and the primitive in N. Then the straight line 
 OQ will be the projection of CQ' ; OQ = r tan 1 CQ' ; 
 KQ = r sec Q'MN; KO = r tan Q[MN. The spherical tri- 
 angle MNQ' is right-angled at N. Whence 
 
 KO 
 
 sin KQO = -- sin KOL = sin Q'MN cos MN = cos MQ'N. 
 KQ 
 
 LO = KO sin MN = r tan Q'MN sin MN = r tan NQ'. 
 
 Hence, when the arc CQ' and the angle MQ'C are given, 
 if we make 
 
 = rtanlCQ'; OL = r cot CQ' ; OQK = 90 - MQ'C, 
 
 and draw LK perpendicular to ON; the circle MQ described 
 round K as a center, with the radius KQ, will be the pro- 
 jection of MQ'. 
 
135 
 
 293. Let SQ be the projection of any other great circle 
 S'Q' passing through Q', R its center. Then 
 
 OQR = 90 - S'Q'C. 
 But 0^ = 90- JJ/Q'C; . 
 
 Therefore the angle between any two great circles is equal 
 to the angle which their projections make with each other 
 at the point in which they intersect. 
 
 294. When a crystal belongs to the octahedral system, 
 the projection of the sphere, upon which its poles are laid 
 down, may have either the zone-circle through two adjacent 
 poles of JO 1 1 j, or the zone-circle through two adjacent poles 
 of {lOO}, for its primitive. When the crystal belongs to 
 the pyramidal or prismatic system, it will be found convenient 
 to take the zone-circle through the poles (l 00), (0 1 0) for the 
 primitive. When the crystal belongs to the rhombohedral 
 system, the primitive should be the zone-circle containing the 
 poles of {oT]J- When the crystal belongs to the oblique- 
 prismatic system, the primitive should be the zone-circle 
 through (001) and (100). When the crystal belongs to 
 the doubly-oblique system, any zone-circle may be taken for 
 the primitive. 
 
 295. To draw the stereographic projection of the poles of 
 a crystal of Axinite (fig. 100), having given mp = 45. 12', 
 pf= 44. 43', mx = 49. 32', my = 79. 24', fas 64. 5?'. 
 
 Let the zone-circle mp (fig. 101) be taken for the primitive; 
 and let O be its center, r its radius. Make mp = 45. 12', pf = 
 44. 53', and draw diameters mOm , pOp', fOf. In Of take 
 OK = r sec 64. 5?', and in Om take OL = r sec 49. 32', and 
 with centers JT, L and radii Kfc-rism 64. 5?', L oo = r tan 49. 32' 
 describe circles intersecting in x. Draw the circles mxm ^ 
 poop, foof- Draw OM perpendicular to Om, meeting mxm 
 in M; m'M meeting the primitive in J/'; take M'N' a quad- 
 rant, and draw m'N' meeting OM in N. In mpm take 
 mY = 79. 24', and draw the straight line ^F meeting 
 mxm in y. Draw pyp', and fyf meeting posp in v. Draw 
 
136 
 
 mvm meeting foof in , and pyp in w ; fw'f meeting mxni 
 in c, and pxp in w, and ptp meeting mxm in s. Draw 
 OT perpendicular to Ot meeting the primitive in T 7 ; draw 
 TU perpendicular to t T meeting Ot in U 9 and draw the 
 circle Uyt meeting mpm in e, and fwf in o. Draw eve' 
 meeting pyp' in g; fsf meeting mvm in Z; pip meeting 
 mxm in r, and mom meeting fyf in g. Then m 9 p, at, &c. 
 will be the projections of the poles of the faces m, p, #, &c. 
 
 296. In the gnomonic projection of the sphere, points 
 upon the surface of the sphere are referred to a plane touching 
 the sphere by lines drawn through them from the center of 
 the sphere. To an eye placed in the center of the sphere, 
 the projection of any point will be the picture of the point 
 upon the plane of projection. 
 
 297. Let O (fig. 146) be the center of a sphere ; C the 
 point in which the sphere touches the plane of projection, and 
 which is termed the center of the projection ; P, Q any two 
 points on the sphere. Draw straight lines OP', OQ' meeting 
 the plane of projection in P, Q. Therefore P, Q are the 
 projections of P', Q'. Let r be the radius of the sphere. 
 Then CP = r tan CP', CQ = r tan CQ', and QCP = Q'CP . 
 
 The plane of every great circle passes through O, and 
 intersects the plane of projection in a straight line. Hence 
 the projection of a great circle is a straight line. 
 
 Let PQ be the projection of a great circle P' Q' having 
 its poles in the great circle CP'. Then, since the planes CPQ, 
 OPQ are perpendicular to the plane CPO, their intersection 
 PQ will be perpendicular to CPO, and, therefore, PQ will be 
 perpendicular to CP. 
 
 298. Let O (fig. 14?) be the center of the sphere; C the 
 center of the projection ; Q'/?' an arc of a great circle ; QR 
 its projection. Draw ECB perpendicular to QR. Make 
 DB = CO, EB = CD. QR is perpendicular to BE and BO, 
 and BE=CD = BO. Whence QER = QOR. Therefore 
 QER measures the arc Q'R'. 
 
137 
 
 Hence we can measure the arc of a great circle projected 
 into a given straight line ; or cut off' a portion of a given 
 straight line, which shall be the projection of a given arc of 
 a great circle. 
 
 299. Let C (fig. 148.) be the center of the projection ; 
 O the center of the sphere ; CQ, PQ the projections of two 
 great circles CQ', P'Q'; CP f perpendicular to CQ', and, there- 
 fore, ECP perpendicular to CQ. 
 
 Take CE = CO. Draw CF perpendicular to QE 9 and 
 make CG = CF. Then, CG = CQ cos CQ', tan PQC = cos CQ' 
 tan CGP. Therefore CGP = CQ'P'. 
 
 Hence we may either find the angle between two great 
 circles of which the projections are given ; or, having given 
 the angle between two great circles, the projection of one of 
 them may be drawn through a given point in the projection 
 of the other. 
 
 When CQ' is small, the above method of comparing the 
 angle between two great circles with the angle between their 
 projections is not so accurate as the following. 
 
 300. Let C (fig. 149) be the center of the projection ; 
 CQ, PQ the projections of two great circles CQ', P'Q'. Draw 
 HQ perpendicular to CQ, and make HQ equal to the radius 
 of the sphere. Make KQ = CH ; bisect HK in Z,, and let 
 PQ meet HK in P. 
 
 HQ KP 
 
 KP HL - LP 
 Hence tan Q = 
 
 .. 
 
 HL I + tan Q' 
 
 18 
 
138 
 no 
 
 If / = Q0 - CO' 
 
 (sin *)*} 
 
 For the purpose of facilitating the above construction 
 sectors are provided with a scale called " inclination of 
 meridians," in which the distance between the divisions 
 0, ra = c{l - tan (45 - m) J, and with a scale called "lati- 
 tudes," in which the distance between the divisions 0, 1 
 
 2c sin Z 
 
 Let rad. sphere = CQ tan 1. Draw QH perpendicular to 
 CQ, making QH = (0, 1) of the line of latitudes. With 
 center H, and radius HK = (O n , 90') of the line of inclination 
 of meridians, describe a circle cutting CQ in K. Then, it will 
 be easily seen that, if KM = (0, m a ) of the line of inclination 
 of meridians, MQ will be the projection of a great circle that 
 makes an angle of m with the great circle projected into CQ. 
 
 301. The gnomonic projection may be used with ad- 
 vantage in projecting the poles of a crystal belonging to either 
 of the three systems in which the three axes make right angles 
 with each other. The plane of projection is most conveniently 
 situated when it meets the three axes at equal distances from 
 their intersection. The axes FZ, ZJf, XY will then be pro- 
 jected into the three sides of an equilateral triangle. 
 
 When the crystal belongs to the rhombohedral system, 
 the plane of projection may be parallel to a face of {l 1 ij, or 
 of {2 IT}- When the crystal belongs to the oblique prismatic 
 system, the plane of projection may be parallel to the faces of 
 {010J ; and when it belongs to the doubly-oblique system, 
 the plane of projection may be parallel to any face. 
 
 302. To draw the gnomonic projection of the poles of 
 
 a crystal of Topaz (fig. 8?) on a plane meeting the axes at 
 
 equal distances from their intersections ; having given ru 
 
 22. 15', rZ = 43.26', rm = 62.10' ? pn = 43. 30', py = 62. 13', 
 
 o = 45. 27'.- 
 
139 
 
 Let/ be the pole (010). Then r,/, p will coincide with 
 X, F, Z, and the arcs joining the poles of r, f, p will be pro- 
 jected into the equilateral triangle r, /, p (fig. 134). Let C be 
 the middle point of the triangle. Draw /C, pC meeting pr, 
 r/in M 9 N. Let O be the centre of the sphere. Then OC 
 will be perpendicular to the plane rfp. Or = O/, and rO/= 
 90, whence ON = Nr. With centre C and radius Nr describe 
 a circle cutting Nr in Q. CN is common to the triangles 
 QNC, OCN 9 QC = N0 9 QNC = OCN, therefore OC = Q^. 
 In Np take NR = Nr, and make uRr = 32.l5 f , IRr = 
 43. 26', ratfr = 62.10'. In J// take MS = Nr 9 and make 
 w ^jo = 43. 30', 2/ > = 62. 1 3'. Draw C T 7 perpendicular to p m. 
 In 7> take TU = OC. In TC take TFrz: C7C, and make 
 o Vp = 45.27'. Let wo meet jZ in x and pf in i. Let r# 
 meet pm in s ; and let/o meet pr in . Then |?, r, m, &c. will 
 be the projections of the poles of the faces p 9 r, m, &c. 
 
 THE END. 
 
PLATE I 
 
J. Grizve, 
 
J. 
 
PLATE VI. 
 
 76. , 
 
 , zincopf 
 
J. Grieve,, Zutcog. T 
 
105. 
 
 PLJLTEVBL. 
 
 JO4-. 
 
 1O6. 
 
PLATE JX. 
 
PLATE X. 
 
14 DAY USE 
 
 RETURN TO DESK FROM WHICH BORROWED 
 
 EARTH SCIENCES LIBRARY 
 
 This book is due on the last date stamped below, or 
 
 on the date to which renewed. 
 Renewed books are subject to immediate recall. 
 
 JA1 
 
 LD 21-40m-4,'64 
 (E4555slO)476 
 
 General Library 
 
 University of California 
 
 Berkeley 
 

 U.C.BERKELEY LIBRARIES