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 A. WENDELL JACKSON, Jr. 
 
 L I B R A R Y 
 
 UNIVERSITY OF 
 
 1 OALIFORNIA. 
 
 S 
 
A TRACT 
 
 CRYSTALLOGRAPHY 
 
 DESIGNED FOR THE USE OF STUDENTS 
 IN THE UNIVERSITY. 
 
 BY 
 
 W. H. MILLER, M.A. FOR. SEC. R.S., F.G.S., 
 
 ^. 
 
 FOREIGN MEMBER OF THE KOYAL SOCIETY OF GOTTINGEN, CORRESPONDING MEMBER OF THE 
 
 ROYAL ACADEMIES OF TURIN, BERLIN AND MUNICH, 31EMEER OF THE IMPERIAL 
 
 MINERALOGICAL SOCIETY OF ST PETERSBURG, HONORARY MEMBER OF 
 
 THE SOCIETY FOR PROMOTING NATURAL KNOWLEDGE IN FREIBURG, 
 
 AND PROFESSOR OP MINERALOGY IN THE UNIVERSITY OF CAMBRIDGE. 
 
 LI iVu AU V 
 
 US1V-EBSITY. OF 
 
 \UF(VKN1 A- 
 
 CAMBRIDGE : 
 DEIGHTON, BELL AND CO. 
 
 LONDON: BELL AND DALDY. 
 1863. 
 
(JTamfcriDge: 
 
 PRINTED BY C. J. CLAY, M.A. 
 AT THE UNIVERSITY PRESS. 
 
INTRODUCTION, 
 
 THE following Tract contains an investigation of the general 
 geometrical properties of the systems of planes by which crys- 
 tals are bounded, and of the formulae for calculating their 
 dihedral angles, indices and elements, given without demonstra- 
 tion in the last edition of Phillips' Mineralogy, or of equivalent 
 expressions in a more convenient shape. To these have been 
 added some theorems which appeared in the Philosophical 
 Magazine for 1857, 1858, and 1859. The last two chapters con- 
 tain concise investigations of the general properties of crystal- 
 line forms by the methods of ordinary and of analytical Geome^- 
 try, These were suggested by a remarkable paper entitled Sulla 
 legge di connessione delle forme cristalline di una stessa sostanza, 
 by the Commendatore Quintino(Sel5j> (Ntiovo Ctmento, Vol. iv.). 
 The Tract, therefore, besides containing all the theorems of 
 Mathematical Crystallography usually required in calculating 
 the angles of crystals, their elements, and the symbols of their 
 faces, will form, it is hoped, a useful supplement to the Mine- 
 ralogy, and also to the Crystallography published by the 
 author in 1839. The reader is referred to either of these works 
 
IV INTRODUCTION. 
 
 for examples, and for an account of the method of using 
 Wollaston's Goniometer. 
 
 The angle made by two faces of a crystal will be measured 
 by the angle between normals to the two faces, drawn towards 
 them, from a point within the crystal. The reasons for ad- 
 hering to this measure of a dihedral angle were given in the 
 Philosophical Magazine for May, 1860. It is needless to offer 
 any reasons for retaining the notation, in addition, to the remarks 
 made by the late Professor Grailich in his KrystallograpJiisch- 
 optiscJie Untersuchungen, p. 6. 
 
 The names used in the Mineralogy to designate two of the 
 hemihedral forms of the Prismatic System, and the hemihedral 
 form of the Oblique System, appeared to be inappropriate, and 
 have, consequently, been changed. 
 
CONTENTS. 
 
 CHAPTER I. 
 
 PAGE 
 PROPERTIES OF A SYSTEM OF PLANES ... 1 
 
 .1. Law according to which a system of planes is constructed. 
 Axes. Parameters. Indices. The symbol of a plane. 2. Angles 
 between the axes and a normal to a plane. Sphere of projection. 
 Poles. 3. Signs of the indices of a pole. 4. Angles which the arcs ->-' 
 joining any pole and two of the poles 100, 010, 001, subtend at 
 the third. 5. Condition that three poles may lie in a great circle. 
 6. A zone. A zone-circle. The axis of a zone. 7. Poles may 
 always exist in the intersections of two zone-circles. 8. Symbol of a 
 zone-circle passing through two poles, and of the poles in which two 
 zone-circles intersect. 9. Condition that a zone-circle may pass 
 through a pole. 10. To find the poles in a given zone-circle, and the 
 zone-circles passing through a given pole. 11, 12. Anharmonic ratio 
 of four poles in one zone-circle. 13, 14. Having given the symbols 
 of the poles P, Q, R, S, in one zone-circle, and the arcs PQ, PR, to 
 find the arc PS. 15. Having given the arcs PQ, PR, PS, where 
 P, Q, R, S are poles in one zone-circle, and the symbols of P, Q, R, 
 to find the symbol of S. 16. Anharmonic ratio of four zone-circles 
 intersecting one another" in the same poles. 17, 18. Having given 
 the symbols of tlie zone-circles KP, KQ, KR, K8, intersecting in the 
 pole K, and the angles PKQ, PKR, to find the angle PKS. 19. 
 Abbreviated notation for the anharmonic ratio of four poles or zone- 
 circles. 20. Having given the symbols of the poles P, Q, R, S, in 
 one zone-circle, and the arcs PQ, RS, to find the arc PR. 21, 22. 
 Change of axes. 23, 24. Change of parameters. 25. The axis of 
 the zone u v w is the diagonal, drawn from the origin, of a parallelo- 
 
VI CONTENTS. 
 
 piped having three of its edges in the axes, and proportional to 
 u#, \b, we. 26. Crystals ; their faces and cleavage planes. Mea- 
 sure of the dihedral angle between two faces. 27. Arrangement of 
 crystals in systems! 28. Forms. Holohedral and hemihedral forms. 
 Combinations. 
 
 CHAPTER II. 
 
 CUBIC SYSTEM 20 
 
 29, Axes and parameters. 30 32. Laws of symmetry of holo- 
 hedral and hemihedral forms. 33. Position of any pole. 34. Arc 
 joining any two poles. 35. Angles subtended by the arcs joining 
 any pole and two of the poles 1 0, 1 0, 1, at the third. 36 
 38. Arrangement of poles of holohedral and hemihedral forms. 
 39 52. Figures and angles of different forms. 53, 54. To find the 
 indices of a form. 
 
 CHAPTER III. 
 
 PYRAMIDAL SYSTEM . + .. . 29 
 
 55. Axes and parameters. 56 60. Laws of symmetry of holo- 
 hedral aiid hemihedral forms. 61 63. To find the position of any 
 pole. 64 67. Arrangement of the poles of holohedral and hemihe- 
 dral forms. 68 83. Figures and angles of different forms. 84 86. 
 To find the arc joining any two poles. 87, 88. To find the indices of 
 a form. 89 91. To find the element of a crystal. 
 
 CHAPTER IV. 
 
 RHOMBOHEDRAL SYSTEM . \- 'Vjj ''"'. 40 
 
 92. Axes and parameters. 93 96. Laws of symmetry of holo- 
 hedral and hemihedral forms. 97 100. To find the position of any 
 pole. 101. Dirhombohedral forms. 102 104. Arrangement of 
 poles of holohedral and hemihedral forms. 105 121. Figures and 
 angles of different forms. 122. To find the arc joining any two 
 poles. 123125. To find the indices of a form. 126128. To find 
 the element of a crystal. 
 
CONTENTS. Vll 
 
 CHAPTER Y. 
 
 PAGE 
 PRISMATIC SYSTEM 53 
 
 129. Axes. 130 133. Laws of symmetry of liolohedral and 
 hemihedral forms. 134 136. To find the position of any pole. 
 137 139. Arrangement of poles of holohedral and hemihedral 
 forms. 140 148. Figures and angles of different forms. 149, 150. 
 To find the arc joining any two poles. 151, 152. To find the indices 
 of a form. 153 156. To find the elements of a crystal. 
 
 CHAPTER VI. 
 
 OBLIQUE SYSTEM 63 
 
 157. Axes. 158, 159. Laws of symmetry of holohedral and 
 hemihedral forms. 160, 161. To find the position of any pole. 162. 
 Arrangement of the poles of the holohedral and hemihedral forms. 
 163166. Figures and angles of the different forms. 167, 168. To 
 find the indices of a form. 169, 170. To find the arc joining any two 
 poles. 171173. To find the elements of a crystal. 
 
 CHAPTER VII. 
 
 ANORTHIC SYSTEM 67 
 
 174. Forms. 175179. To find the position of any pole. 180. 
 To find the indices of a form. 181. To find the arc joining any two 
 poles. 182. To find the elements of a crystal. 
 
 CHAPTER VIII. 
 
 TWIN CRYSTALS . . . . 72 
 
 183. Laws of union of the two individuals constituting a twin 
 crystal. Twin axis. Twin face. 184. Arrangement of the poles of 
 a twin. 185. To find the twin axis. .186. To find the arc joining 
 any poles of each of the two crystals. 
 
Vlll CONTENTS. 
 
 CHAPTER IX. 
 
 PAGE 
 GEOMETRICAL INVESTIGATION OF THE PROPERTIES OF A SYSTEM OF PLANES 74 
 
 187. Equality of the products of the alternate segments of three 
 straight lines intercepted between their mutual intersections, and 
 the points in which they are intersected by a fourth straight line. 
 188. Law according to which a system of planes is constructed. 
 Axes. Parameters. Indices. Edges. 189. Symbol of an edge. 
 190. A zone. The axis of a zone. 191. Condition that a plane 
 may belong to a zone. 192. Plane parallel to two edges. 193 195. 
 Anharmonic ratio of four zone-axes, or four edges, in one plane. 
 196. Anharmonic ratio of four planes in one zone. 197,198. Change 
 of axes. 
 
 ^ CHAPTER X. 
 
 ANALYTICAL INVESTIGATION OF THE PROPERTIES OF A SYSTEM OF PLANES 83 
 
 199. Law according to which a system of planes is constructed. 
 Indices. Parameters. 200. Symbol of the intersection of two planes. 
 201. A zone. The symbol of a zone. The axis of a zone. 202. 
 Condition that a plane may belong to a zone. 203. Symbol of a 
 plane parallel to two zone-axes. 204. Portions of two zone-axes 
 cut off by two planes. 205. Anharmonic ratio of four zone-axes, &c. 
 Change of axes. 
 
1,1 BR AK 
 
 ! UNIVERSITY OL 
 
 GAL1FGE':- 
 
 CRYSTALLOGRAPHY, 
 
 CHAPTEE I. 
 
 PROPERTIES OF A SYSTEM OF PLANES. 
 
 1. LET OX, OY, OZ be any three straight lines not all 
 in one plane, passing through a given point 0; a, bj c any 
 three straight lines given in magnitude ; h t ~k, I any three inte- 
 gers, either positive or negative or zero, one at least being 
 finite. 
 
 Let a plane HKL meet the 
 straight lines X, Y, OZ respec- 
 tively, in the points H, K, L, such 
 that 
 
 ,OH . OK 7 OL 
 
 h K -r~ * 
 
 a b c 
 
 OH, OK, OL being measured 
 along OX, OY, OZ or in the 
 
 opposite directions, according as the corresponding numbers 
 "h, k, I are positive or negative. Suppose a system of planes to 
 be constructed by giving to h, Tc, I different numerical values, 
 the absolute distances of the planes from being perfectly arbi- 
 trary. Let the point be called the origin of the system of 
 
 M. C. 1 
 
2 CRYSTALLOGRAPHY. 
 
 planes ; the straight lines OX, F, OZ its axes ; a, b, c, or 
 any three straight lines in the same ratio, its parameters; 
 h, k, I, or any three integers in the same ratio, and having the 
 same signs, the indices of the plane HKL ; and let this plane be 
 denoted by the symbol h k I. When a numerical index is 
 negative, or a literal index is taken negatively, the negative 
 sign will usually be placed over the index. 
 
 It is evident that when one of the indices of a plane becomes 
 0, the point in which the plane meets the corresponding axis 
 will be indefinitely distant from the origin, and the plane will be 
 parallel to that axis ; also, that when two of the indices become 
 0, the plane will be parallel to the two corresponding axes. 
 
 2. Let the axes meet the surface of a sphere described 
 round as a centre in X, Y 9 Z\ and let OP be a normal to 
 the plane h k I, drawn towards it from 0, meeting the plane in 
 p, and the surface of the sphere in P. Then, if the plane h k I 
 meet the axes in H, K, L, 
 
 ,OH .OK 7 OL 
 
 But h - = k -r- = I - - . Therefore 
 
 a o c 
 
 ? cos XP = \ cos YP = j cosZP. 
 h k I 
 
 When h is positive, OH is measured along OX, and XOP 
 is less than a right angle ; therefore XP is less than a quad- 
 rant. When h is negative, OH is 
 measured in the opposite direction, 
 and XOp is greater than a right 
 angle ; therefore XP is greater than 
 a quadrant. In like manner YP 
 is less or greater than a quad- 
 rant, according as k is positive or 
 negative ; and ZP is less or greater 
 
PROPERTIES OF A SYSTEM OF PLANES. 3 
 
 than a quadrant, according as I is positive or negative. The 
 sphere to the surface of which the planes are referred will be 
 called the sphere of projection. The outer extremity of a radius 
 of the sphere, normal to any plane, will be called the pole of that 
 plane. A plane and its pole will be denoted by the same sym- 
 bol. The points in which the axes meet the surface of the 
 sphere of projection will be invariably denoted by X, Y, Z. 
 
 3. Let A, B, C be the poles 100, 010, 001 respectively ; 
 P the pole h kl. Then (2) 
 
 - cos XA = - cos YA = - cos ZA. 
 
 Therefore YA, ZA are quadrants. 
 
 In like manner it appears that ZB, 
 
 XB, XC, YC are quadrants. Also 
 
 (2) since the symbols of A, B, C 
 
 contain no negative indices, XA, 
 
 YB, ZG are less than quadrants. 
 
 Hence X, Y, Z are the poles of the 
 
 great circles BC, CA, AB adjacent 
 
 to A, B, C respectively ; and A, B, C are the poles of the great 
 
 circles YZ, ZX, XY adjacent to X, Y, ^respectively. Then, 
 
 since h, k, I are positive or negative according as XP, YP, ZP 
 
 are less or greater than quadrants, h will be positive or negative 
 
 according as P and A are on the same side or on opposite sides 
 
 of the great circle BC, k positive or negative according as P 
 
 and B are on the same side or on opposite sides of CA, and 
 
 I positive or negative according as P and C are on the same 
 
 side or on opposite sides of AB. 
 
 When P is in one of the great circles forming the triangle 
 ABC, the cosine of the arc joining P and the pole of the great 
 circle will be 0, and therefore the corresponding index will 
 be 0. 
 
 If a diameter PP' be drawn 
 
 cos XP = - cos XP, cos YF = - cos YP, cos ZP = - cos ZP. 
 
 12 
 
4 CRYSTALLOGRAPHY. 
 
 The ratios of the indices of P will therefore be the same as 
 those of P, but with contrary signs, because P, P are on oppo- 
 site sides of the great circles forming the triangle ABO. 
 
 4. Since X, F, Z are the poles of the great circles BC, 
 CA, AB, the arcs XP, YP, ZP are the complements of arcs 
 which divide each of the triangles BPC, CPA, APB into two 
 right-angled triangles. Therefore 
 
 cos XP = sin CP sin BCP = sin BP sin CEP, 
 cos YP = sin AP sin GAP = sin CP sin A CP, 
 cos ZP = sin BP sin A BP = sin AP sin 
 
 But cosZP=yCosrP = yCosZP. Hence 
 
 A A; 
 
 sin OP sin <?P = sin JBP sin OSP 
 
 sin^Psin CAP=\ sin CPsmAGP 
 k k 
 
 = sin J5P sin A BP = sin AP sin 
 
 From these equations we obtain 
 
 c 
 
 -smCBP = - 
 c a 
 
 a 
 
 5. Let P, R be the poles Jikl, pqr; Q any point in the 
 great circle PR, the arcs PQ, PR being measured in the same 
 direction from P. 
 
PROPERTIES OP A SYSTEM OF PLANES. 
 
 The spherical triangles PQX, EQX give 
 cos XP = cos XQ cos P Q + sin XQ sin PQ cos PQX, 
 cos ZS = cos XQ cosRQ + sin XQ sin It Q cos It QX. 
 
 Multiply both sides of the first equation by sin RQ, both 
 sides of the second by sinPQ, and add, observing that when 
 PQ is less than PR * 
 
 and sin It Q cos PQ + cosRQ sin PQ = sin PR. 
 The resulting equation is 
 
 cos XP sin R Q + cos XR sin PQ = cos XQ sin PS. 
 
 When PQ is greater than PR, we must interchange Q and .5 
 in the preceding equation, which then becomes 
 
 - cos XP sinRQ + cos XR sin PQ = cos XQ sin PS. 
 
 Writing sin (PR - PQ) for sin RQ, in order to reduce the two 
 cases to one, and then substituting Y and Z successively for X, 
 we obtain the following equations : 
 
 cos XP sin (PR - PQ) + cos XR sin PQ = cos 
 cos YP sin (P - P Q) + cos YR sin P Q = cos F Q sin PR, 
 cos ZP sin (P^ - P Q) + cos ZK sin P Q = cos ^Q sin PR. 
 Whence, by elimination, 
 
 (cos YP cos ZR - cos ZP cos 1TJ2) cos XQ 
 + (cos ZP cos XR - cos XP cos ZR) cos FQ 
 + (cos XP cos IB - cos FP cos XR) cos ^Q = 0. 
 
CRYSTALLOGRAPHY. 
 
 2 
 
 But cos XP = cos FP = cos 
 
 and - cos JO == - cos YR = - cos ZB. 
 
 p q r 
 
 Therefore, 
 
 ua cos X$ + v& cos YQ + we cos ZQ = 0, 
 
 where u = Jcrl^ v = lp hr, w = % - Jcp. 
 
 The great circle passing through the poles h Jc I, pqr may 
 be denoted by the symbol u v w. The numbers u, v, w will be 
 called the indices of the great circle PR. Any three integers in 
 the same ratio as u, v, w, satisfy the equation, between cos XQ, 
 cos YQ 9 cos ZQ, when substituted for u, v, w, and therefore may 
 be used as the symbol of the great circle PR. When u, v, w 
 have a common measure it will be convenient to employ as in- 
 dices the lowest integers in the required ratio. In cases where 
 there is reason to apprehend that the great circle u v w may be 
 mistaken for a plane or a pole, it may be distinguished from the 
 latter by the symbol [uvw]. 
 
 6. When three or more planes of the system of planes 
 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 
 planes not parallel to each other, and which, therefore, passes 
 through the pole of any other plane in the same zone with them, 
 will be called a zone-circle. The diameter which joins the poles 
 of the zone-circle will be called the axis of the zone. A zone, 
 its zone-circle, and any line parallel to its axis, will be denoted 
 by the same symbol. Hence, the intersections of the planes of 
 a zone, being obviously parallel to its axis, and to one another, 
 may be denoted by the symbol of the zone. 
 
 The symbol of the zone containing the planes 010, 001, or 
 of a line parallel to the axis OJT, is 100; that of the zone 
 containing the planes 001, 100, or of a line parallel to the 
 axis Y 9 is 1 ; and that of the zone containing the planes 
 1 0, 1 0, or of a line parallel to the axis OZ, is 1. 
 
PROPERTIES OF A SYSTEM OF PLANES. 7 
 
 7. Let h k 1, p q r be the symbols of any two zone-circles 
 intersecting in the points Q, Q'. Then (5), since Q is a point 
 in each of the zone-circles, 
 
 ha cos XQ + k& cos YQ + lc cos ZQ = 0, 
 pa cos X Q + qb cos YQ + re cos ZQ = 0. 
 
 Hence, putting u = kr Iq, v = lp - hr, w = hq kp, 
 
 - cos XQ = - cos YQ cos ZQ. 
 u v w 
 
 The indices of each of the zone-circles are integers, therefore 
 u, v, 10 are integers. Hence, Q, Q' are poles of planes belonging 
 to the system of planes, and common to the zones h k 1, p q r. 
 
 The points Q, Q are the opposite extremities of a diameter 
 of the sphere, therefore (3) the indices of Q being u, v, w, the 
 indices of Q' will be u, v, w. 
 
 8. It appears that when u v w is the symbol of the pole in 
 which the zone-circles h k 1, p q r intersect, the expressions for 
 u, v, w, in terms of h, k, 1, p, q, r, are precisely the same as the 
 expressions for u, v, w, in terms of A, &, I, p, q, r, where u v w is 
 the symbol of the zone-circle passing through the poles h Jc I, 
 p q r. If the symbols h k I, p q r be written twice, as below, 
 one under the other, and the letter X three times in the middle 
 three intervals, it will be seen that each of the indices u, v, w is 
 the product of the indices joined by the thick stroke of the cor- 
 responding letter X, minus the product of the indices joined by 
 the thin stroke, 
 
 h k I h k I 
 
 XXX 
 p q r p q r 
 
 u = kr - ?#, v = lp hr, w = hq- kp. 
 
 It will sometimes be found convenient to use the symbol 
 h k I, p q r to denote either the zone-circle containing the poles 
 
8 CRYSTALLOGRAPHY. 
 
 h Jc I, p q r, or one of the poles in which the zone-circles h k I, 
 p qr intersect, the two cases being distinguished, when requisite, 
 as in (5). 
 
 9. Let u v w be the symbol of the pole Q in the zone- 
 circle p q r. Then (2), (5), 
 
 - cos XQ = - cos YQ = - cos ZQ, 
 u v w 
 
 and pa cos XQ + qZ> cos YQ + re cos ZQ = 0. Hence 
 
 pw + qv + fw = 0. 
 
 This equation expresses the relation between the indices of a 
 zone and those of any one of its planes. Any positive or nega- 
 tive integers, including one or two zeros, which satisfy this 
 equation, when substituted for u, v, w, are the indices of a plane 
 in the zone p q r ; and any positive or negative integers, in- 
 cluding one or two zeros, which satisfy the same equation, when 
 substituted for p, q, r, are the indices of a zone containing the 
 plane u v w. 
 
 10. When the zone-circle p q r passes through the pole 
 u v w, we have, by (9), pw + qv + rw = 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 discover the integral 
 values, in which one or two zeros may be included, of x, y, z 
 which satisfy the equation ax + ly + cz = 0, where a, , c+ are 
 the indices of the given zone-circle in the former case, and of 
 the given pole in the latter, not necessarily arranged in the order 
 in which they stand arranged in the symbol. Let the coeffi- 
 cients c, b be prime to each other. Transform c : b into a con- 
 tinued fraction, and let e : d be the last but one of the resulting 
 converging fractions. Then by the solution of an indeterminate 
 equation of the first degree, y = (eax me), z = (mb dax], 
 where the upper or lower sign is to be taken, according as cd is 
 greater or less than be. The value of x being assumed, the cor- 
 
PROPERTIES OF A SYSTEM OF PLANES. 9 
 
 responding values of y and z may be obtained by substituting 
 different positive or negative integers for m. 
 
 11. Let P t Q, R, 8 be four poles in one zone-circle, PQ, 
 PR, PS being all measured in the same direction from P; e f g, 
 p q r the symbols of any two zone-circles KP, KR passing 
 through P, R respectively, neither of which coincides with PR ; 
 Ji k I, u v w the symbols of Q, S respectively. Then (5) 
 
 cos XP sin (PR - PQ) + cos XR sin PQ = cos XQ sin PR, 
 cos FPsin (PR - PQ} + cos YR sin PQ = cos YQ sin PR, 
 cos ZP sin (PR -PQ) + cos ZR sin PQ = cos ZQ sin PR, 
 
 Multiply both sides of the first, second, third of the preceding 
 equations by ea, f b, gc respectively, and add, observing that P 
 is a pole in the zone-circle e f g, and therefore (5), 
 
 ea cos XP+ fb cos YP+ gc cos ZP= 0. 
 
 Next, multiply by pa, qb, re respectively, and "add, observing 
 that R is a pole in the zone-circle p q r, and therefore 
 
 pa cos XR + qb cos YR + re cos ZR = 0. 
 
 The equations thus obtained are 
 
 (ea cos XR + fb cos YR + gc cos ZR) sin PQ 
 = (ea cos XQ + f cos YQ + gc cos ZQ) sin P#, 
 
 (pa cos XP+ qb cos FP+ re cos ZP) sin (PS - PQ) 
 = (pa cos -3T$ + qb cos F$ + re cos ZQ) sin 
 
 By the substitution of S for Q in the preceding equations, 
 we have 
 
 (ea cos XR 4 f b cos YR + gc cos ZB) sin PS 
 
 = (ea cos XS+ fb cos F + gc cos ZS) sin PS, 
 
 (pa cos XP+qb cos FP+ re cos ZP) sin (PS - PS) 
 = % (pa cos XS + q cos Y# + re cos ZS) sin PR. 
 
1 CRYSTALLOGRAPHY. 
 
 Bat Q, 8 are the poles hkl,uvw respectively, therefore 
 T cos XQ = T cos YQ = j cos ZQ, 
 
 - cos XS= - cos YS = - cos & 
 
 W V W 
 
 Hence 
 
 sinPQ sin (PR - PS) _eh + fk 
 
 sin PS sin (PR - PQ) ~ QU + iv + gw iph 
 
 12. It is easily seen that the left-hand side of the preceding 
 equation is positive, except when one only of the zone-circles 
 KP, KB passes between Q and $; or that the arcs PQ, PS, 
 RQ, RS must be considered positive or negative according as 
 they are measured in the directions PR or RP. If we attend 
 to this rule the equation may be written 
 
 sin PQ sin RS _ eh + f Jc + gl pu + qv -f tw 
 sin PS sin R Q ~~ QU + iv + gw pA + q& + il ' 
 
 in which the correspondence between the poles P, Q, R, S on 
 the left-hand side of the equation, and the symbols e f g, h k I, 
 p q r, u v w on the right-hand side, is more easily perceived 
 than in the original form of the equation. 
 
 13. sin (PR - PQ} = sin PR sin PQ (cot PQ - cot PR) , 
 sin (PE -PS) = sin PR sin PS (cot PS - cot PR). 
 
 Therefore (11), 
 
 cot PS cot PR _eh + fk + gl 
 cot PQ cot PR ~ Qu + fv + gw 
 
 From which, having given the symbols of the zone-circles 
 through the poles P, R, the symbols of the poles Q, S, and the 
 arcs PR, PQ, the arc PS may be found. 
 
PKOPEKTIES OF A SYSTEM OF PLANES. 11 
 
 14. Putting 
 
 ^ _ eh + f Jc + gl pu + qv + rw sin (PR PQ) 
 ~ eu + fy + gw Tph + qk + il sin PQ 
 
 we have -^ - = tan 6. 
 
 sinPff- sin (PR-PS] __ 1 -tan d 
 smPS'+sin(P#-P/S f )~H-tan0' 
 
 sin PS - sin (PR- PS) _ tan (P - 
 sin P + sin (P# - PS) ~ tan JPfl 
 
 l-tan0 
 and 
 
 Therefore tan (P5 - JPB) = tan %PR tan ( JTT - 0) . 
 
 Whence, having given the symbols of the zone-circles 
 through P, E, the symbols of Q, 8, and the arcs PR, PQ, the 
 arc PS may be found. 
 
 15. Let m n o be the symbol of the zone-circle PR. Then 
 from (11) and (9) we have 
 
 pu + qv 4- rw _ p + q& + r? sin Pg sin (PR PS) 
 w ~ eh + ik + gl sirTPtf sin (PR - PQ) ' 
 
 and mu + nv + ow = 0, two equations from which, having given 
 the arcs PR, PQ, PS, and the symbols of P, Q, R, the ratios of 
 u, v, w, the indices of 8 9 may be found. 
 
 16. Let KP, KQ, KR, KS be four zone-circles passing 
 through the pole K-, e f g, p q r the symbols of KP, KR ; hJcl, 
 uvw the symbols of the poles Q, S in the zone-circles KQ, 
 KS. Let the zone-circle QS meet KP in P, and KR in R. 
 Then 
 
12 CRYSTALLOGRAPHY. 
 
 sin KP sin PKQ = sin PQ sin KQP y 
 smKR sinRKQ = sinRQ sin KQR, 
 sin KP sin PKS = sin PS sin KSP, 
 &inKR sinRKS = sin RS sin 7T/S#. 
 
 Hence, observing that smKQP 
 = sin KQR, and si 
 we obtain 
 
 sn 
 
 sinRKS _ sinPQ sinJSff Therefore 
 ~~ ' 
 
 ,, , 
 
 swPKS smRKQ~ eu + fv+gw ph + qk + il' 
 
 As in (12) the left-hand side of the preceding equation is 
 positive, except when one only of the zone-circles KP, KR 
 passes between Q and 8. 
 
 17. It may be proved exactly in the same manner as in 
 (13), that 
 
 cot PKS- cot PKR eh + f k + gl 
 
 cotPKQ-cotPKR~ Qu+fv+gw yh + qk + il ' 
 
 Hence, having given the symbols of KP 9 KR, Q, S, and the 
 angles PKR, PKQ, the angle PKS may be found. 
 
 18. Putting 
 
 eh + f k + gl yu + qv + rw sin (PKR -PKQ) 
 ~ siuPKQ 
 
 we obtain exactly as in (14) 
 
 tan (PK8- %PKR) = tan \PKE tan (JTT - (9). 
 
 Whence, knowing the symbols of KP, KR, Q, S, and the 
 angles PKR, PKQ, the angle PKS may be found. 
 
PROPERTIES OF A SYSTEM OF PLANES. 13 
 
 19. The symbols of the zone-circles KP, KR being efg, 
 p q r, and the symbols of the poles Q, S being h k I, u v w, it is 
 sometimes convenient to denote the expression 
 
 eh + fk + gl pu -f- qv + rw 
 GU + f v + gw iph + q& + il 
 
 by [e f g, h Jc I . p q r, u v w], or by KP,Q.KR,S, either of which 
 suggests the formation of its numerator. The reciprocal of the 
 same expression may be denoted by [e f g, u v w . p q r, h k Z], or 
 by KP,S.KR,Q, either of which suggests the formation of its 
 denominator. 
 
 20. Let e " + f f ? + g ? E*S*?._,-. Then (11), sup- 
 eh + fk + gl pu + qv+rw 
 
 posing PS greater than PR, 
 
 sin PS sin (P Q - PR) = sin PQ sin (PS - PR). 
 
 But 2sinPsin(P<2-P.fl) 
 
 = cos (PS -PQ + PR) - cos (P8+ PQ - PR) 
 = cos (2PR- PQ + RS) - cos (PQ + RS), 
 
 And 2 sin PQ sin (P - PR) 
 
 = cos (PC - PS+ PR) - cos (PQ + P8- PR) 
 = cos (PQ -RS)- cos (PQ + RS). 
 
 Therefore 
 cos (2PR-PQ + RS) = (1 -t) cos (PQ + .&) + *' cos (PQ-RS). 
 
 Whence, having given the symbols of KP 9 KR, Q, S, and 
 the arcs PQ, JS$, the arc PR may be found. 
 
 In one of the most frequent applications of the preceding 
 equation, PQ is a quadrant, and the equation becomes 
 
 sin (2PB H- RS) = (2* - 1) sin RS. 
 
14 
 
 CRYSTALLOGRAPHY. 
 
 21. Let EF, FD, DE be the zone-circles e f g, h k 1, p q r ; 
 the pole m n o ; P the pole u v w. Then (16) 
 
 sin EFO sinDFP em 4- in 4- go hw + kv 4- Iw 
 sin jEFP sinZ^^O ~ ew + f v 4- gw hm + kn + io' 
 
 sin FEO sin 
 
 em + in + go pw + qv + 
 
 in DEO QU + iv + gw 
 
 Let mWo' be the symbol of 0, u'v'w' the symbol of P, when 
 referred to the axes of the zone-circles EF, FD, DE as axes of 
 the system of planes. Then (6) the new symbols of EF, FD, 
 DE will be 100,010, 001. Therefore (16) 
 
 sin EFO smDFP_m f 
 sin EFP siuDFO" w 7 
 
 sin FEO smDEP_m 
 sin FEP sin DEO ~" w 7 
 
 Hence, equating the right-hand sides of equations having iden- 
 tical left-hand terms, we obtain two equations which are satis- 
 fied by making 
 
 w'= em + f n + go, u zu 4 f v + gw, 
 
 o = 
 
 ' = pw 4- qu + 
 
 The coefficients of u, v, w are integers, therefore u, v, w', 
 the indices of P when referred to the axes of the zone-circles 
 e f g, h k 1, p q r as axes of the system of planes, will also be 
 integers. Hence, the planes of the system are subject to the 
 same law when referred to any three zone-axes, as when re- 
 ferred to their original axes. 
 
PROPERTIES OF A SYSTEM OF PLANES. 15 
 
 22. Let D, E, F be the poles efg, hkl, pqr. Let EF, 
 FD, DE meet the zone-circle m n o in M, N, 0, and the zone- 
 circle uvw in U, V, W. Then (12) 
 
 sin OD sin WE _ me + n/+ o>g u^ + v& + wl 
 sin OJSs'm WD ~ mh+nk + ol ue + v/+ w^ ' 
 
 sin ND sin VF me + nf+ og \\p + vq + wr 
 sin NF sin KZ) ~ m^> + n^ + or ue + vf+ wg 
 
 Let m' n' o f be the symbol of the zone-circle M0 t u' v f w' 
 the symbol of the zone-circle UW, when referred to the axes of 
 the zone-circles EF, FD, DE as axes of the system of planes. 
 Then (6), (7) the new symbols of D, E, F will be 1 00, 010, 
 001. Therefore (12) 
 
 sin OP sinJVE_m_ V sin ND sin VF _ m' w' 
 sin OE smWD ~ n' u 7 ' sinNF smVD~!7 u 7 * 
 
 Hence, equating the right-hand sides of the equations 
 having identical left-hand terms, we obtain two equations which 
 are satisfied by making 
 
 m'= em +/h + #o, u' = eu +fv + gw, 
 n' = Tim + Jen + fo, v' = Jin + lev + ?w, 
 o' =pm + qn + ro, w'=^9U + ^v + rw. 
 
 23. Let h Jc I, u v w be the symbols of the poles 0, P, the 
 parameters of the system of planes being a, Z>, c ; h'k'l', uvw 
 the symbols of 0, P when referred to the same axes, but with 
 the parameters a, b', c. Then (2) 
 
16 CRYSTALLOGRAPHY. 
 
 T cos XO T cos YO = -j cos ZO. 
 h K I 
 
 p cos XO = cos FO = cos ZO, 
 
 fl rC l> 
 
 -cosXP = -cos FP = -cosZP, 
 w V w 
 
 ^cos JTP = ~ cos YP= -, cosZP. 
 
 U V W 
 
 Hence Tin' : h'u = &y' : k'v = Iw : I'w. These equations are satis- 
 fled by making 
 
 v = AAj7t; w' = hU'w. 
 
 24. Let A A; I be the symbol of a pole, u v w that of a 
 zone-circle, the parameters being a, Z>, c ; A' k' I', u' v' w' the 
 symbols of the same pole and zone-circle when referred to the 
 same axes, but with the parameters a', Z>', c'. 
 
 Let m n 0, p q r be the symbols of any two poles in the zone- 
 circle, the parameters being a, 5, c ; m' w' <?', p' q r' their symbols, 
 the parameters being a', &', c'. Then (5) 
 
 u = w oq, v = op mr, w = mq np, 
 u'=7iV o'q'j v'=op'm'r, w' = m'q-rip'. 
 
 But (23) m' = Kklm, n=hk'ln, o' = hkl'o, p = h'klp, q' = hk'lq, 
 r hkl'r. Substituting these values of m', ri, o r , p', <?', r' in the 
 expressions for u', v', w', and rejecting the common factor Ti k I, 
 we obtain 
 
 25. Let .5T be the pole of the zone-circle u v w ; P, Q, R 
 poles of the great-circles KX 9 KY, KZ. The great-circles 
 FP, ZP, ZQ, XQ, XR, YR make with the great-circles KX, 
 KY, KZsui right-angled triangles having KX, KY, KZfor the 
 sides opposite to their right angles. Hence, 
 
PROPERTIES OF A SYSTEM OF PLANES. 17 
 
 cos YP = sin XKY sin KY, - cos ZP = sin ZKX sin KZ, 
 cos ZQ = sin YKZ sin /v^, - cos XQ = sin XfiTF sin KX y 
 cos F# = sin YKZ sin KY, - cos XR = sin ZZOT sin KX. 
 
 Since P, ft jft are poles of KX, KY, KZ, cosXP=0, 
 cos YQ = 0, cos ZR = 0. KP, KQ, KR are quadrants, therefore 
 P, $, ^ are points in the zone-circle u v w. Hence (5) 
 
 \b cos FP + we cos ZP 0, 
 ua cos XQ -f we cos ZQ = 0, 
 ua cos XR -f vS cos FZ2 = 0. 
 
 r - sin KX 7 sm K\ 
 
 Therefore ua =v& 
 sm YKZ 
 
 Construct a parallelepiped UVW having OK, the axis of the 
 zone, for a diagonal, and three of its edges OU, 0V, OW 
 coincident with the axes of the system of planes. Let KE, 
 KF, KG be the edges respectively parallel to OU, 0V, OW. 
 The angles GOU, GOV are the segments into which UO V is 
 divided by OG, the intersection of the planes WOK, UOV, 
 and are therefore measured by the arcs NX, NY, N being the 
 intersection of XY and KZ. Hence 
 
 V sin G U_ sin NX _ sin KX sin ZKX _ vj 
 U~ sin G V~ sin ^F ~ sin KY sin YKZ" ua ' 
 
 T ri OW CW 
 
 In like manner u., T = . 
 
 OZ7 ua 
 
 ,, , OT7 OF OTF 
 
 Therefore - = r = -- 
 
 ua vo we 
 
 M. c. 
 
18 CRYSTALLOGRAPHY. 
 
 Or, the axis of the zone u v w is the diagonal of a paral- 
 lelopiped, the edges of which coincide with the axes of the sys- 
 tem of planes, and are equal to ua, v&, we respectively. 
 
 26. Many natural substances, and many of the results of 
 chemical operations, occur in the form of polyhedral solids. 
 These, when broken, frequently separate in the directions of 
 planes passing through any point within the solid, either parallel 
 to certain planes of the solid, or making invariable angles with 
 them. Solids of this description are called crystals; the planes 
 by which they are bounded, their faces; and the planes in which 
 they separate, their cleavage planes. It appears from accurate 
 measurements of the mutual inclinations of the faces of a crystal, 
 including under the term faces, its cleavage planes also, and 
 from calculations founded on those measurements, that the posi- 
 tions of the faces of a crystal are subject to the law according 
 to which the system of planes described in (1) was constructed. 
 Hence, all the geometrical properties which have been esta- 
 blished for such a system of planes, are also properties of the 
 system of planes by which a crystal is bounded. 
 
 The angle between any two of the faces of a crystal will 
 be measured by the plane angle between normals to the two 
 faces, drawn towards the planes of the faces, from any point 
 within the crystal, or by the arc of a great-circle of the sphere 
 of projection joining the poles of the faces. 
 
 27. In many crystals axes may be discovered which make 
 right angles with one another; in others, axes of which one 
 makes right angles with each of the other two ; and in others, 
 axes making equal oblique angles with one another. In the crys- 
 tals with equiangular axes, and in some of the crystals with 
 rectangular axes, the parameters are all equal ; and among the 
 remaining crystals with rectangular axes, some which have two 
 of the parameters equal. Upon these differences in the mutual 
 inclinations of the axes, and in the relation betwen the parame- 
 ters, is founded the arrangement of crystals in systems. The 
 
PROPERTIES OF A SYSTEM OF PLANES. 19 
 
 different systems 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 sym- 
 bol h k I, it will generally be accompanied by the faces having 
 for their symbols certain arrangements of + A, + k, 1 deter- 
 mined by laws peculiar to each system. 
 
 28. The figure consisting of a given face and the faces 
 which* by the law of symmetry of the system of crystalization, 
 are required to coexist with it, is called a form. The form 
 consisting of the face h Jc I and its coexistent faces, may be de- 
 noted by the symbol {hkl}. When, however, there is no 
 danger of mistaking the form for a zone or a face having the 
 same indices, the braces may be omitted. 
 
 Forms possessing all the faces required by the law of sym- 
 metry of the system to which they belong, are sometimes called 
 holokedral, in order to distinguish them from peculiar forms of 
 frequent occurrence, which are derived from holohedral forms by 
 suppressing half of their faces according to certain laws, and are 
 called hemihedral. The figure consisting of the faces of any 
 number of forms is called a combination of those forms. 
 
 22 
 
' UNIVERSITY OF 
 
 PORNIA. 
 
 CHAPTER II. 
 
 CUBIC SYSTEM. 
 
 29. IN the cubic system the axes make right angles with 
 one another, and the parameters are all equal. 
 
 30. The form h Jc I is contained by the faces having for 
 their symbols the different arrangements of + h, Jc, I. 
 These are: 
 
 hkl 
 
 klh 
 
 Ihk 
 
 IJch 
 
 Tchl 
 
 hi k 
 
 hkl 
 
 klh 
 
 III 
 
 III 
 
 kll 
 
 hll 
 
 hkl 
 
 klh 
 
 7 hi 
 
 IJch 
 
 Ihl 
 
 hll 
 
 III 
 
 llh 
 
 Ihk 
 
 llh 
 
 III 
 
 hlk 
 
 III 
 
 llh 
 
 III 
 
 llh 
 
 III 
 
 hll 
 
 hkl 
 
 llh 
 
 Ihk 
 
 Ikh 
 
 khl 
 
 hlk 
 
 hkl 
 
 klh 
 
 IhJt 
 
 Ikh 
 
 khl 
 
 hlk 
 
 hkl 
 
 klh 
 
 Ihl 
 
 Ikh 
 
 khl 
 
 hlk 
 
 When h, k, I are all different, the number of arrangements 
 will be forty-eight ; when any two indices are equal, it will be 
 twenty-four ; when two of the indices are equal, and the third 
 is zero, it will be twelve ; when all three indices are equal, it 
 will be eight; and when two of the indices are .zero, it will 
 be six. 
 
CUBIC SYSTEM. 21 
 
 31. The form contained either by the faces of the form h k I 
 which have an odd number of positive indices, or by the faces 
 which have an odd number of negative indices, is said to 
 be hemihedral with inclined faces. It will be denoted by the 
 symbol /chkl, where hklis the symbol of any one of its faces. 
 The symbols in the upper and lower halves of the table in (30) 
 are those of the two half forms respectively. 
 
 32. The form contained either by the faces of the form 
 h k I having their indices in the order hklhk, or by the faces 
 having their indices in the order Ikhlk, is said to be hemihe- 
 dral with parallel faces. It will be denoted by the symbol 
 TT h k I, where hklis the symbol of any one of its faces. The 
 symbols in the left and right halves of the table in (30) are those 
 of the two half forms respectively. 
 
 33. Let A, B, C be the poles 100, 010, 001 respec- 
 tively ; P the pole h k L The axes make right angles with one 
 another, therefore the sides of the triangle XYZ are quadrants, 
 its angles are right angles, and X, F, Z are poles of the arcs 
 YZ, ZX, XY. But A, B, C are poles of YZ, ZX, XY, and 
 they have no negative indices, therefore (3) A, B, C coincide with 
 X, F, Z respectively. Hence, the sides of the triangle ABC 
 are quadrants, and its angles are right angles. The quadrantal 
 triangles PAB, PEG give 
 
 (cos BPy = (sin AP) 2 (cos BAP}\ 
 (cos OP) 2 = (sin^P) 2 (cos CAP)\ 
 
 Add, observing that (cos BAP)* + (cos CAP)*= 1, and that 
 (cos AP) Z + (sin AP) 2 = 1, and we obtain 
 
 (cos Apy + (cos spy + (cos cpy = i. 
 
 The parameters are all equal, and A, B, C coincide with 
 X, F, Z, therefore (2), 
 
22 CRYSTALLOGRAPHY. 
 
 i cos AP = i cosPP = 7 cos <7P. 
 h k I 
 
 Hence (cos APf = & . + 
 
 
 34. Let P, be the poles h Jc I, p q r respectively. 
 
 cos PQ = cos AP cosAQ+ sin AP sin A Q cos PA Q, 
 cos P4 Q = cos .&4P cos BA Q + sin J5^P sin P^4 Q, 
 sin ^tP cos BAP= cos PP, sin A Q cos JL4 (2 = cos 
 sin ^1P sin BAP = cos OP, sin A Q sin P^4 Q = cos 
 
 Hence 
 cos P$ = cos u4P cos A Q + cos BP cos PQ + cos CP cos (7$. 
 
 Therefore 
 
 hp + kq 
 
 )S ^ - 
 
CUBIC SYSTEM. 23 
 
 35. The quadrantal triangles BPC, CPA, APB give 
 cos AP = sin BP cos ABP = sin CP cos A OP, 
 cos BP = sin OP cos BCP = sin ^P cos BAP, 
 cos CP = sin AP cos O4P = sin BP cos 
 
 But i cos AP = y cos BP = -, cos OP. Hence 
 
 /I K I 
 
 = tan -4 OP- 
 
 36. It appears from the expressions in (33), that if the 
 symbols of two poles of the form h k I differ only in the signs 
 of h, they will be equidistant from the pole 010, and also equi- 
 distant from the pole 001. Therefore the arc joining the two 
 poles will be bisected at right angles by the zone-circle passing 
 through the poles 010, 001. Hence, the poles of the form 
 h k I are symmetrically situated with respect to the zone-circle 
 passing through the poles 010, 001, and the two diametrically 
 opposite poles. In like manner the poles of the form h k I are 
 symmetrically situated with respect to any one of the three zone- 
 circles containing four poles of the form 100. It appears from 
 (34) that, if the symbols of any two poles of the form h k I differ 
 only in the arrangement of the second and third indices, the 
 poles will be equidistant from 111, and also from 111. There- 
 fore the arc joining the two poles will be bisected at right angles 
 by the zone-circle passing through the poles 111, 111, and 
 the two opposite poles. In like manner the poles of the form 
 h k I are symmetrically situated with respect to any one of the 
 six zone-circles containing four poles of the form 111. 
 
 The poles of a hemihedral form with inclined faces are sym- 
 metrically situated with respect to each of the six zone-circles 
 containing the poles of th*e form 111. 
 
 The poles of a hemihedral form with parallel faces are sym- 
 metrically situated with respect to each of the three zone-circles 
 containing the poles of the form 1 0.. 
 
24 CKYSTALLOGKAFIIY. 
 
 37. If h be supposed the greatest, and I the least of three 
 unequal indices h, k, I, the first of the annexed figures will 
 represent the distribution of the poles of the form h k I on one- 
 
 011 .]<i on. .101 
 
 /- : i/.. 7iA/ 
 
 111 111 
 
 W*. Jdh OAA* *AOfc 
 
 fcAfc" "AAfc 
 
 MJ. .AM *** 
 
 c-o IMI oio . . 100 
 
 . HO **0 !i JikO 
 
 eighth of the sphere of projection. The second figure exhibits 
 the poles of the forms obtained by making one of the indices 
 zero, or by making two of them equal. Both figures show the 
 poles of the forms 1 0, 1 1 1, and 110. 
 
 If the surface of the sphere be divided into eight triangles 
 by the three zone-circles passing through the poles of the form 
 1 0, the poles of a hemihedral form with inclined faces will be 
 found in four alternate triangles. 
 
 If the surface of the sphere be divided into twenty-four 
 triangles by the six zone-circles passing through the poles of the 
 form 111, the poles of a hemihedral form with parallel faces 
 will be found in twelve alternate triangles. 
 
 38. The two hemihedral forms either with inclined or with 
 parallel faces, derived from the same holohedral form, differ 
 only in position. For, by turning the sphere of projection 
 through a right angle, round a diameter joining any two oppo- 
 site poles of the form 100, the poles of one of the two hemilie- 
 dral forms derived from the same holohedral form, will change 
 places with those of the other. But a combination of any two 
 hemihedral forms derived from the forms h k I, pqr, when their 
 poles fall in the same triangles formed by the system of zone- 
 
CUBIC SYSTEM. 
 
 25 
 
 circles passing through the poles of the form 100, or of the 
 form 111, is essentially different from a combination of the 
 hemihedral forms when their poles fall in different triangles. 
 
 39. The form 100 has six faces. Let 
 F be the arc joining any two adjacent poles. 
 Then cosJ^=0, therefore F= 90. Hence 
 the faces of the form 100 are respectively 
 parallel to those of a cube. 
 
 40. The form 111 has eight faces. 
 Denoting by D. the arc joining any 
 two adjacent poles, we have cos D = J. 
 Therefore D = 70 31'-7. Hence the faces 
 of the form 111 are parallel to those of 
 a regular octahedron. 
 
 The cosine of the arc joining any 
 pole of the form 111, and each of the adjacent poles of the form 
 1 is \/3. The corresponding arc is 5444''l. 
 
 41. Each of the forms 1 1 1, /till 
 has four faces. Let T be the arc joining 
 any two adjacent poles. Then cos jT= J, 
 therefore T=10928'-3. Hence each of the 
 hemihedral forms is a regular tetrahedron. 
 
 42. The form Oil has twelve faces. 
 two adjacent poles being denoted by G, 
 we have cos Q- -J- . Therefore G = 60. 
 The arc joining the poles of any two 
 alternate faces, meeting at their acute 
 angles, being denoted by D, cos D 0. 
 Therefore D = 90. 
 
 The arcs joining any pole of the form 
 
 The arc joining any 
 
26 
 
 CRYSTALLOGRAPHY. 
 
 Oil, and the two adjacent poles, the two opposite poles and the 
 two remaining poles of the form 100, have for their cosines 
 -|V 2 , |V 2 > 0> respectively. The corresponding angles are 
 45, 135, 90. The arcs joining any pole of the form Oil, and 
 the two adjacent poles, the two opposite poles and the four re- 
 maining poles of the form 111, have for their cosines j\/6, 
 -JV6, 0. The corresponding angles are 3515''85, 14444'-15, 
 90. 
 
 43. The form h k has twelve faces. Let 
 the arc joining any two adjacent poles be F 
 or 6r, according as they differ only in the 
 order of h, k, or in the order of k, 0. Then, 
 h being greater than k, 
 
 2hk * h z 
 
 cosF-. 
 
 44. Each of the forms TrhkQ, TT khis 
 contained by the alternate faces of the form 
 hkQ. Denoting by D the arc joining any 
 two adjacent poles differing only in the signs 
 of kj and by U the arc joining any two adja- 
 cent poles in the symbols of which the indices 
 occupy different places, we have 
 
 hk 
 
 cosZ> 
 
 cos U= 
 
 45. The form hkk has twenty-four 
 faces. Denoting the arc joining any two 
 adjacent poles by D or F, according as the 
 order of their indices is the same or differ- 
 ent, li being greater than &, we have 
 
 cos D T0 . ~, , 
 
CUBIC SYSTEM. 
 
 27 
 
 46. Each of the forms ichkk, K hk k 
 is contained by the alternate triads of faces 
 which meet in the edges F of the form 
 h k k. Let T be the arc joining any two 
 adjacent poles differing only in the signs of 
 k. Then 
 
 cos T = T ., . , x .o . 
 
 47. The form h h k has twenty-four 
 faces. Denoting the arc joining any two 
 adjacent poles by D or Gf, according as 
 the order of the indices is the same or 
 different, h being greater than /, we have 
 
 cosZ) 
 
 cos G = 
 
 48. Each of the forms /ch hk, K h h k is 
 contained by the alternate triads of faces 
 which meet in the edges G of the form h h k. 
 Denoting by T the arc joining any two adja- 
 cent poles differing in the order of the indices, 
 and in the signs of two of them, we have 
 
 cosT = 
 
 49. The form h k I has forty-eight 
 faces. Denoting by D, F, G the arcs join- 
 ing adjacent poles differing only in the 
 signs of Z, in the order of h, k, and in the 
 order of k, I respectively, h being greater, 
 and I less than k, we have 
 
 cosD = 
 
 COS j 
 
 cos Gf = 
 
28 
 
 CRYSTALLOGRAPHY. 
 
 50. Each of the forms /chkl, K hk I 
 is contained by the alternate groups of six 
 faces meeting in the edges F, G of the form 
 h k L Let T be the arc joining any two ad- 
 jacent poles differing only in the order and 
 signs of k, L Then 
 
 cosT^=~ 
 
 51. Each of the forms Trhkl, vrlkh is 
 contained by the alternate pairs of faces 
 meeting in the edges D of the form h k L 
 Denoting by W 9 U the arcs joining any two 
 adjacent poles differing only in the signs of 
 k, and in the places occupied by the several 
 indices, respectively, we have 
 
 U + Ih + hk 
 
 cos IF = 
 
 cos U = 
 
 52. The cleavages are usually parallel to the faces of one 
 or more of the forms 1 0, 1 1 1, 1 1. 
 
 53. If we have given the arc joining any two poles, not 
 opposite to one another, of one of the forms h k 0, h k k, hhk, 
 the expression for its cosine, in terms of the indices of the poles, 
 will supply an equation from which the ratio of the indices may 
 be deduced. 
 
 54. If we have given the arcs joining any pole of the form 
 h k I, and each of two other poles of the same form, no two of 
 the poles being opposite to one another, the expressions for their 
 cosines, in terms of the indices of the poles, will supply two 
 equations from which the ratios of the indices may be found. 
 
CHAPTER III. 
 
 PYRAMIDAL SYSTEM. 
 
 55. IN the pyramidal system the axes make right angles 
 with one another, and the parameters a, b are equal. 
 
 56. The form h k I consists of the faces which have for 
 their symbols the different arrangements of h, k, I, in 
 which I holds the last place. These are : 
 
 hkl 
 
 hkl 
 
 khl 
 
 khl 
 
 hkl 
 
 hkl 
 
 khl 
 
 khl 
 
 khl 
 
 khl 
 
 hkl 
 
 Tiki 
 
 khl 
 
 khl 
 
 hkl 
 
 hkl 
 
 When h and k are different, and I is finite, the number of 
 faces will be sixteen ; when one of the indices is zero, or when 
 h = k, the number will be eight ; when I is zero, and h = A?, or 
 one of the indices h, k is zero, the number of faces will be 
 four ; and when h and k are zero it will be two. 
 
 57. The form contained either by the faces of the form 
 h k I which have an odd number of positive indices, or by the 
 faces which have an odd number of negative indices, is said to 
 be hemihedral with inclined faces, and will be denoted by the 
 
30 CRYSTALLOGRAPHY. 
 
 symbol tchkl where h k I is the symbol of any one of its faces. 
 The left and right halves of the table contain the symbols of the 
 two half forms respectively. 
 
 58. A second hemihedral form with inclined faces, contained 
 by the faces of the form h k I in which the order of h, k changes 
 with the sign of ?, will be denoted by the symbol \hkl, where 
 h k I is the symbol of any one of its faces. The first and 
 fourth columns of the table contain the symbols of the faces 
 of one half form, the second and third columns those of the other 
 half form. 
 
 59. The form consisting of the faces of the form h Tel in 
 which the order of h, k is the same or different according as 
 h, k have the same or different signs, is said to be hemihedral 
 with parallel faces, and will be denoted by the symbol Trhkl, 
 where h k I is the symbol of any one of its faces. The first 
 and third columns of the table contain the symbols of one half 
 form, the second and fourth those of the other half form. 
 
 60. The form contained by the faces of the form h k I, in 
 which the order of the indices h, k is the same or different 
 according as an odd number of the indices are positive or nega- 
 tive, is said to be hemihedral with asymmetric faces, and will be 
 denoted by the symbol ahJcl, where h k I is the symbol of any 
 one of its faces. The upper and lower halves of the table con- 
 tain the symbols of the two half forms respectively. 
 
 61. Let , a, c be the parameters ; A, B, C the poles 100, 
 1 0, 1 respectively ; P the pole h k I. The axes make right 
 angles with one another, therefore the sides of the triangle XYZ 
 are quadrants, its angles are right angles, and X, Y, Z are the 
 poles of YZ, ZX, XY. But A, B, C are poles of YZ, ZX, XY, 
 and they have no negative indices, therefore (3) A, B, C coincide 
 with X, F, Z respectively. Hence, the sides of the triangle 
 ABC are quadrants, and its angles are right angles. The quad- 
 rantal triangles PBC, PGA, PAB give 
 
PYKAMIDAL SYSTEM. 31 
 
 cos AP = sin BP cos ABP = sin CP cos A OP, 
 
 cos BP = sin CP cos BCP= sin JP cos .#^P, : 
 
 cos OP= sin APcos CAP = sin PP cos CBP. 
 
 cot 4P = tan BCP cos P^P = tan CBP cos O^P, 
 cot BP = tan 0^4P cos CBP = tan ^4 OP cos ABP, 
 cot OP = tan ABP cos .4 OP = tan BAP cos POP. 
 
 Also, since A, B, C coincide with X, Y, Z, 
 
 ? COS AP = y COSPP = y COS OP. 
 
 A A; I 
 
 Hence, substituting in the preceding equa- 
 tions the values of cos^P, cosPP, cos OP 
 given above, we obtain 
 
 T -, 
 
 k c ' 
 
 T -, 
 h c ' 
 
 62. Let E be the arc joining the poles 001, 101. Then 
 E measures the angle it subtends at B. Therefore the second of 
 the preceding equations gives tan E = c : a. Hence 
 
 Ilk 
 rcotj, tan^lj5P=^cot^, tan^OP^. 
 
 cot AP = | cos BAP = * tan E cos O^tP, 
 
 cot BP = tan E cos CBP = cos 
 
 cot OP == cotJ^cos JOP= cotEcosBCP 
 
32 CRYSTALLOGRAPHY. 
 
 63. Since tan E = c : a, E may be taken for the element of 
 a crystal belonging to the pyramidal system. 
 
 64. The poles of the form 110 bisect the arcs joining any 
 two adjacent poles of the form 100. For the poles of the forms 
 100, 1 1 are all in one zone-circle ; the arc joining the poles 
 1 0, 1 is a quadrant ; and (62) the arc joining the pole 
 100, and any pole of the form 110, having for its cotangent 
 either 1 or 1, is an odd multiple of 45. 
 
 65. It appears from the expressions in (62) that the arcs 
 joining the poles of the form h k I, and the nearest of the two 
 poles of the form 1, are all equal; and that the angles sub- 
 tended at either pole of the form 1 by the arcs joining any 
 pole of the form h k ?, and the nearest pole of the form 100, are 
 all equal. Hence, the poles of the form k h I are symmetrically 
 situated with respect to each of the five zone-circles containing 
 poles of any two of the three forms 001, 100, 110. 
 
 The poles of the form ichkl are symmetrically situated with 
 respect to each of the two zone-circles drawn through the poles 
 of the form 001, and those of the form 110. 
 
 The poles of the form \hkl are symmetrically situated 
 with respect to each of the two zone-circles through the poles of 
 the form 001, and those of the form 100. 
 
 The poles of the form vrhkl are symmetrically situated with 
 respect to the zone-circle containing the poles of the form 100. 
 
 66. If h be supposed greater than k, the annexed figure 
 will represent the arrangement of the poles of the forms h k l y 
 hhl, hkO, hQ I, 110, 100, 1 on the surface of the sphere 
 of projection. 
 
 If the surface of the sphere be divided into eight triangles 
 by zone-circles passing through the poles of the forms 001, 
 100, the poles of the form /chkl will be found in four alter- 
 nate triangles. 
 
PYRAMIDAL SYSTEM. 33 
 
 If the surface of the sphere be divided into eight trian- 
 gles by zone-circles passing through the poles of the forms 
 001, 1 1 0, the poles of the form X h k I will be found in four 
 alternate triangles. 
 
 If the surface of the sphere be divided into eight lunes by 
 zone-circles passing through the poles of the form 001, and 
 those of the forms 100, 110, the poles of the form Trhkl will 
 be found in four alternate lunes. 
 
 The poles of the form ahkl are eight alternate poles of the 
 form h Jc I. 
 
 *7,0 
 
 LIBRAE' 
 
 UNIVERSITY 
 
 CALIFOMNL 
 
 67. Any two hemihedral forms with inclined or with paral- 
 lel faces, derived from the same holohedral form, differ only in 
 position. For, by making the sphere of projection revolve 
 through a right angle round a diameter joining the poles of the 
 form 001, the poles of /chkl and \hkl will change places 
 with those of /chkl and Xkhl respectively ; and by making 
 the sphere revolve through two right angles round a diameter 
 joining any two opposite poles of the form 100, or of the 
 form 110, the poles of Trhkl will change places with those 
 of TT k h I. The two forms ahkl, akh I are essentially 
 different. 
 
 M. C. 3 
 
34 CRYSTALLOGRAPHY. 
 
 68. The form 001 has the two paral- 
 lel faces 1, 1. 
 
 69. The form 1 has four faces. Let 
 jFbe the arc joining any two adjacent poles. 
 Then F=10 0,0 1 0, and cot F= 0. There- 
 fore F= 90. 
 
 70. The form 110 has four faces. 
 Let K be the arc joining any two ad- 
 jacent poles. Then 1^=100,110, 
 and cot K= I. Therefore K= 90. 
 
 In a combination of the forms 100 
 and 110, all the faces are in one zone, 
 and any face of one form makes angles 
 of 45 with the adjacent faces of the 
 other form. The arc joining a pole of 
 
 the form 001, and any pole of either of the forms 100, 110, 
 is a quadrant. Therefore, in combinations of the form 001 
 with the forms 100, 110, the faces of the form 001 make 
 right angles with those of the forms 100, 110. 
 
 71. The form h k has eight faces in 
 one zone. Let K be the arc joining any 
 two adjacent poles differing in the signs 
 of k ; F the arc joining any two adjacent 
 poles differing in the order of the indices 
 h, Jc. Then %K= I00,hk 0. Whence 
 
 ~, F=W-K. 
 
 The arc joining a pole of the form 001, and any pole of the 
 form h k 0, is a quadrant. Therefore, in a combination of the 
 forms 001, h k 0, the faces of one form make right angles with 
 those of the other form. 
 
PYEAMIDAL SYSTEM. 
 
 35 
 
 72. Each of the forms TrhkO, TrkhQ, consists of the alter- 
 nate faces of the form h k 0. Any two adjacent faces make 
 right angles with one another. 
 
 73. The form h I has eight faces. 
 Let L be the arc joining any two adja- 
 cent poles differing in the signs of I ; F the 
 arc joining any two adjacent poles in the 
 symbols of which I has the same sign. 
 Then 90 -'. = 01,&OZ, and F sub- 
 tends an angle of 90 at the pole 001. 
 Hence 
 
 tan L = -r 
 
 , cos F= (sin 
 
 74. Each of the forms \hOl, X h Z, is contained by the 
 alternate faces of the form h L Let U be the arc joining any 
 two poles in which I has the same sign ; F the arc joining any 
 two poles in which I has different signs. Then 
 
 U= 180 - L, V= 180 - F. 
 
 75. The form h h I has eigjit faces. 
 Let K be the arc joining any two adja- 
 cent poles in which I has the same sign, 
 L the arc joining any two adjacent poles 
 in which I has different signs. Then 
 90 \L = l,h h I, and K subtends an 
 angle of 90 at the pole 001. Hence 
 
 tan \L = T cot E cos 45, cos K = (sin l L) 2 . 
 
 76. Each of the forms /chhl, ichhl, consists of the alter- 
 nate faces of the form h h L Let W be the arc joining any two 
 poles in which I has the same sign; T the arc joining any two 
 poles in which I has different signs. Then 
 TF= 180 - L, T= 180 - K. 
 
 32 
 
36 
 
 CRYSTALLOGRAPHY. 
 
 77. The form h k I has sixteen faces. 
 LelfTT, JTTbe the arcs joining any two ad- 
 jacent poles differing in the signs of Jc, I 
 respectively ; F the arc joining any two 
 adjacent poles differing in the order of the 
 indices h, ~k ; and let < be the angle which 
 the arc joining the poles 1 0, h Jc I, sub- 
 tends at the pole 001. Then 
 
 90 -K=OlQ,hkl. Hence 
 I 
 
 sn = cos 
 
 sn 
 
 78. Each of the forms \hkl, \7chl, consists of the alter- 
 nate pairs of faces of the form h k I which meet in the edges 
 K. Let H be the arc joining any two poles differing only in 
 the signs of h ; F the arc joining any two poles differing only 
 in the order of h, k, and in the signs of I. Then 
 
 90- %H= 1 Q0,hkl, \V= 110,hkl. 
 Hence 
 
 cosily cos<, cosF = cos \L cos (\TT <). 
 
 79. Each of the forms K h k ?, K h k I, consists of the alter- 
 nate pairs of faces of the form h k I which meet in the edges F. 
 Let T be the arc joining any two poles differing only in the 
 signs of k and l\ Gr the arc joining any two poles differing 
 only in the signs and order of h and k. Then 
 
 80. Each of the forms Trhkl,7rkhl, consists of the alter- 
 nate pairs of faces of the form h k I which meet in the edges L. 
 Let M be the arc joining any two alternate poles of the form 
 h k I, equidistant from the pole 001. The angle subtended by 
 M, at the pole 001, will be 90. Hence cos M = (s 
 
 81. Each of the forms ahkl, akhl, consists of the alter- 
 nate faces of the form h k I The arcs joining the adjacent poles 
 
PYRAMIDAL SYSTEM. 37 
 
 in the symbols of which I has the same sign, the signs of k are 
 different, and the order of h, k different, are M, T, V respectively. 
 
 82. The principal cleavages are parallel to the faces of one 
 or more of the forms 1, 1 0, 1 1 0, h I, h h L 
 
 83. Let C be the pole 001; P, Q any two adjacent poles 
 of either of the forms h h I, p r, equidistant from C', and let 
 the arc PQ contain S a pole of the other form. Then CS will 
 bisect the right angle PCQ, and the angle CSP will be a right 
 angle. Whence, tan CS = cos 45 tan CP. 
 
 84. Let A, B, C be the poles 1 0, 1 0, 1 respec- 
 tively ; P the pole h k 1] Q the pole p q r. Then (62), 
 
 cot AP = | cos BAP = * tan E cos CAP, 
 
 Let Q be in the zone-circle AP. Then BAQ = BAP, and 
 CAQ=CAP. Therefore 
 
 h tan AP _ k _ I 
 p tan A Q q r' 
 
 In like manner, when Q is in the zone-circle JSP, 
 k tan BP I h 
 
 q t8m r p 
 
 Also, when Q is in the zone-circle CP, 
 
 l_ tan CP h = k 
 r tan CQ~ p~~ q* 
 
 85. Let C be the pole 1 ; P, Q the poles h k I, p q r 
 respectively. Then (62), 
 
 (tan CQf = -r- (tan E}*. 
 
38 CRYSTALLOGRAPHY. 
 
 Therefore j~^ (tan CP) 2 = -^. i (tan 
 
 86. Let A, B, C be the poles 100, 010, 001 respec- 
 tively; P, Q any two poles the symbols of which are given. 
 Then, knowing E, and the symbols of P, Q, we can find CP, 
 CQ, AGP, ACQ by (62). Hence, knowing OP, CQ and 
 PCQ, the arc PQ can be found. 
 
 Or, having found the angles which CP, CQ subtend at one 
 of the poles A, B, and the arcs joining this pole, and P, Q re- 
 spectively, we have two sides and the included angle, from 
 which the third side PQ may be found. 
 
 87. If the arc joining any two poles of the form hkQ, not 
 being either a quadrant or a semicircle, or the arc joining any 
 two poles not opposite to one another, of either of the forms 
 h I, h h ?, be given ; the given arc, or its supplement, will be 
 one of the arcs F, K, L (71), (73), (75). Hence an equation is 
 obtained from which, knowing E, the ratio of the indices of the 
 form may be found. 
 
 88. If we have given the arcs joining any pole of the form 
 h k I, and each of two other poles of the same form, no two of 
 the three poles being opposite to one another, the given arcs, or 
 their supplements, will be two of the arcs H, K, L, F, V, M (77), 
 (78), (80). Therefore two equations are obtained from which, 
 knowing j&, the ratios of the indices of the form may be found. 
 
 89. When the last index in the symbol of a form is finite, 
 the arc joining any two poles not opposite to one another, or its 
 supplement, is one of the arcs H, K, L, F, F, M. Therefore, if 
 this arc and the symbol of the form be given, tanE may be 
 found from the equations in (77), (78) or (80). 
 
 80. Let A, B, C be the poles 100, 010, 001 respec- 
 tively ; jfr, #any two poles in a zone-circle containing C; Pthe 
 
PYRAMIDAL SYSTEM. 39 
 
 intersection of RS and AB, PR being less than PS', p q r the 
 symbol of any zone-circle, except RS, passing through R- 
 uvw the symbol of S. Then CP is a quadrant, and the sym- 
 bol of AB, a zone-circle passing through P, is 001, therefore 
 (20), 
 
 sin (2PR + RS) = (2 - 1) sin RS, 
 
 YW 
 where ^ = 
 
 pu + QV + YW 
 
 Having found CR or OS by means of this equation, tan E 
 is given by (62). 
 
 91. Let A, B, C be the poles 100, 010, 001 respec- 
 tively ; R, S any two poles not oppo- 
 site to one another. Let RS meet 
 AB in P, PR being less than PS, 
 and let s tQ be the symbol of P. 
 Let M be the pole Us ; Q the in- 
 tersection of RS and CM. Then 
 tan A CM= - cot ^L CP, therefore PM 
 is a quadrant. But CP is a quadrant, 
 
 therefore PQ is a quadrant. The symbol of AB, a zone-circle 
 passing through P, is 00 1. Let pqr be the symbol of any 
 zone-circle passing through R, except RS. The numerical values 
 of the indices of Q can be readily found from those of R and 
 S, and the relation between the indices of P and M. Let h k I 
 be the symbol of Q, uvw that of S. The arc PQ is a quad- 
 rant, therefore (20), 
 
 sin (2PZ2 + RS) = (2t - 1) sin RS, 
 
 where 
 
 w ph + 
 = 
 
 - , - 
 t pw + qv + YW 
 
 Having found PR or PS by means of this equation, and 
 tanPCR or tanPO/S; we have 
 
 cos PR = cos P OB sin OB, cos P$ = cos P OS sin 08. 
 Hence, knowing OB or OS, tanjSJis given by (62). 
 
CHAPTER IV. 
 
 RHOMBOHEDRAL SYSTEM. 
 
 92. IN the rhombohedral system the axes make equal 
 angles with one another, and the parameters are all equal. 
 
 93. The form h k I consists of the faces which have for 
 their symbols the different arrangements of + h, + k, +1, to- 
 gether with those of h, k, I. These are 
 
 hkl 
 
 Ikh 
 
 hkl 
 
 III 
 
 klh 
 
 khl 
 
 klh 
 
 khl 
 
 Ihk 
 
 hlJc 
 
 Ilk 
 
 Ilk 
 
 When h, k, I are all different, the number of faces will be 
 twelve. When two of the indices are equal, or when they are 
 1, 0, 1, it will be six. When all three indices are equal, it 
 will be two. 
 
 94. The form consisting either of the faces having for 
 their symbols the different arrangements of + h, + k, + Z, or 
 of the faces having for their symbols the different arrangements 
 of h, k, I is said to be hemihedral with inclined faces. 
 It will be denoted by the symbol /chkl, where h k Z is the sym- 
 bol of any one of its faces. The left and right halves of the 
 
EHOMBOHEDRAL SYSTEM. 41 
 
 table in (93) contain the symbols of the faces of the two half 
 forms respectively. 
 
 95. The form consisting either of the faces of the form 
 h k I which have their indices in the order hkl hk, or of the 
 faces which have their indices in the order I k h Ik, is said to be 
 hemihedral with parallel faces. It will be denoted by the sym- 
 bol TT h k I, where h Jc I is the symbol of any one of its faces. 
 The symbols of the faces of one half-form are contained in 
 the first and third columns of the table in (93), those of the 
 other in the second and fourth columns. 
 
 96. The form consisting either of the faces of the form 
 h k I having for their symbols the arrangements of + h, + k, 
 + I which stand in the order hklhk, and those of - h, k, I 
 which stand in the order Ikhlk, or of the faces having for their 
 symbols the arrangements of + h, + k, +1 which stand in the 
 order Ikhlk, and those of h, k, I which stand in the 
 order hklhk, is said to be hemihedral with asymmetric faces, 
 and will be denoted by the symbol ahkl, where h k I is the 
 symbol of any one of its faces. The first and fourth columns 
 of the table in (93) contain the symbols of the faces of one 
 half-form; the second and third columns those of the other 
 half-form. 
 
 97. Let be the pole 111; P the pole hkl Since the 
 parameters are equal, and is the pole 1 1 1, we shall have 
 
 cos XO = cos YO = cos ZO, and XO = YO = ZO. 
 The axes make equal angles with one another, therefore 
 
 Hence, YOZ, ZOX, XOYsue each 120. Therefore 
 cos YOP= cos 120 cos XOP + sin 120 sin XOP, 
 cos ZOP = cos 120 cos XOP - sin 120 sin XOP. 
 
42 CRYSTALLOGRAPHY. 
 
 Hence, observing that 2 sin 120 = V3, and 2 cos 120 = 1, 
 cos FOP - cos ^OP = sin XOP^S, 
 cos XOP 4- cos FOP 4- cos ^OP = 0. 
 
 cos XP = cos XO cos OP 4- sin XO sin OP cos XOP, 
 cos FP = cos FO cos OP 4- sin FO sin OP cos FOP, 
 cos ZP = cos ZO cos OP 4- sin ZO sin OP cos ^OP. 
 
 Hence sin XO sin OP sin XOPV3 = cos FP - cos ZP, 
 3 sin XO sin OP cos XOP = 2 cos XP - cos FP - cos ZP, 
 3 cos XO cos OP = cos XP 4- cos FP 4- cos ZP. 
 
 But i cos XP= y cos FP = \ cos ZP. 
 
 h k I 
 
 Hence tan XOP = - 
 
 7 
 2/t k I 
 
 tan XO tan OP cos XOP = 
 
 Similarly tan FOP = 
 
 tan FO tan OP cos FOP = 
 
 And t 
 
 tan ^0 tan OP cos ^OP = 
 
RHOMBOHEDRAL SYSTEM. 
 
 98. Let A, B, G be the poles 100, 
 010, 001 respectively. Then (97) 
 
 tanXOA = 0, tanFOS = 0, tan Z0 (7=0, 
 tan XO tan OA = 2, tan YO tan OB = 2, 
 tan ZO tan 0(7= 2. Hence A, B, C are 
 in the great circles OX, OY, OZ, and 
 OA=OB=OC. Let OA = D. The 
 expressions in (97) become 
 
 tan COP = 
 
 2 tan OP cos <70P 
 
 tan 
 
 99. The great circle OZ' divides the triangle XOY into 
 two right-angled triangles, and bisects the arc XY. In one of 
 these triangles, OX is the side opposite to the right angle, one 
 side is ^XY, and the opposite angle is 60. Therefore 
 
 sin J XY = sin OX sin 60. 
 
 But tan D = 2 cot OX. Therefore the arc D depends upon XY, 
 and may, consequently, be taken for the element of a crystal 
 belonging to the rhombohedral system. 
 
 100. Let be a pole of the form 111, u4_any pole of the 
 form 100, M, N any poles of the forms 211, 101 respec- 
 tively. The expressions in (98) show that OM, ON are quad- 
 rants, that A OM is a multiple of 60, and that A ON is an 
 odd multiple of 30. Hence, the poles of the form 211 lie in 
 one zone-circle, and divide it into six equal arcs ; and the poles 
 
44 CRYST ALLOGRAPH Y. 
 
 of the form 101 bisect the arcs joining the adjacent poles of the 
 form 211. The poles of the form 211 are in the zone-circles 
 containing the poles of the form 111, and those of the form 
 100. Each pair of opposite poles of the form 1 1 is in a 
 zone-circle containing four poles of the form 100. 
 
 101. Let 0, P, Q be the poles 111, hkl, pqr respec- 
 tively ; and let the indices of P, Q be connected by the equa- 
 tions 
 
 l. Then (98), 
 
 and 2tan 
 
 q+r 
 
 7, i 7 07, 
 
 7 7 
 h + k + l 
 
 tan D = - 2 tan OP cos A OP. 
 
 Hence, Q = OP, &u&AOQ = 180 + A OP. Therefore the 
 arc PQ is bisected in 0. The forms h k I, p q r are said to be 
 inverse with respect to each other. A combination of these two 
 forms is called dirhombohedral. It may be denoted by 8 h k I, 
 where h k I is the symbol of any face of either of the two forms. 
 
 102. It appears from the expression for tan OP, that the 
 arcs joining the poles of the form hk I, and the nearest pole of 
 the form 111, are all equal. By interchanging the indices 
 h, k, I, and changing their signs, in the expressions for tan A OP, 
 tan BOP, tan COP, it will be seen that the angles subtended at 
 1 1 1 by the arcs joining any pole of the form h k I, and the 
 nearest pole of the form 100, are all equal. Hence, the poles 
 of the form h k I are symmetrically situated with respect to each 
 of the three zone-circles containing the poles of the form 111, 
 and those of the form 211. The poles of a hemihedral form 
 with inclined faces are symmetrically situated with respect to 
 the same zone-circles. 
 
EHOMBOHEDKAL SYSTEM. 
 
 45 
 
 The poles of a dirhombohedral combination of any two 
 holohedral forms are symmetrically situated with respect to 
 each of seven zone-circles, six of which contain the poles of the 
 form 111, and those of the forms 211 and 101, and the 
 seventh contains the poles of the form 101. The poles of a 
 dirhombohedral combination of any two hemihedral forms with 
 inclined faces, are symmetrically situated with respect to each 
 of the six zone-circles containing the poles of the form 111, 
 and those of the forms 211, 101. The poles of a dirhombo- 
 hedral combination of any two hemihedral forms with parallel 
 faces, are symmetrically situated with respect to the zone-circle 
 containing the poles of the form 101. 
 
 103. The annexed figure represents the arrangement of the 
 poles of the form h k I on the surface of the sphere of projection, 
 h being the greatest, and I algebraically the least, of three un- 
 equal indices. 
 
 If the surface of the sphere be divided into two parts by 
 the zone-circle containing the poles of the form 101, the poles 
 in either hemisphere will be those of a hemihedral form with 
 inclined faces. When the algebraic sum of the indices of a form 
 is zero, the poles of the form h Jc I lie in the zone-circle contain- 
 ing the poles of the form 1 T. The poles in three alternate 
 arcs joining the poles of the form 101, will be those of a hemi- 
 hedral form with inclined faces. 
 
 UNIVEUSITY O 
 
 CALIFORNIA 
 
 oil 
 
46 CRYSTALLOGRAPHY. 
 
 The alternate poles of the form h k I are those of a hemihe- 
 dral form with parallel faces. 
 
 If the surface of the sphere of projection be divided into 
 six lunes by zone-circles through the poles of the form 111, 
 and those of the form 2 1 1, the poles of a hemihedral form with 
 asymmetric faces will be found in three alternate lunes. 
 
 104. The two hemihedral forms, either with . inclined or 
 with parallel faces, derived from the same holohedral form, differ 
 only in position; for, by turning the sphere of projection 
 through two right angles round a diameter joining any two 
 opposite poles of the form 101, the poles of one of the hemi- 
 hedral forms will change places with those of the other. The 
 two hemihedral forms with asymmetric faces are essentially 
 different. 
 
 105. The form 111 has the two parallel faces 111, 111. 
 A normal to these faces is sometimes called the axis of the 
 rhombohedron. It appears from (97) that the angles it makes 
 with the three crystallographic axes are all equal. 
 
 106. The forms Kill, Kill consist of the faces 111,111 
 respectively. 
 
 107. The form 2 1 f has six faces in 
 one zone. Let G be the arc joining any 
 two adjacent poles. Then (100) G = 60. 
 
 G 
 
 108. Each of the forms K 2 1 1, K 2 1 1 
 consists of three alternate faces of the form 
 2 if. 
 
 109. The form 101 has six faces in one zone. Let H be 
 the arc joining any two adjacent poles of the form 101. Then 
 (100), jBT=60. 
 
RHOMBOHEDRAL SYSTEM. 
 
 47 
 
 In a combination of the forms 211, 
 101, all the faces are in one zone the 
 symbol of which is 111, and any face of 
 one form makes angles of 30 with the ad- 
 jacent faces of the other form. In a com- 
 bination of the form 111 with the forms 
 211, 101, it appears from (98) that the 
 faces of the form 111 make right angles with those of the two 
 latter forms, 
 
 110. The form h Ic k, called a rhombohedron, has six 
 faces. Let be either pole of the form 111; A, P any two 
 adjacent poles of the forms 100, h Jc Jc re- 
 spectively; OA = D, OP=T. Let F be 
 the arc joining any two poles of the form 
 h k k, on the same side of the zone-circle 
 111; W the arc joining any two adjacent 
 poles on opposite sides of the zone-circle 
 111. The poles of the form h ~k k are in 
 the zone-circle containing the poles of the forms 111 and 100, 
 therefore (98) the arc .F subtends an angle of 120 at 0. Hence, 
 making I = k in the expression for (tan OP) 2 , we have 
 
 tanJ r 
 
 Ji k 
 r -- 7 
 ft ~\~ 2/c 
 
 tanI>, sin JF= sin 60 sin T, TF=180-F. 
 
 The position of a rhombohedron is said to be direct or in- 
 verse according as tan T is positive or negative, or, according 
 as OP, OA are measured in the same or in opposite directions 
 from 0. 
 
 In a combination of the forms 101, h Jc Jc, each face of the 
 form 1 1 is in a zone containing four faces of the form h Jc k. The 
 arcs joining any pole of the form h Jc Jc and the poles of the form 
 101, are 90- JF, 90, 90 + 1F. 
 
 111. Each of the forms ichkl, /chic I consists of three 
 faces of the form h Jc k, making equal angles with one another. 
 
48 
 
 CRYSTALLOGRAPHY. 
 
 112. The form h 7c ?, where h 4- k + I = 0, has twelve faces 
 in the zone 111. Let H be the arc joining any two adjacent 
 poles, on opposite sides of a pole of the form 211, h being 
 numerically the largest index ; W the arc 
 joining any two adjacent poles, on opposite 
 sides of a pole of the form 101. Then 
 (98), since h + k + l = Q, the arc joining 
 the pole 111, and any pole of the form 
 h k I, is a quadrant. Hence 
 
 In a combination of this form with the form 111, the faces 
 of the two forms make right angles with one another. 
 
 113. Each of the forms /chkl, /chic I, where h + k+ 1 = 0, 
 has the faces of the form h k I, which meet in alternate edges 
 H. The angles between any two adjacent faces are alternately 
 IT and 120 -If. 
 
 114. Each of the forms jrhkl, Trlkh, where h + k + I = 0, 
 consists of alternate faces of the form h k I. The angle between 
 any two adjacent faces is 60. 
 
 115. Each of the forms ah Tel, alkh, where h + k + I 0, 
 consists of the faces of the form h k I, which meet in the alter- 
 nate edges W. The angles between any two adjacent faces are 
 alternately TFand 320- W. 
 
 116. The form hkl has twelve faces. Let D, T be the 
 arcs joining any poles of the forms 100, h k I respectively, and 
 the nearest poles of the form 111; -ET, K, L 
 
 the arcs joining any two poles of the form 
 hkl, equidistant from the pole 111, in the 
 symbols of which h, k, I occupy the same 
 places ; W the arc joining any two adja- 
 cent poles unequally distant from the pole 
 111; 20, 2<, 2>|r the angles subtended at 
 the pole 1 1 1 by the arcs H, K, L. Then (98), 
 
RHOMBOHEDRAL SYSTEM. 49 
 
 In the triangles having their vertex in the pole 111, and 
 the bases H, K, L, the sides which meet in 1 1 1 are each equal 
 to T. Hence 
 
 sin \H= sin 6 sin T, sin \K sin < sin T> 
 sin \L = sin ^ sin T, IF = 180 - K. 
 
 When 2k = h + I, the angles H, L are equal, and the edges 
 IF" are parallel to the faces of the form 111. 
 
 In a combination of the forms 1 1, h k I, each face of the form 
 1 1 is in a zone containing four faces of the form h k L The 
 arcs joining any pole of the form h Jc I and the poles of the form 
 1 T are 90 + \H, 90 + \K, 90 + \L. 
 
 117. Each of the forms ichkl, fchlcl, consists of six faces 
 of the form h k I, the poles of which are equidistant from a pole 
 of the form 111. 
 
 118. Each of the forms Tthkl, IT lie h, is contained by 
 alternate pairs of parallel faces of the form h k L Let V be the 
 arc joining any two alternate poles of the form hkl, equally 
 distant from a pole of the form 111. Then F will subtend 
 an angle of 120 at 1 1 1. Therefore sin JF= sin 60 sin T. 
 
 119. Each of the forms a hkl, a I Jc h, is contained by pairs 
 of faces of the form hkl, which meet in alternate edges W. 
 The arc joining any two poles equidistant from the pole 1 1 1, is 
 F, and the greater of the arcs joining two adjacent poles un- 
 equally distant from 1 1 1, is 180- H. 
 
 120. The principal cleavages are parallel to the faces of one 
 of the forms 111, 1 f , 2 1 T, hick. 
 
 M. c. 4 
 
50 CRYSTALLOGRAPHY. 
 
 121. Let P, Q, E be three poles of a rhombohedron, equi- 
 distant from 0, the pole 111; and let the zone-circle through 
 P, Q, contain S, a pole of another rhombohedron. 8 is in the 
 zone-circle OR which bisects the angle POQ and the arc PQ. 
 The angle POQ = 120, and therefore #6>P=60, 0#P=90, 
 and cos SOP= tan OS cot OP, cos 60 = i, therefore 
 
 tan OP = 2 tan 
 
 122. Let 0, A be the poles 1 1 1, 1 00 respectively; P, Q 
 any two poles the symbols of which are known. Then (98) 
 tan A OP, tan A Q can be found in terms of the indices of P 
 and Q, therefore tan POQ is known in terms of the same in- 
 dices ; also tan OP, tan Q can be expressed in terms of tan D 
 and the indices of P and Q. Therefore, knowing OP, Q, two 
 sides of a spherical triangle, and PO Q the included angle, the 
 third side PQ may be found. 
 
 123. Let F be the arc joining any two of three equidis- 
 tant poles of the form hkk. Then (110), 
 
 sinJF=sin60 smr, 
 
 tan T being positive or negative according as T and D are 
 measured from the pole 1 1 1 in the same or in opposite direc- 
 tions. Hence, when D and V are known, the ratio of h to Jc 
 may be found. 
 
 124. Let H be the arc joining any two poles of the form 
 h Jc I, where h + Jc + I = 0, in which the largest index holds the 
 same place. Then, if the arc, not being a multiple of 60, 
 which joins any two poles of the form, be given, we can find 
 H. The ratios of the indices can then be found by means of the 
 equations 
 
HHOMBOHEDKAL SYSTEM. 51 
 
 125. Suppose the arcs joining any pole of the form h k 7, 
 and each of two other poles of the same form, the three poles 
 not being in the same zone-circle, and the arc D, to be given. 
 The given arcs or their supplements will be two of the arcs 
 R, K, L, V (116), (118). By eliminating T between the equa- 
 tions in (116), (118), observing that < - 0= 60, ifr+ 6 = 60, we 
 obtain 
 
 tan0 tan l(K-L] sin<9 _ si 
 
 tan 60 tan J (K + L) ' sin 60 sin V ' 
 tan d> tan (L 4- H] sin <f> sin K 
 
 tan 60 tan ^(L- H)' sin 60 sin V ' 
 
 tan -^ ta 
 tan 60 5 = tani(JT+Zn ' sin 60 
 
 Two of the arcs H, K, L, V, and D, being known, T and 
 one of the angles 6, <, -^ may be found. The ratios of the 
 indices may then be obtained from two of the equations 
 
 a - -^ A - . n 
 
 tan 6 = ^ r^-j , 2 tan T cos = -y 7 - r tan />, 
 2h k l' h+k + l 
 
 -- 
 
 tan i/r = -^ z-^-i , 2 tan T 7 cos -^ = -= - , - r tan 
 "2l h k h + k + l 
 
 126. When the arc joining two poles of either of the forms 
 hkk,hkl, and the symbols of the poles, are known, the expressions 
 in (110) or (116) enable us to find the angle which the given 
 arc subtends at the pole 111, and T, the arc joining either pole 
 and the pole 111. Then, knowing tan Tand the indices of the 
 form, tan D may be found. 
 
 127. Let be the pole 111; R, S any two poles in a 
 zone-circle passing through ; p q r the symbol of any zone- 
 circle passing through J?, except RS; uvw the symbol of S; 
 and suppose the arc JRS to be given. Let P be the intersection 
 
 42 
 
52 CRYSTALLOGRAPHY. 
 
 of the zone-circle RS and the zone-circle 111, PS being greater 
 than'PR. Then 1 1 1 is the symbol of a zone-circle passing 
 through P; the symbol of is 1 1 1 ; and OP is a quadrant. 
 Therefore (20), 
 
 sin (2P2? + RS) = (2* - 1) sin RS, 
 
 where t = " - . 
 
 3 pw + qv + iw 
 
 Having found OR or OS by means of the preceding equa- 
 tion, tanZ> is given by (98). 
 
 128. Let 0, A be the poles 1 1 1, JL respectively; R, S 
 any two poles ; p q r the symbol of any 
 zone-circle containing R, except RS; 
 uvw the symbol of S, and suppose 
 the arc RS to be given. Let P be the 
 intersection of RS and the zone-circle 
 1 1 1, PS being greater than PR ; Q the 
 intersection of RS and- a zone-circle 
 
 having for its symbol the symbol of P, and therefore pass- 
 ing through 0, for the symbol of a pole in the zone-circle 
 1 1 1 is the symbol of a zone-circle containing the pole 111. 
 It is easily proved that t&nAOQ = - cot A OP. Hence POQ 
 is a right angle, and PQ is a quadrant. Let h k I be the sym- 
 bol of Q, the indices of Q being deduced from those of R and S. 
 Then (20), 
 
 sin (2PZ2 + RS) = (2t - 1) sin RS, 
 
 where i 
 
 h + K + l 
 
 Having found PR or PS by means of the preceding equa- 
 tion, and tan POR or tan POSj we have 
 
 cos PR = cos POR sin <9 cos PS = cos 
 
 Hence, knowing 0$ or OS, tan J9 is given by (98). 
 
CHAPTER V. 
 
 PRISMATIC SYSTEM. 
 
 129. IN the prismatic system the axes make right angles 
 with one another. 
 
 130. The form h k I consists of the faces in the symbols of 
 which each of the indices h, k, I may be either positive or nega- 
 tive, but always occupies the same place. When h, k, I are all 
 finite, the form has the eight faces 
 
 hkl hkl hkl III 
 hkl hkl hkl hkl 
 
 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. 
 
 131. The form contained by the faces of the form hkl, 
 which have an odd number of positive indices, or by the faces 
 of the form h k I, which have an odd number of negative indices, 
 is said to be hemihedral with asymmetric faces, and will be de- 
 noted by the symbol a hkl, where h Jc I is the symbol of any 
 one of its faces. The upper and lower lines of the table in (130) 
 contain the symbols of the two half forms respectively. 
 
 132. The form consisting of the faces of the form hkl, in 
 the symbols of which the sign of one of the indices remains 
 unchanged, is said to be hemihedral with inclined faces, and 
 
54 CRYSTALLOGRAPHY. 
 
 may be denoted by the symbol Khkl, the index which pre- 
 serves its sign unchanged having that sign either prefixed or 
 placed over it. 
 
 133. The form having the faces of h k I, in which two of 
 the indices change their signs together, is said to be hemihedral 
 with parallel faces, and may be denoted by the symbol Trhkl, 
 a dot being placed over the index the sign of which is inde- 
 pendent of the signs of the other two indices. 
 
 134. Let a, b, c be the parameters ; A, B, C the poles 100, 
 1 0, 1 respectively ; P the pole hkl. The axes make right 
 angles with one another, therefore the sides of the triangle 
 
 XYZ are quadrants, its angles are right angles, and X, Y, Z 
 are the poles of YZ, ZX, XY. But A, B, C are the poles of 
 YZ, ZX, XY, and they have no negative indices, therefore (3) 
 A, B, C coincide with X, Y, Z respectively. Hence, the sides 
 of the triangle ABC are quadrants, and its angles are right 
 angles. The quadrantal triangles PB C, PC A, PAB give 
 
 cos AP = sin BP cos ABP = sin CP cos A OP, 
 cos BP = sin CP cos BOP = sin AP cos BAP, 
 cos CP = sin AP cos CAP = sin BP cos GBP. 
 
 cot AP = t&nBCP cos BAP = tan GBP cos CAP, 
 cot BP = tan CAP cos CBP = tan A CP cos ABP, 
 cot CP = tan ABP cos A CP = tan BAP cos BCP. 
 
 Also,, since A, B, C coincide with X, Y, Z, 
 
 \ cos AP = j cosBP = j cos CP. 
 hkl 
 
PRISMATIC SYSTEM. 55 
 
 Hence, substituting in the preceding equations the values of 
 cos AP, cos BP, cos CP given above, and observing that 
 
 cos CAP = sin BAP, cos ABP = sin GBP, cos BCP = sin A CP, 
 we obtain 
 
 tanJ24P=4 b , tan<7PP = y C , tan^OP = ff. 
 k c la ho 
 
 135. Let D be the arc joining the poles 010, Oil; E 
 the arc joining the poles 1, 1 1 ; F the arc joining the 
 poles 100, 110. Then, since the sides of the triangle ABC 
 are quadrants, the arcs D, E, ^ measure the angles they respec- 
 tively subtend at A, B, C. Therefore 
 
 Hence 
 
 |tanI>, tan CBP = * tan E, tan A CP = | tan ^ 
 
 cot JLP = cot F cos A4P = -- tan E cos CL4P, 
 
 cot BP = y cot D cos CBP = | tan F cos ^.P, 
 
 cot CP = r cot ^cos A CP = tan D 
 
 136. Since the ratios of the parameters can be expressed 
 in terms of the tangents of any two of the arcs Z>, E, F, and 
 their product, any two of the arcs Z>, E, F may be taken for the 
 elements of the crystal. The arcs D, E, F are connected by 
 the equation 
 
 tanZ> tan ^ tan ^=1. 
 
 137. It appears from (135) that the arcs joining either pole 
 of the form 100, and the adjacent poles of the form h k I, are 
 all equal ; that the arcs joining either pole of the form 010, 
 
56 
 
 CRYSTALLOGRAPHY. 
 
 and the adjacent poles of the form h k I, are all equal ; and that 
 the arcs joining either pole of the form 001, and the adjacent 
 poles of the form h k I, are all equal. Hence, the poles of the 
 form h k I are symmetrically arranged with respect to each of 
 the zone-circles 1 0, 1 0, 1. 
 
 The poles of a hemihedral form with inclined faces are 
 symmetrically arranged with respect to two of the zone-circles 
 100, 010, 001, the first, second or third being excluded, ac- 
 cording as the first, second or third index preserves its sign 
 unchanged. 
 
 The poles of a hemihedral form with parallel faces are sym- 
 metrically situated with respect to the zone-circles 100, 010, 
 001, according as the sign of the first, second or third index 
 is independent of the signs of the other two indices. 
 
 138. The annexed figure represents the arrangement of 
 the poles of the forms hkl,Qkl, hOl, hkQ, 100, 1 0, 1 
 on the surface of the sphere of projection. 
 
 Ml 100 
 
 010 
 
 A hemihedral form with asymmetric faces has the alternate 
 poles of the form h k L 
 
 The poles of a hemihedral form with inclined faces are 
 contained in one of the two hemispheres, into which the sphere 
 of projection is divided by one of the zone-circles 100, 010, 
 001. 
 
PRISMATIC SYSTEM. 57 
 
 If the surface of the sphere be divided into four lunes by two 
 of the zone-circles 100, 010, 001, the poles of a hemihedral 
 form with parallel faces will be found in two alternate lunes. 
 
 139. The two hemihedral forms with either inclined or 
 parallel faces, derived from the same holohedral form, differ 
 only in position; for, by making the sphere revolve through 
 two right angles round the poles of one of the forms 100, 010, 
 001, the poles of one half-form will change places with those 
 of the other. The two hemihedral forms with asymmetric faces, 
 derived from the same holohedral form, are essentially different. 
 
 140. The three forms 100, 010, 001 have each two 
 parallel faces. The arc joining poles of any two of three forms 
 is a quadrant (134). Hence, in a combination of these forms 
 with one another, the faces of each form make right angles 
 with those of the other two. 
 
 Either of the preceding forms may become hemihedral. 
 
 141. The form k I has four faces. Let L be the arc 
 joining any two adjacent poles differing in the signs of L. 
 Then \L = 1 0,0 Ic I Hence (135), 
 
 tan f = - tan D, K = 180 - L. 
 
 The arc joining either pole of the form r 
 
 100, and any pole of the form k I, is a 
 quadrant. Therefore, in a combination of 
 the forms 100, k I, the faces of the 
 two forms make right angles with one 
 another. 
 
 142. The form h I has four faces. Let H be the arc 
 joining any two adjacent poles differing in the signs of h. Then 
 
 = I 9 h L Hence (135), 
 
58 
 
 CRYSTALLOGRAPHY. 
 
 The arc joining either pole of the form 
 010, and any pole of the form h Z, is a 
 quadrant. Hence, in a combination of the 
 forms 010, h Z, the faces of the two forms 
 make right angles with one another. 
 
 143. The form h k has four faces. Let K be the arc 
 joining any two adjacent poles differing in the signs of k. Then 
 PT= 1 0,h k 0. Hence (135) 
 
 17T-^ 
 
 h 
 
 The arc joining either pole of the 
 form 001, and any pole of the form h k 0, 
 is a quadrant. Hence, in a combina- 
 tion of the forms 001, h k 0, the faces 
 of the two forms make right angles with 
 one another. 
 
 1 44. When either of the forms & Z, A Z, hkO becomes 
 hemihedral with inclined faces, the hemihedral form consists 
 of two adjacent faces. 
 
 When either of them becomes hemihedral with parallel 
 faces, the hemihedral form consists of two opposite faces. 
 
 145. The form hk I has eight faces. Let 
 H, K, L be the arcs joining any two adja- 
 cent poles differing in the signs of h, k, I 
 respectively. Then 90 - %H= 1 0,A k I ; 
 90- ^K=Q10,h k Z; 90 %L = 00 l,hk L 
 Hence (135), (134), </> being the angle which 
 the arc 1 0,h k I subtends at 1, 
 
 I 
 
 tan 
 
 k 
 = r tan F, 
 
 tan %L j-cotJ cos 
 
 sn 
 
 = cos 
 
 sn 
 
 sn 
 
 = cos 
 
 cos </>. 
 
 146. A hemihedral form with asymmetric faces is a four 
 sided figure contained by the alternate faces of the form h k L 
 
PRISMATIC SYSTEM. 59 
 
 147. A hemihedral form with, inclined faces consists of 
 four faces making one of the solid angles of the form h k I. 
 
 148. A hemihedral form with parallel faces has four faces 
 of the form h k Z in one zone. 
 
 149. Let A, B, C be the poles 100, 010, 001 respec- 
 tively; P the pole hkl; Q the pole p gt r. Then as in (84), 
 when Q is in the zone-circle AP, 
 
 h tan AP _ k _ I 
 p tan A Q q r* 
 
 When Q is in the zone-circle BP, 
 
 k tanBP _l _h 
 q tauBQ~ r ~ p ' 
 
 When Q is in the zone-circle CP, 
 
 I tan CP _ h __ k 
 r tan CQ p ~~ q ' 
 
 150. Let U, V be any two of the three poles 100, 010, 
 001; P, Q any two poles the symbols of which are given. 
 Then, knowing two of the arcs D, E t F, and the symbols of P t 
 Q, we can find UP, UQ, VUP, VUQ by (135). Hence know- 
 ing UP, UQ and PUQ, the arc PQ can be found. 
 
 151. If the arc joining any two poles, not opposite to one 
 another, of one of the forms k I, h I, h k 0, be given, the ratio, 
 of the indices may be obtained from (141), (142) or (143). 
 
 152. In the form h k ?, the arcs joining any pole, and each 
 of two others, no two of the poles being opposite to one another, 
 or their supplements, will be two of the arcs H, K, L. There- 
 fore two of the arcs H, K, L being known, we can find <, 
 and thence the ratios of h, k, I, by (145). 
 
 153. The arcs Z>, E, F may be found from the expressions 
 in (141), (142) or (143), having given the arcs joining any two 
 
60 CRYSTALLOGRAPHY. 
 
 poles, not opposite to one another, of any two of the forms k I, 
 hQ I, hkO- or, from the expressions in (145), having given the 
 arcs joining any pole of the form h k I, and each of two other 
 poles, the three poles not being in the same zone-circle. 
 
 154 Let U, V be any two of the three poles 1 0, 1 0, 
 1 ; P, Q any two poles the symbols of which are known ; and 
 suppose the arcs UP, VQ to be given. Then, T being the inter- 
 section of the zone-circles UP, VQ, the symbol of T is known 
 by (5), (7) ; and the arcs UT, VT by (149). The quadrantal 
 triangle UTV gives the angles UVT, VUT, whence the arcs 
 Z>, E, F may be found by (135). 
 
 155. Let U, V be any two of the three poles 100, 010, 
 001; P, R two poles in a zone-circle containing U-, Q, S two 
 poles in a zone-circle containing V. Two of the zone-circles 
 containing every two of the poles 100, 010, 001, will have 
 U, V respectively for poles. Let PR, QS meet these zone- 
 circles in M, N respectively. Then UM, VN will be quad- 
 rants. Hence, if the arcs PR, QS, and the symbols of P, R, 
 Q, 8 be given, the arcs UP, VQ become known by (20), and 
 then the arcs D, E, .Fmay be found by (154). 
 
 156. The arcs D, E, F may also be found from the arcs 
 joining three given poles in one 
 
 zone-circle not passing through any 
 one of the poles 100, 010, 001. 
 Let P, Q, R be the given poles ; 
 A, B, C the poles 1 0, 1 0, 1 
 respectively. Let L, L' be the inter- 
 sections of PR and BC-, M, M' those 
 of PR and CA ; N, N' those of PR and AB-, and let x be the 
 less of the arcs NM, MN' -, y the less of the arcs NL, LN' ', 
 z the less of the arcs ML, LM' . Then, knowing the symbols of 
 P, Q, R, and the arcs joining P, Q, R, the symbols of L, M, N 
 may be found by (5), (7), and the arcs PL, PM, PJVby (13) or 
 
PRISMATIC SYSTEM. 61 
 
 (14). Hence the arcs joining L, M, JVare known. It is easily 
 seen that 
 
 tanBL _ tan LN' tan CM _ tanLM tan AN __ tanMN 
 tan CL ~ tan LM ' tan^lJf ~ tan MN' tanBN' 
 
 and that 
 
 tan CL = cot BL, tan ^ M = cot (7^, tan BN' = 
 
 Hence (tan BL)* = tan y cot z, 
 
 (tan CM)* = tan z cot 07, 
 
 (tan ^L^) 2 = tan x cot y. 
 
 Then, knowing tan.5L, tan CM, tan AN, and the symbols of 
 Z, Jf, ^V, the arcs D, E, Fare given by (135). 
 
CHAPTER VI. 
 
 OBLIQUE SYSTEM. 
 
 157. IN the oblique system one axis (OY) makes right 
 angles with each of the other two axes. 
 
 158. The form h k I consists of the faces in the symbols of 
 which + h, k, I occupy the same places respectively, and h 
 and I change their signs together. When k is finite, the form 
 has the four faces 
 
 hkl 
 
 hid 
 
 hkl 
 
 hkl 
 
 When Jc is zero, or when the symbol of the form is 010, 
 the number of faces will be two. 
 
 159. The hemihedral form has the faces of the form hkl in 
 the symbols of which the sign of k does not change. It may 
 be denoted by K h k I, where h k I is the symbol of either of its 
 faces. The poles of the two half forms are on opposite sides of 
 the zone-circle 010. 
 
 160. Let a, &, c be the parame- 
 ters; A, JB, C the poles 100, 010, 
 001; G the pole 111; P the pole 
 hkl. The axis OY makes right 
 angles with each of the other two 
 axes, therefore YZ, YX are quad- 
 rants. But YA, ZA, ZB, XB, XC, 
 
OBLIQUE SYSTEM. 63 
 
 YC are quadrants (3). Hence, B coincides with Y; the poles 
 (7, A are in the great circle ZX] and SO, BA are quadrants. 
 Let the zone-circle CA meet the zone-circle BP in S, and the 
 zone-circle BG in L. The symbols of S, L will, therefore, be 
 h ? and 101 respectively. 
 
 But cos XP= cos P = cos ZP, 
 
 cos XP = sin BP sin C, and cos ZP = sin #P sin ^ & There- 
 fore 
 
 ? sin (75= | cotPP = y 
 ft A; I 
 
 Hence a sin (7.L = b cot 5(7 = c 
 These equations give 
 
 smCL siu rp. f 
 
 - JT - 77^- = f . Therefore, putting 
 
 smAL sm (7$ h 
 
 ~ h sin (7Z- , ,, sin CS ~ 
 
 tan 6 = j - j-p , and consequently - j-^ = tan 0, 
 I sm AL J sin AS 
 
 we obtain tan (A8-^AG)= tan \A C tan ( JTT - 0) . 
 
 7* sin CT Z sin AL 
 
 . , 
 
 Also 
 
 k sin CS k siuAti ' 
 = sin BP cos A 8, cos CP = sin BP cos 08. 
 
 161. The arc ZZ is the supplement of A C, or of ^ + CL, 
 and the ratios of the parameters are given in terms of sin AL, 
 cotJBGr, sin CL. Hence the arcs AL, B6r, CL may be taken 
 for the elements of a crystal of the oblique system. 
 
 162. The arc joining two poles of the form h k I differing 
 only in the signs of k, is manifestly bisected at right angles by 
 the zone-circle 010. The poles of the form h k I are, therefore, 
 symmetrically situated with respect to the zone-circle 010. 
 
 The arc joining the two poles of the form nhkl is bisected 
 in a pole of the form 010. 
 
64 CRYSTALLOGRAPHY. 
 
 163. The form 1 has two parallel faces. 
 
 164. The form h I has two parallel faces in the zone 010. 
 Let S be the pole h I. Then (160) PS is a quadrant ; the arc 
 AS is given by the equations 
 
 tan 0= 
 
 I sin 
 and CS is either the difference or sum of AC and AS. 
 
 165. The form li k I has four faces. Their poles are in a zone- 
 circle passing through the poles of the form 010. Let K be 
 the arc joining any two adjacent poles differing in the signs of 
 k ; P the pole h k I. Then K= 180 - 2BP, where BP is given 
 by the equations 
 
 tan = j Sm A L T , tan (AS-^AC) = tan \AG tan (JTT - 0), 
 & sin jfLJLj 
 
 tan J5P _ A sin CL _l sin ^4 L 
 tanBa^kmTCS^lS^AS' 
 
 The arcs ^4P, CP are given by the equations 
 
 cos AP = sin BP cos ^4$, cos <7P = sin BP cos (7$. 
 
 166. The form /chic I has two faces of the form h k I, the 
 poles of which are equidistant from a pole of the form 010. 
 The arc joining the two poles is equal to 2BP. 
 
 167. Suppose the arc AS in (164) to be given. Then the 
 ratio of the indices of the form h I can be found from the 
 equation 
 
 I _ sin AL sin CS 
 h sin CL sin AS ' 
 
 168. Suppose any two of the arcs AP, BP, CP in (165) to 
 be given. Then, having found the arcs AS, BP, the ratios of 
 the indices of the form h k I are given by the equations 
 
 h sin ALsin CS k _ sin AL tan BGr 
 1 ~ sin AS tan.BP* 
 
OBLIQUE SYSTEM. 65 
 
 169. Let B, P, Q be the "poles 010, kkl, pqr respec- 
 tively. Then, when Q is in the zone-circle HP, it appears from 
 the equations between cotBP, sin ^4$, sin CS in (160) that 
 
 k tan BP _ I _ h 
 q t&uBQ r p ' 
 
 170. Let A, B, C be the poles 100, 010, 001 respec- 
 tively; P the pole h k I ; Q the pole pqr. Let BP, BQ meet 
 CA in 8, T respectively. Then BP, BQ, AS, AT may be 
 found by (160). Hence, knowing the arcs BP, BQ, and the 
 included angle PBQ, which is measured by 8 T, the difference 
 between AS and AT, the arc PQ can be found by the rules of 
 spherical trigonometry. 
 
 171. Let A, B, C be the poles 1 0, 1 0, 1 ; G, L the 
 poles 1 1 1, 1 1 ; P the pole h k I. Suppose the arcs AP, BP, 
 CP to be given. Let BP meet CA in 8. Then (160), 
 
 cos CP = sin BP cos C8, cos AP = sin BP cos A8, 
 whence CS, AS, A C are known. But 
 
 sin CL I sin C8 ^ I sin CS 
 
 ~ TT L ~= JCY Hence, putting tan = T -^ , 
 sm AL h sin AS h sin AS 
 
 tan (AL -%AC) = tan \A C tan (Jw - 0). 
 
 Having found AL, CL by means of the preceding equations, 
 BG is given by 
 
 tan BG k sin CS k BUI AS 
 tan BP ~ h sin CL ~~ I sin AL ' 
 
 Hence AL, BGr, CL, the angular elements of the crystal, are 
 known. 
 
 172. Let A, B, C be the poles 100, 010, 001; P any 
 pole not in CA U, V, W three poles in CA and suppose the 
 arcs BP, UV, VW, and the symbols of P, U t V, Wto be given. 
 Let BP meet CA in 8. The symbol of S may be found by 
 
 M. G. 5 
 
66 CRYSTALLOGRAPHY. 
 
 (5) and (7) ; and the arcs CU, AU, SUloy (13) or (14). There- 
 fore, knowing AS, BP, CS } the elements of the crystal may be 
 found by (171). 
 
 Let Q be a pole in a zone-circle BP, and suppose that the 
 arc PQ, and the symbol of Q had been given, instead of the arc 
 BP. Then, since BS is a quadrant, the arc BP may be found 
 by (20), and the elements of the crystal by the method given 
 above. 
 
 173. Let A, B, C be the poles 1 0, 1 0, 1 ; P, Q 
 any two poles of different forms, not in CA ; and suppose the 
 
 arcs BP, BQ, PQ, and the symbols of P and Q, to be given. 
 Let BP, BQ, PQ meet GA in 8, T, U respectively. The sym- 
 bols of S, T, U can be found from those of P and Q ; and the 
 arcs SU, TU can be computed from BP, BQ, PQ. The arcs AS, 
 CS are then given by (13) or (14) ; and the elements of the 
 crystal may be found by (171) from A8, BP, CS, or from AT, 
 BQ,CT. 
 
CHAPTER VII. 
 
 ANORTHIC SYSTEM, 
 
 1 74. IN the anorthic system the form h k I has the two parallel 
 faces h k I, hkl. 
 
 175. Let a, b, c be the parameters; A, B, G the poles 
 1 0, 1 0, 1 ; Q the pole 1 1 1 ; D, E, F the points in 
 which AG, EG, CG intersect BC, CA, AB, and, therefore, the 
 poles Oil, 101, 110 respectively ; P the pole hkl. 
 
 Since X, Y, Z are the poles of the great circles BC, CA, 
 AB, the arcs XP, YP, ZP are the complements of the perpen- 
 diculars from P on BC, CA, AB. Therefore 
 
 cos XP = sin CP sin BCP = sin BP sin CBP, 
 cos YP = sin AP sin CAP = sin CP sin A CP, 
 
 cos ZP = sin BP sin ABP = sin JT sin BAP. 
 
 52 
 
68 CRYSTALLOGRAPHY. 
 
 But cos XP = cos YP = cos ZP. Therefore 
 
 | sin CP sin OP = 2 sin BP sin OflP 
 sin^Psin (L4P = sin CPsin^CP 
 
 sin P sin ABP = sin AP sin 
 
 Hence | sin ^ p = - sin 
 
 b c 
 
 -s 
 a 
 
 Ts 
 b 
 
 The symbol of G is 1 1 1, therefore 
 
 -s 
 c 
 
 a 
 
 But sin # sin BVG = sin u4 sin BA G, 
 sin GD sin CDG = sin CZ4 sin O4. #, 
 sin CJ5J sin CJF^ = sin BG sin 
 sin AE sin u4## = sin AB sin 
 sin AF sin ^l^ 7 ^ = sin CA sin 
 sin BF sin ^^(^ = sin BG si 
 sin jBZ) G = sin OZ> ft sin CEG = sin 
 
ANOKTHIC SYSTEM. 69 
 
 Therefore 
 
 c _ sinAB sin CD a _ sin BO sin AE I _ sin CA sin BF 
 ~sin CA sinBD ' c ~ sin AB sin CE* a ~ smBC sm~ZF * 
 
 sin O4P A; sin AB sin 
 
 Hence 
 
 sin BAP I sin CA sin 
 sinABP I si 
 
 sin CBP~h sinAB smCE ' 
 $inBCPh sin (L4. 
 
 : sn 
 Therefore, putting 
 
 . ^ sin AB sin (7Z) 
 tan ^ = -T 77-7 
 sin CA 
 
 sin CA si 
 
 we obtain 
 
 tan (BAP - \EA C) = tan i^^ tan (Jw - ^), 
 tan (GBP - JO^) = tan \GBA tan (JTT - </>), 
 tan ( J. (7P - \A CB] = tan ^ CB tan (JTT - 1)- 
 
 By means of these equations we can find the angles which 
 the arcs AP, BP, CP make with the adjacent sides of the 
 triangle ABC, and then, by the rules of spherical trigonometry, 
 the arcs AP, BP, CP which determine the position of the 
 pole P. 
 
 176. Multiplying together the expression for the ratios 
 of the parameters in terms of the sides of the triangle ABC t 
 and their segments (175), we obtain 
 
 sin BD sin CE sin AF= sin CD sin AE sin BF. 
 
70 CRYSTALLOGRAPHY. 
 
 If we suppose five of the six arcs BD, CD, CE, AE, AF, 
 BF to be. known, the remaining arc will be given by this 
 equation. The sides of the triangle ABC, and, therefore, its 
 angles also are known. Therefore the angles which the axes 
 make with one another, being the supplements of the angles of 
 the triangle ABC, are known, and the ratios of the parameters 
 are given in terms of the sides of ABC and their segments. 
 Hence, any five of the six arcs BD, CD, CE, AF, BF may be 
 taken for the elements of the crystal. 
 
 177. The six segments may also be deduced from one of 
 the sides of ABC, and the segments of the other two sides. 
 Suppose BC and the segments of CA, AB given. Then 
 
 sin BD sin AE sin BF _, 
 
 - - = _ - - _ _. .Therefore, putting 
 
 sin CD sm CE sm AF 
 
 n _ 
 ~ 
 
 we have tan ( CD - %BC] = tan \BQ tan (J?r - 0). 
 Whence CD and BD are known. 
 
 178. The place of a pole in one of the zone-circles BC, 
 CA, AB, or in any zone-circle containing three poles joined by 
 arcs of known length, may be found by (13) or (14). In this 
 manner it is usually possible to determine the places of all the 
 poles of a crystal belonging to the anorthic or any other 
 system. 
 
 179. Let L, M, N, be any four poles of which no three 
 are in one zone-circle; efg, hkl, pqr the symbols of the 
 zone-circles MN, NL, LM respectively ; m n o the symbol of 0; 
 uvw the symbol of P. Suppose five of the six arcs joining 
 every two of the poles L, M, N, to be given. The remaining 
 arc and the angles MLN, MLO, LMN, LMO can be found 
 by the methods of spherical trigonometry. Then (18), 
 
ANORTHIC SYSTEM. 71 
 
 L I B R A R Y 
 
 UNIVERSITY 0] 
 
 CALIFORNIA. 
 
 putting tan-P m + ( ^ + ro htt + kv + lw *(MLN-MLO) 
 pu + qt? + r^ 1m + k^ + lo sin JL 
 
 and tan = 
 
 pu + qv + iw em + fn + go sin LM ' 
 
 we have tan (MLP - \ MLN) = tan \MLN tan (J?r - &), 
 and t8m(LMP- %LMN] = tan %LMN tan (JTT - 0). 
 
 Hence, knowing LM, and the angles MLP, LMP, we can 
 find the arcs LP, MP which determine the position of P. 
 
 180. When the position of any pole P is given with respect 
 to any two of four given poles, no three of which are in one 
 zone-circle, the ratios of the indices of P are given by the equa- 
 tions in (175) or (179). 
 
 181. Let L, M either have the same signification as in 
 (179), or be any two of the poles A, B, G in (175) ; P, Q any 
 two poles, the symbols of which are given. Let the angles 
 MLP, MLQ and the arcs LP, LQ be found by (175) or (179). 
 Then, knowing the sides LP, LQ, and the included angle PLQ, 
 the third side PQ may be found. 
 
 182. When five of the six arcs joining every two of the 
 poles L, M, N, are given, the arc joining any two poles may 
 be found by (179) and (181). Hence we can find the arcs BD, 
 CD, CE, AE, AF, BF, or the angular elements of the crystal. 
 
CHAPTER VIII. 
 
 TWIN CRYSTALS. 
 
 183. A TWIN crystal consists of two crystals joined together 
 in such a manner, that one would come into the position of the 
 other, by revolving through two right angles round an axis 
 which is either normal to a possible face, or parallel to the axis 
 of a possible zone, of each of the two crystals. This axis is 
 called the twin axis. When it is normal to a possible face, the 
 face is called a twin face. It frequently happens that, in twin 
 crystals of any system except the anorthic, the twin axis is 
 normal to a possible face, and also parallel to the axis of a pos- 
 sible zone, of each of the two crystals. 
 
 184. Let Tj T be a diameter of the sphere of projection 
 parallel to the twin axis ; P, p any corresponding poles of the 
 
 two crystals. Since p may be made to coincide with P by 
 turning the crystal to which p belongs through two right angles 
 
TWIN CRYSTALS. 73 
 
 round TT' 9 the arc Tp = arc TP, and the angle PTp = 180, or 
 Pp is an arc of a great circle bisected in T. In like manner 
 Q, q being any other corresponding poles of the two crystals, 
 the arc Qq will be bisected in T. If p', q be the poles opposite 
 to p, q respectively, it is manifest that Pp, Qq are bisected at 
 right angles by the great circle MN having T, T' for its poles. 
 Hence the opposite poles of the two crystals are symmetrically 
 arranged with respect to a great circle having its poles in the 
 twin axis. 
 
 185. In order to find the twin axis in any given twin 
 crystal, when it cannot be found by simple inspection, we must 
 determine by measurement or by the observation of zones, the 
 intersections of two great circles each of which passes through 
 corresponding or opposite poles of the two crystals. If the 
 diameter of the sphere joining the intersections of the two cir- 
 cles be normal to corresponding faces or be the axis of corre- 
 sponding zones of the two crystals, it will be the twin axis. 
 
 Let P, Q be any two poles of one crystal ; p, q the corre- 
 sponding poles of the other ; p, q the poles opposite to p, q ; 
 T, T' the intersections of the great circles pP, q Q. Then, if 
 TT be normal to a possible face or parallel to the axis of 
 a possible zone of each of the two crystals, it will be the twin 
 axis. 
 
 186. When the twin axis, and the angles of one of the 
 crystals are given, the arc joining any pole of one crystal, and 
 any pole of the other, can be readily determined. First let 
 P, p be corresponding poles of the two crystals, p the pole 
 opposite to p. Then Tp TP, and pPp is a semicircle, there- 
 fore Pp = 2TP, and Pp' = 18Q-2TP. When TP is greater 
 than a quadrant, Pp' is negative, and the faces P, p' will form a 
 re-entrant angle. Next let P, Q be any two poles of one crystal ; 
 p, q the corresponding poles of the other. From the given arcs 
 TP, TQ, PQ the angle PTQ is known, andpTQ = 18Q Q -PTQ. 
 Therefore, knowing TQ, Tp and the angle p TQ, the arc p Q 
 may be found. 
 
CHAPTER IX. 
 
 GEOMETRICAL INVESTIGATION OF THE PROPERTIES OF A SYSTEM 
 
 OF PLANES. 
 
 187. LET any three straight lines in one plane, intersecting 
 one another in the points A, B, C, meet any other straight line 
 in the same plane, in Z>, E, F, the points Z>, E, F being in the 
 
 lines respectively opposite to A 9 B, C. From A draw AH 
 parallel to BC, meeting DF'm H. By similar triangles AF : AH 
 = BF: BD, and Ag: AE = C> : C/. Hence 
 
 CD . AE. BF= BD.CE. AF. 
 
 188. Let OX, OYj OZ be any three straight lines passing 
 through a given point 0, and not all in one plane ; a, , c any 
 three straight lines given in magnitude ; h, k, I any three inte- 
 gers, positive or negative or zero, one at least being finite. Let 
 the symbol h k I be used to denote the plane HKL which meets 
 OX, OY, OZin the points H, K, L such that 
 
GEOMETRICAL INVESTIGATION. 75 
 
 ,OH 7 OK ,OL 
 
 h = Jc -j- =1 , 
 
 a o c 
 
 Off, OK, OL being measured along OX, OY, OZ, or in the 
 
 opposite directions, according as the corresponding numbers 
 h, k, I are positive or negative. And suppose a system of such 
 planes to be obtained by giving h, k, I different numerical 
 values. Let the point be called the origin of the system of 
 planes; OX, OY, OZ its axes; a, b, c, or any three straight 
 lines in the same ratio, its parameters ; h, k, I, or any three in- 
 tegers in the same ratio, and with the same signs, the indices of 
 the plane HKL. When an index is taken negatively, the 
 negative sign will be placed over the index usually, but not 
 invariably. It is evident that when one of the indices of a 
 plane becomes 0, the point of intersection of the plane with the 
 corresponding axis will be indefinitely distant from the origin, 
 and the plane will be parallel to that axis ; also, that when two 
 
 of the indices become 0, the plane will be parallel to the plane_ 
 containing the two corresponding axes. The planes Ti k I, hkl 
 are obviously parallel, and on opposite sides of the origin. 
 Either symbol may be used to denote a plane through 0, 
 parallel to the plane HKL. The straight line in which any two 
 planes intersect will be called an edge. 
 
 189. Let be the origin of a system of planes; OX, OY, 
 OZ its axes ; a, b, c its parameters. Let OB b ; and let the 
 planes hkl, pqr, passing through B, intersect one another in 
 
76 CRYSTALLOGRAPHY. 
 
 the edge BM meeting the plane ZOX in M-, and let them meet 
 OZm L, E, and OX in H, P. Then (188) 
 
 - OH=\ OB= l - OL, and OP= j-OB = - OE. 
 a o c a be 
 
 Therefore l.OL = Jcc, h.OH=ka, r.OE = qc, p.OP=qa. 
 Hence, Ir . LE = (fear - lq) c, hp . HP= (hq - kp) a. But (187) 
 HM . OP.LE = HP.OE. LM. Therefore, putting n = kr-lq, 
 \ = lp hr, w = hq Jcp, we have 
 
 wLLM=uh.HM, wLLH=-vk.HM, uh.LH=-vk.LM. 
 
 Draw MD parallel to OZ, meeting OX in D. By similar tri- 
 angles OD : LM= OH : LH 9 and DM : HM= OL : LH. 
 Hence v . OD = ua, and v . DM= we. Draw MF equal 
 and parallel to OB, on the opposite side of the plane LOH. 
 Then - v . MF= v.OB= v. The edge BM is obviously paral- 
 lel to OF, the diagonal of a parallelepiped, the edges of which 
 are respectively coincident with the axes OX, OY, OZ, and 
 equal to OD, MF, DM, and therefore proportional to v . OD, 
 v.MF, v.DM, or to ua, v, we. 
 
 The edge BM, and any straight line parallel to BM, will be 
 denoted by the symbol u v w, or by any whole numbers in the 
 same ratio. The integers u, v, w, or any other integers in the 
 same ratio, will be called the indices of the edge BM, or of 
 any straight line parallel to BM. 
 
GEOMETRICAL INVESTIGATION. 77 
 
 190. Since a plane of the system may be parallel to any 
 given edge, and also to any one of the other edges of the system, 
 it follows that a number of planes may exist parallel to a given 
 edge, and, therefore, intersecting one another in parallel lines. 
 Such an assemblage of planes is called a zone. A straight line 
 through the origin parallel to the edge in which any two of its 
 planes intersect one another, is called the axis of the zone. A 
 zone and its axis will be denoted by the symbol of the edge in 
 which any two of its planes intersect. Hence (189), h Jc I, p q r 
 being the symbols of any two planes of the zone, not parallel to 
 one another, the symbol of the zone will be u v w, where 
 
 u = kr Iq, v = Ip hr, w = hq kp. 
 
 It appears from (188) that the symbols of the planes YOZ, 
 ZOX, XOY are 1 0, 1 0, 1 respectively. Hence, the 
 symbol of OX, the intersection of the planes 010, 001, will 
 be 100; the symbol of Y, the intersection of the planes 
 001, 100, will be 010; and the symbol of OZ, the inter- 
 section of the planes 100, 010, will be 001. 
 
 191. Let the plane u v w, meeting the axes of the system 
 of planes in U, V, W, be parallel to the edge p q r. If VM be 
 drawn parallel to the edge p q r, it will lie in the plane UVW y 
 and its symbol will be p q r. Let VM meet WU in M. Then 
 (189) pu . WU+ qt? . WM= 0, and iw . WU+ qv . UM= ; 
 whence, adding, and observing that WM + UM = WU, we 
 obtain 
 
 pw + qv -f YW = 0. 
 
 This equation expresses the condition which must be satis- 
 fied in order that the plane u v w may belong to the zone p q r. 
 Any three integers either positive or negative or zero, one at 
 least being finite, which satisfy the preceding equation when 
 substituted for u, v, w, are the indices of a plane in the zone 
 p q r ; and any three such integers which satisfy the same 
 equation when substituted for p, q, r, are the indices of a zone 
 containing the plane u v w. 
 
78 CRYSTALLOGRAPHY. 
 
 192. Let h k 1, p q r be the symbols of any two edges. 
 In OF take 0F=&, and through 
 V draw VM, VS parallel to the 
 edges h k 1, p q r respectively, 
 meeting the plane ZOX in M, S. 
 Let M8 meet OZ, OX in W, U. 
 Draw MD, SG parallel to OZ, 
 meeting OX in D, Q. The sym- 
 bols of VM, VS are hkl, pqr, 
 therefore (189) 
 
 -c, q 
 By similar triangles 
 
 OW: OU=DM:DU=DM-GS: OG-OD. 
 
 Hence k (kr - Iq) .DU= 1 (hq kp) a ; also, observing that 
 OU=OD+DU, we obtain (kr -Iq) . OU= (Ip-hr) a, and 
 (hq - kp) . TF = (Ip - hr) c. Therefore 
 
 a b c 
 
 where u kr - Iq, v = lp hr, w = hq kp. 
 
 Since u, v, w are integers, the plane UVW which is parallel 
 to the edges hkl, pqr, is a plane of the system. 
 
 193. Let the plane u v w meet the axes of the system in 
 U, F, TF, and the zone-axis e f g in P. Draw WP meeting UV 
 in N 9 UP meeting FPF in L, and PQ parallel to OU 9 meeting 
 the plane VOW in Q. The symbols of 0TF, OU, OP are 001, 
 100, e f g respectively. Therefore the symbol of the plane 
 TF0P will be f e 0, and that of the plane UOP will be g f. 
 The symbol of the plane UVW is u v w. Hence, the symbol 
 of the edge WN will be QW, fw, eu + ft?, and the symbol of 
 the edge UL will be fv + gw, fu, gu. The edges WN, 
 UL are in the plane UVW, therefore (189) eu . UN= fo . VN 9 
 and fv.VW=(fv + gw).WL. But by (187) UP.WL.VN 
 = PL.VW.UN. Therefore eu. UP= (fv + gw) .PL. There- 
 
GEOMETRICAL INVESTIGATION. 79 
 
 fore eu.UL=(Qu + fv + gw).PL. But QP: OU=PL: UL. 
 Therefore eu. OU (eu + fv + gw) . QP. In like manner, if 
 
 the plane hJc I meet OU in H, and OP in D, and if DE be 
 drawn parallel to U, meeting the plane VO W in E, we shall 
 have eh . OH= (eh + ik + gl) . ED. But OP:OD=QP: ED. 
 Therefore (eu + fv + gw) . OP : (eh + ik +g?) .OD=u.OU:h. OH. 
 Hence, if the zone-axis p q r meet the planes uvw,hkl in J?, 
 F s we shall have 
 
 (pu + qv + YW) . OB : (ph + qk + rt) .OF=u.OU: h.OH. 
 eu + fe + * OP e^ + f^ + ? OD 
 
 Therefor 
 
 ~ OF' 
 
 The preceding equation will still be true, if we suppose OP, 
 OB to be the edges e f g, p q r passing through any point 
 which is not the origin of the system of planes. For OP, OR 
 will be parallel to the zone-axes efg, pqr respectively, and, 
 therefore, the ratios OP : OR, OD : OF will be the same in 
 either case. 
 
 194. If DF, PR intersect in K, we shall have 
 
 KP sinP = KD sinZ>, KB sinB = KFaiu F, 
 and OD sin D=OF sin F, OP sin P = OB sin B. 
 Hence KP.KF : KD . KB = OP . OF : OD . OB. 
 
 Therefore (193), 
 
80 CRYSTALLOGRAPHY. 
 
 195. Let the planes h k I, uvw 
 meet the zone-axis e f g in D, P, and 
 the zone-axis p q r in F, R, being 
 the origin. Draw OQ, OS parallel 
 to DF, PR respectively. Then OQ, 
 OS will be the axes of zones con- 
 taining the planes hkl,uvw, and will 
 be in the plane POR ; 
 
 sinPOQ : smROQ=smD : sinF= OF: OD, and 
 
 : sin P : sin R = OR : OP ; also (193), 
 
 eu + ft? + gw OP e + f k + gl OP T , f 
 rw OR h + k + rl OF' 
 
 sinPOQ sin EOS _zh + f7c + gl pu + qv + rw 
 sinPOS smROQ ~ ew + fv + gw pA + qk + rZ ' 
 
 where OP, , OR, OS are four zone-axes in one plane ; OP, 
 OR the axes of the zones e f g, p q r ; and 0, OS the axes of 
 zones containing the planes hkl,uvw. 
 
 It appears from (13) that the left-hand side of the preceding 
 equation can be put under the form 
 
 (cot POS - cot P OR) : (cot P Q - cot FOR) , 
 
 which is manifestly positive, except when one only of the zone- 
 axes OP, OR lies between OQ and 08. 
 
 196. Let P, Q, R, S be four planes in one zone. Let a 
 plane passing through the origin 0, 
 normal to the axis of the zone, meet 
 the planes Q t S in df, pr ; and planes 
 passing through 0, parallel to the 
 planes P, B, in dp, fr. Let hkl, 
 u v w be the symbols of the planes Q, 
 S-j efg, pqr the symbols of any 
 zones containing the planes P, R respectively, except the zone 
 containing P and R. Then the zone-axes e f g, p q r lie in the 
 planes parallel to the planes P, R respectively; Od, Op are 
 
>M ETHICAL INVESTIGATION. 81 
 
 proportional to the portions of the zone-axis e f g intercepted 
 between and the planes Q, S; and Of, Or are proportional 
 to the portions of the zone-axis pqr intercepted between and 
 the planes Q, 8. Therefore (193), 
 
 ew + fv + gw Op _ eh + f k + gl Od 
 pu + qv + YW Or ~ ph + qk + il Of 
 
 If PQ, PS, RQ, RS be taken to denote the angles which the 
 planes Q, S make with the planes P, R, we shall have 
 sin P Q = sin d, sin PS = sinp, sin R Q = sinf, sin US = sin r. 
 But snip : sinr = Or : Op, and sine? : siny= Of: Od. Hence 
 
 sin P Q sin RS _ eh + f k -f- gl pw + qv + YW. 
 sin PS sin ft Q ~ eu + fo + gw? pA + q& + il ' 
 
 where P, Q, 12, $ are four planes in one zone ; e f g, p q r the 
 symbols of zones containing the planes P, R ; and hkl, uvw 
 the symbols of the planes Q, S. 
 
 It may be shewn, as in (195), that the left-hand side of the 
 preceding equation is positive, except when one only of the 
 planes P, R lies between the planes Q, S. 
 
 197. Let efg, hkl, pqr be the symbols of three zone- 
 axes OP, OQ, OR meeting the plane m n o in the points D, E, F, 
 and the plane u v w in the points P, Q, R. Then (193), 
 QU + ft? + gw OP _ hu + kv + \w OQ _pu + qv + rw OR 
 
 _ 
 em + fn + go OD ~ hm + kn + 10 OE ~ pm + qn + TO OF ' 
 
 But if m n o, u v' w be the symbols of the planes mno, 
 uvw, when referred to the zone-axes efg, hkl, pqr, as axes 
 of the system of planes, we shall have 
 
 _ 
 m' OD~ri OE o OF' 
 
 Hence, comparing identical terms, two equations are obtained 
 which are satisfied by making 
 
 m! QM + in + go, u QU + ft? + gw, 
 
 n = \im + kn + 10, v = hu + ky + \w, 
 
 o pm + qw + r0, w' = pw + qv -f- rw. 
 
 M. c. 6 
 
82 CRYSTALLOGRAPHY. 
 
 198. Let m n o, u v w be the symbols of the zone-axes 
 OG, OP. Through G draw the planes efg, h k I, p q r meet- 
 ing OR in R, S, r respectively. Then (193), 
 
 ne + yf+Wff OR _ uh + vk + w? OS_ _up + vq + wr OT 
 me + n/-j- off G mh + uk + ol OG~ mp + n^ + or OGr ' 
 
 Let m'n'o', u'v'w' be the symbols of OG, OR, when re- 
 ferred to axes parallel to the intersections of the planes ef g, 
 hkl, p q> r. The symbols of these two planes when referred 
 to the new axes will become 100, 010, 001 respectively. 
 Therefore (193), 
 
 m 1 r\ S~Y ' i~\ /~Y ~~~ ' /~i i~i * 
 m (/Or n l/Cr l/Cr 
 
 Hence, comparing identical terms, we obtain two equations 
 which are satisfied by making 
 
 m' = em + /n + go, u' = en +fv+ gw, 
 n' = hm + Arn + Zo, v' = Au -t- kv + Zw, 
 o' pm + qn + ro, w' = pu + gv + rw. 
 
CHAPTER X. 
 
 ANALYTICAL INVESTIGATION OF THE PROPERTIES OF A SYSTEM 
 
 OF PLANES. 
 
 199. As in (188), let OX, OY, O^be any three axes not 
 all in one plane ; a, , c any three straight lines given in mag- 
 nitude ; h, k, I any three integers, positive or negative or zero, 
 one of them at least remaining finite ; H, K, L three points in 
 OX, Y 7 OZ respectively, subject to the condition 
 
 . OH , OK 7 OL 
 
 fl - = K ^ = I - . 
 
 a o c 
 
 Then, d being any positive quantity, the equation to the 
 plane HKL will be 
 
 Let the plane HKL be denoted by the symbol h k I, or by 
 any three integers respectively proportional to h, k, I, and hav- 
 ing the same sign, the numbers A, k, I being called the indices 
 of the plane HKL. A system of planes being formed by giving 
 A, k, I different numerical values, let the straight lines a, b, c 
 be called the parameters of the system of planes. 
 
 200. The equations to the planes h k I, p % r are 
 
84 CRYSTALLOGRAPHY. 
 
 where d, t are positive quantities. The intersection of the 
 planes hkl, pqr will, therefore, be parallel to the line which 
 has for its equations 
 
 j2L.JL.JL 
 
 ua vb we ' 
 where u = &r Iq, v = lp hr, w = hq kp. 
 
 This straight line, or any straight line parallel to it, will be 
 denoted by the symbol u v w, or by any three integers propor- 
 tional to u, v, w. These three numbers will be called the in- 
 dices of the line. 
 
 This straight line is obviously the diagonal OK of a paral- 
 lelepiped having its edges OU, OF, OW coincident with the 
 axes, and equal to ua, v&, we respectively. 
 
 201. Any number of planes intersecting one another in 
 parallel lines are said to constitute a zone. A straight line 
 through the origin, parallel to the intersection of any two planes 
 of a zone, and, therefore, parallel to each of the planes of the 
 zone, will be called the axis of the zone. A zone, and its axis, 
 will be denoted by the symbol of a line parallel to the intersec- 
 tion of any two planes of the zone. Hence (200) the symbol of 
 the zone containing the planes h k I, pqr will be u v w, where 
 u = hr lq, v = Ip hr, w = hq kp. 
 
 202. Let the zone-axis p q r be parallel to the plane u v w. 
 The equations to the zone-axis and plane are 
 
 =^T = , and u-= + vf + w? = d: 
 pa q& re a be 
 
 and the zone-axis is parallel to the plane. Hence 
 
 + rw = 0. 
 
 Any three positive or negative integers, including one or 
 two zeros, which satisfy the preceding equation, when substi- 
 tuted for u, v, w, are the indices of a plane in the zone p q r ; 
 and any three such integers which satisfy the same equation, 
 
ANALYTICAL INVESTIGATION. 85 
 
 when substituted for p, q, r, are the indices of a zone contain- 
 ing the plane u v w. 
 
 203. The equations to the zone-axes h k 1, p q r are 
 
 x y z , x y z 
 
 \- = rT = i and = ^r= . 
 ha K.O ic pa q6 re 
 
 Hence, if a plane be drawn parallel to the zone-axes h k 1, 
 p q r, its equation will be 
 
 x y z 
 -+vf+w- 
 a o c 
 
 where u = kr Iq, v = Ip hr, w = hq kp. 
 
 Therefore, since u, v, w are integers, a plane parallel to any two 
 zone-axes will be a plane of the system. 
 
 204 Let efg, pqr be the symbols of the zone-axes OP, 
 OR meeting the plane h k I in D, F, and the plane u v w in 
 P, R. Let planes be drawn parallel to YOZ, through the 
 points D, P, F, R, meeting OX in the points Z>', P', F', R'. 
 The equations to the zone-axes e f g, p q r are 
 
 , 
 ea 10 gc pa q& re 
 
 and the equations to the planes h k ?, u v w are 
 
86 CEYSTALLOGRAPHY. 
 
 The distances OD', OP, OF', OR are the values of x at 
 the points in which the zone-axes e f g, p q r intersect the planes 
 h Jc I, uvw. Therefore 
 
 (oh + f k + gl) . OD' = ead, (eu + fv + gw) . OF = eat, 
 (ph + qk + rl). OF' = pad, (pu + qv + rw) . OR' = pa*. 
 And by similar triangles 
 
 OD' : OD = OP' : OP, and OF' : OF= OR : OR. Therefore 
 gw OP _ eh + f Jc + gl OI>_ 
 
 _ _ 
 
 pu + qv+xw OM ~~ ph + qk + rl OF ' 
 
 205. From the preceding equation the expressions for the 
 anharmonic ratios of four zone-axes in one plane, and of four 
 planes in one zone, and the indices of a plane or a zone when 
 the axes are changed, can be found as in (195), (196), (197) and 
 (198). 
 
 CAMBRIDGE: PKINTBD AT THE tJNIVEKSITY PKESS. 
 
 
PUBLISHED BY 
 
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 Camfcrfoge, 
 
 AGENTS TO THE UNIVERSITY. 
 
 A Treatise on Crystallography. By W. H. MILLER, M.A. 
 
 8vo. 73. 6d. 
 The Mathematical and other Writings of EGBERT LESLIE 
 
 ELLIS, late Fellow of Trinity College, Cambridge. Edited by WILLIAM 
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 of Ely. [/ the Press. 
 
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