CRYSTALLOGRAPHY. 



469 



Mathema- 

 tical 

 Theory. 



«« + l 2b'+2 ' ,. - 4«'+l 

 ! — — — . From this equation we get «= - — -—■ r» 



4 ;, J 



"""Y""" andn'ss. . Therefore, since the values of n and »' 



2w-f4 



are rational numbers, it follows, that the same form of 

 rhomboid which is possible in consequence of a certain 

 law of decrement in breadth, is possible also by a diffe- 

 rent decrement in height, and vice versa. We may al- 

 ways pass from the one to the other, in consequence of 

 the preceding formulas. 



Suppose, for example, we inquire what would be the 

 decrement in breadth which would produce a seconda- 

 ry rhomboid similar to the cuboid. To resolve this 



question, we have only to take the formula nz=.- — 5— ;j 



and make n'—^. This gives us m=4. Hence the law 

 sought would be a decrement by 8 ranges of mole- 

 cules. 



On Intermediate Decrements relative to the Rhomboid. 

 Interme- The decrements called intermediate, depend upon two 



diate e- variable elements which must enter into their calcula- 

 relaiive to t ' on " The one * s t ^ le rat '° between the number of 

 the rhom- lengths of molecules, subtracted from the two sides of 

 boid. the angle on which the decrement takes place. The 



other is the number of ranges subtracted, or the dis- 

 tance between the same angle and the edge of the first 



v 

 plate of superposition. The fraction — represents the 



ratio between the sides ; and n, as usual, denotes the 

 number of ranges subtracted. 



In proportion as y diminishes in relation to x, the 

 edge of each plate inclines always more towards the 

 edge of which x constitutes a part, and when y vanishes 

 it coincides with that edge. On the other hand, in 

 proportion as y augments in relation to x, the edge of 

 each plate of superposition becomes more nearly paral- 

 lel to the diagonal opposite to the angle on which the 

 decrement takes place ; and when y becomes equal to 

 x, we have the ordinary decrement on the angles. 



Let us consider, in the first place, the effects of an 

 intermediate decrement towards the lateral angles b, u, 



Piate (Fig. 22.) of any rhomboid, one of the superior faces 



f^ X 9? of which is represented by ah du. Let us suppose yX 

 £' to be the edge of the first plate of superposition, so that 



by, b x, measure the respective lengths of molecules on 

 the edges, with this single condition, that b x or x is al- 

 ways greater than by or y. 



In that case, the secondary solid will be in general a 



Fig. 23. dodecahedron, HX, (Fig. 23.) with triangular faces. 



Fig. 2-1. Let agsd (Fig. 24.) be the principal section of the nu- 

 cleus, and h x the axis of the secondaiy crystal. It is 

 easy to see that of two contiguous edges, such as hq, qx, 

 the first passes though the angle d, while the other is 

 formed at a certain distance above the diagonal a d. 

 Draw dp parallel to this last edge; dp will be situated 

 as the oblique diagonal of a rhomboid resulting from a 

 decrement on the angle d, in which the distance of one 

 plate from another, taken in the direction d a, will be 

 the same as in the dodecahedron that we are now con- 

 sidering. Let dkf be the measuring triangle referred 

 to the plane pdr. i/" will represent the length of one 

 molecule. Let us find an expression for d k. 



Draw Xfi (Fig. 22.) parallel to da; yn- perpendicu- 

 lar to x p*. ; dt parallel to y X ; then taking i $ equal to 

 by, draw 3v parallel todt, and v<p parallel to a b. It 

 is evident that d i, v$ will correspond with the edges of 

 the consecutive plates of superposition ; and of course 



tieal 

 Theory. 



Plate 

 CCXXIV. 



fig. 22. 



d > will be the distance of one plate from another, taken Matheraa 

 in the direction da, supposing always only one range sub- 

 tracted. Therefore we have d t x nzzdk (Fig. 24.). The 

 question is therefore reduced to find an algebraic expres- 

 sion forrfy (Fig. 22.) 



But x p measures as many times the oblique semidia- 

 gonal of a molecule, as there are lengths of a molecule 

 contained in bx -f- b pz= 2 x. Therefore denoting by p' 

 the oblique semidiagonal of a molecule, we may repre- 

 sent x p by 2 p'x. On the other side av—ih—byzzy. 

 But the similar triangles aid, ypx give yp : X/a :: 



dvJWfl. Therefore 

 x—y 



d k (Fig. 24.) = — — . Therefore, denoting by 



g' the horizontal semidiagonal of a molecule, we will 



have dk : Jef :: -^ — ~- : Vg^-^p' 1 . Or, since the 



a v : d v ; or, x — y : 2 p'x : : y 



x—y 



dimensions of the molecule are proportional to those of 

 2 p n x y 

 x—y 



the nucleus dk : kf 



Vg* + p\ 



Let us now determine the respective incidences of 

 the neighbouring faces towards the same summit of 

 the dodecahedron HX (Fig. 23.), in which the edge 

 QX is conceived to be the same as qx (Fig. 24.) 



Let us begin with the incidence of CXQ ( Fig. 23.) 

 on NXQ. Let b u be one of the horizontal diagonals of 

 the nucleus, and b a u, one half of the rhomb to which 

 that diagonal belongs. Let br, uy be sections of the 

 same rhomb, prolonged for the purpose on the triangles 

 CXQ, NXQ. If we prolong these sections till they 

 meet in a common point m, the triangle b m o will be 

 similar to the triangle yijr (Fig. 22.), since b m, um 

 (Fig. 23.), or their parts b r, ny, of necessity represent 

 the two decreasing faces of the same plate of superposi- 

 tion. 



Draw b n perpendicular to QX (Fig. 23). Then join 

 the points o, n. The angle /; n o will be the half of 

 that which measures the incidence of CXQ or NXQ. 

 We must therefore find the relation between the sine 

 b o and the cosine o n of the angle b n o. But b ozzg. 

 We have only therefore to find o n. 



From the point o ( Fig. 24. ) in the centre of a d, and 

 from the point a, draw the lines o n, a b perpendicular 

 to qx, and produce ga till it meets dp. The line o n 

 is the same as on (Fig. 23.). Let us find its algebraic 

 expression. 



The triangles m o n, (Fig. 23.) and doz, (Fig. 24.) 

 are similar, since om (Fig. 23.) coincides with od, (Fig. 

 24.) on with oz, and dp, (Fig. 24.) of which dz is a 

 part, is parallel to »»X, (Fig. 23.) on which m n is si- 

 tuated. Therefore om : on (Fig. 23.) : : od : oz (Fig. 

 24. ) : : a d : a I. Hence to find o n, we must find o m, 

 a d, and a I. But a d=z2p. We have only therefore to 

 find o m and a I. 



1. For om. The triangles ywx (Fig. 22.) and bom 

 (Fig. 23.) are similar, as has been already observed. 

 Therefore yir : wX : : bo : om. But *r x measures as ma- 

 ny demidiagonals p as there are lengths contained in b x 

 ■^-byzzx^c-y ; and yx measures as many demidiagonals 

 g as there are lengths contained in ypiz=x — y. Hence yjr 

 : tX : : gx — gy : px+py :: bo (Fig. 23.) : om :: g : 



om. Hence we obtain om— — "" . 



2. For a I (Fig. 24.). The triangles a Ip, drp give 

 usap : al:: dp : dr. But drzzV^ g 1 . We have 

 only then to find ap and dp. 



