464 



M.ithema. 



deal 



Theory. 



Plate 

 U0XXIV. 

 Pigs. 12, 

 IS. 



Let pg (Fig. 12.) be the edge which passes through 

 the points marked by the same letters in Fig. 13. and 

 which is equal to ;;/. Having drawn sy (Fig. 12.) 

 perpendicular to p g produced and through the point t, 

 the centre of sg, another perpendicular t g to the same 

 line, Ave have t g-=e h (Fig. 13.) To obtain the value 



CRYSTALLOGRAPHY. 



us now apply the different expressions thus 



Let 

 found. 



of t, 



let us find that of its double sy. 



triangles png, pys give us pgtgfi; 



The similar 

 p s : sy. 



I. For pg. Wehave|jg=v'(pM) 2 +(gw) 2 , and pnzz 



ap + an=z V9p z — 3# 8 +t \f 9 p 2 —3g* = 



3n- 



—g" V 9 P* — 3 £ 2 - And S n — ^ t S 2 « Hence pg: 



//ra + 2\ 2 „ 



2. g «=*/^g 2 , as has been just observed. 



3. For ps. We have p s—ctp-\-a s=: r V'a 2 + 



re 



./«2 = ./a 2 . 



n — 1 



The proportion p g : g n : : p s : s y, thus becomes 



v \3«— 3y 



re — W / «4-2 \ 2 



_(£!k+^ 



expression for the value of eh (Fig. 13.), we will have 



. n J T« 2 g 2 . . 



+ f < 



v'io- 2 : 



re— 1 T 



a* = *3/ = 



Taking the half of this 



d e : eh : : g 







Let us now find the value of /c and cz. But as 

 fczzg, we have only to find c z. 



From the point a (Fig. 12.), taken at the extremity 

 of the axis, and from the point c, in the centre of a d, 

 draw the lines a x and c z both perpendicular to dp. 

 c z is the same line as in Fig. 13, and a a; is its double. 



But the similar triangles prd, pxa give dp-.dr:: 

 a p : a x. 



As a p and d r are already known, we have only to 

 find dp. 



We have dp=zV , (p r y + (dry. And 



pr-ap^ar--— ^a* + \^ a *J^±}-V a l. Hence 

 «— i 3 M — 3 



1 he proportion above stated of course becomes 



ax. Ta- 



king the half of a x, found from this proportion, we get 

 the value of eg (Fig. 1 3.). Now/c : cz : : g .- 



/ / l v 



V w — ] / 3 * //2»+l\3 " 



\3n—3J TTfc 



If in the equation apz=^—^V 9p l — 3 g z we ma ke 



re=2, it becomes ap= V9p l — 3g z , that is to say, that 

 in this case the part of the axis of the dodecahedron 

 which passes the axis of the nucleus on each side, is 

 equal to the axis ; or which comes to the same thing, that 

 the axis of the dodecahedron is three times the length 

 of the axis of the nucleus. This property is general. 



It may be useful to observe in passing, that the two 

 solids have the same ratio to each other as the axes. 

 We leave the demon stration of this as an exercise to the 

 young crystallographer. 



If we make re=l, we have ap=±^9p 2 — 3g*. This 

 indicates, that in such a case the axis becomes infinite, 

 and of course the planes produced by the decrements 

 are vertical. This case occurs in the corundum. 



Let us resume the hypothesis w=2 ; and let us make 

 likewise g—*/3 and pz=»/2, as in the primitive rhom- 

 boid of calcareous spar. Substituting these values in 

 the expressions for fc and cz (Fig. 13.), we obtain fc : 

 cz : : a/29 : a/3. This gives 144° 20' 26" for the in- 

 clination of fpd to bpd. 



If we substitute the same values in the expressions 

 for de and eh, we obtain de : eh : : a/5 : a/3. This 

 gives 104° 28' 40" for the inclination of fpd to fpq. 

 But this is the angle which measures the inclination of 

 the primitive faces bafd, gafq, which correspond to 

 the secondary faces fpd and jpq. 



Suppose amhl (Fig. 14.) to be the four-sided figure 

 which would be obtained by cutting the rhomboid as 

 (Fig. 7.) by a plane passing through aw and perpen- 

 dicular to abdf. Draw ai (Fig. 14.) perpendicular to 

 Inn, and corresponding to the line ai (Fig. 7.) 



It is easy to see, that the angle ma I (Fig. 14.) mea- 

 sures the inclination of the two faces of the rhomboid 

 taken round the same summit, and consequently it 

 measures that of the rhombs b ofd, g ofq (Fig. 7.) 

 It remains only to prove, that the ratio of m r to a r 

 (Fig. 14.), the sine and cosine of half that angle, is the 

 same as that of de to eh (Fig. 13.) 



We have had already am (Fig. 14.) = / '*S 2 P i _ 



/24 

 5 

 We have likewise had 



ai= Js¥? 



Mathema* 



tical 



Theory. 



Plate 

 CCXXIVV 



Fig. 13. 



Figs. 7, 14. 



-f? 



But by construction mlzz 

 _aiy.mh I %Y. z ~ 



v 2 



\l 



12 



2g=A/12. And 

 — i/f . Therefore 



m r = g : ar : : a/3 : A/f : : a/5 : a/3 : : de (Fig. 13.) 

 : e h, which was the point to be proved. 



Let us now compare the solid angle^ formed by the 

 three plane angles pfd, pfq, dfq, with the solid an- 

 gle a of the nucleus. We have demonstrated, that the 

 inclination of dpf to pfq, is equal to that of b afd to 

 g afq. Farther, the angles pfd, pfq are equal, as 

 also the angles b af g af. And the angle dfq is equal 

 to the angle bag. Hence the two solid angles are 

 equal in every respect ; and since b a g is equal to each 

 of the two other angles b af and g af, it follows, that 

 dfq is equal to each of the angles pfd, pfq. So that 

 not only the inclination of the faces of the secondary 

 crystal, adjacent to the edge pf, is equal to that of the 



