CRYSTALLOGRAPHY. 



465 



Mathema- 



ticat 

 Theory. 



Plate 

 CCXXIV. 



F.gs. 13, 

 *5. 



corresponding faces of the nucleus, but likewise the 

 obtuse plane angle of the faces of the secondary crys- 

 tal is equal to that of the faces of the micleus. 



The preceding residt furnishes us with a very sim- 

 ple method of obtaining the inclination of any one dpf 

 of the faces of the dodecahedron to the adjacent face 

 below the edge df For that inclination is equal to 

 that of dpf to dfqs -f- the difference between this 

 last and that of b afd to dfq s. But the inclination of 

 dpf to dfq s, is 104° 28' 40". That of b afd to dfqs, 

 the supplement to the preceding, is 75° 31' 20". The 

 difference, of course, is 28° 57' 20". Adding this dif- 

 ference to 104° 28' 40", we obtain 133 c 26'. 



Let bafd (Fig. 15.) be the same rhomb as in Fig. 13. 

 Draw by bisecting af. The triangle bay is similar 

 to any one dpf (Fig. 13.) of the triangles of the se- 

 condary dodecahedron, so that the sides of the one are 

 double those of the other. 



Let us, in the first place, find the value of the three 

 sides of the triangle d pf. We have 



1. df=VgT+f=J5. 



2. dp (which is the same line as in Fig. 12.) = 



A 



3- Pg ( Fi g- 12.)=/>gor/>/(Fig. 13.)= 



*/(£S)"*+ }*•=%* 



•94-4 =A /20. 



Let us find, in like manner, the value of the three 

 sides of the triangle bay (Fig. 15.) 



1. ay=> a f=±S5=hdf (Fig. 13.) 



2. Having drawn y m perpendicular to bf (Fig. 15.), 

 we have 



b y= V (bmy + (m yy= */($ bjf + (£ " c) 1 



sV^I2+i.2=V2&=i«*P ( Fi g- 13 



3. a b (Fig. 15.) =</5= ± */20=± pf (Fig. 13.) 



We see likewise, that the mean side pf of the trian- 

 gle dpf is double the small side. All these results 

 take place in the variety of calcareous spar called me- 

 tastique by Hauy. 



This variety of crystal gives an opportunity of re- 

 solving another problem of considerable importance. 

 It is to determine, from certain data, the ratio of the 

 two semidiagonals g and p of the nucleus, from which 

 the angles of the nucleus and of the secondary crystal 

 may be calculated with rigid accuracy. 



Let us employ, as data, the equality observed be- 

 tween the angles pfd and dfq (Fig. 13.), and the law 

 of decrement by two ranges, from which the metastatic 

 crystal results, or, if it is preferred, the equality of that 

 part of the axis of the secondary crystal which passes 

 the axis of the nucleus with that of the axis of the nu- 

 cleus itself. Our object is, from these data, to find the 

 ratio between g and p. 



The angle pfd being equal to the angle dfq, or 

 (which is the same thing) to the angle b af, the an- 

 gles dfk and dfa, which are the supplements to these 

 angles, are likewise equal. Then, since df is equal 

 to af, the sine d k of the angle dfk will be equal to the 

 sine am (Fig. 7.) of the angle dfk (Fig. 13.) But 



am = J±SZ. 

 V g*+P x 



It remains for us to find d k, in order to form an 

 equation with the value of am. 



The triangle d e k is rectangular at e ; for the plane 

 dfs being perpendicular to the plane afs, is perpen- 

 dicular also to the plane pfs, which coincides with 



VOL. VII. PART II. 



afs. Therefore, since d e is at onee situated in the 

 plane dfs and perpendicular to fs, the common sec- 

 tion of this plane with the plane pfs, it must be per- 

 pendicular likewise to this last plane. Therefore k e, 

 situated in the prolongation of the plane pfs, and 

 which falls upon d e, will be perpendicular to this last 

 lihe. Therefore the triangle d e k is rectangular at e. 

 Hence d k=V(d e) x + (ek)\ 



But d e=rg, and e k 



Mathema. 

 tical 



Theory. 



=--v 



ffl» g' 



\3 n—3) 



+ fg* 



And 



because «=2, we have 

 e k 



=2 /_L«v_=y. 



3« 2 g 2 



becomes 



(4) 2fl2+ 4g2- "» 4fl*+3g* 



Equating the values of the squares of d k and a m, we 

 have 7« 2 g 2 +3g*_4g^ 

 4« 2 - r -3g 2 g^+p 1 

 Substituting for a 2 its value 9p 2 — 3g*, it becomes 

 7gH9p 2 — 3g 2 ) +3g 4 _ 4g'V 

 4(9/? 2 — 3g 2 ) + 3g* g'+py 



By making the two denominators disappear, 

 reducing and transposing, the expression 



g* £ p'l g2_ g. p4 # 



This equation gives for the two values of g 2 , g 2 =l f, 

 andg 2 =3p 2 . 



In the first case, g : p : : s/3 : \/2. This is the ra- 

 tio wanted between the semidiagonals of the primitive 

 rhomboid. In the second case, we have g:p'-- \/3 : 1. 

 This corresponds with the hypothesis in which the nu- 

 cleus and secondary crystal would be confounded un- 

 der the same plane, which would be a regular hexa- 

 gon. 



There is another variety of calcareous spar, called as- 

 cending by Hauy, because the different laws on which 

 it depends act from below upwards. Among its bound- 

 ing faces, 12 result from a decrement on the inferior 

 edges, and supposing them prolonged till they cut each 

 other, they would give the form of a dodecahedron si- 

 milar to the metastatic. 



Suppose the triangles bpd, dpffpq represent three 

 of the superior faces of this dodecahedron. If we measure 

 the incidence of dpf or g pf, we find that it is nearly 

 101°. This makes us presume, at first, that it is equal 

 to the great angle of the primitive rhomb, namely 

 101° 32' 13". Let us suppose this equality to hold 

 true, and let us determine from thence the decrement, 

 or the value of n. 



We have by hypothesis de: ek : : g : p: : */3 : v^ 2 - 



But we formerly obtained 



de-.et,: J(£&£+ + 4f 



Putting in place of a* its value 9, and of g 1 its value 3, 



we get • 



/ ., I O \ 2 / / M \ 



3 



* \3n—3j 



9 + 4 



d of g 1 it 



-v/3 



de : ek 



:a/2. 



Taking the products of the extremes and means, and 



suppressing the radical signs, we get 



( ra + 2) 2 18+(3tt-3) 2 8 _/ JL-Vo- or 

 (3«— 3) 2 ~\n—l) 



( w + 2) 2 18-H3m— 3) 2 8 



3 N 



: 9 H s 



