466 



Plate 

 UCXXIV. 



Fig. 12, 13, 



Decre- 

 ments oa 

 the lateral 



angles. 



Ft-?. 1G. 



Fig. 17. 



Fig. 8. 



Tlite 



scxxii. 



rig. 7- 



Getting rid of the denominator (3) 2 , and developing 

 the quantities («-|-2)'-, (8n — 3) 2 , and dividing the 

 whole by 9, the expression becomes (« 2 -f-4»-f-4)2 -f- 

 [ii* — 2«4- 1)S = 9«-. Hence we have /; 2 — 8 ra-f- 16=0. 

 Consequently «— 4=0 and /; = 4. So that the decre- 

 ment takes place by lour ranges. 



With respect to the inclination of dpf to dpb, we 

 obtain it by substituting for g, a, n, .their values in the 

 expression of the ratio of fc to cz. We will haveyb : 



,y/3. This gives for the inclination sought l6'l° 4S' 

 18". 



Another property of this dodecahedron, supposing it 

 complete, is, that the great angle dfp of its faces is a 

 right angle. 



To prove this, let us find the value of the three sides 

 df pf' aucl d P- 



1. df=Vg>+pi= s /5- 



2. dp (Fig ; 12.) = y(^___iy fl 2 -f-4 g - 2 = i/4^p 



3. pf (Fig. 13.) or its equal pg (Fig. 12.) = 



J (3SI) a ' 2 + * «* = $P^V* = V8. Therefore 

 (Fig. 13.) (d P y=(p/y-+(df)\ Of course the tri- 

 angle dfp is rectangular inf. 



4. Decrements on the Lateral Angles. 



The secondary forms produced by these decrements 

 are usually dodecahedrons, in which three of the edges 

 contiguous to each summit are parallel to the oblique 

 diagonals which correspond to them iu the nucleus. 



Let t i (Fig. 16.) be one of these dodecahedrons, and 

 1 one of the edges parallel to the diagonals of the 

 nucleus. Let b be the point of the edge i k which co- 

 incides with the lateral solid angle of the nucleus, or 

 which is the point from which the decrement sets out. 

 Let b c be the horizontal semidiameter of the rhomb 

 upon which the same decrements act. Draw b e per- 

 pendicular to 1 0, and join the points c, e. Let b u m 

 be the measuring triangle ; and let g' be the horizontal 

 semidiameter of a molecule. We will have b «=z2 n g'. 

 As for um, it coincides with the corresponding lateral 

 face of the first plate of superposition, and measures 

 the height of that face. Let as (Fig. 17-) be the nu- 

 cleus represented separately in a position analogous to 

 that which it has in the interior of the dodecahedron. 

 A little attention will enable us to perceive that the 

 lateral face which lias been just mentioned, being con- 

 tiguous to a range of edges of molecules, parallel .to ag 

 and d s, must be itself parallel to the principal section 

 which passes through the points a, d, s, g. And since 

 nm (Fig. 16.) measures the height of that lateral face, 

 it will be equal to the height of a molecule, or to the 

 line ak (Fig. 8.) supposing adsg to represent the 

 principal section of the molecule. Therefore we have 



w ,« ( Fig.7.)=ys^' 



^S::2« : y^S 



g for g', and p for //, because the dimensions of the 

 molecule are proportional to 'iiiise of the nucleus. 



Let us now obtain the ive inclinations of the 



faces of t! e dodecahedron )eginning witii that of p 1 

 to /Wo. (Fig. Id.) 



CRYSTALLOGRAPHY. 



It is easy to perceive, that the angle bee is equal to 

 half that inclination, and since the angle bee is a right 

 angle, the two triangles bum, bee are similar. Hence 



: bn : nm : : 2gn : ^J 



be : 



P 



P* 



Therefore bn: nm :; 



? 



Before calculating the second inclination, or that of 



otktort k, let us determine tiie portion of tile axis of 

 the dodecahedron which exceeds on each side the axis 

 of the nucleus. 



Let adsg (Fig. 18. ) be the principal section of the 

 nucleus, t o an edge of the secondary crystal parallel to 

 the diagonal a d, and i the inferior edge contiguous to 

 the preceding. From the point a, and from c, the 

 centre of ad, draw . a a; and ce both perpendicular to 



Plate 

 U.'XXIV. 

 Fig. 18. 



to. 



The similar 

 a x=.c e : at. 



triangles axt, ales give vis a k : as : 



But a k= J 3 S l P Z -S' 



'V 

 a s= v 9 p 1 



ce, being the same line as in Figure 16, we have, F - 16i 



bc=g 



1 



2n 



2 gn 



j- 



3g : 





Therefore 



J 1 - 



P' 



Therefore the proportion becomes 



J 



Ss 1 



1> 



'-3 e \ 



Let us suppose a plane oyr (Fig. 16.) perpendicular 

 to the axis. Let oty, r.iy (Fig. 19.) be the portions Fig. 1?. 

 of the triangles otk, r.t k (Fig. 16'.) cut oft' by this plane. 

 Let tn be the corresponding part of the axis, which Ave 

 shall suppose equal to t n, (Fig. 18.) Having drawn 

 on, rn, and yn (Fig. 19-)' we wiU nave V n ec l llal to 

 gn (Fig. 18.;, and on or rn (Fig. J 9.) equal to nl 

 (Fig. 18.), or to the prolongation of gn till it meet to. 



Draw oz (Fig. 19-) perpendicular to ty, op perpen- 

 dicular to ny, and join the points z, p. The angle ozp 

 will measure half the inclination of otk (Fig. 10'.; to 

 r t k. Let us find expressions for the sine o p (Fig. 1 9.) 

 and the co-sine p z of the angle ozp. 



1. For op. Because o rip is an angle of 60°, or opn 

 a right angle, opz=on\/\. But we must find the va- 

 lue of on or of its equal nl. (Fig. 18.) The similar 

 triangles adr, tin give us ar : dr : : tn : nl : : at •+- 



an:n I. Or, fyV : V%g 



t : HfetO 



v", 



\i= 



2«-f 3 ,-—: 



X_V-; »- = 0tf. 



2n + 



2n + 3 , — - 

 Therefore op=-^[— Vfg* 



Vi = 



vV 



2. For pz. The similar triangles tny, pzy (Fig. 

 19. ) give ty : tn:: py : pz. Now 



2 n + 3 



Jgn 



J 



substituting ty _ ^ t ^ + y 



/ 1 \ 2n - 



// 2n -f. 3 



v/a 2 . 



py.-y^ 



■ pn ■= yi 



6n 

 on=V^ 



6 v 



« 2 + 4g 2 



2m _ 

 in 



•t7 = ^-(^±- 3 ))^=^V S 



