CRYSTALLOGRAPHY. 



471 



M-ith£ma» 



ti-ai 

 Theor . 



and q r=V^g*. Let us find an expression for v r or 



its equal r t ( Fig. 21. ) The similar triangles p r t, h r d 



nx 1/4-2 x4-ii , — / 



—ZT. TJL vV; -/jg 2 :: 



S nx y — x-f-j/) 



jrive h r : dr :.: » r : r / : or 



2 «X J/-J-X — // 



. — Inx if -l-x — ?/ .- — - . 



Snxy—x+y) 



1.) 



Join b q ; then draw q z perpendicular to v x. Draw 

 also / z. The angle q z I will be the half of that which 

 measures the incidence of (/ t b on (> x v, or of QXC 

 (Fig. 23. ) on BXC. But it is easy to have the ratio 



between q! (Fig. 1.) and Iz in values of q r, 



and 

 v r, by a method fonr.erly explained. By that method 



q I = V\ a/s^ We obtain Iz by means of this proportion 



v r — rl: lz::vr 



)x:xr::vl;tz;or</(xry + (i>ry:xr 



— \ qr: Iz. Hence we get Iz =xr 



Hence q I : Iz 



qr , — xrivr — %qr) 



V(xry 4- (v > y 



1 



V(xry + (;vry 

 qr ^/3J (.rr) I + («»•)* ):xr(2vr — qr). But if we 



make g= ■/:■>, }>—</% nz=l, x=2, y—\, in the alge- 

 braic expressions for q r, x r, and v r given above, we 



get q r=2, xr—5, v r= — . Substituting these values, 



instead of qr, x r, v r, in the ratio of q I to / z, we get 



/ ■/' 100\ „/20 \ ,— .- 



ql: Iz : : 2 J 3 (25 + ^ ) : 5(y-2) :: ^53 Vs. 



This gives us for the incidence of CXB upon CXQ 

 (Fig. 23.) 153° 13' 58". 



These observations and illustrations will give the 

 reader a pretty accurate idea of the method of proceed- 

 ing in cases of intermediate decrements. A complete 

 discussion of the subject would swell to a greater length 

 than would be tolerated in a work of this kind. In- 

 deed, we are afraid that we have already extended this 

 Chapter farther than many readers will be disposed to 

 follow us. Those persons who wish to see the interme- 

 diate decrements explained at full length, with all the 

 requisite examples, may consult Hauy's Mineralogie, 

 vol. i. p. 357. where they will meet with ample satis- 

 faction. 



Of the Compound Secondary Forms relative to the 

 Rhomboid. 



The compound secondary forms, especially those 

 which have a rhomboid for a nucleus, are usually no- 

 thing else than a combination of several simple forms, 

 which have often a separate existence in particular va- 

 rieties of the same substance. When the faces be- 

 longing to each of these simple forms are sufficiently 

 near each other, and of sufficient size to admit the 

 measurement of their mutual incidences, the compound 

 form may be determined by the calculation of these 

 incidences alone, which is always easy ant! simple. But 

 it is sometimes necessary, and often useful, to be able 

 to measure the incidences of the faces of one or.iei on 

 those of another order. There are even ca.ses when it 

 becomes interesting to know die plane angles of these 

 faces. To be able to resolve these probh n -. it is ne- 

 cessary to be accustomed toc>< ve ai 

 the results of the intersections of different planes in- 

 dined in different directions. But in tiie rhomboid we 



have this advantage, that the determination may be 

 made analytically from the ratios between the quanti- 

 ties which represent the system of lines relative to this 

 kind of solid. 



We shall satisfy ourselves with a single example, and 

 we shall select the variety of carbonate of lime called 

 annlogie by Hauy. 



This variety is derived from the prismatic carbonate, 

 by its six vertical faces, from the metastatic carbonate 

 by its twelve faces, situated six and six on each side of 

 the six vertical faces, and from the eqniaxc carbonate 

 by its terminal faces, to the number of three at each 

 extremity. These different faces are situated so advan- 

 tageously, that the knowledge of the angles which the 

 faces of the same order make with each other, (sup- 

 posing the structure of the forms from which they are 

 derived known) is sufficient to verify the laws upon 

 which the crystal depends. But without any regard to 

 this knowledge, let us endeavour, in the first place, to 

 determine the plane angles of the different faces, and 

 then the incidences of the faces of one order upon those 

 of another. 



Let criz (Fig. 2.) be one of the vertical faces, cypr, 

 cisz, two of the faces that belong to the metastatic 

 variety, and yciy* one of the faces derived from the 

 equiaxe. 



Let dvf, qof, duf, q uf be four faces of the me- 

 tastatic variety, supposed complete. Draw the axis 

 o u, the two diagonals c i, r z of the trapezoid criz, 

 the great diagonal y r of the trapezoid cy p r, and the 

 two diagonals c p., y t of the trapezoid y c t f~ 



Let us begin with criz. The points r, z being situ- 

 ated in the middle of the edges df qj which are com- 

 mon to the metastatic crystal and to the nucleus, it is 

 obvious that rz—g=Vs. Of course rh — */±. Let 

 h (Fig. 3.) be the same point as in (Fig. 2.) If we 

 draw Ac (Fig. 3.) parallel to the axis, that line will 

 likewise be the same as the line marked with the same 

 letters in Fig. 2. But the point h ^Fig 3.) is situated 

 at jth the oblique diagonal ft. Hence jf/fcr|jp 'J But 

 fh : ft : : h c : t 6. And t 0—2a—6. So that the pro- 

 portion becomes ^ : 2 : : he : 6. This gives us hc=l. 

 Therefore (Fig. 2.) rh:hc:: Vf: \ : : 1 : ^37 Hence 

 the triangle r c z is equilateral. 



Again, h i (Fig. 3.) is equal to the same line in (Fig. 

 2.) and from comparing the similar triangles fh i, ftu 

 (Fig. '; ) it appears that h iz=~ t K=.j. Hence hi (Fig. 

 2.) = ^hc. From these data it follows, that rcz~(JO°, 

 crior.cz i= 100° 53' 37", r i z=98° 12' 46". 



Let us determine, in the second place, the trapezoid 

 ycifi. Let o u, of, uf, c p. (Fig. $.) be the same lines 

 as in Fig. 2. Through the point £ (Fig. 4.) the same 

 as in Fig. % ; draw o £ (Fig. 4.) and prolong it indefi- 

 nitely." This line is evidently in the plane yos (Fig. 

 2.) or, which comes to the same thing, in the plane 

 d q. Therefore it passes through the middle of the 

 diagonal which joins xl, q. Let $ ( Fig. 4 ) be that mid- 

 dle point. Draw- 3 ep, cir, g * perpendicular to the axis, 

 and £ r perpendicular to c a-. It may be shewn as fol- 

 low:,, that c£— 2 ^(K. 



•The similar triangles ct£, crp, give us c r : r <r : : , •£ : 

 £i«. Hence it is clear, that *c £==££/«, provided it can 

 be proved that ct=2t«-. For this purpose, let us de- 

 e '.lie values of crand tt. 



1. FoT'CT. Ct=CA-{ 7 At, and CXz=Ct trX. 



To obtain a value of «r , let us observe that the line 

 3-f being the demiperpenuicular to tiie axis in relation 



Mathema- 

 tical 

 I'heoi v. 



Plate 

 CCXXV. 

 Fig. 2. 



Fig. 3.: 



H' 



