472 



CRYSTALLOGRAPHY. 



M.uhema. to one of the inferior rhombs of the nucleus, its posi- 



tical tion is the same as gn (Fig. 3.) Therefore o <p 



^Theory^ (Fjg ^ _ Qg (Fig 3.) = a + a g= 3 + 2=5. Far- 



Platb ther, $Q (Fig. 4.) = \Ag l =l. Now the similar tri- 

 CCXXV. angles oa, 0^0, give oo-: e-A :: oip : <p$: : 5 : 1. 

 Fig. S, 4. j3 ut we must fi n( j the value of o<r. If from the point 



c (Fig. 3.) we draw a perpendicular to the axis, it will 

 fall at the extremity a of that axis. For cf=\of. 

 Therefore, since c a is parallel to fr, the distance a r 

 will be \ of o r. Hence a r=f a 0. From this it fol- 

 lows, that the extremity a of the perpendicular coin- 

 cides with that of the axis of the nucleus. Since then 

 cs- (Fig. 4.) corresponds withe a (Fig. 3.), the point 

 <r (Fig. 4.) is so situated, that oo- is the excess of the 

 axis of the metastatic crystal above that of the nucleus. 

 Hence o<r=3. Of consequence the proportion or : s-A : : 

 5 : 1 becomej 3 : o-A : : 5 : 1. So that o-A=:f. 



To obtain a value for c<r or its equal ac (Fig. 3.) we 

 have th ; j proportion, o r :fr : : ao:ac ; or 4 : 2 : : 3 : ac 

 — 3 — J <r. Therefore the equation c A = c <r — s- A becomes 



cx—i ? = 9 



We have still to find the value of at. The triangles 

 Ct£ and co-p give ct. t£: : co-: 57* ; orcA-f-Ar: t£: : c<r: 

 e-fi. But t^=5at; because these quantities are pro- 

 portional to o<r=3 and <rA=f. 



Calling g' and p' the two demidi agonals o f the equi- 

 a xe, we have c<r : rfl, : zVfg 1 *: f</9p' z —3g'* : Vf .12 

 . x */g 5 _37l2 : -. i/& r »/\ : : 2 : 1. We have seen 

 already, that c*= T 9 5 . Therefore the proportion ca + 

 At:t£.::c<t: <r^ becomes ^ + At : 5 At : : 2 : 1. This 

 gives us At = iV 



If we now substitute for ca and At their values thus 

 found in the equation ct=ca + at, we obtain crss^+ 

 _5-=l. 



IO _ , 3 I 5 1 



2. For TO". TS-=C(T CA AT=-| /g - r5 -_ T ^— g. 



Therefore ct=z 2 t<t. Therefore likewise c£=2c>; 

 which was the thing to be proved. Now, since £> 

 (Fig. 2.) : vC : : \/5 : V" 12 ' we have c £ : V? :: ' v/ 20 : \/j2 

 : : \/5 : \/3 ; which is exactly the ratio of the two de- 

 midiagonals of the inverse rhomboid. Hence, of the 

 two triangles spy, icy, the one belongs to the equiaxe, 

 and the other to the inverse variety ; and the two 

 heights c£, p£ of these triangles have to each other the 

 same ratio as the heights ch, ih of the triangles which 

 compose the trapezoid iy, cz. 



Let us now go to the trapezoid Cypr, and find, in the 

 first place, expressions for the three sides of the tri- 

 angle cyr. ■ 



1. Forcy. cysyCcSy-HyQ* and_cg_(Fig. 4.) = 



^(<*-) i +K) s =* / ( cT ) 8 + (5AT)»=y i-+^5=yi +i 



Wf Farther, y? (Fig. 2.) :c?:VSV5. ° r vC : 

 V^ : : a/3: a/5. Of consequence y^—</\ and cy= 



2. For cr. 



3. For yr. 



.y%+$=\f3. 



crzz\/{cky+(hry-. 



If we conceive a plane perpendicular to 

 the axis to'pass through the line ye, it will cut the axis 

 at a point 1 (Fig. 4.) Let us determine the value of 



Ov. We have OvrrOtr — <rv—0<r — T^rrOo- — 5Ar=3 ro = 



4.. But the axis oti=9= r J. Therefore the point y 

 (Fig. 2.) which is at the height of the point » (Fig. 4.) 

 is situated opposite the ^ of the axis. But the point d 

 (Fig. 2. and 3.) is situated opposite the £ of the axis, 

 .since os=5. Hence it follows, that the point y (Fig.2.) 

 is in the middle of the line od. But the point r is in 

 the middle of the line df. Therefore yrzz\ ofz= 

 V(«r)»+C/r) 8 (Fig. 3.)=±\ r L6 + 4 ! = v /5. 



Thus we have found cy~*/2; crst/3; and yrz= 

 a/ 5. Hence we are entitled to conclude, 1. That the 

 angle ycr is a right angle ; 2. That the triangle cyr is 

 similar and equal to the fourth part of one of the faces 

 of the nucleus divided by the two diagonals. 



Having obtained the angle ycr=90°, let us find the 

 angles ypr and cyp. 



1. For the angle ypr. This angle is the supple- 

 ment of dpr. But in the triangle rpd, we know 

 d r= J df= -\\/5. We know p r=zi r= -/\r lif- + (i hy— 

 Vf -\-J- a -= \ V21. Hence d r : pr :: ^20 : ^21. 

 The an "le pdr, which belongs to one of the faces of 

 the metastatic crystal, is supposed known. Its value 

 is 54° 27' 30". From these data we obtain dpr= 

 52° 34' 7". Hence it follows, that y;jr=127° 25' 53". 



2. For the angle cyp. That angle is composed of 

 the two angles cyr, pyr, the first of which is half the 

 obtuse angle of the primitive rhomb, that is to say, it 

 is equal to 50° 46' 6". Now, as ypr=127° 25' 53"; 

 as pr=%*/2l and yr—</5, it is easy to discover, that 

 «yr=24° 0' 24". Adding this value to that of Cyr, 

 we obtain c y p=74° 46' 30". 



The fourth angle crp must, of course, be equal to 



67° 47' 37". 



We have still to determine the incidence of cypr 

 upon czir, and that of c y f* t on c y p r. 



1. For the incidence of cypr on czir. Let rcz 

 (Fig. 5.) be the same triangle as in Fig. 2. Draw 

 en (Fig. 5.) situated as cf (Fig. 2.), and produced 

 in such a manner that the lines r n, z n, drawn to its 

 extremity, are perpendicular to it. Draw likewise c h, 

 the height of the triangle rcz, and n h and n g per- 

 pendicular to cr, n a perpendicular to c h, and lastly 

 a g. The angle n g a, which measures the incidence 

 of n c r or c r z, is the supplement of that which mea- 

 sures the mutual inclination of the planes criz,cypr 

 (Fig. 2.) Hence the problem is reduced to find the 

 angle nga (Fig. 5.) Let us find, in succession, the 

 value of n g and n a. 



(1.) For ?2 a. On account of the right-angled tri- 



, . cnxpn 

 angle c n h, n a= -. — . 



c h= */% ; and r h: k n : : V5 : V 3, because the angle 

 rnz measures the smallest inclination of the faces of 

 the metastatic crystal. But rhzz^j, therefore /i)i= 



Mathema- 

 tical 

 Theory. 



Plate 

 CCXXV. 



Kig. 5. 



= •! 



Hence n a 



a/ 9- X *>- — 

 == — VTT' 



Hg: 



cnY.11 »■ 



c « = \/ » ; 

 = ^/|^ ; and 



Hence n g= 



v/3 - 



(2.) For»g. 



Now we already know that 

 A/(r hf + {h nf = */% + 



\f(chy+(rhy= v^+l= Ss 



Therefore ng-.na:: ^18 : a/Q:: \/2: \/^- Hence 

 it follows, that n g a=45°. Of consequence, the incli- 

 nation of cypr to cz i r is 135°. 



2. For the incidence cy/nt (Fig.2.) c Y pr. Let 

 y c e (Fig. 6.) be the same triangle as in Fig. 2. Draw 

 c v ( Fi°\ 5. ) situated as c o ( Fig. 6. ) and of such a length 

 that the straight lines ■/»,.» (Fig. 6.) drawn to its ex- 

 tremity, are perpendicular to it. Draw likewise v< per- 

 pendicular to y h v A perpendicular to c £, » sr perpendi- 



Fig. e. 



