393 
principal crystallographers for these forms, together with the 
minerals in which faces of them have been found. 
1 
Naumann. 
Miller. 
Brooke, &c. 
Minerals. 
20 
12 2 
cA 
Amalgam. Fluor. Pharmaco- 
Argentite. Franldinite. siderite. 
Blende. Galena. Pyrite. 
Cuprite. Magnetite. Skutterudite. 
Diamond. Perowskite. Spinelle. 
30 
13 3 
a? 
Cuprite. Fluor. Galena. 
40 
2 3 3 
at 
Fahlerz. Garnet. Cuprite. 
40 
14 4 
1 
a-r 
Galena. Kerate. 
~~io~~ 
4 7 7 
4 
(XT 
Galena. 
40 
4 5 5 
a4 
Galena. 
440 
64 65 65 
aul- 
Alum. 
47. To find the ratio of the octahedral axis of the three- 
faced octahedron to that of the circumscribing cube, or of Ao x 
to A0 V 
Fig. 29, Plate IY. By construction 0 1 D 5 = 1 and AD S =^/ 2, 
Therefore tan AO-J) b ~y ^ 2 = 54° 44'; 
And therefore 0 3 AD 5 = 35° 16'. 
AM 
Also tan Mel, A = ^ = m \/ 2 . 
Ad 5 -%%/ ^ 
But Mo^ 5 = 1 8 0 — 14.4^ + AdfrM) = 1 8 0° — 3 5° W-Ad 5 M. 
= 1 44° 41/— Ad r M. 
Hence sin Ao-^— cos (90— Aop7 5 ) = cos (90 — 144° 44'.+ Ad 5 M). 
' = cos {Ad 5 M- 54° 44'). 
-r, , • i • -i a i A. 0-i sm Ad,M 
Put in triangle Ao-m*, -—4==— , 
1 5 Ad. sm AoM. 
Therefore 
. A 7 sin Ad AM . 7 sin Ad,M 
Ao- l =Aa p - — — -- -- = A , n i - 'i/v 
5 sm Ao 1 c? 5 cos (Acl 5 M— 54 44 ) 
__ . , sin Ad 5 M 
~~ C 0 cos Ad 5 M cos 54° 44' + sin Ad 5 M sin 54° 44' 
_ . -j tan Ad 5 M 
~ 6 5 cos 54° 44' + tan Ad 6 Jf sin 54° 44' 
But A(L = ^?== -J_ an( d t an Ad~M=m\/d . 
5 2 V2 5 
