———————— S—<“—éCSéCt 
Crystallographic Examination of Eremite. 73 
As these crystals (especially fig. 2.) afford a fine example of the 
method of crystallographic calculation in the monoclinate system, I 
here subjoin, for the assistance of such as may be interested in the 
study of Mathematical Crystallography, the several steps by which 
the above results have been obtained. In these explanations, I 
must necessarily make frequent references, for principles, to my 
System of Mineralogy. 
We may select for this examination, figure 2. Assuming 4 asa : 
face of the fundamental form, M= wPo,P=amP’m. The gen- 
eral descriptive expressions for the remaining planes are as follow: 
é=mPo re in/P’@ 
e=-—mPo =n a 
e=mP’o e= oPn 
2——- mE e’= oPn’ 
The plane e (mP’o ) forms parallel intersections, with 4 and a, 
which intersections, since they are parallel with the edge a: M 
(Po ) are also parallel with the orthodiagonal edge of 4 (P.) 
Hence 
a= —P (§ 84, 2,) and e=P’m ($ 84, 2.) 
and also since e forms parallel intersections with a and —a, 
e— ai (yet, 12) 
é truncates the edge between 4 and a, which is the clinodiagonal 
edge of P, and therefore, 
é=Po (§84, 4.) 
é in the same manner truncates the clinodiagonal edge of —P. Con- 
sequently, 
é= —Po. (§ 84, 4.) 
The intersection of 6’ (mP’n) with a (P) is parallel to the clino- 
diagonal edge of P, (fig. 2.) consequently n=m (§$.84, 8) and 
0’=mP’m. Again 0/ forms parallel intersections with e (P’c ) and 
° . e : m 
e( oP,) and therefore its sign is of the general form mP/— ’ 
—] 
($ 84,9.) But from the above, n=m, and consequently m=——s; 
from which we find m=2 ae 
ee 
The edge 0’ ; e’ (mP’a i is ‘pare to the orthodiagonal edge of 
6’ (2P’2, fig. 2. b) consequently m=2 (¢ 84,2,) and 
e’=2P’a 
Vou. XXXILI—No. 1. 10 
