36 On the Drawing of Figures of Crystals. 
MA” equal to a, and MC’, MC” equal toc: if c=1, make MB”, 
MB’, equal to 6. By connecting the extremities of the axes, as 
already explained, the rhombic octahedron may be constructed. The 
rectangular prism may be projected, in the same manner as the cube ; 
the rhombic prism in the same manner as the second square prism 
just described; and the rectangular octahedron, in the same manner 
as the second dimetric octahedron explained in the last section. 
10. Monoclinate System.—The axes a and 6 in the monoclinate 
system are inclined to one another at an oblique angle=y. To pro- 
ject this inclination, and thus adapt the monometric axes to a mono- 
clinate form, lay off on the axis MA, 
Ma= MA cosy, and on the axis 
BB' (before or behind M according 
as the inclination of 6 on a, in front, 
is acute or obtuse) Mb=MB xsiny. 
From the points 6 and a, draw lines 
parallel respectively with the axes q 
AA’ and BB’ and from their inter- 
section D’, draw through M, D'D, 
making MD=MD’. The line DD’ 
is the front lateral axis, and the lines 
AA, C’C, DD’ represent the axes 
in a monoclinate solid in which 
a=b=c=1. The points a and b and the position of the axis DD’ 
will vary with the angle y. The relative values of the axes may be 
given them as above coer: that is, if 6=1, lay off in the di- 
rection of MA and MA’ a line equal to a, and in the direction - 
MC and MC’ a line equal to c, &c. 
The right rhomboidal prism may be projected in the same man- 
ner as the cube or right rectangular prism, and the oblique rhombic 
prism, in the same manner as the right rhombic prism. 
11. Diclinate System.—In the diclinate system, the vertical sec- 
tions through the horizontal axes intersect one another at right an- 
gles, as in the preceding system, but the inclination of a to (7) and 
a to c(#) are each oblique. This obliquity may be given the mono- 
metric axes as follows: Lay off on MA, (fig. 4,) Ma=MA xcosy, 
and on the axis BB’ (brachydiagonal), M6= MB’ xsiny. By com- 
pleting the parallelogram MaD’8, the point D’ is determined. Make 
MD=MD’; DD‘ is the projected brachydiagonal. Again lay off 
on MA, Ma’=MA xXcos6, and on MC, to the left, Mc=MC’ x 
sing. Draw lines from a/ and ¢ parallel to MC’ and MA; E’, the 
