On the Drawing of Figures of Crystals. 33 
position for projecting it. In the following demonstration the an- 
gle of revolution is designated 5, and the angle of the elevation of 
the eye, e. Fig. 1. represents the nor- 
mal position of the horizontal axes, sup- 
posing the eye to be in the direction of 
the axis BB; BB is seen as a mere 
point, while CC appears of its actual ¢; 
length. On revolving the whole through ¢ 
a number of degrees equal to BMB’ (°) 
the axes have the position exhibited in 
the dotted lines. The projection of the 
semiaxis MB is now lengthened to BUG 
MN, and that of the semiaxis MC is 
shortened to MH. Since the angle HMC=MB’N=s, MH= 
cos and MN=sin 6; and if the ratio of the projected axes be as 
r 31, MH: MN::cosd: sind::r 31; 
-“.cosd=rsin 0, 
and consequently, cotd=r. 
If the eye be elevated, the lines B/N, BM and C’'H will be pro- 
jected respectively below N, M and H, and the lengths of these pro- 
jections (which we may designate 6’ N, 6M and c’H) will be directly 
proportional to the lengths of the lines B/N, BM and C'H. Now 
B’ N= cos), and 6M, the projection of BM, = tan«; consequently, 
BM(=1) : tane ::cosd (B/N) : b’N (the projection of B/N). 
Hence, b/N= tan € cos, 
In the same manner we find c/ H=tané sind. 
Fig. L. 
If the relation of b’N to MN(=the first projection =sin 5) equals = 
1 
tan e cos d (b’N) =Fsin 9d ; 
consequently, tan cot i=* : 
and finally, since coté=r, 
cotée=7s. 
6. The preceding demonstration affords the following simple 
method of projecting the monometric axes; r is supposed to be 
given equal to 3, and s equal to 2. 
1. Draw two lines 4.4’, H’H (fig. 2.) intersecting one another 
at right angles.) Make MH=MH’=). Divide HH into rr parts, 
and through the points, N, NV’, thus determined, draw perpendicu- 
Vou. XXXITL.—No. 1. 
