IN TRUE PERSPECTIVE. 
497 
A B must always be=CD = DE — FG. In the figures 
of Professor Mohs' Treatise on Mineralogy it is = § A G. 
It is evident, that if W =: 1 R W, also X must be 
1 1 
-PX, and Q^Urr -Q U. The ratio of O N : O G is 
dependent upon that of AB : AG, and is found by draw- 
ing perpendicular lines from all the angles of the hexagon 
upon the line O G, which is situated in a plane parallel to 
the visual ray. If A B : AD = 1 : S, we obtain O N : O K 
=: 1 : 5. 
The apparent length of the lateral edges of the prism, 
the faces being squares, is found in the following way. 
Draw A B perpendicular to B D (Fig. 3.), B D being a 
vertical section of the plane upon which the figure is to be 
represented. Take B C ~ G^ (Fig. 2.), and A C 
= T D, Fig. 1 . Draw C E perpendicular to A C, and 
equal in length to P E (Fig. 1.) ; from the point E draw 
E D perpendicular to B D, the line C D will be the required 
length of the lateral edges of the hexagonal prism, which, 
applied to the projection of the terminal hexagon in Fig. S., 
yields Fig. 4. the projection of the hexagonal prism itself. 
This method of projecting a regular six-sided prism may 
likewise be conceived in analytical terms. 
From the ratio of A B : A D (Fig. 1. Plate XVI.) being 
known, it will be possible to deduce that of O N : O K. 
Let A D be = a, O K =r 5, A B = ^, O N = A; and 
n m 
PE=EI=:IS=:c. We derive the following equations: 
PE^^PBHBE^ EP = EG2h-GP; SP= IN^+NS^ or 
VOL. V. 
