of Crystalline Combinations. 397 
pyramid gRm, and which has the edges of its base passing through 
the points 4, B, C, from which the derivation of gRm proceeds. 
Or, we may reduce the axis and the linear dimensions of the trun- 
cating pyramid to one half of these dimensions, and its form will 
remain the same. Hence, its base being that of the fundamental 
pyramid (in a transverse position, R’), its axis will be half that of 
3m—-1 
the derived pyramid gRm. Now the axis of gRm is qa; 
‘ pp Bias hatred 3 d 2BM—1 
. pa=5——— 9a, and p=——— 
This, .if,.m— 3, p=2¢; if m=\, p=—2 93 if m=2, p= 29. 
II. 5. Let pR and qRm be the forms. In order that pR 
may truncate obtuse edges of gRm, the pyramid pR must be equi- 
valent to one which has the edges of its base passing through the 
_ points S, and its axis equal to that of the truncated pyramid OV’ 
(Fig. 1.). Now, diminish the linear dimensions of pR in the ratio 
b(3m—a) | 
of OS : OK, or of : : id (see H). Its base then becomes 
am+1 2 
3m+1 
2(3m—1) ae 
that of the fundamental form R, and its axis becomes 
3m-—1 
the axis of gRm. But the axis of the pyramid gRm is ga 
3m+1 
4 
3m+1 
A 
m=2,p=iq; if m=3, p= 49; if m=5, p=4g. 
Il. 6. Let pR and gRm be the forms. In order that the 
more acute terminal edges of the pyramid gRm, may coincide 
with the edges of the rhombohedron pR, the axes of the two forms 
must be equal ; 
q. Thus, if 
(see G). Therefore, pa= ga, and p= 
3m—1 
32 wet 
‘ 3m—1 
 pPa=—>— 94, and p= 
3E2 
