On the Drawing of Figures of Crystals. 43 
_parts on the edge €, and ap (4 of é) will equal 1 part on é, agreea- 
bly to the expression 4P2; npd is therefore the plane 4P2. The 
Fig. 9. 
perimeters of the planes npb (4P2) and nmo (P) intersect one anoth- 
er in the points m and «; consequently the line of intersection, be- 
tween these two planes must be situated between these points, and 
therefore the direction of the intersection of P and 4P2 is no. 
The planes nmb (2P) and npb (4P2) intersect in the line nb, 
and therefore the intersection of 2P and 4P2 is in the direction | 
of nb. 
Again, the intersection of P and 2P has the direction mn. 
We may next lay off the plane 2P2, (2: 2: 1,) which may be 
constructed by marking off 2 parts on each e and é and 1 ‘part on e. 
Such a plane.is mro, since aa=2e, am=26é, and ar=1e. There- 
fore the intersection of mro (2P2) with mno (P) has the direction 
of the common line mo. 
The perimeters of the planes mro (2P2) and npb (4P2) intersect 
in the points « and 8. If therefore these planes formed an edge of 
intersection it would have the direction of the line of or ro. 
The line ro of the plane mro (2P2) is parallel to nb, of the plane 
mnb (2P); the intersections of 2P2 and 2P would therefore, be par- 
allel with these lines. In this manner all the mutual intersections 
of these and other planes may be obtained. 
Fig 10 exhibits these planes in their respective positions as above 
determined. ‘The planes may be lettered as in the figure; mno=a, 
mnb=a’, mro=0, npb=0'. The edges a: P anda ; a’ were made 
parallel with mn (fig. 9.) The intersection of P with 6 has the di- 
