74 Crystallographic Examination of Eremite. 
The intersection of 6’ (2P’2) with e’ ( Pn’) is parallel to the 
basal section of 2P’2 (apparent in the crystal, though not in the 
figure, a perspective representation of it;) hence n’=2 (¢ 84, 1) 
and Ci — cogs 
Thus all the expressions for the planes of this crystal have been 
determined without a measurement. If the intersection of 0’ with e 
were not apparent in the crystal, it would be necessary first to de- 
termine e’ by measuring the interfacial angles M:e and Me’; 
these angles 136° 35’ and 117° 51’, diminished by 90° give the 
angles X in the two forms e (Po ) and e’ ( wP/n’;) and then since, 
tan 46° 35/=2 tan 27° 51’, it follows that n’=2 and e/= wP’2. 
Thence since the intersection of 6 with e’ is parallel to the basal 
section of 6 (mP’m), 6=2P’2 (§ 84, 1) as before found. The same 
might have been similarly determined by tneasuring the inclination 
of P on e and e’. 
For the determination of 6 and 0 of fig. 3. we observe that 0, e, 
M, and 0 form parallel intersections with one another, which inter- 
sections are parallel to the orthodiagonal edge of 6 and 0; therefore 
d6=mPm and 6= —m/Pm’. Again, 0’, € and the opposite e, form 
parallel intersections. Introducing therefore in the general equation 
for the parameters of planes forming parallel intersections, (¢ 28.) 
1, », —1 for m,n, r 
m;, Bane Moran? ents 
De beiood for im’ 0/40 
mT m ° 
we obtain 2 ee But we have already determined that n=m, 
1 
hence m= and m=; accordingly 6 =2P2. Inthe same 
m— 1 
manner it is found that o= — 2P2. 
For the calculation of the dimensions and angles of the crystal 
we have as data, aPoa : Po =140° 40’, wmPo : —Po =126° 
8’, oPm : oP=136° 35’. 
180° — 140° 40’=39° 20/=/’ (Min. App. pp. 67. 68.) 
180° — 126° 8’=538° 52/=u 
136° 35’— 90° =46° 35’=X in oP. 
Since Poo and — Poo are coordinate forms, we may detonate ¥ 
2 sin @ sin “/ 
sip (“@— “’) 
7=002 14—=C. 
by the equation, tany="= whence we obtain 
