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 ee @P’2: 
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 (Poo ) and e’ ( mP/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 (4 84, 1) as before found. ‘The same 
might have been similarly determined by measuring the inclination 
of P on e and e’. 
For the determination of 6 and 6 of fig. 3. we observe that 6, e, 
M, and 6 form parallel intersections with, one another, which inter- 
sections are parallel to the orthodiagonal edge of 6 and 6; therefore 
_6=mPm and 6=~-m/Pm’. Again, 6’, é 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 
93/0 Torin’, Wir; 
1, 3, @elor-an’, 2“ sid 
. m . 
we obtain n=~_]: But we have already determined that n=m, 
hence m=— j and m=2; accordingly 6 =2P2. In the same 
manner it is found that 6= —2P2. 
For the calculation of the dimensions and angles of the crystal 
we have as data, mPa : Pw =140° 40’, wPm : —Po =1 
8, owPao : wmP=136° 35’. 
180° — 140° 40’=39° 20’=.’ (Min. App. pp. 67. 68.) 
180° — 126° 8’=53° 52/= 
136° 35’— 90° =46° 35’= Xi in oP. 
Since Po and —Po are coordinate forms, we may determine 7 
ae 2 sin # sin pe! 
by the equation, tany = sin (@=W) 
‘ 7=16° 14’=C. 
whence we obtain 
