( 1043 ) 
The extinetion observed under these circumstances amounts with 
regard to the trace of m,(m,) to w= — 6°. The plane of the optic- 
axes OA lies // (010); be further the angle ¢:¢ = 10°, then the 
axial-angle can be calculated. 
For the poles s, and s, one finds by calculation the following 
coordinates with regard to the plane-systems OA, / (= projection- 
plane) and OA, C (1 bisectrix €): 
— ABG = o'e = 108°15'38" ALI =,= 71°44'22" 
Gs, —=0e—= 44°35'28" ls, =0',—= 44°35'28" 
J Ats, = ple = 116°48'13" Z Acs, =o" = — 81°46'2" 
~ Hs, =0',= 40°44'20" _ Ms, =o".  46°53'35" 
For the pole 7 of the plane m, one finds from u, = ~ ABDT = 
—PBDa ai AT ea bs in 8 by calculation U 2 7 ot 28° 16’ oe 
rv, = —- 8°49'35". The angle between the trace of m, and the fictive 
trace of Cis according to (5) for the secant-plane S, 
cos 6', tg Pe — sin 6'¢ cos (Q'¢ — Ue) 
cot h, == TS 
sin (0 -— Ue) 
cos 40°44'20" tg 8°49'35" — sin 40°44'20' cos 88°31'2 1" 
sin 88°31'2 1" rey 
from which ‚== 84°14'20" 
and for S, by: 
== (as 40. Go aoe ty 8°49'35" — sin 46°53'35" cos 69°57'6" 
ON are a TT <a 
ji sin 69°57'6 
from which h, = — 69°13'20", 
so that according to (6): 
; Uu = Ye —h 
one finds for S, (Ilrd octant): 
Ye =U Ari 02 + 84° 14'20" = 78°14'20" 
and for 5S, ([V octant) 
yp =u dh, = — 6° — 69°13'20" — — 75°13'20". 
From the diagram fig. 4 it can immediately be seen, that y for a 
secant-plane in the [V& octant can never become <0 and conse- 
quently JS, does not correspond. 
Now is according to (2): 
1 — sin? V cos? 9 — (1 — sin® V sin? v) sin? O 
int © we, Ns \ 
coe 2y = = ——— en - = 
sin 2u sin 6 sin? V 
sin? V (cot 2y sin 2 sin 6 + cos? vy — sin? y sin? 6) = 1 — sin? 6 
pit. cos” 6 
sin? a = — en 
cot 2y sin 2 sin 6 + cos* pg — sin® vy sin? 6 
(10) 
