( 1087 ) 
a It 
by the diminution of cos yg >> 0 and becomes with v = 
to 
1 — sin? o (1 — sin? V) 
cot 2y = en == + 2, 
(sin 20 — 0) sin 5 sin® V 
Krom this follows : 
ieee LOO? Or. OF 
EE OUT GOR MG) ir: eae AV OI ae) 
In fig. 2 the a-axis has been chosen as normal to the projection- 
plane; it needs no further elucidation, that the plane that halves 
Z As, here always indicates the direction of the vibration of the 
quicker ray. If the pole of the secant-plane lies in the plane $b’, 
then y,, the angle between the longer ellipse-axis and the secant-line 
Jt 
Sit becomes! == 5» So that the value y=0O= y, relates to 
the short ellipse-axis. 
if consequently the projection-plane is placed 1 to the negative 
bisectrix, one finds with the help of (2) the angle, that the long 
axis of the ray-velocity-ellipse forms with the secant-line of the 
projection-plane and the secant-plane, if with a positive value of 
cot 2y one takes the angle 24, in the 3'¢ quadrant. Then y, is likewise 
an angle in the 2nd or 4% quadrant, and has consequently, if we 
TT . 
take the value te irs a negative sign. 
If in the 1st oetant g diminishes from the point where with a 
definite value of 6 and WW 
1 — sin? o (1 — sin? V) — sin? V (1 + sin? 6) cos? po = 0 
then, in consequence of the increase of cosy, the formula (2) has 
negative values. With v = 0 becomes: 
1 — sin? V — sin? o cos” 6 — sin? V 
0 aoe cot 2y SS 5 TT Ref ns 
(sin 20 = 0) sin 6 sin? V (sin 2y = 9} sin asin? V 
Jt 
If now one assumes for o all values between 0 and zi then it 
appears, that cot 24, becomes indefinite for 
8 Jt ¥ 
cos 6 = sin V = cos 5 EN Yo} 
The pole of the second plane lies here in the optic-axis. For 
T ; : : 
6 rie — V, cos’ 5 — sin? V becomes < 0, by which oot 2y obtains 
a positive value, and 27 must be supposed in the 3rd quadrant; 
68 
Proceedings Royal Acad, Amsterdam. Vol. XIIL 
