430 
ON THE DETERMINATION OE THE 
[476 
Article Nos. 66 to 82. Planogram No. 1, the Meridian 90°—270° (see Plate II.). 
66. Supposing that the orbit-plane rotates about the axis >81 (fig. 6, see No. 64) 
in the plane of the ecliptic, the orbit-pole will describe the meridian 90°—270°, the 
position of the orbit-pole being 6 = 90°, c = 0 C to 90°, or else ¿ = 270°, c=0° to 90\ 
But the same analytical formula extends to the two half meridians, viz., we may take 
h = 90°, and extend c over 180°, in the final results making c an arc between 0° and 
90°, and b = 90°, or =270°, as the case requires. 
67. Assuming then b = 90°, we have 
a , ß , 7=1, 0 , 0 , 
a' , ß', y = 0, cos c, — sin c, 
a , ß , 7=0, sm c, cos c, 
and, moreover, x', y' = x, y: so that instead of (a?/, yi), &c., we may write at once 
(x 1} 2/j), &c. The formulae become 
Xi : y x : 1 = V 3 cos c + sin c : 0 : sin c + V3 cos c, 
x 2 : y 2 : 1 = V3 cos c — 2 sin c : — 3 : sin c — 2 V3 cos c, 
x 3 : y 3 : 1 = V 3 cos c — 2 sin c : 3 : sin c — 2 \/3 cos c, 
that is 
= 1, 2/i = 0, 
(viz. the orbit-plane, as is evident, meets the ray 1 in a fixed point, its intersection 
with the plane of xy)\ 
V 3 cos c — 2 sin c 
-3 
sin c - 2 V3 cos c ’ 
and writing 
2V3 
1 
—7= = Sin CO, 
V13 
—^ = tan co, 
S) A/0 
V13 
= COS <0, 
2 v 3 
(whence &) = 16°6') we find 
and we thence have for the hyperbola, the locus of (x 2 , y 2 ) and {x 3 , y 3 ) 
(«+&>■=*(/-*).