14,2 Dr, Robertson’s demonstrations of the late 
Demonstration of the second rule. 
It is evident that the circle of latitude for any star in EA<? coin- 
cides with EAtf, and therefore in Fig. 1,2,3, (PI. VI. ) in which 
A represents the equinoctial point of aries, the longitude of 
such a star is 0. Now in Fig. 3. let S be a star at the inter- 
section of the arcs Ae, pH, and in this case, SAL in the first 
quadrant is equal to 6=90°. Again in Fig. 3. let S be a star 
at the intersection of the arcs EA, PH, and in this case SAC 
in the fourth quadrant is equal to B= 9 o\ In Fig. 6. (PI. VI.) 
let S be a star at the intersection of the arcs eA,pD, and accord- 
ing to the rule, SAL in the second quadrant is equal to B=90°. 
Lastly, in Fig. 6. let S be a star at the intersection of the arcs 
EA, PH, and according to the rule, SAC in the third quadrant 
is equal to B =90°. It follows from these circumstances, that 
if B be equal to 90°, the star must be in EA<?, and therefore 
that its longitude must be either o or 180 0 . 
Demonstration of the third Rule. 
Let S be a star in Fig. 2. between the arcs Ap, Ae , and then 
it is evident that its right ascension H is in the first, but its 
longitude F is in the fourth quadrant, and that SAL=B is 
greater than e AL or 90°. Again let S be a star in Fig. 2. be- 
tween the arcs EA, PA, and then it is evident that its right 
ascension H is in the fourth quadrant, but its longitude F is in 
the first, and SAC, which is equal to B, is greater than EAC 
or 90°. In Fig. 5. (PI. VI.) let S be a star between the arcs 
A e, Ap. Then H the right ascension is in the second qua- 
drant, but F the longitude is in the third, and SAL, equal to 
B, is greater than e AL or 90°. Again in Fig. 5. let S be a 
star between the arcs EA, PA, and then it is evident that H 
