236 
ON THE BICIRCULAR QUARTIC. 
[667 
as have also R 2 and R 2 , and R 3 ' and R 3 . We may take 8, 8 1} 8 2 , and 8 3 as each of them 
positive: the signs of 
day doy 1 day 2 dco 3 
TBvtr vE^/W,’ VOT vn,v®, are +> +> or +> +: 
hence to make dS, dS 1} dS 2 , dS 3 all positive, 
€ , Cl , € 2 , 6 3 , €1 J € 2 , e 3 > e > 
must have either the signs of 
R', -R', R', ~R 3> -R 1 , R 2 , -R 3> R, 
or else the reverse signs: hence in either case e= — e, e/ = e x , e 2 =e 2) e 3 — e 3 ; or the 
equations are 
dS = — e R' 8 = e.RA-J- 1 — , 
Vil V© Vili V©! 
eA8l v!v?¥,“ e " A V^Vl s ' 
rfSs " A 5 _ 211 la, V@ s ’ 
i ry t) / i day 3 -ir> iZft) 
do 3 = € 3 ti 3 o 3 -p=—7— =eRb —p=- 
Vil, V®, 
Vii V® 
2Vn 
24. But we have R’ — R = ———-, &c.; and hence, putting for shortness 
¿• 2 + y 2 
Si S 2 
8 :i 
(iC 2 + y 2 ) V® ’ («! 2 + 2/a 2 ) V©! ’ (# 2 2 + y 2 2 ) V® 2 ’ (% 3 2 + 2/3 2 ) V©: 
dS + dS 3 = + 2e P day, 
dS 3 — dS = — 2 e 1 P l day 1 , 
dS 2 — = — 2e 2 P. 2 day 2 , 
dS 3 — dS = — 2e 3 P 3 day 3 , 
and consequently 
dS = ePday + e 1 P 1 day 1 + e 2 P 2 day 2 + e 3 P 3 dco 3 , 
dtSj = ePday — e 1 P 1 day 1 4- e 2 P 2 day 2 + e 3 P 3 day 3 , 
dS 2 = ePday — e 1 P 1 day 1 — e 2 P 2 day 2 + e 3 P 3 day 3 , 
dS 3 = ePday — e 1 P l day 1 — e 2 P 2 day 2 — e 3 P 3 day 3 , 
which are the required formulae for the elements of arc. 
= P, P u P 2 , P 3 
25. The determination of the signs has been made by means of the particular 
figure; but it is easy to see that the pairs of terms could not for instance be 
dS—dS 3) dS 1 — dS, dS 2 — dS 1 , dS 3 — dS, or any other pairs such that it would be 
possible to eliminate dS, dS 1} dS 2 , dS 3 , and thus obtain an equation such as 
ePday + e 1 P 1 day 1 + e 2 P 2 day 2 + e 3 P 3 day 3 = 0 ;