“A=(I—) +15 (aa 
RS a aplas 
“y sis e 2 - ve 
on. des ' for 
ily had nearly Completed his. 
Ki lai, whom we have already 
od of so Iving the problem, which 
nd convenient and is capable of con- 
accura acy. 
Let cand h denote the same quantities as before : let 
Rose s)ot the former formula : let 
x denote the assumed difference of meridians and e the 
error ; so that we may always have c<=x+e. Then will 
the apparent time of the moon’s culmination at the west- 
ern observatory, be 
eh A )eO 
nearly. 
= Resting them as equal: and let a and 6 denote, as be- 
fore, the true right ascension of the moon at those assum- 
ea periods respectively. Then if 15A =(a—8,) the value 
of x lias been assumed correctly, and the problem is solved. 
But if not, call the difference, in this last equation, d; 
Weetee we Bball have 
fs, 15A =(a—b)+d 
y d=154—(a-6) 
« Bat di evident! tly a function of the moon’s difference 
in right ascension ; and the time (e) in which it is describ- 
ed (or the variation which it will cause in the value of x) 
will depend on the relative motion of the moon, in right 
ascension, ina true solar hour. Now, since e is generally 
a very small quantity, the relative motion of the moon, du- 
ring that short interval, may be deduced with sufficient 
accuracy from the moon’s motion in 24 hours as shown by 
an ephemeris. Whence the value of e may be expressed 
by the following equation : 
gna = xa 
m 
where ¢ may be — in all cases, equal to 244 4”, 
Von. 1X —No. 15 
