1829 .] 
Occultations and Eclipses. 
275 
Cot. L=0, 27433 sin. L = 9,67161 
Cos. P— 9,59077 tang. P = 0,37347— 
a =36° 14' 30" tang.=9, 86510 sin a = 9,77172 cos. ar. co. a = 0,09333 
A' = 71 58 12 
A ’— a = 35 43 42 sin. ar. co. — a) = 0,23363— cos. (A' — a) = 9,90945 
V = — 67° 19' tang, = 0,37882— A=28° 12'sin.= 9,67444 
For the altitude 28° 12' and horizontal parallax 56' 50", p e= 50’ 28 r . 
log. p = 3,48116 3,48116 
sin. V = 9,96504— cos. V = 9,58618 
p sin. V=— 0°46 34'' =3, 44620— 
5= 79 35 17 
D sin. V= 78 48 43 
* = 79 18 00 
p cos. V= 0°19' 28= 3,06734 
A —71 58 12 
A + p cos. V=72 17 40 
A =72 11 32 
a= + 29 17 
y=— 6 08 
The difference of Polar distance (y) being small, I conclude that there will be 
an occultation ; and proceed to repeat the calculation for an instant an hour later 
than the preceding, the moon’s apparent AR. being less than that of the star. 
©’s AR at 4 1 ' 3 m G. T. = l 11 08"> 54* 
App. time at place of observation = 9 38 00 
M =10 46 54 
2) = 5 20 38 
P— — 81° 34' = — 5 26 16 
Col L = 0-27433 
Cos. P = 9- 1 6631 tang. P= 0,82897— 
a= 15° 25' 20" tang. = 9 44064 sin. a = 9,42476 
A '=71 57 00 
A'— a = 56 31 40 
sin. co. =0,07876 
sin. L = 9,67161 
cos. co. a = 0,01592 
cos. (a'— a) = 9, 741 57 
log. p=3, 51825 
sin. V =9, 95745— 
» sin. V= — 0° 49' 50"=3, 47570— 
])= 80 09 25 — 
D+p.sin.V= 79 19 35 
*= 79 18 00 
s>= — 1 35 
V=— 65° 03' tang = 0,33249— A= 15° 35' sin. =9,42910 
p = 54' 58" 
3,51825 
cos. V=9,62513 
«. cos. V= 0°23' ll''=3, 14338 
A'= 71 57 00 
A +p. cos.V = 72 20 11 
A =72 11 32 
y=— 8 39 
which IT = horizontal parallax and N = true zenith distance. Two terms of the 
series are amply sufficient for the present purpose. 
