376 Ohservntions on the value of the ciphen, ^c. [Oct. 



the place, to find the angle 0 opposite to ~. By trigo- 

 nometry 



cos^ — = sm \—--~^) sin [—-^~- z ) 



sin P. sin c. 



when 0 H T' is the true time of rising. — Let y de- 

 note the time from the approximate to the true time of 

 rising: now the variation of the Moon's polar distance 

 in the time y causes an alteration in the angle 0, there- 

 fore the ^loon does not rise when the hour angle is 

 equal to 0 hut when it becomes 6 ^ d. O. hence we 

 have to determine from the variation of P. the change 

 in 6 or the value d. 0. d representing the dill"erential. 



From Trigonometry cos 0. sin c. sin P = cos z — 

 cos c. cos P. differentiating this expression we get 

 — d 0. sin 0. sin c. sinP d. P. cos. P. cos. 0. sin c = 



P sin P. cos c. 



^ d e. sin 0 ^ d P. cot P. cos 0 = d.F. cot c- 

 ^ sin 0 = P cot P cos ^ — dF cot c. 

 '.' d 0 = V. cot P. cot 0 — P. cot c. cose c 0, As- 

 suming dV = V. Make g ~ d 0. 

 Now in the interval y, the Moon's polar distance will 

 be altered by A y, and the corresponding change in the 

 angle is. : as P : g : Ai/ : : g A ?/; but the Moon ap- 

 proaches the meridian in this space by (Ih. Om. 9s, 

 86 — ^) 2/ therefore 



H — (1 0 9.86— m) y = 0 + g. a, y 

 U — 0=g. Ay-i-(l — 0 — 9.86 — m) y 

 U--0 



Hence y = 



g A -f (1. 0. 9.86 •— m) 



T' -f ?/ =^ R the exact time of rising. 

 An example wull sufficiently exemplify this method. 

 Required time of Moon rising on 24th October, 1836, at 

 Madras. 



