160 Astronomical and Nautical Coikctions. 



V SI 11" 27^ 43^87 AR. N. P. D. 



Or, 171° 58' r,2 Decl. 9 20,61^ 



iii. Find by inspection of the moon's right ascension, the 

 hour in which the occultation is likely to happen, and com- 

 pute, by means of first and second differences, the moon's 

 change of right ascension and of declination for that hour. 



Example. In the Connaissance des Terns for 1822, the right 

 ascensions stand thus : 



7 M 161° 10' 12" 1 Diff. a Diff. Mean. 



8 N 166 52 41 "^ ^° ^^' ^^" — 6' 13" 



M 172 28 57 ^ 36 16 ^ ^^ —5' 13" 



9 N 178 59 5 32 2 



Hence it appears, that the occultation will happen about 

 11 o'clock on the 8th : the change in an hour is 28' 1",3 be- 

 sides the second difference, which at 11" is 12", leaving 27' 

 49",3 for the change from 11 to 12. The declinations are, 



7 M 6° 30' 22" * ^^^' ' ^^^' *'**" 



8N 3 32 35 N 2° 57' 47" --0' 5"^^^,^ 



M 34 53 2 57 42 ^ ^^ 



9 N— 2 21 6 ^ ^^ ^^ 



Consequently we have 14' 48",5 for each hour, and 2",1 for 

 the second difference at 11", leaving 14' 46",4 for the change 

 in the last hour. 



iv. To the proportional logarithm of the hourly change of 

 right ascension, add the logarithmic secant of the declination ; 

 subtract the sum from the proportional logarithm of the 

 change of declination, and find the angle of which the dif- 

 ference is the tangent, for the polar orbital angle, agreeing 

 nearly with the angle appropriate to the star in the second table. 



