THE ORBITS OF THE EIGHT PRINCIPAL PLANETS. 183 



Rigorous expressions for the difference of the declinations depending on the 

 precession are not readily deducible from equations (580) and (581) of sufficient 

 simplicity to possess any advantage in computation over the original formulae, since 

 is not necessarily a small angle. 



10. Having given the necessary formulae for reducing the position of a star which 

 is given in longitude and latitude, or right ascension and declination, with refer- 

 ence to the equinox of 1850, to that of any other date, and the reverse, we shall 

 now give some examples by way of illustration. 



Example I. The mean place of polaris, at the beginning of the year 1850, was, 

 a=16 15' 22".815, 5=88 30' 34".889; it is required to find its right ascension and 

 declination at the beginning of 1950, 2050, and 2150, and also the maximum decli- 

 nation of the star. 



In 1850 the planetary precession was equal to nothing, therefore 3 disappears 

 from the second member of equations (580). For <=+100, Table X, gives 

 z=0 38' 23".482, z'=0 38' 36".527, 0=0 33' 24".811, and $= 0' 12".433. 

 Whence a-|-z=16 53' 46".297 ; and the computation is as follows, using the first 

 and second of equations (580): 



a+z 



8 



a+z 











sin 9.4633534 



cos 8.4151046 



cos 9.9808361 



cos 9.9999795 



sin 7.9876415 



sin 9.9998531rc 



cos 8 cos (a+z) cos =+0.02488400 

 - sin 8 sin =0.009716 16 



cos 8' cos (a' z' S')l=+0.01516784 

 cos 8' sin (a' z' 3') =cos 8 sin (a+z) 

 a' z' 3' =26 29' 2V.70 

 38 24.10 

 a' =27 7' 45".80 



log. 8.3959202 



cos 5' cos (a' z' 



=89 1' 44".277 



" 8.1809237 

 " 7.8784580 

 tan^.6975343 



cos 9.9518314 

 log. 8.1809237 

 cos 8.229092T 



Computing in the same way for (!=-)- 200 and <=-{-300, we shall find the values 

 of a! and <$' as in the following table : 



89 



89 



89 



1' 



27 

 28 



44". 277 

 20.654 

 13.594 



1950 27 7' 45".80 



2050 56 51 57.27 



2150 120 32 36.31 



The declination evidently attains a maximum value some time between 2050 

 and 2150. If we compute its place for (1=250, we shall find z=l 35' 59".93, 

 ^=1 36' 26".63, 0=1 23' 29".80, and 3'= 21".80. 



