3.5*2 Mr. Smith on finding tin-Latitude by the Altitudes of 'two Stars. 

 Example. 1. 



Given. The altitude of Aliath =53° 25' 



The altitude of y Cassiopeia? =32 41 

 Required. The latitude of the place. 



Zen. dist. of Aliath 36° 35' Decl. Aliath... = 56° 55 



Zen. dist. of Cassiopeia? 57 19 Alt. Aliath ... = 53 25 



Constant arc 63 20 2 ) 3 30 



2)157 1* Half difference = 1 4-5 

 Half sum 78 37 



First remainder 4-2 2 sine = 9*82579 



Second remainder 15 17 sine = 9*42093 



Log. in table =19*78592 



2)3903264 - 



Half sum of the three logarithms 19*51632 



Half diff. decl. and alt. of Aliath 1° 45' sin e 8*48485 



Tangent of an arc 84° 41' =11*03147 



Half sum of the three logarithms =19*51632 



Sine of the above arc 84° 41' = 9*99813 



Sine of half co-latitude 19° 15 = 9*51819 



2_ 



Co-latitude ... 38 30 

 Latitude 51 30 north. 



Example 2. 



Given. The altitude of Aliath 42° 20' 



The altitude of y Cassiopeia? 57 48 

 Required. The latitude of the place. 



Answer. Latitude 63° 42' north. 



Note. — The constant arc 63° 20', is the distance between 

 the two stars, and the logarithm in the table is the sum of 

 the log. cosecant of that arc, and the log. cosine of the decli- 

 nation of Aliath. 



It may be necessary to add, that the star Aliath is the first 

 star in the tail of the Great Bear ; and that y Cassiopeia? is the 

 centre star of five bright ones in that constellation, arranged 

 in the form of the letter W. Their right ascensions and de- 

 clinations for 1825, are as follows: 



Aliath M 12 h 46 m 17 s Decl. 56° 54' 42" N. 



y Cassiopeia? M. 46 12 Decl. 59 46 3 N. 



1 remain, gentlemen, 



Your most obedient servant, 

 Nov. 9, 1824. M. Smith. 



P.S. 



