MATHEMATICAL PAPERS. 33 



may easily find N ^ and Ns, and NM and Nx; and con- 

 sequently, obtain the spaces which Mercury would have run 

 through between the external and internal contacts, had his 

 semidiameter been equal to that assumed. These spaces are 

 5 s and Mx, each ^ 11". 242; but the spaces run through 

 by observation, were not so great j therefore, 5's semidiam- 

 eter by observation, was less than that which we have as- 

 sumed, and may now be exactly obtained, i?'s by observa- 

 tion, was 10". 594, therefore, 11". 242 ; 5" : : 10". 594 : 4". 

 712. Mx by observation, was 11". 061 ; therefore, 11". 

 242 ;. 5" : ; 11". 061 : 4". 919. Erom these results it nppear 

 that 5's semidiameter by the observation of the two first 

 contacts, was 4". 712 ; and by the observation of the two last, 



9 



4". 919 J the mean of which is i'\ 815; which we may use 

 as the semidiameter of 5 , in the calculations which follow. 

 For 2's geocentrick latitude at the time of the external 



contacts : 



We have already found 5 GU = N OD=i8°19' 54". 5, 

 and ^ N, the visible transit line = 1732''. 39, the half of 

 which, viz. 866". 195, is = 5? N or M N ; also $ 's semi- 

 diameter by observation = 4". 815, which added to G's re- 

 duced semidiameter, 967''. 665, will give 972". 48 == O 5 : 

 Therefore, 5 N 972". 48 : Radius : : ?^ N 866". 195 : Sine < 

 SON 62° 57' 44". 2. From this substract < N O D, 

 8° 19' 54". 5, and the remainder, viz. 54° 37' 49". 7, will be 



< i^ O D ; the complement of which angle, viz. 35° 22' 

 10". 3 will be = < 5 O C; because D O C is a right angle. 

 Having found < 5 O C, we may say, Radius : O 5 , 972". 

 48 : : Sine < 5 O .C, 35° 22' 10". 3 : C S , 562". 917 



E 



9' 22". 917 



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