50 U. S. COAST AND GEODETIC SURVEY. 



we know that the center hes upon YY\ The circle 

 which represents the parallel of latitude — a has an infinite 

 radius with center at infinity on the line YY'; it is there- 

 fore a straight line perpendicular to YY\ The lower 

 point at which the parallel crosses the central meridian is 

 given by 



_ g(cos g — COS <p) 

 ^^ ^ sin a+sin (p 



This takes the form 0/0 for (p= —a, and the limit must 

 be determined for this point. 



,. a (cos a — cos ^) T a sin 0? 



lim -^^ ; — -. — ^= lim ^=— a tan a, 



^=-a sma-f-sm^ ^=_aCosv> 



or, otherwise, 



a(cos a — cos ^) ^ 1 / x 



— : ; — : = atano \<P^<^)i 



sm a + sm <p 2 ^ ^^ 



which for <p= —a becomes — a tan a. 

 The straight line parallel, therefore, conicides with the 

 line of centers for the meridians; and hence must be the 

 perpendicular bisector of pp\ It is the line RB^ drawn 

 in the figure. 



In figure 13 the details of the construction of the merid- 

 ians are given, p and p^ are determined in the same way 

 as in figure 12. To determine the coordinates of p and 

 of p', we set;; a; = in the equation of the meridian and 

 solve for y. We thus find that 



y = —a tan a ±a sec a; 

 therefore 



Ep= —a tan a + a sec a 

 and 



Ep' = —a tan a — a sec a. 



The middle point of pp^ is given by 



^{Ep + Ef)=-a tan a. 



The perpendicular bisector of pp^ is, of course, the line of 

 centers of the meridians, since they must all pass through 

 the poiQts y and p^ and they thus have j>2?' as a common 

 chord. This liae of centers is the hne RR' in the figure. 



