20 rr. s. coast and geodetic survey. 



at the points M and M', respectively, since the projection 

 is rectangular. This angle is evidently equal to dd. 

 By differentiation we obtain 



jde \ , 



sec^ ~2~ ~ 2 ^^^ ^ *^' 



since X is a constant for a given meridian. 

 Hence 



,„ X cos (p dip 

 dO= :^ 



1+^smV 



The radius of curvature of the meridian, denoted by pg 

 is given in the form 



^^ dd X cos (f 



By substituting the values oi-j-y -j-i and cos B and reduo 



ing, we find 



X^ X^ 



a [1 - e2 + (1 - e2)-j sin2 <^ + -j cos^ <^ (1 - e^ sin^ ^)] 



Pg^ - . 



X cos (p (1 — e^ sin^ ipY'^ 



The magnification of area becomes 



Z= 



/cosec^v? eMl + cos2(^] l-e^sin^(^ \ sm d 



( -^^ — ^ —^ — ?— ^ ^ — ^ — cot^ (p cos ^ JrTT;, ~ • 



V^ 1 — e^ 1 — £2 1 — e^ ^ /X sm ^ 



But 



l--^^m^ip 

 cos = 



and 



4 



. ^ X sm ip 

 sin = 



l+^%in2^ 



4 



