From the first and last of either sets of equations (2) find 



A , e AT . r. / e sincS cos 4> /„\ 



sm A^ = Nsin2^= ^ -I— . (3) 



2a Vl-e'sin'^S 



To find the maximum value of A<;6 and the value of ci at which the maximum occurs, one 



e^sin <f) cos 

 differentiates A(^ = arc sin ~ to obtain 



\/ 1 — e^ sin^^ 



dAoS e^ cos"" 24> + 2(2 -e'') cos 2oS + e"" 



d0 (2- e' + e' cos 2(f)) /2(2^^^^)^^^^T2^^1?^^'^^T^^~^^^^^ 



neither factor of the denominator of (4) is zero for Q <^(f) <^90°. Hence to find the maximum from 

 (4), place the numerator equal to zero and solve for cos 2<p to obtain 



cos 20 = 1 + 2 ( V 1 - e' - 1 ) /e\ (5) 



The flattening, f, of the reference ellipsoid is given by f = (a — b)/a = 1 — b/a = 1 —\J 1 — e , 

 whence e^ = 2f-f^, we can write 



cos 2<^ = 1 - 2(1 - V 1 - e' )/e' = 1 - 2f/(2f - f^) = - f/(2-f) 



sin '20 = 1 - cos'20 = 1 - fV(2 - f)' = 4(1 - f)/(2 - f)' 



• 2^ 1 1 If 1 



sin = cos 20= — + — = . 



2 2 2 2(2-f) 2-f 



l-e'sinV= 1 - f(2 - f)/(2 - f) = 1 - f. 



e" sin'20 f'(2-f)' 4(1 -f) 1 



from (3) sin 'A0 = — = 



4 l-e'sin^ 4 (2-f)' 1-f 



sin'A0 = f' 

 hence sin A0j„j,^ = f = 0.0033900753 (Clarke 1866 ellipsoid), 

 cos 20 = - 0.001697914 

 = 45°O2'55:'1O6 , 

 and A0j„a^ = 0° 11' 39':255, (6) 



(9 = 0- A0 = 44° 51' 15':851. 

 Now from (3) and = 0- A0 a complete table for corresponding latitudes can be computed 

 readily since complete tables for N to 0.001 meter have been computed for most reference 

 ellipsoids. [2] 



To develop sin A0 is a series for computation without the necessity of tables of N, write 

 n the form sin A0 = 

 binominal formula to get 



-1/2 

 (3) in the form sin A0 = e' sin cos (1 - e' sin'0) , then expand the radical by the 



e 3 5 



sin A0 = e' sin cos (1 h sin' + — e sin + — e sin 



^ ^^2 8 16 



14 



