Thus it is found that the first four terms of (24) are 



e' sin 61 cos (9 + Vie" sin' cos + {3/8)e' sin= 6 cos + (5/16)e' sin' cos d; 



e" sin cos 6*+ (2e'- 26") sin' 6 cos 6 + (3e' - 3e' ) sin= (9 cos - 4e ' sin' d cos 6 ; 



c" sin cos + (5e'-"/6 e') sin' (9 cos 6 + CVee' - '' /,e^) sin' cos (9 + "/.e' sin'0 cos 6; 



e' sin cos - 12e' sin' cos 6* + 30e° sin' d cos 6 - 206° sin' 6 cos d. 

 Adding corresponding terms of these we have 



A(?i = vS - ^ = (e' + e" + e' + e') sin 61 cos 6 - [(3/2)e'' + (23/6)e' + 7e' ] sin '0 cos d (31) 



+[(77/24)e' + (55/4)e'] sin' d cos 6* - (127/16)e" sin' 6 cos (9. 

 Now sin 61 cos = V2 sin 2d 



sin' cos = % sin 20 - (1/8) sin 40 



sin' cos = (5/32) sin 26 - (1/8) sin 40 + (1/32) sin 60 (32) 



sin' cos = (7/64) sin 20 - (7/64) sin 40 + (3/64) sin 60 - (1/128) sin 80. 



The values from (32) placed in (31) give finally 



(f) = (p - 6 = C^ sin 20 + Cj sin 40 + C3 sin 60 + C4 sin 80 

 where C^ = 'Ae' + (1/8) e" + (ll/256)e' + (31/1024)e' (33) 



Cj = (3/16)e'' + (5/64)e' + (25/1024)e' 

 C3 = (77/768)e' + (59/1024)e% C, = (127/2048)e\ 

 Again for the Clarke 1866 spheroid 



e' = 0.006768657997, e" = 0.00004581473108, (34) 



e' = 0.0000003101042459, e' = 0.000000002098989584, whence from (33) 

 Ci = 3.390069228 x 10"', C^ = 8.614540216 x 10"% (35) 



C3 = 3.12121 X 10-% C4 = 1.302 X 10"°. 

 We now check (33) directly from the maximum value of A(f>, the assumption being that if it 

 holds for the maximum it will hold for all A<^. 



From (6) = 44° 51' 151' 851, whence 

 sin 20 = 0.99998708, sin 40 = 0.01016441, sin 60 = - 0.99988377, sin 80 = - 0.02032777. (36) 



With the values from (35) and (36) find 



C, sin 20 = 0.0033900254283 C3 sin 60 = - 0.0000000312085 



Cj sin 40 = 0.0000000875617 C4 sin 80 = - 0.0000000000026 



0.0033901129900 -0.0000000312111 



AqS (radians) = 0.0033900817789 



A0 (seconds) = (0.0033900817789) (206,264.8062) = 699."2545611, 

 or Av!)jjj^^= 11' 39:'255 which checks (6). 



18 



