In these formulas, Oo is the complement of the parametric latitude of the vertex of the great 

 elliptic arc. To see this, find on page 119 of Forsyth, the expression 



sin oo = (tan (;6o)/[(p sec ^(^0- 1) (p'sec ^(^0 + l)]''^ 

 where p = sin' Viid^ + d^/ sin 0^ sin 6^ (112) 



p'= cos^ Viidi + 0^/s'va. 6^ sin Q^ 

 Now replace Forsyth's 6^ and 62 by 90 — 0,, 90 - O^ respectively and his cjig by A A/2. 

 Then find: 



tan <fia = tan (AA/2) = (1 - cos A A)/ sin AA 



p sec ''^So - 1 = [(1 - cos AA)/sin^AA] (1 + sec d^ sec $2 - tan d^ tan O^) - I (113) 



p'sec ^(jig + 1 = [(1 - cos AA)/sin'AA] (- 1 + sec 6^ sec 62 + tan di tan 0^) + 1 

 The values from (113) placed in (112) give 



sin Oo = sin AA/(tan'0, + tan ^62 - 2 tan d^ tan 62 cos A A + sin 'A A)'/' (114) 



Now the right member of (114) is cos do where do is the parametric latitude of the vertex 

 of the great elliptic arc [17] . (See also GEODESICS AND PLANE ARCS ON AN OBLATE 

 SPHEROID, L. E. Ward, American Mathematical Monthly, Aug.-Sept., 1943 page 427). 

 From Ilia, 111b, 111c, 111m, llln we have, retaining terms to c^ inclusive: 



$ = ^'+ c (n +— sec Qotan Oo) (115) 



+ c'' [W +V2 sec Oo tan Oo! U + Af'(1 + csc^'ao) + (1/8) v'{l - 6 tan^ ao)W 

 If R, S are the coefficients respectively of c and c' in (115), then 



tan = tan <;6'+ c sec ^<;i' R + c' sec '^'(S + R' tan ^') (116) 



With the values of R and S from (115) and the values of ft + (i/V2) sec ajtan Ooand U 

 from lilt, cot ^'from Ills, we can write (116) as 



tan $ = tan <^ '— c A cot i/'cscaosec'c^' (117) 



+ c sec 



W + A' cot v' CSC^Co 



+ 72 sin Oosec Op 



A[i/'(1 + csc'oo) - cot v'\ 



- (1/8) sin 2v' + - (1-6 tan 'oo) 



83 



