From the first of (30), placing No = a/(l - e^ sin^^o) '/^ and solving for sec^(/)o find 



sec^^o = (1 - e' + k')/(l - e') (1 - kVe'). 

 With the value of Ng sin 00 from the first of (30) placed in the second find 



N sin (fi = (ak/e\/l - e^) cos d and with N = a./y\- e^ sin^0, solving for sec^cj) find 



1 - e' + k^ cos'd 

 sec^0 



(1 -e^)[l-(kVe')cos^d] 



By differentiating N sin cji = (ak/e\/l - e^) cos d obtain 



(N sin <^)'d<^ = -(ak/eVl -e') sin d 5d 



R cos (f) / \ 1 . 



Since (N sin 0) '= , equation (33) may be written 



1-e^ 



R cos <^ 

 1 -e^ 



- (ak/e\/l - e^) sin d Sd or finally 



(R^ cos' 0d0') = (1 - e')a' (kVe^) sinM Sd^ 

 low from (31) and (32) find 



(kVe^) sin'd 



sec 00 ~ sec 



(1 - e') (1 - kVe') [1 - (kVe^) cosM] 

 and the numerator of (29) becomes 



(1 - e') sec'<;6[0- " e') sec'0o + 2e'] + e* 



1 - k' + k' cosM 



(1 -kVe^') [1 -kVe') cosM] 

 With the values from (34), (35), (36) the linear element (29) becomes 



1-k' + k'cosM 



(1 -e')(l -kVe')[l-(kVe')cosM] . ^ _ ^,^ 



(1 - kVe') [1 - (kVe^) cosM] (kVe^) sinM (1 - e')' 



a='(k7e') sinM Sd' = a'(l - k' + k' cos'd) 8A\ 

 ds' = a'(l - k' sinM) Sd'. 

 Now equation (37) is the usual elliptic integral form with modulus k, and we write 



d. 



f 



(1 - k' sin'd) '/' Sd, 



where k = (e yj \ - e'/a) Nq sin 0(, , the modulus of the elliptic integral, and 



dj = cos"~^ (Nj sin 0i/No sin <jSo), dj = cos^' (Nj sin c^j/Nq sin <f)^. (k is equal to Co the 



eccentricity of the great elliptic arc — see Figure 15). 



The integrand of (38) may be expanded by the binomial formula and integrated term by 

 term to obtain an approximation formula for direct computation. To 6th order terms in 

 k: (1 - k' sin'd)'/' = 1 - Kk' sin'd - (l/8)k' sin-d - (l/16)k' sinM -. 



(31) 



(32) 



(33) 



(34) 



(35) 



(36) 



(37) 



(38) 



(39) 



51 



