From (23) one has then 



- A sin A A. + B cos AA = [(A' + B') - C (1 - e')' tan' 0] '''' and this value placed 



(22) gives 



C(l -e') sec' (hid> 



(dAA)= ■ , 



[(A'+B^)-C'(l-eTtan'<^] V' 



whence, by means of relations (20) and trigonometric identities, 

 C'(l -e')' secVd^' sec^^d,^' 



(dAA)' 



A'+B'-CMl -eTtan'cS A' + B' 



C'(l -eT 

 sec^c^d^ sec''(^d0' 



tan'(;6 



tan 0n-tan 



sec cDn ~sec 



Now the linear element of the spheroid is, [8] page 62, 



sec'vSd^'+ [—] (dAA)' 



R' cos'^, 



1-e' 



vhere R = a(l - e')/(l - e' sin'0)^/' = ^^ N^ N = a/(l - e' sin',^)"/' 



(24) 



(25) 



(26) 



Now from (25) and (26) it is seen that we will be able to express the quantity in brackets 

 in terms of sec <f> and sec (^^ since 



/N\' (l-e'sin'<A)' [(1-e') sec'(i + e'? 



,R/ (1 - e')' 



(l-e')'sec 



(27) 



^N\ 



With the values of (dAA)' and ( — j from (25) and (27), the linear element (26) may be 



\R/ 



be written 

 ds' 



[(1 -e') sec' <b + e 



(R'cos'<jSd0'). 



(1 - e')' (sec'<j!)|j - sec 



If the quantity in brackets is given a common denominator, then (28) may be written as 



J , (l-e')sec'0[(l-e')3ecV„+2e'] + e^ 



ds = — (.n cos 0d0 ) . 



(1 -e')'(sec'0„-sec'0) 



e^l-e' 

 To bring (29) into manageable form we place k = N,, sin cj)^ , and 



d = 



N sin (f) 



No sin 00 

 (Note that k = Co, is the eccentricity of the great elliptic arc. See Figure 15.) 



(28) 



(29) 

 (30) 



49 



