s = 



d -Ce78) (1 + e72 + 3eV8) (Xd - Y sin d) 



(lOd + sin 2d) X' + 8(sin d) XY 



+ 2(sin 2d) Y' 



3(22d + 3 sin 2d) X' - 3(15 sin d - sin 3d) X'Y ' 



- 18(sin 2d) XY' - 4(sin 3d)Y- 



- (eV512) (1 + 3e72) 



- (eVl2,288) 



(87) 



Again if all terms above first order in f (e' = 2f) in (87) are ignored then the first two terms of 

 (87) represent the Andoyer-Lambert form as given by the first of equations (80). 



For the case where geographic latitudes, 0, are not first converted to parametric, but are 

 considered spherical, the corresponding differential right triangles are: 



a cos (^dX. 



id(f> 



N cos 0d A. 



d0 

 D5d 



We have for the approximation 

 Rd^ = ds cos a 



d(P 

 or Rd^ = ds , placing cos a = cos a 



ds = RDSd = a(l-e^ (1 -e'sin'0)-'^' DSd. (88) 



If (88) represents the equator, then when = 0, ds = aDSd. Hence add e' cos'<;6o to the 

 integrand of (88), to obtain 



(ds/a) = [1 - e') (1 - e' sin^)"'^' + e' cos^ol DSd. (89) 



Note the following for (89): When = <;6o = 0, ds = aDSd; when 9S0 = rr/2, DSd = d^S, 

 equation (89) will represent the meridian. 



Expanding (89) to 6th order terms in e get 



(ds/a) = 1 + (3/2)e' sin'0 + (15/8)6" sin^ + (35/16)e' sin^) 



- e'[ 1 + (3/2)6== sinV + (15/8)6" sin^] + 6=^(1 -sin^o) 



_ 



which may be written in the integral form 



DSd 



(90) 



72 



