Great Elliptic Section Azimuths in terms of 6 



(tan 6 J cos A A - tan 62) (cos ^i ) (1 - e^ cos'' 0, )'/^ 



cot a^B = + — — 



sin IS A 



(tan 6i - tan 6^ cosAA) (cos d^) (1 - e^ cos^^J'/^ 



cot OoA = + — 



t>A sin A A 



(18) 



GREAT ELLIPTIC ARC DISTANCE 



Referring to Figure 9, it is seen that the great elliptic arc is orthogonal to a meridian at 



a point Po((^o> ^0) vvhich is the vertex of the great elliptic arc determined by the points 



P,(<;6,, Aj), P2(02' '^2) on th^ ellipsoid. The equation of the great elliptic plane through Pj 



and P2 is given by equations (14). Now a meridional plane orthogonal to (14) has an equation 



of the form Bx - Ay = and the rectangular coordinates of P^ic/iQ, \f) must satisfy both planes. 



From (1), the rectangular coordinates of Pq i(f>o, Ag) are Xq = No cos 0|,cos A A,,, 



Xo = No cos 00 sin AAo, z = No(l - e^) sin 0o and these placed in Bx - Ay = and (14) give 



B cos AAo-AsinAAo=0, 



(19) 

 A cos AAo+B sin AAo = C (1 — e^) tan <pg . 



From the first of (19) find tan AAo= B/A, whence sin AAo= B/(AV B^)'/^and these values 



placed in the second of (19) give tan 0o= (A^ + B^)vVC (1 - e^), 



, w, 2 ^0 / A^ +B^ \ '/' 



sin cba = tan 0„/ (1 + tan 0o) = ) . 



VA^ + B^ + C^d-e^)^ / 



(20) 



tan AAo = B/A. 



With the values of A, B, C from (14), equations (20) may be written 



/ tan^ J — 2 tan 0i tan 02cos AA + tan ^ 0j \ 



sin0„= — , (21) 



\tan^0 1- 2 tan 0, tan 02 cos A A + tan^02+ sin ^AA/ 



tan A Ao = (cot 0, tan 02 - cos AA)/sin A A, 



tan 00 = ( tan ^0j + tan^02 - 2tan 0, tan 02 cos A A)''7sin A A . 



From the second of equations (19), dropping the subscript zero and differentiating we obtain 



(- A sin AA+B cos A A) (d A A) = C (1 - e') sec' d 0. (22) 



By solving A cos AA+B sin A A = C (1 - e') tan with the identity sin 'A A+ cos' A A = 1, find 



BC (1 - e') tan + A [ (A' + B') - C (1 - e')' tan' 0] '/' 



sinAA = — ^— , (23) 



A' + B' 



- AC (1 - e') tan + B [(A' - B') - C'(l - e')' tan'0] '/' 



cos AA = 



A' + B' 



48 



