A SPHERICAL RECTANGULAR 



COORDINATE SYSTEM WITH A GREAT 



CIRCLE BASE LINE AS AN AXIS 



Figure 5 is a further elaboration of Figures 2 and 3. M is the midpoint of the spherical 

 segment QiQj . The section MPT" is perpendicular to the base line at M. The general point 

 Q {6, \) has for the foot of the perpendicular from Q upon the base line, the point Q' {d',\') as 

 shown in figure 2. The great circle arc QQ 'passes through P,'and QQ'is taken for spherical 

 rectangular coordinate y. The great circle perpendicular to the section MP'P^and passing 

 through Q meets MP'P" in T. The distance OQ'is S as shown in Figure 5. Note that the s of 

 Figure 3 in the y of Figure 5. The great circle arc QT is taken for x. That is the spherical 

 rectangular system chosen is x = QT, y = QQ '. Spherical polar coordinates are then r and a as 

 shown in Figure 5, where r = MQ, and a is the angle between r and MQ '. 



From the right spherical triangles MQT, MQQ ' one finds 



sin X = sin r cos a 



(14) 



(15) 



sin y = sin r sin a 

 whence 



sin r = (sin^x + sin^y) ''^ 



tan a = sin y/sin x, 

 that is (14) and (15) represent the conversion formulas between the spherical rectangular and 

 spherical polar systems as given. 



We now develop the coordinates x and y as functions of S and of 6 and A. Also 6 and A 

 as functions of x and y. 



COMPUTATION OF S, x, y, FROM d AND A 



Assume that the base line has been established, that is the coordinates do , Aj, of the 



vertex, 0, of the great circle base line have been computed from the coordinates of the two given 



points Qi((?i, Aj), 02(^2, Aj) by means of the equations as given on page 23. Then referring to 



Figure 5, find in spherical triangles: 



PP'Q: cos y sin S = cos 0sin (Ao- A), (16) 



: sin y = cos 60 sin 6 — sin d^ cos 6 cos (Ao — A), (17) 



OPQ: cos f= sin 00 sin + cos 00 cos 9 cos (Ao- A), (18) 



OQQ': cos y cos S = cos f, d") 



TP "Q: sin X = sin d cos y. (20) 



28 



