58 U. S. COAST AND GEODETIC SURVEY. 



The triangle WP'c is isosceles, and the angle P' W'p equals 

 the angle TFP'/S', which is measured by kIo"^^^^ PN\ 



= ^SiTcPN W; that is, the arc REp = arc PN W. Hence lay 



off the arc PEp = arc PNW and draw Wcp. This is the 

 same as laying off a polar distance PiVTf from P; thus the 

 line of centers is the projection of a small circle passing 

 through the line of sight and having the polar distance 

 PNW=Tr — ^y where ^ denotes the inclination of the circle. 



From figure 19 WQ=PE; QSp = T- (pE+WQ) =Tr- 

 PEp = Tr-PNW=WQ; hence lay off WQp = 2PE, and 

 draw Wp, thus locating c. Wp is evidently perpendicular 

 to PQ, so that c can be located in that way. 



Z WEp= lPOE= Z WOQ; hence a line joining E and p 

 is parallel to PQ; this gives another method for locating c. 



Problem 5. — To draw a great circle through P, making a 

 given angle with N8: 



In figure 19 the tangent to the required circle at P makes 

 the given angle (m)^with P'OS; the perpendicular to the 



IT 



tangent makes with P^OS the angle o"^- Hence con- 

 struct SP'R = ^ — m with P'P intersecting the line of cen- 

 ters at P, the center of the required cirple. 



The projection of a great circle always meets the primi- 

 tive circle at the extremities of a diameter as MM^ in 

 figure 19. 



Problem 6, — To find the projection of a pole of a given 

 circle : 



In figure 18 let Wp^E be a great circle; draw the per- 

 pendiciSar diameters WE and NS, and draw Wp'p; lay off 



pP equal to - and draw WP, thus locating P', the required 



pole. 



In figure 16 let p'q' be a g^^en small circle; through its 

 center c draw N8 and draw WEQ,t right angles; draw Wp' 

 to locate p and Wq' to locate g; bisect the arc qNEp, locat- 

 ing P, and draw WP, thus locating P', the projection of 

 the required pole. 



Problem 7. — To construct the projection of a great circle 

 passing through the projections of two given points: 



