clearing a Lunar Distance. 



281 



Take CS, representing the apparent zenith distance ^"1 of Sj, 



CS. 



& 



and with radius CSg describe a circular arc EF. CSj and CSg 

 are the natural tangents of fj, and fg to radius D. 



From ^1 draw any Une S^P, and lay down upon it SjK repre- 

 senting the angular distance Sj Sg by the first rule. In this 

 way one point is obtained of the conic section which represents 

 the small circle described about Sg, whose intersection with the 

 circle EF described with radius ZSg determines the place of Sg. 

 In the same manner any number of points in the conic section 

 may be found ; but as it is easy to see very nearly where Sg is 

 situated, it will generally be sufficient to find two points on the 

 conic section in the presumed vicinity of 85, and the intersection 

 of a straight line joining them with the circle will give the place 

 of 83 very nearly. The locus of the point P is a circle described 

 with radius |S|C, and the locus of the point Y is a circle de- 

 scribed with radius S,V, SiV^zrSiC^ + D^. If R is near the true 

 place of 83, Sg will be found nearly by the intersection of a cir- 

 cular arc described with radius SjU and the circular arc EF. 

 SjCSg is the accurate delineation in perspective of the spherical 

 triangle SjZSg. 



To measure the angle ZSjSg, produce SjC to Q, making 

 SiQ = 90°, by making S^BQ aright angle, and CB perpendicular 

 to CQ=D. 



Draw QT perpendicular to S^Q, and in SjQ take 



QV= ^D2 + CQ^=BQ. 

 Produce SjSg cutting QT in the point T and join VT, the 

 original angle ZSjS^ is equal to QVT. 



