280 G. H. KNIBBS. 



direction L'N) is alone taken as the conjugate. The fact that 

 the intersection antipodal to D' must be ignored, shews that the 

 interpretation is purely formal and not without inconsistency. 



The other method of resolving the discontinuity in determining 

 the conjugate points C and D is illustrated in Figs. 10, 11 and 12. 

 Let the circle OB be of unit diameter measured on the surface of 

 an infinite sphere of which 00' is the radius and upon which it 

 lies. Then since OB is finite and 00' infinite, B will be only 

 infinitesimally distant from a sphere of which 00' is the diameter. 1 

 If the plane O'OB, Fig. 10, or O'O A" Fig. 11, be made perpen- 

 dicular to the plane O'AOB in either figure, the point C will be 

 at O, and D at O" Fig. 12; O'O" being perpendicular to O'O. 

 But the quadrant OEO" is equal to the semicircle ODO', since 

 any projection from O' through any point D to E makes the arcs 

 OD, OE equal, 2 hence instead of regarding D as at 0", we may 

 treat it as at O'. Similarly for any rotation 9 from the vertical 

 position, A'OA", see Figs. 8, 10, 11, the real conjugate points 

 should be C and E, but D may be substituted so long as the angle 

 is measured at E, not at D. 3 For finite distances the error of 

 this representation will be infinitesimal, 4 both in respect of the 



1 And will be the greatest distance. 



2 We may call the curvilinear triangle ODE a curved isosceles triangle. 



3 The inclination of the planes may of course be measured at D : that 

 inclination will be 6, see next footnote. 



* Let the several angles be denoted by letters as follows : — 



EOA." = ^7r-0; OA"E=i7r; OEA"=<£; 00'A" = tt; 



OC^y, measured on the sphere of which O' is the centre ; OE = & E being 



on the same sphere : then by spherical trigonometry 



a 2 

 cos cf> = cos a cos 6 = (1 \-) cos d°. 



But a is a first order zero, hence cos <£ differs only infinitesimally from 

 cos for infinite distances. Again 



tan 8 - tan a cosec 6 



tan £y = tan \<x sin 6. 

 hence 2 tan \y tan 8 = 2 tan \a tan a. 



Consequently y 8 = a 2 +^ (5a 4 - 8y 3 - 4yS 3 ) + etc. 



Taking the radius as infinity, this last is equivalent to 



^ = a 2 {l + -°- 2 (5a 2 -c 2 -d 2 ) + etc.) 



which so long as d is finite can be only infinitesimally in error. The 

 neglected terms are smaller still. Consequently the error of the scheme 

 is infinitesimal. 



