730 



Sir G. Greenhill on Pendulum 



in Mr. Rose-Innes's manner on a sphere, by lines drawn 

 from the centre ; corresponding points on the sphere in 

 fig. 4 and the plane AQD of fig. 1 may be represented 

 without confusion by the same letter, in most cases. Then 



(i) 



k- = 



AD 

 AE 



^j 2 = snr AOD ; 



so that AOD is the modular angle, denoted by c 

 EO = EL, and OL bisects the angle AOD. 



Invert] with respect to 0, making OQ.OQ' = OD 5 



Fier. 3. 



and 

 the 



vertical plane AQD inverts into a sphere on the diameter 

 OD, and the circle AQD into another circle A'Q'D in a 

 plane perpendicular to OA, so that these circles are circular 

 sections of the cone, vertex and base AQD. 



Now AQ is perpendicular to the plane ODQ, so that the 

 planes OAQ, ODQ are at right angles, and the angle AQD 

 on the sphere in fig. 4 is a right angle. 



If DX is the perpendicular on the tangent at Q, QDX = 

 QDA = </>in figs. 1 and 4 ; so also in fig. 4, if AY is the per- 

 pendicular from A on the tangent at Q, QAY=Q AD = (/>', 

 but </>' is the angle Q'A'D in fig. 2, or ADQ' in fig. 1. 



3. In fig. 4, by Spherical Trigonometry, 



(1) sin AQ = sin AD sin ADQ = k sin </>, cos AQ = A<£, 



(2) sin DQ = k sin </>', cos DQ = A</>', 



(3) cos AQ cos DQ = cos AD = cos c, A0 Acf> / = k', 



(4) <j)' = am (K— u), with <f> = am u. 



