858 Mr. J. Rose-Innes on the Motion of a Pendulum 



Then the intersection of CD with its neighbouring position 

 will be ultimately the point G, halfway between C and D. 

 By means of the theorem in Spherical Trigonometry, Part I., 

 we deduce that CF=DE, and since GC = GD we must 

 have GF = GE. It follows that the small change in the 

 value of the perpendicular BF owing to the displacement of 

 the arc CD inusr, be equal and opposite to the small change 

 in the value of the perpendicular AE ; so that the sum 

 AE + BF will remain constant during the displacement. 

 By means of the theorem in Spheiical Trigonometry, Part III. 



A. A 



we deduce that the sum ADE + BCF will also remain constant 

 during the displacement. 



We can easily show that the spherical excess of the 



pentagon LMDCN is equal to KLH-(ADC + BCD) ; 

 hence, when the arc CD moves as described above, the 

 area of the pentagon will remain unaltered. From the 

 triangle LKH subtract half the sphero-conic, add the 

 segment of the sphero-conic cut off by CD, and subtract 

 the pentagon LMDCN ; the remainder consists of two 

 elliptic areas AKNC and BHM, whose sum is therefore 

 proved constant when CD moves under the prescribed 

 conditions. 



Let u and v denote the elliptic integrals determined by the 

 angles ABO and BAD respectively, so that 



am u = ABC = FBD, 



am v = BAD = CAE. 



Suppose that C and D move clockwise along the sphero- 

 conic, subject to the condition already mentioned. Then, 

 by letting D approach indefinitely close to B, we see that the 

 elliptic area BHMD and the angle BCD can be made 

 simultaneously as small as we please ; we obtain in the 

 limit 



am (m+») = ADC + BCD. 



Take the two triangles ACE and BDF, and place them 

 alongside each other so that the side DF coincides with the 

 side CE, D coinciding with C, and F coinciding with E. 

 This is possible, since the side DF can be proved equal 

 to the side CE by the theorem in Spherical Trigonometry, 

 Part I. Also, since the angles AEC and BFD are both 

 right angles, the sides AE and BF in the new positions 

 would be portions of the same great circle. Hence the 

 two triangles so placed would make up one triangle, whose 

 angles would -be respectively equal to am u, am r, and 



