C. B. WARRING. I 17 



same is true of B and of Bb. The new tangents will be 

 Be and Am. 



By another resolution, we get the triangles Bbc and 

 Amx. From them we have Ax = Am - 1 : cos m Ax, or 

 cos 2a ; 1 or, since Ax = cos a, we have cos a : m A •■■■ 1 : 

 cos 2a. 



Hence, Am = cos a cos 2a. The other component, mx, 

 expends itself in pulling against 0. 



Another reversal gives Am : Am' ■■■■ 1 ■■ cos 2a. And a 

 third Am' = Am" ■■•■ 1 : cos 2a, and so on. 



We have therefore, 



1st tang = cos a. 



After 1st reversal — 2d tang = cos a cos 2a. 

 ' ' 2d " — 3d tang = cos a cos 2 2a. 

 " 3d " — 4th tang = cos a cos 3 2a. 

 " 4th " — (n + l)th tang = cos a cos n 2a. 



It will be seen that at each reversal, a part of the 

 effect produced by the one impulse of gravity is ex- 

 pended on the point O. 



As in case of light reflected between parallel mirrors, 

 the intensity grows less and less, and gradually fades 

 away. 



The above formula gives some curious results. 



If the wheel is so large that 2a=90°, then the re- 

 mainder becomes after the first reversal, for cos. 2a (or 

 90°) =0 ; therefore, all the terms disappear except the 

 first, for in that cos 2a is not a factor. 



That this is true is easily seen by the diagrams 10 and 

 11. When B, of fig. 10, comes to the top, B b (fig. 11), 

 which was perpendicular to O B, will keep its direction, 

 and hence will be perpendicular to A O, because the 

 latter is now in OB's place ; and as A O B is a right 

 angle, its direction will coincide with B O. Hence, the 



1 To prove m Ax = 2a ; draw O w parallel to Bb, and, of course, perpendicular to A. 

 wOB (i. e., 90°- 3a) =OBb = Bbc. .-. Bbc = 90°- 2a and bBc = 2a. 



But Bbc and m A x are similar, having their sides perpendicular each to each, therefore . 

 m Ax is also equal to 2a. 



lOl 



