Vibrations of Heavy Bodies, 71 



16 V. 2* J 16 h 



But the time of descent down the arc of a cycloid, or through 

 any small circular arc, is inversely as the square root of the mov- 

 ing force. 



Let gravity, as acting in a vacuum, be represented by unity. 



Let S = the specific gravity of the pendulum, 



s = the specific gravity of air, at the height (A) of the 

 barometer. 



Then will 1 — — represent the moving force, 

 S 



and t = 



v/'-i 



S 

 But t, as compared with h, will be 



t = 1 L The first term 



J\- s x hVh xA~- 



v l "S x T" v s 



Is h 



of which expression, being expanded, is -r- x "g" x ~r as f. 



By making this equal to the variation in the time occasioned by 

 buoyancy, 



JL X -1 X A = 1 z* X A V ** = 8 ± 

 2 1 h h 



and z ss 



Then if air be taken at the specific gravity of 



JL its log 7.0819697 



828 fe 



The spec. grav. of brass 8.8, 4 its log. 0.9242793 > 9 0757 o 07 



Ar. Com S 



8 its log. . 0.9030900 



2)7.0607804 



8.5303902 



Degrees in radius 57°.2958 the log 1.7581226 



.2885128 

 which gives 1°.9432, or 1° 56' 35" for the vibration on each side 



