Force of Gravity. 249 



370. It is not necessary to go through the same process in 

 order to find the length of a pendulum required to vibrate in any 

 other proposed time, as half a second, or half a minute. The 

 principles we have investigated will enable us to solve all prob- 

 lems of this kind with the greatest facility and exactness, when 

 the length and time of vibration of one pendulum is known. Thus 

 if it is proposed to find the length of a pendulum required to vi- 

 brate half minutes, the proportion 



t 2 : t' 2 : : a : a', 



by substituting for J, t', \" and 30", and for a 39,1386, the length 

 of the seconds pendulum, we have 



I 2 :: (30) 3 :: 39,1386 : a' = 39,1386 X 900 = 35224,74 

 inches, or 2935,39 feet. 



In like manner, the length and time of vibration of one pen- 

 dulum being known, the time of vibration, in the same place, of 

 any other pendulum whose length is given, may be determined. 

 Suppose, for example, that it is required to find the time in which 

 a pendulum of 20 feet, or 240 inches in length, would perform its 

 vibrations ; by substituting the known quantities in the general 

 proportion, 



Va : \/a f : : t : t', 

 we have 



V39,1386 : V240" : : 1" : C = _?!2_- = 2",5 nearly. 



\39,1386 



Measure of the Force of Gravity. 



371. It will be easy now to determine through what space a 

 liravy body must pass in the first second of its fall, the air and 

 all other obstacles being removed. For the equation 



g 



gives, by squaring both members and transposing^ 



2r 2 a 

 = -jTT', 



Mech. 32 



