34 
0 = angular deflection in circular measure produced by p. 
9 = gravity. 
If 0 = 0 we have the position of equilibrium given by. 
e = 2 MW* < 2 > 
The semi-periodic time is 
/MI 2 + 2Pa 2 
,= V mV+Mgl ^ 
From equations (2) and (3) we can eliminate 2P/&-f- Mgh, 
obtaining 
M^P + 2Pa 2 0 
p = id— 35 (4) 
M ag r w 
From this expression it appears that if we know the 
moment of inertia of the beam, its length, and the weight 
at each end, we can find the excess p from the time of 
vibration and deflection. 
The results given in this paper were obtained with a 
16-inch chemical balance by Oe idling. The exact length of 
the half beam (a) measured by a dividing engine is 20*2484 
centimetres. 
To find the moment of inertia MI 2 of the beam — The 
simplest way theoretically would appear to be this. Find 
the times of vibration t x t 2 and the deflections 0i 02 due to 
the same excess p with two different loads Pi P 2 in each 
pan. Equating the values of p given for each by equation 
(4) we have 
MyP 4 - 2 Pi (I 0 2 h 2 
MyP + 2P 2 n 2_ 0p 2 2 
An equation which will give M(/I 2 in terms of known 
quantities. But on trial it was found that a very small 
proportional error in the observed time made a large error 
in the value of Myl 2 , and the following method, that usually 
adopted in magnetic observations was employed in prefer- 
ence. A stirrup was suspended by a platinum wire and its 
