Intelligence and Miscellaneous Articles. 149 
A mass m on the upper right pan is counterpoised by a weight 
My placed in the lower left pan. The same mass m is then placed 
in the lower right pan, and is counterpoised by weights m, placed 
in the upper left pan. For simplifying the subsequent scheme of 
calculation, let us assume that the absolute value of the vertical 
components of the accelerating forces which the block exerts on 
the upper and lower scale is the same; let it be denoted by &. If 
Jo and gy represent the values of gravity at the upper and lower 
scales respectively, we have for the two weighings the equations 
mM Jo +k)=mMy(Ju—k), 
M(Ju—k) = Mo(Jo+k) ; 
m(Jo+k)'—(gu—k)* 
(Jo +k)\(Gu—k) 
If now we put gu=go+y, then, seeing that & and y are very 
small compared with g,, 
from which we get 
on=M, — Mp= 
Qn 
Om= — (2k—y). 
Jo y 
The magnitude y is to be determined by weighings, which are 
to be made in the same balance in the same way before building up 
the lead block. If m stands for the same mass as above, and 
m', and m', correspond to the above values m, and mo, we have 
fortwo such weighings 
Mo = M'vJus 
My = MoJo 
From this follows 
nt Ie. 
/ / , 
Om= Mo — My, = 
Guo 
If, now, we put again 
Ju =9o Ur Y 
then 
Om = “Ym, 
o 
and therefore 
Aid OO 
LA an - 
If we introduce this value into the expression for 6, we get 
