AT 4 Professor STEVELLY on a@ new 
the pillar or stem of the hydrometer; s* the excess of the cross section of the vessel 
A, at the surface of the fluid it contains, above s’’. All these quantities are sup- 
posed to have one inch for their linear unit. Let w denote the weight of one cubic 
inch of mercury, and if the hydrometer float in any other fluid than mercury, let w’ 
denote the weight of one cubic inch of that other fluid. 
Then 8h” —dh’” is the part of the lower portion of the tube B which emerges 
from the mercury in the cistern, or becomes more immersed in it, in consequence 
of the oscillations of the instrument. And 6h’+ 0h" is the portion of the stem or 
pillar of the hydrometer which becomes immersed in the fluid in the vessel 4, or 
emerges from it in consequence of the same oscillation. 
Again dh.’ s. w—(sh —oh”) s. w is the alteration of weight of the cistern, and 
(8h’ + 8h") s.” w' is the alteration of the buoyant force of the hydrometer, and that 
the equilibrium may continue these must be equal ; hence 
m ww 
dh.'s —(8h" —8h'") s' = (8h + 8h ) s. (D) 
Also because the perpendicular distance of the surface of the mercury in the tube B 
from the surface of that in the cistern C is at all times equal to the height of the 
common barometer, therefore 
oh = bh! — bh" + oh'" . (1) 
And from considering the cause of the rising or falling of the surface of the mercury 
in the cistern, it will appear that . 
dh'.s — (8h —8h'") s'=8h" 5", hence 
oh's—oh''s! 
Ai = fo) 
oh a (2) 
and by elimination from (1) and (2) we get 
_ 8h C's!) —$h"' (s!—s') +8's—8h''s! 
oh _ ee ence 
cS ‘ 
yf BAG SD TEMS” (3) 
ae co 
and by substitution of this in (2) we get - 
onl! — dh s+ 6h's—dh's (4) 
s+s'—s 
Also by considering the cause of the rising or falling of the surface of the fluid in 
the vessel 4 it will be seen that 
(8h" + 8h’) s\’ =8h.” 8’.) Hence— 
3h! sl" 
SY +3" 
(5) 
w= 
oa 
pe 
