_ © Temperature yin a Siphon Barometer, 259 
placing the body of the column below it. Suppose then the parts 
of the column in the vicinity of the circle HI to have such a hor- 
izontal tendency towards the axis of the tube, without disturbing 
the colamn below, as to cause the meniscus HBI to take the form. 
of EAF. We shall then have this equation : 
Meniscus EAF =meniscus HBI+cylinder HEFL. (11.) 
Put height AD of meniscus EAF=H. 
Height BC of meniscus HBI=H’. 
Radius of tube ED=r. 
Correction AB=z. 
Height of cylinder CD=y. 
Circumference of a circle whose diameter is unity =z. | 
And let the capacity of any meniscus EAF be represented by (ar?) 
multiplied by a function F (H) of its height. Equation (11) 
therefore becomes ar?. F(H)=ar?. F(H’)+-2r?y. 
F'rom the figure we have H=H/+y+2. -— 
Dividing the first of these equations by (7r?,) and eliminating (y,) 
Wehave e=H—H’+F(H’)-F(H.) (12.) 
It remains only to determine F (H,) F (H’,) which depend 
Upon the form of the meniscus. oe HO 
If the tube is small, the meniscus differs insensibly from the, | 
segment ofa sphere, (Mécanique Céleste, 9334.) And generally, 
since the meniscus in barometer tubes isa small portion of the 
solid of revolution of which it isa segment, every practical pur- 
pose would be answered by considering it a common paraboloid, 
This being so, we have F(H)=4H, and F(H’)=3H’; and con- 
} H— 
sequently (12) becomes t=~—5—, (13.) Wherefore (a’) is cor- 
ay Hi! 
tected to a’+- 3 (14.) 
In like manner, denoting the corresponding heights of the me- 
hiscuses in the lower branch by (/,) (h’,) we shall find (2’) to be- 
Ah—h! 
come gated (15;) and consequently (a’—8’) becomes, after 
Correction, a’ — Ett ee, (16.) Hence, to find the cor- 
tection for the difference of readings (a’ — b’,} we subtract the 
half sum of the heights of the meniscuses which terminate this 
column from the half sum of the heights of the meniscuses which 
terminate the column of which the readings are (a,) (b.) 
