92 New experimental Researches 
by substituting the value of c in one of these equations, we get 0. 
Lastly, putting 2 and c in one of the two first equations, we have ~ 
a. Thus we find” 
a. = —0°01537419550 
b = —0-00006742735 
¢ = + 0:00000003381 
Whence the whole formula Log. F, = Log. 30+ an+ bn? + cn3 
is completely determined, and may serve for calculating F,,, re- 
lative to any proposed value of 7. 
If we.make, for.example, 2= 100, we shall have the elastic 
force at 100 degrees below the boiling point, or at the tempera+ 
ture of melting ice. We thus obtain 
Log. F, = 1°4771213—2-1778831=—0-°7007618. 
Or employing negative indices in order to make use of the or- 
dinary logarithmic tables, 
Log. F, = 12992382, whence 
F =0°19917 inches ; and observation 
gives us 0-200. 
The error is obviously insensible ; and we may adopt, says 
M. Biot, our formula as representing the experiments of Mr. | 
Dalton. To introduce the Fahrenheit degrees into the formula, 
calling them /, and counting from 212°, we have 3 f=n; and 
substituting this valueof 7 in the preceding formula, we obtain 
a =—0:00854121972 
b = —0:00002081091 
c = +0-00000000580, 
whence Log. FY =1:4771213 + af + lf? + cf3, f being the 
number of degrees of Fahrenheit, reckoning. them from 212°, 
positive below and negative above this point of departure. 
By the above formula, thus elaborately investigated by M. Biot, 
I have computed the elastic forces of steam at the three succes- 
sive temperatures of 232°, 262°, and 312°, or 20°, 50° and 
100°, above the boiling point of Fahrenheit’s scale. 
In the first case we have f= —20 and af + bf? + cf?=20 
+ 400 —8000 c; f is negative, being above the point of de- 
parture 212°, and, consequently, the products af and ‘cf? are 
positive, while Jf? becomes negative. . 
20a= 0:170824 
400 4 =—0-008324 
8000 c = +0:000046 
0:162546 +4 log. 30 or 1477121 
1:477121 
Log. of 43:62 =1+639667 By 
