68 
Proceedings of the Royal Society 
of the cylinder, and putting AB = 6, AO = c, and l for the length of 
the correspondihg simple pendulum, we have 
3 7 _ b 2 + be + c 2 . 
2 b + c • 
Let now (3 be the rate of expansion of the supporting rod AB, y 
that of the cylinder BC, we have for a change of temperature t. 
b' = b(l + !3t), c =c(l +yt). 
and therefore, in order that there be no change in the value of b, 
we must have 
b 2 + bc + c 2 3 7 V 2 + b’c + c 2 (3 b 2 + Sbc)(3 + (2c 2 — be — £ 2 )y 
b + c 2 b' + c' 2b(3 — by + cy 
whence the equation 
f3 . b 3 + (4/? - 2y)6 2 C + ((3 + y)bc 2 + yc 3 = 0, . . . (2.) 
by help of which the ratio of b to c may be found. Now, unless 
y exceed the double of (3 , all the terms of this equation have the 
same sign, and therefore there can be no positive root ; so that, on 
this account, there can he no such pendulum as we are thinking 
of, unless the one rate of expansion be more than double of the 
other. In order to discover the least possible disparity of expan- 
sion, let us put b = cx, y = n/3, and our equation becomes 
x 3 + (4 — 2n)x 2 + (1 +n) x + n = 0, .... (3.) 
When our solution just begins to be possible, this equation must 
have two equal roots, and therefore must have a divisor common to 
it and its derivative 
3sc 3 + (8 — 4:n)x 2 + (l +w) = 0, (4.) 
On eliminating x from the equations (3) and (4), we obtain an 
equation of the fifth degree; of which n= 2 is one solution, and on 
dividing there results the biquadratic 
12 n* - 80rc 3 + 123 n 2 - 60n + 4 = 0, (5.) 
which has two possible roots, viz., 
n = 0 . 07873 06113 75, 
and n = 4. 71418 14416 13. 
*» 
The first of these can have no application to our present problem 
