10 SIXTH REPORT. — 1836. 



be led to use a less z, or a less length of metallic tube than that 

 which will truly compensate the changes of the suspending rod. 



But to proceed with the investigation, — if d s and dji respectively 

 denote the alterations of the unit length of steel and of white deal 

 for the same change of temperature that causes dm in the unit length 

 of the metal used for the compensating tube, then the measures 

 being expressed in inches we shall have 



2ds-\-{\-2 + z)dp=\z- ^ , , '! . \dm 



^ /I o / 



3 



Since we have supposed the steel spring to be two inches long, 

 and the length of the deal rod is X — 2 + z inches ; or, 



(r2±r^ z2\^ 

 \ 4 ^ 3 / dm... 

 2{ds-dp) +Xdp-z(dm-dp) = -2 ■.,,„„ .0 



4 3 ....(b) 



But from (a) it will appear that A' — — — = 2X^ — l\ 



And solving (a) for A, we get A = — + \ /i! _ ^'"'" ^'" _ f! 



2 V 4 4 3 • 



the value of A, found with the negative radical, being the intercept 

 between the centre of gravity and centre of oscillation. Now if we 

 denote by R the radical in the value of A, and substitute these 

 quantities in {b), we shall have 



l \ 4 



2(ds—d2:i)+-^dp -\-'B.dp —z(dm—dp) = — 



or, by reducing, substituting for R* its value, and dividing by 

 dm — dp 



/'2(ds-dp) +ldp ^ ,2 7> J 



\ dm — dp y 3 A dm — dp 4 



According to the table given by Kater, rf* = -0000063596, 

 dp= -0000022685, and if the material of the tube be lead, dm 

 = •0000159259; also, Z = 39-13929. If we assume the dia- 

 meter of the leaden tube to be an inch and a half, and the hollow 

 along its axis to be six tenths of an inch, then write r = "75, and 

 r' = -3, and substituting these respective values in the foregoing 

 equation, and changing the signs of all the terms, we shall have 



(X- 3-84963) (382-807880426025 - ^y -j= 63-774774 ..(c) 



