CALIBRATION OF A GLASS TUBE. 309 



zero of the scale. The point is to determine the correction a 

 for each division. 



A column of mercury is taken which occupies the nth part of the 

 tube, and its length is measured in n successive parts of the scale. 

 Let a Q and a l9 a' and a 2 , a' 2 and a 3 ..... a' n _ l and a n be the divisions 

 corresponding to the end of the column, <x , a lf a 2 , a 3 ..... a n 

 the various terms of correction. As a first approximation we may 

 assume that these corrections are respectively the same for adjacent 

 divisions a^ and a\ , a 2 and a' 2 ..... The corrected length / of the 

 column that is to say, the length which it would occupy in a 

 tube of mean section is expressed by a series of values such that 



If 



are the various lengths observed, and assuming that the volume of 

 mercury is the same that is, that the temperature has not varied 



a -a 1 +/=8 1 , 

 (2) a 2 -a 3 +/= 3 , 



As the divisions a and a n correspond to the ends of the scale, 

 corrections a and a n 

 these n equations, we have 



the corrections a and a n are null. If we eliminate / between 



(3) a 1 - 4 

 a l + a 



Adding these latter, we get 



(4) na 



