344 Hydrostatics. 



In the case of a cylindrical tube, let n represent the ratio 

 of the circumference to the diameter, h the height of the fluid 

 column, reckoned from the lowest point of the meniscus, h' the 

 mean height to which the fluid rises, or the height at which the 

 fluid would stand if the meniscus were to settle down and assume 

 a level surface; then we have n R 3 for the solid contents of a 

 555* 1 "' cylinder of the same height and radius as the meniscus ; and as 

 the meniscus, added to the solid contents of a hemisphere of 

 the same radius, must be equal to n R 3 , (or in other words, the 

 cylinder TfR 3 , diminished by the hemisphere f n R 3 , is equal to 



*U \ U 2 TfR 3 7TR2 , .. , 



the meniscus,) we have n R 3 -- - , or ^ 5 for the solid con- 



3 o 



tents of the meniscus. But since = n R 2 X ;-, it follows 



3 o 



71 R 3 



that the meniscus - - is equal to a cylinder whose base is 



O 

 jy T> 



. n R 2 , and altitude -. Hence, we have h' = h -{- -\ or, which 



O O 



is the same thing, the mean altitude h' is always equal to (he al- 

 titude h of the lower point of the concavity ot the meniscus, in_ 

 creased by 'one third of the radius, or one sixth of the diameter 

 of the capillary tube. Now, since the contour c of the tube 

 = 2 n R, and since the bulk b of water raised is equal to h' X n R 2 , 

 we have, by substituting these values in the general formula (i.) 



g A h' n R 2 = 2 n R (2p /) 

 and, dividing by n R a; d g A, we obtain, 



x 



g& gA R 



In applying this formula to M. Gay Lussac's experiments, we have 



2 JP^iJL = R .V = 0,647205x(23,1634-f-0,215735)=15,1311, 



o 



or 0,023454 of an inch for the first experiment; and, since the 

 heights are inversely as the radii or diameters, 0,023454 or its double 

 0,046908 is a constant quantity. In order to find the height of 

 the fluid in thfe second tube by means of this constant quantity, 

 we have 



