372 Mr. A. M. Worthington. [June 16, 



due to curvature at the base is wrong, and that the pressure is in reality 

 negative, and that each value of the tension requires to be diminished 

 by a correction which, being proportional in each case to the area of 

 the section, will affect the higher values most, the smaller least. If 

 we call the unknown pressure due to curvature at the base c, and 

 introduce this in the equation, it becomes 



T cos 6 x Zttt = ( V + Trrm) D - 7rr% 



or T _(V+^H)D_ rc 



27rrcos9 2cos#' 



so that we have now four equations for determining T and c, viz., 



T = T T + — ^ — , 

 2 cos X 



T = T + 



2 cos 6c, 



T = T„ + 



2 cos 3 



T=T 4 + -^i- c 



2 cos 4 



any two of which are sufficient. 



Combining these equations by the method of least squares so as 

 to obtain the most probable values of the unknown quantities, we 

 get 



c=— '09943 grm. per square centim., 



T= '03373 grm. per linear centim., 



and using these values to calculate the most probable values of the 

 observed quantities T : . To. &c, we have 



Calculated. Observed. 



T 1= -04935 .... -0-4940 



T 2 = '05121 .... '05181 



Tg= -05777 .... -05688 



T 4 = -06506 .... -06538 



It remains to observe that, as will appear in the table of results,, 

 the pressure due to curvature at the base was found to be negative 

 even where the drop hung from a plane surface of indefinite extent 

 which it wetted entirely. 



It must be admitted that the surface here (see fig. 1, Plate 7) really 

 passes into a continuous plane where the curvature, and consequently 

 the nressure due to curvature, is zero ; and that, nevertheless, at an 

 indefinitely small distance below this, the pressure due to curvature 

 is undoubtedly negative. 



