Connected with the Earth's Field of Force. 155 



The resultant X, of all forces along the ^r-axis, is 

 p. + g., or 



X = 2 h {av^ - pv, z,) (11) 



It is easy to show that ^x is numerically greater than 

 ^x, or that the resultant force is towards the equator, 

 Bs and C being essentially negative. 



The effect of the lack of symmetry with respect to the 

 ^5;-plane, which we have so far ignored, is easily seen to be 

 negligible. The equation of the free surface is found by 

 putting the right-hand side of (1) equal to zero. If x or 

 y be taken as a small quantity of the first order, z is easily 

 seen to be of the second and s- of the fourth order, so that 

 we may write for the point A, whose coordinates are 

 {x, 0, ^i), 



^ _ Oj ^ _ a x^ / 2 h X \ 



''' ~ go — '^ hx~ g^ \ "^ g^ J 



For the point B, approximately symmetrical, with 

 coordinates (a;, 0,^2) 



_ ax^ / 2 h x\ 



^' ~ ^ \ ^ ) 



(12) 

 Jf ah x^ 



If in (12) we take x equal to r, the radius of the sphere, 

 then. 



Jj. ah r^ 



9o 



f3r' 



is the maximum possible depth of the strip on which the 

 pressure is unbalanced ; the maximum width in computing 

 (6) or (8) is 2r^ and the maximum pressure, since the strip 

 is at the '^water-line,'' is evidently less than J^ ( depth) ^ 

 X width X (go p), i. e., less than 



4 /3^ (^)' go P r' (13) 



The quantity (13) is seen to be quite negligible when 

 compared with (9) or (10), both on account of the addi- 

 tional factor 13 and a factor of the order (^\ . This 



unbalanced pressure, such as it is, is furthermore toward 

 the equator. 



