WALLACE O. FENN 375 



express x and 5 in equation (1) in terms of h when the radius, r, of the 

 cell (when spherical as at a) is equal to 1. In order to do this it is 

 necessary to neglect the effect of gravity and assume that C always has 

 the shape of a sphere or spherical segment. This is quite legitimate 

 since the effect of gravity would be practically negHgible where 

 the dimensions of C are those of leucocytes and where the difference 

 in density between C and P is small.^ By familiar formulas we may 

 write 



x = Tr{a^ + h^) (2) 



^ = 7ra2 (3) 



where a is the radius of the base of the cell. 



The volume of the spherical segment at h equals the volume of the 

 sphere at a (Fig. 1) or 



1 4 



-Th U2 4-3^2) =--,rr2 

 6 3 



Putting r = 1 and solving 



(4) 



ISA 3 



Substituting the value of a from equation (4) in (2) and (3) and in- 

 troducing the resulting values of x and s in equation (1) we have 



E = cTp[irh^ + -- — \ + (gTc -gTp) {-- — ] (5) 



/ , Stt A^A . . /Stt 7rh^\ 



Putting ti = cTp and m = gTc — gTp, differentiating with respect 

 to h, and simplifying, we find 



dh 3/j2 



(6) 



dE 

 At equiUbrium y- = 0. Hence, putting the right hand member of 



equation (6) equal to and solving for h, we have 



^ Gravity would merely shift the equilibrium point without altering the princi- 

 ples involved. 



