474 THE FORMS OF TISSUES [ch. 



partition lies midway between the two centres; and the diameter 



OP a/3 



of the partition-wall, PP\ is -^^ = sin 60° = — - = 0-866 times the 



diameter of each of the two cells. This gives us, then, the form 

 of a combination of two co-equal spherical cells under uniform 

 conditions. 



By integrating between the known values of the meridian section and the 

 plane partition, we should find each half of the double cell (or soap-bubble) 

 to be equal to 27/32 of a complete sphere. Therefore the radius of curvature 

 of each half of the divided bubble is greater than that of a sphere of equal 

 volume in the ratio of: 



\^32 : \^27 = 2 . ^4 : 3 = 1058 : 1 = 1 : 0-945. 



And the radius of the original sphere, before division, is to the radius of 

 each half, or each product of cell-division, as 



^54:^^32 = 3.^^2:2.^^4 = 1191: 1 = 1:0-84. 



In the case of three co-equal and united bubbles (to which case we shall 

 presently return), each is approximately five-sevenths of a whole sphere: 

 and their radii, therefore, are to the radius of the whole sphere as 



^7 : \^5 = 1 : 0-893 = 1 : (0-945)2. 



When two co-equal bubbles coalesce, the internal pressure, du5 to the 

 tension of the wall and varying inversely as its radius of curvature, is 

 diminished in the ratio of 1 : 0-945, or say 5i per cent. And we begin to see, 

 in the case of three bubbles, that the process proceeds in a geometrical 

 progression, each new coalescence increasing the radius of curvature and 

 diminishing the internal pressure, by a constant fraction of the whole. This 

 and other simple corollaries may perchance, some day, be found useful to the 

 biologist. 



In the case of unequal bubbles, the curvature of their partition- 

 wall is easily determined, and is shewn in Fig. 159. The three 

 films meeting in P being (as before) identical films, the three 

 tangents, PQ, PR, PS, meet at co-equal angles of 120°, and PS 

 produced bisects the angle QPR. PQ, PR are tangents perpen- 

 dicular to the radii CP, C'P\ and C"P, the radius of the spherical 

 partition PP\ is found by drawing a perpendicular to PS in P. 

 The centre C" is, by the symmetry of the figure, in a straight line 

 with C, C. 



Whether the partition be or be not a plane surface, it is obvious 

 that its line of junction with the rest of the system lies in a plane, 



