VII 



OF CELL-PARTITIONS 



473 



proportional to their radii: that is to say, p: p' :: Ijr : l/r'; and the 

 partition-wall must, for equilibrium, exert a pressure (P) which is 

 equal to the difference between these two pressures, that is to say, 

 P = l/R = Ijr' — Ijr = (r — r')lrr'. It follows that the curvature of 

 the partition must be just such as is capable of exerting this pressure, 

 that is to say, R = rr'j(r — /). The partition, then, is a portion of 

 a spherical surface, whose radius is equal to the product, divided 

 by the difference, of the radii of the two bubbles; if the two bubbles 

 be equal, the radius of curvature of the partition is infinitely great, 

 that is to say the partition is (as we have already seen) a plane 

 surface. 



Fig. 158. 



In the typical case of an evenly divided cell, such as a double 

 and co-equal soap-bubble (Fig. 158), where partition- wall and outer 

 walls are identical with one another and the same air is in contact 

 with them all, we can easily determine the form of the system. 

 For, at any point of the boundary of the partition, P, the tensions 

 being equal, the angles QPP', RPP', QPR are all equal, and each 

 is, therefore, an angle of 120°. But PQ, PR being tangents, the 

 centres of the two spheres (or circular arcs in the figure) lie on lines 

 perpendicular to them; therefore the radii CP, C'P meet at an 

 angle of 60°, and CPC is an equilateral triangle. That is to say, 

 the centre of each circle Hes on the circumference of the other; the 



