Now the surfaces of the two bubbles exert a pressure inwards which is inversely proportional to their radii: that is to say pã₤:ã₤pÿ£¢ãýã₤::ã₤1ã₤ãã₤rÿ£¢ãýã₤:ã₤1ã₤ãã₤r; and the partition wall must, for equiôÙlibôÙrium, exert a pressure (P) which is equal to the difference between these two pressures, that is to say, P =ã₤1ã₤ãã₤R =ã₤1ã₤ãã₤rÿ£¢ãýã₤ãã₤1ã₤ãã₤r
=ã₤(rã₤ãã₤rÿ£¢ãý)ã₤ãã₤rãrÿ£¢ãý. It follows that the curvature of the partition wall must be just such a curvature as is capable of exerting this pressure, that is to say, R =ã₤rãrÿ£¢ãýã₤ãã₤(rã₤ãã₤rÿ£¢ãý). The partition wall, then, is always a portion of a spherical surface, whose radius is equal to the product, divided by the difference, of the radii of the two vesicles. It follows at once from this that 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.
The geometrical construction by which we obtain the position of the centres of the two spheres and also of the partition surface is very simple, always provided that the surface tensions are uniform throughout the system. If p be a point of contact between the two spheres, and cp be a radius of one of them, then make the angle cpm =ã₤60ô¯, and mark off on pm, pcÿ£¢ãý equal to the {300} radius of the other sphere; in like manner, make the angle cÿ£¢ãýpn =ã₤60ô¯, cutting the line ccÿ£¢ãý in cÿ£¢ã°; then cÿ£¢ãý will be the centre of the second sphere, and cÿ£¢ã° that of the spherical partition.
Fig. 105. Fig. 106.
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, and is at right angles to the axis of symmetry. The actual curvature of the partition-wall is easily seen in optical section; but in surface view, the line of junction is projected as a plane (Fig. 106), perpendicular to the axis, and this appearance has also helped to lend support and authority to ãSachsãs Rule.ã