in this way the air drag was calculated by Green for an ellipsoidal pendulum. Similarly, the inertia parallel to Oy and Oz is B0 C0 [beta]W´ = ------ W´, [gamma]W´ = ------ W´, (26) 1 - B0 1 - C0 _ / [oo] abc d[lambda] B_[lambda], C_[lambda] = | -------------------------------; (27) _/[lambda] (b² + [lambda], c² + [lambda])P and A + B + C = abc/½P, A0 + B0 + C0 = 1. (28) For a sphere a = b = c, A0 = B0 = C0 = 1/3, [alpha] = [beta] = [gamma] = ½, (29) so that the effective inertia of a sphere is increased by half the weight of liquid displaced; and in frictionless air or liquid the sphere, of weight W, will describe a parabola with vertical acceleration W - W´ ------- g. (30) W + ½W´ Thus a spherical air bubble, in which W/W´ is insensible, will begin to rise in water with acceleration 2g. 45. When the liquid is bounded externally by the fixed ellipsoid [lambda] = [lambda]1, a slight extension will give the velocity function [phi] of the liquid in the interspace as the ellipsoid [lambda] = 0 is passing with velocity U through the confocal position; [phi] must now take the form x([psi] + N), and will satisfy the conditions in the shape _ abc /[lambda]1 abcd[lambda] ------ + | ---------------- A + B1 + C1 a1b1c1 _/ [lambda] (a² + [lambda])P [phi] = Ux ----------------- = Ux ------------------------------------------, (1) B0 + C0 - B1 - C1 abc /[lambda]1 abcd[lambda] 1 - ------ - | ---------------- a1b1c1 _/0 (a² + [lambda])P and any confocal ellipsoid defined by [lambda], internal or external to [lambda] = [lambda]1, may be supposed to swim with the liquid for an instant, without distortion or rotation, with velocity along Ox B_[lambda] + C_[lambda] - B1 - C1 U ---------------------------------- B0 + C0 - B1 - C1 Since - Ux is the velocity function for the liquid W´ filling the ellipsoid [lambda] = 0, and moving bodily with it, the effective inertia of the liquid in the interspace is A0 + B1 + C1 ----------------- W´. (2) B0 + C0 - B1 - C1 If the ellipsoid is of revolution, with b = c, A + 2B1 [phi] = ½Ux -------, (3) B0 - B1 and the Stokes' current function [psi] can be written down B - B1 [psi] = - ½Uy² -------; (4) B0 - B1