Thus with g = 0, the cylinder will describe a circle with angular velocity 2[rho][omega]/([sigma] + [rho]), so that the radius is ([sigma] + [rho])v/2[rho][omega], if the velocity is v. With [sigma] = 0, the angular velocity of the cylinder is 2[omega]; in this way the velocity may be calculated of the propagation of ripples and waves on the surface of a vertical whirlpool in a sink. Restoring [sigma] will make the path of the cylinder a trochoid; and so the swerve can be explained of the ball in tennis, cricket, baseball, or golf. Another explanation may be given of the sidelong force, arising from the velocity of liquid past a cylinder, which is encircled by a vortex. Taking two planes x = ± b, and considering the increase of momentum in the liquid between them, due to the entry and exit of liquid momentum, the increase across dy in the direction Oy, due to elements at P and P´ at opposite ends of the diameter PP´, is [rho]dy (U - Ua²r^(-2) cos 2[theta] + mr^(-1) sin [theta])(Ua²r^(-2) sin 2[theta] + mr^(-1) cos [theta]) + [rho]dy (- U + Ua²r^(-2) cos 2[theta] + mr^(-1) sin [theta])(Ua²r^(-2) sin 2[theta] - mr^(-1) cos [theta]) = 2[rho]dymUr^(-1)(cos [theta] - a^2r^(-2)cos 3[theta]), (8) and with y = b tan [theta], r = b sec [theta], this is 2[rho]mUd[theta] (1 - a²b^(-2) cos 3[theta] cos [theta]), (9) and integrating between the limits [theta] = ±½[pi], the resultant, as before, is 2[pi][rho]mU. 31. _Example 2.--Confocal Elliptic Cylinders._--Employ the elliptic coordinates [eta], [xi], and [zeta] = [eta] + [xi]i, such that z = c ch[zeta], x = c ch [eta] cos [xi], y = c sh [eta] sin [zeta]; (1) then the curves for which [eta] and [xi] are constant are confocal ellipses and hyperbolas, and d(x, y) J = -------, [xi]) = c²(ch²[eta] - cos² [xi]) d([eta] = ½c²(ch 2[eta] - cos 2[xi]) = r1r2 = OD², (2) if OD is the semi-diameter conjugate to OP, and r1, r2 the focal distances, r1, r2 = c(ch[eta] ± cos [xi]); (3) r² = x² + y² = c²(ch²[eta] - sin² [xi]) = ½c²(ch 2[eta] + cos 2[xi]). (4) Consider the streaming motion given by w = m ch([zeta] - [gamma]), [gamma] = [alpha] + [beta]i, (5) [phi] = m ch([eta] - [alpha]) cos ([xi] - [beta]), [psi] = m sh([eta] - [alpha]) sin ([xi] - [beta]). (6) Then [psi] = 0 over the ellipse [eta] = [alpha], and the hyperbola [xi] = [beta], so that these may be taken as fixed boundaries; and [psi] is a constant on a C4. Over any ellipse [eta], moving with components U and V of velocity, [psi]´ = [psi] + Uy - Vx = [m sh([eta] - [alpha]) cos [beta] + Uc sh[eta]] sin [xi] -[m sh ([eta] - [alpha]) sin [beta] + Vc ch [eta] cos [xi]; (7) so that [psi]´ = 0, if