...terms of complete elliptic integrals, gives (1.135) where K, _ I -4? sin2 (9 E(q) = \-tC-sm26de (1.136) are the complete elliptic integrals of the first and second kind, respectively. The kernel (1.135) has a logarithmic singularity at p = p. For a half-full tank, the kernel function...

...the reduced coupling constant, K given by Equation (2.147b), «' - 2tanh 2 2K - 1 and F(K) and E(K) are the complete elliptic integrals of the first and second kind, respectively: F( K ) = r \ dtp, (2.155a) Jo v 1 — K 2 sin (f (a) 0 (b) 3 T/T C Note that at T = T c the variable...

...functions of k, ab = -K(k)(E(k) - K(k)). be — -K(k)(E(k)-(l k2)K(k)), ac = -K(k)E(k), and K(k), E(k) are the complete elliptic integrals of the first and second kind, respectively, with 0 < k < 1 . The coordinate t is given by the change of variables / = — ~ ^K^k) ' up to an additive...

...of k, ab = -K(k}(E(k) -K(k)), bc = -K(k)(E(k) - (1 k2)K(k)), ac — —K(k)E(k), and £X&), £(fc) are the complete elliptic integrals of the first and second kind, respectively, with 0 < k < 1. The coordinate t is given by the change of variables t — — ^K~(k) UP to an additive...

...by the mirror image is (5.1.2) where r'± = \/ (a T R')2 + (z T Z')2, A' = (r_ - r'+)/(r'_ + r'+). and F and E are the complete elliptic integrals of the first and second kind, respectively. The vortex ring is assumed to have a similar core structure with initial core size 6Q. Because of symmetry,...

...considerations give us x(u) = -a(\-e)[F(sn(u,k),k)+K(k)] - a (1+ e) [E(sn(w, k), k) + £(*)] where K(k) and E(k) are the complete elliptic integrals of the first and second kind, respectively. We must also convert y(<j>) to y(u) by the transformation sin0 = 1 — 2sn2(w). This gives y(u) = av/(l...

...are the radii of the two circles, d is the distance between their centers, and k = <P sin 7. fff ji F and E are the complete elliptic integrals of the first and second kind, respectively, to modulus k. He also obtained an expression for the mutual inductance between coaxial circles in the...

...constant and y 2 cos2<f>d<f> .(5-51) .'oVl - sin2 \y sin2 <£ = E — K cos* fa> (5-52), A' and E being the complete elliptic integrals of the first and second kind respectively to the modulus sin Jy. By considering the integral in equation (5-51) we can show that r/X or y^ steadily increases...

...fa = where (8) an= (l + p)2 + £2/i'n, (9a) /32 = (1 -p)* + £*/»',«, (9b) and where Kn and A',, are the complete elliptic integrals of the first and second kind, respectively, with modulus mn given by (10) A similar procedure is used to calculate the displacement component n-...

...is found from 1 ra i; = i = 4 / z2 ds a2 -~ b2 /14 b2y2 a2(a2 -y2) dy = 4ab2 %-EJ 3e2 (A18) and EJJ are the complete elliptic integrals of the first and second kind, respectively, icre (A19) eccentricity. The following approximate expressions for P and Iy were found to give good...