Do you share my annoyance that the formula for the a-by-b ellipse, √(1-(b/a)²), only works for a ≥ b? Like turning an ellipse 90º turns it into a hyperbola? Is there a formula for √(1-(Min[a,b]/Max[a,b])²) nicer than √(1 - E^(-2 Abs@Log[b/a]))? Some arctrig(trig)? True|False puzzle: An ellipse is a special case of an ellipsoid. My answer: .si espille derevoc-elbuod a tub ,oN The three eccentricities of the axis-aligned cross-sections of an a by b by c ellipsoid are not independent, and the asymmetry of the eccentricity formula makes things weird. If a ≥ b ≥ c, the eccentricities are ba := √(1-(b/a)²) cb := √(1-(c/b)²) ca := √(1-(c/a)²) with ba^2 = (ca^2 - cb^2)/(1 - cb^2) i.e. cb^2 = (ca^2 - ba^2)/(1 - ba^2) but then ca^2 = ba^2 + cb^2 - ba^2 cb^2 For each ellipse shape, we can imagine a different π: π(eccentricity) = circumference/major axis, π(0) = π, π(1) = 2, π(𝜀) = 2 EllipticE(𝜀²). This would make a nice alternative quantification of ellipticity if EllipticE weren't such a pain. It is possible to choose semiaxes a > b > c (the interesting spinning body!) so that at least two of the elliptic integrals come out in closed form. E.g., eccentricities ca = (√2 - 1)², cb = 4^⅝ (√2 - 1), and hence ba = √(6√2 - 3)/4^⅝ determine π ratios π( (√2 - 1)²) = 2 EllipticE( (√2 - 1)⁴) = (2-√2) π^(3/2)/(8 ¼!²) + 4√(2/π) ¼!² π(4^⅝ (√2 - 1)) = 2 EllipticE(8 - 4 √2) = (2-√2)√π³/(4 ¼!²) + 4 ¼!²/√π π(√(6√2 - 3)/4^⅝) = 2 EllipticE[⅜(4 - √2)] = ? I haven't looked very hard for a nonspheroidal ellipsoid with all three (π)s in closed form. —rwg What kind of prigs would deprive us of vulgar fractions 3/2 and 5/4? 5/4 is traditional in carpentry. And 3/2 dates back millennia as "sesqui". (I know of a school that still marks kids off for "improper fractions".) Not that anyone has said otherwise, but there are infinitely many "non singular-value" k for which EllipticK(k) (and EllipticE(k)) come out in closed form. E.g., EllipticK[1/2 (1 - Sqrt[2])]/EllipticK[1/2 (1 + Sqrt[2])] == 1/3 (I + Sqrt[2]) (instead of the square root of an integer) EllipticK[1/2 (1 - Sqrt[2])] == (8 2^(1/4) (1/8)! (3/8)!)/(3 Sqrt[\[Pi]]) EllipticK[1/2 (1 + Sqrt[2])] == (8 2^(1/4) (1/8)! (3/8)!)/((I + Sqrt[2]) Sqrt[\[Pi]]) (These would be more interesting if they were real. Unless they can't be. Which would be interesting.)