Response to In which Matt Parker introduces equations for the perimeter of an ellipse. My solution is a very good approximation (I am curious if the exponent e/2 is entirely accidental? Is there a clever math theorist out there who can explain if there's a reason this term improves the approximation?). c ≈ 2πr*sin(Θ)^(e/2) + 4r - 4r*sin(Θ)^(e/2) Where sin(Θ) is the viewing angle. Deriving a generalized equation using the a and b terms (because sin(Θ) is simply b/a) to make it more useful: c ≈ 2πa*(b/a)^(e/2) + 4a - 4a*(b/a)^(e/2) @4:55 yes, zero to a power is zero. . . . . . Introducing a logarithm to the exponent term so it varies from .5 to .62 as the angle drops to zero produces perimeter values that are accurate to several decimals as well as giving you a null result at 0°, which it should, since technically that is no longer an ellipse. Special thank you to tim1328 who asked the same questions others have asked, but in a way that helped me see where I'd made an error in deriving the general form.