Explanation of sin(ω/2) = |a-b|/(a+b) on an ellipse.

Question

In the paper "Map projections - a working manual" by John P. Snyder, I've come across result (4-1), but there's is no explanation given as to where it came from. The omega is defined in the second paragraph. Is there a geometrical explanation?

Answer 1

I think this is only an approximate formula, valid if $a-b$ is small (I'll suppose $a>b$), let's see why.

If two diameters of the circle on the left are perpendicular, after the deformation they become a couple of conjugate diameters of the ellipse. If we denote by $\alpha$, $\beta$ their lengths, and by $\theta$ the angle between them ($\theta\ge90°$), then we have by the theorems of Apollonius: $$ \sin\theta={ab\over\alpha\beta},\quad \alpha^2+\beta^2=a^2+b^2. $$ The maximum value of $\theta$ is attained for the minimum value of $\sin\theta$: by AM-GM inequality we immediately get: $$ \sin\theta_\max={2ab\over a^2+b^2}. $$ Now I must do some guesswork: I suppose $\omega=\theta_\max-90°$ (this is not so clear from the given paper), which gives: $$ \cos\omega={2ab\over a^2+b^2}. $$ From the trig identity $\cos\omega=1-2\sin^2(\omega/2)$ we then get: $$ \sin{\omega\over2}={a-b\over\sqrt{2(a^2+b^2)}}. $$ This is an exact formula. But, to first order in ${a-b\over a+b}$, we have $$ \sqrt{2(a^2+b^2)}=(a+b)\sqrt{1+\left({a-b\over a+b}\right)^2}\approx a+b, $$ which gives your formula: $$ \sin{\omega\over2}\approx{a-b\over a+b}. $$