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Given the semi-major axis and a flattening factor, is it possible to calculate the semi-minor axis?

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    @Katie also, I was being guided by the joint top most on-topic question at http://area51.stackexchange.com/proposals/3355?phase=definition which relates quite strongly, and is possibly easier to find an algorithm for2010-07-20

2 Answers 2

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Possibly something like this. Correct me if I'm wrong.

$j$ = semi-major
$n$ = semi-minor
$e$ = eccentricity

$n = \sqrt{(j\sqrt{1 - e^{2}}) \times (j(1 - e^{2}))}$

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Where, $a$ = transverse radius = semi-major axis (for ellipse/oblate spheroid);
$b$ = conjugate radius = semi-minor axis (" " ");
$oe$ = angular eccentricity = $\arccos(\frac{b}{a})$;
$f$ = flattening = $\frac{a-b}{a} = 1 - \frac{b}{a} = 1 - \cos(oe) = 2\sin(\frac{oe}{2})^{2}$;

then $b = a\cos(oe) = a(1-f)$.