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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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  • 1
    $\begingroup$ Define 'flattening factor' please. $\endgroup$ – Noldorin Jul 20 '10 at 19:37
  • $\begingroup$ "versine of the spheroid's angular eccentricity" $\endgroup$ – Rowland Shaw Jul 20 '10 at 19:47
  • $\begingroup$ do we want to allow/encourage questions that could easily be homework questions? I know we're not MathOverflow and we don't need research-level questions, but to me it still seems like there are better places for homework help (the line being much thinner here than it is on MO). (Rowland, I'm not looking to disparage this as a homework question, it's just asked a lot like one). $\endgroup$ – Jamie Banks Jul 20 '10 at 20:16
  • $\begingroup$ Please use meta.math.stackexchange.com to discuss what's approprate. Thanks! :) $\endgroup$ – Jon Bringhurst Jul 20 '10 at 20:21
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    $\begingroup$ I think this particular case is useful and acceptable. I'm waiting to flag homework questions that have specific values in mind. And a great answer could go into some good detail here, not just provide a formula. $\endgroup$ – Nick Jul 20 '10 at 20:24
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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 $b = a\cos(oe) = a(1-f)$.

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