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. $\endgroup$– NoldorinCommented Jul 20, 2010 at 19:37
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$\begingroup$ "versine of the spheroid's angular eccentricity" $\endgroup$– Rowland ShawCommented Jul 20, 2010 at 19:47
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$\begingroup$ 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 BanksCommented Jul 20, 2010 at 20:16
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$\begingroup$ Please use meta.math.stackexchange.com to discuss what's approprate. Thanks! :) $\endgroup$– Jon BringhurstCommented Jul 20, 2010 at 20:21
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1$\begingroup$ 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. $\endgroup$– NickCommented Jul 20, 2010 at 20:24
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2 Answers
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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 $b = a\cos(oe) = a(1-f)$.