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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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Rowland Shaw
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    Define 'flattening factor' please. – Noldorin Jul 20 '10 at 19:37
  • "versine of the spheroid's angular eccentricity" – Rowland Shaw Jul 20 '10 at 19:47
  • 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). – Jamie Banks Jul 20 '10 at 20:16
  • Please use http://meta.math.stackexchange.com/ to discuss what's approprate. Thanks! :) – Jon Bringhurst Jul 20 '10 at 20:21
  • @Katie I'd be flattered if you think I was of homework age - as it happens, this was inspired from a software engineering problem I was having, and having never been taught geometry beyond spherical geometry, I was struggling to get my head around the terms. For reference, I wanted this is to generate the bits I need to feed in to do a Helmert transformation. – Rowland Shaw Jul 20 '10 at 20:23
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    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.

     – Nick Jul 20 '10 at 20:24
  • @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 for – Rowland Shaw Jul 20 '10 at 20:26

2 Answers2

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

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

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

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