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Position angle of Moon's bright limb (formula)

  1. Mar 4, 2009 #1
    Hi everyone !

    I have a question about the position angle of Moon's bright limb.
    In "Astronomical Algorithms" (Jean Meeus), one can find the formula to calculate this angle (formula is tagged "46.5"), but there is no explanation about the derivation of this formula.

    This is the formula "46.5" :

    [tex]tan \chi = \frac{cos \delta_0.sin(\alpha_0 - \alpha)}{sin \delta_0.cos \delta - cos \delta_0.sin \delta.cos(\alpha_0 - \alpha)}[/tex]

    where [tex]\alpha[/tex], [tex]\delta[/tex] are the geocentric right ascension and declination of the Moon and
    [tex]\alpha_0[/tex], [tex]\delta_0[/tex] are the geocentric right ascension and declination of the Sun.

    I think the derivation is maybe explained in the "Practical astronomy with your calculator" (Peter Duffett-Smith) but, unfortunately, I don't have this book at home.

    Does anybody can explain the derivation of this formula 46.5 ?

    Thanks (... and sorry for my approximative english)
    Jeff (from France)
     
  2. jcsd
  3. Mar 4, 2009 #2
    Of course [tex]\chi[/tex] is the so-called "position angle of the Moon's bright limb"...

    Jeff from France
     
  4. Mar 14, 2009 #3
    No one ?
     
  5. Mar 14, 2009 #4

    tiny-tim

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    Welcome to PF!

    Hi Jeff! Welcome to PF! :wink:

    Without a diagram, it's difficult to see what's what …

    but the denominator is the standard spherical trig formula for cos of the side of a triangle if the other two sides are δ and 90º - δ0, and the opposite angle is α0 - α :smile:
     
  6. Jul 9, 2010 #5
    Re: Welcome to PF!

    Does anyone capable of tracing a diagram of this spherical triangle in order to understqnd the relation given above ?
     
  7. Jul 16, 2011 #6
    Last edited by a moderator: Apr 26, 2017
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