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    Re: Azimuth Formula Questions
    From: Bill B
    Date: 2005 Oct 31, 18:18 -0500

    George wrote:
    
    > But even after those corrections, is that now correct, I wonder? Here, the
    > mystery deepens. I don't think it gives the right answers.
    
    Paul is correct in his note as to which table values are taken from.
    Latitude vs. declination is the key.  They are either both north or both
    south (same) or have opposing signs (contrary).
    
    That being said, 229 claims to derive values via the formula given.
    
    EXCEPT WHEN THE ELEVATION OF THE BODY IS WITHIN 1.5 d OF THE ZENITH.
    
    Which it would be in your example.  Then 229 uses a formula something like
    (my keyboarding skills are, as you know, poor):
    
    sin^2 1/2z = cos^2 1/2LHA sin^2 1/2(L-d) + sin^2 1/2LHA cos^2 1/2(L+d),
    where z is the zenith distance.
    
    I doubt rounding errors by using the primary 229 formula are causing the
    difference you see.
    
    How they are adjusted for LHA 0 to 90 & 270 to 360 Same tables,
    LHA 90 to 270 Same tables, and LHA 0 to 90 & 270 to 360 Contrary tables I do
    not know. This is, I suspect, is of the utmost importance. The values may
    have been adjusted for each table, therefore their rules apply.
    
    Unfortunately I do not have the 229 volume with tables for the 0-15 latitude
    range to inspect the relationship.  I would welcome values for LHA 1, Dec 1,
    Latitude 1 from the Same and Contrary tables if anyone has that volume.
    
    Bill
    
    
    

       
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