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    Re: Basics of computing sunrise/sunset
    From: Brad Morris
    Date: 2009 Jun 17, 16:58 -0400

    Please note that you must consider the Equation of Time to convert from Apparent Time to Mean Time.
    Be careful about what you mean by sunset, there are many
    1) Upper limb of sun touches the horizon
    2) sun 6 degrees below horizon
    3) sun 12 degrees below horizon
    4) sun 18 degrees below horizon
    You must consider your offset in longitude from the center longitude of your 
    time zone.  That is, sunset occurs earlier in the eastern section of a given 
    time zone than a sunset in the same time zone, but on the western edge!
    That should pretty much do it!
    Best Regards
    -----Original Message-----
    From: NavList@fer3.com [mailto:NavList@fer3.com] On Behalf Of Christian Scheele
    Sent: Wednesday, June 17, 2009 4:41 PM
    To: NavList@fer3.com
    Subject: [NavList 8673] Basics of computing sunrise/sunset
    I'm trying to solve the following sunrise/sunset astronomical triangle and
    am encountering unexpected pitfalls.
    cos(LHA) = [sin (a) - sin (lat) sin(dec)]/ cos(lat)cos(dec)
    where I am assuming a = -0.83 degrees
    It is of course something very basic and I'm almost embarassed to say that
    although I keep rechecking what I'm doing, I remain with an error of about
    15 minutes by the time I get to the final result in mean zone time. Before I
    drag everybody into it, could somebody please tell me whether there are any
    snags that I should look out for?
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