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    Re: Basics of computing sunrise/sunset
    From: Christian Scheele
    Date: 2009 Jun 17, 23:46 +0200
    Okay. I appreciate your help. I think I'll go back to the equation with substituted values again.
     
    Regards,
     
    CS
    ----- Original Message -----
    Sent: Wednesday, June 17, 2009 11:39 PM
    Subject: [NavList 8679] Re: Basics of computing sunrise/sunset

    The equation looks right

    On Wed, Jun 17, 2009 at 5:35 PM, Christian Scheele <scheele@telkomsa.net> wrote:

    I've considered EoT. I'm using the upper limb on horizon definition of
    sunset and am therefore applying the semi-diameter of 16 minutes. I've also
    corrected my result for error within time zone. Am still checking, it must
    be a trig error.

    Thanks for your time.

    The very best regards,

    Christian Scheele




    ----- Original Message -----
    From: "Brad Morris" <bmorris@tactronics.com>
    To: <NavList@fer3.com>
    Sent: Wednesday, June 17, 2009 10:58 PM
    Subject: [NavList 8676] Re: Basics of computing sunrise/sunset



    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
    Brad


    -----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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