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    Azimuths
    From: Brad Morris
    Date: 2019 Jun 30, 14:59 -0400
    Hello Brian

    You properly corrected me regarding azimuths of a celestial object when in the horizon.  After the second time this occured to me in recent memory, it caused me to review and see where my understanding was at issue.

    In both cases, I relied upon "Azimuths of the Sun", HO 71, substituting a star for the sun.  It is clear to me that I did not look up the azimuth correctly.   

    In review of the table, I found that the tables were derived from

    Azimuth = acos((sin(declination)/cos(latitude)))

    and that there was no need to fiddle about with the table.  I should directly compute the azimuth instead.

    A=acos((sin(-17°)/cos(-17))) = 107.8°
    For  a star at -17° declination, in the horizon, when observed from Tahiti.

    A=acos((sin(-17°)/cos(17))) = 107.8°
    For a star at -17° declination, in the horizon, when observed from Hawaii.

    I'd like to thank you, Brian, for helping me to broaden my horizons (heh) a bit.  I had an intuitive expectation that there would be a curve.  There was, just not at the curvature I expected. 

    A=acos((sin(-17°)/cos(0))) = 107.0°
    For a star at -17° declination, in the horizon, when observed from the equator.

    I attempted to see how the curve would look when extended from pole to pole.  In doing so, I found that the equation exploded when 
    Abs(latitude) > 90-Abs(declination)

    For the instant case of declination = -17°  the equation does not yield an azimuth for latitudes higher than +/-73°, as the acos of a number greater than 1 is not permitted.

    In between -73° and +73°, the azimuth varied from 180° (at either extreme) to 107° at the equator, in a long, gradual curve.  In between -/+20°, it behaved as you stated!

    Brad










    On Sun, Jun 30, 2019, 2:48 AM Brian Walton <NoReply_Walton@fer3.com> wrote:

    Brad,

       “Assume a star at 17°S”.   A star with a declination of 17°S ?

        A star with a declination of 17°S will rise on an azimuth of 090+17=107° anywhere within Lat 20N to 20S,  and set on an azimuth of 270-17=253° anywhere from 20N to 20S.  I am just using the rules as a sailor would. It has nothing to do with seasons, or the Sun.

        That star will also go over all places with a latitude of 17°S, every day. and if it is dark, and it goes overhead, you are on Lat 17°S. It has nothing to do with seasons, or the Sun.

    Brian Walton

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