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    Re: The Great Circle Challenge
    From: Jonathan M Bresler
    Date: 2014 Dec 26, 12:35 -0800

    Start: Cook Strait {at} 41d 30’S, 174d 30’E           End:   Waimea Bay  {at} 21d 57’N, 159d 40’W

     

    Using HO-229:

    Case II -- Latitudes of departure and destination of contrary name and great-circle distance less than 90°. Enter the tables with latitude of departure as latitude argument (Contrary Name) and, latitude of destination as declination argument, and with the difference of longitude as local hour angle argument. If the respondents do not lie above the C-S Line on the right-hand page, Case IV is applicable. Extract the tabular altitude which subtracted from 90° is the desired great-circle distance. The tabular azimuth angle is the initial great-circle course angle.

    Must use an assumed position for the start that results in a whole degree latitude for the start and a whole degree LHA from start to destination.

    Assumed Position for start: 41d S, 174d 20'E

    End longitude                               159d 40'W

    LHA                                             334d 00'

    HO-229 Vol 3 Page 237: Hc 23d 27.8'  (tabular altitude Ht); d -56.0; Z 153.5

     

    S. Lat LHA > 180d Zn = 180d - Z

                      Zn = 180d - 153.5d

                      Zn = 026.5d

     

    Declination Increment 57' yields -47.5 for 50' and -5.7 for 06', total -53.2

    Hc (Ht) 23d 27.8'

               22d 87.8'

                    -53.2'

    Hc       22d 34.6'

     

            89d 59.10'

    Hc    22d 34.6'

            67d 25.4'

     

    Distance = 4045.4nm = 67 * 60nm + 25.4 Actually short due to assumed position being 30nm north of actual.

     

    Course   = 026.5

     

    The time was far too long, about 30 minutes, as there was some reading instructions, learning, re-reading instructions involved.

     

    Jonathan

       
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