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    Re: Exercise #13 Azimuth of a planet
    From: Andrés Ruiz
    Date: 2008 Jun 13, 10:49 +0200

    gyro error = 231.9-232.7 = -0.8º

     

    Fecha y hora UT1

    12/06/2008

    19:29:00

     

    SITUACION ESTIMADA

    Latitud: Be = 15.233333 º =  15º 14.0'

    Longitud: Le = 145.116667 º = 145º  7.0'

     

    ALMANAQUE NAUTICO

    Cuerpo celeste: Planeta Jupiter

    Polo de iluminacion del astro: Coordenadas Ecuatoriales

    GHA = 261.445097 º = 261º 26.7'

    Dec = -21.988427 º = -21º 59.3'

    Correcciones altura:

    SD  = 0.384586 '

    HP  = 0.034311 '

     

    DETERMINANTE DE LA RECTA DE ALTURA

    Angulo Horario Local: LHA = 46.561764 º =  46º 33.7'

    Coordenadas Horizontales: Hc, Z

    Altura Calculada: Hc = 31.1163 º =  31º  7.0'

    Azimut: Z = 231.9 º

     

     

    Andrés

     

    -----Mensaje original-----
    De: NavList@fer3.com [mailto:NavList@fer3.com] En nombre de Anabasis
    Enviado el: viernes, 13 de junio de 2008 10:35
    Para: NavList
    Asunto: [NavList 5414] Exercise #13 Azimuth of a planet

     

     

    Exercise #13  Azimuth of a planet

     

    On 13 June 2008, a ship is approaching anchorage.  On the gear test

    check off list, there is a box stating “gyro error calculated and

    posted.”  The sky is mostly overcast and the sun looks like it will be

    obscured by clouds for most of the morning.  Fortunately abaft of the

    starboard beam, Jupiter is shining brightly in the late morning

    twilight.  The alidade was set up on the starboard gyro compass

    repeater and a bearing was taken at 05h 29m 00s ZT.  (ZD -10).  The

    bearing was 232.7 degrees.  The ship is in position Latitude 15deg 14’

    North, Longitude 145deg 07’East.  Determine the gyro error for

    arrival.

     

    Jeremy

     


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