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    Re: Diego Garcia Lunar
    From: Andrés Ruiz
    Date: 2010 Dec 17, 11:44 +0100

    Antoine wrote:

    2 - If we assume that both your position and your Observed Sextant values are correct, then your UT would have been 14h41m16.5s , (or LT = 8h41m16.5s), which translates into a difference of 2m29.5s (+6h) between UT and Watch.

    Could you be wrong with the time UT/LT? At 14:43:46 the Sun is not be seen at Diego Garcia!

     

     

     

     

     

    Here is my solution.

     

     

     

     

     

    The input file is added. The solution is:

     

    12/12/2010

    08:43:46 UT1

    Geocentric equatorial coordinates

    Moon:

    GHA = 234.950673 º = 234º 57.0'

    Dec = -4.529893 º =  -4º 31.8'

    Phase: 39% (+)

    Sun:

    GHA = 312.539490 º = 312º 32.4'

    Dec = -23.076579 º = -23º  4.6'

    Geocentric lunar distance

    LD = 76.816700 º = +76º 49' 0.1"

     

    DR:

    B = -7.271667 =  -7º 16.3'

    L = 72.446667 =  72º 26.8'

     

    IE = -1.200000 '

    air T = 10.0 ºC

    air P = 1010.0 hPa

    h Eye = 32.31 m

    Dip = 0.166733

     

    Time by lunar distances

    Moon:

    Hs = 37.115425 =  37º  6.9'

    Limb = 0

    SD = 14.798223 '

    HP = 54.308845 '

    R  = 0.021554 º

    OB = -0.000037 º

    PA = 0.723512 º

    AG = 0.002363 º

    SDag = 14.940022 '

    Sun:

    Hs = 61.451472 =  61º 27.1'

    Limb = 0

    SD = 16.243914 '

    HP = 0.148858 '

    R  = 0.008882 º

    OB = 0.000001 º

    PA = 0.001194 º

    AG = 0.000010 º

    SDag = 16.244531 '

    Lunar observation:

    LDs = 76.966667 =  76º 58.0'

    Moon Limb = 1

    body Limb = 1

    Clearing Lunar Distance:

    m = 36.928692  =  36º 55.7'

    M = 37.630650375632321  =  37º 37.8'

    s = 61.264739  =  61º 15.9'

    S = 61.257050084970373  =  61º 15.4'

    d = 77.466410  =  77º 28.0'

     

    T1 = 7.000000 LD1 = 76.034602

    Tc = 8.696442 LDo = 76.801740

    T2 = 9.000000 LD2 = 76.939010

     

    Error:

    Ta = 08:43:46 LDc =  76º 49.0'

    Tc = 08:41:47 LDo =  76º 48.1'

    Tc = 08:41:47 LDc =  76º 48.1'

    |LDo-LDc(Ta)| = 0.014960º = 0.897612'

    |LDo-LDc(Tc)| = 0.000040º = 0.002397'

    |Ta-Tc| = 1.980155 min = 118.809318 s

     

    Regards,

    --
    Andrés Ruiz
    Navigational Algorithms
    http://sites.google.com/site/navigationalalgorithms/

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