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    Re: time of meridian passage accuracy
    From: Andrés Ruiz
    Date: 2009 Sep 29, 09:41 +0200

    Interesting points of view about of subject.

     

    Gary LaPook in [NavList 9968] is right, but calculating the position at LAN has the taste of old times, when navigation was more art and less science. Do you remember the film “master & commander” taking shoots at noon aboard the HMS Surprise?.

     

    George Huxtable, [NavList 9958], has put his finger on the pulse, the problem is longitude, because the rate of change of GHA.

     

    My opinion is the same as Joseph Schultz, [NavList 9964]. This fact is problem for the beginners when take a prestigious book as a reference.

     

     

    Recapitulating:

     

    Avoid the EoT, BOWDITCH says: [NavList 9914]

    “To calculate latitude and longitude at LAN, the navigator seldom requires the time of meridian passage to accuracies greater than one minute”

     

    At time of LAN:

    LHA = 0 for upper culmination and 180º for lower.

    Then |L| = |GHA|

    Sin H = sin Dec sin B +/- cos Dec cos B

    Where B is the latitude

    B = f( Dec, H )

    L = f( GHA )

    And with some simplifications the error e is function of the rates of change of the variables:

    eB = f( dDec/dt ) since at culmination dH/dt = 0 and is near transit.

    eL = f( dGHA/dt )

     

    This is the academic point. Douglas Denny, [NavList 9939], compare practical navigation and theoretical astronomy.

    For practical navigation, the question is: are you near the coast or shallow waters? Is the error of 1 min acceptable? Are you in blue water with no dangers around? Of course, the position obtained in such a way is better than nothing.

     

    A numerical example: (Position from observation of a single body. Jim N. Wilson)

     

    30/12/1982  19:55:45

    Sun

    Number of shoots = 26

     

    a0 = -664.781491

    a1 = 69.933785

    a2 = -1.752351

    Mean Square Error

    mse = 0.000257

    Maximun Altitude:

    t (Hmax) = 19.954279 h = 19:57:15

    Hmax = 32.957642 º

     

    LAN (James Wilson): (full equation)

    t LAN = 19.929029 h = 19:55:45

    Dec = -23.149089 º

    GHA = 118.285394 º

    Checks:

    LHA = 0.000000 º

    Hc = 32.956525 º

    Z = 180.000004 º

     

    heye = 1.86 m

    IC = 1.5

     

    Latitude

    Dec = -23.149088 = -23º  8.9'

    Hs = 32.956525 =  32º 57.4'

    Ho = 33.189697 =  33º 11.4'

    Culmination = S

    B LAN = 33.661215 =  N33º 39.7'

     

    Longitude

    30/12/1982  19:55:45

    GHA = 118.287461

    L = W118.287461 = W118º 17.25’

     

     

    For other UT  - Peter Hakel has  provided his solution: [NavList 9961]

    30/12/1982  19:55:42

    Dec = -23.149091 = -23º  8.9'

    B = 33.661212 =  33º 39.7'

    GHA = 118.274965

    L = W118.274965 = W118º 16.50’

     

    30/12/1982  19:55:44

    Dec = -23.149089 = -23º  8.9'

    B = 33.661214 =  33º 39.7'

    GHA = 118.283296

    L = W118.283296 = W118º 17.00’

     

     

    Introducing an error of 1 min:

    30/12/1982  19:56:45

    Dec = -23.149041 = -23º  8.9'

    B = 33.661262 =  33º 39.7'

    GHA = 118.537378

    L = W118.537378 = W118º 32.24’

     

     

    In my humble opinion, 1 min of error in time of LAN is unacceptable for longitude calculations, but is not a problem for latitude.

     

     

    Andrés Ruiz

    Navigational Algorithms

    http://sites.google.com/site/navigationalalgorithms/

     


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