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    Fw: Re: A simple three-body fix puzzle
    From: George Huxtable
    Date: 2010 Dec 9, 10:15 -0000

    Frank wrote-
    
    "I'm also attaching the scan of the relevant paragraph. Note that in the
    last line, L and B with subscripts refer to longitude and latitude, but all
    we need here is the difference in these in miles which is given by
    (AE-BD)/G and (CD-BE)/G".
    
    ===============
    
    No, not in miles.
    
    Frank has taken the result of that last step in the iteration, which gave a
    difference in degrees of long and lat, and omitted the cos Bf term in the
    denominator of the long to provide the longitudinal distance instead. But
    those distances, in lat and long, being converted degrees, will be in units
    of 60 miles, not of miles.
    
    The Almanac doesn't explain well, and Frank hasn't either, that this is
    just one step in an iteration, which starts from a guessed initial
    position, such as the DR position. The next step is to test whether the
    resulting change in position is outside some preset limit, and the Almanac
    suggests 20 miles. Depending on how close the DR position was, a single
    iteration may well suffice. It depends.
    
    If another iteration is called for (and this is the bit that isn't
    explained well) then all the azimuths and intercepts have to be
    recalculated, using the new values of assumed long and lat, and the
    summations repeated using those new values, and these are used to provide a
    better estimate still. Convergence is usually rapid.
    
    If only two observations are being worked, there will of course be two
    solutions to the problem.. If you happened to choose an initial position
    that was on the great circle of points that are equidistant between those
    two solutions, the program is likely to get into a mess, not knowing which
    way to turn, and infinities may crop up. Otherwise, the algorithm seems to
    find its way to whichever solution is in the same hemisphere as the initial
    position was placed in, in relation to that equidistant great circle.
    Having found one solution from some chosen starting point, you can then be
    sure to flush out its partner by starting from the antipode.
    
    The same procedure is described also in the publication by the British
    Nautical Almanac Office, "AstronavPC and compact data 2001-2005", which is
    the edition that I have. It then goes on to deduce procedures for obtaining
    the error-ellipse parameters, but there are errors, which have been
    discussed here before.. I don't know whether later editions have put things
    right. I hope so.
    
    George.
    
    contact George Huxtable, at  george@hux.me.uk
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    
    
    

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