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    Re: No Two-Body Fix Problems
    From: George Huxtable
    Date: 2009 Nov 6, 17:25 -0000

    John Karl wrote, in [10464]
    | As I posted on the other thread, It seems to me that if the computer
    | code simply calculates the answers for the two possibles fixes, using A
    | +B and A-B in Eq. 7.5d, and then lets the navigator pick the closest
    | fix, all possible combinations of the locations of the GPs and ship
    | are covered.
    | And I don't see any problem with the ship being on, or near, GP1's
    | meridian.  This is a well defined problem with no ambiguity in the
    | equations, and the navigator doesn't need to know his position
    | relative to the meridian.
    Those pages in that book "Celestial Navigation in the GPS age", are John's
    copyright, but it could be helpful to Navlist readers who wish to follow
    this matter if he would permit me to post copies of pages 78-9, or better,
    did so himself; with updates, if any, from a later edition.
    Thie continued discussion still leaves me puzzled. Despite all the maths and
    spreadsheets, I think something is being missed.
    The example shown in fig 7.3 has the ship somehere between GHA1 and GHA2, in
    longitude, and then everything works as described. But what if it isn't?
    What if the ship is actually to the West of GP1?
    That happens in two cases; either if A-B is less than zero, or if A+B is
    greater than 180. Then that needs to change the sign of LHA1, in 7.5f. It's
    the same thing as taking the alternative value for arc-cos LHA1, when
    obtaining the answer to 7.5e. That step, in getting the answer from 7.5e to
    plug in to 7.5f, is non-trivial, and needs spelling out more carefully.
    John is correct in maintaining that there's no actual "ambiguity" about
    this, that needs to be resolved. Only one of the two alternative solutions
    for arc-cos LHA1  can actually apply, but to discover which, the
    corresponding value of (A-B) or (A+B), whichever is the chosen one, has to
    be inspected.
    John's equations may provide what looks like a perfect mathematical
    solution. But to get the right answer, it seems to me that a real-life
    navigator needs a bit more.
    Or am I misunderstanding the matter, rather seriously? Could be...
    contact George Huxtable, at  george{at}hux.me.uk
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    NavList message boards: www.fer3.com/arc
    Or post by email to: NavList@fer3.com
    To unsubscribe, email NavList+unsubscribe@fer3.com

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