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    Re: No Two-Body Fix Problems
    From: Peter Hakel
    Date: 2009 Nov 6, 10:52 -0800
    I think the simplest thing anyone on NavList can do is to use the latest version of that spreadsheet, enter numbers defining all kinds of arrangements and see if the results make sense.  If something breaks (I am confident now that this can happen only in highly academic cases), then we can dig deeper and fix it.  The most recent numerical problems I saw had nothing to do with John's equations.  They are caused by Excel's frustrating ability to return NaN instead of zero as a result of ACOS(1.00000....).  Go figure...

    In my original artificial example the ship was in fact west to both GP1 and GP2 along the equator, with osculating LOPs.  Once the LHA1 caveat of going from Eq. 7.5e to 7.5f was explicitly addressed, John's equations passed that test (and all others I did) with flying colors.

    I think that computers and trig equations give us a benefit that old-fashioned printed sight-reduction tables do not - the ability to use negative numbers in the same formulae without any special treatment.  Nowhere in this debate do we need to worry about terms like "same" and "contrary."  Nor has this discussion been complicated by the notion that in the southern hemisphere the celestial triangle is "upside down" with the South Pole as one of its vertices.  (My apologies to our Anzac and South African friends. :-) ).  Negative latitude values seem to take adequate care of all that.  And hour angles do not have to always conform (certainly for intermediate results) to the 0<=HA<360 convention; LHA=-10 works just as well as LHA=350.  I convert longitudes to their usual signage and intervals only at the very end on output.

    Peter Hakel

    From: George Huxtable <george@hux.me.uk>
    To: navlist@fer3.com
    Sent: Fri, November 6, 2009 9:25:57 AM
    Subject: [NavList 10473] Re: No Two-Body Fix Problems

    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.

    [rest deleted by PH]

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