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    Re: Resume of "Averaging"
    From: Chuck Taylor
    Date: 2004 Nov 6, 15:47 -0800

    Alexandre Eremenko and Herbert Prinz have been
    discussing the averaging of celestial sights.  The
    following is an attempt to summarize:
    
    Consider the following hypothetical situation (the
    actual numbers here are made up):
    
    Body       Time        Altitude
    
    Spica      20-00-00    25d 10.5'
    Spica      20-01-00    25d 05.1'
    
    Arcturus   20-03-00    35d 30.0'
    Arcturus   20-04-00    35d 25.0'
    Arcturus   20-05-00    35d 20.0'
    Arcturus   20-06-00    35d 15.0'
    Arcturus   20-07-00    35d 10.0'
    
    Dubhe      20-09-00    42d 00.0'
    
    We have 2 observations of Spica, 5 of Arcturus, and 1
    of Dubhe, for a total of 8 observations.  How should
    we proceed in obtaining a fix?
    
    One obvious solution is to simply choose one
    observation of each body, then reduce and plot.  The
    fix would be plotted somewhere within the "cocked
    hat".
    
    Alex would argue that we could do better than that by
    averaging the two sights of Spica (treating the
    average time and the average altitude as a one sight),
    averaging the 5 sights of Arcturus, and taking the
    single sight of Dubhe. (The motivation for averaging
    is the expection that random errors will cancel out in
    the process of averaging.)
    
    If I understand Herbert's objection correctly, he
    would claim, "Now wait a minute.  You have 5 sights of
    Arcturus, but only 2 of Spica and 1 of Dubhe.  How can
    you give the same amount of weight to the single sight
    of Dubhe as you do to the 5 sights of Arcturus?  There
    is more information in the 5 sights of Arcturus than
    in the 1 sight of Dubhe, yet you are giving them the
    same weights."  (I hope Herbert will forgive me for
    putting words in his mouth.  If I have misunderstood
    him, I apologize in advance.)
    
    In theory you can only justify giving equal weights to
    each set of observations if each set contains an equal
    number of sights.  In practice, you will probably not
    go far wrong anyway.  Nor will you go far wrong in
    practice by selecting one sight from each set ("run")
    of sights.
    
    I happen to have a copy of the paper, "On the
    Overdetermined Celestial Fix", by Thomas and Frederic
    Metcalf.  (Tom was once a contributor to this list.
    The last I heard, he was associated with the Institute
    of Astronomy, University of Hawaii, Honolulu.) I also
    have a copy of a follow-on paper, "An Extension to the
    Overdetermined Celestial Fix" by Tom Metcalf (alone).
    I obtained my copies of these papers some years ago
    directly from Tom.
    
    At the risk of gross oversimplification, I present the
    following analogy:
    
    Suppose you have two equations in two unknowns, each
    representing a line in a plane.  Assuming the two
    lines are not parallel, the two lines intersect at a
    single point, and that point can be found by solving
    those two equations simultaneously. The solution can
    be by one of several methods, including Gaussian
    elimination, Gauss-Jordon elimination, or by the use
    of matrix algebra.
    
    Now suppose that we have 3 equations in two unknowns.
    We observed above that a unique solution is determined
    by two equations in two unknowns.  Now we have an
    "overdetermined" set of equations with no unique
    solution.  This is where the method of least squares
    enters in.  We find the point such that the squared
    distance between that point and each of the lines in
    minimized.  As Herbert pointed out, we can do this
    using matrix algebra.
    
    Now, let us return to the example of our sights of
    Spica, Arcturus, and Dubhe.  Herbert's argument is
    that we ought to compute 8 lines of position (2 for
    Spica, 5 for Arcturus, and 1 for Dubhe) and find that
    point which minimizes the sum of the squared distances
    between it and each of the 8 lines of position.  That
    would be solving an overdetermined celestial fix by
    the method of least squares.  With this method, each
    observation from each body is given equal weight.
    
    Herbert, have I stated your argument fairly?
    
    The Metcalf papers actually address a slightly
    different form of overdetermined celestial fix.
    Quoting from the earlier of the two papers, "... using
    these methods, an accurate fix can quite often be
    determined from 10 to 20 observations of a single
    celestial body spanning only 10 to 20 min of time."
    The algorithm uses a Lagrange multiplier and the
    simultaneous solution of a nonlinear equation for the
    Lagrange multiplier and a number of linear equations.
    The solution is iterative.  It's an interesting
    concept.  Tom tried it out with "real observations of
    the Moon taken on land with an inexpensive plastic
    sextant, and with a pan of vegetable oil serving as an
    artifical horizon."  He took 18 sights over
    approximately 35 minutes, and was able to confirm his
    known position within about 2.5 nautical miles. He
    programmed his algorithm on an HP-48SX calculator.
    The paper offers to make the code available to
    interested parties who send a stamped, self-addressed
    envelope.  I'm not sure whether there was any
    expiration date on the offer, as it appeared in
    "Navigation: Journal of The Institute of Navigation",
    Vol 38, No. 1, Spring 1991. The follow-on paper
    appeared in Vol. 39, No. 4 (Winter 1992-1993).
    
    Best regards to all,
    
    Chuck Taylor
    Everett, WA, USA
    
    
    
    
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