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    Re: R: Problems with AstronavPC
    From: George Huxtable
    Date: 2004 Feb 16, 15:52 +0000

    Federico Rossi asked-
    >I don't have this book (though I've often read about it on the internet)
    >but I'm very interested in the matter of error ellipse and even if I
    >found the equation to plot it, I've never been able to find the formula
    >giving its orientation with respect to xy axis.
    >I would greatly appreciate if you wrote it down for me.
    With pleasure. Below is what the booklet AstronavPC 2001-2005 (which is
    accompanied by a cdrom) has to say.
    In emailese, I don't have the Greek letters or subscripts, but I will do my
    best to substitute so that you can make sense of it. It follows directly
    after 7.4, which is equivalent to section 11 in the Nautical Almanac,
    "Position from intercept and azimuth by calculation", and uses the same
    symbols. I presume you have a copy of that. My own comments will be in
    [square brackets].
    "7.5. Estimated position error.
    If three or more position lines are obtained an estimate of the error may
    be calculated. The standard deviation of the estimated position sigma in
    nautical miles is given by
    sigma = 60 * square root of (S/ (n-2))
    where S = F - DdB - EdL cos B(F)  [the F is a subscript, in the original]
    The standard deviations sigma(L) and sigma(B) in longitude and latitude are
    given by
    sigma(L) = sigma * square root of (A/G)  [this is in error by a factor of
    cos lat, I suggest, if it's to be expressed in units of longitude rather
    than units of distance)]
    sigma(B) = sigma * square root of (C/G)
    In general as the number of observations increases the error in the
    estimated position decreases. Statistical theory shows that the estimated
    position has a probability P of lying within a confidence ellipse which is
    specified by the lengths of its axes a and b and the azimuth theta of the
    a-axis, where
    Tan 2*theta = 2B/(A - C)
    a = sigma*k/(square root of (n/2 + B/sin 2*theta))
    b = sigma*k/(square root of (n/2 - B/sin 2*theta))
    and k = square root of(-2 * (log to the base e of(1-P))) is a scale factor.
    Values of the scale factor for selected values of P are given in the table below
    Probability  P   0.39  0.50  0.75  0.90  0.95
    Scale factor k   1.0   1.2   1.7   2.1   2.4
    Normally a confidence level of 95% is chosen, that is P = 0.95
    The shape of the confidence ellipse depends only upon n and the
    distribution of the observations in azimuth; while the size of the ellipse,
    apart from the scale factor, depends on the errors of observation. The
    method assumes that the observations have equal weight. The ideal solution
    is to produce a circular distribution of errors, with A = B and C = 0, so
    that the errors are the same in all directions. [This is another passage
    that I question; I suggest it should be "A = C and B = 0"] This situation
    is approached when the bodies are equally spaced in azimuth. In addition,
    if the bodies are observed at similar altitudes, this has the effect of
    minimising the systematic errors of the final calculated position."
    A comment from George.
    What makes me a bit uneasy about the resulting error ellipse is this-
    Take a simple measurement of three stars at 120 deg apart, with a random
    distribution of errors (i.e. no systematic errors). You will get a cocked
    hat with the sides all angled 120 deg apart, but the size of the cocked hat
    is unpredictable. Sometimes it will be a big triangle, sometimes a small
    one, occasionally (rarely) it will be tiny. For each observation of a set
    of 3 stars, the program has to base its assessment of the error ellipse on
    the size of that cocked hat: it has nothing else to go on. So the error
    ellipses will vary wildly from one observed set of three, to the next, just
    because of these statistical fluctuations.
    It's a bit like asking someone to throw a single dart at the bull's eye on
    a dartboard, and then assessing his skill on the basis of how close that
    one dart got. Sometimes, just by chance, he will hit the target even if he
    is quite unskilled. If you were assessing his skill on the basis of that
    one lucky dart, your judgment would be seriously at fault.
    For that reason, I suggest that an error ellipse from a single set of 3
    star altitudes is not taken too seriously. It's not due to any defect in
    the program, which is doing the best it can with the limited information
    available. Because a nice error-ellipse has been plotted out by the
    computer, it's tempting to give it undue weight. Resist that temptation.
    contact George Huxtable by email at george@huxtable.u-net.com, by phone at
    01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
    Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.

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