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    Re: On LOPs
    From: Trevor Kenchington
    Date: 2002 Apr 17, 18:08 -0300

    Many thanks to Bill Murdoch for posting that mass of trigonometry!
    
    He wrote:
    
    > These are the formulas. (Any mistakes in copying and changing this to
    > survive being typed and sent over the internet are mine.  I typed it
    > in Courier.)
    >
    > From B.D. Yallop and C.Y. Hohenkerk, Compact Data for Navigation and
    > Astronomy for the Years 1991-1995.  [snip]
    
    
    Rearranging Bill's material somewhat, he gave the formulae for the
    probability ellipse as:
    
    > 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 = 2 B / (A - C)
    >   a = sigma k / sqrt(n / 2 + B / sin 2 theta)
    >   b = sigma k / sqrt(n / 2 - B / sin 2 theta)
    >
    > where
    >
    >   k = sqrt(-2 ln(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
    >
    > The usual confidence limit is 95%, that is P = 0.95.
    >
    > The shape of the confidence ellipse depends only upon n and the
    > distribution of the observations in azimuth; whilst the size of the
    > ellipse, apart from the scale factor, depends upon the errors of
    > observation.  The method assumes that the observations have equal
    > weight.  The best results will be obtained when the observations are
    > equally spaced in azimuth.  In such cases the effect of systematic
    > errors on the final calculated position will be minimized.
    
    
    I am still not convinced that an elliptical shape for the boundary is
    anything more than a convenient approximation when there are more than
    three LOPs but, that aside, the above equations make a lot of sense,
    even to this non-statistician. The length of the major and minor axes of
    the ellipse depend only on the sines and cosines of the azimuths, plus
    the chosen value of P. If each LOP is given equal weight (i.e. is judged
    to have equal accuracy), then those are the variable that should affect
    the confidence interval, while the equations use sums of squares and
    sums of cross products of the sines and cosines, which is what one would
    expect in a calculation of a confidence interval. I don't pretend to be
    competent to judge whether the equations are or are not correct but they
    certainly don't look to be off the wall.
    
    
    But Bill also wrote:
    
    > 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 sqrt(S / (n-2))
    >
    > where
    >
    >   S = F - D dB - E dL cos BF
    >
    >   sigma sub L = sigma sqrt(A / G)
    >   sigma sub B = sigma sqrt(C / G)
    
    
    That I don't understand. D, E and F all include the intercepts of the
    sights as terms in their calculation yet the size of the intercept is,
    in part, a result of the arbitrary choice of an assumed position. Unless
    this is supposed to be the standard deviation of that part of the error
    in the estimated position which arises from projecting an azimuth
    (calculated onto the the nearest full degree) and an LOP from the
    assumed position to the estimated position, I don't see why the
    calculation includes the azimuth at all. Can anyone explain that?
    
    
    The first part of Bill's posting was:
    
    > If p1 and Z1 are the intercept and azimuth of the first observation,
    > if p2 and Z2 are the intercept and azimuth of the second observation,
    > and so on; form the summations
    >
    >   A = cos^2 Z1 +cos^2 Z2 + ...
    >   B = cos Z1 sin Z1 + cos Z2 sin Z2 + ...
    >   C = sin^2 Z1 + sin^2 Z2 + ...
    >   D = p1 cos Z1 + p2 cos Z2 + ...
    >   E = p1 sin Z1 + p2 sin Z2 + ...
    >   F = p1^2 + p2^2 + ...
    >
    > As a check A + C should equal n, the number of sights.
    >
    >   G = A C - B^2
    >
    > The improved estimated position is (Lat, Long), (LI, BI) made by
    > correcting the first estimated position (LF, BF)
    >
    > LI = LF + dL
    > BI = BF + dB
    >
    > where
    >
    > dL = (A E - B D) / (G cos BF)
    >
    > and
    >
    > dB = (C D - B E) / G
    >
    > The distance between the first estimated position and the improved
    > estimated position is
    >
    >   d = 60 sqrt(dL^2 cos^2 BF + dB^2)
    >
    > If d exceeds about 20 nautical miles, set LF = LI and BF = BI then
    > repeat the calculation until d, the distance between the previous
    > estimate and the improved estimate, is less than about 20 nautical miles.
    
    
    I think that I follow most of that, without being able to confirm that
    it is correct.
    
    But shouldn't "BF" be the assumed latitude and "LF" the assumed
    longitude? If those were that way around, the change in longitude, "dL"
    would be multiplied by the cosine of the (assumed) latitude to get a
    departure in miles for use in the Pythagorean calculation of the
    distance between the two positions. Likewise, division by Cos"BF" (_not_
    the cosine of the product of "B" and "F" from the first set of
    equations) would convert the departure calculated as "(A E - B D) / G"
    into a difference of longitude "dL".
    
    
    Trevor Kenchington
    
    
    

       
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