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    Re: Still on LOP's
    From: JC Sutherland
    Date: 2002 Apr 20, 02:20 +0100

    Quoting George Huxtable :
    
    > The reason why I haven't yet responded to Martin Gardner is because I
    > have
    > been thinking hard about what he said, and have not (yet) come up with
    > a
    > reply that I feel happy with. Trevor Kenchington has expanded on
    > Martin's
    > contribution, in a way that causes me to think harder still.
    >
    > It's not at all clear to me whether I am going to be able to convince
    > them
    > (that the probability of the true position being embraced by the cocked
    > hat
    > is just 0.25), or whether they will persuade me otherwise. A bit more
    > discussion is called for yet.
    >
    > It's gratifying, anyway, that neither of them is arguing the
    > "traditional"
    > view, that the true position must lie within the cocked hat. Indeed,
    > they
    > both seem to accept that the probability of the true position lying
    > inside
    > the triangle may be even lower than my figure of 0.25.
    >
    > Let me pick up a few points that have been raised, while recognising
    > that
    > these do not demolish their arguments.
    >
    > Martin said-
    >
    > >When I drew this, RRR and LLL lay outside my triangle.  I was able to
    > >>redraw it so that RRR lay inside, but then there is no area LLL.
    >
    > This puzzles me. Are we perhaps at cross-purposes about the words Left
    > and
    > Right and how they relate to the diagram of the cocked hat? If we used
    > the
    > words Clockwise and Anticlockwise instead, for the changes in bearing,
    > would that remove some ambiguity? It's just that it seems to me that on
    > any
    > picture of a cocked hat, RRR or LLL must always represent the triangle
    > at
    > its centre (but never both at the same time).
    >
    > Next, Martin said-
    >
    > >I think something went awry in your reasoning:  for any given three
    > lines
    > >intersecting to form a triangle, there is only one area  'inside'.  If
    > two
    > >different patterns of R and L both described some of the 'inside' then
    > the
    > >'inside' would have to be partitioned, which it is not.
    >
    > There are 8 possible combinations, as I said. But they can not all be
    > illustrated on the same diagram. Each combination needs a different
    > diagram
    > showing how it comes about. That is why LLL and RRR do not appear on
    > the
    > same picture. First, show all 3 bearings, through the 3 landmarks,
    > displaced to the left (or anticlockwise) of the line that leads
    > exactly
    > from the true position. This shows a triangle LLL which contains the
    > true
    > position. Now draw a new diagram in which those three bearings are
    > displaced to the right, which shows a new triangle RRR, and again,
    > that
    > triangle contains the true position. Draw another diagram showing two
    > bearings to the left and one to the right, and then the true position
    > is
    > outside the triangle formed by those bearings. And so on, for each of
    > the 8
    > conbinations.
    >
    > So 8 possibiliies (not 7) remain, and according to my view each of them
    > has
    > an equal probability.
    >
    > Martin's final words, and Trevor's endorsement of them, are what has
    > caused
    > me to ponder most, when he said-
    >
    > >My worry is that I don't think the three bearings are actually
    > independent.
    > >It's not as if we were dropping three long straws at random on a chart
    > -
    > >we're in some sense measuring the same thing all three times.
    >
    > I am still thinking about that point.
    >
    > =============================
    >
    > Perhaps it might help this discussion if, to give another view, I quote
    > JED
    > Williams' explanation of this matter, in "From Sails to Satellites",
    > which
    > is somewhat different to mine, and perhaps more elegant.
    >
    > On page 273 he provides figure 14.6, entitled "The probability of
    > being
    > within the cocked hat is 1/4". This shows a cocked hat, showing the
    > value
    > "1/4" in the central triangle, "1/6" in the areas adjacent to the
    > sides,
    > and "1/12" in the areas outside each corner.
    >
    > Williams says-
    >
    > "Fig 14.6 is the cocked hat formed by three position lines. If no
    > information is available on the distribution of error the probability
    > of
    > being in each of the seven zones defined by the cocked hat is as shown
    > in
    > the figure. The proof is so simple that an outline of it is given in
    > note
    > 13; but the writer is unaware of any text likely to be read by mariners
    > or
    > airmen of the period which even suggests that the true position is
    > probably
    > outside the cocked hat...."
    >
    > Note 13 to that chapter, on page 282, reads as follows-
    >
    > "If there is no information on the distribution of probable error, a
    > position line of any shape simply divides the Earth's surface into two
    > domains, and the position is as likely to be in one as in the other. We
    > may
    > arbitrarily denominate them the + domain and the - domain, and we choose
    > to
    > do so by nominating the enclosed area of the cocked hat the 3+ zone.
    > Every
    > zone bounded by a side of the triangle is a 2+,1- zone, and each zone
    > at
    > the vertex of the triangle is a 1+,2- zone. The 3- zone is cospicuous
    > by
    > its absence. If the true position were in the the minus zone of each
    > and
    > every position line as they have severally been labelled, the cocked
    > hat
    > would necessarily be shaped in the form of its mirror image, and the
    > position would be within it. The fact that the cocked hat has taken
    > the
    > form it has tells us that the position is in the + domain of at least
    > one
    > position line. Given that we are in the + domain of one position line
    > and
    > may equally be in either the + domain or the - domain of two other
    > position
    > lines,out of four there is one chance that the domain is in the 3+
    > zone,
    > two chances that it is in a 2+, 1- zone, and one chance that it is in
    > a
    > 1+,2- zone. The probability of being in each zone is therefore as shown
    > in
    > fig 14.6."
    >
    > (end of quotation from Williams)
    >
    > The above is offered on an "if it helps" basis, and I am not intending
    > to
    > defend or explain its logic.
    >
    > ===========================
    >
    > My thanks to all who have given a stir to this interesting pot, which
    > is
    > not yet fully cooked. Sorry if any list-members find it all boring.
    >
    > Meanwhile, I will continue to chew over the food for thought that
    > Martin
    > and Trevor have provided, and would welcome any further contributions
    > to
    > this discussion.
    >
    > Perhaps an answer might arise from a "Monte Carlo" type simulation
    > devised
    > by one of our computer experts, though its parameters would need to be
    > very
    > carefully specified beforehand, to ensure it was answering the right
    > question.
    >
    > George Huxtable.
    >
    >
    >
    >
    >
    >
    > ------------------------------
    >
    > george---.u-net.com
    > George Huxtable, 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    > Tel. 01865 820222 or (int.) +44 1865 820222.
    > ------------------------------
    >
    
    Hi All;
    The arguments that I have read so for on this subject are to my mind  rather
    obscure and un-convincing
    
    George�s presentation , for example, of three intersecting position lines
    resulting in only 25% probability of the position being within the cocked hat
    has long been the argument in textbooks on this subject, but it doesn�t gel
    with me. To support this contention we are supposed to agree that for each one
    of these position lines the true position lies equally likely to the left or to
    the right but not actually on the line! This is where I get off!
    
    Let�s imagine for a moment that the earth is still and many observations of the
    same LOP can be taken. If these are analysed statistically ( assuming that all
    errors are random) the most probable value will be represented by the average
    and the further off this average  line you inspect the less likely you are to
    find the true  position line.
    
     If we assume that the line is the actually the peak of a Gaussian
    distribution, yes, it has an equal probability either side, but this is  a
    diminishing probability value the further you move away from it and the MOST
    PROBABLE VALUE MUST BE  THE LINE ITSELF.
    
    If we take a three dimensional view, our observations could be better
    represented by a solid figure rather like a length of wood cut so that it has
    Gaussian cross-section the same size and shape all along its length. We could
    then lay this on the chart to represent our LOP.
    
    If you imagine two such position lines intersecting , merging and adding. The
    probability curve at the point  of intersection would become a Bell shape. It
    would have circular contours only if the error distributions of both LOPs are
    the same , but would have elliptical contours if one LOP group is tighter  than
    the other.
    
    Each contour will represent an uncertainty area at a particular confidence
    level.(the smaller the ellipse the less confident you are of being inside it).
    For simplicity it is usual to show only one contour for the confidence level
    (usually 95%) the navigator prefers.
    
    Now it is possible to extend this theory to include three or more LOPs, but my
    maths is not up to this, however I can convince myself that the true position
    lies inside the cocked hat if I look at this plot in a more colouful way.
    This is my way of imagination. On maps contours are often coloured. If we
    wereto use dark transparent colour for high probability and  a light colour for
    low,  this would reveal the shape of the probability terrain. The darkest area
    would represent the highest confidence in your position. For example, if we
    plot the Gaussian distribution of a single group of LOPs  as a ridge line, the
    center would be a dark  line and the further off either side of this line the
    lighter the shading  would be.
    
    For two position lines, at the point of intersection, the shading would add and
    this would then produce a single darker peak with  the shading getting lighter
    in all directions away from this point. This would be consistent with the
    circular Bell shape,  and would reveal a ellipsoid bell, if this were in fact
    the case as above.
    
    If you use this trick to imagine what three or more curves intersecting as a
    cocked hat would look like, personally I am convinced that the darkest
    cumulative shading would be a plateau inside the triangle. This would show that
    the further you go from the �Center of gravity� of the triangle the lighter the
    shading would be. Therefore by inference the highest probability of the true
    position fix would be inside  the triangle.
    
    PS.  In surveying it is common to find �the most probable position� in a
    triangle of error produced by three sight lines,  by drawing lines inside this
    plotted  triangle, parallel to each side, at a distance from the side
    proportional to the length of the observed ray so as to produce a smaller
    triangle inside the first. This is applied successively until the smallest
    acceptable triangle is found and this then becomes the surveyed point.
    
    Clive Sutherland.
    
    
    

       
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