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    Re: Still on LOP's
    From: George Huxtable
    Date: 2002 Apr 20, 11:18 +0100

    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.
    ------------------------------
    
    
    

       
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