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    Re: Dip uncertainty
    From: Trevor Kenchington
    Date: 2004 Dec 8, 20:58 -0400

    George,
    
    It's not like you to be unable to either grasp my argument or else blow
    it out of the water! (More usually the latter, I'm afraid.)
    
    
    You wrote:
    
    > I still
    > have difficulty with drawing out these ray-diagrams, in regard to the
    > following text.
    
    
    That is, perhaps, because you still insist on:
    
    
    > It is true that if the observer measures the angle POL, between the
    > direction the light appears to be coming from (OP) and his direct view of
    > the lamp OL (not through the prism), then THAT angle will change
    > considerably, depending on where he is on the parade ground, on the
    > distances LP and PO. But so what? That's an angle we have no interest in.
    > It's a red-herring that has distracted from the truth.
    >
    > The only thing that's relevant is the direction of the incident light-beam
    > and the constant angle that it's deflected through by the prism. No matter
    > where the prism is placed.
    
    
    Imagine that your observer on the parade ground seeks to establish an
    LOP for himself by using a horizontal sextant angle, with his two
    landmarks being the light at the north end and a flag pole on the east
    side. The flag pole can be observed directly but all he can see of the
    light is the beam after its refraction by the prism.
    
    As a result of that refraction, the observer's measurement of the
    horizontal sextant angle would be too small by an amount equal to the
    angle POL which, as you say, will vary in magnitude depending on how far
    from the prism the observer is standing.
    
    I would argue that that is analogous to a sextant observation of the
    altitude of some body when, as a result of anomalous refraction very
    close to the sea surface, the observer's perceived horizon does not
    correspond to the expectation of the dip tables.
    
    
    You also wrote:
    
    > You can see that the standard refraction has
    > just twice the effect on reducing dip, from a 24 ft. viewpoint, than it
    > does viewed from 6 ft. In the same way, non-standard refraction, which is
    > "anomalous dip", will have twice the effect at 24 ft height than it does at
    > 6 ft. This is presuming that the air density gradients that cause the
    > refraction remain uniform up to a height of at least 24 ft.
    
    Interesting. On its face, that is persuasive: The further from the
    horizon, the greater the curvature of the light and so the larger the
    effect of refraction -- both standard and any anomaly (assuming that the
    refraction is constant across the volume of air of concern). So why does
    this lead to the opposite conclusion from your parade-ground model and
    Bruce's prism-on-a-buoy model?
    
    I am not certain of the answer but I think it may be because you are
    placing the two observers in different positions relative to each other.
    In all of our discussion thus far, the yachtsman with his eye 6 feet off
    the water and the coaster captain with his at 24 feet have had their
    sextants on the same anomalously-refracted ray. Now you are placing two
    observers on the same straight line, extending from the same point on
    their mutual geometrical horizon. I haven't convinced myself of that but
    it seems like the most likely explanation for the apparent contradiction.
    
    
    Trevor Kenchington
    
    
    
    --
    Trevor J. Kenchington PhD                         Gadus{at}iStar.ca
    Gadus Associates,                                 Office(902) 889-9250
    R.R.#1, Musquodoboit Harbour,                     Fax   (902) 889-9251
    Nova Scotia  B0J 2L0, CANADA                      Home  (902) 889-3555
    
                         Science Serving the Fisheries
                          http://home.istar.ca/~gadus
    
    
    

       
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