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Re: Anomalous dip. was: [NAV-L] Testing pocket sextant.

From: George Huxtable (no email)
Date: Sat Jun 17 2006 - 18:01:58 EDT

  • Next message: George Huxtable: "Re: Advancing a position circle."

    Bill has asked

    | George et al
    |
    | Now that I have the basic concept, what happens on a larger scale?
    We could
    | use Frank's beach shots as an example (Indiana to Chicago, approx.
    22 nm).
    | Assume a thermal inversion as Frank stated. What will that do to
    the
    | horizon relative to a horizon with well-mixed air (shift it up or
    down) and
    | what will it do to building tops around 1100-1500 feet above water
    level
    | (raise or lower them)?

    Well, although I was aware of those Chicago skyline observations being
    noted on Nav-l at the time, I didn't follow all the details, so am
    unable to respond fully to Bill's request.

    However, I can say that where there is a thermal inversion near sea
    level, the dip of the horizon is reduced, and may become zero, or even
    reversed in sign, compared with its predicted value. You correct a
    sextant observation by subtracting the dip, so if there happened to be
    an inversion you would be subtracting too large a number (from the
    book) than you should, and arriving at an answer for altitude that was
    too small. That seems to be opposite to Bill's experience, as
    described below.

    In the unlikely event that the inversion (cooler below, warmer above)
    extended upwards as far as the tops of the Chicago buildings, then
    they, too would appear higher than you would expect, raised in a
    similar way to the way the horizon gets raised. But more likely, such
    a temperature inversion will extend up from sea level, only to a
    fraction of that height. And so for light rays from the top of the
    building to the eye of the observer, only a fraction of their path
    would be through that inversion layer, and the rest of the path would
    be above it. That would result in a complicated picture for the
    refraction, and I would not know how to predict the details.

    | The problem I was having using the current Bowditch Table 15 formula
    was
    | that the distance kept falling short, meaning if the T15 constants
    are
    | correct, the angle measured was too large. Therefore if the horizon
    was
    | shifted up, the building tops must have been shifted up to a greater
    extent.
    |
    | I understand the problem as stated may be too complex for a simple
    answer,
    | but try a if-then.
    |
    | I also strongly suspect another problem area is the Bowditch
    constants.

    George

    contact George Huxtable at
    or at +44 1865 820222 (from UK, 01865 820222)
    or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.


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