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Re: dip, dip short, distance off with buildings, etc.

From: Bill (no email)
Date: Thu Jan 05 2006 - 22:41:19 EST

  • Next message: Marcel E. Tschudin: "Re: dip, dip short, distance off with buildings, etc."

    > In Bowditch and elsewhere, there are formulas for dip, dip short, Table XV
    > for distance based on measured height, maximum visibility distance, etc., and
    > they all have various mysterious corrections for "mean refraction". I've got
    > this stuff all figured out pretty well now, and it turns out that there is a
    > really easy, though somewhat bizarre (!), way of thinking about the effect of
    > refraction in terrestrial, or coastal navigation, situations.

    Frank

    I let go of your Chicago observations as there seemed to be little interest
    and focused on Bowditch vs. the web article you mentioned.

    Bowditch's table (now 15) indicates that an object of 1400 ft with an
    observed angle of 0d 30' would be 20.5 nm away. Using the formula the table
    is derived from, the distance is 20.55 nm.

    Using trig and accounting for the curvature of the earth, in the absence of
    air 373 feet would be hidden below the horizon at 20.55 nm.

    Using Bowditch (0 height of eye) 20.55 = 1.169 * square root height of
    object, it would take 309 feet of height to appear to touch the observers
    horizon at a distance of 20.55 nm.

    Therefore Bowditch predicts a lift of 64 feet due to refraction (373-309),
    hence 1091 feet visible to the observer.

    Working backwards from an oblique triangle, distance of 20.55 nm along the
    surface of the earth and an observed angle of 0d 30', visible height should
    be 1089 ft, or 62 feet lift due to refraction.

    That is certainly close enough to indicate the Bowditch system is internally
    in agreement.

    But does it hold true in practice? In the link you provided refraction of a
    terrestrial object was stated as: Lift from refraction in meters = distance
    in meters^2 * 2.55 10^-8. Adjusting that to nautical miles and feet: Lift
    from refraction in feet = distance from object in nm^2 * .287.

    Predicted lift from refraction at 20.55nm = 121 ft using the latter formula,
    vs Bowditch's predicted 62 to 64 feet. That is a pretty significant
    difference.

    NOTE: All the above assume height of eye as 0, and barometric pressure
    sea-level. No adjustment for BP.

    Hard to tell what to put stock in. Can you, or any of the other learned
    list members point me in the right direction?

    Thanks

    Bill


  • Next message: Marcel E. Tschudin: "Re: dip, dip short, distance off with buildings, etc."



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