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Re: Distance off with Chicago buildings-corrections

From: Bill (no email)
Date: Mon Dec 05 2005 - 16:50:05 EST

  • Next message: Alexandre E Eremenko: "Re: Russian Sextants"

    > Adding dip back into your 30.8' figure to correct it to 0 ft height of eye,
    > we have 31.1', compared to calculated (with generous refraction lift as the
    > base is approx. 595 ft above sea level) of 28.6', we find your observation
    > 2.5' high.

    The above was a blunder. With height of eye of 11 ft, dip is 03.2'

    30.8' - 3.2' dip = 27.6'

    NOTE: Frank stated the horizon was 4 to 5 miles away and stated distance in
    statute miles, which is SOP for Great Lakes charts. Using the mean of 4.5
    sm and converting to nautical miles, dip was calculated as 3.2'. This seemed
    reasonable for a beach shot. Using both sm and nm, the dip range could be
    from 2.9' to 4.1'

    And now the fatal flaw. Working from memory, I did not check the table
    explanations, and assumed the H-h accounted for dip. Frank's 30.8' was not
    adjusted for dip. IC *and* dip should be accounted for.

    Below are revised results from the first of the latest group of posts:

    Following are results of my calculations using calculated angles and your
    observations with the Bowditch formula:

                  ANGLES
    Frank Observed Observed - 3.2' Calculated Diff
    Sears 30.8' 27.6' 26.6' 1.0'
    Hancock 22.1' 18.9 17.8' 1.1'
             ----- -----
    Diff 8.7 8.8

    NOTE: As distance was known I flipped the Bowditch formula D = 1.17 sqr rt
    (H-h) with height of eye as 11 ft. to establish how much of the buildings
    were hidden and used the visible portion to calculate angles.

    DISTANCE FROM OBSERVED & CALCULATED ANGLES (nm)
    Frank Observed Calculated GPS
     Sears 22.54 23.04 23.08
     Hancock 22.83 23.51 23.53

    ====================================

    In the next post the portion of the structure (Sears) below the horizon was
    calculated with trig from 0 ft height of eye (471 ft). Then 153 ft lift
    from refraction was added to the visible portion and the angle was
    calculated at 0 ft height of eye.

    ANGLE: 0d 28.6'
    DISTANCE (from Bowditch formula): 22.20 nm (GPS target 23.08 nm)

    I note Bowditch was written for sea-level, not Great Lakes sailors. I
    understand comparing Frank's observations at approx. 595 ft above sea level
    to sea-level calculations is a bit of apples-to-oranges.

    Even so, I remain confused about Bowditch predictions of 389 ft hidden (sea
    level) vs. trig calculated (sea level) hidden of 318 ft (471 hidden - 153
    lift).

    Any insights would be appreciated.

    Bill


  • Next message: Alexandre E Eremenko: "Re: Russian Sextants"



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