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    Re: Bowditch (2002) Table 17
    From: Bill B
    Date: 2012 May 01, 03:29 -0400

    As the math, engineering and physics gurus have passed on your question
    to date, I'll attempt a layperson's explanation. Once I build a straw
    man, I'll wager they chime to correct my blunders :-)
    (It's been a long time since I reverse engineered the equations).
    
    To over simplify, Table 15 is a case of measuring the portion of the
    object still visible and adjusting for the unseen portion below the
    horizon due to curvature of the earth, then adding to the observed angle
    for refraction.
    
    Way way over simplifying, Table 17 is similar to measuring the height of
    an nearby object above ground level. I surmise it then elegantly  uses
    height of eye to divide it into to right triangles. (Table 16 says that
    directly.) Throw in a few tweaks for curvature of the earth and
    refraction...
    
    In table 15 you enter height of object minus eye and and the angle
    (height) of the exposed portion of an object of known height over the
    horizon. Similar to taking an angle from a celestial body, dip (height
    of eye) comes into play. Refraction also comes into play. As does the
    curvature of the earth.
    
    If memory serves,there is indeed an editorial error in the equation
    given in 2002. (H-h)/0.7349 should be included under the square root
    symbol.
    
    As a side bar, Frank Reed did some observations of Chicago, IL
    skyscrapers from a beach in Indiana. We were both of the opinion the
    Bowditch constant of 0.0002419 is a bit heavy.  Frank suggested an
    alternative and I derived my own. Both were close to spot on 17 miles
    off Chicago when I tested them from a sailing vessel. I cannot say that
    of the Bowditch constants.  If you are interested I can dig into my
    archives and find our proposed constants.
    
    Table 17 is a bit more up close and personal. (Although on page 560
    there is definitely a typo, the table does explain it correctly.)  Given
    the same height of eye, the distance to the horizon in nm is 1.169 *
    square root of height of eye (in feet).  In table 15 the base of the
    object is below the horizon so a good distance away.  In table 17 the
    base is visible and the horizon is above the base. We still have
    curvature of the earth and refraction to deal with.
    
    I understand I have not fully answered your questions. As the dimmest
    bulb on the NavList marquee, it's my best shot.
    
    Bill B
    
    
    On 4/30/2012 4:46 PM, slk1000---.com wrote:
    > It has been a while since we discussed distances by vertical angles
    > (2005 and 2009).  I just ran into something I hope someone can clear up
    > for me.  In the Explanation of Tables on page 560 of Bowditch 2002, it
    > says that for Table 15 the sextant angle is corrected for index error
    > and dip, but for Table 17 it is only corrected for index error.  Why no
    > dip if height of eye is part of the formula?
    
    
    
    
    

       
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