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    Re: Bowditch Table 15
    From: Bill B
    Date: 2005 Feb 3, 14:28 -0500

    Jim wrote:
    
    In another section he shows how to derive the constants 0.0002439 and
    0.7409.  Note that 0.7409 is not 0.7349, the constant used in the current
    2002 Bowditch.
    
    Ah, the .7349 constant has been driving me nuts.
    
    Horizon is 1.169 (SQRT height of eye, ft).
    
    Geographical range, with height of eye = 0 is 1.17 * SQRT height of object.
    Note the difference between 1.169 and 1.17 is most likely because the object
    has to be a smidge over the horizon to be seen.
    
    Both the above are flip sides of the same coin.
    
    In the table 15 formula if I set the the observed angle, corrected for IC
    and dip to 0, and height of eye to 0, I get
    
    D = SQRT (H/.7349) = SQRT (H * 1.36184) = SQRT H * SQRT 1.36184 = 1.167*
    SQRT H
    
    Beginning to look familiar?  By why the difference between 1.167 and 1.169
    or 1.17?
    
    Using .7409 and running through the above exercise, it becomes 1.162 * SQRT
    H
    
    It is puzzling to me.
    
    Bill
    
    > This is for nit-pickers only, since for practical purposes we have already
    > answered our questions about Table 15 (was 9).
    >
    > I obtained a reprint of Guier's paper from the Journal of the Institute of
    > Navigation by joining the ION, so that I can download reprints from their
    > members'-only website.
    >
    > In his Introduction he wrote that the 1958 edition of Bowditch had
    > "neglected a correction for observer height in the Explanation of Tables.
    > This error was removed in the Corrected Reprint, 1962 Edition.  The
    > discovery of that error together with a desire to extend the tables to
    > smaller angles and shorter objects motivated the author to examine the
    > accuracy and application of the tables."  He developed the current tables
    > "having in mind the particular application of small craft piloting."
    >
    > His Figure 1. shows the complex curves, lines and angles behind the
    > equations.  He shows the derivation of the equation that we see today in
    > Bowditch.  The solution is not trivial -- lots of calculus and assumptions.
    > No wonder I could not reverse-engineer the solution.
    >
    > The equation that he derives, which he wrote was the one given in the 1962
    > Corrected Edition of Bowditch, is not written precisely the same as in the
    > current 2002 Bowditch.  Since I am not a mathematician, I will reproduce
    > Guier's equation here:
    >
    > R =(approx) rR x [-tan(a) + sqrt{tan^2(a) + 2 x ((hs - ho)/rR)}]
    > Where
    > R = "Range to base of sighted object"
    > rR = "refracted equivalent of earth radius"
    > a = "vertical angle at observer"
    > hs = "height of sighted object"
    > ho = "height of observer"
    >
    > In another section he shows how to derive the constants 0.0002439 and
    > 0.7409.  Note that 0.7409 is not 0.7349, the constant used in the current
    > 2002 Bowditch.  In that section he gives this equation:
    >
    > R =(approx) cos(Hs - D) {-(sin(Hs - D)/0.0002439) + [(sin(Hs - D) /
    > 0.0002439)^2 + ((hs - ho)/0.7409)^0.5.
    > Where
    > D = "dip of horizon of observer (min. of arc)".
    >
    > The equation shown in the current 2002 Bowditch, which Bill found produces
    > the values in the current tables, is:
    > D = sqrt[(tan^2a/.0002419^2) + ((H-h)/0.7349)] - (tan a/.0002419)
    >
    > Guier said that the tables are accurate to about 0.1 NM in non-standard
    > refraction. Interestingly his table values are 0.1 NM less than the table
    > values in the 2002 Bowditch for long ranges, but they are equivalent for
    > short ranges.  So somewhere after his publicaton there must have been more
    > work that we have not found reference to.
    >
    > Not that any of this would be significant to a cruiser estimating range from
    > a distant object over the horizon by using a sextant to measure the vertical
    > angle.  Table 15 in Bowditch would work fine, since the Table's precision of
    > 0.1 NM is more than enough for such purposes, for example to check that GPS
    > datum is accurate for a remote Pacific island (in my dreams).
    >
    > Jim Thompson
    > www.jimthompson.net
    > --------------------
    > Outgoing email scanned by Norton Antivirus
    
    
    

       
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