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    Re: Bowditch Table 15
    From: Bill B
    Date: 2005 Jan 25, 16:19 -0500

    George
    
    I have the 1995 edition
    
    > I transcribed into emailese, writing A instead of Alpha, the expression in
    > my edition (1981 vol 2) of Bowditch as-
    >
    > d = sqr { (tan A / .0002419 )^2 + ((H-h) / .7349) - (tan A / .002419) }
    
    > Is the expression and the text the same in later editions? Somebody please
    > state, for my benefit, the expression for d as given in a more recent
    > edition.
    
    The formula as presented in the 1995 edition has only the first term, tan
    a/0.0002429^2 as radicand under the radical symbol.  This must be incorrect
    for reasons previously noted.
    
    If I read your emailese version correctly, all three terms are treated as
    radicands in your version. I cannot make that match the tabular values when
    using that formula in Excel.  Happy to send you the spreadsheet if you
    request.
    
    The formula that seems to match the tabular values as well as the published
    formula are attached as a jpeg.
    
     > I wrote- To me, it seems a bit suspicious that that text refers to "The
    > constants .0002419 and 0.7349", although a third, and different, constant
    > of .002419 also appears to be used in the expression. Is that a misprint, I
    > wonder?
    
    I do not see a value of 0.002429 in the explanation of Table 15 (Table 9 in
    your addition?) in the 1995 version.
    
    > But it seems me there's more wrong with the expression I quoted above than
    > a simple matter of the number of decimal places in a constant term. When
    > the angle =0, then the expression gives the same results as does the table,
    > but the two seem quite inconsistent in the way they vary with the angle.
    
    Have not played with that as of yet.
    >
    > Just to check that my Table 9 and the later Table 15 are the same, here are
    > two spot values to compare-
    > angle = 0',  H-h = 100 ft., distance 11.7 miles
    > angle = 1 deg 50', H-h = 450 ft., distance = 2.3 miles
    >
    > Does Table 15 give those same values?
    
    Yes.
    
    
    
    

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