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    Summary of Bowditch Table 15
    From: Trevor Kenchington
    Date: 2005 Jan 25, 12:19 -0400

    Thank you all for your responses to my initial question. I will try to
    summarize where we have got to.
    
    1: When I complained that Table 9/15 doesn't work as advertised, I was
    thinking of dip as a minor correction (in the sense that refraction is)
    and hence that, after correction for dip, an observer should still see
    more of a distant object as he increases his height of eye. As George
    has pointed out, the reason that objects appear to rise over the horizon
    as height of eye increases is that dip increases. Correcting for dip
    changes the whole thing.
    
    Specifically, correcting for dip changes the reference plane from the
    apparent horizon to the sensible horizon. As height of eye increases,
    the sensible horizon obviously moves further from the centre of the
    Earth, which means that it cuts the masts of a distant ship closer to
    the truck. Hence, the angle subtended at the observer's eye by the bit
    of mast projecting above the sensible horizon is reduced as height of
    eye increases.
    
    That is what Table 9/15 shows and so I have no reason to question the
    meaning of the table.
    
    [Jim: Does that explanation make sense to you? In your last posting, you
    said that you still thought the direction of change in the angles was
    wrong in the Table. I previously thought that but George has, I think,
    explained the confusion.]
    
    2: Specifically, I withdraw my suggestion that it shows the distance
    from the observer's horizon to the distant object. My reply to Nels
    should have disposed of that one anyway.
    
    3: However, I cannot confirm numerically that Table 9/15 is correct.
    
    Consider a 100ft mast seen from 5 miles away by an observer with 10ft
    height of eye. The 90ft of the mast extending above his sensible horizon
    should subtend an angle of about 10' at his eye, yet entering the table
    with (H-h)=90ft and angle 10' gives a range of 4.3 miles.
    
    Let the observer climb a further 60ft up his own mast and the remaining
    30ft of the observed mast showing above the sensible horizon should
    subtend about 3.4'. However, Table 9/15 would then reduce the range to
    3.3 miles.
    
    I suspect that these differences can be accounted for by refraction but
    the way to be sure of that is to have someone derive the Bowditch
    formula as a formula for range from observer to object, based on the
    height of the object above the sensible horizon. Maybe one of the
    mathematically-gifted list members could do that. I cannot.
    
    4: Any such derivation requires, of course, that we can agree on what
    the Bowditch formula is.
    
    George has reported that the 1981 Bowditch makes it:
    
    d = sqr{(tan A / .0002419 )^2 + ((H-h) / .7349) - (tan A / .002419)}
    
    My 1995 Bowditch has seen the attention of some over-zealous editors,
    who "corrected" that to:
    
    d = sqr{(tan A / .0002419 )^2} + ((H-h) / .7349) - (tan A / .0002419)
    
    (Note the extra "0" as well as the shift ion "}".) Jim says that the
    2002 edition has the same.
    
    In contrast, Bill's numerical analysis has shown that the only version
    which makes sense is:
    
    d = sqr{(tan A / .0002419 )^2 + ((H-h) / .7349)} - (tan A / .0002419)
    
    So, Bowditch has had this wrong for at least 25 years and still has it
    wrong. I wonder when it was last correctly quoted. [I'm guessing but I
    suspect the actions of generations of editors since the last attentions
    of a compiler who actually thought about the meaning of the material,
    instead of just carrying forward the last edition, with amendments.]
    
    Jim has pointed out that this issue was discussed on the list
    previously, when George correctly deduced what the formula should be. (I
    don't feel so bad about having forgotten that exchange, since George
    evidently did too!)
    
    5. That leaves, by my count, only two outstanding yarns within this
    thread (to stretch a metaphor): Jim asked:
    
    "How should the Dip correction be applied?  Add?  Subtract?"
    
    It should clearly be subtracted from H[s], just as with a celestial sight.
    
    "See my new Figure 1 and tentative explanation of the Dip correction for
    Table 15 at
    http://jimthompson.net/boating/CelestialNav/CelestNotes/Bowditch15.htm"
    
    Jim: I think you have complicated things by drawing a distant object
    which doesn't rise above the observer's sensible horizon. For such a
    case, you have the angles correctly labelled.
    
    However, your numerical example (as is true of most of the numbers in
    Table 9/15) is for an object which does rise above it. In trying to
    reconcile those two different cases, you have put a double negative into
    your numerical example. Take that out and draw yourself a new, higher
    object, extending above the sensible horizon, and all should fall into
    place.
    
    "How is H-h calculated?  I had assumed that one subtracted the
    observer's height of eye from the known height of the object above its
    sea level"
    
    Yes.
    
    
    
    Trevor Kenchington
    
    
    
    --
    Trevor J. Kenchington PhD                         Gadus{at}iStar.ca
    Gadus Associates,                                 Office(902) 889-9250
    R.R.#1, Musquodoboit Harbour,                     Fax   (902) 889-9251
    Nova Scotia  B0J 2L0, CANADA                      Home  (902) 889-3555
    
                          Science Serving the Fisheries
                           http://home.istar.ca/~gadus
    
    
    

       
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