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    Re: Summary of Bowditch Table 15
    From: Jim Thompson
    Date: 2005 Jan 26, 11:13 -0400

    > -----Original Message-----
    > From: Trevor J. Kenchington
    > 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.
    
    Yes, I've got that now.  Being a visual learner, I think that I have the
    right images on my page, I hope:
    http://jimthompson.net/boating/CelestialNav/CelestNotes/Bowditch15.htm.
    Getting the sensible horizon below the top of the object is key to
    understanding this table.  If the sensible horizon shoots into space above
    the top of the object, then the relationships in Table 15 are "wrong".  In
    real life the sensible horizon is unlikely to be above the object, I
    presume.
    
    The "correct" (consistent with Table 15) relationships are these, for an
    object with constant height, stationary distance between ship and object
    over the horizon, and decreasing height of eye:
    
    - Measured vertical angle decreases (observer sees less of the lighthouse).
    - Dip (angle between sensible and visible horizons) decreases (sea horizon
    gets closer to observer).
    - Corrected vertical angle increases (consistent with the table, I know see,
    but not intuitive).
    - D increases, if D is the distance from the observer's horizon to the
    object.
    
    > 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.
    
    Doh!  So what is "D" then, and how does one use the table?
    
    My height of eye is 30 ft.
    My horizon is 6.4 NM away.
    (Distance to Horizon for h in feet = 1.17 * sqrt(h)).
    And my Dip is - 5.31'.
    (Dip in minutes of arc for h in feet = -0.97 * sqrt(h)).
    The lighthouse over the horizon is 100 ft high from its sea level.
    
    I measure a vertical angle of 00d 05.40' with my sextant (not that I can
    measure to a precision of hundredth of minute of arc).
    The corrected vertical angle is 5.40' - 5.31' = 0.09'.
    (H-h) = 100-30 = 70 ft.
    From the table then, D = 3.8 Miles.
    
    D is, I thought, the distance from the sea horizon to the top of the object.
    I presumed that I should add my distance to my horizon (6.4 NM) and the
    distance from the horizon to the top of the object (3.8 NM) to get a total
    distance of 10.2 NM, with a 0slight vertical dogleg two thirds of the way
    along.  Is that incorrect?
    
    As a rough check I know that:
    My horizon is 6.4 NM away from an eye height of 30 ft.
    The top of 100 ft lightouse sees a horizon 11.7 NM away.
    So when I can just see the top of the lighthouse at my visible horizon, we
    are separated by about 18.1 NM.
    At least the magnitude seems reasonable.  The lighthouse must be closer than
    18.1 NM when I can measure a greater vertical angle than 00.00'.
    
    >> Jim wrote,
    >> "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.
    
    Aha!  I replaced that figure with one drawn so that the sensible horizon
    does not fly over the top of the object (increased the radius of my model
    earth).  Now I think that I am catching up to the gazelles.
    
    Jim Thompson
    
    
    

       
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