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    Re: Distance by Vertical Angle Adaptation
    From: Greg Rudzinski
    Date: 2009 Jul 27, 07:35 -0700

    Hi Mike,
    
    You have brought up an important point for regions that experience
    extreme tides. The tide height measured relative to mean low or lower
    water must be figured relative to mean high water to derive a
    correction to the object height. Dip correction is often forgotten
    when using The Bowditch table so it is a good idea to pencil in a note
    of caution at the top of that table.
    
    Greg
    
    On Jul 27, 3:25�am, Mike  wrote:
    > Thanks for the interesting post on the use of vertical sextant angles.
    > I examine Masters and Mates here in N.Australia and inevitably
    > candidates get the calculation wrong even using the tables in
    > Nories,probably similar in Bowditch, what they seem to forget is that
    > charted heights are in metres above Mean High Water Springs and
    > inevitably their observations are never taken at one of those magical
    > tidal moments! We have 7 metre tides here so the height differences
    > are appreciable, I wonder if US charts observe the same 'worst case'
    > scenario for heights and do observers take tidal differences into
    > account?
    >
    > Cheers
    > Mike Bowman
    > Darwin NT
    > Australia
    >
    > On Jul 27, 2:04�pm, Greg Rudzinski  wrote:
    >
    > > Example of vertical distance short of horizon for:
    >
    > > Height of eye 6ft.
    > > Height of Object 250ft.
    > > M.O.A. by sextant 60'
    >
    > > .567(250)/60 = 2.36 NM
    >
    > > Example of vertical distance beyond horizon for:
    >
    > > Height of eye 9ft.
    > > Height of object 250ft.
    > > M.O.A. by sextant 18'
    > > Approximate distance 8 NM
    >
    > > (8/1.144 - 3)Sq = 15.9ft. �(height correction to be subtracted from
    > > object height)
    >
    > > .567(234.1)/18 = 7.37 NM
    >
    > > Run calculations again using 7.37 NM as approximate distance
    >
    > > (7.37/1.144 - 3)Sq = 11.8ft.
    >
    > > .567(238.2)/18 = 7.5 NM �(more accurate distance)
    >
    > > note - Nav.Mark is replaced by Object in original post.
    >
    > > On Jul 26, 4:00�pm, Greg Rudzinski  wrote:
    >
    > > > An interesting adaptation of the distance by vertical angle short of the horizon formula:
    >
    > > > Dist.(NM)=(.567 x Height of Nav.Mark in ft.)/Observed Minutes of Arc
    >
    > > > To get distance by vertical angle beyond the horizon - correct the 
    height of the navigation mark (subtract ft.) using formula:
    >
    > > > ((Approx.Dist.Nav.Mark/1.144)-Root Height of Eye)Squared
    >
    > > > Now try this out using a slide rule.
    >
    > > > Greg- Hide quoted text -
    >
    > > - Show quoted text -
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