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    Re: Distance by Vertical Angle Adaptation
    From: Mike Bowman
    Date: 2009 Jul 27, 03:25 -0700

    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
    Mike Bowman
    Darwin NT
    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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