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    Re: Distance by Vertical Angle Adaptation
    From: Greg Rudzinski
    Date: 2009 Jul 26, 22:04 -0700

    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
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