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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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