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> 3: However, I cannot confirm numerically that Table 9/15 is correct.
>
> Consider a 100ft mast seen from 5 miles away by an observer with 10ft
> height of eye. The 90ft of the mast extending above his sensible horizon
> should subtend an angle of about 10' at his eye, yet entering the table
> with (H-h)=90ft and angle 10' gives a range of 4.3 miles.

You forgot to subtract dip at 10 ft. of 3.1' from your calculated 10' before
entering the tables.  That would make the entering argument 6.9' tabular
value approx. 6 nm.
>
> Let the observer climb a further 60ft up his own mast and the remaining
> 30ft of the observed mast showing above the sensible horizon should
> subtend about 3.4'. However, Table 9/15 would then reduce the range to
> 3.3 miles.

If the observer was at 10 ft, then climbed an additional 60 ft, he/she would
be at 70 ft above water level, correct?

Using the simplified formula for distance to horizon, 1.169(sqrt height of
eye) the observer's horizon would be 9.8 nm away.  If the observed vessel is
5 nm away, as stated above, the observer should be able to see the hull and
waterline, in which case Table 16 would apply.  (Note that the geographical
range would be 21.5 nm from 70 ft aloft.)

I also do not understand how you arrived at the 3.4' angle.  Roughly
speaking we have 2 right triangles with opposite sides of 30 and 70 feet,
with approx. 5 nm (30380 ft) adjacent sides.  Using the tangent, find and
add the angles of both. I get approx. 11.3'.  Do a rough sanity check with
the law of cosines, (c^2 = a^2 + b^2 -2ab cos c) where c=100, a and b =
30380, and again we get approx. 11.3'.

That's my story and I'm sticking to it. 

Bill

   

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