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Couple of days away on a business trip and I get back to a score or more
of messages on this thread! Hopefully, this one won't repeat anything
posted by others in the interim.

Bill wrote:

 >> 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.

I did not forget dip. I calculated the 10' for the 90ft part of the mast
projecting above the observer's sensible horizon. Dip is the angle below
the sensible horizon, down to the visible horizon.

 >> 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?

Yes

 > 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.)

As George has noted, despite its title and explanatory note, Table 15
works perfectly well with objects closer than the observer's horizon
distance.

If the distant vessel's waterline is visible, Tables 16 or 17 would be
alternative ways of finding its distance but Table 15 is still an option.

Geographic range is something quite different: Not the actual distance
to the vessel but the maximum distance at which its truck could just be
seen, if it was that far away.

 > 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'.

Once again, I was working with only the part of the mast above the
observer's sensible horizon, hence only the right triangle 30ft in the
perpendicular and 5 miles in the base. My calculator gives 3.39468...
arc minutes. The almanac gives a dip of -8.1' for a height of eye of
70ft, so the entire angle from visible horizon to mast truck would be
3.4+8.1=11.5'. I'd guess that the extra 0.2' is the result of
refraction, which is included in the almanac's value for dip but ignored
in the above tangent calculations.

 > Getting down to short strokes, I am assuming 100 ft of mast visible
 > above the deck (as opposed to a 100 ft mast keel stepped, showing
 > maybe 92 ft above deck).

My example assumed that the 100ft mast was measured above the waterline
of the vessel, since that is the reference level for the observer's
height of eye. A 100ft mast would not extend 90ft above the sensible
horizon of an observer with height of eye of 10ft unless both heights
were taken from a surface with constant geopotential.




It was a bit off the topic of the thread but Jim included in one of his
posts, as an explanation for Bill:

 > 1. The Visible Horizon is the one you see with your naked eye when
you look
 > at the apparent bounday between the sky and sea (or earth).  A line from
 > your eye to that horizon is called the Visible Horizon.
 > 2. The Geometric Horizon is the real line between your eye and the actual
 > boundary between sky and sea/earth.  Since light bends between the
horizon
 > and your eye, then the Visibile Horizon is rarely coincident with the
 > Geometric Horizon.  The difference is accounted for mainly by terrestrial
 > refraction.
 > 3. The Sensible Horizon is part of the Horizon Coordinate System.  The
 > Sensible Horizon is a line from your eye that runs out into space
parallel
 > to the Celestial Horizon.  The angle between the Sensible Horizon and the
 > Visible Horizon is called the Dip.  The Sensible Horizon is a very
 > non-intuitive concept for learners.


The sensible horizon might be better understood as a plane,
perpendicular to the direction of gravity acting on the observer and
drawn through the observer's eye. It is parallel to the celestial
horizon because that too is a plane perpendicular to the direction of
gravity acting on the observer but drawn through the centre of the Earth.

The visible horizon isn't a line drawn from the observer's eye to the
apparent sea/sky boundary. For piloting purposes, it is a circle drawn
on the Earth which coincides with that apparent boundary but for
celestial navigation it is another plane perpendicular to the direction
of gravity (as are all the horizons). Specifically, the visible horizon
is a plane drawn through the points where rays of light from the
observer's eye which just graze the sea/sky boundary cut the celestial
sphere. But that is a really weird concept since the celestial sphere
doesn't actually exist.

The geometric horizon is closely similar, in concept, to the visible
horizon. But I don't think it is right to say that the difference is a
matter of the geometric horizon being defined by the "actual boundary
between sky and sea". The actual boundary between sky and sea is the
curved (and often rough) surface of the water. The geometric horizon is
defined in terms of a cone, centred on the observer's eye and tangential
to the water surface, in contrast to the visible horizon's definition in
terms of curved (by refraction) light rays.

Bill responded to Bill Noyce's explanation of part of this with:

 > Is the sensible horizon is a plane perpendicular to the vertical (line
 > through center of the Earth and viewer), AT THE VIEWER'S HEIGHT OF
EYE, as I
 > understand Jim's drawing?

At the observer's height of eye, yes. But not perpendicular to a line
drawn from the observer to the centre of the Earth. It is perpendicular
to the direction of gravity, which means very nearly tangential to the
surface of the geoid. But living here at close to 45 degrees North, a
line perpendicular to the geoid is appreciably different to one directed
towards the centre of the Earth.

 > General instructions for a pan of oil etc. instruct the user to place
it on
 > the ground.  No dip correction.  If it relates to the sensible
horizon and
 > my understanding of the definition is true, then dip correction would be
 > required if the pan were placed on a stool?  That doesn't seem right
to me.

If your oil pan were big enough to reach to your horizon, then the oil
surface would be curved as the ocean surface is and you might have to
worry about that curvature if you set the reflected image of the Sun
near the far side of the pan. However, realistic artificial horizons are
so small that their surfaces can be assumed to be flat -- hence no
reason to worry about dip.



Trevor Kenchington



--
Trevor J. Kenchington PhD                         Gadus@iStar.ca
Gadus Associates,                                 Office(902) 889-9250
R.R.#1, Musquodoboit Harbour,                     Fax   (902) 889-9251
Nova Scotia  B0J 2L0, CANADA                      Home  (902) 889-3555

                     Science Serving the Fisheries
                      http://home.istar.ca/~gadus

   

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