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I think Bruce has got it wrong, about anomalous dip affecting observations
from small vessels rather than from large ones.

On my small boat, the cockpit sole is no more than inches above sea-level:
it's only just self-draining. So my height of eye is no more than 6 feet,
which puts my visible horizon 2.8 miles away. Let's say there's a marker
North of me, just 2.8 miles away, just there to mark my horizon, appearing
to float right on it.

When I see the horizon to the North, the light-ray marking that horizon has
skimmed just over the sea-surface, at a tangent to it, close to that marker
vessel. The dip that I see is the angle between that light ray as it passes
my head, and the direction of my own local horizon.

You can think of that dip as being made up of three parts. The major part,
by far, being simply the curvature of the sea surface, the "geometrical"
component of dip, which is easy to calculate precisely.

As you ascend above sea-level, the air gets less dense, according to a
well-known law for a "standard" atmosphere. This density-gradient refracts
the light from the horizon into a predictably curved path. The curvature is
in a downward direction, as is the curvature of the surface, and the dip,
depending on the difference between the two, is reduced by about one part
in 12 as a result. So far all is predictable, and given in the standard
"dip tables".

But the region of the atmosphere, within a few feet of the sea-surface,
through which the light has been passing, is rather special, because of
exchange of heat with the surface. Depending on some complex factors, such
as wind (or lack of it), temperature difference, surface roughness, complex
layers may exist at different levels. The light, in passing through those
layers, is refracted in an unpredictable way. That unpredictable component
is "anomalous dip"; it can add to or subtract from the predicted dip.
Without a dipmeter, nobody knows of its existence.

What happens to the light which has passed my head, from the horizon 2.8
miles away? It still carries on rising, and may be seen from the bridge of
a larger vessel, from 24ft high, which is 5.6 miles from that marker on his
horizon (because the horizon distance increases with the square root of
eye-height) and 2.8 miles to my South.

And what dip affects observations from that vessel? The standard dip, from
the almanac, is doubled. The anomalous part of the dip will be, initially,
the same as affected me at 2.8 miles, but now the light goes on, still
close (from 6 to 24 ft) to the sea surface, so there's extra curvature
added as a result of that additional 2.8 miles of its path.

What Bruce is arguing, it seems to me, is that in the latter part of its
path, that light is  sufficiently far (6 to 24 ft) from the sea-surface,
that no anomalies of temperature-gradient will occur to affect the dip.
This seems to me an unlikely proposition, but even if it were true, the
anomalous dip would be no less than it was for me at 6 ft height-of-eye. So
I don't accept his argument that anomalous dip will be more of a problem to
smaller vessels; it's the other way about, as I see it.

George.


>Alex,
>
>This IS interesting. I've had the notion you only want a lower height of eye
>when there's fog, or for some other reason you have to get the horizon close
>to see it clearly.
>
>My thinking goes this way: Ocean temperature would change the density of air
>near the surface. Updrafts and downdrafts from clouds would tend to keep that
>air in flux, so it would vary in density from place to place. Coming through
>that air, the light from the horizon is bent more than it normally would be.
>But, as seen from the bridge of a ship, the light has been coming through
>"normal" air for miles. The "abnormal" bending is a long way off. So
>there's only a
>slight difference between where the navigator sees the horizon and where he
>would see it with normal refraction.
>
>But the small boat navigator is right there in the midst of the abnormal
>refraction. He sees the horizon in line with the tangent of the bend the light
>makes just before entering his eye.
>
>Set a prism between you and a line on the wall. Put the prism near the wall
>and look at the line from across the room. You'll see the line not far from its
>true position. Bring the prism close to your eye and you'll see the line a
>considerable distance out of place.
>
>Bruce

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