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Re: Dip uncertainty
From: George Huxtable
Date: 2004 Dec 7, 22:37 +0000

```Yesterday I wrote-

"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."

====================

Since then, there have been so many messages on this topic that I am left
quite confused.

Would one or more opponents of my view, expressed above, kindly summarise
what's wrong with it, in simple terms that I can understand?

I would hope that any explanation would deal with the influence of
air-layers on the dip as seen by our two mariners, at heights of eye 6 ft
and 24 ft. Comparing three situations, as follows-

Normal atmosphere, right down to sea level.
A layer with abnormal temperature gradient, confined to within 6 feet of
sea level.
A layer with similar abnormal temperature gradient, confined to within 24
feet of sea level.

If we discuss it in such terms, then we don't need to make the impossible
study that Bruce proposes, to discover what those temperature gradients
actually are.

Bruce wrote-

"I still think my simplistic logic is nearer the truth. Put normal refraction
aside, as Alex does in his thought experiment, and suppose light should come
straight from the horizon. Now put a bend in the light somewhere along its
path. For simplicity, let's say the bend is caused by a prism planted on
top of a
five-foot tall buoy. Climb on the buoy and look at the horizon through the
prism. The angle between the horizon and where you see it through the prism will
be huge.

Now go twenty miles away, and from a height where you can see both the prism
and the true horizon, look (you'll need a powerful telescope) at the prism.
The horizon you see through the   prism will be only slightly out of line with
the true horizon.

That's because, from twenty miles away, the angle between top of the buoy and
the horizon is small."

"I realized, too late, that instead of a prism I should have specified a plate
of glass with just enough difference in parallelism of surfaces so that a ray
coming from the horizon (say 2' below horizontal) would be bent down 2' in
passing through it, and come out of the glass horizontal to the earth at that
point. At the buoy, looking through the glass, you'd see the horizon 2' above
its true place. Looking through the glass from 20 miles away you'd see the
horizon very close to its true place."

Well, that's a prism too, if a narrow one, that (within limits) bends all
light through 2', no matter where it is in the light path.

I've been trying, but failing, to draw out the picture that Bruce wishes me
to visualise. I wonder if he would kindly help me a bit further to put that
geometry together.

And if simplifying assumptions will help to make the point that I seem to
be missing, (e.g. flat-earth?, no-atmosphere?, bending in horizontal
plane?) then simplify, do!

George.

================================================================
contact George Huxtable by email at george---.u-net.com, by phone at
01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
================================================================

```
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