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Thanks to Trevor for taking his explanation rather further, but I still
have difficulty with drawing out these ray-diagrams, in regard to the
following text.

>However, the horizon is not at an infinite distance from observers with
>heights of eye of 6 or 24 feet. Thus, rays of light from the horizon to
>their eyes will not be parallel to one another -- except in the case
>where they are both placed on the same ray. You, however, have defined
>the model as one in which they are both on the same
>anomalously-refracted ray. That being so, they cannot both be on the
>same ray that would be expected with standard refraction.
>
>If the actual light ray suffers anomalous refraction primarily when very
>close to the sea's surface, then the further the observer backs away
>from the horizon and hence the higher his eye, the smaller than angular
>anomaly will become. I don't know how to prove that mathematically but
>it is easy to see if you draw out the light rays and easy to understand
>if you follow Bruce's argument and reduce the anomalous refraction to
>the effects of a single prism placed in the light path.

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

The diagrams are hard for me to visualise. If only we could gather round a
blackboard, I think we would be able to resolve the matter rather quickly.

Let's try to isolate this prism question, eliminating the Earth's curvature
and any refraction by switching it into the horizontal plane. Take a large
flat parade ground, and put a light L with a narrow directional beam at its
the North end, pointing South so that it defines a precise North-South
line. Somewhere South of L, put a prism P. Let's say it bends that
North-South light ray 5 degrees towards the West into a direction 185
degrees. No matter how far from L the prism was placed, that would be the
case.

If an observer O looked into that prism in a direction of 005 degrees,
placing himself somewhere on the parade-ground to do so, he would be
looking right into the light-beam, no matter how far back from the prism he
stood.

It is true that if the observer measures the angle POL, between the
direction the light appears to be coming from (OP) and his direct view of
the lamp OL (not through the prism), then THAT angle will change
considerably, depending on where he is on the parade ground, on the
distances LP and PO. But so what? That's an angle we have no interest in.
It's a red-herring that has distracted from the truth.

The only thing that's relevant is the direction of the incident light-beam
and the constant angle that it's deflected through by the prism. No matter
where the prism is placed.

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

Now let's revert to that vertical section through the Earth, to see the
analogy. When an observer looks at a point on his horizon, he sees it by a
light-ray that leaves it tangential to the surface. If not, it wouldn't be
on his horizon. So its direction at the horizon is precisely defined, just
as on the parade ground it was defined by the North-South light-beam. And
what matters is its final direction, because that's the apparent direction,
in the horizon mirror, that the sextant measures up from. And the angle
between those two directions is constant, just depending on the bend-angle
of the prism, or the refraction in the air path, quite independent of where
the bend occurs.

I hope that this will dispose of the argument about "Bruce's prism" (but
fear that it may not...)

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

Let me suggest another tack.

Chauvenet, vol I deals with this matter, explaining the "geometrical" dip
just due to the spherical Earth, and the"standard" dip that's caused by the
combination of standard refraction with the geometrical dip. Chauvenet
doesn't consider "anomalous dip" as such, but never mind; we may be able to
get something out of it.

If there was no atmosphere (or if there was no refraction near the
sea-surface), so that the light at low levels travelled in precisely
straight lines, then the "geometrical" dip
is obtained from
tan dip = square-root of (2 h / a)
where h is the height above sea-level and a is the Earth's radius in the
same units (in feet, say).

I make a to be 20,900,000 feet. Hope nobody disagrees. In which case the
geometrical dip for h = 6ft would be 2.605' and for 24 ft would be 5.210'.
That's not what's shown in the dip tables, however. They take into account
the effects of refraction in a "standard" atmosphere. Chauvenet gives, in
his table XI, 2.4' for 6 ft., which is a reduction of 0.205', and 4.8' at
24 ft., a reduction of 0.410'. You can see that the standard refraction has
just twice the effect on reducing dip, from a 24 ft. viewpoint, than it
does viewed from 6 ft. In the same way, non-standard refraction, which is
"anomalous dip", will have twice the effect at 24 ft height than it does at
6 ft. This is presuming that the air density gradients that cause the
refraction remain uniform up to a height of at least 24 ft.

George.

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