You can see from elementary geometry that the
*dip* *a*
of the lower line of sight **OPM** is the same as
the angle **OCP** at the center of the Earth, because both are complements
of the angle **COP**. So the angle **OCM** at the center of the Earth,
subtended by the arc between the observer at **O** and the point **M**
where the lower ray re-crosses the observer's
(curved) level surface, is 2*a*.
And that's the same as the angle between the upper and lower rays at the
observer.

The ray **OPM** below the observer's horizon
has the same altitude *a* at **M** as
its dip at the observer; so
whatever refraction it might have *above* eye level
(to the right of **M**)
is the same as that of the ray above the horizon at **O**.
In fact, the whole course of the lower ray to the right of **M** has exactly
the same shape as the upper ray has to the right of **O**, because both
rays have the same altitude at the same height above the surface of the
Earth.
So, no matter what refraction these rays suffer *above* eye
level, their angular separation outside the atmosphere is the same as
their angular separation at the observer, namely, 2*a*.
Consequently, the sky at the horizon is not distorted: the magnification
is unity.

As an aside, I note that Cassini's uniform atmosphere is a model of this kind. Consequently, his model produces unit magnification at the astronomical horizon, and very little distortion of the setting Sun.

These angles are very small — a few minutes of arc —
so the curved ray **OPM** is well approximated by an arc
of a circle; let **S** be the center of that arc.
Then its central angle θ is also the change in direction of the bent
ray within the layer below the observer (because the ray is perpendicular to
the dashed radii **OS** and **MS**, at **O** and **M**,
respectively).
So the ray separation 2*b* at the observer is less than the value of
2*a* we had without refraction by just this bending, θ; that is,
2*b* = (2*a* − θ).
Notice that θ is the contribution to the refraction of the lower ray
made by the air below eye level.

But on every level surface above the observer (including the top
of the atmosphere), the two rays are still separated by 2*a*, the
angle between the local verticals at **O** and **M**
(because both rays make the
same angle, *b*, with their local horizons).
Therefore the angular magnification of a small celestial object at the
astronomical horizon is just

i.e., the ratio of the apparent to the true angular separation of the rays.

It is most convenient to write the angular magnification *m* in terms of
the ratio of curvatures *k* of the ray and the Earth,
which we found on the
bending
page. As the
curvature is just the reciprocal of the radius of curvature,
*k* = **OC**/**OS**.

Now, the length of the curved ray **OPM** is very nearly the same as
the arc length **OM** measured along a level surface.
But (using the fact that the arc length is just the angle in radians times
the radius of the circle)
**OM** = 2*a* × **OC** ; and similarly,
**OPM** = θ × **OS**.
If we equate these two expressions for the distance from **O** to **M**,
we have

Sometimes it's convenient to work with the reciprocal *K* of
the ratio *k* ; then *m* = (*K* + 1)/*K*.

On the
bending page,
we found that *k* is usually about 1/6 (which means *K* is about 6).
This value makes *m* = 5/6, which is a typical value for the ratio of
vertical and horizontal diameters of the setting Sun.
(See the
simulation
of a sunset in the Standard Atmosphere, for comparison.)

However, when the observer is in a thermal inversion, the value of *k* can
be much larger — approaching, or even exceeding, unity.
What does this do to the flattening?

As *k* → 1, *m* → 0. That means the Sun flattens down
into a thin line at the horizon. This condition is always seen from just
above a duct.
However, such strong inversions are usually accompanied by
complex thermal structures
that make it difficult to assign a well-defined flattening to the observed
images.

When *k* exceeds unity, the observer is inside a duct. Then the Sun
cannot be seen at the astronomical horizon, which is blocked by Wegener's
blank strip.

As each value of *m* corresponds to a value of *k*,
and each value of *k* corresponds to a particular lapse rate just
below eye level, the observed flattening of the Sun at the horizon
provides a good estimate of the lapse rate near eye level.

Copyright © 2006 – 2009, 2012 Andrew T. Young

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