Wrong. (Although that's a reasonable question.)

What do we need to know?
Sunset occurs when the apparent position of the Sun's upper limb reaches the
*apparent horizon* ;
so we need to know the positions of both the upper
limb (as a function of time) and the apparent horizon.

Well, we certainly do know the Sun's geometric position accurately.
And we know its diameter, too; so we can compute the *true*
zenith distance of its upper limb accurately at any time. That's the easy
part (i.e., the astronomical part) of the problem.

However, the *apparent* positions of both the
Sun's upper limb and the visible horizon are affected by atmospheric
*refraction*
— but in somewhat
different ways.
And, to calculate refraction accurately, we need to know the distribution
of temperature and pressure in the atmosphere accurately. That's the
hardest part (i.e., the meteorological part) of the problem.

Another difficult part has to do with just where the apparent horizon is.
If it's a sea horizon, the apparent horizon depends not only on the
observer's position, but also on the height of the waves.
If it's a land horizon, irregularities in the topography determine the
geometric position where the Sun disappears.
Given the heights of the eye and the physical horizon, we can
calculate
the
*dip*
of the apparent horizon — but we must take refraction into account.

Thus, refraction near the horizon itself — i.e., near the sea
surface — hardly affects the dip; but it affects the distance to
the horizon, and the Sun's refraction, with full force. As a result,
the dip turns out to be affected by the mean curvature of the ray; while
the distance to the horizon, and the solar refraction, are affected
by the (reciprocal of ) the mean *reciprocal* of the curvature.
So, in the real world, where the lapse rate varies with height, the
dip depends on just the (arithmetic) mean lapse rate, while the Sun's
refraction depends on the reciprocal of the mean reciprocal —
i.e., the *harmonic* mean — of the lapse rate. (See my
1998 paper
with George Kattawar on the
dip diagram
for technical details.)

Therefore, while we can find the dip from just the temperatures at eye
level and at the sea horizon, without knowing the detailed temperature
profile in between, the refraction of the setting Sun depends on all the
details of the temperature profile. In general, the factor of four
discussed
above
is a **lower limit**; if the ray curvature is similar to
the Earth's curvature near the sea surface, the solar refraction can
become very large, even though the dip of the sea horizon is hardly
affected. (This actually happens in
**Fisher**'s
Type B sunsets.)

As the variations in astronomical refraction at the horizon are larger
than the dip variations
by a factor of at least four, they clearly amount to a few tenths of
a degree ordinarily, and sometimes reach a few degrees. These angles
correspond to a minute of time or so at low latitudes, and several
minutes at higher ones. In the polar regions, the observed times of
sunrise and sunset can differ from predictions by several *days *;
see the section of the bibliography devoted to the
**Novaya Zemlya**
effect.

So it is hardly surprising that the standard tables of sunrise and sunset
times are given only to the nearest minute. That's about as accurate
as we can expect a prediction to be. As G. M. Clemence,
the Director of the Nautical Almanac Office, wrote in
**1951**,

… the indeterminacy is a geometrical one, and the refraction at any instant may differ by several minutes of arc from the most accurate value that can be calculated.

Evidently, the variability of refraction is the big problem. But can't we just use the atmospheric profiles measured by radiosondes twice a day to calculate the refraction accurately?

No. It turns out that standard meteorological instrumentation isn't
accurate enough to allow the calculation of refraction very precisely,
even at the
*astronomical
horizon *.
And *below * it, at the apparent
horizon, the refraction is still more sensitive to the details of
atmospheric structure that are missed by the radiosondes,
as Bruton and Kattawar explained in their papers of
**1997**
and
**1998**.
In any case,
the radiosondes aren't launched from sea level, but from (usually)
airports, at locations many meters above the sea; so even if the
instruments had perfect accuracy and unlimited resolution, they couldn't
provide data for the most important part of the profile: where the sunset ray
is horizontal — i.e., the part **below** eye level. (See my
2004 paper
on low-altitude refraction for more details.)

For an excellent and thorough discussion of refraction near the horizon,
see
**Fletcher's
great article**.

This subject is discussed further in
**Lutz Hasse's thesis**
(in German; but summarized in English in
**1964**.)

Hasse's unique contribution was to point out that a strong thermal
inversion can produce a *Kimmfläche* or
“horizon surface” above the horizontal.
This false horizon
can best be thought of as the upper edge of Wegener's
blank strip
associated with a superior mirage. If that mirage contains only an
inverted image of the distant sea surface, it may not be noticed.
However, as
**Wegener (1918)**
pointed out, this strip blocks the observer's view of the Sun (and other
celestial objects); so its upper edge will be where the setting Sun
disappears, if the strip's lower edge is hidden by the physical horizon.
Because of the very great refraction at this false horizon, the sunset
will appear to be greatly delayed.

A similar elevation of the apparent horizon can sometimes be produced by a
distant layer of coastal stratus. If its upper surface is flat enough, it
can be mistaken for the sea horizon. This may account for some reports of
anomalously *small* refraction at sunset (i.e., early sunsets).

Also, if there is an
inferior mirage
at the horizon, the
“last glimpse” of the Sun occurs a few minutes of arc
*above* the apparent horizon. (See, for example, an
animated simulation
of an inferior-mirage sunset — or the same sunset at
high resolution.)
As the diurnal motion carries the Sun through 1 minute of arc in 4
seconds of time, that alone would make the sunset appear several seconds before
the expected time; however, the
fold line
of the inferior mirage is always below the astronomical horizon, so the
mirage actually makes the sunset occur a little later than expected.

Normally, the place to start is the USNO website — but it's being re-done, and is not available.

For arbitrary places on Earth, you can get similar results from either the FAA calculator or NOAA's calculator.

Sarah Wahlberg has pointed out another page with many links to solar positional and other information. It leads to nominal predictions for moonrise and moonset, as well as twilight. Just be aware that these values are as arbitrary (and as inaccurate) as predictions of sunrise and sunset times.

© 2010, 2012, 2018, 2020 Andrew T. Young

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