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>> I'm trying to solve the following sunrise/sunset astronomical triangle and
>> am encountering unexpected pitfalls.

When I am playing that game with myself just for the fun of it, I recall the
counsel of the list when I first joined and wanted to establish an LOP using
HO229 tables, and binoculars to spot the moment of sunrise/sunset.  The fly
in the ointment ("problem" for non US of A readers) as counseled by the list
gurus is that the actual refraction from the tables in the front of the
Nautical Almanac (NA) can be *quite* different from the published values of
-0d 49!6 to -0d 49!8 for the center of the Sun (even if corrected for
temperature and barometric pressure) due to the amount of atmosphere the
light passes through and other gremlins.

To your quest, if longitude and latitude are known, the variables are
declination and refraction.  Not too much we mortal layman can do about
refraction unless living on a celestial body without atmosphere (in which
case there are other problems to deal with, but the exact moment of sunrise
may be quite important if Hollywood movies are to be believed). Therefore
declination becomes the variable you can nail down.

When I play (jump starting my brain in the AM by calculating sunset--and the
moment the sun is due west--I hate mornings) I look in the NA for sunset on
the daily page for my latitude, then correct for any difference in latitude
using the "Table for Interpolating Sunrise, Moonrise, Etc." on page xxxii in
the back of my NA.  That adjusted time is corrected for longitude before or
after the 15d increments defining my time zone with the usual method (arc to
time). Then I have a time within a couple of minutes (refraction ignored)
for my known position. For calculated time I can interpolate a declination
that is more than good enough for practical purposes.

Now the fun begins. When doing sight reductions I use the sine formula for
elevation of the body and the cosine formula for azimuth. As a sidebar, if I
recall George H. has tangent formulas for azimuth--one if memory serves,
that can derive azimuth without first establishing elevation of the body.
Please see the archives for these.

Back on track, if you have a longitude, elevation (given as -0d 49!6 to -0d
49!8 for the center of the Sun) and declination you can flip the cosine
formula normally used for determining azimuth when elevation has already
been solved. This gives you the Local Hour Angle (LHA) of the body, and as I
use it is a derivation of the classic time sight (as used by Sumner et al)
and explained by Frank in the archives.

Specifically:

cosine LHA = (sin elevation - (sin lat * sin dec)) / (cos lat * cos dec)
Hope I have that right--I have not posted in a long while.

Combine the computed LHA with the longitude in the usual manner, and you
have your Greenwich Hour Angle (GHA).

Take the computed GHA and find the value just below it on the daily page for
your date.  Subtract that tabular hour's  GHA value from your computed GHA
and convert the difference from arc to time.  That time conversion plus the
UT/GMT for the hour used is the UT of your sunset or sunrise.

The above allows you to escape messy Equation of Time adjustments and
interpolations for your local time.

It may sound complicated. A few passes with a TI-30Xa (three memory banks)
and I can knock it out in few minutes even with the NA interpolations. With
the hourly rate of change of declination now (mid June) interpolating
declination is pretty easy ;-)

Bill B.




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