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From: Frank Reed (no email)
Date: Thu Jul 22 2004 - 15:53:24 EDT
Jim wrote:
"My idea was simply that for any observation you make with watch and sextant
where the object appears on the horizon, the same observation could be made
with watch only (or watch and binoculars). Then you run the computations, using
using zero for the sextant reading, making all the appropriate corrections
(but using zero for the sextant index error
correction!), and you should get identical results. Possibly that's what
Bill was getting at."
Yes, this is all certainly do-able in principle. A minor problem is that the
refraction tables are a little "iffy" for very low altitudes. This could
result in an error of 2 miles or maybe a little more, but this isn't huge
(especially considering you can do it without a sextant).
And wrote:
"Furthermore, it seems that on a moonless night you might be able to get
a LOP from the setting of a star or planet more easily using this method,
since you don't have to see the horizon, you only need to note the instant
at which the star or planet blinks out."
It's a great idea in principle. But I've never seen a star or planet rise or
set on a sea horizon (without optical aid and excepting the Sun as a star for
the moment), and nor has any other celestial navigator in history more likely
than not. Surprising, isn't it? The problem is not refraction, but extinction.
Atmospheric extinction is the dimming of a star's light as it passes through
the atmosphere. Near the zenith, the atmosphere typically dims a star by about
0.2 to 0.3 magnitudes (side note: this means that stars in "outer space" only
look very slightly brighter than they do at sea level on a dark night). Like
refraction, atmospheric extinction depends directly on how much "air mass"
there is between you and the star. And just as refraction increases dramatically
close to the horizon, do does extinction. For an object at 2 degrees altitude,
the average extinction at sea level is over five magnitudes. That means that
all first magnitude stars would be rendered completely invisible to the
unaided eye at that height. Even the planets Venus and Jupiter which can be
extremely bright would be invisible right at the horizon. Typical extinction at zero
degrees altitude is 11 or 12 magnitudes, so even an object that has a normal
magnitude of -4.5 (e.g. Venus when brightest) would have an apparent magnitude
fainter than 6 right at the horizon. If you think about it, that makes sense.
The magnitude of the Sun at high altitude is around -26.5, and it's impossible
to look at it for more than a moment. But when it's sitting on the horizon,
it seems comparable to the Full Moon in brightness and you can look at it
easily. And sure enough, that observational evidence is consistent with some 12
magnitudes of extinction at zero degrees altitude. Note that the technique you
and Bill described DOES work with the Sun and Moon. If you note that time when
the upper limb just disappears on the horizon, you've got a Sun UL sight with
the altitude equal to zero. Also, you could catch Venus at it brightest right
on the horizon if you're using binoculars.
Again though, these observations are subject to some inaccuracy because the
refraction near the horizon in the usual tables is approximate. It's not hard
to see where errors might arise. Take a look at the temperature/pressure
correction tables for zero altitude. Just one step left or right in the table
changes the refraction by 1 minute of arc.
By the way, speaking of looking at the Sun, there is presently a very large
sunspot group near the center of the Sun's face. It is visible with the unaided
eye at sunset and an easy mark through a sextant with the shades in place.
Frank R
[ ] Mystic, Connecticut
[X] Chicago, Illinois
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