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Subject: Calculated Altitudes for Lunars
From: Bruce Stark (Stark4677@XXX.XXX)
Date: Mon Oct 21 2002 - 09:02:32 EDT
Before going further with the problem of calculating altitudes to clear a
lunar distance when GMT can only be guessed at, it may be worthwhile to look
into the question of how accurate the altitudes have to be.
Since a lunar distance has to be measured accurately it's natural to suppose
the altitudes used to clear it have to be accurate also. But experience
suggests otherwise. A look at how the altitudes are used will, I hope,
explain why.
Suppose you've taken a lunar distance and adjusted for index and instrument
error, and for semidiameter. That gives you the distance between the places
where YOU saw the moon and other body. But, unfortunately, those are not the
spots the Almanac has assigned them. Refraction and parallax made you see the
moon too low, the other body too high. You can't use your distance to get
Greenwich time from the Almanac until you've adjusted for the shift in
altitudes.
First you find how much the altitudes were shifted by refraction and
parallax. Then, using the shape of the triangle formed by your zenith, the
moon, and the other body, you adjust the distance for that shift.
The primary use of the altitudes is in finding the vertical shift: the
refraction and parallax corrections. But, unless you use one below 7°, you
only have to have the altitude within 6' of the truth to get the correction
within 0.'1. Furthermore, only about half of that 0.'1 will, on average, show
up in the distance. There's only one case in which the whole 0.'1 could show
up. That's when the two bodies and your zenith are in one, straight, vertical
line.
Normally the zenith, moon, and other body form a triangle, the shape of which
determines what fraction of the vertical refraction and parallax shows up in
the distance. But to put that triangle out of shape enough to cause trouble,
the altitudes would have to be amazingly inaccurate.
I don't see how errors of 6' or so in the altitudes could be a serious
problem.
Bruce
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