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Jan Kalivoda said, about the errors in finding GMT by Sun and Moon
altitudes rather than by lunar distance-

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>The errors of dip and refraction in both altitudes of the Moon and a star
>can be each enormous for finding the GMT by this way - but they could
>neutralize each other, not double in this method. From these altitudes the
>local hour angle of Moon and other body is computed. Star's LHA is added
>to its Right Ascension to obtain the Local Sidereal Time. The Moon's LHA
>is then subtracted from this LST to obtain the Right Ascension of the Moon
>in the moment of observation. You can then obtain the GMT by interpolating
>the time according to the gained Moon's RA in the almanac backwards.
>
>So if one LHA is added, the other LHA subtracted during the procedure,
>their errors from the wrong dip and refraction can neutralize each other,
>if they are roughly the same in the both altitude values, not reinforce,
>however great they are. The old authors sought the conditions, when these
>errors can be compared in amount - similar azimuths and altitudes of both
>bodies observed in the same time, above all.
>
>Please, send me the Sadler's paper, George. Thank you very much in advance.
>
>
>Jan Kalivoda

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Comment from George-

I agree with Jan, but only in part.

He is right, that in some circumstances the possible altitude errors,
inherent in using the horizon as a reference, can cancel out (for the Moon
and the other-body) rather than add. But he and I seem to differ somewhat
in our view of what those circumstances are.

Let's consider a really simple state of affairs. The observer is on the
Equator, and observes Sun and Moon on a day when both have a declination of
zero (or nearly so).

In that case the Sun and the Moon will follow the same path, rising due
East, passing exactly overhead, and setting due West. One will be ahead of
the other, and the angle between them gives a measure of GMT. A
lunar-distance observer measures that angle in the sky directly, without
involving the horizon.

Instead, our observer measures their altitudes up from the horizon. Let's
imagine that on that day, because of unusual refraction within a few feet
of the sea-surface, the horizon dip differs from its expected value for the
observer's height-of-eye. Let's presume that dip is actually 1 arc-minute
greater than the dip table in the almanac predicts. This would not be an
uncommon state of affairs.

When our observer corrects his sextant altitude by subtracting the dip
taken from the table, he won't subtract quite enough. If the corrected
altitude ought truly to be 30deg 00', our observer will make it 30deg 01'.
And exactly that same error would apply to any altitude of any body
measured at that time, whatever its altitude and whatever its azimuth.

The observer needs to deduce the angle-in-the-sky between these bodies from
those two altitudes. Because they are both in a straight East-West line
through the zenith, the arithmetic is simple and obvious. There are just
three relevant possibilities-

1. They are both East of the observer, rising toward the zenith, and at the
same azimuth of 90deg. In that case the errors in the altitudes of the two
bodies will both be the same amount and in the same direction. The angle
between the two bodies will be taken by subtracting their altitudes, so the
dip error will cancel out, exactly. It doesn't matter a fig what those
amplitudes are.

2. They are both West of the observer, setting at an azimuth of 270deg.
Exactly the same arguments apply.

3. One is East of the zenith, at azimuth 90deg, and the other is West, at
270deg. Now the angle between the two bodies is given by taking 180deg -
(alt1 + alt2). In this case, it's obvious, isn't it, that the dip error
will now give rise to a 2 arc-min error in the resulting angle, roughly
corresponding to a 4-minute error in GMT and therefore to a 1deg error in
the longitude (or 60 miles at low latitudes). This would be true for any
combination of altitudes of the two bodies.
[Note. As long as one of the bodies is at an altitude greater that 60deg,
there's a trick the observer could play here to get around this problem. He
could measure one of the altitudes by a back-observation, using the fact
that the sextant can measure altitudes up to 120deg. This involves facing
away from the body being observed, so that it's somewhere up behind his
head, and measuring up from the horizon in the opposite direction to the
azimuth of the body: then subtract the two altitudes.]

The argument above illustrates the principle in a simple situation, but it
remains valid in more complex cases. The effect of dip errors will be
greatest when the azimuths of the two bodies are furthest apart, and will
tend to zero in cases where the azimuths converge, irrespective of the
values of the two altitudes.

So I doubt that there is any virtue at all in waiting for a situation where
the two altitudes are the same. It's the azimuths that matter.

Is that argument convincing? If not, please argue back.

George.

================================================================
contact George Huxtable by email at george@huxtable.u-net.com, by phone at
01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
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