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On Thu, 15 Jan 2004 17:14:38 +0000, George Huxtable wrote:
>rmpisko wrote:
>>On the other hand, an observer at (for example) one
>>quarter of the Earth's circumference away at that same latitude, would
>>see the moon hanging against a different part of the sky; so the
>>longitudes might still be determined as accurately as they are now, if
>>the latitude is known.
>
>No, at a different latitude the retardation would be less, so the Moon
>wouldn't be "hanging" stationary at all, observed from there..
>
I see that I failed to make myself clear in my "On the other hand..."
paragraph about the earth with the doubled rotational speed.  I was
*not* referring to a simultaneous observation, but to an observation
taken six hours later when the earth had rotated and the moon was
again apparently stationary against the stellar background.  Same
latitude, six (or rather three) hours later, 90 degrees longitude more
westerly, and the moon again seems stationary to the second observer;
but in a *different* position on that background.

>So as a result, the overall precision of a lunar does NOT depend on the
>"parallactic retardation" of the apparant Moon. That's my new view, anyway.
>Unshaken, as yet...
>
I took this out of order, because I think the precision of a lunar,
whether on our earth or one with a doubled rotational speed, is not
affected by the "parallactic retardation" of the moon if you are
determining the _longitude_, but it does affect the ability to
determine the _time_ of that observation.

By thinking of this extreme case (doubling the rotational speed in the
analogy), the implications to me would seem to be that "clearing" the
elliptical path, (which would project to sine wave motion on both the
vertical and horizontal axes) from the apparent path (in order to
calculate the true position of the moon against the stars as seen from
the center of the earth) would hardly be necessary to give a very
accurate _difference_ in longitude between the two observers on a
single day.

Incidentally, the track of this apparent moon on the sky background,
the sum of a sine motion of the earth and a constantly moving moon
against the background, would be what I think is called a "Cycloid".
As least it would approximate the trace of an illuminated tire valve
on a bicycle wheel traveling on almost level ground at night, but it
would be upside down and mostly hidden for the moon.  The valve
appears stationary relative to the ground, the moon stationary
relative to the stars; both bobbing up and down a bit.  You can tell
_where_ the valve is on the ground, but it is difficult to tell from a
stopwatch when the invisible axle of the bicycle wheel passes over the
valve stem.

The next day both stations would see a changed apparent position of
the moon, but the difference between the two apparent positions would
be the same; and so on from day to day.   The actual (GMT equivalent)
time doesn't matter very much, and maybe not at all if you happen to
be at a latitude where there is a near hang or slight retrograde
motion.

But; to get the GM time onto a chronometer I believe you could not use
the lunar observation, which would be indefinite long.  I think you
would be better off measuring the interval between equal altitudes of
the sun, and thus determine the chronograph time of the local solar
noon at your known latitude.  Work back from your longitude to
determine GMT; and you should have very good accuracy of your time.

I rather doubt the standard current formulas and tables are set up in
this way, but didn't I read of a way for the longitude to be
determined directly from the lunars not long ago in this list?  It
could have been in the Slocum thread.

Thank you for your patience.


--
Richard ...

   

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