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    Re: Real accuracy of the method of lunar distances
    From: George Huxtable
    Date: 2004 Jan 17, 22:36 +0000

    I need to respond to a couple of messages from Richard Pisko.
    
    First, I should quote a bit of dialogue between us-
    
    ==============
    
    >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.
    
    ================
    
    Richard is kind when he says "I failed to make myself clear..." In fact,
    what he said was entirely clear, but I failed to read it properly, or
    failed to take it in. Sorry about that.
    
    He said-
    
    >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.
    
    The whole point of a lunar is to determine the almanac-time (Greenwich
    time) from the position of the Moon in the sky. Knowing that time , you can
    determine how much the Earth has rotated since the previous noon. From a
    time-sight of some sort you determine local time. From the difference
    between these times, you can determine longitude. So unless you can
    determine Greenwich time accurately, you can't determine longitude
    accurately. So I don't follow your statement above.
    >
    >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.
    
    =======================
    
    Richard, I like your analogy with the bike wheel and the torch and I would
    like to develop it firther, if you don't mind.
    
    It's dark, you are standing on a high bridge, looking down on a road
    passing below. A bike is passing along that road at a rather steady speed,
    and you want to determine the exact position of that bike at a particular
    moment. However, being dark, you can't see the bike, all you can see below
    you is a lamp which happens to be attached to the spokes of one of the
    wheels, about halfway out to the rim. That lamp will be moving unsteadily,
    faster or slower than the bike, at different moments, so any position you
    record for the lamp will be a very inaccurate measure of where the bike is.
    
    But now, imagine that on board the bike is a sensor, which can measure
    exactly the angular position of the wheel (and the lamp, with respect to
    the wheel hub), and can transmit that information by radio to you. Now, you
    can use that information to add or subtract the appropriate displacement to
    your observed lamp position, and the result will be that you will know the
    steadily-changing position of the bike, precisely, at any time..
    
    Of course the bike represents the steadily-moving true-Moon, the road
    represents the path of the Moon across the sky, unrolled. And the lamp,
    with its variable speed, represents the apparent Moon.
    
    Even if the lamp was mounted right at the wheel's rim, so that it came to
    an actual stop at the bottom of its path, the method would still work, as
    long as you knew the position of the lamp, and its angle and radius from
    the hub.
    
    No matter how big the perturbation is, caused by the lamp's motion on the
    rim, as long as its angle and radius are precisely known at any time. the
    true position of the bike can be deduced, to the same accuracy as the
    apparent position of the lamp was measured.
    
    The analogy isn't a perfect one. It would be better if the lamp was
    obscured from the observer during the upper half-turn when it appeared to
    be going faster than the bike, so it was only visible when going slower.
    Then it would correspond to parallactic "retardation", as happens with the
    Moon, and never "acceleration".
    
    =================
    >
    >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.
    
    I don't follow this well. Are you referring to two observers, about 90
    degrees apart at the same latitude, at the same moment, or to the same
    observer moving 90 degrees in latitude? And if the former, how do they
    commumicate their observations to each other to establish the longitude
    between them?
    
    >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.
    
    This would be (and was) a good way to establish chronometer error, but you
    needed to know your longitude: perhaps when passing a known cape.
    
    In mid-ocean, if you only had a lunar distance to check your chronometer
    against, you could check that it hadn't developed a gross error of more
    than a minute or two of time, but it wasn't possible to get much more
    precise than that.
    
    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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