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    Re: Venus, Jupiter, and the Moon
    From: Frank Reed
    Date: 2008 Dec 16, 22:44 -0800

    Two weeks ago, Dave wrote:
    "If you grabbed your sextent (or your octant, or your quadrant, or your 
    protractor and string) and made the following near limb distance observations 
    at 20:00:00 UT 2008 Dec 1: Venus 1 deg 16.9 min, Jupiter 1 deg 35.3 min, what 
    island would you be near?"
    This is a case where we can get a complete position fix by lunar distances 
    [NOTE: this was not how lunar distances were used historically]. The measured 
    lunar distances at known GMT will determine both latitude and longitude 
    without any measured altitudes --a fix without a horizon. There is a brute 
    force approach to the problem, but I've been trying to think of ways of 
    explaining how this works in a graphical fashion. And that's what you'll see 
    in the attached diagram.
    We know from any nautical almanac that the Moon, at this instant of time, is 
    located directly above a point off the west coast of South America. Its GHA 
    was 74d 09.0' and its Dec was S22d 35.7'. If you were observing the Moon from 
    that location its apparent position in the heavens would be exactly the same 
    as its geocentric location. In other words, the given GHA and Dec would 
    correspond exactly to the GHA and Dec of the center of the Moon as observed 
    (except for a very small oblateness correction that we don't have to worry 
    about quite yet). The Moon, Jupiter, and Venus would make a nice triangle in 
    the sky as seen from this spot. The top half of the attached diagram shows 
    how this would look. But if we observe from any other spot on the Earth, the 
    Moon's position would be shifted relative to Jupiter and Venus. This is where 
    the measured lunar distances come into play...
    The lunar distances from Venus and Jupiter allow us to fix the actual observed 
    position of the Moon's center in the sky. In the upper half of the diagram 
    below, I've drawn circles around Jupiter and Venus with radii equal to the 
    observed distances plus the Moon's semi-diameter (the scale is 0.5 minutes of 
    arc per pixel). And where those circles cross is where we would find the 
    center of the Moon as seen from our observing location. This observing 
    location can be graphically represented by taking the Moon out of the sky and 
    replacing it with a "ghost image" of the Earth with its continents and lines 
    of latitude and longitude. This image represents all of the possible 
    locations of the Moon's center as seen from every observing point on the 
    surface of the Earth. Note that the center of the projected Earth is exactly 
    that spot in the Pacific where the Moon would be straight overhead. Of 
    course, the Earth is reversed because if we move north, the Sun drops down to 
    the south as a result of parallax. Incidentally, the semi-diameter of this 
    ghost Earth is exactly equal to the Moon's horizontal parallax at this 
    moment: 54.2 minutes of arc. We can now examine the diagram and directly read 
    off our position, at least roughly. Eyeballing it, I estimate that the 
    position is in the Atlantic around 8 degrees South and 10 degrees West. 
    That's near Ascension Island, which is the answer to the puzzle, asssuming 
    I've done nothing wrong.
    This analysis, projecting the circles and finding their intersection on the 
    projected face of the Earth, can of course be done rigorously, and I believe 
    that the math Dave posted a couple of years ago (when I first brought up this 
    idea of a position fix by lunar distances) does just that. We can also take 
    this approximate position from the graphical solution, 8 South and 10 West, 
    and use it as a starting point for a straight-forward clearing process. You 
    use this position as a "DR" in your favorite lunars clearing software (e.g. 
    this one: www.HistoricalAtlas.com/lunars) and then just walk your way to the 
    point where the calculated "errors" in the observed distances go to zero. 
    When I do that, I find that the observer's actual position was closer to 8d 
    South and 13d West --not far from that rough, graphical estimate.
    Notice also that the diagram makes it easier to see the relative accuracy of 
    the position fix. Just by examining that "ghost Earth" you can see that the 
    resulting fix in latitude, in this particular case, is more accurate than the 
    longitude. Indeed the fix in latitude would be about as accurate as if the 
    Moon were in the zenith. That is, assuming measured distances accurate to +/- 
    0.1 minutes of arc, the latitude would be accurate to +/- 6 nautical miles, 
    So that there is no misunderstanding, I am not proposing this "graphical 
    solution" as a general method for solving this particular problem. It's a 
    nice means of visualizing what's going on, and that's all.
    Navigation List archive: www.fer3.com/arc
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