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A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Venus, Jupiter, and the Moon
From: Frank Reed
Date: 2008 Dec 16, 22:44 -0800
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, approximately. 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. -FER http://www.HistoricalAtlas.com/lunars/ --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---