NavList:

Antoine C, you wrote:
"would a typo not explain..."

Yes, thanks. April 30, not April 29. Of course I have made a common celestial navigation mistake here, and it's one that I regularly warn my students about! The local date was April 29, but switching into the early hours of UT, I forgot to advance the date (and this was in my paper notes, too, so I can't say it was "just a typo"). My mistake does also narrow the possible longitudes, right? I previously said that the observer was in the western hemisphere, and that was the only clue. This knocks out a further fraction of the western hemisphere, before we even simulate the astronomical circumstances.

I mentioned the issue of error in the fix. Each "lunar distance", observed at known UT, places us on a cone of position. The apex of the cone is at the center of the Moon. If we observe two stars, the cones have the same apex at the Moon's center, and they intersect along a ray (formally infinite in length) emanating from there. The spot where that ray intersects the Earth is our fix. 

Looking at just one cone of position, its "opening angle" is equal to the measured lunar distance. The cone opens wider by one minute of arc if our observation changes by one minute of arc. But down here on terra firma we don't care about some giant cone in three-dimensional space. We only need that small portion that is near us, and that small portion is nearly a flat plane, The observation places us on a plane, and the uncertainty in the observation shifts that plane towards/away from the Moon's center. If the Moon is high in the sky, a shift in the measured distance by 0.1' of arc (at the Moon's center) does not correspond to a position error of 0.1 nautical mile, as it would on the Earth. Instead our surface is sixty times beyond the Earth's radius (Radius of Earth about 3438 n.m., distance Moon-Earth about 206,000 n.m. *), so an error of 0.1' shifts the "plane of position" by 6 nautical miles. Direction has minimal impact when the Moon is near the zenith.

Suppose the Moon is low in the sky, as it is in this scenario, The portion of the cone of position that I am referring to as a plane of position can now cut the Earth's surface at various angles. Make a little vector from the Moon's center toward the star being observed. The plane is perpendicular to that. If our star is aligned above or below the Moon, then the plane reaches the Earth's surface low and flat, almost parallel with the ground. Thus a small change in its angular position produces a big shift in the intersection with the ground, considerably worse than when the Moon was near the zenith. Meanwhile if our star is aligned left or right of the Moon, then a shift of 0.1' once again shifts the intersection by about 6 n.m.

Frank Reed

* The mean radius (meaning of "mean" left open)  of the Earth in nautical miles is, more or less by definition, identical to the number of minutes of arc in a unit angle (one radian): 3438 = 180·60/pi. And the approximate distance to the Moon in nautical miles is, just by chance, equal to the number of seconds of arc in a unit angle: 206265 = 180·3600/pi. A simple corollary: physical sizes at/on the Moon (of mountains, craters, moonbases, etc.) measured in nautical miles are very nearly identical to seconds of arc as seen by observers here on Earth.