NavList:

Antoine Couëtte, you wrote:
"While we all acknowledge that sextant observations are limited to 0.1' in the very best cases, setting up software with much higher accuracy is not detrimental to Lunars as long as we remember all the essential points you indicate Frank, including instrument accuracy and Lunar limb shape effects and their magnitude. Within our usual CelNav applications publishing software with overkill accuracy can do absolutely no harm but... "

Absolutely it can do harm when such accuracy is described in a misleading way. When asked about the accuracy of his software, Paul explained that he calculates the positions of the stars (and in truth this applies only to certain easy cases, roughly 50% of the stars) to "mas" precision --milliarcseconds. He has doubled-down and posted more examples. I have no complaint with anyone finding entertainment and pleasure in these sorts of calculations. And if Paul is working for a team developing pointing software for a new telescope in the Atacama in Chile, that would be useful. But these "mas" stellar coordinates have zero relevance to common celestial navigation (for celestial altitudes 0.1' is all we ever need and that's 6000x larger than 1 mas) and zero relevance to lunars (even for lunars 0.05' is all we can realistically ask for and that's still 3000x larger than 1 mas). Talking about milliarcseconds when the real uncertainty in the results can be no better than a few arcseconds is misleading. I certainly don't think Paul is doing this fraudulently or with an intention to mislead. He's proud of his project, and perhaps he naively does believe that it is has some significance to celestial navigation. Regardless of good intentions or naivete, the end result is a misleading diversion. 

Let's stop for a minute, too, to consider the limitations of these sorts of milliarcsecond computations. What would it take for a milliarsecond to matter in a lunar distance? We can start with a genuinely useful rule for lunars: one arcsecond seen from the Earth is almost exactly a nautical mile on the Moon (within less than a third of 1%). This follows because an arcsecond is an angular ratio of distance/size equal to 206265 (this is the arcsecond version of my usual 3438 rule for minutes of arc). That is, if you have a 1cm object at a distance of 206265cm (equal to just over 2km), then it subtends an angle of one arcsecond. The Moon meanwhile has an average distance that is customarily quoted as 238000 (US statute) miles which happens to be 206820 nautical miles. Thus a nautical mile subtends an arcsecond at the Moon's distance, to within a fraction of one percent. Divide both of these numbers by a thousand, and you can see that one milliarcsecond corresponds to just about six feet at the Moon's distance. If you want milliarcseconds to matter for lunars, then you better have a limb model that picks up every damn rock! 

And I mentioned that milliarcsecond positional accuracy for stars only applies to the easier 50% of stars. So what's not easy, anyway? Just for a few examples among the standard navigation stars: 

• Consider Betelgeuse. It's a physically enormous star. Its angular diameter is about 50 mas, and its brightness is known to vary dramatically across the stellar disk, so its visual position (its "center of light") is necessarily always uncertain at a minimum of about 10 mas.
• Or Capella. It's a spectroscopic binary in which two components with different brightnesses and masses revolve around each other every hundred days. The center of light of that system oscillates by about 10mas on that same time scale. Unlike Betelgeuse, this one is predictable and open to calculation, but every case like this is unique. 
• Or Sirius, Procyon, and alpha Centauri (Rigil K.). These double stars systems require special solutions, and it's not easy. The photometric (center of light) orbit of Sirius, for example is on the order of 2 arcseconds. 

Accuracy in stellar coordinates at the milliarcsecond level can, no doubt, provide a certain amount of entertainment when you're done with everything else. But its value to traditional celestial navigation is exactly zero.

Frank Reed
Clockwork Mapping / ReedNavigation.com
Conanicut Island USA