NavList:

RE 1 - https://NavList.net/m2.aspx/Zubenelgenubi-occultation-Walden-jan-2022-g51921

RE 2 - https://NavList.net/m2.aspx/Zubenelgenubi-occultation-Couëtte-jan-2022-g51944

Dear Dave, dear All,

Thanks again for submitting this quiz (RE 1).

After publishing a first solution (RE 2) I decided to further study this problem.

What I needed was to accurately compute an approximate position rather than more or less 100% guessing the "early/initial starting points" (See RE 2  §2.1.1 and 2.1.2).

So I derived the attached solution. It works well, quite well indeed, by it does takes time to use ... It took time to narrow it down onto just 2 pages.

That's all I can say.

And by the way, I have used very accurate Ephemeris this time - directly from a most recent version of INPOP - , setting aside my own computed one.

Going from (5) to (6) in the attachment can be done :

(a) - Either through hand-drawing on paper : this is what I displayed in one of the attachments of RE 2.

(b) - Or through direct computation, which is greatly simplified here since it is possible to easily solve it geometrically (Thales theorem) and this is what I did here.

Other than that, I think that everything else in this method is sufficiently clear for somebody with an Engineer's degree and fully familiar with 3D computations : you need to "3D see" in space if you want to fully understand it. If questions, please ask.

Final result is within less than 1 mile from N32°/W109° which is the "exact" solution used by Dave using a recent software (private communication).

I would have thought that I would have come closer. A few possible explanations here : a difference of just 1km in the Moon Diameter will exactly translate into - at least - a 1 km offset in the curve because the Moon image is stamped from infinity onto the Earth. Maybe it already explains all. I did not correct for Observer's height. Certainly this not required for the sea position. The usual Official Ephemeris give Moon Centre of Gravity Coordinates and not Moon Center of Figure and that alone could explain most of the difference. What else ???

So, thanks again Dave, you forced me into taking a close look at accurately computing the intersection of a cylinder and a sphere. I had delayed it for years. At least, this computation is extremely accurate, but it is only an "approximate" computation for our current aim to-day: Spherical Earth, Spherical Moon and no refraction.

Very few of you will see any interest in such work because you are so much "modern software addict". But for the possible remaining ones, I am offering this research unpublished anywhere so far to the best of my knowledge.

Thanks again also to Frank for offering such an awesome publishing platform.

Antoine M. "Kermit" Couëtte

PS : And as usal : hopefully not too many typos …