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Moon meridian passage and its Tilt
From: Igor S.
Date: 2019 Sep 17, 16:14 -0700
From: Igor S.
Date: 2019 Sep 17, 16:14 -0700
Hello everyone,
Recently I've met an idea that moon is upright (Tilt = 0) during the meridian passage (see the schematic picture attached). To test this idea I've wrote a short simulation (in Octave) based on the data from NA 2010. The report with all calculations, results and references one can find in the attached pdf file.
The lunar tilt can be expressed by surprisingly simple but rare formula:
Tan(Tilt) = [ cos(h_m)*sin(h_s) - sin(h_m)*cos(h_s)*cos(AZ_m - AZ_s) ] /(cos(h_s)* sin(AZ_m - AZ_s) )
where index m - moon and s - sun. The references to this formula are given in my report.
Upon quick inspection of this formula at the moment of moon meridian passage (AZ_m = 0 or AZ_m = 180) I don't see the aforementioned statement to be necessarily valid.
I calculated lunar Tilt during it's meridian passage (in it's last quarter phase only) during the whole year 2010 for several latitudes (see attached report) and found a non-zero tilt which agrees well with that of shown by Stellarium program. Stellarium shows the picture which is supposed to be seen by the observer at specified time and position.
Please comment.
kind regards,
Igor
File:
