The illustration in the middle is not 95% luminosity.
*** I apologize for my rather approximate English. I'll try my best! ***
Why is it not 95% luminosity? the best way to know is to calculate! And it's so easy I wonder why you did not do it yourself already.
If
alpha is the angle between the directions Moon-Observer and Moon-Sun, the illumination ratio of the moon is given by :
r = (1+cos(alpha))/2 (this is an approximate formula, assuming the light rays coming from the moon are parallel, but the error is very small, and in any case in the right direction...)
Taking into account the data from the Naval Observatory you quoted on your post of Aug. 10, 01:16:53 AM (reply#10), we find that
alpha = 26°. This gives an illumination ratio of...
0.9494. Hence the 0.95 given by the National Obervatory.
So I think everything is ok : contrary to your feeling,
the moon phase in the video at the begining of the thread is perfectly compatible with Round Earth Theory.Now the next question from me would be :
is it compatible with Flat Earth Theory?...I would very much appreciate if you - or any other authoritative flatearther - post a similar demo for the case of FET, of course with the same parameters (sun and moon altitudes and azimuts, same place and date,...). Is it possible? I don't know enough FE theory to be able to do it myself (in fact I know nearly nothing concerning FET...)
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