the problem with my missing post was kindly sorted (it had been marked as spam!) and it now appears in the thread above.
I have had a look at the situation where the sun and the moon are different apparent sizes and I don’t think the situation is too bad. You have to take into consideration that the two “lens” shaped areas either side of the chord PP’ in my diagram are not the same size, but the formulae for their areas have similar forms. I think that the percentage obscuration in this instance is given by (50/pi)(w – sin w + f^2(W – sin W)). As before w is the angle subtended (in radians) by the chord of length l at the centre of the sun. However, now there is also an angle W subtended by the chord at the centre of the moon. You can see that the w – sin w part is the same as before, but now there is an additional part that comes from the area of the “lens” to the right of the chord which is governed by the shape of the moon. The factor f is the ratio of the moon’s diameter to that of the sun’s. Now as before w = 2 arcsin(l/d) where l is the length of the chord and d is the diameter of the sun. But we also have that W = 2 arcsin ((1/f)(l/d)). I hope this make sense.
I now calculate that if f=0.99253 (assuming that the moon was smaller than the sun at the time of your measurement) then the measured obscuration is now 18.08 +/- 0.13. So it does make a small difference. I guess that this formula works up until W or w is less than or equal to pi. I haven’t figured out what happens otherwise!
- This reply was modified 1 month, 4 weeks ago by Duncan Hale-Sutton.