After the partial solar eclipse on the 25th October I came up with a formula for how to calculate the percentage area of the sun that is covered by the moon (the percentage obscuration). I now want to show how I derived this formula.
I have assumed that the sun and the moon have the same apparent diameter during an eclipse. This won't necessarily be true but will be a reasonable approximation. So in the above diagram we will assume the moon is the circle to the left and the sun to the right. Where the two circles intersect represents the area where the moon eclipses the sun. If we can calculate the area of the sun to the left of the chord AB then the area of the intersection of the two circles is just twice this by symmetry.
Firstly, as ADC is a right-angled triangle we have that AD/AC is sin t. Let AC be the radius r and let AD be x. So (x/r) = sin t. If the chord length AB is l then l = 2x.
We can see that the area of the sun to the left of the chord AB is the area of the sector of the circle AEBC (the piece of cake) minus the area of the isosceles triangle ABC. The ratio of the angle 2t to the angle 2pi (pi being the constant 3.14...) is the same ratio as the area of the sector to the area of the circle (t being measured in radians). Thus the area of the sector is tr^2 (where r^2 represents r squared).
The area of the triangle ABC is half the base AB times the height DC. Half AB is x and DC is rcos t, so the area of the triangle is xrcos t. So the area of the sun to the left of the chord AB is tr^2 - xrcos t and the area of the intersection of the two circles is 2(tr^2 - xrcos t).
We can now calculate the obscuration fraction as the ratio of this intersection area to the area of the sun. We can see this is (2/pi)(t - (x/r)cos t). But (x/r) = sin t, so the obscuration fraction is (2/pi)(t - sin t cos t). But sin t cos t = (1/2)sin 2t. So the obscuration fraction becomes (1/pi)(2t - sin 2t). If we let 2t = w, then the obscuration fraction is (1/pi)(w - sin w) or as a percentage (100/pi)(w - sin w) as we stated before. Note that to find w we use that w = 2t = 2arcsin (x/r). But x = l/2 and r = d/2 where d is the diameter of the sun. So w = 2arcsin (l/d). Thus measurements of the chord l and diameter d give the angle w and hence the obscuration.
There is another measure of the depth of the partial eclipse and that is the magnitude. This is the fraction of the sun's diameter that is occulted by the moon. In the above diagram you can see that the portion of the diameter of the sun that is covered by the moon is twice the distance ED. Now ED is EC - DC which is r - rcos t = r(1 - cos t). So twice ED is 2r(1 - cos t) and so the magnitude is just (1 - cos t) which in terms of w is (1 - cos (w/2)). By convention the magnitude is usually quoted at maximum eclipse.
I now want to consider the more complicated obscuration calculation when the moon and the sun are not the same apparent diameter on the sky. The above figure shows the exaggerated situation. We again assume that the disc on the left is the moon and that on the right is the sun. It can be seen that the obscured area of the sun AEBF consists of two 'lens' shaped areas either side of the chord AB. We have already calculated the first AEBD. This is (r^2/2)(w - sin w) with w = 2arcsin (l/d). By a similar argument the lens shaped area AFBD is given by (R^2/2)(W - sin W) where W = 2arcsin(l/D) (D=2R). So the total area eclipsed is (1/2)(r^2(w - sin w) + R^2(W - sin W)) and as a percentage of the area of the sun this is (50/pi)(w - sin w + (D/d)^2(W - sin W)). Here W = 2arcsin((d/D)(l/d)).
All text and images © Duncan Hale-Sutton 2022
