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Area of a Crescent

Date: 06/18/2001 at 10:34:37
From: John Dixon
Subject: Area of a crescent - as in an eclipse

When observing a total eclipse of the sun we need to determine the 
area of the sun that has not been covered by the moon. The important 
point is that the eclipse magnitude (the ratio of the apparent size of 
the moon to that of the sun) can change from 1 to 1.18. The most 
critical time is when the eclipse is 90% complete and the change in 
area is no longer linear. Is there a formula for the relative area 
of a circle that is not covered by a larger circle as a function of 
the distance between the centers?

Date: 06/18/2001 at 12:47:25
From: Doctor Peterson
Subject: Re: Area of a crescent - as in an eclipse

Hi, John.

Here's your eclipsed sun:

                 *********** C
             ****           **oooooooooo
          ***            ooo/|\ ***.....oooo
        **             oo  / | \   **.......oo
       *             oo   /  |  \    *........oo
      *             o    /r1 |  \r2   *.........o
      *             o   /    |   \    *.........o
     *             o   /     |    \    *.........o
     *             o  A\    d|    /B   *.........o
      *             o   \    |   /    *.........o
      *             o    \   |  /     *.........o
       *             oo   \  |  /    *........oo
        **             oo  \ | /   **.......oo
          ***            ooo\|/ ***.....oooo
             ****           **oooooooooo
                 *********** D

The shaded crescent CEDF is bounded by arcs of a circle with center A 
and radius r1 (the moon), and another circle with center B and radius 
r2 (the sun), where the distance AB is d.

You can see this area as the difference between the areas of two 
segments of circles, CFD - CED. According to the Dr. Math FAQ on 
Circle Formulas,   

this area is

    K_segment = r^2[theta-sin(theta)]/2

where theta is the central angle subtended by the chord. Applying this 
to each of our segments,

  K_crescent = K_CFD - K_CED
             = r2^2[2<CBE - sin(2<CBE)]/2 - r1^2[2<CAE - sin(2<CAE)]/2
             = r2^2[2pi - 2<CBA - sin(2pi - 2<CBA)]/2 -
               r1^2[2<CAB - sin(2<CAB)]/2
             = pi r2^2 - (r1^2<CAB + r2^2<CBA ) +
               [r1^2 sin(2<CAB) + r2^2 sin(2<CAB)]/2

You can find angles <CAB and <CBA by solving triangle ABC, whose sides 
you know. I don't think this will simplify into a nice formula, but 
you can calculate it.

Let me know if you need more help, or find an interesting formula.

- Doctor Peterson, The Math Forum   
Associated Topics:
College Euclidean Geometry

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