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--+------+-----+---*E--------oF * 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, http://mathforum.org/dr.math/faq/formulas/faq.circle.html#segment 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 http://mathforum.org/dr.math/
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