Simplifying Circle Formulas from the Dr. Math FAQ
Date: 04/30/2004 at 14:47:14 From: Jim Subject: faq.circle.segment.html In your FAQ on circle formulas, in the sections where the other five values are derived from any two known values, could you write each formula in terms of only the two known values, instead of using the intermediate steps? For example, in Case 10 I know c and d, so can all the formulas be expressed simply in terms of c and d? Can the trig function be represented more concisely? It's even more of an issue for K, where I would need to take the sin of an arcsin to get to the answer.
Date: 04/30/2004 at 15:54:41 From: Doctor Peterson Subject: Re: faq.circle.segment.html Hi, Jim. Certainly you can combine formulas to get an individual result directly from the givens, if you don't need the other values, and sometimes that will give neat results. In Case 10, you are told r = sqrt(c^2 + 4d^2)/2 h = r - d theta = 2 arcsin(c/[2r]) s = r theta K = r^2[theta - sin(theta)]/2 As you suggest, combining to get s directly gives the unwieldy s = sqrt(c^2 + 4d^2) * arcsin(c/sqrt(c^2 + 4d^2)) To simplify the arcsin, look at this triangle: + sqrt(c^2 + 4d^2) / | / |c / | +-----------+ What is the other side? sqrt[c^2 + 4d^2 - c^2] = sqrt(4d^2) = 2d So the tangent of the angle is c/(2d), and our formula simplifies to s = sqrt(c^2 + 4d^2) * arctan(c/(2d)) I think that helps a bit. Now, for K, we are taking the sine of twice the angle whose sine is c/sqrt(c^2 + 4d^2). We can use the double-angle identity sin(2x) = 2 sin(x) cos(x) = 2 sin(x) sqrt(1 - sin^2(x)) to get, for the sine, 2 c/sqrt(c^2 + 4d^2) sqrt[1 - c^2/(c^2 + 4d^2)] which simplifies nicely to 4cd ---------- c^2 + 4d^2 (check that and make sure I'm right!), so the formula becomes K = (c^2 + 4d^2)/4 [2 arctan(c/(2d)) - 4cd/(c^2 + 4d^2)]/2 = (c^2 + 4d^2)/2 arctan(c/(2d)) - cd/2 Still not simple, but a lot nicer that it would have been without the simplifications. Now, you can also get these formulas directly by going back to the geometry. You have this figure: ooooooooo oooooo....|....oooooo ooo..........|..........ooo +--------------+--------------+ oo \ | c/2 / oo o \ | / o o \ d| / o o \ | / o o \ | / o o \ |t/2/ o o \ | / o o + o o | o o | o o | o o | o o | o o | o oo | oo oo | oo ooo | ooo oooooo | oooooo ooooooooo The area will be the area of the sector minus the area of the isosceles triangle. The latter is just cd/2 (sounds familiar!), and the former is r^2 theta/2, where theta/2 (shown as t/2 in the figure) has tangent (c/2)/d. So K = r^2/2 arctan(c/(2d)) - cd/2 which is just what we had above, once we replace r with its value. I wouldn't mind seeing more of these formulas in the FAQ, since it takes a bit of work to get them one way or the other. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 04/30/2004 at 16:05:41 From: Jim Subject: Thank you (faq.circle.segment.html) Thanks! That was fast, thorough, and it even included an ASCII graphic! - Jim
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum