How do I calculate the diameter of a circle that has n evenly-spaced points on its circumference where adjacent points are m distance apart?
Diameter of circle with n points where adjacent points are m distance apart
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0It seems like this should be trivial, but I must not be using the right terms when I google $f$or it. – 2012-11-01
2 Answers
If you have $n$ points uniformly distributed on a circle then the angle between the points is $\frac{2\pi}{n}$. If the distance between adjacent points is $m$ then the chord of the sector with angle $\frac{2\pi}{n}$ has length $m$.
Trigonometry then yields $r = \frac{m\sin\left(\frac{(n-2)\pi}{2n}\right)}{\sin\left(\frac{2\pi}{n}\right)}$
Alternatively, this formula will probably be simpler $r = \frac{m}{2\sin\left(\frac{\pi}{n}\right)}$
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0beautiful. I probably could have gotten there in a few days, and maybe gotten a little smarter in the process, but I would much rather be writing logic, so thanks! – 2012-11-01
The circle is divided into $n$ wedges with angle $ \theta = \frac{2 \pi}{n}. $ Consider the triangle formed by the center of the circle (point $A$), one of the two adjacent points (point $B$), and the midpoint of the line joining the two adjacent points (point $C$). Now, $\angle BAC = \theta / 2 = \pi / n$, $AB = d/2$, and $BC = m / 2$, where $d$ is the diameter of the circle, so $ AB \sin \left(\angle BAC\right) = BC \Rightarrow \frac{d}{2} \sin\left(\frac{\pi}{n}\right) = \frac{m}{2} \Rightarrow d = \frac{m}{\sin \left(\pi / n\right)}. $
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0ooohhh, pretty. – 2012-11-01