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I'm looking for a radius formula. Assume a disc with evenly spaced holes, say 10, just inside the disk perimeter. The distance between hole centers is known, say 0.500". No further data available.

I'm looking to calculate a radius for: the inner edge, the center point, and the outer edge of the holes.

Thanks,

Alex

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  • $\begingroup$ Are you certain that there's no other information given? Perhaps in the form of a picture? If nothing else is given, see my answer below. $\endgroup$ – Cameron Buie Aug 10 '13 at 0:24
  • $\begingroup$ Ok, this is a machining problem, where the # of holes, the distance between hole centers, and the degrees between hole centers (360 / # of holes) is known. The author gave this formula as the solution: D = C / sin(180 M / N) where C = chord length, M is the # of spaces between holes, and N = # of holes. He made 2 disks, one with 20 holes, one with 40. He said this formula would give the size disk needed. I pluged it into Excel but didn't get disk sizes anywhere near his. He did not say how large his were, but the largest was about 1 foot, the smaller a couple of inches. $\endgroup$ – Alex Aug 10 '13 at 14:41
  • $\begingroup$ Ops! Ran out of characters in my comment above. Maybe I made an Excel mistake? $\endgroup$ – Alex Aug 10 '13 at 14:42
  • $\begingroup$ Hmmm.... I'm not at all sure what "number of spaces between holes" means. I'll update my answer to clarify what I've got so far (similar to what the author has, but not the same). Can you explain what $M$ means in more detail? $\endgroup$ – Cameron Buie Aug 10 '13 at 18:42
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There doesn't seem to be enough information to answer all of that (unless I'm misunderstanding what you mean by "inner edge" and "outer edge"). However, if I understand correctly, the centers of the holes will (in this case) be the vertices of a regular decagon with side length $0.5$ inches. Some quick right-triangle trigonometry will tell you the distance of the hole centers from the disc's center.

In particular, if you consider the triangle formed with a vertex in the centers of two adjacent holes and another vertex in the center of the disc, you'll find that it is an isosceles triangle whose matching side lengths are the distance from the center of the disc to the center of the holes, which I will call $R$; whose third side length is the distance between the centers of adjacent holes, which I will call $r$ ($0.5$ inches in your example); and whose angle at the disc's center is $\dfrac{360^\circ}N$ ($N=10$ in your example).

Cutting the triangle in half from the center of the disk, we have a right triangle whose hypotenuse is $R$, whose angle at the center is $\dfrac{180^\circ}N,$ and whose side length opposite this angle is $\dfrac r2.$ By right triangle trigonometry, we have $$\sin\left(\frac{180^\circ}N\right)=\frac{r/2}{R},$$ so that $$R=\cfrac{r}{2\sin\left(\frac{180^\circ}N\right)}.$$

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  • $\begingroup$ Thanks Cameron Buie, I'll go try that. I'm not much of a math person which is what brought me to posting. Thanks again! $\endgroup$ – Alex Aug 10 '13 at 21:22
  • $\begingroup$ I just remembered one more thing. The author of the formula I posted said that any 2 holes could be used to calculate the radius. The # of spaces are used when using non-adjacent holes. $\endgroup$ – Alex Aug 10 '13 at 23:15
  • $\begingroup$ Ah! That makes sense, and it's easily adaptable. I'm still not sure why the author doesn't have the 2 in the denominator, though? $\endgroup$ – Cameron Buie Aug 11 '13 at 1:07

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