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For a simulation, I need to be able to calculate the radius $r$ of a circle $C$, knowing only two points on its circumference, $P_1$ and $P_2$, as well as the distance between them ($a$) and how much of the whole circumference $c$ is in the arc between those two points ($\frac{c}{x}$, where $x$ is known and $\geq 1$). Neither the arc itself nor its angle is known, but the arc should be equal to $\frac{2\pi r}{x}$.

The following image should illustrate this:

example of a circle, $P_1$, $P_2$, $a$ and $x$ are known

While being closely related to questions just as this one, it's not quite the same, as I don't know the angles.

Should this not be possible, what else would I need? Would a third point suffice?

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Big-Blue
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  • You're missing one thing: which way is the arc curving? In your diagram you could have the center directly below $p_1$ (as shown), or directly above $p_2$. – Martin Kealey Dec 10 '23 at 20:55

2 Answers2

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We know that the arclength $s$ between the two points is given by $s = 2\pi r/x$, where $x$ is known. So, we know the angle $\alpha$ of the arc between the two points -- it's just $\alpha = s/r = 2\pi/x$. Then, using the formula from the first answer, we have:

$$r \sin\left(\frac{\alpha}{2}\right) = \frac{a}{2} $$

and so

$$r = \frac{\tfrac{1}{2}a} {\sin\tfrac{1}{2}\alpha } = \tfrac{1}{2}a\,\mathrm{cosec}\tfrac{1}{2}\alpha $$

Or, in summary

$$r = \frac{1}{2}a\,\mathrm{cosec}\left(\frac{\pi}{x}\right)$$

bubba
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  • I added an additional sentence about the arc in the question. The arc itself is not known, only the distance between the two points, but it is known that the arc equals $\frac{2\pi r}{x}$ with $x$ being known. – Big-Blue Mar 09 '13 at 13:32
  • @Big-Blue, then you know $arc \over circumference$. The needed formula is in my answer. – gukoff Mar 09 '13 at 13:50
  • Thanks for providing a formula that is usable on-the-fly! But somehow, the results I get with this are far off. This circle has a radius of $5$, but letting WolframAlpha calculate the radius using the formula you provided, I get a result around 28, which resembles the circumference. – Big-Blue Mar 10 '13 at 08:59
  • Tell us the $P_1$, $P_2$, and $x$ that you used in your example test. – bubba Mar 10 '13 at 09:05
  • $P_1 = [0,5]$, $P_2 = [5,0]$, $x = 4$ – Big-Blue Mar 10 '13 at 09:17
  • My formula for \alpha was wrong. My apologies. So, in your example, $\alpha = 2\pi/4 = \pi/2$, and everything else should work. – bubba Mar 10 '13 at 09:28
  • Your picture shows $\alpha = 1/4$, which is not correct. – bubba Mar 10 '13 at 09:30
  • Thanks, with the correct $\alpha$, everything works as intended. – Big-Blue Mar 10 '13 at 12:39
  • I'm sorry to dig this up again, but I tried using this formula again on a circle with a higher radius and it was way off. This circle has $r = 220$, $B = [220,0]$, $B' = [190.53,110]$, $x = 12$ and $a = 113.87\ldots$. But while giving me the correct radius for the circle with the radius $5$, using it on this circle gets me $58.94\ldots$, which is way different from the desired $220$. Any ideas on how to fix this? – Big-Blue Mar 13 '13 at 19:38
  • $\sin(\pi/12) = 0.259$, so $\cosec(\pi/12) = 3.864$. Multiply this by $a/2$, which is 56.9, and you get 220. – bubba Mar 14 '13 at 04:24
  • I'm very sorry, this was indeed an error in my implementation, as I implemented $\csc(x)$ as $\frac{1}{\cos(x)}$ instead of $\frac{1}{\sin(x)}$, which works as intended for $r=5$. – Big-Blue Mar 14 '13 at 14:05
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Law of cosines: $a^2 = 2R^{2}(1-2cos(\alpha))$, where $\alpha$ is the angle measure of an arc, and $a$ is the distance between points. $\alpha = 2\pi ({arc \over circumference})$.

Also $R \cdot sin({\alpha \over 2}) = {a \over 2}$, it is also pretty obviously.

If you only know $arc$ and $distance$, then $distance = (2R)\cdot sin({arc \over (2R)})$. The inverse function of $sin(x)/x$ you need here can be sure approximated.

gukoff
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