# Get radius of circle given arc length and chord length

I have to program a software to get radius of a circle with arc and chord

Let's say arc = $$9.27$$ and chord = $$8$$, here is what I tried so far:

from the arc formula, I know that:

$$\pi r * \frac{\theta}{180} = 9.27$$

And base on the cosine law:

$$2r^2 - 2r^2\cos\theta = 8^2$$

Currently I'm stuck at how to get the value of r from this two equation, I'm not really sure how to deal with the $$\cos()$$ here

Or there's easier way to calculate the radius from arc length and chord length?

• Try rearranging the second equation so that $\theta$ is the subject. Then you can plug it into the first equation. Sep 3, 2021 at 14:17

If you have radius $$r$$ and angle $$\theta$$ in rad, then $$A = r\theta$$ and $$C = 2r\sin(\theta/2)$$. To get this independent of $$r$$ lets consider $$C/A = \frac{2\sin(\theta/2)} \theta$$ This is a function that is falling montonously for $$0\leq \theta \leq 2\pi$$, so while we cannot invert this exactly we can easily invert this numerically, for example by simple bisection:

1. Set $$l = 0,u = 2\pi$$
2. Set $$m = (l+u)/2$$
3. If $$(2*sin(m/2))/m = C/A$$ we’re done
4. If $$\ldots < C/A$$ set $$u=m$$
5. Else set $$l=m$$
6. If $$u-l<\epsilon$$ for some $$\epsilon$$ return $$m$$, else jump to 2.

So now you know some (numerical) value for $$\theta$$. Use $$A=r\theta$$ to get $$r$$.

EXAMPLE: A=9.27, C=8. Then $$C/A=0.86...$$.

Set $$l=0, u=2\pi,m=\pi$$. Then $$f(m)$$ for $$f(\theta)=2\sin(\theta/2)/\theta$$ is about $$0.64. So we set $$u=\pi$$. So $$m=\pi/2$$, then $$f(m)\approx 0.9>C/A$$. So we set $$l=\pi/2,m=3/4\pi$$. Then $$f(m)\approx0.74, so we set $$u=3/4\pi,m=5/8\pi$$, $$f(m)\approx0.85$$.

We could continue this way to improve this value, but let’s go with $$\theta = m = 5/8\pi$$. Then $$9.27\approx r\theta$$, so $$r\approx 9.27/(5/8\pi)\approx 4.72$$.

So for checking the error we check: $$2r\sin(\theta/2)\approx 7.85 \approx C=8$$

So we are somewhat off, mainly because we calculated $$\theta$$ only with a precision of $$3/4\pi-1/2\pi = \pi/4$$, which is kind of large. But you get the point.

EDIT: Of course, instead of bisection we might also use Newton: The derivative of $$2\sin(\theta/2)/\theta$$ is $$\cos(\theta/2)/\theta - 2\sin(\theta/2)/\theta^2$$.

So for $$Y=C/A$$ set $$x_0 =\pi$$, and set

$$x_{n+1} = x_n - \frac{\frac{2\sin(x_n/2)}{x_n} - Y}{\frac{\cos(x_n/2)}{x_n} - \frac{2\sin(x_n/2)}{x_n^2}}$$ $$= x_n - \frac{2x_n\sin(x_n/2) - Yx_n^2}{x_n\cos(x_n/2) - 2\sin(x_n/2)}$$

This should converge faster. Example with $$Y=C/A\approx0.86$$:

We have $$x_0=\pi$$. Then $$x_1\approx 2.02$$ according to the formula above. Then $$x_2\approx 1.859$$. Then $$x_3\approx 1.85266$$. With using this for $$\theta$$ we get $$r\approx 5.004$$ in only three iterations. With two additional iterations the error is smaller than the precision of a double float.

• If we use $\epsilon=0.001$ we get a precision of $\pi/4096$ with $\theta\approx 2415/4096\pi\approx 1.85$ and thus $r\approx 5$. Then we get $C\approx 2\cdot 5\sin(1.85/2) \approx 7.986$ or with the unrounded values we get $C\approx 8.0005$, which is not that much off.
– Lazy
Sep 3, 2021 at 14:35
• Where does B come from in step 3? Sep 3, 2021 at 19:30
• @jialeko Sorry, that should have been a $C$. That comes from writing $A$ for arc and $C$ for chord, and then just continuing using $A$ and $B$ instead.
– Lazy
Sep 3, 2021 at 21:07
• Tested in program, looks like if $m=l+(l+u)/2$ where $\epsilon=0.001$ will have infinite loop, but $m=(l+u)/2$ work fine, may I know why $+l$ is used here? Sep 3, 2021 at 21:29
• I’ve just added a note about using newton method to approximate $\theta$, which converges much faster. Also I’m sorry, that was again a mistake. It should of course be $l+(u-l)/2$ or $(u+l)/2$ (which is the same thing).
– Lazy
Sep 3, 2021 at 21:32

I think there's no elementary formula for this.

Denote arc and chord as $$a$$, $$b$$, then $$r\theta=a$$, $$2r\sin{(\theta/2)}=b$$. Divide the first equation by the second, we get $$\frac{\theta}{2\sin{(\theta/2)}}=\frac{a}{b} \\ \theta=\frac{2a}{b}\sin{(\theta/2)}.$$ Subsitute $$\theta/2$$ with $$u$$, we get $$u=\frac{a}{b}\sin{u}$$ which is an equation you can't give simple solution to.

I was attempting the same thing today as part of developing a calculator program. I posted my solution under a different thread here. I think the mods may frown on me reposting the same answer, but I'll summarize it here:

Working in degrees (not radians), and given the chord length ($$C$$) and the arc length ($$A$$), I reduced the formula for the radius ($$R$$) to:

$$R=\frac{C}{2\sin\left(\frac{360A}{4\pi R}\right)}$$

But there's the problem of having $$R$$ on both sides, with no straightforward way to simplify. 🤔

I discovered that my TI-84 Plus calculator has an iterative solver built-in, so I can modify the formula slightly so that it equals zero, and plug it in to the solver.

I believe you can do similar iterative solving in Excel or in other programming languages.