geometry
I have a $56.1$" diameter cheese vat. I need to build two supports to lay across the top. Each support is $8$" from center on either side. I found the chord lengths by $$2\sqrt{r^2-d^2}$$
where
So chord lengths on both sides of center are $53.77$".
How can I find the arc length between these two points with this information?
Using the notation of the figure you have linked to, we have
\begin{equation} R \sin \frac{\theta}{2} = \frac{a}{2} \end{equation}
we can also write
\begin{equation} \theta = \frac{s}{R} = \frac{2 s \sin \theta/2}{a} \end{equation}
From this equation, you can solve for $\theta$.
Once you have solved for $\theta$, you have
\begin{equation} h = R - R \cos(\theta/2) \end{equation}
Since $R = a/(2 \sin \theta/2)$, we have
\begin{equation} h = \frac{a}{2 \sin \theta/2} \left( 1 - \cos\left(\frac{s \sin\theta/2}{a}\right)\right) \end{equation}
If I understand correctly, you are given $s$ and $h$ below, and you are looking for $r$, $\theta$ and $c$.
This means solving the following equations
$$\begin{align} s &= r \theta \\ h &= r - r\cos\theta \\ c &= 2 r \sin\left( \frac{\theta}{2} \right) \end{align}$$
If you knew the radius of curvature $r$, then $$c = \frac{h}{\sin\left( \frac{s}{2 r}\right)}$$
Unfortunately, there isn't an analytical solution to the equation $$h = r -r \cos\left( \frac{s}{r} \right)$$ for $r$ in terms of $h$ and $s$.
A second order estimate, for small $\frac{h}{s}$ is $$r \approx \frac{s}{4 \sin\left( \frac{1}{3} \sin^{-1} \left( \frac{3 h}{2 s} \right) \right)}$$
This was found using a Taylor series expansion on $ \frac{h}{s} = \frac{1}{\theta} (1-\cos\theta)$
A higher order estimate is $$\frac{r}{s} \approx \frac{ \sqrt{\frac{5}{18}-\frac{\sqrt{13}}{18}}}{ \sin^{-1} \left(2 (\sqrt{13}-3) \sin \left( \frac{1}{3} \sin^{-1}\left( \frac{h}{s} \sqrt{ \frac{\sqrt{13}}{4}+1} \right) \right) \right)}$$
This last one with $\theta < \frac{\pi}{2}$ the error is less than $0.007\%$.
Best Answer
If you assume the center of the circle to be the origin of a coordinate system, then the coordinates of one of your points (point $B$ in the illustration I added to your question) has coordinates $\left(d, \sqrt{r^2-d^2}\right)$. From that you can deduce the angle $\varphi$ between the horizontal $x$ axis and the line connecting that point to the origin. That angle will satisfy
$$\tan\varphi = \frac{\sqrt{r^2-d^2}}{d}$$
I've marked $\varphi$ in the above illustration. Now you can use that to compute the angle, and from the angle the arc length:
\begin{align*} \varphi &= \arctan\frac{\sqrt{r^2-d^2}}{d} \\ a_{AB} &= 2r\varphi = 2r\arctan\frac{\sqrt{r^2-d^2}}{d} \end{align*}
If you compute this using a pocket calculator, make sure you have its angle measurement mode set to radians, since a result in degrees won't work for the conversion from angle to arc length.