Please imagine that we have a circular segment with some arc length 's' and chord length 'a' (using notation from http://mathworld.wolfram.com/CircularSegment.html):

circular segment

Provided only 'a' and 's', and placing the left-hand-side point of the chord at the origin of the Euclidean plane (or a more convenient point), is there sufficient information to write an expression for the height of the circular segment (i.e. the y-axis/"vertical" distance between the chord on the x-axis and the circular arc) as a function of a position on the chord?

It's a simple matter to express the chord length in terms of the arc length and theta: $a = (s) * 2\frac{\sin(\frac{\theta}{2})}{\theta}$, or an expression for the arc length in terms of the chord length and theta: $s = \frac{a\theta}{2\sin\frac{\theta}{2}}$. And one can write an express for the maximum height as: $h = R - \frac{1}{2}\sqrt{(-a)^2+4R^2}$, where the radius of the circle, 'R' is related to theta as: $R = \frac{1}{2} \sqrt{\frac{a^2}{\cos^2\frac{\theta}{2}-1}}$.

If there is insufficient information to accomplish the above, I would love to have an intuitive explanation for why this is so.

  • $\begingroup$ With just $s$ and $a$, the problem is ill-posed. There's no way to determine the radius or other such useful info from just those two. $\endgroup$ Nov 1, 2010 at 2:35
  • $\begingroup$ I feel that we should be able to derive the curvature of the line segment from 's' and 'a'? And with that, reconstruct the circle? $\endgroup$
    – David
    Nov 1, 2010 at 2:39
  • $\begingroup$ Consider the further constraint that the arc segment must have constant curvature, and that we have two points we know the segment of the circular curve passes through. I feel intuitively the information is there for this calculation. $\endgroup$
    – David
    Nov 1, 2010 at 2:44
  • 1
    $\begingroup$ Remember that a chord divides a circle into a "small arc" and a "big arc" (tacitly excluding the case where the chord is a diameter). It should not be hard to find small arc-big arc pairs (the two implied circles are of different radii, of course) that have the same length. $\endgroup$ Nov 1, 2010 at 2:50
  • $\begingroup$ Circles have constant curvature, so that restriction is moot. $\endgroup$ Nov 1, 2010 at 2:51

1 Answer 1


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}

  • $\begingroup$ ok. I just saw J.M's comment about small and big arcs. My solution implicitly assumes that the arc length $s$ is the length of smaller segment of the circle. If you consider the other possibility (that the length s is the length of the larger segment), the solution you will get is $ h = \frac{a}{2 \sin \theta/2} \left( 1 + \cos\left(\frac{s \sin\theta/2}{a}\right)\right)$ $\endgroup$
    – svenkatr
    Nov 1, 2010 at 2:58
  • $\begingroup$ Yes, apologies, we should assume that it is the smaller length of the circle! $\endgroup$
    – David
    Nov 1, 2010 at 2:59
  • $\begingroup$ I need this too. But in this solution, you give a height calculation including theta. Given only an arch length and a chord length (no radius) I don't know how to calculate theta. I tried deriving it myself, but arrived at an equality containing both cosine theta and theta squared, and I don't know how to solve it. $\endgroup$
    – rumtscho
    Feb 8, 2011 at 21:39
  • $\begingroup$ @rumtscho - You would have to solve for $\theta$ using the equation $\theta = \frac{2s \sin \theta/2}{a}$. $s$ and $a$ are known. You would solve for $\theta$ numerically in general. $\endgroup$
    – svenkatr
    Feb 9, 2011 at 4:57
  • $\begingroup$ With $\phi = \theta/2$ we get $h = \dfrac{a(1 - \cos \phi)}{2 \sin \phi} = \dfrac{a \sin \phi}{2 (1+\cos \phi)}$. $\endgroup$
    – Paul Frost
    Jan 3, 2020 at 13:03

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