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Circle Radius
10
same units as the inputs
Diameter 20

What this calculator does

This tool finds the radius of a circle when you only know the length of a chord and the sagitta — the perpendicular height of the arc measured from the midpoint of the chord to the arc itself. It is a universal geometry tool that works in any consistent unit (mm, cm, inches, feet), so no country or jurisdiction applies.

How to use it

Measure the straight-line distance across the arc (the chord, c) and the height of the arc above that line (the sagitta, h). Enter both values in the same unit and press calculate. The calculator returns the radius and the diameter of the full circle the arc belongs to.

The formula explained

The relationship comes from the right-triangle formed by the radius, half the chord, and the distance from the center to the chord:

$$r = \frac{\text{Chord }(c)^{2}}{8 \cdot \text{Height }(h)} + \frac{\text{Height }(h)}{2}$$

Here \(c^{2}/(8h)\) accounts for the spread of the chord while \(h/2\) corrects for the curvature of the arc. As long as \(h > 0\) the formula gives an exact radius. The diameter is then \(d = 2r\).

Circle showing chord c, sagitta height h, and radius r
The radius r relates to the chord length c and the sagitta height h of the circular segment.

Worked example

Suppose a curved doorway top has a chord of 12 inches and rises 2 inches at its peak. Then $$r = \frac{12^{2}}{8 \cdot 2} + \frac{2}{2} = \frac{144}{16} + 1 = 9 + 1 = 10 \text{ inches},$$ giving a diameter of 20 inches. So you would set a compass to a 10-inch radius to draw the matching arc.

FAQ

What is the sagitta? It is the height of the arc — the maximum vertical distance between the chord and the curve, measured at the chord's midpoint.

Why does the height have to be greater than zero? If \(h = 0\) the points are collinear (a straight line), so there is no finite circle and the formula would divide by zero.

Can I use any units? Yes — just keep the chord and height in the same unit, and the radius comes out in that same unit.

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