Layout and shop math
Circle segment calculator
Chord 4.000 in and height 0.500 in give a 4.2500 in radius; enter any two values to get the rest.
Technical review pending. Formulas and values are cited. This notice comes down after a machinist review.
Radius
in
R = (c^2 + 4 x h^2) / (8 x h) = (4.0000^2 + 4 x 0.5000^2) / (8 x 0.5000) = 4.2500 in
4.2500 in (107.95 mm). Included angle 56.1450°.
| Value | Result | Formula with your numbers |
|---|---|---|
| Radius R | 4.2500 in | R = (c^2 + 4 x h^2) / (8 x h) = (4.0000^2 + 4 x 0.5000^2) / (8 x 0.5000) = 4.2500 in |
| Chord c | 4.0000 in | Given |
| Height h | 0.5000 in | Given |
| Included angle theta | 56.1450° (0.97991 rad) | theta = 2 x asin(c / (2 x R)) = 2 x asin(4.0000 / (2 x 4.2500)) = 56.1450° |
| Arc length s | 4.1646 in | s = R x theta = 4.2500 x 0.97991 = 4.1646 in |
| Segment area A | 1.3499 sq in | A = (R^2 / 2) x (theta - sin(theta)) = (4.2500^2 / 2) x (0.97991 - sin(0.97991)) = 1.3499 sq in |
Source: Machinery's Handbook (segments of circles); standard plane geometry
How it works
R = (c^2 + 4 x h^2) / (8 x h); theta = 2 x asin(c / (2 x R)); s = R x theta; A = (R^2 / 2) x (theta - sin(theta))
| Symbol | Meaning | Unit |
|---|---|---|
| R | Radius of the circle | in or mm |
| c | Chord: straight-line width of the segment | in or mm |
| h | Height (sagitta): chord to the top of the arc | in or mm |
| theta | Included angle of the segment; in radians in s and A | ° or rad |
| s | Arc length | in or mm |
| A | Segment area, between the chord and the arc | sq in or sq mm |
From radius and angle: c = 2 x R x sin(theta / 2) and h = R x (1 - cos(theta / 2)). From radius and chord: h = R - sqrt(R^2 - c^2 / 4). From radius and height: c = 2 x sqrt(h x (2 x R - h)) and theta = 2 x acos((R - h) / R). From radius and arc length: theta = s / R in radians. Chord with arc length, and height with arc length, need an iterative solve and are not offered here.
Worked example: radius from chord and height
A curved part measures 4.000 in across the chord and 0.500 in high.
- R = (4.000^2 + 4 x 0.500^2) / (8 x 0.500) = (16.000 + 1.000) / 4.000 = 4.2500 in.
- theta = 2 x asin(4.000 / (2 x 4.2500)) = 2 x asin(0.47059) = 2 x 28.0725° = 56.1450° (0.97991 rad).
- s = 4.2500 x 0.97991 = 4.1646 in.
- A = (4.2500^2 / 2) x (0.97991 - sin(0.97991)) = 9.03125 x (0.97991 - 0.83045) = 9.03125 x 0.14947 = 1.3499 sq in.
Result: radius 4.2500 in (107.95 mm), angle 56.1450°, arc 4.1646 in, area 1.3499 sq in.
Worked example: radius and angle
A 60° segment on a 3.000 in radius.
- c = 2 x 3.000 x sin(30°) = 6.000 x 0.50000 = 3.0000 in.
- h = 3.000 x (1 - cos(30°)) = 3.000 x 0.13397 = 0.4019 in.
- s = 3.000 x 1.04720 = 3.1416 in.
- A = (3.000^2 / 2) x (1.04720 - 0.86603) = 4.5 x 0.18117 = 0.8153 sq in.
Result: chord 3.0000 in, height 0.4019 in, arc 3.1416 in, area 0.8153 sq in.
Shop notes
- Radius of an arc from a part you can measure: take the chord with calipers and the height with a caliper depth rod or over a pin, then use chord and height.
- Flats on round bar: the flat width is the chord and the depth of cut is h. Use radius and chord.
- Metric check: R = 50 mm and c = 60 mm give h = 50 - sqrt(2,500 - 900) = 50 - 40 = 10.00 mm and theta = 2 x asin(0.6) = 73.7398°.
- A height larger than the radius means the segment is more than half the circle. The calculator flags it and uses 360° minus the small angle.
- Radius and chord gives the smaller of the two segments on that chord.
FAQ
How do you find a radius from a chord and height?
R = (c^2 + 4 x h^2) / (8 x h). A chord of 4.000 in and a height of 0.500 in give (16.000 + 1.000) / 4.000 = 4.2500 in.
How deep is a flat 0.500 in wide on 1 in round bar?
Use radius and chord: R = 0.500 in, c = 0.500 in. h = 0.500 - sqrt(0.2500 - 0.0625) = 0.500 - 0.4330 = 0.0670 in.
What is the arc length for a 60° segment of 3 in radius?
3.1416 in: s = R x theta = 3 x 1.0472 rad.
Why can one height give two answers?
A height above the radius means the segment is larger than half the circle, and the included angle is 360° minus 2 x asin(c / (2 x R)). The calculator flags it and uses the larger angle.
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