Annulus calculator
Area of the ring between two circles, its width and both circumferences.
Fill in the fields and the result will appear here automatically.
An annulus between concentric circles has area π(R² − r²). The expression π(R − r)² describes a different shape, a disc of radius R − r, and can look plausible for a narrow ring while giving the wrong answer. This tool handles positive width, R > r ≥ 0. Equal radii are a zero-area degenerate case excluded here; r = 0 is allowed and gives a solid disc.
How it works
Formula and logic
S = π(R²−r²) = π(R−r)(R+r). The factored form avoids subtracting nearly equal squares. Width is R−r, circumferences are 2πR and 2πr, and mean radius is (R+r)/2. At r = 0 the inner circumference is zero.
Example
A ring with radii of 10 and 6 cm has an area of 201.06 cm² and a width of 4 cm.
Fields and units
- Length unit — list option
- Outer radius — cm
- Inner radius — cm
How to use
- — Choose the length unit.
- — Enter the outer radius.
- — Enter the inner radius — it must be smaller than the outer one.
- — Leave the inner radius at zero for a solid disc.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax: circle area and circumference
- Limitation
- The circles share one centre and require R > r ≥ 0. Selecting a unit does not convert existing numbers. An off-centre hole or oval cross-section needs a different geometric model.
FAQ
Why can't the area be π(R − r)²?
Because that is the area of a disc of radius R − r, not of the ring. For radii 10 and 6 the correct answer is 201.06 cm² while the mistaken one is 50.27 cm² — a factor of four apart, though both look plausible.
What if the inner radius is zero?
You get a solid disc, and the calculation allows it: the area becomes πR² and the inner circumference is zero.
Why can't the inner radius equal the outer one?
This page requires positive width R − r. At R = r the mathematical area is zero, but this degenerate case is rejected here. Enter r = 0 for a solid disc.
What is the mean radius for?
The mean radius (R+r)/2 gives the midline circumference. Circumference times width equals the annulus area exactly: 2πR_mean(R−r) = π(R²−r²). The identity does not require a narrow ring; physically flattening material into a straight strip is a separate question.
How do I find the cross-section of a pipe?
It is exactly this problem: the outer radius of the pipe and the inner radius of the bore. The difference gives the area of metal in cross-section.