An annulus is the flat ring between two circles that share a center: the face of a flat washer, the wall of a pipe seen end-on, a rubber gasket, or a path laid around a round pond. This annulus calculator finds the ring’s area, width, both diameters and both circumferences from whichever pair of measurements you have, and it can also work backwards from a required area to the size of the hole.
How to use the annulus calculator
- In I know the, pick the pair you have measured: Outer radius R and inner radius r, Outer and inner diameters, Outer radius R and ring width w, Inner radius r and ring width w, or Area and outer radius R.
- Type the values into the fields that appear. The ring width is R − r, the thickness of the ring measured straight across.
- Choose the Units. Lengths come back in that unit and areas in the matching square unit.
- Read the area of the ring at the top of the results. In area mode the headline is the inner radius instead. Below it you get the radii, width, diameters, outer and inner circumference, the mean radius and the area of the hole.
Annulus formulas
| Quantity | Formula |
|---|---|
| Ring width | w = R − r |
| Area from diameters | A = π(D² − d²) ÷ 4 |
| Area from mean radius and width | A = 2π × (R + r) ÷ 2 × w |
| Inner radius from area | r = √(R² − A ÷ π) |
| Edge lengths | 2πR outside, 2πr inside |
The third line has a neat explanation. Cut the ring along one radius and straighten it: you get a trapezoid whose parallel sides are the two circumferences and whose height is w. Its area is the average edge length times w, which is the circumference of the mean circle times the width.
Worked example: the wall of a steel pipe
A 2-inch Schedule 40 steel pipe has an outside diameter of 2.375 in and an inside diameter of 2.067 in. How much metal is in its cross-section?
- Choose Outer and inner diameters, enter 2.375 and 2.067, and set the units to inches.
- The calculator halves them: R = 1.1875 in and r = 1.0335 in, so the wall is 0.154 in thick.
- Area of the ring: π × (1.1875² − 1.0335²) = π × 0.342034 ≈ 1.075 in² of steel.
- The area of the hole, about 3.356 in², is the open flow area inside the pipe.
Multiply the ring area by the pipe's length for the volume of steel, or the hole area by the length for the volume of water it holds.
“2-inch” is a nominal size: neither diameter is actually 2 in, so always work from the real outside and inside diameters.
A path around a round pond
A round pond 16 ft across is getting a gravel path 3 ft wide. Choose Inner radius r and ring width w, enter r = 8 and w = 3 in feet, and the calculator returns R = 11 ft and an area of π × 57 ≈ 179.07 ft². Spread 3 in (0.25 ft) deep, that is about 44.8 ft³, or roughly 1.66 yd³ of gravel. The edging runs about 50.27 ft along the inner edge and 69.12 ft along the outer edge.
Why ring width matters more than size
For a fixed width, the area of a ring grows in step with its mean radius, not with its square. Here is a path 1 m wide around circles of different sizes:
| Inner radius r | Outer radius R | Ring area |
|---|---|---|
| 1 m | 2 m | 9.42 m² |
| 2 m | 3 m | 15.71 m² |
| 4 m | 5 m | 28.27 m² |
| 8 m | 9 m | 53.41 m² |
| 16 m | 17 m | 103.67 m² |
Doubling the pond roughly doubles the path, while the pond’s own area quadruples.
The edges behave even more simply: the outer circumference is always 2π × w longer than the inner one, whatever the radius. In the pond example, 69.12 − 50.27 ≈ 18.85 ft, exactly 6π. That is the old rope-around-the-Earth puzzle: add 1 m of rope and it rises only 1 ÷ 2π ≈ 0.16 m, just as it would around a basketball. It is also why races on a bend use staggered starts.
Common mistakes
- Subtracting before squaring. π(R − r)² is not the ring area. Square each radius, then subtract.
- Mixing radii and diameters. Halve any diameter before using the radius formula.
- Converting area like length. To turn in² into ft², divide by 144, not 12.
For a solid disc use the circle calculator; for the volume of a hollow cylinder, try the tube calculator.
Frequently asked questions
What is the formula for the area of an annulus?
Subtract the area of the hole from the area of the outer circle: A = π(R² − r²), where R is the outer radius and r the inner radius. A ring with R = 10 cm and r = 6 cm has an area of π × 64 ≈ 201.06 cm².
Can I use diameters instead of radii?
Yes. Choose 'Outer and inner diameters' and the calculator halves them for you. If you work by hand with diameters, use A = π(D² − d²) ÷ 4; leaving out the division by 4 makes the answer four times too large.
Why can't I just square the ring width?
π(R − r)² is the area of a small circle whose radius equals the ring width, not the area of the ring. For R = 10 and r = 6 it gives about 50.27 instead of the correct 201.06. Square each radius first, then subtract.
How do I find the size of the hole from the area?
Choose 'Area and outer radius R'. The calculator rearranges the formula to r = √(R² − A ÷ π). The area you enter must be smaller than πR², the area of the solid disc, or there is no room left for a hole.
Does the hole have to be centered?
For the area, no. A disc of radius R with a hole of radius r anywhere fully inside it still has an area of πR² − πr². Only a centered hole gives a true annulus with an even ring width, though, so the width and mean radius in the results assume the circles share a center.