An annulus is the ring left when a smaller concentric circle is removed from a larger one, and its area is simply the difference of the two circle areas. Factorising gives pi times the sum of the radii times their difference, which is often quicker to evaluate by hand and makes the dependence on wall thickness obvious.
This is the cross-section of every pipe, washer, tube and bearing race, so the annulus area is what determines the material in a length of pipe while the inner circle alone determines its flow capacity. A thin-walled tube of large diameter can have far more metal in it than a thick-walled narrow one, which is why the ring area rather than the wall thickness is the number to compare.
The formula
R- Outer radius
r- Inner radius
R − r- Wall thickness
(R + r) ÷ 2- Mean radius, used for the mean circumference
How to use this calculator
- Enter the outer radius.
- Enter the inner radius — it must be smaller.
- The gauge shows the ring area.
- Wall thickness and the share of the disc removed appear below.
Worked examples
Radii of 10 and 7
The area is pi × (100 − 49) = 160.2212. Using the factorised form, pi × 17 × 3 gives the same 160.2212. The inner hole removes 49% of the outer disc.
A thin-walled tube
Radii of 30 and 29 give a ring area of 185.3540 with a wall only 1 unit thick. That is more material than the 10 and 7 example, despite the far thinner wall.
How to read your score
Frequently asked questions
Do I use radius or diameter?
Radius. If you have outer and inner diameters, halve both before entering them, or the area comes out four times too large.
How do I get the volume of a pipe wall?
Multiply the annulus area by the length of the pipe. That gives the material volume, which times density gives the mass.
Why does the factorised form help?
Because pi times the sum of the radii times their difference avoids squaring large numbers by hand, and it shows directly that the area is proportional to wall thickness for a given mean radius.
What happens if the inner radius is zero?
The annulus becomes a full disc and the formula reduces to pi R squared, which is the correct answer.
Ring areas for an outer radius of 10
| Inner radius | Wall thickness | Ring area | Share of disc removed | Mean circumference |
|---|---|---|---|---|
| 0 | 10 | 314.1593 | 0.00% | 31.4159 |
| 1 | 9 | 311.0177 | 1.00% | 34.5575 |
| 2 | 8 | 301.5929 | 4.00% | 37.6991 |
| 3 | 7 | 285.8849 | 9.00% | 40.8407 |
| 5 | 5 | 235.6194 | 25.00% | 47.1239 |
| 7 | 3 | 160.2212 | 49.00% | 53.4071 |
| 8 | 2 | 113.0973 | 64.00% | 56.5487 |
| 9 | 1 | 59.6903 | 81.00% | 59.6903 |
| 9.5 | 0.5 | 30.6305 | 90.25% | 61.2611 |
The removed share depends on the square of the inner radius, so small holes remove very little.