About the Annulus (Ring) Calculator
An annulus is the flat ring between two circles that share a centre — the shape of a washer, a gasket, a pipe cross-section, a donut seen from above, or a circular path around a pond. This annulus calculator finds the ring’s area by subtracting the inner circle from the outer one, and also reports the ring width, both circumferences, the mean radius and how much of the outer disc the ring covers.
You can enter the outer and inner radii, the outer and inner diameters (how washers and O-rings are usually specified), or the outer diameter plus wall thickness (how pipes and tubes are specified). The cross-sectional area of a pipe wall is exactly an annulus, so the result can be multiplied by length to get the volume of material, or by density to estimate weight.
All lengths must use the same unit; the area comes out in that unit squared. If you enter the inner size larger than the outer, the two are swapped.
With the default inputs, the ring area π(r² − r²) is 201.0619. Change any value above to recalculate instantly.
How to use the annulus (ring) calculator
- 1Choose whether you have radii, diameters, or an outer diameter with wall thickness.
- 2Enter the outer size.
- 3Enter the inner size or the wall thickness.
- 4Read the ring area, width and both circumferences.
Formula and method
The ring is what is left of the outer disc once the inner disc is removed, so its area is πR² − πr² = π(R² − r²). Factoring gives π(R + r)(R − r), which reads as the mean circumference 2π × (R + r)/2 multiplied by the ring width R − r — a handy mental check.
With diameters the same formula becomes (π/4)(D² − d²). For a pipe or tube given by outer diameter D and wall thickness t, the inner radius is D/2 − t, and the annulus area is the cross-sectional area of the material.
- R, r
- Outer and inner radius
- D, d
- Outer and inner diameter
- t
- Wall thickness (R − r)
Worked examples
Ring with radii 10 and 6
π(10² − 6²) = 64π ≈ 201.06 square units. The ring is 4 units wide and covers 64% of the outer disc; the hole takes the other 36%.
Washer: 20 mm outside, 10.5 mm hole
With diameters, A = (π/4)(20² − 10.5²) = (π/4)(400 − 110.25) ≈ 227.57 mm² of steel on each face of the washer.
4-inch Schedule 40 steel pipe (OD 114.3 mm, wall 6.02 mm)
The inner radius is 57.15 − 6.02 = 51.13 mm, so the steel cross-section is π(57.15² − 51.13²) ≈ 2,047.8 mm², while the bore offers about 8,213 mm² of flow area.
Frequently asked questions
What is the formula for the area of an annulus?+
A = π(R² − r²), where R is the outer radius and r is the inner radius. For R = 5 and r = 3, A = π(25 − 9) = 16π ≈ 50.27.
How do I find the area of a ring from diameters?+
Use A = (π/4)(D² − d²). A washer 30 mm across with a 12 mm hole has area (π/4)(900 − 144) ≈ 593.8 mm².
How do I find the cross-sectional area of a pipe?+
Treat the wall as an annulus: inner radius = outer diameter ÷ 2 − wall thickness, then A = π(R² − r²). Multiply by length for the volume of material.
Is the area of a ring the same as its width times its average circumference?+
Yes, exactly. π(R² − r²) = 2π × ((R + r)/2) × (R − r), so mean circumference × width gives the same answer, which is useful for estimating paths and borders.