The Annulus Calculator computes the area, perimeter, and width of an annulus — the ring-shaped region between two concentric circles — from the outer and inner radii. Used in engineering, architecture, and mathematics for pipe cross-sections, washers, ring foundations, and hollow cylinder analysis.
235.6194
sq units
62.8319
units
31.4159
units
5
units
235.6194
sq units
62.8319
units
31.4159
units
5
units
A washer, a pipe cross-section, a circular fountain basin, a ring beam foundation — all of these are annuli: the flat region trapped between two concentric circles. The calculator for annulus geometry finds area, perimeter, and width from any two of the three defining measurements (outer radius, inner radius, or width), handling the geometry that the basic circle calculator cannot.
For an annulus with outer radius R and inner radius r (where R > r ≥ 0):
Area = π(R² − r²) = π(R + r)(R − r)
Perimeter = 2π(R + r) — the sum of the outer and inner circumferences
Width (ring thickness) = R − r
The area formula is simply the outer circle area minus the inner circle area. Note that the annulus perimeter includes both the outer circumference (2πR) and the inner circumference (2πr) — the full boundary of the shape consists of two separate circles. For a washer with outer diameter 30 mm and inner diameter 20 mm: R = 15 mm, r = 10 mm; Area = π(225 − 100) = 392.7 mm²; Perimeter = 2π(15 + 10) = 157.1 mm. Use this online calculator for any annulus dimensions. The ellipse calculator covers the analogous elliptical ring geometry.
The annulus is the fundamental shape in piping and hollow structural analysis:
The regular polygon calculator and plane geometry calculators cover related geometric shapes for engineering and mathematical applications.
Ring-shaped structures appear throughout architecture: circular foundations, donut-shaped reflecting pools, ring roads, and stadium track cross-sections all involve annulus geometry. A circular swimming pool with inner radius 5 m (pool area) and outer radius 6 m (including deck) has a deck annulus area = π(36 − 25) = 34.56 m² — useful for material quantity estimation and cost calculation. Ring beam foundations for circular tanks and silos distribute load around the perimeter; the annulus cross-sectional area determines the concrete volume per linear meter of beam.
Two special cases of the annulus reveal its relationship to other shapes. When inner radius r → 0, the annulus becomes a full disk with area πR². When inner radius r → R (ring width → 0), the annulus becomes a thin circle (circumference) with area → 0. The annulus is therefore a generalization of both the disk and the circle, and its area formula π(R² − r²) interpolates continuously between these extremes. This makes the annulus a pedagogically useful shape for demonstrating the difference between area and perimeter as limiting cases of a parameterized family.
The Annulus Calculator uses the difference of two circle areas:
$$A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$$
Where:
This can be factored as:
$$A = \pi(R + r)(R - r)$$
This factored form is useful: it shows that the annulus area equals $$\pi$$ times the average diameter times the width of the ring.
Circumferences:
Steps:
The Annulus Area represents the ring-shaped region between the two circles. This is the amount of material in a washer, the cross-sectional area of a pipe wall, or the surface area of a ring-shaped region. If the outer radius is in meters, the area is in square meters.
The Outer Circumference and Inner Circumference give the lengths of the two boundary edges. The outer edge is always longer. The difference $$C_R - C_r = 2\pi(R - r)$$ depends only on the width of the annulus, not on the absolute radii — a surprising and useful property.
Inputs
Results
A = π(100 - 25) = 75π ≈ 235.62 sq units. The annulus has three-quarters of the area of the full outer circle (which would be 100π ≈ 314.16).
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Results
A = π(1600 - 1332.25) = π × 267.75 ≈ 845.09 m². The track surface area is about 845 m². The outer lane is 251.33 m around vs. 229.34 m for the inner — a difference of nearly 22 m per lap.
If $$r = 0$$, the annulus becomes a full circle with area $$\pi R^2$$. There is no hole, so the formula reduces to the standard circle area formula.
This is physically impossible — you cannot have a hole larger than the outer boundary. The formula would give a negative area, indicating an error. Always ensure $$R > r$$.
The width (radial thickness) is simply $$w = R - r$$. For a pipe with outer diameter 10 cm and inner diameter 8 cm, the wall thickness is $$(10 - 8)/2 = 1$$ cm (using diameters, divide by 2 to get radii first).
The cross-sectional area of the pipe wall (annulus area) determines the material quantity and structural strength. The inner circle area determines the flow capacity. Together, they define the pipe's mechanical and hydraulic properties.
The outer circumference is longer than the inner circumference. On a 400 m track with standard dimensions, each lane outward adds approximately $$2\pi \times 1.22 \approx 7.67$$ meters to the lap distance. Staggered starts compensate for this difference to ensure all runners cover the same total distance.
Yes. If $$R_m = (R + r)/2$$ is the mean radius and $$w = R - r$$ is the width, then $$A = 2\pi R_m w$$. This form is often more convenient in engineering applications where width and mean radius are directly specified.
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