AJ Designer

Annulus Calculator

Area equals pi times the quantity R squared minus r squared

Solution

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Calculate Annulus Area from Outer and Inner Radii

Use this form when both radii of the annulus are known. The annulus is the flat ring between two concentric circles, so its area is just the outer disk's area minus the inner disk's.

A = π(R² − r²)

Calculate Outer Radius from Area and Inner Radius

Use this rearrangement when the ring area and the hole radius (inner) are known and you need the outer radius — for example, sizing a washer to a required bearing surface area around a known bolt hole.

R = √(A/π + r²)

Calculate Inner Radius from Area and Outer Radius

Use this rearrangement when the outer radius and ring area are known and you need the inner radius. The expression under the root must be non-negative — if the requested area exceeds the outer disk's area, no real inner radius exists.

r = √(R² − A/π)

Calculate Ring Width from Outer and Inner Radii

Use this form when you need the radial thickness of the ring — useful for gasket sizing, pipe wall thickness, or washer face width.

w = R − r

How It Works

An annulus is the flat 2D ring between two concentric circles. Its area is the area of the outer disk (πR²) minus the area of the inner disk (πr²), which factors to A = π(R² − r²). Pick the unknown with the solve-for toggle, enter the remaining values, and the calculator shows the area, outer and inner circumference (2πR and 2πr), and the radial width w = R − r together. The outer radius must be strictly greater than the inner radius — when R = r the ring collapses to a circle of zero area, and when R < r there is no annulus.

Example Problem

A flat steel washer has an outer radius of 5 cm and an inner radius of 3 cm (a 3 cm bolt hole). Compute the face area of the washer, its ring width, and the inner and outer circumferences.

  1. Knowns: R = 5 cm, r = 3 cm
  2. Area: A = π(R² − r²) = π(25 − 9) = 16π ≈ 50.265 cm²
  3. Width: w = R − r = 5 − 3 = 2 cm
  4. Outer circumference: C_outer = 2πR = 10π ≈ 31.416 cm
  5. Inner circumference: C_inner = 2πr = 6π ≈ 18.850 cm
  6. Sanity check (inverse): R = √(A/π + r²) = √(16 + 9) = √25 = 5 cm, which recovers the outer radius.

This R > r case is the only valid annulus. When R = r the ring has zero area; when R < r the geometry has no annulus interpretation.

When to Use Each Variable

  • Solve for Area — when both radii are known — washer face area, pipe cross-section, ring-shaped sign or gasket coverage.
  • Solve for Outer Radius — when the ring area and inner radius are fixed and you need the outer radius — sizing a washer around a known bolt hole.
  • Solve for Inner Radius — when the ring area and outer radius are fixed and you need the hole radius — gasket inner cutout sized to a target sealing area.
  • Solve for Width — when you need the radial thickness w = R − r — gasket width, pipe wall thickness, washer face width.

Key Concepts

An annulus is defined by two concentric circles with radii R (outer) and r (inner). The area is most easily derived as the difference of the two disk areas: πR² − πr² = π(R² − r²). The ring has two circumferences — an outer one at 2πR and an inner one at 2πr — and a radial width w = R − r. The factored form π(R² − r²) = π(R − r)(R + r) is sometimes useful: it shows that for thin rings (R ≈ r), the area is approximately the width times the mean circumference, A ≈ w · 2π(R + r)/2 = πw(R + r). For a true annulus, R must be strictly greater than r.

Applications

  • Mechanical washers: compute face area, ring width, and outer/inner circumferences for plain washers, fender washers, and shoulder washers
  • Pipe cross-section: solid pipe wall area for stress, mass, and thermal-conduction calculations (outer radius minus bore radius)
  • Gaskets and O-ring seals: flat ring gaskets between flanges, where the sealing area is the annulus face
  • Race tracks and ring roads: track surface area between an outer kerb and an inner kerb
  • Astronomy: planetary rings and circumstellar disks, where the disk's projected area is annular

Common Mistakes

  • Computing π(R − r)² instead of π(R² − r²) — the correct formula is the difference of squares, not the square of the difference
  • Forgetting that R must be strictly greater than r; entering R ≤ r gives no valid annulus (zero or negative area)
  • Confusing the outer circumference (2πR) with the inner circumference (2πr); they are different unless the ring is infinitely thin
  • Treating the width w as a chord or diameter — it's the radial distance between the two circles, not the gap measured at any angle other than radially
  • Solving for r when the requested area exceeds the outer disk's area (A > πR²); in that case the inside of the square root is negative and no real inner radius exists

Frequently Asked Questions

What is an annulus?

An annulus is the flat, two-dimensional ring-shaped region between two concentric circles — like the face of a washer, a CD viewed flat, or a ring road. It is defined by an outer radius R and an inner radius r, with R > r.

How do you calculate the area of an annulus?

Use A = π(R² − r²), where R is the outer radius and r is the inner radius. For R = 5 cm and r = 3 cm, A = π(25 − 9) = 16π ≈ 50.27 cm². You can also factor it as A = π(R − r)(R + r).

What is the formula for an annulus?

Area: A = π(R² − r²). Outer circumference: C_outer = 2πR. Inner circumference: C_inner = 2πr. Ring width: w = R − r. Inverting the area formula gives R = √(A/π + r²) and r = √(R² − A/π).

What is the difference between an annulus and a circle?

A circle (or disk) is a filled round region defined by one radius. An annulus is the ring left over when you remove a smaller concentric disk from a larger one. An annulus has a hole in the middle; a disk does not.

How is an annulus used in engineering?

Annular geometry appears in washers (the flat ring between the hole and the outer edge), pipe cross-sections (the solid wall between the bore and the outer diameter), flat gaskets, thrust bearings, and any ring-shaped seal or contact surface. Stress, mass, and heat-transfer calculations all start with the annular area.

What is the ring width of an annulus?

The ring width is w = R − r — the radial distance between the inner and outer circles. For a washer with R = 5 cm and r = 3 cm, the ring width is 2 cm. This is the dimension you measure across the face of the washer with calipers held radially.

Can the inner radius be larger than the outer radius?

No. By definition R > r for an annulus. If you accidentally swap them, π(R² − r²) becomes negative, which has no geometric meaning for a ring. This calculator rejects inputs where R ≤ r.

How do you find the inner radius from the area and outer radius?

Rearrange A = π(R² − r²) to r = √(R² − A/π). The expression under the root must be non-negative — that is, A ≤ πR². If A exceeds the outer disk's area, no real inner radius exists.

Reference: Weisstein, Eric W. "Annulus." MathWorld — A Wolfram Web Resource. https://mathworld.wolfram.com/Annulus.html

Worked Examples

Washer

What is the face area of a flat steel washer with a 5 cm outer radius and 3 cm hole?

A plain washer has an outer radius of 5 cm (10 cm OD) and an inner radius of 3 cm (6 cm bolt hole). Compute the contact face area between bolt head and clamped surface.

  • Knowns: R = 5 cm, r = 3 cm
  • Formula: A = π(R² − r²)
  • A = π(25 − 9) = 16π ≈ 50.27 cm²
  • Width: w = R − r = 2 cm

Face area ≈ 50.27 cm² (≈ 7.79 in²)

This is the canonical example in the Example Problem section above. Bearing-pressure calculations divide the bolt's clamp load by this area.

Pipe Cross-Section

What is the wall cross-section area of a steel pipe with 75 mm OD and 60 mm ID?

A circular steel pipe has an outer diameter of 75 mm and an inner bore diameter of 60 mm. Compute the solid wall cross-sectional area for stress and mass calculations (use radii, so R = 37.5 mm and r = 30 mm).

  • Knowns: R = 37.5 mm = 0.0375 m, r = 30 mm = 0.030 m
  • Formula: A = π(R² − r²)
  • A = π(0.0014063 − 0.0009000) = π · 0.0005063 ≈ 1.5904 × 10⁻³ m²
  • Equivalently: A ≈ 1590.4 mm² ≈ 15.904 cm²
  • Wall thickness: w = R − r = 7.5 mm

Wall area ≈ 1590 mm² (≈ 15.9 cm²)

Multiply by steel density (≈ 7850 kg/m³) and pipe length to get mass per length. Divide axial load by this area for axial stress.

Race Track

What is the surface area of a circular race track with a 50 m outer radius and 40 m inner radius?

A circular running track has an outer kerb at 50 m radius and an inner kerb at 40 m radius. How much surface area does the track itself cover (for resurfacing material estimates)?

  • Knowns: R = 50 m, r = 40 m
  • Formula: A = π(R² − r²)
  • A = π(2500 − 1600) = π · 900 = 900π ≈ 2827.4 m²
  • Width (lane span): w = R − r = 10 m
  • Outer circumference: 2π · 50 = 100π ≈ 314.2 m

Track surface ≈ 2,827 m² (≈ 30,434 ft²)

Tartan and asphalt resurfacing is often priced per m², so this is the headline number. For a single 1.2 m lane on the same track, recompute with R = 41.2 m and r = 40 m.

Annulus Formulas

An annulus is the flat 2D ring between two concentric circles with outer radius R and inner radius r (with R > r).

A = π(R² − r²)

R = √(A/π + r²)

r = √(R² − A/π)

w = R − r

C_outer = 2πR  ·  C_inner = 2πr

Annulus with outer radius R and inner radius rRr

Where:

  • A — annulus area (m², cm², mm², in², ft²)
  • R — outer radius (length)
  • r — inner radius (length); must be less than R
  • w — ring width = R − r (radial thickness)
  • C_outer — circumference of the outer circle = 2πR
  • C_inner — circumference of the inner circle = 2πr

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