Torus Calculator: How to Find the Volume and Surface Area of a Doughnut Shape
About the Author
Marko Šinko
Co-Founder & Lead Developer, AI Math Calculator
A torus calculator needs exactly two numbers to pin down a doughnut shape completely: the major radius R, measured from the centre of the hole to the middle of the tube, and the minor radius r, the radius of the tube itself. Hand over those two and everything else follows — volume, surface area, the size of the hole, the overall width. The formulas are among the shortest in solid geometry: V = 2π²Rr² and A = 4π²Rr. No integrals, no angle tables, nothing conditional.
Short doesn’t mean obvious, though. Where does that 2π² come from? What happens when the hole closes up? And why does the doughnut you measured with a ruler refuse to match the R and r the formula wants? This article walks through the 1,700-year-old shortcut behind the formulas, the three torus species and where the equations stop telling the truth, and two fully worked examples — one pastry, one 37-litre fuel tank.
Two Radii, Four Diameters, One Shape
Slice a torus through its axis and you see the tube’s cross-section twice: a circle of radius r sitting a distance R from the centre line on either side. Every other measurement is built from those two lengths. The outer diameter — the widest span of the whole ring — is 2(R + r). The inner diameter, the width of the hole, is 2(R − r). Add those two equations together and you get the conversions this calculator uses behind the scenes:
R = (outer diameter + inner diameter) ÷ 4
r = (outer diameter − inner diameter) ÷ 4
That divide-by-four trips people up constantly, but it’s just two halvings in a row: averaging the two diameters gives a diameter-like quantity, and halving again turns it into a radius. It matters in practice because real parts are almost never specified by R and r. O-ring catalogues list inner diameter and cross-section width. A doughnut gets measured across the outside with a ruler. An engineering drawing dimensions the bore and the outside. All three routes lead to the same (R, r) pair, which is why the calculator offers all three as input modes.
Pappus Beat Calculus to This by 1,300 Years
The volume formula looks like it should need an integral. It doesn’t, because of a result written down by Pappus of Alexandria around 300 AD — thirteen centuries before Newton and Leibniz. Pappus’s centroid theorem says: the volume of any solid of revolution equals the area of the shape being revolved, multiplied by the distance its centroid travels. For a torus, the shape is a disc of area πr², and its centroid rides a circle of circumference 2πR. Multiply them:
V = (πr²) × (2πR) = 2π²Rr²
A = (2πr) × (2πR) = 4π²Rr
The surface area line is the same trick applied to the circle’s boundary instead of its interior. There’s an equivalent mental picture: cut the doughnut once and straighten it out. You get a cylinder of radius r and length 2πR, and the cylinder formulas give identical answers. The straightening isn’t free — material on the inside of the bend compresses while the outside stretches — but the two distortions cancel exactly, which is precisely what Pappus proved.
Two consequences are worth keeping in your head. Volume scales with r² but surface area only with r, so doubling the tube thickness quadruples the volume while merely doubling the area. And dividing the formulas gives V ÷ A = r ÷ 2 — a torus with a 2-unit tube radius has the same numeric volume as area, whatever R is.
Ring, Horn, Spindle: One Formula, Three Shapes
Shrink R while holding r fixed and the torus passes through three distinct species. The formulas keep producing numbers the whole way down — but they stop meaning the same thing.
| Condition | Name | What it looks like | Are V and A exact? |
|---|---|---|---|
| R > r | Ring torus | A doughnut with an open hole | Yes, both |
| R = r | Horn torus | The hole has pinched shut to a single point | Yes, both |
| 0 < R < r | Spindle torus | The tube overlaps itself through the middle | No — swept values only |
| R = 0 | Sphere (degenerate) | The tube closes into a ball of radius r | No — both collapse to 0 |
The spindle case deserves the warning label this calculator gives it. Once R drops below r, part of the tube sweeps through space the opposite side already claimed, so 2π²Rr² is a net swept volume rather than the capacity inside the outer skin. The R = 0 endpoint fails more loudly: the real shape is a solid ball of volume (4/3)πr³, but the torus formulas return zero because Pappus’s theorem requires the centroid to stay off the axis. If your shape has no hole and no waist, you want the sphere calculator instead.
How Much Dough Is in a 9 cm Doughnut?
Time to use the machinery as a doughnut volume calculator, on an actual doughnut. A typical ring doughnut measures about 9 cm across the outside, and the tube — the fried part you’d pinch — is roughly 3.6 cm thick. That’s the “overall diameter & tube diameter” input mode: r = 3.6 ÷ 2 = 1.8 cm, and R = (9 − 3.6) ÷ 2 = 2.7 cm.
Volume: V = 2π² × 2.7 × 1.8² = 2π² × 8.748 ≈ 172.7 cm³. At a fried-dough density of about 0.35 g/cm³, that’s roughly 60 g of dough, which matches what a bakery scale says a plain ring doughnut weighs. The hole, for the record, is 2 × (2.7 − 1.8) = 1.8 cm wide.
The surface area is where it gets commercially interesting: A = 4π² × 2.7 × 1.8 ≈ 191.9 cm². A glaze coat about 1 mm thick therefore uses 191.9 × 0.1 ≈ 19 cm³ of glaze — around 25 g at glaze density. Nearly a third of a glazed doughnut’s sugar rides on that surface term, which is why V ∝ r² but A ∝ r matters to a bakery: shrink the tube by 20% and the dough cost falls 36% while the glaze cost falls only 20%.
A 37-Litre Fuel Tank Hiding in a Spare-Wheel Well
The most common engineered torus you’ll ever sit above is the toroidal LPG tank, shaped to drop into a car’s spare-wheel well with the filler hardware through the hole. Idealise one as a true torus with a 58 cm overall diameter and a 20 cm tube: r = 10 cm, R = (58 − 20) ÷ 2 = 19 cm.
V = 2π² × 19 × 10² = 37,505 cm³ ≈ 37.5 litres to the brim. LPG tanks are legally filled to 80%, so the usable capacity is about 30 litres. Compare that against the cylindrical envelope the tank occupies — a 58 cm wide, 20 cm tall drum holds π × 29² × 20 ≈ 52.8 litres — and the torus recovers 71% of the space while leaving the hub clear. The missing 15 litres are the price of the hole and the curved shoulders. For odd shapes beyond cylinders and tori, a general volume calculator covers the standard solids side by side.
Numbers People Feed a Torus Calculator by Mistake
Because V = 2π²Rr² multiplies three lengths together, input errors don’t add — they multiply. These are the four that show up over and over, with the exact damage each one does:
- Entering diameters in the radius fields. Both lengths double, and since volume carries one factor of R and two of r, the result is 2 × 2² = 8 times too large. Surface area comes out 4 times too large.
- Using the outer radius as R. The distance from the centre to the outer edge is R + r, not R. On the doughnut above that substitutes 4.5 for 2.7 and inflates the volume by 67%.
- Treating the tube diameter as r. A “6 mm cord” O-ring has r = 3 mm. Skip the halving and any torus volume calculator will report 4 times the rubber that’s actually there.
- Measuring a squashed part. A used O-ring or an inflated inner tube isn’t a perfect circle in cross-section. The formulas assume it is; a cross-section flattened by 10% changes πr² and shifts the volume by roughly the same 10%. Measure the relaxed part, or average two perpendicular thickness readings.
A quick sanity check catches most of these: the outer diameter shown in the results table should match what your ruler says the part actually spans. If the calculator’s 2(R + r) disagrees with the physical object, one of the inputs is wrong — and for a deeper check on the area side, the surface area calculator handles the comparison solids.
Five Real Toruses, From O-Ring to Tokamak
The same two-number recipe spans about nine orders of magnitude in volume. Each row below was converted from the measurements you’d realistically have — catalogue ID and cord for the O-ring, outside measurements for the rest.
| Object | R | r | Volume | Surface area |
|---|---|---|---|---|
| O-ring, 30 × 3.5 mm | 16.75 mm | 1.75 mm | 1.01 cm³ (≈1.2 g of rubber) | 11.6 cm² |
| Ring doughnut | 2.7 cm | 1.8 cm | 173 cm³ | 192 cm² |
| Road-bike inner tube (700×25) | 32.4 cm | 1.25 cm | ≈1.0 litre of air | 1,599 cm² |
| Toroidal LPG tank | 19 cm | 10 cm | 37.5 L | 0.75 m² |
| ITER plasma (circular equivalent) | 6.2 m | 2.0 m | ≈490 m³ | ≈490 m² |
Two footnotes on that last row. The matching 490s aren’t a typo — with r = 2 the identity V ÷ A = r ÷ 2 forces volume and area to share a numeral. And the real ITER plasma chamber holds about 840 m³, not 490, because fusion engineers stretch the cross-section into a tall D-shape rather than a circle. That gap is the honest limit of any torus surface area calculator: the formulas are exact for circular cross-sections and silently wrong for everything else. For doughnuts, O-rings, inner tubes, and tank estimates, circular is exactly what you have — measure two lengths, pick the matching input mode, and the other six numbers are already on screen.
