The Napkin Ring Problem: Why the Volume of a Drilled Sphere Depends Only on the Hole’s Height

Here is one of the most counterintuitive results in elementary geometry: when you drill a cylindrical hole through a sphere, the volume of what remains depends on just one number — the height of the hole.

What Is the Napkin Ring Problem?

Take a sphere and drill a cylindrical hole straight through its center. What remains looks like a bead, or a chunky ring with a large gap running through the middle. This shape is known as a napkin ring, and it sits at the heart of a beautiful geometric puzzle.

The napkin ring problem asks: what determines the volume of this leftover solid? The answer turns out to be far simpler — and far stranger — than you might expect.

The Surprising Result: Only the Height Matters in the Napkin Ring Problem

The volume of the napkin ring depends only on the height of the hole — that is, the height of the ring if you set it down with the hole pointing downward.

The size of the original sphere is completely irrelevant. Whether you drill through a tennis ball or through the entire planet Earth, if the remaining ring has a height of 1 cm in both cases, the two rings will have exactly the same volume. A tiny sphere with a tiny hole and a planet-sized sphere with an enormous hole can yield identical napkin rings — as long as the height matches.

Why the Napkin Ring Volume Is Independent of the Sphere’s Size

Intuitively, this makes sense once you think about what drilling actually removes. To produce a ring of a given height from a large sphere, you must drill a very wide hole, cutting away a great deal of material. For a small sphere, the hole is narrow and removes much less. These two effects — the size of the sphere and the size of the hole — cancel each other out perfectly.

The geometric relationship between the hole’s radius, the sphere’s radius, and the ring’s height follows directly from the Pythagorean theorem:

R² − a² = h²/4

Here R is the radius of the original sphere, a is the radius of the cylindrical hole, and h is the height of the ring. This single equation encodes the exact balance between what you keep and what you remove.

Deriving the Napkin Ring Problem Formula

From there, the derivation is straightforward. Slice the napkin ring into infinitely thin horizontal discs and integrate their areas over the height of the ring. Because the Pythagorean relation eliminates R from the calculation entirely, the final result contains only h:

V = πh³/6

The sphere’s original radius vanishes from the formula completely. The napkin ring problem is a rare case where a seemingly complex geometric object has a volume determined by a single, elegant measurement.