Thursday, June 20, 2024

How big will a carpet be when you roll or unroll it?

Computer scienceHow big will a carpet be when you roll or unroll it?

If you know the dimensions of a carpet, what will the dimensions be when you roll it up into a cylinder?

If you know the dimensions of a rolled-up carpet, what will the dimensions be when you unroll it?

This post answers both questions.

Flexible carpet: solid cylinder

The edge of a rolled-up carpet can be described as an Archimedian spiral. In polar coordinates, this spiral has the equation

r = hθ / 2π

where h is the thickness of the carpet. The previous post gave an exact formula for the length L of the spiral, given the maximum value of θ which we denoted T. It also gave a simple approximation that is quite accurate when T is large, namely

L = hT² / 4π

If r1 is the radius of the carpet as a rolled up cylinder, r1 = hT / 2π and so T = 2π r1 / h. So when we unroll the carpet

L = hT² / 4π = πr1² / h.

Now suppose we know the length L and want to find the radius r when we roll up the carpet.

T = √(hL/π).

Stiff carpet: hollow cylinder

The discussion so far has assumed that the spiral starts from the origin, i.e. the carpet is rolled up so tightly that there’s no hole in the middle. This may be a reasonable assumption for a very flexible carpet. But if the carpet is stiff, the rolled up carpet will not be a solid cylinder but a cylinder with a cylindrical hole in the middle.

In the case of a hollow cylinder, there is an inner radius r0 and an outer radius r1. This means θ runs from T0 = 2π r0/h to T1 = 2πr1/h.

To find the length of the spiral running from T to T1 we find the length of a spiral that runs from 0 to T1 and subtract the length of a spiral from 0 to T0

L = πr1² / h − πr0² / h = π(r1² − r0²)/h.

This approximation is even better because the approximation is least accurate for small T, and we’ve subtracted off that part.

Now let’s go the other way around and find the outer radius r1 when we know the length L. We also need to know the inner radius r0. So suppose we are wrapping the carpet around a cardboard tube of radius r0. Then

r1 = √(r0² + hL/π).

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