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 Archimedean spiral. In polar coordinates, this spiral has the equation

*r* = *h*θ / 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* = *hT*² / 4π

If *r*_{1} is the radius of the carpet as a rolled up cylinder, *r*_{1} = *hT* / 2π and so *T* = 2π *r*_{1} / *h*. So when we **unroll** the carpet

*L* = *hT*² / 4π = π*r*_{1}² / *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 *r*_{0} and an outer radius *r*_{1}. This means θ runs from *T*_{0} = 2π *r*_{0}/*h* to *T*_{1} = 2π*r*_{1}/*h*.

To find the length of the spiral running from *T*_{0 } to *T*_{1} we find the length of a spiral that runs from 0 to *T*_{1} and subtract the length of a spiral from 0 to *T*_{0}

*L* = π*r*_{1}² / *h* − π*r*_{0}² / *h* = π(*r*_{1}² − *r*_{0}²)/*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 *r*_{1} when we know the length *L*. We also need to know the inner radius *r*_{0}. So suppose we are wrapping the carpet around a cardboard tube of radius *r*_{0}. Then

*r*_{1} = √(*r*_{0}² + *hL*/π).

Years ago someone in the Contracts Dept asked me almost the same question. Part of his job was to affix labels to equipment, thousands of pieces of equipment over the course of a year. The labels were the “sticky” kind that come in rolls (like packing tape). He needed to buy more rolls, but the number of labels per roll was not given. How many labels are in a roll?, he asked, My very first thought was that I must be doing something right that he would approach me with such a question. My next was to think of Archimedean spirals. I found an answer along the lines of this post, details long forgotten. He didn’t know anything about Archimedes, and he couldn’t accept it without testing. So he counted the labels in a partial roll. He became a believer. Here is a non-Archimedean spiral explanation that he liked: Find the cross-sectional area of the annulus of the roll, and divide by h. (He measured the thickness h with a caliper, making sure it was not over tightened). Then, of course, divide by the length of a label. This gives an excellent approximation.