The problem is that the toilet roll’s depth (i.e., thickness of paper on the roll) doesn’t diminish in a linear fashion. With an unused roll, you can get about three sheets in one rotation. But getting the same three sheets toward the end of the roll can take almost three rotations. A constant rate of use leads to an increasing rate of shrinkage in the remaining depth.

Using basic geometry and a measuring tape, I found the following:

• The diameter of a typical roll of toilet paper is about 10 cm, of which about 4 cm is space inside the tube and 6 cm is paper.

• The paper’s starting depth is about 3 cm (on each side of the tube), with 100% of the paper remaining.

• When the paper

*appears*half gone, with 1.5 cm of depth, the actual fraction of paper remaining is only 39%.

• When the paper

*appears*three-quarters gone, with .75 cm of depth, the actual fraction of paper remaining is only 17%.

But there’s more! I’ve derived formulae for the relationship between apparent fraction remaining (as measured by paper depth) and actual fraction remaining. Let X = apparent fraction remaining and Y = actual fraction remaining. Then:

• Y = [(3X + 2)^2 – 4]/21

• X = [sqroot(21Y + 4) – 2]/3

The first formula is probably the more useful, since you can use it to calculate the actual fraction left when you know the apparent fraction. The formulae still work for larger-than-usual rolls,

*if*the tube is enlarged by the same proportion as the whole roll. (Otherwise, the formulae are a bit more complex.)

Now there’s news you can use!

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