This (123-112)/123 = 1-0.91 = 9% reduction in diameter doesn't look much for a single roll:

123mm v 112mm, to scale overlaid for comparison |

the difference is clear with 40 rolls |

^{2}= 1-0.83 = 17% reduction in volume. Hence more rolls can be packed in each delivery lorry, with a claimed saving of 500 lorry trips, or 140 tonnes of CO

_{2}, per year.

But we could do better, surely? I've shown the rolls on a square grid, because that's how they are packed:

But what about a hexagonal packing?

smaller rolls, and now hexagonally packed |

*r*, not a square of half-height

*r*:

square boundary hexagonal boundary equilateral triangle |

What space does this take? The hexagon is made of 6 equilateral triangles, each of height r. If the side is of length h, then we have

*r*=

*h*sin 60 =

*h*√3 / 2. The area of the triangle = 1/2 x base x perpendicular height =

*h*/2 x

*r*=

*r*

^{2}/ √3. The area of the hexagon is 6 times this, or 2√3

*r*

^{2 }= 3.464

*r*

^{2}. The square, meanwhile, has area 4

*r*

^{2}. (And not coincidentally, 3.464 is a better approximation to π than is 4.) Hence hexagonal packing is (4 - 3.464)/4 = 0.134, or a little over 13%, better than square packing.

Doing the sums to combine the percentages properly [viz, 1-(1-17%)(1-13%)=0.28], this means hexagonal packing

*and*smaller rolls combined gives about 28% improvement over the original large, square packed rolls.

So Sainsbury's could save a

*further*300-odd lorry trips, and a further 90 tonnes CO

_{2}, by packing the rolls hexagonally.

Maybe I should write them a letter?

*UPDATE*(19 Jan 2014)

A commenter queried the correctness of the QI calculations. I just so happen to have an old and new roll size cardboard tube. Here's the difference:

the smaller tube has a diameter of ~35mm, the larger a diameter of ~50mm; also the cardboard is much flimsier in the smaller tube |

This QI "fact" just doesn't seem right. If the toilet roll diameter shrinks from 123mm to 112mm, the inner cardboard tube must shrink by a lot more than 11mm. To keep the same amount of paper per roll, the cross-sectional area of the paper annulus (no pun intended) must remain constant. In particular, the original diameter of the inner tube must be greater than 50.84mm (since 123^2 - 112^2 = 2585 = 50.84^2). And since, in practice, the final diameter of the inner tube would have to be at least 30mm, the original diameter would have to be at least √(2585 + 900) = 59mm. Surely the original tubes couldn't have been that big?

ReplyDeleteYour calculations are correct. I've provided a few photos of the before and after central cardboard tubes, along with their measurements. They are a little different from the exact figures claimed (as your calculations demonstrate they must be), but there is still a large difference between the two tube sizes, enough to make the claimed savings.

Delete