Crates of Bottles

The other day I was looking at a crate of 12 water bottles, and looked at the wasted space between them. Why not use a honeycomb? or even other bottle shapes?

Well, let us have a mathematical look at bottle packing. A round bottle occupies a surface (at the bottom of the box) of $S=\pi r^2$ (for some radius $r$ ) in a square of $4r^2$ . The occupancy is therefore

$\displaystyle \frac{\pi r^2}{4r^2}=\frac{\pi}{4}\approx 0.785$

So the round bottle in a square grid uses less than 80% of the available space. What if we packed them in a honeycomb grid? The grid would look something like this:

If we are to make crates with (approximately) the same number of bottles, we can either have 11 or go up with larger crates. Let’s say we stay with 11. Then the arrangement is a row of four, a row of three, and another row of four. The length of the box is 4 bottles, and the depth is $2(h+r)$ , where $h$ is the half-height of the hexagon above.

However, we’re not exactly interested in the hexagons, but to know how wide are three rows. To do so, we need to consider the following arrangement:

If the bottles have radius $r$ , then

$\displaystyle h=r_h\frac{\sqrt{3}}{2}$

with $r_h=2r$ (which is obvious from the diagram), is the height of the equilateral triangle (shown dashed), which gives the half-height of the hexagon. However, $H=2(r_h+r)$ because $h+r$ is the distance between the top of the hexagon to the bottom side of the bottle. Therefore, the depth of three rows is $\approx 5.46r$ (rather than $6r$ with a regular grid.

The box is therefore $\approx 10\%$ smaller, and the packing efficiency only went up a few percent to just a wee bit short of 80%. (Because the original box has area $48 r^2$ , the hex-grid box has an area of $5.46\times{}8r^2\approx{}43.7r^2$ , and $\frac{43.7}{48}\approx{}0.91$ . Also, because an area of $11\pi{}r^2$ is actually bottles, we find that $\frac{}{}\approx{}79\%$ , we see that the gain is negligible.)

However, hexagonal grid have an asymptotic regime of $\frac{\pi}{\sqrt{12}}\approx 0.9$ , therefore, an infinite (and infinitely inconvenient) box would have only 90% of its surface devoted to actually supporting bottles, with 10% unusable space.

What if we had triangular bottles? Well to be fair, one would need to consider triangles with the same surface as the circles, so the side $a$ of such a triangle is related to $r$ by:

$\displaystyle a=2r\sqrt{\frac{\pi}{\sqrt{3}}}$

which we obtain by isolating $a$ in

$\displaystyle a^2\frac{\sqrt{3}}{4}=\pi r^2$

where the left hand side corresponds to the area of an equilateral triangle and the right hand side is the area of the circle.

From there on, we could just plug this and compute the surface using this complicated expression, but this is irrelevant, in many ways, to determine if triangular packing is more efficient. Let us look at a row of triangular bottles:

We see that the two unused spaces at the end of the row for exactly one (complete) triangle, and that, therefore, there will be one unused triangle-area unused for any number of triangles in the row, with total area being $n+1$ triangles for $n$ bottles. The larger the box, the smaller (relative) unused space; the unused space is $O(\frac{1}{n})$ which can be made arbitrarily small.

For reasonable values of $n$ , say, $n=4$ , we find that the wasted space in a row is $\frac{1}{5}$ , and we break even with the hexagonal grid and classical round bottles. Larger rows will bring the efficiency as close to 1 as you wish, although for real bottles, you may not want to go much beyond 6, or 8, or 12. Or maybe have triangular bottles in a triangular crate? And triangular crates transported by triangular trucks?

Of course, the real idea is to use square bottles for rectangular boxes. Neglecting the necessarily rounded edges, square bottles would occupy essentially all of the box (which itself would be about 20% smaller), with no waste, and without asking for an infinitely large box either.

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This entry can be understood as a cautionary tale about overthinking/overengineering a solution. Initially, the triangular bottles seems to be a good consequence of the hexagonal packing strategy (and some makers of wine do have triangular bottles) which itself seems better than the regular grid of bottles and quite a natural extension to it.

We do, I think, overthink/overengineer solutions too often. Sometimes by not asking the right question, sometimes by not going in the right direction. Why have an hexagonal grid of triangular bottles when just square bottles solves the problem… better?

This entry was posted on Tuesday, February 5th, 2013 at 10:03 am and is filed under Mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to Crates of Bottles

Yann Collet (@Cyan4973) says:

February 5, 2013 at 10:41 am

How much cost a square bottle, as compared to a regular round bottle ?

Reply
- Steven Pigeon says:
  
  February 6, 2013 at 0:08 am
  
  I suppose we could compare surface area at equal volume? Let’s oversimplify: A round bottle is a cylinder, a square bottle some kind of boxoid:
  
  The volume of the cylinder of radius $r$ and height $h$ is
  
  $\displaystyle V_c=\pi r^2 h$ ,
  
  and its surface $\displaystyle S_c=2\pi r(r+h)$ .
  
  For a box, we would have volume $V_r=4r^2 h'$ , for some height $h'$ ,
  and surface $8r(r+h')$ .
  
  Solving for $h'$ such that $V_c=V_r$ , we find $h'=\frac{\pi}{4}h$ .
  
  Now, is $S_c$ or $S_r$ larger?
  
  Plugging back $h'=\frac{\pi}{4}h$ in $S_r$ , we find that $S_r=8r(r+\frac{\pi}{4}h)$ , and that:
  
  $8r(r+\frac{\pi}{4}h) \overset{?}{\geqslant} 2 \pi r(r+h)$ ,
  
  $4(r+\frac{\pi}{4}h) \overset{?}{\geqslant} \pi(r+h)$ ,
  
  and yes,
  
  $4r+\pi h \geqslant \pi r+ \pi h$ ,
  
  so that, at equal volume, the rectangular bottle costs more than the round bottle.
  
  Reply