… Y’see, now, y’see, I’m looking at this, thinking, squares fit together better than circles, so, say, if you wanted a box of donuts, a full box, you could probably fit more square donuts in than circle donuts if the circumference of the circle touched the each of the corners of the square donut.

So you might end up with more donuts.

But then I also think… Does the square or round donut have a greater donut volume? Is the number of donuts better than the entire donut mass as a whole?



A round donut with radius R1 occupies the same space as a square donut with side 2R1. If the center circle of a round donut has a radius R2 and the hole of a square donut has a side 2R2, then the area of a round donut is πR12 - πr22. The area of a square donut would be then 4R12 - 4R22. This doesn’t say much, but in general and  throwing numbers, a full box of square donuts has more donut per donut than a full box of round donuts.

The interesting thing is knowing exactly how much more donut per donut we have. Assuming first a small center hole (R2 = R1/4) and replacing in the proper expressions, we have a 27,6% more donut in the square one (Round: 15πR12/16 ≃ 2,94R12, square: 15R12/4 = 3,75R12). Now, assuming a large center hole (R2 = 3R1/4) we have a 27,7% more donut in the square one (Round: 7πR12/16 ≃ 1,37R12, square: 7R12/4 = 1,75R12). This tells us that, approximately, we’ll have a 27% bigger donut if it’s square than if it’s round.

tl;dr: Square donuts have a 27% more donut per donut in the same space as a round one.

All I can think now is that I want donuts

It’s a good stuarting point, but this discussion of donut packing neglects several crucial details.

(a quick note about the extruded annular donut case: (R12-R22) factors out so the area ratio is always 4:π ≃ 1.273:1, regardless of the two radii.)

  • Donuts are 3D objects, and have a volume rather than a cross-sectional area. This wouldn’t matter if ordinary circle donuts were extruded annuli, but they’re tori.
    • The volume of a toroidal donut of major radius (distance from centre to the centre of the tube) of R and a minor radius (radius of the tube) of r is 2π2Rr2.
    • We can convert to have this in terms of the inner radius R2 and outer radius R1 using R=(R2+R1)/2, and r=(R2-R1)/2<s
    • In terms of these values, the volume of the toroidal donut is π2(R12-R22)(R1-R2)/4.
    • The volume of the corresponding square donut is 4(R12-R22)(R1-R2).
    • So… all the factors of R1 and R2 just factor out again! The volume ratio per donut turns out to be 16:π2 ≃ 1.62:1, even more in favour of the square donuts.
  • Tiling in a square grid is not the most efficient way to pack identical circles into the plane.
    • If we’re putting our donuts into an infinitely large box, we should use hexagonal packing.
    • More likely we have a square box, in which case we could use this data on the best known packings to figure out how many circles we can fit in a given size of square donuts we can fit in a given sized box.
    • We’re never going to do quite as well as the square donuts in terms of volume, but we might be able to get an extra donut or two in and make things a bit better for the toroidal donuts. In this picture, we can get one extra round donut per layer…
    Donuts laid out flat in layers. The illustrated box can fit five square donuts in a layer, and six round donuts.
    • [Admittedly we’ll probably have a square flat box and stand our donuts on their sides only one layer deep, in which case we’re pretty much guaranteed to have the same number of square and round donuts.]
    Donuts are laid on their sides, one layer deep. In this layout, the box contains the same number of square donuts and round donuts.
    • Using a fairly naive packing scheme of ‘put it in a grid I guess’, the number of square donuts we can fit in a square layer of side L is at least floor(L/(2R1))2. We can sometimes fit an extra donut in, iff (L mod 2R1) ≥ R1√2. There are probably more efficient packing schemes - apparently packing squares inside a bigger square is still an open problem.
    • Let’s plot the number of donuts we can fit in a flat layer with the two schemes…
    • So it varies, but yeah, you can often get a lot more circular donuts in your box. To apply this to the volume ratio, we need the ratio between the number of circular donuts and the number of square donuts…
    A plot of the ratio N_circle/N_square for various sizes of box. It's always possible to fit at least as many circular donuts as square donuts in the box, and you can usually fit around 1.15 times as many circular donuts.
    • If we call this number ratio q, the volume ratio becomes 16:2 ≃ 1.62/q:1. Let’s plot 1.62/q
    A plot of the V_square/V_circle for various sizes of box. Except for one value, the square donuts always contain a larger volume, which seems to asymptote towards 1.4 times as much.
    • From the graph, we see that for most box sizes, square donuts give you about 1.4 to 1.6 times as much donut per box. However, intriguingly, there is one size of the box for which square donuts are no good…
    • At this point, Mathematica crashed and I hadn’t saved my worksheet. The mysterious value where circular donuts are better will have to remain a mystery.
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