> It feels like the smaller marbles are denser but obviously they actually pack the same efficiency as the tennis balls

That's not obvious to me. Won't the marbles pack more efficiently due to the edges of the container? As a limiting case, consider a container slightly smaller than 1 tennis ball: the packing efficiency of tennis balls will be 0%, while marbles will be something like 60-70%.

Packing efficiency is usually defined for an infinite space because finite containers don’t always favor the same size object.

Consider a marble with a diameter of 2inch and a tennis ball with diameter if 3 inches. A cube of length 9.01 inches now favors the tennis balls where one of 8.01 inches favors the marbles.

According to http://hydra.nat.uni-magdeburg.de/packing/scu/scu.html a cube of side 9.01 can fit 27 balls of diameter 3, but 100 balls of diameter 2. Since 27×3^3 = 729 and 8×2^3 = 800, the balls of diameter 2 are more efficient in this case.

In fact it seems that the diameter 3 balls are most efficient only for cubes of edge-length between 3 and 2+sqrt(2) = 3.414.... These are the sizes for which you can fit in 1 diameter 3 ball, but can't fit in 4 diameter 2 balls. Above that, diameter 2 balls are more efficient (except that a cube of edge-length 6 can fit in the same volume either way; 27 diameter 2 balls or 8 diameter 3 balls).

> Since 27×3^3 = 729 and 8×2^3 = 800

I think you mean 100 x 2^3 = 800. But, yer correct that’s what I get from using a half remembered example without checking.

A "barrel" though is way larger than a tennis ball, so we're not close to the limiting case.

The mixture seems like the intuitive answer to me. A single tennis ball packs the volume of a tennis ball more efficiently than the equivalent volume of marbles.

The problem is ambiguously stated. A "volume of X" isn't well-defined when the point of the problem is to reason about packing efficiency.

Yes. I really like the metallurgy version: An alloy can be denser than pure elements.

I was just talking to a friend about how 1 part water plus 1 part ethanol becomes 1.92 parts solution.

This raised a question for me that I have yet to research/answer. Maybe one of you knows... if the above solution is 50% ABV, what happens if I add one more part alcohol to the solution? Is it now 66.6% ABV? More? Less? How does ABV take into account the fact that this solution is packed together tighter than its constituent parts?

The wikipedia article has all the charts and formulas you need. https://en.wikipedia.org/wiki/Alcohol_by_volume

(To make the above solution 50% abv, btw, you need to keep adding water until it's 2 (liters, whatever) total volume.)

This is a common issue for people doing liquors at home: you mix 1l of alcohol which you used to extract some flavour with 1l of water and don't get back 2l.

Thank you for being the best of HN.

The muderation might be awful, but the users are sharp.

Learned something new, and I'm a chem grad.

Assuming the barrel is an infinite space, and the marbles are small enough to fit in the gaps from an optimal tennis-ball sphere packing:

Not just tennis balls, because after adding all the tennis balls you can, there's still room for marbles to fill in the gaps

Not just marbles, because the optimal sphere packing density is the same whether it's all marbles or all tennis balls. Since all tennis balls wasn't optimal, neither is all marbles.

I figure things get interesting once the ratio between the two spheres gets closer to 1.

(And checking my work, this argument is described in a fancier sciencey way by https://en.wikipedia.org/wiki/Sphere_packing#Unequal_sphere_... )

Define efficiency - marbles are more dense and will fill the barrel with more mass.