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0 pointscubefox6mo ago0 comments

Yeah, though they are somewhat similar. We could perhaps say the volume of the unit ball is (in some sense) "larger" than the area of the unit disk, because both are measured relative to a radius of 1. So the area of the disk is fewer "units" than the volume of the ball.

Anyway, it seems independently interesting that this value peaks for the 5-ball, or the ~5.2569-ball. The non-fractional difference between fractional dimension of peak hyper volume and peak hyper surface area seems also interesting. (I assume there is some trivial explanation for this though.)

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srean6mo ago

No it doesn't work that way because the units of hyper-volume and length are different.

However, once you take appropriate roots of hypervolume to get same units you can safely compare. Or the otherway round take appropriate powers of length to get same units as hypervolume.

evanb6mo ago

I agree; the fair comparison is the nth root of the hypervolume in n dimensions, (V(n))^(1/n). This monotonically decreases from n=0, which shows the counterintuitive point that people often want to make anyway: an n-sphere takes up less and less of an n-hypercube. The peak at n~=5 is illusory.

Another fair comparison is between dimension-dependent lengths is the ratio of the (hyper)volume to the surface (hyper)area V(n)/A(n). This monotonically decreases from n=1.

cubefoxOP6mo ago

Imagine you want to compare sizes of video game levels, where some are 2D, made from pixels, and some are 3D, made from voxels. You could stipulate that n pixels are equivalent to sqrt(n) voxels. But you could also stipulate that n pixels are equivalent to n voxels. I don't think either is more correct than the other.

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cubefoxOP6mo ago

With that it would probably mean the units monotonically approach 0, rather than first increasing and then decreasing. At least for whole dimensions. I'm not sure about monotonicity with fractional dimensions.

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0 pointscubefox6mo ago0 comments

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srean6mo ago

No it doesn't work that way because the units of hyper-volume and length are different.

However, once you take appropriate roots of hypervolume to get same units you can safely compare. Or the otherway round take appropriate powers of length to get same units as hypervolume.

evanb6mo ago

Another fair comparison is between dimension-dependent lengths is the ratio of the (hyper)volume to the surface (hyper)area V(n)/A(n). This monotonically decreases from n=1.

cubefoxOP6mo ago

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cubefoxOP6mo ago

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