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If you consider the regular packing in an infinite plane, tri/hex packing is the most space efficient (least wasted space), so I’d assume larger packings would tend towards that. But in smaller packings, the efficiency loss from the extra size needed to offset the circles outweighs the efficiency gained by hex packing.
7x7 is the boundary where those efficiency tradeoffs switch.
Want a hint? Think about a circle bound by an n-sided polygon. What happens to the space between the bounding polygon and the circle as n increases? And when n is infinite?
So of three possible regular tilings, which will be most and least efficient?
(Btw, strictly speaking, I shouldn’t have said tri/hex before, as it’s really just hex tiling.)
You could also use some fancy trig to calculate the efficiency %, but that’s way too much work for me. :)
Because it’s a smaller area than 7x7.
If you consider the regular packing in an infinite plane, tri/hex packing is the most space efficient (least wasted space), so I’d assume larger packings would tend towards that. But in smaller packings, the efficiency loss from the extra size needed to offset the circles outweighs the efficiency gained by hex packing.
7x7 is the boundary where those efficiency tradeoffs switch.
Thank you for this explanation!
Trying to think how tri/hex is more efficient than any regular tiling, say squares.
Want a hint? Think about a circle bound by an n-sided polygon. What happens to the space between the bounding polygon and the circle as n increases? And when n is infinite?
So of three possible regular tilings, which will be most and least efficient?
(Btw, strictly speaking, I shouldn’t have said tri/hex before, as it’s really just hex tiling.)
You could also use some fancy trig to calculate the efficiency %, but that’s way too much work for me. :)