2
$\begingroup$

Reference: https://en.wikipedia.org/wiki/Smoothed_octagon

Background: The smoothed octagon is conjectured to have the lowest maximum packing density of the plane of all centrally symmetric convex shapes. If central symmetry is not required, the regular heptagon has even lower packing density, but its optimality is also unproven. A circular segment is a region cut off from a circular disk by a chord.

Question: Circular segments that are larger than semidisks of same radius appear good candidates for low packing fraction among non-centrally symmetric convex regions (segments smaller than semidisks can be paired into 'convex lens' shapes which are centrally symmetric and convex). Are the packing densities of all such large circular segments known to be strictly greater than that of the regular heptagon? If "yes", how does the packing densities of such large segments compare with that of the disk itself?

Additional Question: Is the constant width curve with least packing fraction the circle - or say, an irregular figure formed by unequal length arcs (for a construction of the latter, see 'Colossal Book' by Martin Gardner)?
Note: Messnikoff (https://arxiv.org/pdf/1504.06733.pdf) has conjectured the constant width curve with highest packing fraction is the Reuleaux triangle.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Have you tried calculating the density for the obvious double-lattice candidate for circular segments? I'm pretty sure it should easily beat the the circle-packing density. Even the obvious lattice candidate probably beats it, no? $\endgroup$
    – Yoav Kallus
    Commented Oct 6, 2021 at 21:17
  • $\begingroup$ Thank you. For various circular segments bigger than semidisks, I tried welding 2 of them along the straight edge and then to naively tile the plane with all these pairs having same orientation. Note: The segment was determined by a, distance from the center of the full disk to the midpoint of its straight edge (as a goes from 0 to 1, the segment goes from semidisk to full unit disk). It was found that for v low values of a, the packing fraction is slightly more than 0.906 (full disk optimal value); as a increases this fraction rises gradually, then reduces and for a > ~.71, falls below 0.906. $\endgroup$
    – Nandakumar R
    Commented Oct 8, 2021 at 13:55
  • 1
    $\begingroup$ I think you made a mistake in your calculation. I get densities above 0.906 for all cirular segments (see my answer below) $\endgroup$
    – Yoav Kallus
    Commented Oct 10, 2021 at 2:01
  • $\begingroup$ Thanks very much! yes, I considered the possibility of pushing the paired units close only in the y direction; had simply put them side by side in x direction; mistake! $\endgroup$
    – Nandakumar R
    Commented Oct 10, 2021 at 5:37

1 Answer 1

Reset to default
4
$\begingroup$

Circular segments all pack at a higher density than disks:enter image description here

The density is

$$\frac{a\sqrt{1-a^2}+\pi-\cos^{-1} a}{\sqrt{3}+a\sqrt{4-a^2}}$$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.