Lionel Levine suggests trying the octahedron in hope of finding some interesting duality phenomenon (maybe random planes cutting an octahedron have something to do with random lines cutting a cube while random lines cutting an octahedron have something to do with random planes cutting a cube). Can someone try this too? Thanks, Jim Propp On Sat, Aug 10, 2019 at 11:50 AM James Propp <jamespropp@gmail.com> wrote:
Thanks, Tom!
Does anyone recognize these numbers?
Jim
On Sat, Aug 10, 2019 at 11:02 AM Tomas Rokicki <rokicki@gmail.com> wrote:
And here are the results after an overnight run.
At 34359738368 pts 3.28853583722957 area 0.129186219302995 perim 1.49118840781298
On Fri, Aug 9, 2019 at 10:49 PM Tomas Rokicki <rokicki@gmail.com> wrote:
It's easy to do, and I think I did it correctly. Here are the results I'm seeing. They aren't as nicely round. These values are for a unit tetrahedron. (I'm calculating them for a tetrahedron with edge length 2 sqrt 2, and then scaling the results down.)
At 268435456 pts 3.28851559385657 area 0.129189154723324 perim 1.49120946841637
On Fri, Aug 9, 2019 at 5:04 PM James Propp <jamespropp@gmail.com> wrote:
Can the software that gave accurate predictions for the cube be repurposed to predict properties (expected area, perimeter, and number of sides) for a random cross-section of a regular tetrahedron?
We can derive a prediction for the expected area by dividing the volume by the mean width, but I don’t know how to compute the expected perimeter; the method I described yesterday (for the cube) shows that it’s equal to pi / (4 sqrt(3)) times the expected number of sides, but I don’t know how to compute the expected number of sides.
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