3 ms·
Haven't watched yet but an obvious lower bound is 2 times it's shadow (one for each side) - in fact a coin achieves this.
by quickthrower2 8y ago
Haven't watched yet but an obvious lower bound is 2 times it's shadow (one for each side) - in fact a coin achieves this.
- theunamedguy 8y agoI don't see why this is getting downvoted. It's a perfectly valid argument that's very much in the spirit of 3b1b.
- lightedman 8y agoBecause a huge chunk of people on this site aren't as educated and as intelligent as you'd like them to be.
- dmurray 8y agoAnd 6x is a good upper bound (achieved by a cube). It's nice that the correct answer is halfway between these.
- fred256 8y agoI guess that only works if you are considering convex shapes (which sphere and cube obviously are)
- empath75 8y agoThe cube surface area is 4 times the average size of the shadow across all rotations, according to the video.
- rocqua 8y agoIn fact, the video states this for all convex shapes. I've been trying something similar for 2D, but there it doesn't quite seem to hold. Consider a very thin rectangle of size 1 by epsilon. Then it has circumference 2 (ignoring the epsilon). The shadow it casts at angle phi has size |sin phi|. Now, if we average |sin phi| from 0 to 180 degrees, (or 0 to 360 or 0 to 90) we get (2 / pi). I haven't checked whether this average holds for things other than thin rectangles, but I'd imagine so. I then find it weird we get a trancendental number in 2D but an integer in 3D.
- pervycreeper 8y agoNot a surprise to see pi in there really, since we're averaging over "surfaces" of circles/spheres. In general (spoiler), it does generalize to arbitrary dimensions. We get a rational factor for odd dimensions, and 1/pi * rational for even. See sections 6-9 here for demonstration: https://arxiv.org/pdf/1109.0595.pdf https://arxiv.org/pdf/1109.0595.pdf
- 00N8 8y agoand Graham's number is a correct but very weak upper bound -- as usual.