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Apr 14, 2021 at 16:17 vote accept April
Apr 14, 2021 at 3:07 answer added Willie Wong timeline score: 5
Apr 13, 2021 at 21:45 comment added Mateusz Kwaśnicki As long as the curvature of $\gamma$ is less than something of the order $1/r$, and there are no "overlaps", the area of the "sausage" $\gamma_{+r}$ will be equal to $2 r L + \pi r^2$, and it is intuitively clear this is the maximal value. I guess this is standard, I vaguely remember having read that somewhere, but unfortunately I do not have a reference.
Apr 13, 2021 at 19:29 history edited April
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Apr 13, 2021 at 19:24 history asked April CC BY-SA 4.0