Timeline for Closed curve whose neighborhood is as large as possible
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 13, 2015 at 5:07 | comment | added | Thomas Richard | Basically, you need $\varepsilon$ to be smaller than the injectivity radius of the normal exponential map. | |
| Oct 12, 2015 at 23:26 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Added another illustration.
|
| Oct 12, 2015 at 20:24 | comment | added | Joseph O'Rourke | @TomSolberg: Good point. Any sharp turn w.r.t. $\epsilon$ will cause "overlap" of the neighborhood and so lose area. I believe the radius of curvature should exceed $\epsilon$ to achieve $2 \epsilon L$. So that's a hypothesis: Any smooth curve (not necessarily convex) such that its radius of curvature exceeds $\epsilon$ (at every point) leads to area $2 \epsilon L$. | |
| Oct 12, 2015 at 20:19 | comment | added | Tom Solberg | But couldn't I approximate this with a smooth curve by rounding the corners by a tiny, tiny amount, and still retain the size of the neighborhood (in other words, does this really have anything to do with smoothness?) | |
| Oct 12, 2015 at 20:02 | history | answered | Joseph O'Rourke | CC BY-SA 3.0 |