Timeline for Interpolating points with minimum curvature constraint
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 10, 2017 at 10:58 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Image links broken; now fixed.
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| Feb 28, 2012 at 12:51 | comment | added | Joseph O'Rourke | Concerning the box constraint: It seems three points close to three distinct corners forces high curvature, as only two of those points can be curve endpoints. | |
| Feb 28, 2012 at 12:50 | comment | added | Joseph O'Rourke | @Scott: You are right! Added a corresponding image. @Ryan: Remaining inside $R$ eliminates many options. @Will: Thanks for the relevant link! Their focus is aesthetics, but still there are many ideas in there I might use. | |
| Feb 28, 2012 at 12:46 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 244 characters in body
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| Feb 28, 2012 at 9:39 | comment | added | Ryan Budney | Ah, I missed your non-self-intersecting constraint. | |
| Feb 28, 2012 at 9:38 | comment | added | Ryan Budney | Is it really pointwise curvature you want to minimize, or total curvature (the integral of the pointwise curvature wrt arc length)? If it's just pointwise curvature, why not just connect up pairs of points on great circles passing through infinity on the complex plane, and then use very gradual spirals to merge one great circle into another? Do you want the curve to be embedded or is immersed enough? Because I think if you're okay with immersed curves and are only constraining the pointwise max curvature, you should be able to approximate 0 curvature. | |
| Feb 28, 2012 at 5:56 | comment | added | Will Jagy | take a look at levien.com/phd/LevienSequinCAD09_014.pdf | |
| Feb 28, 2012 at 3:48 | comment | added | S. Carnahan♦ | It looks to me like the ordering 5-4-2-1-3 could yield a lower curvature path. In particular, it might be best to avoid points of inflection whenever possible. | |
| Feb 28, 2012 at 2:17 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Illustrating Anton's annulus idea.
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| Feb 28, 2012 at 2:14 | comment | added | Joseph O'Rourke | @Garabed: Thanks for those related MO questions. Yes, it is the rectangle that is spoiling that approach. | |
| Feb 27, 2012 at 19:44 | comment | added | Garabed Gulbenkian | You might be interested in questions 19074 and 22601 as well as some of the answers given to them even though the sort of minimizing curve sought for in these questions is of the "y=f(x)" type. With regard to your question, it occurs to me that if your curve C was the arc of a spiral enclosing a sufficiently large portion of the Euclidean plane, the maximum absolute value of its curvature at any point might be made arbitrarily small while all your n points could still be points of C. Of course this would violate your condition (c) unless your rectangle R was large enough. | |
| Feb 27, 2012 at 16:31 | comment | added | Joseph O'Rourke | Oh, OK--Nice! $\mbox{}$ | |
| Feb 27, 2012 at 14:50 | comment | added | Anton Petrunin | Ups, I wanted to say $\kappa_\max\le \tfrac1r$. | |
| Feb 27, 2012 at 12:29 | comment | added | Joseph O'Rourke | @Anton: Thanks! However, if the points in the annulus happen to lie on a nearly collinear arc, they could be captured by that low-curvature arc. | |
| Feb 26, 2012 at 21:21 | comment | added | Anton Petrunin | Note that if your points lie in an annulus $B_x(R)\backslash B_y(r)$ then $\kappa_{{\max}}\ge \tfrac1r$. | |
| Feb 26, 2012 at 19:05 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |