Timeline for Two-dimensional catenary in Mathematica
Current License: CC BY-SA 4.0
32 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 11 at 9:00 | history | edited | user21 | edited tags | |
| Mar 30, 2019 at 5:02 | history | edited | J. M.'s missing motivation | CC BY-SA 4.0 | don't call it an "improvement" if you are going to introduce a typo |
| Mar 30, 2019 at 4:57 | history | edited | user64494 | CC BY-SA 4.0 | The title is improved. |
| Mar 30, 2019 at 4:51 | history | edited | J. M.'s missing motivation | edited tags | |
| Nov 17, 2018 at 5:56 | history | edited | Brandon | CC BY-SA 4.0 | added 78 characters in body |
| Aug 21, 2018 at 17:53 | comment | added | Brandon | @xzczd uploaded pic | |
| Aug 21, 2018 at 17:52 | vote | accept | Brandon | ||
| Aug 21, 2018 at 17:52 | history | edited | Brandon | CC BY-SA 4.0 | added 655 characters in body |
| Aug 16, 2018 at 14:17 | comment | added | Brandon | @Henrik, I couldn’t come up with a 2D analog of what a 1D chain does. A gold necklace is slinky, and the changes in distance between hinge fulcrums is generally fixed as compared to those of a sheet of chain mail, local portions of which will go from square to trapezoidal under shear. No shear should occur in the weight loading of a 2D catenary some (i.e., by magically bringing it from outer space to the surface of a planet flatly). The chamois cloth then stretches, which a necklace (or other chain) doesn’t do when it’s used to trace a catenary for a masonry arch. | |
| Aug 16, 2018 at 7:51 | history | edited | xzczd♦ | CC BY-SA 4.0 | The picture is formatted properly now, I think :) |
| Aug 15, 2018 at 7:32 | comment | added | Henrik Schumacher | IMHO, one has to add a stretching term to the total energy of the system and the gravitational energy term should read $\int_\varOmega u \, \operatorname{d} \!x$ (it has to involve only the area element of the surface in rest state; it must not dependend of the area element of the deformed surface). | |
| Aug 15, 2018 at 7:30 | comment | added | Henrik Schumacher | The 2D case is more sophisticated as the 1D case because one cannot work with isometric deformations. I don't think that the model elaborated in mathoverflow.net/a/69837 is physically meaningful. In particular, the area constraint is nonsense: It is precisely the (local!) change of area that produces the stretch forces that counter the gravitational forces. So it's no wonder that the physical experiments do not fill well to simulated data. | |
| Aug 15, 2018 at 6:01 | history | tweeted | twitter.com/StackMma/status/1029609030346567680 | ||
| Aug 14, 2018 at 17:05 | history | edited | Brandon | CC BY-SA 4.0 | added 255 characters in body |
| Jul 3, 2018 at 18:05 | history | edited | Brandon | CC BY-SA 4.0 | update on fabrication |
| Jun 21, 2018 at 21:52 | comment | added | Brandon | @mariusz - RegularPolygon works in other problems - documentation here reference.wolfram.com/language/ref/RegularPolygon.html | |
| Jun 21, 2018 at 21:05 | history | edited | Brandon | CC BY-SA 4.0 | make words sound more smarter |
| Jun 21, 2018 at 20:52 | history | edited | Brandon | CC BY-SA 4.0 | Response added |
| Jun 21, 2018 at 20:46 | history | edited | Brandon | CC BY-SA 4.0 | image of chamois cloth added |
| Jun 21, 2018 at 20:40 | comment | added | Brandon | @xzczd I did mean it and took a stab at it - see edited post; looks like more folks have chimed in with more skill, but at least I learned how to properly use Div and Grad. | |
| Jun 21, 2018 at 17:33 | history | edited | xzczd♦ | edited tags | |
| Jun 21, 2018 at 17:17 | comment | added | xzczd♦ | @mariusz probably means there's no built-in function named as RectangularPolygon… do you mean RegularPolygon at that position? | |
| Jun 21, 2018 at 17:16 | answer | added | xzczd♦ | timeline score: 8 | |
| Jun 21, 2018 at 15:36 | history | edited | Brandon | CC BY-SA 4.0 | Translation added that is closer to the problem statement originally found. |
| Jun 21, 2018 at 15:32 | comment | added | xzczd♦ | Do you mean that, you just want to solve this equation as a simplified one at the moment? | |
| Jun 21, 2018 at 15:15 | comment | added | Brandon | @Mariusz: Lambda is a Lagrange multiplier; rectangular polygon herein specified with 4 vertices is a cute way to require a square domain (similar to Disk). | |
| Jun 21, 2018 at 15:11 | comment | added | Brandon | @xzczd: I'll work to transcribe the PDE as-written to make this more fun to follow in the future. | |
| Jun 21, 2018 at 11:26 | comment | added | xzczd♦ | Your translation for the PDE in the link is wrong, the $\sqrt{1+|\nabla u|^2}$ term cannot be cancelled, notice one of them is inside $\nabla\cdot{}$. | |
| Jun 21, 2018 at 8:16 | history | edited | Kuba | CC BY-SA 4.0 | deleted 4 characters in body |
| Jun 21, 2018 at 8:04 | history | edited | xzczd♦ | edited tags | |
| Jun 21, 2018 at 7:56 | history | migrated | from math.stackexchange.com (revisions) | ||
| Jun 21, 2018 at 0:56 | history | asked | Brandon | CC BY-SA 4.0 |