Timeline for Plotting implicitly-defined space curves
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 17, 2015 at 4:57 | history | edited | m_goldberg | CC BY-SA 3.0 |
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| Dec 30, 2013 at 19:30 | history | made wiki | Post Made Community Wiki by Daniel Lichtblau | ||
| Sep 17, 2012 at 8:33 | comment | added | Yves Klett | As rightly you should be keeping all the geometric goodies to yourself. There should be a spelunking tag… | |
| Aug 30, 2012 at 0:45 | history | edited | J. M.'s missing motivation♦ | CC BY-SA 3.0 |
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| May 25, 2012 at 4:29 | vote | accept | J. M.'s missing motivation♦ | ||
| May 24, 2012 at 14:22 | comment | added | Daniel Lichtblau | @J. M. Thanks for the updates. I seem to be getting an awful lot of upvotes for code that was never of my own devising, and not notably better (that I can tell) than that in the other response. Embarrassed am I. | |
| May 24, 2012 at 5:49 | comment | added | J. M.'s missing motivation♦ |
I'll just note that if you take a look at the InputForm[] of the output of Daniel's version, the Polygon[] objects representing the isosurfaces are still there, just transparent. One could of course do something like ContourPlot3D[{(x^2 + y^2 + z^2 + 8)^2 - 36 (x^2 + y^2), y^2 + (z - 2)^2 - 4}, {x, -4, 4}, {y, -4, 4}, {z, -2, 2}, BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> {}}, ContourStyle -> None, Mesh -> None] (as Vitaliy comments) if one just wants the curves themselves. So many unused points, though...
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| May 24, 2012 at 3:01 | comment | added | rcollyer |
Nice! I find it interesting, though, that the docs do not say that the Filling format specification can be applied to BoundaryStyle. Very useful.
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| May 24, 2012 at 1:28 | history | edited | rm -rf♦ | CC BY-SA 3.0 |
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| May 23, 2012 at 22:28 | comment | added | Vitaliy Kaurov |
+1 Magical ;-) {1,2}->Green means intersection (boundary) between surface 1 and 2 will be green. Here is a minimal set of options to make it work: ContourPlot3D[{(x^2 + y^2 + z^2 + 8)^2 - 36 (x^2 + y^2), y^2 + (z - 2)^2 - 4}, {x, -4, 4}, {y, -4, 4}, {z, -2, 2}, ContourStyle -> None, Mesh -> None, BoundaryStyle -> {1 -> None, 2 -> None, {1, 2} -> Green}]
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| May 23, 2012 at 20:47 | history | answered | Daniel Lichtblau | CC BY-SA 3.0 |