| bio | website | |
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| location | ||
| age | 27 | |
| visits | member for | 1 year |
| seen | 2 days ago | |
| stats | profile views | 326 |
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Dec 12 |
awarded | Nice Answer |
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Dec 12 |
comment |
Movable text on a curve @AlexeyPopkov If you know what to do with the kerning, using text shouldn't be a problem. However you might need to convert to filled curves and do some cleaver integration between characters to figure out the kerning. Independent of which you use to draw the symbols, I don't think you'll be able to select the text without extensive workarounds. Since they are individually positioned pieces of text. |
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Dec 12 |
awarded | Civic Duty |
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Dec 12 |
comment |
Movable text on a curve @Mr.Wizard Kerning, the bane of all. Though in seriousness, it might not even be enough to find equidistant spacings unless you are dealing with monospaced fonts, and even if the normal kerning of the font could be accurately captured, it might not look good when curvature is taken into account. |
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Dec 12 |
revised |
Movable text on a curve added 46 characters in body |
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Dec 12 |
answered | Movable text on a curve |
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Dec 12 |
comment |
Mathematica not simplifying square root expressions even with assumptions @murray doesn't Surd[x,2] fit that description? Or am I misunderstanding you. |
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Dec 12 |
revised |
Quirk in VertexColors interpolation when displaying Polygon added 43 characters in body |
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Dec 12 |
comment |
Quirk in VertexColors interpolation when displaying Polygon @MarkMcClure Strictly speaking polygon triangulation is done without adding vertices. I think methods adding points are refereed to as meshing, or equivalently a triangulation but of a modified set of points which is then confusingly called constrained triangulation. Besides this, it looks like triangle does indeed implement useful routines. Definitely a program worth playing around with. Thanks |
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Dec 12 |
comment |
Quirk in VertexColors interpolation when displaying Polygon @MarkMcClure A triangulation shouldn't introduce new vertices, unless you mean to resolve self intersections simultaneously. In such cases I would suspect it would be trivial to just do a linear interpolation at the intersection point. |
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Dec 12 |
revised |
Quirk in VertexColors interpolation when displaying Polygon added 269 characters in body |
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Dec 12 |
answered | Mathematica not simplifying square root expressions even with assumptions |
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Dec 12 |
revised |
Is there a way to continuously update only *some* variables in Manipulate? added 200 characters in body |
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Dec 12 |
comment |
Is there a way to continuously update only *some* variables in Manipulate? @Mr.Wizard Indeed, and it seems like that was also what I used in my testing. When under the same Dynamic slow[b] is called again because it updates both and not just the one depending on a. Thanks for the catch. |
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Dec 12 |
answered | Quirk in VertexColors interpolation when displaying Polygon |
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Dec 12 |
comment |
Quirk in VertexColors interpolation when displaying Polygon So really the problem is that it's doing a less then optimal triangulation which actually seems to result in zero area triangles at times. Is there any documentation or information about which method is used to break it down? Perhaps someone knowledgeable in triangulation algorithms would know, Afaik Delaunay triangulation avoids these effects. |
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Dec 12 |
comment |
Quirk in VertexColors interpolation when displaying Polygon @ruebenko There is only one polygon. It happens to be oriented counter clockwise, but the same behavior is present independent of orientation. |
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Dec 12 |
answered | Is there a way to continuously update only *some* variables in Manipulate? |
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Dec 12 |
asked | Quirk in VertexColors interpolation when displaying Polygon |
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Dec 11 |
comment |
Implementing Position In the last solution you should use MatchQ[#1, x] not #1==x. Position works on patterns not just strict equality. |