Timeline for Compute the Centroid of a 3D Planar Polygon
Current License: CC BY-SA 2.5
20 events
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| Mar 10, 2017 at 9:42 | history | edited | CommunityBot |
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| Oct 5, 2016 at 6:28 | answer | added | Max Mammel | timeline score: 0 | |
| Nov 21, 2014 at 12:34 | history | edited | user9072 |
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| Nov 7, 2011 at 19:12 | comment | added | Joseph O'Rourke | You need to define what "this hexagon" is. If you triangulate it, then you could compute the centroid of each triangle, and then the centroid of those area-weighted centroids. | |
| Nov 7, 2011 at 17:53 | comment | added | user19087 | Related to this topic: what if the points do not lie all in the same plane ? I need to calculate the center of an hexagon. However, the hexagon in not planar (it is a 3D object, a distorted hexagon). Can the above formulas be exended to calculate the centroid of this hexagon ? | |
| Aug 1, 2010 at 11:26 | vote | accept | Graviton | ||
| Jun 16, 2010 at 12:49 | answer | added | Joseph O'Rourke | timeline score: 6 | |
| Jun 15, 2010 at 22:26 | comment | added | JBL | Perhaps one of the people who have given the complete answer in comments could give the answer as an answer, so it can be accepted and this can stop being bounced up to the top? | |
| May 4, 2010 at 20:30 | comment | added | TonyK | To subtract from what Suresh says, you don't need to rotate anything. Just project onto the (x,y)-plane to get the x- and y-coordinates, and project onto the (x,z)-plane to get the z-coordinate. | |
| Apr 6, 2010 at 19:07 | answer | added | Faken | timeline score: 0 | |
| Mar 18, 2010 at 20:18 | comment | added | Suresh Venkat | to add to what Douglas says, since the points are coplanar, just find that plane and rotate the object so it's in the x,y plane, and proceed as before | |
| Mar 4, 2010 at 18:35 | answer | added | Quimey | timeline score: 2 | |
| Mar 4, 2010 at 6:43 | comment | added | Douglas Zare | Then the centroid sits over the centroid of its projection to any plane. So, you can use the formulas for two coordinates to compute each coordinate for the centroid. | |
| Mar 4, 2010 at 3:47 | history | edited | Graviton | CC BY-SA 2.5 |
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| Mar 4, 2010 at 3:33 | history | edited | Graviton | CC BY-SA 2.5 |
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| Mar 4, 2010 at 3:33 | comment | added | Graviton | @Michael, all the points are coplanar | |
| Mar 4, 2010 at 3:33 | comment | added | Graviton | @David, yes, all vertices are required to lie on the same plane | |
| Mar 4, 2010 at 3:16 | comment | added | Michael Lugo | I don't think this question is well-defined unless all your points are coplanar. In that case you'd probably just want to project down onto R^2. | |
| Mar 4, 2010 at 3:14 | comment | added | David Eppstein | Before asking for the centroid of a surface bounded by a polygonal cycle, you should define more clearly what the surface is. Are all the vertices required to lie on the same plane as each other, so you really just have a 2d polygon embedded into 3d? Or are you trying to describe non-flat 3d surfaces? | |
| Mar 4, 2010 at 3:07 | history | asked | Graviton | CC BY-SA 2.5 |