Timeline for Characterizing a tumbling convex polytope from the surface areas of its two-dimensional projections
Current License: CC BY-SA 2.5
27 events
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| Dec 6, 2013 at 15:13 | comment | added | Olga | I looked more attentively at the answers and I got the volume question was asnwered negatively by Sergei Ivanov who found two prisms with different volumes but the same projections. | |
| Dec 6, 2013 at 14:14 | comment | added | Olga | Yesterday I discussed this question with my friend who has a big chevelure. He said that he likes to brush himself in a way that it occupies more volume. I remarked that of course the volume doesn't change. He answered that of course we look at the volume of convex hull. We are both mathematicians. Then it got started... So we were speaking if I can count the volume of the shock of hair by looking at the 2D proections. I'm happyI found this post.I see that we can have different haircuts with the same projections.But the question about the volume: it's still open, isn't it? Even for polytopes? | |
| Jul 9, 2010 at 12:45 | answer | added | Markus K | timeline score: 6 | |
| May 15, 2010 at 11:56 | vote | accept | Rob Grey | ||
| May 15, 2010 at 7:54 | answer | added | Sergei Ivanov | timeline score: 10 | |
| May 14, 2010 at 23:17 | comment | added | Sergei Ivanov | Sorry, of course multiply by 4 (as with the round sphere). | |
| May 14, 2010 at 22:53 | history | edited | Rob Grey | CC BY-SA 2.5 |
Added update on surface area calculation for convex polytope
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| May 14, 2010 at 22:41 | comment | added | Rob Grey | Sergei, apologies, but am I correct in believing that you would multiple the average projection area by 4$pi$ and not 4 to obtain the surface area of the convex polytope? | |
| May 14, 2010 at 21:23 | comment | added | Sergei Ivanov | If interpret your comment correctly, you mean randomness in the sense of probability theory (not in the sense that it is just unknown). If so, the average area over time equals (approximately, after a sufficiently long time) the average area over the direction of the projection in the space. Then you can find the surface area as in Nurdin's answer (multiply the average projection area by $4\pi$). | |
| May 14, 2010 at 20:49 | comment | added | Rob Grey | Given a max and min projective surface area, and given that the polytope is convex, I feel that the mean projective surface area could possibly be informative, but I would have to think more about this... | |
| May 14, 2010 at 20:42 | comment | added | Rob Grey | Sergei, 'random tumbling' means that no particular orientation of the polytope is favored and/or more likely than any other. Furthermore, I do not know the initial orientation of the polytope. Do you think this is too little information for extracting properties like surface area? | |
| May 14, 2010 at 20:27 | comment | added | Sergei Ivanov | The "random tumbling" part is confusing. Do you know which rotations of the polytope gave you particular projection areas? If not, do you know some probability distribution or you just know the set of possible areas? (In the latter case, you will only learn what is the maximum and minimum projection area, and this too little information). | |
| May 14, 2010 at 19:52 | history | edited | Rob Grey | CC BY-SA 2.5 |
deleted 9 characters in body; Post Made Community Wiki
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| May 14, 2010 at 19:41 | history | edited | Rob Grey | CC BY-SA 2.5 |
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| May 14, 2010 at 18:24 | history | edited | Rob Grey | CC BY-SA 2.5 |
Added further justification for physical example, made point about watching evolution of projections rather than unordered collection.
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| May 14, 2010 at 17:13 | history | edited | Rob Grey | CC BY-SA 2.5 |
Specified that the polytope should be convex
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| May 14, 2010 at 17:12 | comment | added | Rob Grey | Cam, thanks - I agree with you, and will update the question accordingly. | |
| May 14, 2010 at 17:06 | comment | added | Cam McLeman | I think it's likely best to restrict to convex polytopes. If I'm thinking about this correctly, you could symmetrically indent two opposite sides of a cube (two so as not to move the center of mass) and be unable to distinguish the resulting object from the cube based only on its shadows. | |
| May 14, 2010 at 15:32 | answer | added | Nurdin Takenov | timeline score: 3 | |
| May 14, 2010 at 15:19 | answer | added | Joseph O'Rourke | timeline score: 2 | |
| May 14, 2010 at 14:53 | history | edited | Rob Grey | CC BY-SA 2.5 |
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| May 14, 2010 at 14:48 | history | edited | Rob Grey | CC BY-SA 2.5 |
Noted that the polytope will not stabilize in a particular orientation and will continue to tumble randomly; edited body
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| May 14, 2010 at 14:42 | comment | added | Rob Grey | Joseph, thanks for your commment! But may I ask you to clarify what you mean by 'such polyhedra'? I'm thinking about simple characterizations / what you can learn about arbitrary polytopes (or perhaps, arbitrary convex polytopes to simplify things). | |
| May 14, 2010 at 14:30 | comment | added | Joseph Malkevitch | Such polyhedra are called unistable or monostatic, and there are some references here: en.wikipedia.org/wiki/Monostatic_polytope | |
| May 14, 2010 at 14:16 | history | edited | Rob Grey | CC BY-SA 2.5 |
Title change; added 20 characters in body
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| May 14, 2010 at 14:11 | history | edited | Rob Grey | CC BY-SA 2.5 |
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| May 14, 2010 at 14:05 | history | asked | Rob Grey | CC BY-SA 2.5 |