Timeline for Light inside a polyhedron
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Feb 22, 2020 at 17:05 | comment | added | Dr. Richard Klitzing | That sentence "Each edge (boundary between faces) on S corresponds to an edge of P, not necessarily between the relevant faces (possibly between the nearer face and an invisible face)" of @user25199 is wrong, cf. the examples porvided in my answer: There might be edges of P both of its incident faces are visible, but their Images on S will not be connected at an common boundary (on S)! | |
| Jun 11, 2014 at 17:02 | comment | added | user25199 | @Mostafa, Yes, but in that case, the edge in $P$ has four ends in $S$, each of which must correspond to a different edge in $P$ as they are not part of the same great circle(s) in $S$. We have still not divided $S$ into parts with maximum angle less than $\pi$ so a sixth edge (at least) is required. | |
| Jun 11, 2014 at 9:33 | comment | added | Mostafa - Free Palestine | A gap in your proof: There might be two edges of $S$ which are projections of parts of the same edge in $P$. | |
| Jun 11, 2014 at 6:39 | history | answered | user25199 | CC BY-SA 3.0 |