Timeline for Thales' semicircle theorem in higher dimensions
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 7, 2022 at 10:06 | answer | added | memorial | timeline score: 1 | |
| Apr 2, 2015 at 19:34 | answer | added | The Masked Avenger | timeline score: 1 | |
| Mar 31, 2015 at 12:17 | answer | added | report | timeline score: 2 | |
| Mar 31, 2015 at 6:22 | answer | added | echinodermata | timeline score: 6 | |
| Mar 28, 2015 at 23:37 | comment | added | Joseph O'Rourke | @DouglasZare: Indeed, this was a tough one for me, maybe because the solid angle is "nearly" constant; and in any case a nuanced geometric situation and computation. | |
| Mar 28, 2015 at 21:31 | comment | added | Douglas Zare | When people ask how to visualize things in higher dimensions, I think it is good to mention problems like this showing that we struggle to visualize things in $3$ dimensions. | |
| Mar 28, 2015 at 21:17 | answer | added | Douglas Zare | timeline score: 10 | |
| Mar 28, 2015 at 20:46 | vote | accept | Joseph O'Rourke | ||
| Mar 28, 2015 at 20:46 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Finish up, answered by Will.
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| Mar 28, 2015 at 14:16 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Added requested series of images.
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| Mar 28, 2015 at 14:13 | comment | added | Joseph O'Rourke | @TheMaskedAvenger: I added a "series that shows the projection onto the steradian sphere as the vertex goes from 90 degrees down to" $5^\circ$. | |
| Mar 28, 2015 at 14:10 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Added requested series of images.
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| Mar 28, 2015 at 2:43 | answer | added | Will Sawin | timeline score: 9 | |
| Mar 28, 2015 at 1:41 | comment | added | The Masked Avenger | I am trying a mental simulation where I attach a semicircle to a sphere of same radius, and then (while having the diameter of the semicircle fixed and tangent to the sphere) folding this like a flap and imagining the change in steradial projection. I am not getting a quadrant filling shadow but something else. I am withdrawing my earlier vote. | |
| Mar 28, 2015 at 0:57 | comment | added | The Masked Avenger | I think an enlightening picture would be a series that shows the projection onto the steradian sphere as the vertex goes from 90 degrees down to zero. You should see a circle elongate into an ellipse, and almost become but stay within a 1/4 wedge of the steradian sphere. | |
| Mar 28, 2015 at 0:53 | comment | added | The Masked Avenger | I vote no to Q1. Imagine the steradian sphere the same diameter as the sphere circumscribing the cone, with the cone rays extended to cut both spheres. As I bring the steradian sphere very close to the equator, I see the steradian increase to close to pi, since the cone vertex approximates the edge of a cube. If you can, do some computation with the vertex at 1 or 0.1 degrees off the horizon/equator. | |
| Mar 27, 2015 at 23:55 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |