Timeline for Advanced view of the napkin ring problem?
Current License: CC BY-SA 2.5
13 events
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| Sep 30, 2013 at 22:47 | comment | added | Timothy Chow | This is a really nice argument. So if I understand correctly, instead of a circle, one could take any smooth simple closed curve whose tangent vector rotates monotonically, and sweep out a solid by sliding a semicircular disk around the curve, always keeping the semicircular disk perpendicular to the plane of the curve and in the plane of the tangent vector. | |
| Sep 1, 2010 at 22:24 | comment | added | Michael Hardy | Of course, if you really wanted to know how much gold is in a wedding ring, you might start by doing the two things Archimedes did: weight it, and see how much water it displaces. | |
| Sep 1, 2010 at 21:43 | history | edited | David Eppstein | CC BY-SA 2.5 |
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| Sep 1, 2010 at 17:29 | comment | added | sleepless in beantown | The half-disc of radius $h$ is the limiting area swept per $d\theta$ in both the sphere of radius $r$ and in David Eppstein's elegant construction. | |
| Sep 1, 2010 at 17:27 | comment | added | sleepless in beantown | @David, Beautiful construction! @Victor, take the limit as $\Delta \theta \to 0$, for either David's construction illustrated above, or for the volume of a sphere of radius $h$ $$V=\int_{\theta=0}^{\theta=2\pi} \frac{1}{2}\pi r^2$$ and notice that when the integration is done over the angle $\theta$, then the inner radius does not play a role in the integration. | |
| Sep 1, 2010 at 17:08 | comment | added | Michael Hardy | @David: Of course all half-disks have the same shape, but isn't the relevant fact here that in this case they have the same size? | |
| Sep 1, 2010 at 15:23 | comment | added | David Eppstein | The instantaneous movement of the center of the disk is parallel to the disk and doesn't affect it's volume. The only remaining component of the instantaneous motion is the rotation of the plane of the disk, and that's the only component that affects the volume, but it only depends on the turning angle and not the sphere radius. | |
| Sep 1, 2010 at 8:07 | comment | added | Victor Protsak | Even after all the clarifications, I have a hard time understanding why is the instantaneous volume a spherical wedge, since both the diameter of the half-disk and the plane of the disk are rotated simultaneously. | |
| Sep 1, 2010 at 0:40 | comment | added | David Eppstein | Draw a stick connecting the center of the sphere to a point halfway from top to bottom on the inner surface of the hole. Attach the disk perpendicularly to the stick. Rotate the stick around the equator keeping the disk attached to it. By "the outer sphere" I mean the sphere that the hole is drilled through. By "the shape of the half-disk is independent" I meant, up to congruence. Obviously all half-disks have the same shape up to similarity. | |
| Aug 31, 2010 at 23:54 | comment | added | Michael Hardy | You wrote "The shape of the half-disk is independent of the outer radius." Could it be that you meant "The size of the half-disk is independent of the outer radius."? | |
| Aug 31, 2010 at 23:53 | comment | added | Michael Hardy | To speak of "the outer sphere" seems to presuppose that one sphere is inside another, so the latter is "the outer sphere". Is that what you meant? It took me a while to reach the point where I felt I knew what you're (probably?) saying. If I were going to present it to a class of high-school students, I'd expect them to find the version in the Wikipedia article clearer. | |
| Aug 31, 2010 at 22:46 | comment | added | Joseph O'Rourke | @David: May I ask you to explain the sweep more precisely? Are the half-disks swept parallel to themselves, or are they rotating on a fixed diameter? Perhaps I am having difficulty with the phrase "bounded by semicircles on the outer sphere," but I am not seeing the picture you are painting. No doubt my fault... | |
| Aug 31, 2010 at 21:04 | history | answered | David Eppstein | CC BY-SA 2.5 |