### funky donut

Posted: **Fri Jun 01, 2007 6:58 pm UTC**

by **4=5**

so, I was sitting in class yesterday (which happens to be in the attic of school on a hot summer day) and I noticed the infectual ocilateing fan

it got me to thinking that if you had a circle of x diameter and you mounted a line n long at the center, at pi/2 radians that would happen if you spun the thingymajigger around the unattached tip of the the line so the line would be come a flat disk, now ignoring the line, what would the space that the circle had travled through look like, would it be a doughnut made of elipses? or would it be something diffrent because the circle was rotated through it instead of just being passed through it at an angle

Posted: **Fri Jun 01, 2007 7:09 pm UTC**

by **Woxor**

That's an interesting question. If I understand your construction correctly, you're talking about tying a string to the center of a 2-D circular disk and then spinning it around by the string, to where the circle's face is pointed towards the center at all times. In that case, I think the volumetric shape that you would get is the same as if you took an ellipse, cut it in half down the vertical, and rotated that. So it would be a doughnut whose cross-section was half of an ellipse, with the flat part on the inside. I'm not 100% sure whether it would be an elliptical shape exactly rather than a similar shape, though.

To see this, just fix a cross-section and mentally trace what parts of the cross-section end up intersecting the disc as it passes through. It becomes obvious that the inner-most part of the cross section is flat, since that's where the disk passes through perpendicularly.

Posted: **Fri Jun 01, 2007 7:13 pm UTC**

by **4=5**

Woxor wrote:That's an interesting question. If I understand your construction correctly, you're talking about tying a string to the center of a 2-D circular disk and then spinning it around by the string, to where the circle's face is pointed towards the center at all times. In that case, I think the volumetric shape that you would get is the same as if you took an ellipse, cut it in half down the vertical, and rotated that. So it would be a doughnut whose cross-section was half of an ellipse, with the flat part on the inside. I'm not 100% sure whether it would be an elliptical shape exactly rather than a similar shape, though.

To see this, just fix a cross-section and mentally trace what parts of the cross-section end up intersecting the disc as it passes through. It becomes obvious that the inner-most part of the cross section is flat, since that's where the disk passes through perpendicularly.

ok thanks that makes perfect sense,