Timeline for Uninteresting questions with interesting answers
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 26, 2015 at 8:52 | comment | added | Fred Daniel Kline | @bubba, Carpenters call it kerf and they must account for it to build things. Sawmills do the same to determine board-feet of a log. The volume of this kerf is the sawdust. | |
| Mar 25, 2015 at 20:30 | comment | added | Michael | @Wojowu: Step 1: The result for $C$ being a circle follows directly from Pythagoras Theorem: Basically, $L/2$ is one side of the triangle, the ray from the center of $C$ to the middle of $L$ is the other side, and the radius is the hypotenuse. Step 2: "morally conclude" the case of a general closed curve $C$ by considering at every point $P$ of $C$ the tangent circle $C'$ of the same curvature as $C$. The infinitesimal amount of area swept by $L$ at $P$ is virtually the same for $C$ and for $C'$. Since that amount does not depend on the radius of $C'$ it's straightforward to conclude the rest | |
| Mar 25, 2015 at 0:04 | comment | added | zeb | This looks like Mamikon's "visual calculus": its.caltech.edu/~mamikon/VisualCalc.html | |
| Mar 24, 2015 at 17:24 | comment | added | Joe Silverman | @Wojowu It was assigned as a homework problem when I took undergrad differential geometry class, at which time I knew who it was named after. But I'm afraid I've forgotten even that piece of information. And I don't know any references. Sorry. | |
| Mar 24, 2015 at 16:55 | comment | added | Wojowu | What are some references for this result? | |
| Mar 24, 2015 at 16:46 | comment | added | Joe Silverman | @RobinSaunders Probably it's fine for non-convex curves with signed area, as you say. It's also okay for piecewise smooth curves. In any case, the proof is a lovely application of Green's theorem, and from the proof one should be able to see how to formulate it more generally than I did in the answer that I gave. But note, for example, if $L$ is too long and $C$ is too small, it may not be possible to slide $L$ around inside $C$, so some restrictions are necessary. | |
| Mar 24, 2015 at 16:44 | comment | added | Joe Silverman | @WillieWong Yes, $C$ is supposed to be closed. | |
| Mar 24, 2015 at 16:17 | comment | added | Robin Saunders | Is there a formulation involving signed area for nonconvex curves, or is the convexity really essential? | |
| Mar 24, 2015 at 16:04 | comment | added | Willie Wong | I am guessing that $C$ is supposed to be closed? | |
| Mar 24, 2015 at 16:03 | comment | added | Michael Hardy | There seem to be lots of problems in geometry, including this one, and the one in my posted question and the napkin ring problem, and others that escape me at the moment, that seem essentially the same as each other in all respects except the specifics. ${}\qquad{}$ | |
| Mar 24, 2015 at 12:17 | comment | added | bubba | Interesting, indeed. Though I know it might not matter much to folks here, this has a real-world application, too. If the line segment represents a cutting tool, the area is the material removed by the tool motion. Understanding rates of material removal is very important in manufacturing. References, please?? | |
| Mar 24, 2015 at 11:57 | comment | added | Hachino | That looks very surprising. Is anyone aware of any 'fundamental' reason which could explain this invariance phenomenon ? | |
| S Mar 24, 2015 at 11:45 | history | answered | Joe Silverman | CC BY-SA 3.0 | |
| S Mar 24, 2015 at 11:45 | history | made wiki | Post Made Community Wiki by Joe Silverman |