Timeline for A lady and a monster
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Apr 5, 2011 at 21:57 | history | edited | user8268 | CC BY-SA 2.5 |
added 2 characters in body; edited body
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| Apr 5, 2011 at 21:45 | comment | added | joriki | Nice solution, by the way -- a monster as a source of fictitious forces acting in the moving frame of reference of a lady :-) | |
| Apr 5, 2011 at 21:38 | history | edited | user8268 | CC BY-SA 2.5 |
added 4 characters in body; deleted 82 characters in body
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| Apr 5, 2011 at 21:36 | comment | added | user8268 | @joriki: ah, silly me; of course you are right | |
| Apr 5, 2011 at 21:28 | comment | added | joriki | Yes, it's wrong -- the slope is correct, but $\mathrm{d}r/\mathrm{d}\phi$ is $r$ times that slope. $dr/d\phi=r dx/dy$. | |
| Apr 5, 2011 at 21:21 | comment | added | user8268 | @joriki: I added a comment why the constraint is as I claimed - is it wrong? | |
| Apr 5, 2011 at 21:19 | history | edited | user8268 | CC BY-SA 2.5 |
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| Apr 5, 2011 at 21:04 | comment | added | joriki | I take that back -- Henry just happened to switch two digits in his answer -- the correct answer from his method is 4.6033, and that's also what you get from your method if you include the factor of $r$. That yields $\int_{1/k}^1\sqrt{k^2-1/r^2}\mathrm{d}r=\sqrt{k^2-1}+\arctan(1/\sqrt{k^2-1})-\pi/2<\pi$. | |
| Apr 5, 2011 at 20:50 | comment | added | joriki | @user8268: You seem to be missing a factor $r$ in your calculation of the constraint for $\mathrm{d}r/\mathrm{d}\phi$? However, that doesn't quite bring your result into agreement with Henry's, so it may just be a mistake on my part. | |
| Apr 5, 2011 at 20:09 | history | edited | joriki | CC BY-SA 2.5 |
typo
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| Apr 5, 2011 at 19:58 | comment | added | joriki | Wow, this did get quite interesting after all! I thought I'd checked Henry's answer; now I'm quite keen to see who turns out to be right... | |
| Apr 5, 2011 at 19:44 | history | answered | user8268 | CC BY-SA 2.5 |