Timeline for Billiard dynamics with angle of reflection a fraction of angle of incidence
Current License: CC BY-SA 3.0
11 events
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| Oct 24, 2017 at 11:12 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 110 characters in body
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| Jun 9, 2014 at 12:24 | vote | accept | Joseph O'Rourke | ||
| Jun 9, 2014 at 7:39 | answer | added | user25199 | timeline score: 6 | |
| Jun 8, 2014 at 1:27 | comment | added | Joseph O'Rourke | @BenCrowell: Nice question! Maybe the right model is that, for incident $\theta \in [0,\pi/2]$, the reflected ray has $\theta' = \alpha \theta + (1-\alpha) 0$ for smaller reflected angles, and $\theta' = \alpha \theta + (1-\alpha) \pi/2$ for larger reflected angles. So $\theta=88^\circ$ could reflect to $\theta' = 89^\circ$ with $\alpha=\frac{1}{2}$. | |
| Jun 8, 2014 at 0:50 | comment | added | user21349 | When you time-reverse the dynamical law governing this system, you get $\lambda\rightarrow 1/\lambda$, in Lucia's notation. What happens when you extrapolate backwards? Lucia's analysis only applies to $\lambda<1$. For $\lambda>1$, does the ball get trapped sliding along two walls? It's not clear to me how the time-reversed dynamics are to be defined if, say, $\lambda=2$ and $\theta_n=\pi/2$; in this example, the ball has to break a symmetry. | |
| Jun 7, 2014 at 21:44 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 21 characters in body
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| Jun 7, 2014 at 19:15 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Added image to show it is not just the angles that are different, but the path itself converges.
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| Jun 7, 2014 at 15:41 | comment | added | Carlo Beenakker | an entire Ph.D. thesis devoted to the dynamics in a rotating circular billiard: etheses.dur.ac.uk/6228/1/6228_3583.PDF?UkUDh:CyT | |
| Jun 7, 2014 at 14:53 | comment | added | Carlo Beenakker | reflection from a rotating billiard would capture some the effect you are seeking; see, for example, diegofregolent.com/wp-content/uploads/2014/04/… | |
| Jun 7, 2014 at 14:32 | comment | added | Lucia | For any rectangular table, if the fraction is $\lambda<1$ then the angles tend to $\frac{\lambda}{\lambda+1}\frac{\pi}{2}$ and its complement alternately. This follows upon noting that $|\theta_n-\frac{\lambda}{\lambda+1}\frac{\pi}{2}|$ becomes the fraction $\lambda$ of its value at the next step. Note that the usual billiards for a rectangular table similarly conserves the first given angle and its complement. | |
| Jun 7, 2014 at 13:17 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |