Timeline for How far can water rise above the edge of a glass?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 19, 2014 at 15:30 | comment | added | Eph | 1. Shouldn't your second boundary condition read $z'(r)=tan\;\alpha$ as $z$ is a distance and $tan$ is a ratio/slope? 2. The water at the top of the glass generally goes vertical before overflowing. This would correspond to $z'=\infty$ so I don't thing the simplification $z'\approx 0$ is valid. | |
| Nov 26, 2012 at 17:53 | history | edited | Jaime | CC BY-SA 3.0 |
missing square root
|
| Nov 26, 2012 at 17:52 | comment | added | Jaime | Yup, square root missing everywhere, John... | |
| Nov 26, 2012 at 16:38 | comment | added | John Rennie | @jaime: I can't claim any credit for the height equation in my answer because I got it off Wikipedia. However I can claim credit for pointing out your equation for $\Delta h$ isn't dimensionally consistent :-) | |
| Nov 26, 2012 at 16:32 | comment | added | learner | Thanks, that's a great starting point! I don't have time to read your post in detail right now but I think your boundary conditions are incorrect, you should have $z'(0)=0$ by symmetry and therefore you can't impose $z(0)=0$ when adopting the convention $p_1=p_0$. So that would make the interface a $\cosh$ curve instead of a $\sinh$ curve. The exponential growth of $\cosh$ means we can't pretend $z(0)$ vanishes even if it is very small w.r.t. $\Delta h$! | |
| Nov 26, 2012 at 4:37 | history | answered | Jaime | CC BY-SA 3.0 |