Timeline for Why do clouds have well-defined boundaries?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 13, 2022 at 14:59 | comment | added | user66043 | @StevenSagona I was thinking in terms of entrainment and detrainment. Then NS equations as applied to cloud dynamics. | |
| Aug 13, 2022 at 14:04 | comment | added | Steven Sagona | @gansub, sad truth is that most of highly accepted answers are usually like this... a fancy looking equation and literal nonsense that backs it up -- most people just skim it and are like "yeah okay that makes sense" | |
| Aug 13, 2022 at 8:52 | comment | added | user66043 | @StevenSagona Exactly. I am a researcher in meteorology and this answer is so wrong and not even sure how it got accepted. | |
| Jul 15, 2022 at 13:09 | comment | added | Steven Sagona | I cannot follow the argument at all. Like fine, the scatting scales to the 6th power of the droplet diameter, but what does that have to do with shapes of clouds at all? I think for this argument to make sense, you have to say that clouds have some decreasing droplet size as you are further away from the cloud. But why would this be true? I think it seems more sensible that the density (number of droplets per area) would be the thing that will change across the area of the cloud, not the droplet size. | |
| Jul 13, 2022 at 14:56 | vote | accept | Roger V. | ||
| Jun 20, 2022 at 18:27 | comment | added | nanoman | ... Since total incoherent scattering from droplets of any given size is directly proportional to their density, the brightness of a small volume should be your Rayleigh scattering formula convolved with the droplet size distribution. Now, your observation about the $d^6$ factor implies that this is dominated by the largest droplets. But this just reduces the question to: How does the density of the largest droplets vary over space? And I'm not clear how you have reached the conclusion that this variation is sudden. | |
| Jun 20, 2022 at 18:26 | comment | added | nanoman | This argument isn't completely clear. You refer to a "smooth gradient of droplet sizes". However, the general situation is that in any given small (mesoscopic) volume, there is a distribution of droplet sizes, and this distribution can expected to vary smoothly over space. You seem to imply that the distribution is narrow (sharply peaked at a particular size), and varies over space mainly by shifting the "location" of its peak (causing the distribution itself to rise and fall rapidly) -- in which case, why is that? ... | |
| Jun 20, 2022 at 18:23 | comment | added | Anders Sandberg | @Ruslan - Note that this is the meaning of the previous sentence. Given that the question wasn't about the finer points of physical approximation theory, I think this update would be overly pedantic and only confuse the reader. | |
| Jun 20, 2022 at 8:59 | comment | added | Ruslan | "for smaller droplets Rayleigh scattering is the best approximation" — this is misleading. Such claims should always note that Mie solution is the universal solution for scattering of waves by spherical particles, regardless of particle size. It works fine (with full precision) in Rayleigh regime, as well as for liquid-water clouds and raindrops. The only reason why Rayleigh description is preferred in its domain is that it's much simpler and works well enough (even though it's not "exact") in case of very small scattering particles such as air molecules. | |
| Jun 19, 2022 at 23:13 | comment | added | uhoh | you may want to have a look at How much blue sky comes from each altitude; dependence of atmosphere's dielectric constant and isothermal compressibility with altitude? as well | |
| Jun 19, 2022 at 8:50 | history | answered | Anders Sandberg | CC BY-SA 4.0 |