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April 17

Crowley's ridge loess

How deep is the loess on Crowley's ridge and what kind of rock is under the loess?Hoover12345! (talk) 00:32, 17 April 2018 (UTC)[reply]

Our article (Crowley's Ridge) does not say, and the only ref that looks promising has a malformed URL. The authors fo that paper are Google Muhs and Bettis. a google on these names show them to be USGS scientists that have published extensively about last-glacial loess formations. -Arch dude (talk) 00:52, 17 April 2018 (UTC)[reply]

I have updated two refs in our article to have working open access URLs for the content. One is an archive, the other is at a different location. Although I didn't check the URL for which I found a new location for the content which is not behind a paywall, I strongly suspect it too was not malformed simply a dead but correctly formatted URL for its time, especially since it is a very similar USGS URL. (I have also added DOIs so the content can hopefully be more easily found, at least behind a paywall, should either URL stop working again.) Nil Einne (talk) 12:49, 17 April 2018 (UTC)[reply]

The map on page 26 of [1] shows a thickness of 5-10 meters. -Arch dude (talk) 00:59, 17 April 2018 (UTC)[reply]
It is underlain by the Lafayette gravel. See [2]. Found by googling for <geology of crowley's ridge>. If you can understand this stuff, please expand our article. -Arch dude (talk) 01:10, 17 April 2018 (UTC)[reply]

How much surface can you plate with one gram of gold

Using modern gold plating, how much surface (of glass or jewels) can you cover per gram of gold? I understand that gold can be spread over a really large surface. I just don't know how large it is. --Doroletho (talk) 13:21, 17 April 2018 (UTC)[reply]

According to this, gold leaf can be as thin as about 0.1 microns. Gold has a density of 19.33 g/cm³. That means that 1 gram of gold has a volume of 1/19.33 = 0.05173 cm³ since 0.1 microns = 1 x 10^-7 meters, that's 1 x 10^-5 cm, so 0.05173/10^-5 = 5173 cm² for the area covered by that leaf. --Jayron 32 13:42, 17 April 2018 (UTC)[reply]

For people as dumb as me: the comma in "5,173cm2" above is a thousands separators, not a decimal point. So that is roughly half a square meter. Tigraan^{Click here to contact me} 13:55, 17 April 2018 (UTC)[reply]

This site says that 1000 leaves of gold leaf will cover 79 square feet and has a mass of 18 to 23 grams. 79 square feet is about 7.3 square meters, so 18 grams per 1000 leaves gives a coverage of 0.4 square meters per gram - very close to User:Jayron32's estimate. Gandalf61 (talk) 15:57, 17 April 2018 (UTC)[reply]

However, when electroplating, much thinner coatings are possible. This site [3] sells microscope slides with 100 Å (i.e., 10 nm or 0.01 micron) coatings, so that would cover ten times the area computed above. Sufficiently thin gold coatings are transparent,. -Arch dude (talk) 16:11, 17 April 2018 (UTC)[reply]

Slightly thicker coatings are transparent red. LongHairedFop (talk) 18:03, 17 April 2018 (UTC)[reply]

I think we ought to distinguish an important practical detail:

The total amount of surface-area covered by one gram of gold is not identical to the surface area you would cover if you added one gram of gold to a real process or machine. There are inefficiencies and losses to consider. Not all the gold ends up where you want it to go!

For every gram of gold input to an electroplating process machine, or to a gold sputtering machine, only a fraction of that gold ends up on the final product. In the case of a sputtering machine, a very huge percentage of the gold ends up as waste-product.

Here's a commonly-cited paper, from the Journal of Vacuum Science and Technology: Ion Sputtering Yield Measurements for Submicrometer Thin Films (1988), which you'll surely find in the archives of any great physical sciences research library.

I'm not very familiar with electroplating - maybe you can get some answers from our regular contributors who are chemists - but I bet electroplating makes more efficient use of the input material than vacuum sputtering! Even still, there are some cases where sputtering is the best and only option - for example, if you want to shoot an scanning electron microscope at a gummy bear, you've got to sputter! The unique preparation of the gold-plated gummy-bear to make it suitable for the SEM is a true art form, a sort of rite-of-passage in the vacuum chamber that one must learn if one wishes to truly master the microscope.

Nimur (talk) 18:26, 17 April 2018 (UTC)[reply]

I know nothing about gold sputtering for electron microscopy, but in other processes using gold, the wast stream is reproicessed to recover the gold. Thus,at the system level, (almost) all of the gold is eventually used. Surely you guys don't jut flush this down the drain? -Arch dude (talk) 22:24, 17 April 2018 (UTC)[reply]

Using the density of 19.3 g/cm^3 and the atomic weight of 197 u, it follows that the surface density of gold is 4.96*10^(-7) g/cm^2, therefore the maximum possible area a gram of gold can cover is about 202 squared meters. Count Iblis (talk) 00:34, 18 April 2018 (UTC)[reply]

Drinking water and hormones

When a human drinks water, does the body release a hormone with diuretic effect? Does it store less water? I imagined that it could exist a similar mechanism to the mechanism of eating/secreting insulin, but related to hydration. Smelling food stimulates release of insulin (which stimulates glucose uptake), since the body anticipates more of it is on its way. --Hofhof (talk) 19:48, 17 April 2018 (UTC)[reply]

You may find some information at one of the items in Water retention. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:03, 17 April 2018 (UTC)[reply]

When humans drink, and that water is absorbed, we produce less of an anti-diuretic hormone called, unsurprisingly, antidiuretic hormone (ADH) and otherwise known as vasopressin. That does imply that under normal conditions we have a constant low rate of release of this hormone which is switched off, or decreased, after drinking. Klbrain (talk) 23:33, 19 April 2018 (UTC)[reply]

Blood volume is controlled by a variety of feedback loops. You may be interested in Atrial natriuretic peptide. -Nunh-huh 03:55, 22 April 2018 (UTC)[reply]

Crowley's ridge

I think my real question is how does something made out of loess and gravel not get eroded away by the Mississippi River over all its meanderings?Hoover12345! (talk) 21:18, 17 April 2018 (UTC)[reply]

How do you know it hasn't already been eroded some? ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:36, 17 April 2018 (UTC)[reply]

Because in that part of the river, deposition rather than erosion is the predominant process. 2601:646:8E01:7E0B:792F:2CDD:A29B:FC67 (talk) 03:08, 18 April 2018 (UTC)[reply]

Looking at meander maps going back eons, the Mississippi has been on both sides of the ridge and Crowley's ridge is never under water. — Preceding unsigned comment added by Hoover12345! (talk • contribs) 14:35, 18 April 2018 (UTC)[reply]

Looking at the aerial map I think that the ridge of today is the slim rest of a much thicker ridge, possibly comprising the whole valley of today and having been eroded alternatively from one side an then from the other. As the valleys on both sides become larger, the eroding action should decline but eventually the ridge will disappear, I suppose. So your answer could be, it gets eroded indeed, just slower and slower. 194.174.76.21 (talk) 10:54, 9 May 2018 (UTC) Marco Pagliero Berlin[reply]