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December 27

Grohe Eurodisk

OK, so I've been trying to remove the handle of a Grohe Eurodisk kitchen faucet in order to clean out a clogged cartridge -- but despite me doing everything right, and despite half an hour of effort, it refuses to come off! Is the handle in this faucet really non-removable, or is it just jammed in place for some reason? 2601:646:8E01:7E0B:8115:EFB1:83C0:5101 (talk) 06:27, 27 December 2017 (UTC)

OR: These cartridges fit extremely tightly, whatever the manufacturer. Use a rubber mallet, some suitable levers and a choice selection of 4-letter words. Even then it will take quite a bit of effort until the cartridge slides off. --Cookatoo.ergo.ZooM (talk) 11:43, 27 December 2017 (UTC)

In my case it's not the cartridge, it's the handle -- I removed the set screw as required, but the handle won't come off! 2601:646:8E01:7E0B:8115:EFB1:83C0:5101 (talk) 14:07, 27 December 2017 (UTC)

Grohe has a section on their website "for professionals" that includes video tutorials, etc. (here). —2606:A000:4C0C:E200:7150:E613:78D3:42AE (talk) 18:55, 27 December 2017 (UTC)

I don't see no video tutorials on repairing kitchen faucets here! 2601:646:8E01:7E0B:8115:EFB1:83C0:5101 (talk) 06:14, 28 December 2017 (UTC)

• If all else fails, use a gear puller. -Arch dude (talk) 03:42, 28 December 2017 (UTC)

December 28

Exact value of the strong interaction (constant)?

For the electric and gravitational field constants we have the equations ~8.98x10^9 N m^2 C^-2 and ~6.67x10^-11 m^3 kg^-1 s^-2, respectively. But what about the strong nuclear force? I can't find a reference anywhere as to it's exact value or dimensions. As of yet unknown or am I just looking for it in the wrong places? Earl of Arundel (talk) 01:01, 28 December 2017 (UTC)

Does Fundamental interaction help you? Else it will surely help to learn all the "basics" first that we humans always think we can skip when we "only want to understand that detail, not trying to become a professional". --Kharon (talk) 19:46, 28 December 2017 (UTC)

I'm simply asking if the strong nuclear force does or does not have an analog to the gravitation proportionality constant (G), and if so what its value and dimensions are. Earl of Arundel (talk) 01:05, 29 December 2017 (UTC)

Until proven otherwise I'll assume that any field can be reduced to individual factual statements. Either the strong force isn't a force or else it has a value. My first search turned up a Stack Exchange set of answers on this at [1] (no two the same) which say the following. I don't pretend to be able to identify errors reliably...

• The exact formula for the interaction is a Lagrangian in quantum chromodynamics.

• In quantum hadron dynamics the strong force can be approximated by the derivative in $r$ of the Yukawa potential, said to be (I quote)

$V(r)=-{\frac {g^{2}}{4\pi c^{2}}}{\frac {e^{-mr}}{r}}$

where $m$ is roughly the pion mass and $g$ is an effective coupling constant. If I am not mistaken that should be

${\frac {g^{2}}{4\pi c^{2}}}{\frac {e^{-mr}}{r^{2}}}+{\frac {mg^{2}}{4\pi c^{2}}}{\frac {e^{-mr}}{r}}$

by a simple application of the chain rule.

• The most-upvoted answer says (quote) $V(r)=-{\dfrac {4}{3}}{\dfrac {\alpha _{s}(r)\hbar c}{r}}+kr$ where the constant $k$ determines the field energy per unit length and is called string tension. For short distances this resembles the Coulomb law, while for large distances the $k\,r$ factor dominates (confinement). It is important to note that the coupling $\alpha _{s}$ also depends on the distance between the quarks. (unquote) This gives us a derivative that depends on $\alpha _{s}(r)$ plus a constant k (force at any distance).

I can't help but observe some difference between the form of these equations... A comment notes that the "breaking of the flux tube has no classical counterpart" and calls the calculation handwaving (by "flux tube", read QCD string). Another comment suggests the equations are invalid because they imply action at a distance, faster than light, and action on a quantum scale that cannot be modelled classically. Nonetheless, there seems here a useful starting point for coming to some sort of understanding. Wnt (talk) 01:22, 29 December 2017 (UTC)

Our article on strong force provides some illuminating specifics. It says the strong force is 137 times more powerful than electromagnetic force at distances of 1 fm or less -- which is consistent with an inverse r^2 term, for small r. It also says the force is 10,000 newtons at any distance, which is consistent with a k term. I assume that ratio has something to do with the fine structure constant, and suggests you might simply multiply the universal gravitational constant by that for your answer, where short ranges are concerned. But that's perhaps a mistaken inference on my part. Wnt (talk) 01:34, 29 December 2017 (UTC)

Spot on, that's exactly what I was looking for. Can't comment specifically on much else though...it'll take some time to digest all of that! Anyhow, thanks so much for the most excellent effort. Cheers! Earl of Arundel (talk) 01:56, 29 December 2017 (UTC)

It depends on what you mean by nuclear force. If you mean the force that acts on nucleons in a nucleus, its depends on distance. On small distances <1 fm it is strongly repulsive and is actually strong enough to keep neutron stars from collapsing. At the intermediate distances of a few fm it is strongly attractive -- around 100 times stronger than the electrostatic repulsion of protons at the same distance. It keeps nuclei bounded. At larger distances it vanishes completely. As to interaction of quarks inside nucleons, it is difficult to express its force in easily understandable numbers. However at very small distances (or high energies) it vanishes. This is called asymptotic freedom. At large distances (small energies) it becomes in some sense infinitely strong. Ruslik_Zero 19:54, 29 December 2017 (UTC)

What do aircraft throttle numbers above 100 mean?

In a flight simulator I've seen the throttle go to 108 and 114 on a newer plane (the Boeing 777?) Is 100 maximum sustainable setting or maximum cruising or what? How much above 100 has an airliner or aircraft in general gone? Sagittarian Milky Way (talk) 02:23, 28 December 2017 (UTC)

There are no numbers on the throttle of the 777. The closest numbers are the flap angle. What numbers are you referring to? 71.85.51.150 (talk) 02:44, 28 December 2017 (UTC)

I seem to remember the throttles going from the idle number (0?) to 108 or 114 (not on the throttle levers themselves nor analogous to the tick marks on a mercury thermometer). That was BS? Or maybe it was on a green heads-up/projected onto windshield display or omniscient on-monitor stats display that wouldn't be there in the real aircraft (there were so many stats they'd intrude into the windscreen if you showed them all, things like odometer to the meter, thrust to the newton, mass to the pound and other stuff better then the real instruments). The Hindenburg's altimeter had a clock face from 0 to 9 (10=0), two hands for hundreds and thousands respectively and the face behind the hands would rotate 10 times slower than the slow hand (the ceiling was low enough that the letters couldn't get upside down). Was that not historically accurate too? Sagittarian Milky Way (talk) 04:24, 28 December 2017 (UTC)

It depends on the exact setting or gauge you're looking at, but it's almost certainly one of the turbine speeds (most likely N1, the turbofan speed). Instead of displaying a raw RPM value, the display is scaled against some 'nominal' speed, and the readout is expressed as a percentage of that value. Generally, the nominal speed is chosen to be somewhere near or at an engine's rated or designed maximum.

What subsequently happens is that a manufacturer will determine that certain components are more (or less) durable than projected, or make minor tweaks to their design. Heck, sometimes this happens before the first jet is delivered to a customer. Instead of changing that nominal calibration number for converting RPMs to percent (which can cause all kinds of paperwork and maintenance issues), the manufacturer and certifying authorities say, "Okay, you can take N1 up to 105% with these engines on this aircraft."

See here for a definition of N1 and N2; here and here for various discussions of N1 greater than 100% and how it comes about. TenOfAllTrades(talk) 05:24, 28 December 2017 (UTC)

Many jet engined-aircraft use engine pressure ratio (EPR) as an indicator of engine output. It is rare for the limiting EPR for take-off to be exactly 1.00 or 100%. It is likely that the numbers in question were EPR. Dolphin (t) 06:00, 28 December 2017 (UTC)

See also Airbreathing jet engine#Terminology. --47.157.122.192 (talk) 07:34, 28 December 2017 (UTC)

Many correct answers are already posted, but here's a few references!

Consider reading Chapter 15 of the Airplane Flying Handbook: Transition to Jet-Powered Airplanes.

That's not a throttle on a 777 - it's a thrust lever, because jets don't have a throttle!

Piston pilots control for RPM, and monitor the resulting power using other instruments (or computing the power by hand). Piston planes have a throttle control, and sometimes have other controls for manifold pressure, fuel rate, and so on.

Jet pilots usually control for "percent power," or "percent thrust," or "engine pressure ratio." They use RPM and temperature as an indicator of engine health and operating condition.

So - if you saw a lever in a simulator for a large jet, you were looking at a thrust lever and it is probably marked in either "percent power" or "EPR" (engine pressure ratio). On some aircraft, those values can exceed 100%.

If you were looking at an accurate 777 simulator, there should be no markings on the thrust levers. Pilots of 777 must monitor engine instruments. Here's Boeing's internal magazine article on the 767-400 and 777 flight deck: Aero Magazine, (described); here's a photo of the actual thrust lever used on 767-400 and 777. Here's the assembled control stand. Here's another issue all about the 777 flight deck (unfortunately, featuring fewer close-up photos than the very similar 767-400 deck). And here's Aviation Week's sneak-peek at the 777X flight deck, replete with five full 15.1-inch Rockwell-Collins touchscreens, and still no marking on the thrust lever or control stand.

A few weeks ago I had the privilege of visiting the flight-deck of an MD-11. The thrust lever on that aircraft is a lot if fun: if you throw it full-forward (for maximum thrust), you even get an audible voice-alert in the cockpit that reminds you if full thrust is inappropriate for the airplane's present configuration. Intelligent!

Nimur (talk) 08:15, 28 December 2017 (UTC)

Flying is special because of all the security needed. So the technology in play is never maxed out in practice. If you have a car that the producer sells as 240Km/h fast, that is really its max speed. When the engine becomes to hot you simply stop on the sideway. That is never an option for the pilot of a 777. He will always try to fly in the technical save zone of maybe 70% of the max the plane could do. In some planes that is probably the "100" marker. --Kharon (talk) 19:18, 28 December 2017 (UTC)

Perhaps I am unadventurous, but I've never taken my car anywhere near the nominal limits. The closest I get might be one or two trips to the higher end of the RPM scale, and as it happens, when that happens I am never moving anywhere at all! And it is invariably winter. You do make me wonder whether the MD-11 has a fancy steering wheel that complains if you aim it at the new World Trade Center. ;) Wnt (talk) 01:42, 29 December 2017 (UTC)

Köppen climate classification in the US

This map, ultimately derived from [2], is in line with the Humid subtropical climate article's statement that this climate zone extends only as far north as "far southern portions of Illinois, Indiana and Ohio." However, sources such as Weatherbase, used in tons of locality articles to determine climate, disagree; Bloomington is hardly far south, and Indianapolis is almost exactly in the center (well north of what you see highlighted on this map), but as you can see at [3] and [4], Weatherbase lists both of them in humid subtropical. What's the difference? Is it a matter of varying definitions? Nyttend (talk) 13:30, 28 December 2017 (UTC)

Well, the article gives the definitions used for the current map, and it has been revised historically. I have linkified the title of this section to increase its usefulness. I find of specific interest the fact that southwestern New Jersey is classified as humid subtropical, and this area corresponds very closely with a change in the dominant climax forest vegetation that is quite obvious (at least to those of us who've studied plant ecology) as one travels from the Pine Barrens westward into lowlands that drain into the Delaware Bay. In other words, that map does seem quite relevant and accurate for that local boundary in New Jersey. μηδείς (talk) 18:05, 28 December 2017 (UTC)

The article mentions (cite) However, while some climatologists have opted to describe this climate type as a "humid subtropical climate", Köppen himself never used this term.. In the end its just a silly category and that is where some science communities go nuts on Scientific formalism. Rock solid Pluto is no Planet anymore since some Astronomers Assembly decided so in 2006 but Saturn, which is believed to be only gases, is? Just take it as it is. Discussing or arguing about disputed scientific categorization is a waste of time unless you also like to discuss or argue about Roman Catholic Dogmatics (which some argue is a science too!). --Kharon (talk) 18:55, 28 December 2017 (UTC)

Saturn only has a thin skin of gas, it's average density is as substantial as 0.7x water and the density barely inside the gas is only an order of magnitude from water but two from air. Sagittarian Milky Way (talk) 20:26, 28 December 2017 (UTC)

I prefere planets to be solid rock. :) --Kharon (talk) 14:36, 29 December 2017 (UTC)

So if aliens added water till the ocean covered everything would Earth stop being a planet? Sagittarian Milky Way (talk) 18:19, 29 December 2017 (UTC)

The temperature threshold for dividing the C and D Koeppen primary classifications (i.e., mean temperature of the coldest month) sometimes is taken as 0 C and sometimes as -3 C. This can produce noticeable differences in geographic extent of the classifications. The article you linked states that for Bloomington "the coolest month on average is January, with an average temperature of 30.0°F (-1.1°C)." Thus if a threshold of -3 C is used Bloomington will be Koeppen type Cf (so-called "humid subtropical") and if 0 C is used the type will be Df (sometimes called severe mid-latitude or other names). Shock Brigade Harvester Boris (talk) 20:01, 28 December 2017 (UTC)

Didn't the -3 climatologists choose it to try to follow the limit of semi-persistent snowcover on the coldest month? At least for Earth and the time it was chosen? (who knows about other geological epoches and planets) Maybe non-zero snow depth 50% of the time?. While 0 might be the best Celsius integer for some other line(s) (50% liquid precip? sharpest cutoff in snowfall?) there's just too much time >0, rain and sun there to avoid a fairly high no snowdepth percentage. Sagittarian Milky Way (talk) 21:43, 28 December 2017 (UTC)

• To address a question about the earth being covered in water, were it not due plate tectonics, the continents would not have formed and once plate tectonics grinds to a halt the ocean, if it still exists, will erode most of the existing elevated areas of the earth below see level. One exception would be volcanic hot spots like Hawaii which would still poke above the waves as long as there were mantle upwellings to produce them. According to this study there was little continental crust before 1.4 billion years, and a google search for "early earth little land surface" provides such things as Quora answers saying the early earth was mostly ocean, as well as the NIH study I first linked to. This paper on visualizing the land surface compares the relief profiles of Venus, Earth, Luna, and Mars, and shows how the earth has much higher and lower peaks due to plate tectonics, although Olympus Mons, another hot spot volcanic formation skews the Martian relief map away from flat. μηδείς (talk) 17:32, 1 January 2018 (UTC)

Modern equivalents of chief engineers such as Brunel

I know these days major infrastructure projects involve hundreds of people from several companies but would the closest thing to Brunel these days, on a major project, be a consultant who is employed by a client to design and develop their project? Clover345 (talk) 20:04, 28 December 2017 (UTC)

Many large companies bestow the title "senior engineer," "distinguished engineer," "engineering director," and other such accolades, with various career implications. University scholars are sometimes awarded a "named chair" or similar accolade. Depending on the company or institution, this type of title can be a serious privilege and honor. For example, I have often observed that a named chair at a prestigious university commands more ears and eyes than a corporate executive.

It can be instructive to have a look at the actual job titles held by such greats who pioneered and transitioned engineering into its modern era: individuals such as Claude Shannon, Alan Turing, Frederick Terman; Douglas Engelbart, Paul Allen, William Boeing; Cecil Howard Green, William Shockley, Gordon Moore... Edwin Aldrin, Vint Cerf, Grace Hopper.

Among such individuals, we have pilots, admirals, entrepreneurs; scientists, engineers, businesspeople; CEOs, professors, individual contributors; capitalists, innovators, and heirs.

Perhaps the hardest thing to recognize is that modern technology is all about miniaturization - so today's great engineers often build invisible things - things that we can't point at and attribute. Contrast our understanding of global navigational satellite technology, and the great engineers who built it - shall we point at the vague direction of space? ... or a cheap commodity hand-held electronic device? ...and compare to pointing at the Brooklyn Bridge, and saying it was "built" by Roebling ("with the help of others"). Both narratives are a summary-abstraction.

One of my favorite documentary series, Connections (1978), incisively explores whether it is appropriate to look at history as the well-orchestrated accomplishments of a few geniuses. Many episodes study famous scientists and engineers, and surrounds each historical story with much-needed context. In particular, historian James Burke explores the transition to modern science and engineering, where every discovery is widely-regarded to be the cumulative effort of many thousand-individuals, rather than a single genius; and he analyzes whether this is because engineering is different today, or if it is only a difference in the narrative that we construct about engineering. By now, surely all educated individuals know about "great man theory," and its strengths and serious weaknesses; any actual survey of the history of science and engineering will deconstruct the conventional narrative; and reconstruct it, or replace it.

So, do we even have a 21st-century equivalent of a 19th-century "Chief Engineer"? And if you can forgive the academic question, which story-book concoction is the more fictitious plot device: the 19th-century Chief Engineer, or the 24th-century version?

Nimur (talk) 21:17, 28 December 2017 (UTC)

Paul MacCready springs to mind, and perhaps more like Brunel (some failures, well quite a lot really) Jim Bede, Colin Chapman Greglocock (talk) 05:18, 30 December 2017 (UTC)

December 29

Absorption (skin) article

The article let's me wondering: why is the skin impermeable to water? How would absorption differ for seawater and fresh water? Does the lining of the mouth or intestine count as skin (that is, should this go into the article)? Do they behave differently? --Hofhof (talk) 11:54, 29 December 2017 (UTC)

It is not impermeable, it is simply already saturated, given the high amount of water our body keeps in every cell. Cells also have the ability to selectively take up, filter, throw out and keep an inner balance of chemicals, so it should not make a difference how much salt water contains that comes in contact with the skin. --Kharon (talk) 14:29, 29 December 2017 (UTC)

What? -165.234.252.11 (talk) 19:34, 29 December 2017 (UTC)

To actually answer this guy's questions: the impermeability of the skin is thanks to a layer of keratin that composes the outermost part of the skin (see e.g. keratin, human skin, epidermis, stratum corneum), and no, the lining of the mouth and intestine are not skin, they're mucous membranes, and they don't belong in that article. -165.234.252.11 (talk) 19:56, 29 December 2017 (UTC)

• Would the mucous membrane in the intestine absorb water? For example, if there's only seawater available, would pumping this water down there hydrate a human? Isn't that the same as a vodka-soaked tampons (aka Alcohol enema? Is a Seawater enema a thing?--Hofhof (talk) 22:41, 29 December 2017 (UTC)

Sea water is hypertonic compared to human blood, and pumping it into the intestine would suck water out of the body, causing dehydration, diarrhea, and death. μηδείς (talk) 02:23, 30 December 2017 (UTC)

See osmosis - to desalinate water requires the input of energy in some well engineered process. The body can do it to a limited degree (and does so in the loop of Henle for example) but is not evolved to be able to do it to the extent of being able to use seawater. Wnt (talk) 11:39, 30 December 2017 (UTC)

Year of description of a new species

Good morning/afternoon/evening. This article describes Varalphadon janetae sp. nov. As you can see, it is due to be published in the April 2018 issue of Cretaceous Research, but it has been available on ScienceDirect since last month.

Since the ICZN Code recognises electronic publication for species described after 1 January 2012, should the authority of this taxon be cited as "Carneiro, 2017" or "Carneiro, 2018"? In other words, should we use the date on which the journal carried the article or the actual date when it was published online? Thanks.--Leptictidium (mt) 12:49, 29 December 2017 (UTC)

Articles should not use commercial sources like Sciencedirect, Elsevier and alike, which are definitely not freely available, not even in university libraries. See WP:Identifying reliable sources#Vendor and e-commerce sources The journals from these companies usually are tho. So best use the journal reference. --Kharon (talk) 14:14, 29 December 2017 (UTC)

Neither Sciencedirect nor Elsevier are vendor and e-commerce sources and the guideline that you cited is not applicable to them. Ruslik_Zero 19:40, 29 December 2017 (UTC)

I don't know what you mean by 'not even in university'. Lots of universities have access to Elsevier. In the same way that Wikipedia editors have access to it: [5].--Hofhof (talk) 22:44, 29 December 2017 (UTC)

• The answer to the original question is that you should use the date assigned to the article by the journal, because any other approach would lead to unacceptable confusion. Looie496 (talk) 20:24, 29 December 2017 (UTC)

Sorry, but this is incorrect in general. Often the date printed in the journal predates when a journal was printed (because of publication delays), or may postdate when an author distributed reprints. The date of publication to be used is "the earliest day on which the work is demonstrated to be in existence as a published work". For that reason the library stamps on journal issues indicating when they were received can be important historical evidence about priority in taxonomy. For zoology, the relevant part of the ICZN code is Chapter 5 (http://www.iczn.org/iczn/index.jsp). Jmchutchinson (talk) 16:04, 30 December 2017 (UTC)

• It only took a moment to search out the ICZN rule change you mentioned at [6] - article 9.9 explicitly says that electronic pre-publication doesn't count. This is an arbitrary decision; however understandable the choice may be, there was no way to decide it from first principles. Wnt (talk) 11:36, 30 December 2017 (UTC)

• It is not quite that simple, but depends on whether the electronic pre-publication is the “version of record”, that is the final version, which will be identical to the printed version. If the electronic version may be updated prior to printing on paper, the rules are clear according to the ICZN: "21.8.3. Some works are accessible online in preliminary versions before the publication date of the final version. Such advance electronic access does not advance the date of publication of a work, as preliminary versions are not published (Article 9.9)." For the botanical code, the rules are very similar (http://www.iapt-taxon.org/nomen/main.php?page=art30).

However, if the electronic version is identical to the later printed version, and especially if the publisher designates it as the version of record, then my understanding is that the date of publication is that of this electronic version. I think to remember that it is not considered important whether an electronic version of record includes the page numbers of the printed version. For instance, the respected taxonomic journal ZooKeys writes (https://zookeys.pensoft.net/about): "The journal publishes electronic versions of the articles when these are ready to publish, without delays that might be caused by completion of an issue. These electronic versions are not "pre-prints" but final and immutable (Version of Record), hence available for the purposes of biological nomenclature. The date indicated on the electronic version is to be considered the actual publication date." Jmchutchinson (talk) 14:29, 30 December 2017 (UTC)

Good catch! I am more accustomed to sections like this, which I think are clearly excluded by the policy, but you give a perfect example to the contrary. Wnt (talk) 15:14, 30 December 2017 (UTC)

• Wikipedia is a tertiary source. We don't need to worry about establishing priority, etc. That is someone else's problem. In so far as we actually need to discuss disputes about priority that ought to be done by citing secondary sources that discuss the issue. However, that does not appear to be the crux of the question being asked. The OP seems to be asking what is the correct way to cite: [7]. Our goal is to provide a clear citation to allow others, including those without internet access, to find the paper. For that purpose, I agree with the comment above that using the date provided by the publisher as it appears on the paper is the only reasonable choice when writing out the citation since that is the natural way that someone would look up the paper. So, I would write the full citation as :

Carneiro, Leonardo M. (April 2018). "A new species of Varalphadon (Mammalia, Metatheria, Sparassodonta) from the upper Cenomanian of southern Utah, North America: Phylogenetic and biogeographic insights". Cretaceous Research. 84: 88–96. doi:10.1016/j.cretres.2017.11.004. Retrieved December 30, 2017.

Giving any other date for the paper than the publisher's date of April 2018 seems likely to make it harder to find. Then, if one is giving a short reference to this citation, I would say "Carneiro, 2018" is the way to go. Dragons flight (talk) 21:44, 30 December 2017 (UTC)

You misinterpret the original question. The issue is not how to cite an article in Wikipedia or in another publication but how to write the "taxonomic authority". You will be familiar with the Latin binomial name given to every species, but the full scientific name includes a reference to the original description; for instance "Rattus norvegicus (Berkenhout, 1769)". See Author citation (botany) and Author citation (zoology). It is often a requirement to quote the name in full including the taxonomic authority (author and date) at the first mention of a species in a scientific article, so that readers can be sure which species is being referred to (another author, or even the same one, might later have used the same name to refer to another species). Except in some taxonomic papers, the full bibliographic details of such works are not routinely listed in the reference section. But it is important to follow the formal rules of which date to use for the taxonomic authority. One reason is so that everybody uses the same date in this context; otherwise it appears that two different species might have been described using the same name. A second reason is that this date is the criterion deciding what name has priority: if two publications described the same species, it is the one with the earlier date that is valid. You would be doing nobody any favours by writing a non-standard date here, even if it might make it easier to find the original publication. And according to the formal internationally agreed rules of scientific nomenclature, it is simply wrong.

In other situations when citing publications, cases of dates written in publications not matching their actual date of publication are often dealt with by writing both dates with one in square brackets, and perhaps using inverted commas. There seems to be considerable variation between journals in how this is done. But, to re-emphasise the main point, including both dates like this and using square brackets are explicitly not allowed when including the taxonomic authority in the formal name. Jmchutchinson (talk) 09:24, 31 December 2017 (UTC)

Perhaps Leptictidium would like to clarify. I agree the question is ambiguous, but the context still suggests to me that he is asking how "we" (as Wikipedians) should state the citation and taxonomic authority. If instead he is asking how he should state it in some other context (a journal article he is writing perhaps?), then the answer might be different. I stand by my answer as it comes to Wikipedia. I believe that a Wikipeidian who is citing this paper when referencing the creation of this species should also give the paper's corresponding date in any statement of authority (i.e. Carneiro, 2018). To be explicit, to the extent there is an interpretation of ICZN rules that might be read to give a different statement of authority, then in my opinion that should be ignored on Wikipedia as long as the only source is the paper itself. A) It is likely to create confusion if the full citation and the short authority statement disagree, B) it isn't Wikipedia's job to decide such issues of priority, and C) the interpretation of ICZN rules arguably requires original research. If some later review article or other source assigned a different taxonomic authority (e.g. "Carneiro 2017"), then Wikipedia could cite the later source when stating the authority. In the absence of such a third-party clarification though, and when the only source provided is the paper itself, I believe the reasonable course is for Wikipedia to treat the information as it would other citations and use the publisher's date. Dragons flight (talk) 19:21, 31 December 2017 (UTC)

Sorry for not making the question clear enough. Indeed, I am not asking how to cite the paper, but how to write the taxonomic authority. Based on the above answers, since the version available on Sciencedirect seems not to be the full article, merely a short advance extract, I'm inclined to go for "Carneiro, 2018". Thanks everyone for your valuable feedback.--Leptictidium (mt) 19:35, 31 December 2017 (UTC)

• The pdf of the full version of this particular article is behind a paywall, to which I have no free access. However, it does appear to be available. With reference to the issue in which this article appears, the journal web site (https://www.sciencedirect.com/science/journal/01956671/84/supp/C) states that "This issue is In Progress but contains articles that are final and fully citable." That implies that the electronic version is the version of record, and so the authority should be Carneiro 2017. My opinion is that there is here no freedom of choice about how to write the taxonomic authority correctly. But if you cite the article, you can write Carneiro (2017 [“2018”]), or something similar, to clarify the situation. I agree that this is an unfortunate circumstance, but presumably not such an unusual one nowadays. However, it is routine in old literature that printed publication dates do not reflect reality, and taxonomists are very used to dealing with that situation appropriately, so that the consequent mismatch of dates is thus not so surprising to professionals. Nowadays in some journals the final printed pdfs make clear the date of online publication, which is good practice. Jmchutchinson (talk) 21:16, 31 December 2017 (UTC)

December 30

Feynman Lectures. Exercises. Exercise 16-5 JPG

. .

...

The Berkeley "bevatron" was designed to accelerate protons to sufficiently high energy to produce proton-anti-proton pairs by the reactio

p + p --> p + p + (p + p)

The so-called threshold energy of this reaction corresponds to the situation when the four particles on the right move along together as a single particle of rest mass M = 4 m_p. If the target proton is at rest before the collision, what kinetic energy must be bombarding proton have at threshold?

—  R. B. Leighton , Feynman Lectures on Physics. Exercises

${\begin{cases}m_{\text{pv1}}+m_{\text{p0}}=4m_{\text{pv2}}\\m_{\text{pv1}}v_{1}=4m_{\text{pv2}}v_{2}\\m_{\text{pv1}}={\tfrac {m_{\text{p0}}}{\sqrt {1-v_{1}^{2}/c^{2}}}}\\m_{\text{pv2}}={\tfrac {m_{\text{p0}}}{\sqrt {1-v_{2}^{2}/c^{2}}}}\end{cases}}$

${\begin{cases}v_{1}={\tfrac {4{\sqrt {3}}}{7}}c\\v_{2}={\tfrac {\sqrt {3}}{2}}c\\m_{\text{pv1}}=7m_{\text{p0}}\\m_{\text{pv2}}=2m_{\text{p0}}\end{cases}}$

So I got $0.5m_{\text{pv1}}v_{1}^{2}={\tfrac {24}{7}}m_{\text{p0}}c^{2}$

But Feynman answer is $6m_{\text{p0}}c^{2}$.

Why?

Username160611000000 (talk) 12:55, 30 December 2017 (UTC)

It seems Feynman has used a formula $T=(m_{\text{pv1}}-m_{\text{p0}})c^{2}$. Username160611000000 (talk) 15:45, 30 December 2017 (UTC)

Suppose that the full 4D momentum of the 4 resulting particles is $u$. Since at the threshold it behaves like a single particle with mass $4m_{p}$ we have $u^{2}=16m_{p}^{2}c^{2}$. On the other hand the 4D momentum of initial proton at rest is $u_{1}=(m_{p}c,0)$ and that of the moving initial proton $u_{2}=(m_{p}c+K/c,\mathbf {p} )$ where $K$ is its kinetic energy and $\mathbf {p}$ its 3D momentum. So, we have $(u_{1}+u_{2})^{2}=2m_{p}^{2}c^{2}+2m_{p}c(m_{p}c+K/c)=u^{2}=16m_{p}^{2}c^{2}$. Quite obvious that $K=6m_{p}c^{2}$. Ruslik_Zero 20:57, 30 December 2017 (UTC)

@Ruslik0: That's a great derivation, except ... aren't you saying that the four resulting particles are at rest? If they're at rest, they should be in the frame where the two colliding protons have equal and opposite velocities. Wnt (talk) 17:05, 31 December 2017 (UTC)

Yes, it is true that they are at rest in the reference frame of their center of mass (at the threshold). In the laboratory reference frame they have exactly the same 4d momentums. You can easily understand why: in the center of mass frame two initial protons have the same energy and momentum but move in opposite directions towards each other. So, their total 3d momentum is zero. If we are at threshold of the reaction, all kinetic energy of both moving protons is spent to create a proton-antiproton pair. Therefore in the final state all four particles are at rest in the center of mass frame. When we transform the reference frame back to the laboratory frame, 4d momentums of the final four particles remain equal to each other. So, if $u'$ is the 4d momentum of one of those four particles, then their total momentum is $u=4u'$ and $u^{2}=16u'^{2}=16m_{p}^{2}c^{2}$. Ruslik_Zero 17:34, 31 December 2017 (UTC)

Momentum and energy is conserved, not momentum and inertial mass. So your first equation is the one that is wrong. Dragons flight (talk) 22:30, 30 December 2017 (UTC)

I feel like an idiot when I try to do this, but hopefully it is, by definition, a method any idiot can follow. Looking at the system in the final frame of the four proton masses, we need two proton masses of mass + two proton masses of energy to begin with, so each proton has a relativistic mass twice that of its rest mass. Actually that just lets us directly set gamma = 2 = 1/sqrt(1-v^2/c^2). So v = sqrt(3)/2 c. Now using the velocity-addition formula I get that if one of the initial moving protons is set at v=0, the other (with twice the speed) is now at sqrt(3) c / (1 + 3/4) = 4 sqrt(3)/7 c. Taking that gamma formula, that gets 1/sqrt(49) i.e. gamma=7. One mc^2 of that is actual rest mass and 6mc^2 is kinetic energy. Double checking with the energy-momentum relation and saying that E^2 = (m c^2)^2 + (p c)^2, we first use p = gamma m v with the two sets of gamma and v values to get 2 * m * sqrt(3)/2 c = sqrt(3) m c and 7 * m * 4 sqrt(3)/7 c = 4 sqrt(3) m c. Then we put that into the formula and we have E = 2 m c^2 and E = 7 m c^2 respectively. So we do indeed have 4 m c^2 on two protons with equal speed in the final frame and 7 m c^2 on one proton moving + 1 m c^2 on one proton at rest in the lab frame. In the lab frame we know that all the protons have 2 m c^2 total energy after the collision x 4 so this energy is properly conserved. And the 4 sqrt(3) m c value for p on one proton in the lab frame allows all the protons coming out to end up moving with momentum sqrt(3) m c.

I will admit, however, that I am having a very hard time following anyone else's calculation above, and I fear the same may be true here. :( Wnt (talk) 02:42, 1 January 2018 (UTC)

Your derivation is correct but it can be done much simpler as I showed above. Ruslik_Zero 17:13, 1 January 2018 (UTC)

If you can lead me to some background on how to use 4D momentum as you are, I'd welcome it. Our offering on four-momentum seems more mysterious than useful to me at present. I was fiddling about on my own and derived a way to add gamma factors directly: "g1+g2" (by which I really mean, the gamma factor of something with gamma factor g1 viewed from a frame at a velocity in the same direction that would impose gamma factor g2) works out to be g1*g2 + sqrt(g1^2*g2^2-g1^2-g2^2+1). There is probably some way to make that more concise and to make it into a vector result to boot, but it is already an improvement for this problem: knowing that each proton has to have gamma factor = 2 to carry the extra relativistic mass to make a new proton, I can then shift the frame and calculate directly the gamma on the moving proton is 2*2 + sqrt(4*4-4-4+1) = 7, meaning it carries 6 proton masses of kinetic energy plus its rest mass. Wnt (talk) 18:35, 1 January 2018 (UTC)

P.S. when g=g1=g2 this reduces to 2*g^2 - 1, so repeatedly "doubling velocity" should get gamma factors of 2, 7, 97, 18817, 708158977... all apparently numerators of diophantine approximations to the square root of 3,[8] if that means anything. ...Actually, if g(n) is 1,2,7,26,97,362,1351,5042,18817,70226,262087,978122 for n=0,1,2,3,4... then viewing g(n1) from a frame with g(n2) seems to yield an integral gamma value g(n1+n2). Wnt (talk) 20:32, 1 January 2018 (UTC)

PP.S. Actually those diophantine approximations seem relevant -- the velocities for those gamma values are the inverse of the diophantine approximations * sqrt(3) * c: 0, sqrt(3)/2, 4*sqrt(3)/7, 15*sqrt(3)/26, 56*sqrt(3)/97 etc. Each velocity is reached from the preceding by shifting the frame by a velocity of sqrt(3)/2 c. (the momenta are thus 0,1,4,15,56... * m * sqrt(3) * c), appearing here as the numerator, with gamma as the denominator of each speed) Wnt (talk) 03:21, 2 January 2018 (UTC)

Writ, I agree that all these maths can be really hard to follow... and the required gamma factor is 2 as you determined above from considering the center-of-mass frame, a factor that fixes the final kinetic energy of the proton accelerated from rest, so in the lab frame it is $m_{p}c^{2}$ (since their kinetic energies are given by $(m-m_{0})c^{2}$ or $(gamma-1)m_{0}c^{2}$). The total energy of the four protons (which I've assumed behave as a single particle at the threshold as pointed out by Ruslik0) is thus 4*(rest energy + kinetic energy) or $8m_{p}c^{2}$. Given that the energy of the rest masses of the initial two protons are $m_{p}c^{2}$ each, by energy conservation, the additional kinetic energy of the initial bombarding proton is $6m_{p}c^{2}$. This is a simpler derivation, if I'm not mistaken and implied by your double check above, that doesn't mess around with the velocity-addition formula or the four-momentum with this particular problem. -Modocc (talk) 03:34, 2 January 2018 (UTC)

Simply put, bring the 4-momentum you want to eliminate to one side of the equation and square it, as the square of the 4-momentum yields the known mass and thereby indeed eliminates the variables (energy and momentum) you want to get rid of. Count Iblis (talk) 16:08, 2 January 2018 (UTC)

December 31

How many of a 100 year old's atoms have been in their body their whole adult life?

Which elements are these atoms most likely to be? Atoms of which elements are most likely to stay? (not necessarily the same since they could be rare)

Do weight changes or yo-yo dieting affect this or is lipid storage an unlikely place for atoms that don't leave the body for decades to have been?

2. Could their body at this instant have an atom that has spent at least 1/2 continuous centuries outside and came back? Well of course it can but is this almost certain, almost certainly false or in between? How is such an atom most likely to leave and return? (i.e exhalation, food, water) Sagittarian Milky Way (talk) 01:00, 31 December 2017 (UTC)

Seriously, SMW, this is the sort of nonsense that made User:Floquenbeam suggest that you might be a candidate for a topic ban. We don't do predictions or debate. This is not a forum. I suggest you do with this what you did with your enquiry on the Klingon word for diaper. μηδείς (talk) 01:07, 31 December 2017 (UTC)

When did I ask about Klingon? I've never asked about Klingon. Sagittarian Milky Way (talk) 01:13, 31 December 2017 (UTC)

Medeis might have been thinking of this, which was posed by a 140 IP. Floq's comment came in reference to SMW's Muzak question.[9] ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:10, 31 December 2017 (UTC)

Oh, my. I think the question speaks for itself μηδείς (talk) 17:22, 31 December 2017 (UTC)

I would say virtually none remain. Cells die, even fat cells. Cells respirate. It's hard to imagine that a cell lives without exchanging atoms. --DHeyward (talk) 02:34, 31 December 2017 (UTC)

Exhalation is the dominant effect here, you can use the information in this article to calculate the exhaled body mass per unit time for some given metabolic rate. This is the same as the absorbed body mass from carbs and fats in food for someone who has a stable body weight. If you then assume that the exhaled CO2 mixes with the atmosphere, then it's easy to compute the fraction of the carbon that comes back later from the food you eat. Count Iblis (talk) 02:37, 31 December 2017 (UTC)

User:Sagittarian Milky Way is not asking about that. He is interested in turnover time. For example, collagen molecules in cartilage have a 400 year median turnover time. So some of them remain in your body for your whole life. Abductive (reasoning) 05:45, 31 December 2017 (UTC)

The question as posed is meaningless. All atoms of the same sort are identical to each other. Therefore one can not distinguish "old atoms" from "new atoms". On the hand complex molecules can be "old" as they can carry some defects which can be used to distinguish them from other molecules of the same sort. Please, see this. Ruslik_Zero 17:15, 31 December 2017 (UTC)

Fundamental particles are identical, yet the ways they are assembled are not - which includes atoms, since there are isotopes. Count Iblis gave an excellent example above -- the amount of carbon-14 in hippocampus DNA depends on date of birth, even though there is some turnover. Wnt (talk) 17:50, 31 December 2017 (UTC)

Note: the excellent answer was given by User:Fgf10 and promptly removed by some drama asshole with nothing to contribute ([10]). Here it is: Another possibility for long lived atoms is in the DNA of non-dividing cells, such as neurons. This has been used to carbon date cells, and see here for an example of cells with DNA containing carbon incorporated well before the onset of atmospheric nuclear testing. Fgf10 (talk) 15:09, 31 December 2017 (UTC)

See the Biology section of the Bomb pulse article --catslash (talk) 19:20, 31 December 2017 (UTC)

Ruslik is correct, the question as posed is rambling and meaningless. We have no way of determining how many of the atoms of you are born with will last over any period, or especially whether "their body at this instant have an atom that has spent at least 1/2 continuous centuries outside and came back?" Iblis hasn't addressed that by pointing out that people born at certain times have more or less of certain isotopes, it doesn't generalize into anything other than a sophomore bullshit session. μηδείς (talk) 00:56, 1 January 2018 (UTC)

I'm not sure I understand the hostility to the question. Carbon dating is based on the exchange of atoms. How close the body and its parts resemble natural radioisotopes is very meaningful. Radioisotopes are also not chemically identical to each other as carbon uptake by plants is different for different carbon isotopes. --DHeyward (talk) 01:12, 1 January 2018 (UTC)

I apologize for inadvertently deleting FGF's comment when I reverted his busy-body hatting.

":Another possibility for long lived atoms is in the DNA of non-dividing cells, such as neurons. This has been used to carbon date cells, and see here for an example of cells with DNA containing carbon incorporated well before the onset of atmospheric nuclear testing. Fgf10 (talk) 15:09, 31 December 2017 (UTC)"

Thankfully, I am not inclined to sink to his level of arrogance and low-life chatter. ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:14, 1 January 2018 (UTC)

Lest there be any confusion, this is my level of annoyance, expressed before I quoted his posting above. We do not need trolls who don't answer the questions and do destroy the good answers to the questions while waving bogus administrative claims against people who asked good questions; people who win every debate on what gets "hatted" and one doesn't according to one simple policy, namely that the race goes not to the swift nor the strong but to him that edit-wars unto the end. This is a classic good question, asked fairly often in biological circles and answered with varying degrees of persuasiveness, and Fgf10 provided a decent example. Wnt (talk) 04:51, 1 January 2018 (UTC)

No one is stopping you, Wnt. Please do answer "How many of a 100 year old's atoms have been in their body their whole adult life" and whether "their body at this instant have an atom that has spent at least 1/2 continuous centuries outside and came back?" μηδείς (talk) 05:14, 1 January 2018 (UTC)

Just because you lack to knowledge to awnser the question doesn't mean everybody else does. Please don't troll. Thank you! Fgf10 (talk) 10:15, 1 January 2018 (UTC)

Just because a question is difficult to answer does not mean that it should not be asked. I've learned a lot about neurons and collagen reading this debate and it's possible that others might contribute further insights. Someone might post an approximate answer or be able to answer a different but related question. It was a perfectly sensible question (to which I don't know the answer). Robinh (talk) 07:31, 1 January 2018 (UTC)

The question about atoms leaving and returning is harder to answer, but we can certainly lay down some approximations. If we suppose that products leaving the body are perfectly mixed into the rest of the world, then we can say that people take about 9 kilograms of nitrogen gas into their body daily (promptly expelling it again) [11] and release about 1 kg of CO2 daily [12] and drink about 3 liters = 3 kg of water a day [13]. These are all very approximate figures, but for reasons we'll see below even an order of magnitude is more accuracy than we need. The global carbon pool in the atmosphere is 750 Pg (that's petagram, x 10^15 g) [14]; obviously there's some exchange out of that to the oceans and crust but again, order of magnitude argument here. The pool of water is 1,338,000,000 km^3 = 1.3 x 10^18 m^3 = 1.3 x 10^21 dm^3 = 1.3 x 10^21 kg. [15] for the first. And Earth's atmosphere should have about 4 x 10^18 kg of nitrogen. (see atmosphere of Earth, multiply by 78% or so - also confirmed here) Now taking each of these dilutions, we see that the water is diluted 3 kg in 1.3 x 10^21 kg is about 1 in 2 x 10^21, the CO2 is diluted 1 kg in (7.5 x 10^17 g C = 7.5 x 10^14 kg C x (48 g CO2/12 g C) = 3 x 10^15 kg CO2) or 1 in 3 x 10^15, the nitrogen is diluted 9 kg in 4 x 10^18 kg is about 1 in 5 x 10^17. Now Avogadro's number is 6.022 x 10^23 atoms per mol; a mol of water is 18 g, a mol of CO2 is 48 g, a mol of nitrogen is 28 g. So 9 kg of nitrogen contains about 300 mol, which means it has 1.8 x 10^26 molecules (multiplying by Avogadro's number). If these are perfectly diluted into the nitrogen of the atmosphere and rebreathed, you get that diluted by a factor of 5x10^17, which means you're still breathing in, oh, close to 4 x 10^8 = 400,000,000 molecules that you breathed on some day in the distant past; double that for atoms. (For this purpose it doesn't really matter if they've found new partners in the meanwhile). Likewise you're breathing that many atoms breathed by Jesus on the day of the Sermon on the Mount or whatever, because according to [16] all the other non-geologic nitrogen pools are pretty small compared to atmosphere. I'll leave the others as an exercise, though tracing the CO2 involves some more figuring since so much of the carbon and oxygen end up in other forms.

The catch to all this though is that the mixing isn't random, especially with non-nitrogen elements that can more readily react close to the site of production. If you end up living in the same house after 50 years somehow, a drop of boyhood pee on the floor may have combined with some calcium in grout or cement or something, to release countless billions of oxygen atoms on a warm day fifty years later. Seems absurd, but think of the levels of dilution we're talking about here! You can release billions of atoms from a spot on a floor and not notice *any* difference in what is left. As a result, the non-random flows seem likely, by idiosyncratic and bizarre means, to swamp the smaller random flows. Wnt (talk) 16:52, 1 January 2018 (UTC)

• It is precisely because I do know the absurd number of variables involved in this arbitrary question with so many undefined parameters, (I majored in biology as an undergrad, did the air Jesus breathed calculation four decades ago, tested out of Chemistry 101 & 102, having gotten a 5 on AP chem, and understand the conservation of DNA molecules versus the recycling of red blood cells) and the fact that not a single RS is going to answer it, that I am closing this bullshit session as a request for prediction and debate. μηδείς (talk) 17:08, 1 January 2018 (UTC)

Stop your disruptive behaviour or I will be forced to take this to ANI. Several people have provided partial answers, something you have neglected to do. The vendetta you and BB have against Sagittarian is ridiculous. Fgf10 (talk) 18:23, 1 January 2018 (UTC)

Earth science not biology might be the biggest problem with 2. for water. Most of the industrialized world doesn't fertilize their food with their waste or drink their recycled pee and breath but flushes it far away so if rebreathing your own water vapor isn't considered becoming part of the body if it doesn't reenter the blood then perhaps random mixing isn't so far off for them. Then the chance of ~10³⁰ water atoms/decade shuffling between pools of ~10⁴⁷ and ~10^27.2 water atoms causing at least one of them to be in the smaller one now and the bigger one for at least 50 contiguous years of the last 100 seems possible to estimate with mathematics. The worldwide tropospheric mixing time is only about a year but most water is sea and the thermohaline circulation takes a millennium so even if that calculation showed a high probability 2. is true it'd require more earth science than I know to know to see if that answer isn't bullshit. Sagittarian Milky Way (talk) 20:17, 2 January 2018 (UTC)

January 1

sanding/filing/grinding stainless steel

I have some stainless steel parts[17] with burbs and burn marks on them that I want to clean up. Googling around I found three ways of accomplishing this easily at home without any heavy power tools:

1. Sand with sandpaper

2. File with a file

3. Grind it with a grinding attachment on a dremel

Cost wise all three are about the same (I have a dremel already, just no grinding tools). Which of these three methods is the least time consuming? Which produces the best results? Mũeller (talk) 03:15, 1 January 2018 (UTC)

For an professional a file is the fastest because there is not need for any tool preparation. The result depend on your skills in every case, especially with a file, which needs allot of practice for best results. For an amateur i would recommend sand paper and i would not, no matter what tool, expect "best results". In case of some "hidden skills" the result may end up being ok ofcourse. --Kharon (talk) 05:33, 1 January 2018 (UTC)

Once you have the equipment, a dremel will almost certainly provide the quickest execution. However, it is also the most prone to accidentally removing more material than you intended. It takes a fair amount of skill to handle a dremel with precision. A small metal file is likely to be the most precise approach. Professionals can use a file in a manner that is both fast and precise; however, I generally find that even amateurs can do a decent job with a file if they take it slow. I'm not a fan of using sandpaper for removing large imperfections on metal (e.g. burrs), though very fine sandpaper may be appropriate for cleaning / polishing. Dragons flight (talk) 08:45, 1 January 2018 (UTC)

Agreed. In particular, it seems that cleaning up the holes with a dremel would lead to unwanted beveling unless great care was taken. Matt Deres (talk) 16:52, 1 January 2018 (UTC)

• I do a lot of this. I would usually polish it with a Garryflex block - a rubber block impregnated with abrasive powder. Several grades are produced, colour coded, and they're produced worldwide under several different brands. As usual, start coarse and work through to fine.

If it's too discoloured or damaged to start with these hand tools, smooth it first by machine, using a foam-backed flap disc on an angle grinder. Use a foam-supported one, to avoid making more machine facets.

If you have the equipment, and especially for many small parts that are mostly corners, then the best way could be to grit blast it.

Stainless steel should be passivated after finishing, to avoid it self-passivating naturally and possibly discolouring in blotches. This can be done by wiping it with citric acid, or even just lemon juice. Andy Dingley (talk) 20:51, 1 January 2018 (UTC)

The best and most used method was not mentioned: Sandblasting! The results are usually superior and you dont have to worry about overdoing or doing it wrong. Its like painting with airbrush, actually a very satisfying, fun task to do. And economical and fast aswell.

In case you count yourself proud member of the Do it yourself-club and do all sorts of work, you may even want to check out how easy it is to build your own little sandblasing cabinet. There are many videos on youtube unter "sandblasting DIY". --Kharon (talk) 12:59, 2 January 2018 (UTC)

Of course it was already mentioned.

Don't use sand for this though, use an abrasive grit, manufactured for the task - probably a silicon carbide grit, maybe glass beads, depending on the grade of stainless, the dirt to be removed and the amount of edge rounding that's acceptable. Sand has very few appropriate uses left for literal sand blasting, as it's not the best medium to work with and it also has safety hazards (the dust hazard from sand blasting is far worse than grit blasting). Andy Dingley (talk) 13:07, 2 January 2018 (UTC)

Oh sorry, i oversaw you mentioned "grid blast". Regarding the hazards was why i mentioned building a cabinet for that. Most professional shops have one and if build and handled right it is very save work. --Kharon (talk) 13:22, 2 January 2018 (UTC)

Most grit blasting cabinets can't be used for sand (if you're bothering to follow the appropriate regs). The exhaust air needs filtering too. You can't allow sand-shard-laden air into a workshop. Andy Dingley (talk) 20:36, 2 January 2018 (UTC)

Some even use glass perls and good cabinets have a circular system where the abrasive is reused. A vacuum cleaner needs a filter too. Never heard that kept anyone from using one in a workshop. --Kharon (talk) 04:37, 3 January 2018 (UTC)

Split rocks on Theodore Roosevelt Island

I am wondering if anyone might shed light on these three rocks that were photographed on Theodore Roosevelt Island? Is this splitting possibly the result of a natural process? If so, please describe ... Many thanks in advance -- P999 (talk) 18:29, 1 January 2018 (UTC)

By definition it is the result of a natural process. ;) The question is which. What I know is that the description given of gneiss and schist doesn't obviously match with the basalt of the most dramatic formation with this appearance I can think of, which is Giant's Causeway. On the other hand, they are metamorphic rocks, which raises a bit of a question of how metamorphic and metamorphed from what ... I wish I could say more, but I certainly don't know presently. Wnt (talk) 19:03, 1 January 2018 (UTC) Sorry, I realize the dramatic top photo on that site probably is just a bannerhead and not about this location at all. Though List of places with columnar jointed volcanics does list a site in Virginia not too far from the Potamac. Wnt (talk) 19:11, 1 January 2018 (UTC)

Such "erratic" rocks were certainly carried by glaciers. Ruslik_Zero 19:08, 1 January 2018 (UTC)

I regret that I was unable to provide a link directly to the photograph of the split rocks in question, but the website where it resides, being a blog, does not seem to allow individual links to the images; however, I have tried to improve the description of the rocks to which I am referring in my query. Thank you very much for your responses, Users Wnt and Ruslik. These rocks are of particular interest to me because they are either on, or a short distance from, the eastern limit of the Atlantic Fall Line which passes through the southern tip of TR Island ... P999 (talk) 20:11, 1 January 2018 (UTC)

You mean this link to these rocks? --76.69.117.217 (talk) 20:54, 1 January 2018 (UTC)

Yes, thank you very much! I will change my query to include it.

Yes, it's likely a glacial erratic, but that doesn't explain the breakage. My guess would be frost weathering; if you check out the photo in that article, the broken rock there is quite similar to the one in question. Matt Deres (talk) 02:38, 2 January 2018 (UTC)

I find it very hard to believe that it is a glacial erratic. The last glacial maximum did not reach as far south as the Roosevelt Island . (Great Falls is a few miles upstream) — Preceding unsigned comment added by Addisnog (talk • contribs) 04:34, 2 January 2018 (UTC)

Much more likely to be a boulder carried down by the river, but I agree about the frost shattering as the likely explanation for the splitting. Mikenorton (talk) 11:11, 2 January 2018 (UTC)

It may not be the last maximum. Ruslik_Zero 13:24, 2 January 2018 (UTC)

Fixed your confusing typo. --76.69.117.217 (talk) 21:26, 2 January 2018 (UTC)

For a simple explanation of the process, see Frost Wedging. Alansplodge (talk) 11:43, 2 January 2018 (UTC)

References

January 3

Does Social Peer Pressure prevent people from stating the obvious?

If everyone is walking upside down on the south pole, does the social peer pressure from the scientific community which lives there in the south pole, prevent a good scientist from mentioning that they are all walking upside down? Perhaps the reason they do not know that they are walking upside down is because if everyone is walking upside down then from their point of view, they seems to be walking the right side up. 110.22.20.252 (talk) 06:11, 3 January 2018 (UTC)