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February 28

Question about a particular deadly condition for old people

What is the condition called where an old person's blood circulation stops working properly and this old person begins needing blood thinner and this old person's various organs and body parts also gradually begin to give out as a result of their poor blood circulation? Futurist110 (talk) 05:27, 28 February 2021 (UTC)[reply]

Are you talking about Congestive heart failure? ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:01, 28 February 2021 (UTC)[reply]

Probably Atherosclerosis.--Wikimedes (talk) 07:19, 28 February 2021 (UTC)[reply]

Or perhaps Peripheral artery disease? If properly treated (possibly including a change in lifestyle), the prognosis is not particularly bad, though. Poor blood circulation can arise as a result of several medical conditions, including (obviously) the already mentioned congestive heart failure, which in turn can have a variety of underlying causes. While statins may be prescribed, they are not blood thinners. Blood thinners may be prescribed if the underlying cause is coronary artery disease – not so much to improve circulation, but specifically to counter the formation of blood clots in the coronary arteries. --Lambiam 08:18, 28 February 2021 (UTC)[reply]

Quantum Entanglement

I am clearly getting in the area where I don't belong, however, the issue bothers me. A couple of years ago an experiment was conducted to prove the quantum entanglement "once and for all." I first ran into the description in the Amazon documentary "Einstein's quantum puzzle" or similar. It is in their Prime Video section. Then there are two descriptions, one by the Austrians [1] who did the experiment, the other by MIT people [2]. The Prime Video channel provides a nice animated diagram, very helpful. In short, they needed two random numbers generators that must be totally independent. They found them in signals from two quasars on opposite sides of the sky, as they said in the opposite sides of the Universe. The distances to the quasars were enormous. They have their reasoning about it, which you can read there. The problem is that they say that those two quasars, to have been created in a very young Universe, never shook hands with one another physically because of the enormous distance between them. They implicitly assume that the history of quasars began at the time of their creation.

I don't think it is correct. In the very early Universe in the first billionth of a second or so there were quantum fluctuations that deformed the universe or perhaps parts thereof which eventually resulted in appearance of those two quasars. It is possible that both areas which much later became those quasars were subjected to the same pressure fluctuations because of the small size of the Universe at that time.

I also can offer a solution for the random number generators for this experiment. They got the computer numbers by some conversion of light signals which are constantly in a flux. They should have selected a small sequence of numbers e.g. 100 from one quasar and tried to find a match (or better a simple correlation) of this sequel in another quasar. Then they had to do such comparison with a slightly larger sequence, perhaps 110 numbers from the first quasar, etc. After repeating this numerous times they could prove total independence of the sources.

I wonder it all makes sense? AboutFace 22 (talk) 21:19, 28 February 2021 (UTC)[reply]

You're asking good questions. Bell's inequalities rely on a "no-conspiracies" assumption that's difficult to pin down exactly. See [3], [4] for some discussion. The version of the experiment that uses quasars as a source of random numbers can't completely eliminate this loophole. However, it can say that the "conspiracy" that could fool you would have to be on the scale of the entire universe and lasting many billions of years. From an MIT press release [5]:

“If some conspiracy is happening to simulate quantum mechanics by a mechanism that is actually classical, that mechanism would have had to begin its operations — somehow knowing exactly when, where, and how this experiment was going to be done — at least 7.8 billion years ago. That seems incredibly implausible, so we have very strong evidence that quantum mechanics is the right explanation,” says co-author Alan Guth, the Victor F. Weisskopf Professor of Physics at MIT.

The version you describe, going back to the quantum fluctuations that seeded the structures that went on to form the quasars, is something like superdeterminism. It may be impossible even in principle to rule this out. Your idea about comparing 100 random numbers doesn't help, because the kind of correlation relevant to the Bell-type experiment is a three-way one that would involve the two quasars and the experimental results on Earth. It would not necessarily produce a correlation between the two quasars alone that could be detected by any possible statistical test. --Amble (talk) 18:19, 1 March 2021 (UTC)[reply]

A little more about looking for correlations in the quasar signals... No statistical test of the quasar signals could eliminate the loophole, even in principle. That’s because it’s not about whether the quasar signals themselves are correlated. They could be perfectly good, completely independent random numbers. The loophole is about whether the experimental apparatus on Earth and the entangled particles could somehow already “know” those numbers before they arrive. It’s like a one-time pad in cryptography. There is no test you can perform on the secret key that can tell you whether your enemy has somehow managed to get a copy. —Amble (talk) 17:01, 2 March 2021 (UTC)[reply]

Slightly off-topic AboutFace 22 but I think you will be interested in the Free will theorem article, if you are not already aware of it. There is a fascinating set of 6 lectures by Conway (linked at the bottom of the article) which cover this related subject in detail and in an entertaining way (IMHO). Mike Turnbull (talk) 13:42, 6 March 2021 (UTC)[reply]

March 1

A stationary object in outer space

Is it meaningful to refer to "a stationary object in outer space" in our buoy article? Is it nonsense or is there a better way to express the idea?--Shantavira|^{feed me} 11:12, 1 March 2021 (UTC)[reply]

This appears under Buoy#Other:

The space buoy is a common element in science fiction that refers to a stationary object in outer space that provides navigation data or warnings about that particular area.

Whether or not it belongs in our article, it is a common element of science fiction.

And you needn't get hung up on "stationary" as suggesting some privileged frame, as it is instead shorthand for a an object which propulsively (via whatever science-fictiony method) maintains a relatively fixed position with respect of some other bodies.

Here's a NOAA page which speaks of DSCOVR (The Deep Space Climate Observatory which maintains position at the Sun-Earth L₁ Lagrange point) as, "acting as a solar storm buoy in deep space." -- ToE 12:56, 1 March 2021 (UTC)[reply]

Thanks Thinking of England. So there really is a "space buoy". That's quite interesting and I'll give some thought to working that into the article.--Shantavira|^{feed me} 13:46, 1 March 2021 (UTC)[reply]

I don't believe it's a commonly used phrase for Lagrange point satellites, but here is NOAA again, in their "50 Years of Science, Service & Stewardship" 2020 publication.

Most of the definitions for the term space buoy read as follows: a common element in science fiction that refers to a stationary object in outer space that provides navigation data or warnings.

This, while true, leaves out an important fact: space buoys are real. Since 2016, NOAA's Space Weather Prediction Center (SWPC), one of the National Weather Service's (NWS) National Centers for Environmental Prediction, has been receiving data from its own space buoy -- the Deep Space Climate Observatory, or DSCOVR -- to help the agency monitor space weather.

...

See page 72 of this document. (Page 76 of the PDF itself.)

Funny, as I assume that their "most of the definitions" definition originated with WP! -- ToE 17:23, 1 March 2021 (UTC)[reply]

Here's a link directly to that one article. -- ToE 17:24, 1 March 2021 (UTC)[reply]

As far as the science fiction usage, its a "In popular culture" type reference, which could be hard to source.

Here's the Memory Alpha entry: Space buoy (which I assume we do not accept as RS). Interestingly, it does not mention Spectre of the Gun, where our very own article uses the phrase in question. -- ToE 17:30, 1 March 2021 (UTC)[reply]

Why can't snow melt here?

Recently USA Texas people posting videos about ice can't melt instead it turned black:

“You’ll see it’s not melting and it’s going to burn. Snow don’t burn. Snow f***ing melts. No water, no dripping, no nothing.

Source

Why can't snow melt here? Rizosome (talk) 13:34, 1 March 2021 (UTC)[reply]

The article you linked explained the effect; there's a combination of soot forming from the fire source, and that they aren't waiting long enough for significant melting to occur. Also, there are some faked videos as well. All of that is already explained in the page you linked. --Jayron 32 13:39, 1 March 2021 (UTC)[reply]

Ice/Snow is surprisingly hard to melt quickly. "XKCD-What If" discusses the math behind why that is. The small drips that do form are probably either absorbed by the snow or boiled off by the flame.

As for the blackness, that's just soot from the lighter. Lighters not not the most efficient flame source in the world, and some of the fuel is only half-burnt. Try pointing it at sheet of glass. Same result. A hotter, cleaner flame wouldn't leave a mark. ApLundell (talk) 04:59, 3 March 2021 (UTC)[reply]

March 2

Colliding EM waves and energy conservation

Please refer to the image.

I create 2 EM waves by the simplest method - moving 2 charged objects back and forth once.

Both E and B fields values are opposite to each other. That is, at any given time, if in wave 1 B=a and E=b, than for wave 2 at the same time B=-a and E=-b.

Problem - At the instance both waves overlap they cancel each other completely.

This cannot be - It breaks conservation of energy.

What is answer to this seemingly contradiction?

I asked this question a while ago, got a horrible (and partially wrong) answer that I didn't think through. The false claim in the answer was that the drawings are irrelevant. This is wrong: at any given moment E and B fields have a distinct value at any point in space, and this is what the drawing depicts.

אילן שמעוני (talk) 05:32, 2 March 2021 (UTC)[reply]

The drawing is not "irrelevant", but it's underspecified. Taken literally, it seems that you're giving the E and B fields at one instant in time, and only on two line segments. In that case, the total energy is zero, and nothing measurable happens.

Probably that's not what you really mean, but I can think of at least two other things you might mean. If you'd be more precise about exactly what you do mean, someone might be able to answer your question, or you might find out you've answered it yourself. --Trovatore (talk) 05:58, 2 March 2021 (UTC)[reply]

I fail to understand what you mean by "...and only on two line segments.". If you refer to that the B and V values are only supplied for the waves, just assume we do this in intergalactical space where the background B and V values are negligible. If you refer to that I only measure values along this single axis and ignore that the waves are in fact 3D - this is true but I can't see why it's relevant.

Please note that the test is conducted after the situation depicted in the image, where both waves advanced to the point they overlap.

I'll rephrase - what data regarding B and V values you seek that is absent from the drawing?אילן שמעוני (talk) 07:27, 2 March 2021 (UTC)[reply]

I'm not sure where V came from; before you were talking about E and B.

The reason it's relevant that you only measure values along a single axis is that Maxwell's equations are three-dimensional. How are you going to compute a curl in one dimension?

So the information I'm missing is, what are the E and B values at all other points in space? Let's say the direction you're taking to be the direction of propagation is the z-axis. What is the dependency on the x- and y-coordinates, and what are the values of E and B at z-coordinates other than those of the two wave packets you show? --Trovatore (talk) 07:44, 2 March 2021 (UTC)[reply]

My bad, it's E, not V.

Mmmmm, I think I understand your question and can not answer it (last time I dealt with cross-products and calculus was 30 years ago), BUT I can't see why this matters to energy conservation.

Energy is supposed to be conserved for any volume without flow of energy in or out.

This true for a volume as small as you wish around the overlapping segment on the z-axis. We can reduce the x and y coordinates around this segment to our liking, and the E and B values will converge to zero.

In the same manner - flow of energy in and out of given volume requires time. Since we talk abot the instance of overlapping, we can again reduce the interval around measurement, and the flow will converge to zero.

I suspect that somehow the cross product that dictates E and B is such that they cannot both be opposites - that if for given time (for wave 1 E values are positive while wave 2 are negative) than (at the same time, B values for both waves must be equal). Buy I lack the knowledge how to check this. I only found how to check the values for moving charged particle, not for EM wave.

אילן שמעוני (talk) 12:14, 2 March 2021 (UTC)[reply]

Link to previous discussion: WP:Reference_desk/Archives/Science/2019_October_31#Destructive_interference_and_conservation_of_energy. You're still assuming you can choose a direction for the E field, choose a direction for the B field, and also independently choose a direction of propagation for the wave. This is not correct. They are related by a cross product as described in Poynting vector. Therefore, the direction of propagation shown in the right-hand half of your drawing is not consistent with the indicated E and B fields. --Amble (talk) 17:29, 2 March 2021 (UTC)[reply]

Ah, thanks, I missed that too. I was trying to figure out how the energy would get in and out of a standing-wave region — I think there's no speed-of-light problem per se, but the energy flow would increase without bound as the region got bigger, which seemed at the very least odd. But it's just the wrong picture. The energy doesn't need to go anywhere; it just sloshes back and forth between electric and magnetic fields. --Trovatore (talk) 20:58, 2 March 2021 (UTC)[reply]

So my suspicion was correct. Thanks. אילן שמעוני (talk) 19:24, 2 March 2021 (UTC)[reply]

Steel bollards on road - higher risk of accidents

Are there any scientific studies to what extent steel bollards on the streets increase the risk of accidents?

The type of bollards I'm talking about: https://www.bz-berlin.de/data/uploads/2019/05/radweg_1558535427-1024x576.jpg --88.78.14.45 (talk) 15:51, 2 March 2021 (UTC)[reply]

Probably decrease the risk of accidents at the expense of average vehicle speed. Abductive (reasoning) 16:02, 2 March 2021 (UTC)[reply]

It's worth considering what type of accident we are talking about. As positioned in the photo, it would appear that they are designed to prevent accidents involving a automotive vehicle hitting a bicycle or hitting a pedestrian. A moving chunk of metal vehicle, even at a gentle 40 km/h, hitting a pedestrian has a pretty large chance of killing said pedestrian. A vehicle moving at 40 km/h and hitting one of those bollards is unlikely to kill the occupants, especially if they are wearing seat-belts. On a risk/benefit analysis, I'd say that even if they somewhat increase the frequency of accidents as cars hit the things (cost), the benefit in eliminating a large portion of fatal accidents and replacing them with non-fatal accidents is high. --OuroborosCobra (talk) 17:14, 2 March 2021 (UTC)[reply]

Different types of barriers have certainly been studied. I don't know if you'll be able to find a paper on exactly the kind of barrier in that picture. They might not have been common enough long enough.

Here are some good starting points, though

An insight into the performance of road barriers − redistribution of barrier-relevant crashes You'll need journal access for this one. Or you'll need to fork out $41.95
Performance Assessment of Road Barriers in Indiana This one is free.

I really expected the NTSB to have something, but I could find anything. ApLundell (talk) 04:49, 3 March 2021 (UTC)[reply]

As an aside, does anyone know the American English word for "bollard"? I'm quite sure that word was not in my active vocabulary, but I'm familiar with objects of that sort and I don't know what I would have called them. Maybe just "barriers". I don't see anything at Wiktionary. --Trovatore (talk) 01:09, 6 March 2021 (UTC)[reply]

physical properties of octane isomers

I have searched for the following physical properties of octane isomers, "molar volume, total surface area and octanol-water partition coefficient", on internet but could not find anywhere, can any body help me to find it. — Preceding unsigned comment added by 182.187.18.182 (talk) 19:26, 2 March 2021 (UTC)[reply]

Before we run into an XY problem-type situation, can you elaborate the purpose for this information? If we know what larger problem you need this information for, perhaps we can better help you find the information that will help you. --Jayron 32 19:30, 2 March 2021 (UTC)[reply]

This may be helpful for the octanol-water partition coefficient... I found n-octane and 2,2,4, trimethyl pentane. There are a lot of C8H18 compounds though. There's n-octane, 3 different methylheptanes, 2 different ethylhexanes, 6 different dimethylhexanes, and a bunch of pentane derivatives I don't feel like counting, plus your 2,2,4,4 tetraethylbutane. --Jayron 32 19:42, 2 March 2021 (UTC)[reply]

The octanol-water partition coefficient article is too technical, in my opinion, and duplicates much of the information at partition coefficient (a much better article overall). The concept is very important in drug and pesticide design and chemists doing such work normally use the LogP as a vital parameter when comparing compounds within a given analogue series. Very many of Wikipedia's articles on commercial products of these types includes the LogP in the Chembox and there are online collections at repositories such as "The Pesticide Properties DataBase".. I have no idea why the OP wants the value for octane isomers but such values can be pretty reliably calculated when they have not been experimentally determined (see cLogP in the partition coeffficient article). Mike Turnbull (talk) 10:24, 4 March 2021 (UTC)[reply]

I have made a proposal to merge the articles and added the appropriate templates. Please comment at Talk:Partition coefficient#Merger proposal: Merge Octanol-water partition coefficient into this article if you wish. Mike Turnbull (talk) 10:37, 4 March 2021 (UTC)[reply]

To answer the OP's original question, many compounds in Pubchem have recorded data for their physical properties, so the the logP (octanol-water partition coefficient) is 5.18 [6], although our octane article says 4.783 without giving a specific source: it is probably a predicted value, since Chemspider agrees with Pubchem (the ultimate experimental source is MM Miller). No doubt Pubchem will have other isomers. Mike Turnbull (talk) 12:34, 4 March 2021 (UTC)[reply]

For molar volume, you can calculate it from the more easily accessible density and molar mass (which will be about 12×8+18=114 g) ie 114/ρ cc. Graeme Bartlett (talk) 07:31, 6 March 2021 (UTC)[reply]

For molecular total surface area, if you care about it, we probably need an article. For polar surface area, your answer will be 0Å². Graeme Bartlett (talk) 07:37, 6 March 2021 (UTC)[reply]

On the article front I have discovered Accessible surface area, Relative accessible surface area and Van der Waals surface. Are any of these describing the measurement you want 182? Graeme Bartlett (talk) 10:18, 6 March 2021 (UTC)[reply]

Higher taxa named after people

What higher taxa (from genus up to a family, order or higher, like Smithatris after Rosemary M. Smith) are named after people? Preferably full list, if available somewhere. 212.180.235.46 (talk) 23:58, 2 March 2021 (UTC)[reply]

Here's a query on WikiData: [7] that is as complete as the information on WikiData. It returns every taxon that's not at a low level (species, subspecies, variety, strain, or cultivar) and is listed as being named for something (not necessarily a person). There are quite a few genuses and some families named for people. There's a class arachnida named for the mythological person Arachne, and a phylum lokiarchaeota named for Loki's castle, so you can decide whether those count. This query doesn't return a fungal subdivision pucciniomycotina, which seems to be named for a Tommaso Puccini [8], although I'm not sure this is the right Puccini. --Amble (talk) 02:14, 3 March 2021 (UTC)[reply]

Ultimately, within taxonomic nomenclature, Pucciniomycotina gets its name from the genus Puccinia, which was established in 1729 by Micheli, named "in honor of Puccini, a Florentine professor".^[9] Its eponym cannot have been the art historian (not "professor of anatomy") Tommaso Puccini (1749–1811). Merriam–Webster states: "from Tommaso Puccini †1735 Italian anatomist".^[10] The bust shown in our article Tommaso Puccini is dated as being from 1718, and its sculptor died in 1725, so it also cannot depict the future art historian. --Lambiam 09:17, 3 March 2021 (UTC)[reply]

Other sources give 1666–1726 for the anatomist, who reportedly studied with Lorenzo Bellini and was connected to the Hospital of Santa Maria Nuova.^[11]^[12] --Lambiam 09:32, 3 March 2021 (UTC)[reply]

Well done. Do you think the anatomist is notable enough for an article? —Amble (talk) 15:16, 3 March 2021 (UTC)[reply]

He obviously was notable in his days, not only having a fungus named after himself (what was the connection there?) but also having a bust in marble by a famous sculptor as well as a medal with his effigy. The bust is in a museum in Pistoia and the inscription patrit(ivs) pistorien(sis) on the medal also connects him to Pistoia, which makes it likely he was a relative – perhaps the grandfather – of the art historian, a native of Pistoia. I also saw that the anatomist was included in the circle of friends of the painter Pietro Dandini.^[13] Nevertheless, altogether I did not spot enough material for an article. --Lambiam 23:54, 3 March 2021 (UTC)[reply]

One more snippet: Villone Puccini aka "Villa Puccini" in Pistoia was built by the anatomist.^[14] --Lambiam 00:26, 4 March 2021 (UTC)[reply]

Our page list of fungal orders lists kickxellomycota as a division; but kickxellomycotina is a subdivision and that page doesn’t list a division (“incertae sedis”). Whether division or subdivision, it’s named for Jean Kickx. I wonder whether this could be the highest example of taxonomic “promotion through the ranks,” where originally a genus was named after him, and the genus name eventually filtered up to (maybe) a division... —Amble (talk) 16:08, 3 March 2021 (UTC)[reply]

March 3

Is there a sea in europe of which its color is green rather than blue?

ThePupil (talk) 09:31, 3 March 2021 (UTC)[reply]

The apparent color of a body of water will depend on various factors, including what's in it, and what the sky looks like that day. ←Baseball Bugs ^{What's up, Doc?} carrots→ 09:51, 3 March 2021 (UTC)[reply]

To expand on that, if the water is clear, then it will look bluish (especially if the sky is blue too), as water absorbs red light and scatters and transmits blue. Sediment or algae in the water tend to absorb other colours, making the water look more green (or brown, or grey). See Ocean colour for a summary, or here for more detail. The sea around the UK often looks greenish (or greyish) due to sediment in the water and an overcast sky. Iapetus (talk) 10:35, 3 March 2021 (UTC)[reply]

Colour variations are probably better expressed in smaller bodies of water with particular conditions, such as at the Blue Pool in southern England, where the colour varies between "red-brown, through grey and green, to the more typical turquoise". PaleCloudedWhite (talk) 12:31, 3 March 2021 (UTC)[reply]

Why coronavirus Indian vaccine “Covishield” is more demanding than “pfizer” which has more efficacy?

India has undertaken contractual supplies of coronavirus vaccines to Saudi Arabia, South Africa, Brazil, Morocco, Bangladesh and Myanmar, the Ministry of External Affairs (MEA) said in New Delhi on Friday.

Source Why coronavirus Indian vaccine "Covishield" is more demanding than "pfizer" which has more efficacy ? Rizosome (talk) 12:05, 3 March 2021 (UTC)[reply]

@Rizosome: What do you mean by "more demanding"? If you saw the expression somewhere then please link to it. It's not used in your link. PrimeHunter (talk) 13:05, 3 March 2021 (UTC)[reply]

I'd guess that by "more demanding" what's meant is "in greater demand". Deor (talk) 18:51, 3 March 2021 (UTC)[reply]

Maybe, but a story about some poor countries getting a cheap vaccine doesn't indicate a greater demand. Looking at other posts by Rizosome, I guess we shouldn't expect much. PrimeHunter (talk) 19:16, 3 March 2021 (UTC)[reply]

Covishield is the Indian name for the Oxford–AstraZeneca COVID-19 vaccine. It is easier to store and distribute than the Pfizer–BioNTech COVID-19 vaccine which needs to be stored at very low temperatures. The Indian-developed Covaxin has been criticised by some because it was approved for emergency use by the Indian regulator before the clinical trials had been completed. See BBC - Covaxin and Covishield: What we know about India's Covid vaccines. Initial doubts about the effectiveness of the AstraZeneca vaccine in the elderly have been dispelled by recent data from the UK where it has been widely used; see Pfizer, AstraZeneca COVID-19 vaccines may offer high efficacy in elderly. Alansplodge (talk) 15:01, 3 March 2021 (UTC)[reply]

March 4

Why is Nyiragongo a stratovolcano?

Given the fluidity of its lava, why isn't it a shield volcano? ZFT (talk) 04:27, 4 March 2021 (UTC)[reply]

That's a good question, the neighbouring Nyamuragira is a shield volcano, although it erupts high potassium basalts. I've looked through this thorough review and I can't find a clear explanation - apart from the earliest stage eruptions, the volcano is dominated by lava flows. Mikenorton (talk) 09:04, 4 March 2021 (UTC)[reply]

I've seen a discussion a while ago on another volcano that the eventual slope of a volcano depends on the median lava flow length, which in turn is a function of the median volume of each eruption. So even a volcano with very liquid lavas will grow steep if the median lava flow is short. Jo-Jo Eumerus (talk) 11:19, 5 March 2021 (UTC)[reply]

Do astronomical observatories really shoot laser into sky like this?

Do astronomical observatories really shoot laser into sky like this? Here is the picture taken from Optical amplifier Rizosome (talk) 06:28, 4 March 2021 (UTC)[reply]

See laser guide star for an explanation of the technique - why would you doubt it? Mikenorton (talk) 08:44, 4 March 2021 (UTC)[reply]

In usual terminology, space observatories are observatories located in outer space. Observatories observing space (usually from the ground, but sometimes from space or even a high-altitude balloon, aircraft or from below the Earth's surface) are called astronomical observatories. And yes, sometimes they shoot a laser into the sky. Apart from lunar ranging, it's sometimes done for adaptive optics. Adaptive optics uses a deformable mirror to correct for diffraction of light in moving bubbles of air of different density. To measure the required corrections, the system has to observe a sufficiently bright reference star (much brighter than the object you're interested in). If no sufficiently bright star is available close enough to the object you want to observe, a laser can be used to project one in the higher parts of the atmosphere. Of course, they have to make sure they don't hit an aircraft passing over, as those lasers can be very annoying to pilots (just as aircraft can be annoying to astronomers). PiusImpavidus (talk) 09:02, 4 March 2021 (UTC) @PiusImpavidus: I changed my title, still your answer supports my question? Rizosome (talk) 13:46, 4 March 2021 (UTC)[reply]

Yes, that was already the question I answered. The laser-equipped observatories are always ground-based, not in the air, in space or below ground. Except gravitational wave observatories, which are laser-equipped and can be below ground, but don't shoot their laser beams into the sky. PiusImpavidus (talk) 16:35, 4 March 2021 (UTC)[reply]

The Royal Observatory, Greenwich has a similar laser, The Meridian Laser, which only serves to show the location of the Greenwich Meridian and has no scientific purpose beyond education. It's more of a museum than a working observatory, but one of their historic telescopes has been brought back into use recently. Alansplodge (talk) 15:30, 4 March 2021 (UTC)[reply]

Yes, I once saw that, when leaving the Maritime Museum on a winter afternoon/evening (after the fire alarm went off). PiusImpavidus (talk) 16:35, 4 March 2021 (UTC)[reply]

March 5

Seabird IDs

In this video, what are the cliff-nesting birds first seen at 23 seconds, and the tern-like birds at 3:30? --24.43.123.79 (talk) 00:19, 5 March 2021 (UTC)[reply]

The birds at 23 seconds are probably Bermuda petrels, a deduction based on the fact that the list of CornellLab's bird cams shows they have a couple of cams covering this species. PaleCloudedWhite (talk) 07:46, 5 March 2021 (UTC)[reply]

My best guess for the tern-like birds - based not on personal knowledge, but on perusing lots of google searches - is red-billed tropicbird. The images at our article seem to have very red bills - maybe too red for the birds in the video - though the images of the same species at this google image search seem to fit better. PaleCloudedWhite (talk) 09:10, 5 March 2021 (UTC)[reply]

Why Space X didn't reveal the reason behind SN 10 failure?

Why Space X still didn't reveal the reason behind SN 10 failure? Rizosome (talk) 02:49, 5 March 2021 (UTC)[reply]

Is that the one that blew up within the last day or so? ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:26, 5 March 2021 (UTC)[reply]

Maybe because they aren't yet completely sure what the reason was? Our article already says "due to a suspected methane leak." In any case, as a privately owned company conducting tests on prototypes, they are under no obligation to reveal their findings immediately, and will doubtless only do so when and if it becomes advantageous for them. {The poster formerly known as 87.81.230.195} 2.125.75.168 (talk) 06:20, 5 March 2021 (UTC)[reply]

It's been a day and a half, what did you expect? It could be easily observed that at least some of the landing legs didn't lock into place, so that might have something to do with it. However, SpaceX (Elon Musk's twitter probably) will let us know when they worked it out, they always have in the past (cf the Starlink 19 landing failure already having been explained). Fgf10 (talk) 09:34, 5 March 2021 (UTC)[reply]
(ec) Usually they provide a full explanation for such failures, but the investigation needed for a full explanation takes time. If there was a methane leak, then the question is, how come? A leak can perhaps explain the explosion, but there should have been no leak in the first place. Did something rupture? If so what, how, and why? There is also some chatter about the landing legs not properly deploying,^[15] which may have been a factor in a chain of mishaps. It is wonderful how the Space X people (that is, people of Space X, not X-Men from space) describe the incident in a self-congratulory text on their website:

As if the flight test was not exciting enough, SN10 experienced a rapid unscheduled disassembly shortly after landing.

Fortunately there was no one on board, or they too would have experienced a rapid unscheduled disassembly. --Lambiam 09:45, 5 March 2021 (UTC)[reply]

Reports yesterday said initially it's suspected there was a leak. Hence the kaboom. That term "unscheduled disassembly" reminds me of when Challenger exploded in mid-air, and the NASA announcer describing things while they were happening called it "a major malfunction". ←Baseball Bugs ^{What's up, Doc?} carrots→ 13:57, 5 March 2021 (UTC)[reply]

First, the announcer who said that didn't have access to live video of the flight. Second, there was no explosion; quoting from memory, what people misperceived as an explosion was only "a fast, almost explosive fire". Third, if by the Challenger you mean the orbiter, it not only didn't explode, it came away from that event intact, only to be destroyed on impact with the sea surface. --142.112.149.107 (talk) 21:54, 5 March 2021 (UTC)[reply]

How do you know he didn't have the live video feed? ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:11, 5 March 2021 (UTC)[reply]

He "was reading routine values off gauges at his console". 39 seconds later he reported that the flight dynamics officer had reported an explosion. --142.112.149.107 (talk) 23:44, 6 March 2021 (UTC)[reply]

It can take months of investigation to understand the causes of such disasters. The Challenger disaster investigation took a long time; the Rogers Commission Report was released in June of 1986, the Challenger disaster happened in January of that year. The Columbia disaster investigation took a similar amount of time, the Columbia Accident Investigation Board released a report in August, 2003 when the accident happened in February. The VSS Enterprise crash investigation took even longer; the final findings were released in July 2015, nine months after the October 2014 crash. I wouldn't expect anything for some time. --Jayron 32 12:58, 5 March 2021 (UTC)[reply]
SpaceX doesn't operate like that in most cases. If the cause is obvious either they or Elon Musk tends to release it within days. That's what has happened with previous Starship tests and Falcon 9 landing failures. The exceptions being the two failures on launches for external customers . Fgf10 (talk) 19:24, 5 March 2021 (UTC)[reply]
But what if they don't know? The OPs question of "Why hasn't SpaceX said why the rocket failed?" has an obvious answer. "They don't know yet". --Jayron 32 20:39, 5 March 2021 (UTC)[reply]
These are tests that are conducted with the express goal of finding out what failures may happen so that the design can be improved. Everything is heavily instrumented and recorded to make it possible to quickly get at the bottom. --Lambiam 20:46, 5 March 2021 (UTC)[reply]

Elon Musk:

Thrust was low despite being commanded high for reasons unknown at present, hence hard touchdown. We’ve never seen this before.

Next time, min two engines all the way to the ground & restart engine 3 if engine 1 or 2 have issues.

Tweeted 2010-02-06 01:51 UTC.

Remember that the SpaceX Raptor engine is still being developed, tested, and refined. -- ToE 02:18, 6 March 2021 (UTC)[reply]

Role of anaerobic metabolism in the global carbon cycle

Has anyone ever tried to quantitatively estimate the role of anaerobic metabolism in the global carbon cycle? Jo-Jo Eumerus (talk) 11:17, 5 March 2021 (UTC)[reply]

The citation quoted in anaerobic metabolism that's most likely to have a good estimate is (on Pubmed) [16] which has a "simliar articles" section that could also point to useful items. Ironically, at first sight one might think that organisms which take carbon dioxide out of the atmosphere, as some anaerobes do, would be good to mitigate global warming. Unfortunately, they turn it into methane, which is an even worse greenhouse gas! Mike Turnbull (talk) 14:10, 5 March 2021 (UTC)[reply]

Unfortunately, that source is focused on man-made emissions, but I need information on all anaerobic throughput. Jo-Jo Eumerus (talk) 10:51, 6 March 2021 (UTC)[reply]

@Jo-Jo Eumerus:: I know of one book that covers the GCC pretty comprehensively, and does mention anaerobes. The book is Mannion, A. M. (12 January 2006). Carbon and Its Domestication. ISBN 1402039565., with e-book at ISBN=1402039581. Looking at my copy, on page 53 it has a carbon budget in units of 10⁹ tC yr^-1 that talks about a "missing sink" — maybe that could be where anaerobes have a role? The book refers to the IPCC's "Climate Change 2001. The Scientific Basis", which has its preface/foreword "here" (PDF).. Later reports may be even more relevant. Mike Turnbull (talk) 13:14, 6 March 2021 (UTC)[reply]

March 6

Equation regarding amount of heat in a certain amount of matter.

Hi, I was looking for an equation which tells me how much a certain amount of heat in a certain amount of matter will translate into.

As a reference, it has been said that the head of a pin as hot as the surface of the sun (10,000 Fahrenheit) could kill a person from 90 miles away.

I was looking for a specific equation though - for example, assuming no other variables - If I was standing 90 miles away from a pinhead (let's say 1.5 mm diameter) as hot as the sun's surface (10,000 F) - how hot would it be where I'm standing? Thanks.--IBBishops (talk) 00:04, 6 March 2021 (UTC)[reply]

Simple answer: The Sun subtends an angle of approximately 0.52° as observed from here on Earth. Since a pinhead at 90 miles will appear much, much smaller, you would feel much, much less radiant heat from it than you feel from the Sun.

If your pinhead has a diameter of 1.5 mm, then at 90 miles it subtends only 5.9×10^-7 °. Since you are interested in the solid angle of the heat source, you want the square of the ratio of the subtended angles: Thus the received radiant heat from the pinhead will be (5.9×10^-7 ° / 0.52°)² = 1.3×10^-12 that of the sun. Inversely, the sun will feel 770 billion (7.7×10¹¹) times warmer to you. -- ToE 03:50, 6 March 2021 (UTC)[reply]

Nearly as simple answer: Per Solar radius: 1 R_☉ = 695,700 km, so your pinhead is 0.75 mm / 695,700 (km/R_☉) = 1.1×10^-12 R_☉ in radius.

Per Astronomical unit: 1 AU = 92955807 miles, so your pinhead is 90 mi / 92955807 (mi/AU) = 9.7×10^-7 AU away.

Given its temperature, the amount of heat radiated by the sun (or your pinhead-sun) is a function of its surface area, which varies with its radius squared. The portion of that heat your body intercepts is inversely proportional to your distance squared from the heat source. So the heat you receive from your pinhead will be (1.1×10^-12 / 9.7×10^-7)² = 1.3×10^-12 that of the sun. Inversely, the sun will feel 770 billion (7.7×10¹¹) times warmer to you. -- ToE 04:43, 6 March 2021 (UTC)[reply]

These two back-of-the-envelope calculations match because they are in essence the same geometrical argument.

But they both ignore the practical issue that if you were to be in 90 mi line-of-site of your pinhead, its radiant heat would be further attenuated by passing through much, much more atmosphere than that of the mid-day sun. -- ToE 04:43, 6 March 2021 (UTC)[reply]

But you want a black-body radiation calculation? Stefan–Boltzmann will oblige.

The total power emitted by your pinhead is 4πr²σT⁴ = 4π(0.00075 m)²(5.67×10−8 W⋅m⁻²⋅K⁻⁴)(5800 K)⁴ = 454 watts.

(That's slightly less than 1/3 that of a typical 1500 W electric space heater!)

Divide that be the area of a 90 mi radius sphere = 4π(90 mi)² = 2.63×10¹¹ m² to determine the irradiance of 1.72×10^-9 W/m².

The solar constant is 1361 W/m2, which is 791 billion times greater than the calculated irradiance of the pin, which is a close enough match given the precision of these calculations.

(Both the solar constant and the calculated irradiance of the pin do not consider any atmospheric attenuation.) -- ToE 05:48, 6 March 2021 (UTC)[reply]

But you asked "how hot would it be where I'm standing?" and I instead answered what the radiant flux density would be.

We could instead calculate your planetary equilibrium temperature as if you were a rotating (or heat conductive) black body in a universe empty of everything else except your 5800 K pinhead 90 miles away. How warm would it keep you?

T_{\rm {you}}={\sqrt[{4}]{{\frac {A_{\rm {abs}}}{A_{\rm {rad}}}}{\frac {1}{\sigma }}{\frac {L}{4\pi D^{2}}}}}

Where:

A_abs/A_rad is the ratio of your absorption area to your radiation area. In this case it is 1/4, which is the ratio of the area of the disk, πr², you present perpendicular to the pinheadlight to that of your entire surface area, 4πr², with which you reradiate that energy via your own black body radiation. (We are determining your equilibrium temperature where those energy fluxes balance.)

L/(4πD²) is the irradiance we calculated above, with L being the total luminosity of your pinhead and D the 90 mi between you two.

The 1/σ and fourth root are from us solving the Stefan–Boltzmann law for temperature.

And for a black body albedo a = 0 & emissivity ε = 1 so those terms don't appear.

Thus T_you = ( (1/4)(1/(5.67×10−8 W⋅m−2⋅K−4))(1.72×10-9 W/m2) )^1/4 = 0.295 K.

So as hot as that pinhead is, it does almost nothing to keep you warm. (Though there are colder spots in Florida.)

If you wish to recalculate for objects of different sizes and distance, its much better to substitute in our luminance formula and simplify.

T_{\rm {you}}={\sqrt[{4}]{{\frac {1}{4}}{\frac {1}{\sigma }}{\frac {4\pi R^{2}{\sigma }T^{4}}{4\pi D^{2}}}}}={\sqrt[{4}]{{\frac {1}{4}}{\frac {R^{2}}{D^{2}}}T^{4}}}={\sqrt {\frac {R}{2D}}}T

Redoing our calculation:

T_you = ( 0.00075 m / (2 ⋅ 144841 m) )^1/2 ⋅ 5800 K = 0.295 K.

Más sabe el diablo por viejo que por diablo, and the parallel moral here is that the power of the sun comes more from its enormous size than simply from its temperature. -- ToE 14:53, 6 March 2021 (UTC)[reply]

None of the above calculations consider how your pinhead manages to hold together at such a temperature.:

Nor do they consider the 454 watt power source which maintains it temperature.

Without power input it will quickly dim due to its tiny surface-to-volume ratio.

V = (4/3)π(0.075 cm)³ = 1.77×10^-3 cm³.

If we assign it the density and specific heat capacity of steel, roughly ρ = 7.87 g/cm³ and c = 0.466 J/(g⋅K),

and we assume that specific heat capacity remains constant across a wide range of temperatures and there are no phase changes to deal with,

then your pinhead's heat capacity will be 1.77×10^-3 cm³ ⋅ 7.87 g/cm³ ⋅ 0.466 J/(g⋅K) = 6.48×10^-3 J/K, and even multiplying that by the full 5800 K gives us only 37.6 J. So it clearly can't maintain it's 454 watt luminous output for but a fraction of a second, and it will rapidly dim.

This is no different than a pinhead of steel being blasted by the 40,000° F jet from a plasma cutter. It will be bright but ephemeral. If you are in the workshop, then you should be wearing welding goggles, but 90 miles away, you won't even be aware of its existence.

Compare this to the sun, where other than the sudden drop in neutrino flux, we wouldn't notice anything for 10,000 years were its fusion to somehow suddenly stop. -- ToE 15:58, 6 March 2021 (UTC)[reply]

As pointed out by ToE, the solid angle has to be similar to that of the Sun. This means that at a large distance the object needs to be huge to lead to enough radiant flux. Take e.g. a fireball from a large impact. Or large gas fires: "The fires from the wells and the oil and gas lines (all of which ruptured, one by one) had produced flames with a height of about 200 metres and a peak rate of energy consumption of ~100 gigawatts, three times the rate of UK total energy consumption." In this case the radiant heat was so large that it was felt inside rescue helicopters at a distance of 1 km away. Count Iblis (talk) 02:59, 7 March 2021 (UTC)[reply]

RBMK part 2

So I've read that paper about the Chernobyl disaster which was linked in response to my previous questions about the RBMK reactor (where the hell did the article go?!), and while its main hypothesis is not plausible (it claims there was an actual nuclear explosion during the disaster, among other implausible claims), it does bring up at least one salient point: it says that the RBMK main circulation pumps have a built-in low-voltage/low-frequency trip (which would explain why they wound down so fast all 3 times after the turbine was tripped), and that this, rather than the graphite tips of the control rods, was the reason for the thermal runaway at Chernobyl. So my question is: (1) is it true that the pumps trip automatically due to low voltage and/or frequency, and (2) if so, is it plausible that at least part of the role which the INSAG-7 report attributes to the graphite tips in causing the reactivity spike was actually played by the abrupt stoppage of the circulation pumps (as the paper alleges)? 2601:646:8A01:B180:7446:DB1C:3033:BF8E (talk) 03:22, 6 March 2021 (UTC)[reply]

To answer your first question, see RBMK. Since the "R" part of the acronym stands for "Reaktor", I guess calling it the "RBMK reactor" or "High-Power Channel-type Reactor reactor" is considered tautological. {The poster formerly known as 87.81.230.195} 2.125.75.168 (talk) 05:32, 6 March 2021 (UTC)[reply]

But it would make sense for RBMK reactor to be a redirect to RBMK. Anyone feel like creating it? --142.112.149.107 (talk) 23:57, 6 March 2021 (UTC)[reply]

We have an article on that: RAS syndrome. -- ToE 06:08, 6 March 2021 (UTC)[reply]

Sorry, the article does not answer my first question -- it says that the coolant pumps are prone to cavitation (which I already knew from reading said article, and which was specifically mentioned in both INSAG-1 and INSAG-7 as a contributing factor to the disaster), but it says nothing about the power being cut to said pumps! 2601:646:8A01:B180:7446:DB1C:3033:BF8E (talk) 07:57, 6 March 2021 (UTC)[reply]

The first question you asked in this thread seems to be "where the hell did the article go?!". It may not have been numbered, but since it was the first question you asked, it seems reasonable to call it "your first question". The article is called RBMK not RBMK reactor for the reasons stated which is an answer to that question. (It doesn't look like it was ever called anything else [17] although I don't know if any redirects were deleted. It looks like it was only ever wikilinked under RMBK in Wikipedia:Reference desk/Archives/Science/2021 February 22#RBMK reactor.) No one said the article answers any of your other questions AFAICT. To avoid confusion, it might be better to ensure you number all your questions, or don't number any. Nil Einne (talk) 14:28, 6 March 2021 (UTC)[reply]

March 7

Falcon 9 launch

In the wee hours of March 4, I was photographing the launch of a Falcon 9 from Cape Canaveral, from 180 miles north of there. About 5-6 minutes after launch, it became visible to me again and I took this 39-second exposure. I could see only a faint dot of the flame from the engines, but this photo shows some sort of purplish gas or something below it. I've gotten something like it once before and I've seen other photos showing it. What is it? Bubba73 ^{You talkin' to me?} 02:25, 7 March 2021 (UTC)[reply]

Twilight phenomenon perhaps? --jpgordon^{𝄢𝄆 𝄐𝄇} 06:01, 7 March 2021 (UTC)[reply]

Perhaps, but this happens when it is not too long after sunset or too long before sunrise (30-60 minutes), when sunlight is shining on it. This launch was at 3:24AM and sunrise was at 6:45AM.

I'm guessing that it is (1) dumping fuel or some gas, or (2) it is ionizing the gas in the ionosphere. But I'd like to know. Bubba73 ^{You talkin' to me?} 06:36, 7 March 2021 (UTC)[reply]

Falcon 9 launches (and first stage landings) are continuously filmed from various cameras on the ground and several mounted on/in the first and second stages, and shown (initially live) on SpaceX's YouTube Channel, where they remain available thereafter (as well as on other space-related channels). Ordinarily it would be easy for you to watch the appropriate video and see what events and activities correlate with your exposure, but as it happens the Stage 1 downlink apparently failed this time around: however, you could watch some earlier Starlink missions to see similar Stage 1 events.

As a personal observation: the exhaust gases from such rocket launches spread more broadly as the rockets get higher because of the reducing atmospheric pressures they are expelled into, and around dawn or sunset as seen from those altitudes (per Jpgorden), the Sun's oblique illumination combined with a perspective unusual to ground watchers can give rise to some otherwise rarely seen phenomena.

Falcon 9 lanches do not generally involve 'dumping fuel', but after the Second stage has separated (around 2:40 from lift-off and 70km altitude on this occasion) and its single engine started (a few seconds later at about 75km altitude), the now-ballistic First stage adjusts its attitude using inert pressurised nitrogen jets and later (6:29 from lift-off at 55km altitude this time) re-ignites three of its nine engines in a re-entry burn (of about 20 seconds) to slow it down (and again for a low-altitude landing burn later). {The poster formerly known as 87.81.230.195} 2.125.75.168 (talk) 08:09, 7 March 2021 (UTC)[reply]

Can heavy water be used as a food preservative?

The stronger hydrogen bonds of deuterium compared to hydrogen derails biochemical processes, causing organism that take in too much heavy water to die. Can this be exploited to inhibit bacterial and fungal growth after harvesting of crops by watering crops with heavy water some time before harvesting? Count Iblis (talk) 03:08, 7 March 2021 (UTC)[reply]

Charge a black hole

I understand that having a black hole with a significant charge is p~0 in practice. But suppose that you could dump somehow a few stellar masses of electrons into a black hole. Could you by adding enough charge overcome the black hole's gravitational cohesion? 93.136.167.229 (talk) 05:17, 7 March 2021 (UTC)[reply]