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June 9

Weigeltisaurus species clarification

Is this Weigeltisaurus reptile in this family Rhynchocephalia. Would that statement be true to say. Its for this article Johannes Weigelt. There is source that states it but I don't know how accurate it is. scope_creep^Talk 21:23, 9 June 2024 (UTC)

While Weigeltisaurus flourished in the Late Permian, our article on the order Rhynchocephalia states that the oldest record of the group is dated to the Middle Triassic. According to this chronology, a gap of several million years separates them.  --Lambiam 07:21, 10 June 2024 (UTC)

Right. I'll leave the Rhynchocephalia bit out and only mention the Weigeltisaurus bit, since I don't understand it. Thanks @Lambiam: scope_creep^Talk 08:52, 10 June 2024 (UTC)

June 12

Grignard Reagent with Haloalkane

The reaction between a Grignard reagent (${\ce {RMgX}}$) and an alkyl halide (${\ce {R'-X}}$) gives what product? I see some sources saying there would be substitution reaction thus forming ${\ce {R-R'+MgX2}}$, whereas other source says there would be transmetallation thus forming ${\ce {R'MgX +R-X}}$. Thanks for your time, ExclusiveEditor _{Notify Me!} 12:19, 12 June 2024 (UTC)

The WP article on Grignard compounds itself states that "Grignard reagents do not typically react with organic halides, in contrast with their high reactivity with other main group halides. In the presence of metal catalysts, however, Grignard reagents participate in C-C coupling reactions." Also in this article there is a diagram of different reactions that show the reaction between R'-X and RMgX to yield the C-C coupling product (where R' should be a carbonyl-like moiety such as benzyl or allyl). Logically, the MgX⁺ cation should strongly bond with the halogen atom of R'-X during the reaction, resulting in a very stable salt MgX₂. The more stable the products are, the more favorable a reaction pathway is. 2402:800:639D:9B0A:C9C4:BA57:2B8:DC89 (talk) 14:01, 16 June 2024 (UTC)

Yes, but as you state, it happens in presence of metal catalyst (transition metal). My question doesn't mention our reaction having any such catalyst. ExclusiveEditor _{Notify Me!} 19:11, 21 June 2024 (UTC)

On otherhand, the transmetallation reaction requires that both two of the reactants must be organometallic compounds (i.e, contains at least a metal-carbon bond), which is not the case with one of your reactants R'-X.Vanadium-3065 (talk) 14:08, 16 June 2024 (UTC)

When forming a Grignard reagent by the standard method of adding an alkyl halide like EtBr to Magnesium turnings suspended in ether, controlling the exotherm by the rate of addition, there will be a point where the halide is being added to material that contains substantial amounts of EtMgBr. If coupling (i.e. the substitution reaction) were fast, the main product would be butane and it would be impossible to form a good yield of the Grignard reagent! My trusty "Advanced Organic Chemistry" by Jerry March says that you can deliberately encourage the coupling reaction by adding thallium (I) bromide, Cr, Co or Cu chlorides or various other reagents, depending on the type of Grignard reagent you have already formed. The mechanism is via the R-Metal intermediate which decomposes via free radicals. Mike Turnbull (talk) 15:08, 17 June 2024 (UTC)

@Vanadium-3065: You may be right. But then the question is still half answered. It still does not say if R-R' will be formed or not. If not transmetalation, there could be a metal-halogen exchange as stated by DMacks below. ExclusiveEditor _{Notify Me!} 19:13, 21 June 2024 (UTC)

@Vanadium-3065, Michael D. Turnbull, and 2402:800:639D:9B0A:C9C4:BA57:2B8:DC89: To clarify, I am not an advanced chemistry student, so I don't realize things which really require deep insights. However one thing I would mention is that the statement that R-R' would not be formed but transmetallation would happen is written in Clayden (available at pg 189 at pdfs available online.). Maybe this would help clear any confusion, or if the publisher should be informed that the strict language that R-R' would not be formed is wrong. Reply appreciated, ExclusiveEditor _{Notify Me!} 16:13, 21 June 2024 (UTC)

The Clayden statement is correct and corresponds to what I said above. In order to get R-R' you need to add a transition metal. Mike Turnbull (talk) 17:22, 21 June 2024 (UTC)

I just saw Kumada coupling and read that page of Clayden again. So we need some transition metals like Ni/Pd to act as catalyst to form carbon-carbon bond (R-R'). If it is not available, then transmetalation takes place in equilibrium? ExclusiveEditor _{Notify Me!} 18:18, 21 June 2024 (UTC)

The logic of that Clayden statement seems incorrect (or circular at best). It says that substitution "does not work because of transmetallation. The two alkyl bromides and their Grignard reagents will be in equilibrium with each other so that, even if the coupling were successful, three coupled products will be formed." (my underlining). That doesn't sound like a reason why it doesn't work, just potential evidence that it doesn't work and not even stating that the evidence is true. And isn't this really metal–halogen exchange (swapping M and X on the R groups) rather than transmetallation (M and M', like the Schlenk equilibrium)? DMacks (talk) 18:19, 21 June 2024 (UTC)

So I think my question could be summarized as, what product do we get, if we conducted Kumada coupling without any catalyst, and the answer supposedly is that we get transmetalation (or metal–halogen exchange as said by DMacks) per Clayden. It is only if we use some transition metal like Ni/Pd as catalyst, that we get a carbon-carbon bond (R-R'). ExclusiveEditor _{Notify Me!} 18:41, 21 June 2024 (UTC)

Yes but.... like many things in practical organic chemistry, the issue is not simply whether a particular product forms. Considerations of side-products, rate of reaction and hazard (among others) also come into play. The Wurtz reaction takes R-Hal with sodium and gives coupled products R-R but as our article says is "of little value". That's because in general side-reactions like elimination and rearrangement lead to low yields. The free-radical intermediates are too difficult to control in practice. Hence chemists have developed milder and more specific methods for coupling, of which Kumada coupling is one and there are many others e.g. listed in the "see also" section of that article. These are now a key group of important C-C bond-forming reactions finding extensive general use. I agree with DMacks that Clayden could have expressed himself better. He was arguing that even if the coupling were successful, three coupled products will be formed as if it were the main reason not to try the reaction R₁CH₂MgBr + R₂CH₂Br -> R₁CH₂CH₂R₂ (i.e. because you would also get R₁CH₂CH₂R₁ and R₂CH₂CH₂R₂) whereas in fact the main reason this sort of coupling is unsuccessful is that the rate of any useful reaction is too low. Mike Turnbull (talk) 11:16, 22 June 2024 (UTC)

Check a SpringerLink reference

http://doi.org/10.1007/978-94-017-9861-7_22

Does anyone have access to this? If so, does it make any reference to the diet or hunting behaviour of Euthyrhynchus floridanus? Such claims were added without reference to this insect's article four years ago ([1]), and later someone dumped in a reference to this book. Nyttend (talk) 21:15, 12 June 2024 (UTC)

I don't think so. The closest I can find is "However, only a few notes about its life history on the field are known (Avila-Núñez et al. 2009 )" The reference is "Avila-Núñez JL, Ortega LDO, Pisarelli MPC (2009) Un caso de depredación de adulto de Gonodonta pyrgo Cramer 1777 (Lepidoptera: Noctuidae) por Euthyrhynchus floridanus(Linnaeus 1767)(Heteroptera: Pentatomidae: Asopinae). Entomotropica 23:173–175". (Btw, you should be eligible for Wikipedia Library; that's how I could access the book).--Wrongfilter (talk) 21:46, 12 June 2024 (UTC)

@Nyttend: You can get its access at Wikipedia Library, publications at springer are available there. However, for this one there is no mention of things that were claimed, in that chapter at least, so there removal is legitimate. ExclusiveEditor _{Notify Me!} 16:41, 21 June 2024 (UTC)

June 13

Synchronous orbit

Diagram showing my possibly flawed understanding of astronomic poles of a body with a synchronous orbit.

I think I understand the different poles (near pole, far pole, leading pole, trailing pole) and associated hemispheres of a body with a synchronous orbit, but I'd like to double-check. I've drawn (badly) a diagram showing my understanding of these things (see right), and I've given a shot at explaining it at Rhea (moon):

Rhea is tidally locked and rotates synchronously; that is, it rotates at the same speed it revolves (orbits), so one hemisphere is always facing towards Saturn. This is called the near pole. Equally, one hemisphere always faces forward, relative to the direction of movement; this is called the leading hemisphere; the other side is the trailing hemisphere, which faces backwards relative to the moon's motion.

Is this right, or am I missing something? Cheers, Cremastra (talk) 21:35, 13 June 2024 (UTC)

I have the impression some assumptions are missing from the definitions in our articles Poles of astronomical bodies and/or Synchronous rotation. Without additional assumptions, the aspect of the satellite as seen from the body it circles is subject to possibly significant libration. One assumption is that the orbit of the satellite is circular. The other is that its axis of rotation is aligned with the normal to the orbital plane – in other words, its poles of rotation and orbital poles coincide. Under these assumptions, your diagram corresponds with the definitions given. The intersection of the green–yellow band and the red–blue band is a pole of rotation and the three axes defined by the various kinds of pairs of opposite poles are at right angles to each other.  --Lambiam 18:35, 14 June 2024 (UTC)

Because of the curve of the orbit, the line between the true rearmost and forward-most points is inward from the geometric center of the satellite. The article implies that this difference is ignored in practice. —Tamfang (talk) 20:23, 18 June 2024 (UTC)

June 14

Australian fizzy cans

What is the average mass of an (empty) Australian fizzy can of the typical size? It's easy to find pages talking about the amount of liquid a typical can holds (375 ml), but I can't find anything about the mass of the aluminium. Drink can gives a US figure, but that's not useful because cans here are a different size, and also it's a really rough estimate, being derived from a US-EPA source that says "assume 34 cans per pound". Nyttend (talk) 02:49, 14 June 2024 (UTC)

14.9 grams, according to the Australian Aluminium Council. [2] Figure is from 2001 and the trend seems to be downward so perhaps they're lighter today.  Card Zero  (talk) 04:41, 14 June 2024 (UTC)

An empty 330 ml beer can in New Zealand is 14-15 g. I expect an Australian can would be the same.-Gadfium (talk) 05:13, 14 June 2024 (UTC)

Card Zero, thank you! Information added to the drink can article. If I had a large quantity, I suppose I could just weigh them and divide by the number, but I don't have a good scale, and I don't have room to hold a huge number for a representative sample anyway. Plus, I'd rather take them (and little cans, and plastic bottles, etc.) to redeem the deposit without amassing a huge quantity. Nyttend (talk) 05:21, 14 June 2024 (UTC)

About how much would a full can of Foster's weigh? ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:23, 14 June 2024 (UTC)

Don't ask an Australian that question. They never drink Fosters. HiLo48 (talk) 06:09, 14 June 2024 (UTC)

TV ads in America used to say that Foster's is Australian for "beer". Evidently that is not altogether true? ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:51, 14 June 2024 (UTC)

Not in the last 40 years. The same goes for throwing another shrimp on the barbie. We call them prawns, not shrimps. HiLo48 (talk) 23:31, 14 June 2024 (UTC)

Some Foster's ads, for your possible amusement. The comments are telling. [3] Also, Americans may not know shrimps from prawns, but we know all about barbie. There was a major movie about it last year. :) ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:09, 15 June 2024 (UTC)

Google says it's the typical size of 375ml, so add in the can and we're at about 390ml, just like Coca-Cola or Carlton Draught or Victoria Bitter. I know much more about alcohol containers than I did before Victoria introduced container deposit last year...never had an idea that Coke-diluted whiskey existed (what's the point of diluting a distilled beverage?), let alone was sold in cans with US-specific branding like Bourbon County, or was sold in 200ml cans. But I don't remember seeing Foster's cans in any of the neighbourhoods where I go on foot and collect discarded containers. Nyttend (talk) 05:37, 14 June 2024 (UTC)

Regarding your question, Nyttend. (i) Many people find the taste/mouthfeel of some undiluted spirits over-strong. (ii) Sometimes one wants to drink a larger volume and a weaker ABV than the undiluted spirit; with less intensely flavoured spirits, such as vodka, this can effectively produce 'alcolohic dilutant' (I used to be fond of screwdrivers, for example). (iii) The combination of the spirit and dilutant can result in quite different new and desirable tastes: this is partly chemical – for example, a single drop of water added to a glass of single-malt whisky can have a very noticable effect, I'm told. {The poster formerly known as 87.81.230.195} 151.227.226.178 (talk) 17:47, 18 June 2024 (UTC)

Thank you! As a non-drinker, this is all strange to me. In particular, I assumed that the point of drinking distilled beverages was experiencing the alcohol intensely. Also wouldn't have guessed that the diluted beverage would taste anything other than a proportionate mix of the flavours, e.g. in the water-in-single-malt-whiskey example, I figured it would merely weaken the taste, like putting water into milk or fruit juice. Nyttend (talk) 19:49, 18 June 2024 (UTC)

The reason I ask is that in Monty Python at the Hollywood Bowl, they were throwing what appeared to be full Foster's cans into the crowd. It occurred to me that if they were actually full, that could be dangerous. ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:51, 14 June 2024 (UTC)

Who is "Professor Cleveland Abbe" (Meteorology)

I might as well ask here as someone may be bored and can help solve a mystery. I recently added information published by "Professor Cleveland Abbe" to History of tornado research#19th century (specifically this article). He was the editor of Volume 25, Issue 6 of Monthly Weather Review, published in June 1897. However, after reading the "General Characteristics" (basically the introduction) to the academic paper, it doesn't actually say where Mr. Abbe is a professor, just that his title is "professor". At the time, Monthly Weather Review was run by the U.S. federal government (United States Weather Bureau), so he had to be some meteorological scientist.

It's a long shot, but if someone wants to try to check around for a scientist/professor named "Cleveland Abbe" in the late 1800s/early 1900s, maybe the university he is a professor of could be located. I honestly, don't expect an answer to this, but asking others never hurts. The Weather Event Writer (Talk Page) 05:10, 14 June 2024 (UTC)

Cleveland Abbe has an article. :D  Card Zero  (talk) 05:16, 14 June 2024 (UTC)

Ultimate facepalm moment. LOL. Thank you Card_Zero! The Weather Event Writer (Talk Page) 05:16, 14 June 2024 (UTC)

Scientists disappointed by a shared Nobel prize

Are there cases known when a scientist longed for the Nobel prize so much that he was disappointed by even a Nobel prize he had to share? --KnightMove (talk) 07:39, 14 June 2024 (UTC)

George Santos, but he got over it by the time he got his fourth one. Clarityfiend (talk) 10:05, 14 June 2024 (UTC)

Art for blind people

I have just read the Fantastic Four comic and, before the usual superhero stuff gets started (an invasion of vampires, go figure), there's a brief slice of life intro in a museum. Alicia Masters, one of the characters, is blind, and the museum has reproductions of famous fine arts that can be touched, so that blind people can enjoy them as if reading braille (see here). Is that really a thing, or was the author making it up like the rest of the fantasy and sci-fi stuff? Cambalachero (talk) 13:30, 14 June 2024 (UTC)

Here's a couple of articles [4][5]. Sean.hoyland (talk) 15:30, 14 June 2024 (UTC)

A tactile reproduction of Roy Lichtenstein's Whaam! for the Tate Modern gallery in London, which offers "touch tours" for the visually impaired. Alansplodge (talk) 12:35, 16 June 2024 (UTC)

""Angry Small Boy" is a bronze statue of a nude infant boy modelled by Norwegian sculptor Gustav Vigeland ca. 1928."

Sculptures are sometimes exhibited in public places where they can be touched by blind people. The example shown stands in an Oslo park and its bronze surface is kept polished by visitors who often stroke it (either for art appreciation or just "good luck"). Philvoids (talk) 16:21, 17 June 2024 (UTC)

No wonder the boy's angry. Clarityfiend (talk) 08:49, 18 June 2024 (UTC)

June 17

Unit usage

1. Is the height of clouds and flying animals from earth's surface measured in meters or feet in most countries? I have seen using feet as primary units in this wiki's articles.
2. Are distances on association football pitch, such as length of goal kick, measured in meters in most continental European countries?
3. Are rink and equipment sizes in ice hockey measured in metric units in most continental European countries?
4. Is there any country that measures American football field sizes in metric units?

--40bus (talk) 21:01, 17 June 2024 (UTC)

In aviation, height is often measured in feet, even in countries that otherwise make little use of Imperial or US customary units. This would apply to clouds, but flying animals don't come up too much in aviation, unless your name is Sully Sullenberger. Jc3s5h (talk) 21:07, 17 June 2024 (UTC)

In aviation, altitude is usually measured in flight levels, with one flight level corresponding to the barometric pressure difference of 100 feet under normal conditions. Using barometric altitude is due to the fact that it is easier to measure well, and good enough to ensure vertical separation of aircraft. That on FL corresponds to 100ft is a historical relic, but air traffic is extremely conservative. --Stephan Schulz (talk) 21:19, 17 June 2024 (UTC)

Regarding this, is there any English-speaking country that has metric flight levels? --40bus (talk) 21:34, 17 June 2024 (UTC)

According to the FAA's Aeronautical Information Manual (page 231 of the PDF) in the US feet are used up to but not including 18,000 feet MSL (that is, above mean sea level). At and above 18,000 MSL flight level is used. Moving on to page 601 of the PDF, there is a complex discussion of setting the aircraft's altimeter, but essentially instructs to use the current actual barometric pressure if below 18,000 feet MSL. But at or above 18,000 feet MSL all operators will set 29.92 inches of mercury, which is considered the standard setting. The manual doesn't say so, but I've been taught that below 18,000 feet the chief concern is colliding with terrain, and above that the chief concern is colliding with other aircraft. The altimeter setting is chosen to minimize the risk in the respective altitude ranges.

Other countries may have a transition altitude other than 18,000 MSL. Jc3s5h (talk) 21:53, 17 June 2024 (UTC)

Outside the US, American football is virtually nonexistent, and supporters are likely to be Americans. NFL Europe says that the league gave an additional point for field goals of 50 yards, not 45 metres. Nyttend (talk) 08:03, 18 June 2024 (UTC)

User:40bus, when you say American football field sizes, do you mean measuring American football fields specifically, or anything roughly that size? Rugby league and rugby union, which are related to American football, are played on fields a bit larger than American football, and their dimensions are measured in metres. Nyttend (talk) 19:39, 18 June 2024 (UTC)

Heights of clouds, birds, buildings, anything are measured in metric in most of the world. The only exception is aviation, which shifted in most of the world post-World War 2 from metric to British/American units. It had something to do with the post-war dominance of the British and American aviation industry. Historical aircraft (pre-1945), but also many modern gliders, from Europe typically have altimeters in metres and airspeed indicators in km/h. PiusImpavidus (talk) 10:48, 18 June 2024 (UTC)

Meteorology tends to use the same units as aviation (or at least they did when I worked in the field, over 20 years ago), resulting in a weird mess of systems. So heights and altitudes in feet (unless we were dealing with things more than about 15-20km up, when we switched to km). Visibility in metres, but long distances in nautical miles. Speed in knots. Temperature in Celsius. Pressure in millibars / hectopascals. Iapetus (talk) 14:45, 18 June 2024 (UTC)

Of course, the nautical mile is a very different matter from anything else non-metric, since it's based on the distance of a minute of latitude, because such a figure is very easy to measure on maps. The nautical mile article notes that it was sanctioned for use by France and that other metric countries have defined it in metric terms (and makes an uncited claim that many metric countries approved its use), which is rare or unique for pre-metric units. Nyttend (talk) 19:39, 18 June 2024 (UTC)

I have seen using feet as primary units in this wiki's articles.

We have unit conversion templates that should make it possible to see the units of your choice in any given article. They haven't yet been applied consistently to all articles, but you can update them yourself if you encounter places where they're missing. -- Avocado (talk) 22:26, 19 June 2024 (UTC)

Football

I found that in articles of major association football matches use yards first in distances and lengths of shots, even macthes played in completely metric countries. Should they use meters first? --40bus (talk) 13:47, 19 June 2024 (UTC)

How do the news media cover it? ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:44, 19 June 2024 (UTC)

I meant that this wiki's articles do that. --40bus (talk) 18:41, 19 June 2024 (UTC)

Do Wikipedia's articles conform to the way the news media cover it? ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:56, 19 June 2024 (UTC)

Extracting energy from the Earth's rotation

I've been thinking about the Focault pendulum, and that gave me an idea. For simplicity, assume I put a gyro on a freely rotating platform, exactly at the North Pole, with the axis of the gyro parallel to the ground (and hence the equatorial plane). As I understand it, the gyro-mount would rotate once every 24h relative to the Earth - the gyro remains stable, the Earth is rotating under it. What would now happen we connect a generator to the platform (possibly via a gearbox to get a bit above 1rpd). Would it be possible to extract useful amounts of energy from this construct? I actually have no idea how to calculate this... --Stephan Schulz (talk) 21:11, 17 June 2024 (UTC)

If you wish to extract energy from the Earth's rotation, there are simpler ways to do it. AndyTheGrump (talk) 21:16, 17 June 2024 (UTC)

Tidal power draws mainly on the energy of the Moon's orbit. A windmill blown by a hurricane is an example of drawing power via Coriolis effect from the Earth's rotation. Philvoids (talk) 21:25, 18 June 2024 (UTC)

Yes, but those are boring. I want the mad scientist solution. --Stephan Schulz (talk) 21:20, 17 June 2024 (UTC)

I'm not sure, but aren't the present day Focault Pendulum systems powered using an electromagnet so that they keep going indefinetely? Getting energy from this would never be more than the energy inserted in it using the magnet. Rmvandijk (talk) 08:37, 18 June 2024 (UTC)

The lazy answer is: thermodynamics says no. Without transporting angular momentum from the Earth to anywhere off Earth, the only energy that can be extracted is the rotational energy of the flywheel and you won't get more than what you used to spin up your gyroscope.

Now it's clearly possible to extract energy from the gyro-mount in the way you described. The gearbox will apply a torque to the gyro-mount, trying to put the gyro's spin axis vertical, but it's restricted in its movement, so there will be another torque perpendicular to both the gyro's spin axis and the vertical, causing the gyroscope to precess with the Earth's rotation. But it's possible to tune this torque such that the gyro's precession rate will be half a revolution per day, so that neither the torque nor the angular speed of the mount relative to Earth will be zero, so that power can be extracted. It's not immediately obvious how this energy can be extracted from the rotational energy of the spinning gyroscope, as that requires a torque along the gyro's spin axis. I suspect it has something to do with precession of the Earth itself. PiusImpavidus (talk) 11:53, 18 June 2024 (UTC)

Certainly, the gyro's spin axis is mechanically restricted to a flat plane, then the gyro and the Earth are torquing the gearbox and doing work in that plane. Operating it at low speed on a magnetic suspension reduces windage and friction. Hmmm, of course, for it to be an effective energy source the rate of work extracted from their rotations must be greater than the rate of energy needed in maintaining its operation. Yet whenever I stand up or lay back down, the Earth's rotational energy and momentum are redistributed between my body and the Earth. Thus I don't see how or why energy and momentum conservation is an issue with this clever invention. Modocc (talk) 23:27, 18 June 2024 (UTC)

In other words, the angular momentum and mass-energy of the entire Earth is conserved, but the energized gyro(s), the rotating platform and its gearboxes are redistributing it, hence the Earth is both the source and repository of the energy. Moreover, this idea is related to the Coriolis effect which is very small, yet it would be simple enough to demonstrate its principles on a much smaller scale: simply substitute a spinning wheel for the planet. A second free-spinning wheel placed on the axle serving as the platform per the OP's specs, should act as a regenerative brake. Modocc (talk) 22:11, 19 June 2024 (UTC)

June 18

If atomic number was known in 1869, would the unknown elements (as of that time) be discovered earlier?

In 1869, Mendeleev predicted that there exists elements with atomic masses 44, 68, 72, 100, and they corresponding to atomic numbers 21, 31, 32, 43, unfortunately, at that time, only atomic mass was known and atomic number was not known, and according to the article Discovery of chemical elements, the unknown elements as of 1850 are the elements with atomic numbers 2, 10, 18, 21, 31, 32, 36, 37, 43, 49, 54, 55, 59, 61, 62, 63, 64, 66, 67, 69, 70, 71, 72, 75, 81, …, if atomic number was known in 1869 (or 1850), would these elements be discovered earlier (by finding the elements with missing atomic numbers)? 220.132.216.52 (talk) 06:01, 18 June 2024 (UTC)

You seem to be asking us to speculate, which we don't do here. However, I don't see how that information would lead to any elements being discovered earlier. Shantavira|^{feed me} 09:24, 18 June 2024 (UTC)

Mendeleev only correctly predicted elements 21, 31, 32, 43 and did not predict elements 2, 10, 18, 36, if atomic number was known in 1869, then maybe Mendeleev would also predict elements 2, 10, 18, 36? 220.132.216.52 (talk) 10:38, 18 June 2024 (UTC)

Imagine a document is sent to Mendeleev through a time machine, saying, "All elements have an atomic number. For hydrogen it is 1, for lithium it is 3, ..., and for thorium it is 90. Your four predicted elements have atomic numbers 21, 31, 32 and 43." How would this information have been useful to Mendeleev, or any scientist of his time, in making predictions?  --Lambiam 20:00, 18 June 2024 (UTC)

"Any scientist of Mendeleev's time" will include Bohr (16), Rutherford (2) and van der Broek (1) where their ages at Mendeleev's prediction are in brackets. It's unlikely that any of them will benefit from a magical receipt of numbers without explanation that do not fit then-known atomic weights. There is just a tiny probability that your tipoff causes baby van der Broek who is destined to arrange the periodic table by proton numbers at age 41 in 1911 (4 years after Mendeleev's death) to do so earlier. Philvoids (talk) 21:13, 18 June 2024 (UTC)

OP must be thinking about a Bell number for the combinations and arrangements leading to atomic numbers. Yet Mendeleev would simply have not been Mendeleev himself (and more than probably someone else would have grabbed the document instead). Here is a view described more similar to Medeleev own private domain of thinking. And yet ions themselves... --Askedonty (talk) 21:27, 18 June 2024 (UTC)

Mendeleev should find the missing atomic numbers (2, 10, 18, 21, 31, 32, 36, 37, 43, 49, 54, 55, …) and try to predict the properties of the elements with these atomic numbers. 42.76.70.59 (talk) 05:29, 19 June 2024 (UTC)

Atomic mass are not consecutive numbers, but atomic numbers are. 42.76.70.59 (talk) 05:29, 19 June 2024 (UTC)

What you really need for many of these discoveries is to push the invention of the spectroscope earlier. Otherwise, it is very difficult to find elements like Ga, which don't have any common high-content minerals. Note that element 43 remained elusive for a while even after the discovery of atomic numbers: this is a similar case where having the theory doesn't mean you'll find the element, since you need technology that isn't there yet (in this case, artificial transmutation).

There are a few cases that could get away without it. Henry Cavendish came pretty close to discovering argon, and if he really did it, I can see the noble gases as already being known by Mendeleev's time. Among the lanthanides, europium in particular has a +2 oxidation state that could have been used to find it much earlier. Double sharp (talk) 07:18, 19 June 2024 (UTC)

Gas non-giants

Planet says that planets are classified as "a giant planet, an ice giant, or a terrestrial planet". This doesn't appear to leave room for Earth-size gas planets. Why can't they exist? Or can they exist theoretically, and we just don't know of any? Nyttend (talk) 08:01, 18 June 2024 (UTC)

Could the gas dwarf form in the first place? One theory of formation is that solids have to coalesce first to make a planet that can attract gas to make a gas giant. SO small objects will not form. But if a ball of gas the size of Earth existed would it be stable and survive? Probably not as the density would be lower and so gravity is lower. It would have to be far from the Sun to stop solar wind erosion. Graeme Bartlett (talk) 09:28, 18 June 2024 (UTC)

e/c You need a lot of mass (gravity) to keep that gas together, and with too little mass any atmosphere would disappear (Mercury, Venus, etc). Earth is just the right size. Shantavira|^{feed me} 09:31, 18 June 2024 (UTC)

Though Venus has a significant carbon dioxide atmosphere. So perhaps you could have a planet made from carbon dioxide. Though with pressure this would transform into diamond and metallic oxygen in the core. Graeme Bartlett (talk) 09:58, 18 June 2024 (UTC)

Between ice giants and terrestrial planets, there's room for ocean worlds. PiusImpavidus (talk) 12:17, 18 June 2024 (UTC)

See also Super-Earths. {The poster formerly known as 87.81.230.195} 151.227.226.178 (talk) 17:51, 18 June 2024 (UTC)

Based on the data we have, planets generally obtain a Neptune-like atmosphere once they cross two Earth masses.

On the other hand, in a way Venus is close to what you want. Its atmosphere even goes supercritical as it approaches the surface. :) Double sharp (talk) 07:22, 19 June 2024 (UTC)

June 19

Carnivore diet

(Disclaimer: I don't want to put my cat on any kind of diet, in fact I don't have a cat, so this is purely a science question). I was surprised to find a ton of online search hits about putting cats on vegan diets This one mentions that you at minimum have to include some supplements like taurine that come only from animal sources. Thus, the diet is not 100% vegan.

My question is whether there are at least potential workarounds such as protein synthesis, that would allow putting obligate carnivores like cats on purely plant diets and keeping them healthy? Question is inspired by a discussion elsewhere about "impossible burgers", which supposedly give meat-like nutrition to humans. There must be something missing though, that humans can do without but cats cannot. Thanks. 2601:644:8501:AAF0:0:0:0:78AE (talk) 09:28, 19 June 2024 (UTC)

Most mammals, including humans, can biosynthesize taurine. Felids lack this ability. Taurine can be synthesized in a chemical production process not involving animal sources; see Taurine § Synthesis.  --Lambiam 06:09, 21 June 2024 (UTC)

Buoyancy question

What's the minimum concentration of an aqueous table salt solution which will ensure that a human being will float regardless of his/her body type, even with all his/her muscles fully contracted (assuming standard temperature and pressure)? Disclaimer: this is not a homework question. 2601:646:8082:BA0:649B:7753:3C84:C70D (talk) 10:28, 19 June 2024 (UTC)

0%. Since water is 997kg per cubic meter and the average density of the human body is 985kg per cubic meter. 41.23.55.195 (talk) 11:37, 19 June 2024 (UTC)

Obviously you haven't been to the Dead Sea or the Great Salt Lake. Clarityfiend (talk) 13:53, 19 June 2024 (UTC)

If you are in a swimming pool you can descend to the bottom by contracting your muscles, which I guess decreases your body's volume. I guess that is what OP was referring to. Question is how much salt it takes to make this impossible. I can say 3.5% or so is not enough, since you can also do that in the ocean. 2601:644:8501:AAF0:0:0:0:78AE (talk) 16:45, 19 June 2024 (UTC)

I have never been able to float. --User:Khajidha (talk) (contributions) 17:58, 19 June 2024 (UTC)

Assume the "Jesus position" with an arched back. It works for me, and I'm no swimmer. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:55, 19 June 2024 (UTC)

That's not what I was asking -- the context here is a panic attack causing a person to sink (think something similar to the floundering in the water scene in Titanic, but minus the hypothermia), so obviously voluntary techniques won't work in this context. 2601:646:8082:BA0:649B:7753:3C84:C70D (talk) 23:05, 19 June 2024 (UTC)

Sure. I was talking to Khajidha. But knowing the technique can help conquer a panic situation. ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:01, 20 June 2024 (UTC)

I don't have panic attacks in water (my one and only severe phobia has to do with something else entirely) -- I was asking this simply out of curiosity. So, at what salinity level would a person not sink even if he/she panics and suffers a muscular spasm all over his/her body? 2601:646:8082:BA0:649B:7753:3C84:C70D (talk) 02:30, 20 June 2024 (UTC)

According to this,[6] the Dead Sea has the highest salinity in the world, at 34 percent, making it virtually impossible to sink. So that at least puts a frame on your question. And this[7] suggests an experiment you could do. ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:26, 20 June 2024 (UTC)

I'm aware of the Dead Sea -- however, my question was about the lowest salinity level at which a person would float under all circumstances, which is probably quite a bit lower than that in the Dead Sea. And does an egg have the same maximum density as a person with all his/her (I'd go with his, since a woman's body probably has a lower density than a man's) muscles fully clenched? 2601:646:8082:BA0:649B:7753:3C84:C70D (talk) 05:57, 20 June 2024 (UTC)

Well if you want this to apply to all people you will need to know the density of the most dense human body. That is probably one with deflated lungs, no gas in the intestines and a very low proportion of fat. And do we allow someone who at a barium meal? Graeme Bartlett (talk) 07:04, 20 June 2024 (UTC)

Google has various opinions. Here's one:[8] ←Baseball Bugs ^{What's up, Doc?} carrots→ 08:03, 20 June 2024 (UTC)

IP editor: you may be interested in this calculator which provides the density of saline solutions. Note that the answer you seek depends on the temperature of the water. By setting various concentrations of salt, you can find out the density of the resulting solution. You still have to decide what "floating in all circumstances" means to you. Mike Turnbull (talk) 10:12, 20 June 2024 (UTC)

Most of the short-term volume variation in humans comes from the lungs. Depending on which muscles exactly contract, the volume of the lungs may increase or decrease. Most people have a reflex to fill their lungs when they unexpectedly enter the water, making them positively buoyant.

Also, people who unexpectedly enter the water usually wear clothes. Clothes tend to be negatively buoyant when wet, but can trap a lot of air. When panicking, you can quickly squeeze that air out or your clothes, making you sink. And some clothes wake it very difficult to swim. That's why we teach our children in basic swimming lessons to swim with light clothes on, but also to get their clothes off if too heavy. Panicking people might forget to get their clothes off. PiusImpavidus (talk) 12:39, 20 June 2024 (UTC)

OK, User:Baseball Bugs FTW with the most helpful answer this time around -- thanks! 2601:646:8082:BA0:649B:7753:3C84:C70D (talk) 03:48, 21 June 2024 (UTC)

If it's any help I can sink and lie flat on the bottom of a swimming pool by just letting some air out but I can't do that in the sea. NadVolum (talk) 10:26, 21 June 2024 (UTC)

See isolation tank. 2A00:23A8:1:D801:1484:1471:392F:9B25 (talk) 15:43, 23 June 2024 (UTC)

How can DHA be a primary structural component of the brain, cerebral cortex, skin & retina?

Antioxidants? Do brain lipids ever go rancid while someone's still alive? Sagittarian Milky Way (talk) 23:52, 19 June 2024 (UTC)

Fats may go rancid, usually by exposure to air and light. Fats are esters. DHA is not an ester, it is a (fatty) acid.  --Lambiam 04:55, 20 June 2024 (UTC)

How would brains be a delicacy if the fat was free acids? 12% lipids. And even if it was free acid the main problem's double bonds oxidizing probably eventually making it gum like oil paints' triglyceride molecules tangling and polymerizing. Sagittarian Milky Way (talk) 11:33, 20 June 2024 (UTC)

Whose brains are you eating? ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:02, 20 June 2024 (UTC)

I don't eat any flesh but I've heard you can even buy brains 'n' beans in a can in the South (I've seen a pint or quart of pig blood in a container in a NYC grocery store so could doesn't necessarily mean common). Brains 'n' beans is calorie-dense, if you wolf it down cause it's delicious it becomes junk food. Sagittarian Milky Way (talk) 12:26, 20 June 2024 (UTC)

Rose brand pork brains in milk gravy are readily available at grocery stores 'round my neck of the woods (Hickory, NC). Oh! WHAAOE! eggs and brains. I also see sites selling them fresh or freeze dried.-- User:Khajidha (talk) (contributions) 14:42, 20 June 2024 (UTC)

Sounds about as appetizing as the Beverly Hillbillies' staple: possum innards. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:03, 20 June 2024 (UTC)

That's probably it, eggs 'n' brains, not beans. Sagittarian Milky Way (talk) 17:58, 20 June 2024 (UTC)

Is the question about DHA or about lipids?  --Lambiam 19:08, 20 June 2024 (UTC)

June 20

Are scientists sure that agriculture only started with the Holocene?

Could there have been agriculture hundreds of thousands of years ago before the last ice age?Rich (talk) 08:12, 20 June 2024 (UTC)

Some group of dinosaurs could have had agriculture millions of years ago before they were all wiped out for all we know. But see the big box at he top of this page, one of the things it says is "We don't answer requests for opinions, predictions or debate" NadVolum (talk) 08:19, 20 June 2024 (UTC)

Very unlikely. I suggest you read Mannion, A. M. (5 December 1995). Agriculture and Environmental Change: Temporal and Spatial Dimensions. Wiley. ISBN 0471954780. or similar. Agriculture depends on the domestication of crops and animals and its main centres were in areas of the world that were not covered in ice. Hence we would see archaeological traces whenever that had occurred. Mike Turnbull (talk) 09:29, 20 June 2024 (UTC)

Ant–fungus mutualism is often considered a form of agriculture, and it developed millions of year ago. --Amble (talk) 17:23, 20 June 2024 (UTC)

Humans were slow starters. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:28, 21 June 2024 (UTC)

Scientists can never be sure about their theories. A good theory explains the patterns they observe. But errors can creep in, both in the process of obtaining observations and in recording them. Also, in a laboratory experiment one can perhaps make direct uninterpreted measurements, but in the field observations require interpreting the raw data, and the interpretations themselves depend on theories whose validity can also be subject to doubt. A pattern may be manifest while not being due to some underlying process but emerging by pure chance. And, finally, there is always the possibility of new observations uprooting a generally accepted theory. As to human agriculture, the currently best available explanation of the observations is that sedentary agriculture was an innovation that emerged in a few places and spread out from these during the Neolithic.  --Lambiam 19:46, 20 June 2024 (UTC)

There may have been some earlier domestication attempts in the warmer periods on either side of the Last Glacial Maximum. I recall hearing that Zohary wondered if wheat didn't go back 20,000+ years, and lately there's been some talk about olives and pine nuts, which of course wouldn't show much in genetic studies as they are long-lived trees. I myself wonder about Vicia palaestina (whose stub I created). What makes this hard to tease out is that if it occurred, it occurred in the same area as the successful domestication events, thus masking the clues. Abductive (reasoning) 06:16, 21 June 2024 (UTC)

Please see

If you are interested in genetics, please see Wikipedia:External links/Noticeboard#Human mitochondrial genetics and share an opinion about whether the proposed ==External links== would be interesting or valuable to readers. WhatamIdoing (talk) 22:50, 20 June 2024 (UTC)

June 21

Noctilucent clouds

How high (in angle) do noctilucent clouds have to be to be seen? Do they have to appear outside the lighter area of the sky during nautical twilight? I've been watching clouds before last few sunrises but I'm not sure if i'm just observing high altitude cirruses, they are whitish but not bright at all like in the pictures. Other (lower) clouds stay dark for a long time after that, almost until sunrise. I can't tell if they disappear after sunrise because there's a lot of humidity haze or maybe still Sahara's smog, and my phone is terrible at taking pictures at twilight. Please don't just tell me to read the page, it hasn't helped me with this. 31.217.31.107 (talk) 02:21, 21 June 2024 (UTC) They were visible from around 3.30 to 4.30 here (30min ago) in Zagreb, whatever they are. They weren't that sharp like on the pictures. 31.217.31.107 (talk) 02:57, 21 June 2024 (UTC)

Under the right conditions you should be able to see them at any angle, wherever they are hanging out, from the horizon to directly overhead.  --Lambiam 05:48, 21 June 2024 (UTC)

I saw them at up to 45 degrees up (the 'height' of the polar star here). They didn't really look like the pictures in the article, they were more faint and blurry although they were lighter than the sky all along while low clouds are darker than the sky that close to sunrise. I think they could look that way because of heavy light pollution lately and high humidity but I'm still not sure if they weren't just cirruses because low clouds are always lighter than the sky near midnight, there's enough light pollution. Is there any way to be sure? 31.217.31.107 (talk) 05:58, 21 June 2024 (UTC)

On the only occasion I've seen noctilucent clouds, they were up to 4 degrees above the horizon (half the altitude of Capella at that time), almost straight north. At the same time, the sun was also almost straight north, 14 degrees below the horizon. Not nautical twilight, but the sky was still fairly bright and the noctilucent clouds were in the bright part. With those angles, the radius of the Earth and some high-school geometry, you should be able to calculate the minimum altitude of the clouds to be in sunlight. If that's more than about 12 kilometres, they can't have been ordinary cirrus clouds. Use a planetarium program like Stellarium (open source) to find the position of the sun at the time you observed the clouds. PiusImpavidus (talk) 08:11, 21 June 2024 (UTC)

Lunar Standstill

A lunar standstill is supposed to happen tonight. I've read the article's simplified description and I'm still confused. So, is this correct? In my own words...

During a lunar standstill, the Moon does not actually stand still, nor does it appear to. All this means is that tonight the Moon will rise at its most northeastern point and set at its most northwestern point. Period. It's not something you can go out at a specific time to observe, see something different, the event ends, and you go about your evening. It's an event that takes "all" (air quotes) night (the time the moon is out).

Do I have that right?

Thanks! †dismas†|^(talk) 12:14, 21 June 2024 (UTC)

I wasn't aware of this, this is quite interesting (if you're into that sort of thing). "Standstill" refers to (the lack of) motion in declination, i.e. the north-south coordinate (equivalent to latitude on earth), and is equivalent to the solstices. Summer solstice was yesterday (sun at its northern-most point), we also have full moon tonight, and since the full moon is opposite to the sun this means that moon is at its southern-most point (not northern!). In addition the moon is at its lowest point below the ecliptic, so that in sum we have a major lunistice. In principle these are all point events but the changes over the course of a night are so small that it is effectively an all-nighter. Where I am, the moon will only be 12 degrees above the horizon when it culminates in the south around midnight. --Wrongfilter (talk) 13:31, 21 June 2024 (UTC)

According to this article, we're not at the absolute extreme tonight, though. The situation is quite complicated and the extremes differ for moonrise and moonset, compare the graphics in that article. --Wrongfilter (talk) 13:44, 21 June 2024 (UTC)

For about one year every 18.61 years your hometown Moon rise/peak/set paths are more extreme than any other time (north/south, rises and sets unusually left/starboard, also low/high if you're between latitudes 29 and 62 or so. North or south of 29 you'll never see a shorter hometown Moon shadow than one of these fortnightly standstills in the next year or so probably near one of the standstills near a half moon near one of the next 2 equinoxes (the Full Moons can be up to 1.2 Moon radii less extreme but the c. 19 year cycle is 20 radii to minus 20 radii) Sagittarian Milky Way (talk) 14:01, 21 June 2024 (UTC)

The cycle peaks early 2025 I believe but cause complexities records can be up to seasons away. Records like rightmost Full Moon rise for your location between about 2006 and 2043 or lowest Full Moon path of a night in that timeframe, might be tonight or about 355 days in the future. High Full Moon records are probably December 2024, some other records are probably near an equinox in 2025 or late 2024. If you want I can get each major record to the nearest millisecond (false precision?) for any location on Earth. From what might be the most accurate possible way (https://ssd dot jpl dot nasa dot gov/horizons/app.html#/, there's also other interfaces like email and telnet but for this you don't need the relatively few email/telnet-only tools) Sagittarian Milky Way (talk) 14:27, 21 June 2024 (UTC)

From 74W 40¾N 0ft the highest Moon center in the future of anyone alive is 77.753639° the year 2025 April 3rd 6:05:02.857pm (ellipsoidal coordinates). The highest before that was 2006 Mar 7th 18:56:52.857 77.753848°. The sky was pretty dark for this one as it's always near half moon so can't be too close to midnight. The highest before that was 1950 October 3 6:07:53.010am 77.782823°. These follow the obliquity Milankovitch cycle which is slow enough that a new record doesn't happen every 18.61 year cycle as you go back in time. So the highest since about 3X thousand BC was about 10,000 years ago. Sagittarian Milky Way (talk) 21:19, 21 June 2024 (UTC)

From Earth's center the northernmost Moon of Oct 2006 to Aug 2043 inclusive is March 7 2025 15:56 UTC 28°43.0'N. Thus longitudes of about 45E will get a slightly more northern Moon record than any other place on Earth. Sagittarian Milky Way (talk) 17:25, 22 June 2024 (UTC)

The cycle is dependent on the longitude of the ascending node. This book [9] says the longitude of the ascending node (☊) = ☊_o - 0.0529539D_e, where ☊_o is the ecliptic longitude of the ascending node at the standard epoch J2000 and ☊ is the ecliptic longitude of the ascending node after D_e (an interval measured in days). The ecliptic longitude of the ascending node at the standard epoch (12h 1 January 2000) was 125.044522°.

Now, the major standstill occurs when the longitude of the ascending node is 0°, which means that the total movement since J2000 is 360° + 125.044522° = 485.044522° (remember the movement is backwards). The number of days since J2000 is therefore 485.0422522/0.0529539 = 9159.750689 days. 24 years contain 8766 days, so the standstill occurs (9160 - 8766) = 394 days after 1 January 2024, or around 29 January 2025. I have just found out that the calendar of golden numbers (mentioned on the Reference desk from time to time) is (or was) mentioned elsewhere in Wikipedia, and the text given there has been augmented to embrace the matters raised here:

LUNAR CALENDAR 1 MARCH 1900 - 28 FEBRUARY 2200

The lunar date for 29 February of a leap year is normally the same as that of the preceding day - thus the lunar date for 28 and 29 February 2028 is 3 Ronan. For use of the letters A - g to find the day of the week see Dominical letter. The months are: (1) Harriet, (2) Ronan, (3) Miri, (4) James, (5) Eloise, (6) Thomas, vii, (8) Nicholas, (9) Catherine, (10) Richard, (11) Emma, (12) Paul. Paul II, a 30-day month, is added between Paul and Harriet 7 times in 19 years. When the golden number is 19, Richard has 29 days instead of 30. See Saltus#Latin (third bullet point).

	JAN Paul 30	FEB Harr 29	MAR Ron 30	APR Miri 29	MAY Jame 30	JUN Eloi 29	JUL Thom 30	AUG vii 29	SEPT Nich 30	OCT Cath 29	NOV Rich 30	DEC Emma 29
1	A 12	d 1	d 12	g 1	b	e 9	g	c 17	f	A	d 3	f 3
2	b 1	e	e 1	A	c 9	f	A 17	d 6	g 14	b 14	e	g
3	c	f 9	f	b 9	d	g 17	b 6	e	A 3	c 3	f 11	A 11
4	d 9	g	g 9	c	e 17	A 6	c	f 14	b	d	g	b 19
5	e P2	A 17	A	d 17	f 6	b	d 14	g 3	c 11	e 11	A 19	c
6	f 17	b 6	b 17	e 6	g	c 14	e 3	A	d	f	b 8	d 8
7	g 6	c	c 6	f	A 14	d 3	f	b 11	e 19	g 19	c Em	e 16
8	A	d 14	d	g 14	b 3	e	g 11	c	f 8	A 8	d 16	f 5
9	b 14	e 3	e 14	A 3	c	f 11	A	d 19	g Ca	b 16	e 5	g
10	c 3	f	f 3	b	d 11	g	b 19	e 8	A 16	c 5	f	A 13
11	d	g 11	g	c 11	e	A 19	c 8	f 16	b 5	d	g 13	b 2
12	e 11	A	A 11	d	f 19	b 8	d vii	g 5	c	e 13	A 2	c
13	f	b 19	b	e 19	g 8	c 16	e 16	A	d 13	f 2	b	d 10
14	g 19	c 8	c 19	f 8	A El	d 5	f 5	b 13	e 2	g	c 10	e
15	A 8	d 16	d 8	g 16	b 16	e	g	c 2	f	A 10	d	f 18
16	b Ha	e 5	e Mi	A 5	c 5	f 13	A 13	d	g 10	b	e 18	g 7
17	c 16	f	f 16	b	d	g 2	b 2	e 10	A	c 18	f 7	A
18	d 5	g 13	g 5	c 13	e 13	A	c	f	b 18	d 7	g	b 15
19	e	A 2	A	d 2	f 2	b 10	d 10	g 18	c 7	e	A 15	c 4
20	f 13	b	b 13	e	g	c	e	A 7	d	f 15	b 4	d
21	g 2	c 10	c 2	f 10	A 10	d 18	f 18	b	e 15	g 4	c	e 12
22	A	d	d	g	b	e 7	g 7	c 15	f 4	A	d 12	f 1
23	b 10	e 18	e 10	A 18	c 18	f	A	d 4	g	b 12	e 1	g
24	c	f 7	f	b 7	d 7	g 15	b 15	e	A 12	c 1	f	A 9
25	d 18	g	g 18	c	e	A 4	c 4	f 12	b 1	d	g 9	b
26	e 7	A 15	A 7	d 15	f 15	b	d	g 1	c	e 9	A	c 17
27	f	b 4	b	e 4	g 4	c 12	e 12	A	d 9	f	b 17	d 6
28	g 15	c	c 15	f	A	d 1	f 1	b 9	e	g 17	c 6	e
29	A 4		d 4	g 12	b 12	e	g	c	f 17	A 6	d	f 14
30	b		e	A 1	c 1	f 9	A 9	d 17	g 6	b	e 14	g 3
31	c 12		f 12		d		b	e 6		c 14		A
	Harr	Ron	Miri	Jame	Eloi	Thom	vii	Nich	Cath	Rich	Emma	Paul

The numbers move down one day on 1 March of years which, although divisible by four without remainder, are not leap years. They move up one day eight times in 2500 years. The next eight movements will be on 1 March of 2100, 2400, 2700, 3000, 3300, 3600, 3900 and 4300. Sometimes the movements cancel out - thus in 2100 the numbers stay where they are. The numbers mark the first days of the lunar month. Each year's golden number is found by adding 1, dividing by 19 and taking the remainder. If the remainder is 0, the golden number is 19. Until 2099 Orthodox Easter falls from 19 - 25 Miri, on whichever day is Sunday. From 2100 to 2399 the range is 20 - 26 Miri, and so on. So to find Orthodox Easter in 2025:

2025 + 1 = 2026. 2026/19 = 106 remainder 12. From table, 1 Miri is 31 March. The Sunday letter is E. 20 April (21 Miri) is an "E" day. Orthodox Easter is 20 April.

To calculate the date of occidental Easter, proceed as follows:

1. In the calendar, locate the date of 14 Miri

2. If 14 Miri falls on or before 17 April, Easter is the Sunday following. If 14 Miri falls on 18 April and no golden number is marked against 6 April, again Easter is the Sunday following. If 14 Miri falls on 18 April and a golden number is marked against 6 April, Easter falls on 18 April (if Sunday), and if 18 April is not Sunday Easter falls on the following Sunday.

3. If 14 Miri falls on 19 April, Easter falls on 19 April (if Sunday), and if 19 April is not Sunday Easter falls on the following Sunday.

4. If 14 Miri falls on 20 April or later, the date is to be treated as a day of March, and Easter falls on the day after the Saturday following that date.

The calendar may be used to locate the moon at any given time.

Example

It is 9 PM Greenwich Mean Time in London on 15 February 2024. The golden number is 11, which is printed against 11 February. This is the lunar new year (1 Harriet) and 15 February is therefore 5 Harriet. At the end of the previous month (as at the end of every lunar month) the sun and moon are together in the sky, but the lunar day is on average 4/5 hour longer than the solar day. Thus at 9 PM it is only 5 PM by the moon. Whether the moon is visible at that time may be determined using the fact that the moon moves through the zodiac at the rate of 13.2° per day (compared to 1° per day for the sun). On 15 February (5 Harriet) the moon will have advanced (5 x (13.2 - 1)) = 5 x 12.2 = about 61° ahead of the sun. So it will be where the sun will be about 61 days later, i.e. around 16 April.

But there is another factor. The moon's ascending node (where the plane of its orbit crosses the ecliptic in a northerly direction) moves backwards, completing a circuit relative to the equinox in 18.6 years. When the longitude is 0° (which it will reach around 29 January 2025) it reaches a maximum of 5° further from the celestial equator than does the sun (the major standstill). Half a revolution later (the minor standstill) it reaches a maximum of 5° nearer to the celestial equator than does the sun.

So considering the moon's position at 9 PM on 15 February 2024 we look whewre the sun would be at 5 PM on 16 April and (since the date is fairly close to the major standstill) a little higher in the sky. The moon was thus looked for (and was seen) high in the west.

The "establishment" of a port is the number of hours high tide is reached there after the moon crosses the meridian (i.e. 12 noon or midnight by the moon). The state of the tides may thus be predicted using the method above, remembering that tides are highest at the middle and the end of the lunar month ("spring tides") and their amplitude is greatest at the equinoxes. For accurate predictions consult specialist tables.

June 22

Boeing 707

Several questions: (1) Can a Boeing 707-120B take off at MTOW from Runway 04/22 at Greater Rochester International Airport? (2) How much payload (if any) would have to be taken off for the same jet to take off from Runway 10/28 at the same airport, with a 45-knot headwind and below-freezing temperatures as would be the case in a severe blizzard (and would this be possible at all)? (3) How much payload (if any) would have to be taken off for said Dash-120B to fly nonstop from ROC to Rome Fiumicino Airport with acceptable reserve fuel remaining, assuming no tailwinds along the route (and would this be possible at all)? (Questions inspired by the original Airport movie.) 2601:646:8082:BA0:649B:7753:3C84:C70D (talk) 12:27, 22 June 2024 (UTC)

1: According to the official documentation from Boeing, if it's equipped with the original JT3D-1 engines, the takeoff runway length requirement at 500 feet above sea level, standard temperature and maximum take-off weight is 8500 feet. Those Boeing documents are in US units. Your runway is 8001 feet, so it won't fit. At least, not with the proper safety margin. If equipped with the more modern JT3D-3 engines, the takeoff runway length requirement is 7800 feet, so that fits.

2: Under normal circumstances, assuming JT3D-3 engines, maximum weight on a 6400 foot runway is 232,000 pounds, 26,000 pounds below MTOW. Freezing temperature would take the density altitude down to below sea level, which the documentation doesn't tell about, so I'll assume sea level. The 45 knot headwind reduces takeoff groundspeed to about 70% and required runway length to a bit more than 50% of the original, so about 4000 feet should be enough, even at MTOW. Unless there's snow on the runway. Ploughing through snow may increase drag enough that it can't take off at all, but it also reduces braking performance, so you need more length for an aborted takeoff.

3: The distance from ROC to Fiumicino is 3780 nautical miles. Taking off at MTOW, it can carry a payload of 25,000 pounds, assuming JT3D-3 engines and a standard cabin configuration. This is slightly less than a full payload of 137 passengers and baggage, 28,000 pounds. Maximum payload for short flights is 42,000 pounds, passengers, baggage and cargo. PiusImpavidus (talk) 09:54, 23 June 2024 (UTC)

OP: Why did you choose Rochester? Airport_(1970_film) is set in Chicago, so wouldn't O'Hare International Airport or Midway International Airport be a better reference? RudolfRed (talk) 18:24, 23 June 2024 (UTC)

Because I'm doing some very early work on a rescue simulator game, so for this mission I wanted to move the departure airport further east so as to speed things up (it wouldn't do to have some of the players, in particular the one playing the medic, sit around for 2 hours or more doing nothing) -- but I see this probably won't work, so I guess I'll go with Boston Logan Airport instead (I had rejected it at first because I thought it would have sped things up too much, but on second thought it wouldn't be the case because the flight would be flying parallel to the shore at first). Not happening any time soon, though -- for one thing, a simulator would be way over my head at this point (and I'd have to put together a whole studio, because it's far too much work for just one person, no matter how highly skilled), and also for this particular mission I'd have to first get permission to use the likenesses of Dean Martin, Jacqueline Bisset, Helen Hayes and Van Heflin as Vernon Demerest, Gwen Meighen, Ada Quonsett and D.O. Guerrero respectively! 2601:646:8082:BA0:649B:7753:3C84:C70D (talk) 21:34, 23 June 2024 (UTC)

Stepper motor driver and power saving

Hi. Some motor drivers have an energy saving feature; for example, in the TMC2209[10] it's called "CoolStep".

I'm guess that this is accomplished via some very clever control circuitry inside. For example:

1. if a stepper is holding its position, and there's 0 load on it (just a bare stepper motor sitting on a desk connected to a TMC2209), then 0 current is needed

2. if I use my finger to try to turn the stepper shaft, a back EMF is generated in the stepper. The TMC2209 senses this voltage, and supplies a current in the opposite direction to counteract it

3. as a result, my finger will feel a "torque" from the stepper, and the stepper will hold its position, despite my applied force

(This is my best understanding of how it works.)

Is it possible to apply this same technique, or an analogous technique, to drive a solenoid?

I'm asking because the two situation are almost analogous to each other:

A. a solenoid is connected to a solenoid driver. A permanent magnet is stuck to the solenoid via magnetic force. There is 0 movement, so 0 induced voltage. There is 0 force, so 0 current is needed to hold this position

B. if I use my finger to try to move the magnet, a back EMF is generated in the solenoid. The solenoid driver senses this voltage, and supplies a current in the opposite direction to counteract it

C. as a result, my finger will feel a "force" from the solenoid, and the magnet will hold its position, despite my applied force

Is this kind of control actually possible in reality? In my head, I'm imagining something like this is possible in theory, but I don't know enough electronics to know whether it's actually possible or not.

Is there any commercial solenoid driver that can accomplish the above described power-saving feature? I could not find any myself. Since manufacturers use different marketing terms (such as "CoolStep") to describe their proprietary technology, it's possible that such a driver exists, but I don't know the right keyword to search for so I cannot find it. OptoFidelty (talk) 21:24, 22 June 2024 (UTC)

The TMC2209 Datasheet says thatCoolStep's operation relates to the StallGuard4 feature, which it describes as being based on back-EMF. But "CoolStep is not able to measure the motor load in standstill and at very low RPM". One application of sensing linear motion and counteracting it (effect: holding something in approximate position with minimal required force) is damping...lots of ways of implementing it. DMacks (talk) 21:53, 22 June 2024 (UTC)

Is there any commercially available IC that can achieve this damping effect?

I'm guessing that there are at least a dozen solenoids any new car, so presumably they have some very smart solonoid drivers inside them to control the drive current.

Currently I'm just using a constant current to drive the solonoid, which is very wasteful, since 99% of the time, 0 current is needed. I don't know how to achieve this back-EMF sensing feature on my own.

It'd be helpful if I can find a commercially available solonoid drivers IC that has this feature built-in. OptoFidelty (talk) 22:48, 22 June 2024 (UTC)

Not my field (ha!), sorry. DMacks (talk) 02:39, 23 June 2024 (UTC)

No worries. Thank you for the help so far.

At least I know it's physically possible now. Just not sure where to find any commercial solutions for it, if such a thing exists. OptoFidelty (talk) 06:40, 23 June 2024 (UTC)

The back-EMF gives a velocity sensor. Holding a position with only a velocity sensor is possible with a perfect sensor, but in reality sensors aren't perfect and some creep will happen – which is why the example mentioned above doesn't work in standstill. You need a position sensor, or apply a brake. PiusImpavidus (talk) 10:17, 23 June 2024 (UTC)

June 23

I'm looking for useful physical formulas of the form:

$X_{total}=X_{initial}+X_{additional}$

Where the $Xs$ are physical properties of the same type (e.g. of energy, or of electric charge, and likewise).

Those physical properties don't have to be denoted by the same letter in the formula.

The words "total", "initial", "additional", don't have to be mentioned in the formula, either. They should be understood, though, from the standard meaning of the letters mentioned in the formula. Therefore, formulas of the type $E_{total}=E_{potential}+E_{kinetic}$ should be ignored, because none of those Es (=energies) is usually interpreted as "initial".

"Initial" can also mean "basic".

HOTmag (talk) 10:59, 23 June 2024 (UTC)

What about integrations, eg the amount of electric charge on an object = integration over time of the current flowing through a surface that encloses the object.? (+ initial charge) Graeme Bartlett (talk) 11:21, 23 June 2024 (UTC)

I even avoid simpler formulas, e.g. of the type $X_{total}=X_{initial}+YT,$ not to mention $X_{total}=X_{initial}+\int Ydt.$ HOTmag (talk) 12:05, 23 June 2024 (UTC)

Is Mass–energy equivalence#Low-speed approximation one of the "useful" formulas by your definition, as applied to the Parker Solar Probe? Mike Turnbull (talk) 14:19, 23 June 2024 (UTC)

Your example uses too many letters. I must use three only (the Xs), and all of them should be of the same type, as indicated above. HOTmag (talk) 16:03, 23 June 2024 (UTC)

That's a distinction without a difference. Just call it X_displacement if the letters Y and T offend you. DMacks (talk) 21:30, 23 June 2024 (UTC)

"Where the $Xs$ are physical properties of the same type" - they have to be for the equation to make sense. see Dimensional Analysis. i can't off hand think of a non trivial equation that is simple enough for your requirements. weight_today=weight_yesterday+change_in_weight_per_day? Greglocock (talk) 22:16, 23 June 2024 (UTC)