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April 27

Is all hypochlorite salts of transition metals are colorless?

I'm don't sure that all hypochlorite salts of transition metals (like Fe(ClO)₃, Cu(ClO)₂, Rh(ClO)₃, Pt(ClO)₄,...) will colorless. So what do you think for the color of them? Thanks for much (Sorry if you don't understand, because my English is not good).--Ccv2020 (talk) 07:02, 27 April 2020 (UTC)

Might be difficult to make some of those. Adding Fe³⁺ to NaClO solution creates a giant cloud of chlorine gas, as seen in multiple industrial accidents such as [1]. DMacks (talk) 07:19, 27 April 2020 (UTC)

As noted above, I can't find any evidence the above compounds exist at all. Any reaction that would putatively create them is likely outcompeted by some kind of redox reaction that would reduce the ClO^- to Cl₂. I can't find an MSDS on any of them, which again, is a good sign that they don't exist. I'll repeat what I said last time: just because you can write a correct formula for a compound doesn't mean it can actually exist. Describing the color of a non-existent compound is a futile thing. --Jayron32 12:53, 27 April 2020 (UTC)

For attempts at making copper hypochlorite, a precipitate of cupric hydroxide / copper oxychloride forms. Quite a few of the oxychloride compounds are coloured eg TiOCl VOCl VOCl₂ VOCl₃ and FeOCl. These oxychlorides do not contain O-Cl groups though. Graeme Bartlett (talk) 23:12, 27 April 2020 (UTC)

Yes, just to clarify that, those are compounds that contain oxygen atoms and chlorine atoms as separate moeities, not as hypochlorite ions. That's a very different animal.--Jayron32 13:29, 28 April 2020 (UTC)

Address of a medical machine

Hello, I would like the address of the machine, which is in the USA, which can maintain in life the head despite the body is ceasing to function , please ? Thank you in advance. — Preceding unsigned comment added by 93.23.124.43 (talk) 15:48, 27 April 2020 (UTC)

Are you asking about Advanced Life Support, about Cryonics, or about something else entirely? -- ToE 16:26, 27 April 2020 (UTC)

Outside of Futurama, I don't think such a machine exists. Even cryogenically frozen heads are still dead. --OuroborosCobra (talk) 16:34, 27 April 2020 (UTC)

The OP might be talking about Alcor, which is where the remains of Ted Williams are on ice. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:50, 27 April 2020 (UTC)

And (iirc) most of Alcor's customers are head-only ("neurosuspension"), because it's cheaper and because recovery is thought easier that way. If technology can someday revive the corpsicles at all, cloning a new body is an easier job. —Tamfang (talk) 02:43, 2 May 2020 (UTC)

(52768) 1998 OR2 Asteroid

Hi, just wanted to ask questions concerning the subject above - What will be the impact of the asteroid on the earth? - Where it will hit hard?

Please explain it in less 'Scientific terms', my field is quite different

I'd be grateful Thanks --RazorTheDJ (talk) 18:09, 27 April 2020 (UTC)

Wikipedia has an article titled (52768) 1998 OR2 that should answer your questions. The most relevant sentence I can quote to answer your question by reading that article is "On 16 April 2079, this asteroid will make a near-Earth encounter at a safe distance of 0.0118 AU (4.59 LD), and pass the Moon at 0.0092 AU (3.6 LD)" Which is to say the next closest approach to the earth by this asteroid will be a distance of 4.59 times the distance the moon is from the Earth. In your lifetime, there is zero chance of it having any impact. --Jayron32 18:18, 27 April 2020 (UTC)

Hi @Jayron32, thank you for your response

What's all about the April 29th 2020 frenzy of the asteroid 'crushing' the earth? Please explain

P.S I'll probably be alive in 2079 at a proud age of 76!!😁--RazorTheDJ (talk) 19:52, 27 April 2020 (UTC)

It is also making a "close" approach on April 29th. It is not "crushing" the Earth. It will miss us, at its closest it will be 16 times further away from the Earth than the Moon, and we don't generally consider the Moon to be crushing us. It will not hit us. We know its closest approach to an accuracy of +/- 7km, which is incredibly accurate. --OuroborosCobra (talk) 19:55, 27 April 2020 (UTC)

Where can I see this "April 29th 2020 frenzy"? HiLo48 (talk) 03:17, 28 April 2020 (UTC)

See Snopes.  --Lambiam 04:50, 28 April 2020 (UTC)

There are lots of rocks that get much closer to the Earth than this one. See e.g. spaceweather.com table of Near Earth Asteroids. Dark red entries indicate rocks that come closer than the Earth-Moon distance, lighter red lines are less than five lunar distances. All of that happens several times every month. NASA has a Near Earth Object database where you can dig for more orbital elements of rocks. 85.76.44.116 (talk) 06:51, 28 April 2020 (UTC)

Thanks so much for your responses,all of you. At least there's nothing gonna annihilate us, for now. So what's all the fuss of it being 'potentially dangerous to earth' ? Why should some websites spread fear if at all it will miss us?--RazorTheDJ (talk) 12:13, 28 April 2020 (UTC)

Maybe because they are sensationalist and have some way of benefiting from the attention. That's existed for a long time, in the form of "buy my doomsday pamphlet" or "visit my website as clicks get me revenue," etc. That, or people who genuinely do not understand the science and spread hysteria. That also has existed for a long, long time. --OuroborosCobra (talk) 14:47, 28 April 2020 (UTC)

Will it be seen by the naked eyes or with the aid of a telescope as it passes?--RazorTheDJ (talk) 16:51, 28 April 2020 (UTC)

If I did my maths right, the rock will have an angular size of 0.0002 degrees. Which is like looking at a car tail light at a distance of 200 kilometers. So pretty much negatory on the naked eye. Telescope - sure, though probably you'll want to set a longish exposure time on your telescope camera. 85.76.44.116 (talk) 17:03, 28 April 2020 (UTC)

Wait, the end of the world was two days ago and I missed it?! Ultimate FOMO! —Tamfang (talk) 02:54, 2 May 2020 (UTC)

April 28

Some bird

What's this bird? Yellow beak and yellowish (yellow-greenish) patches on the upper side, spotted in Warsaw, Poland. Thanks. Brandmeister^talk 08:57, 28 April 2020 (UTC)

Common starling. Mikenorton (talk) 09:02, 28 April 2020 (UTC)

Duplicate Wikidata items?

I came across two Wikidata items, Q23015579 and Q132641 which are both called "Glyptodontidae" where one has the description "subfamily of mammals (fossil)" but the other has the description "family of mammals." However, otherwise they appear to be about the same thing. Are these items duplicates? DraconicDark (talk) 15:42, 28 April 2020 (UTC)

There's a fairly consistent latin naming system for various Taxonomic ranks, with consistent use of suffixes for specific ranks. In this case, the -idae suffix is for a zoological family, and the -inae suffix is for zoological subfamily. Other than that, I don't know how to solve any knock-on problems from the double naming, but the -idae suffix indicates that that specific word is for a family and not a subfamily. --Jayron32 15:48, 28 April 2020 (UTC)

@Jayron32: Upon further inspection, it turns out Q132641 is called "Glyptodontinae" which would fit the naming convention you stated. However, there still appears to be significant overlap between the subjects, and apparently the glyptodonts' classification isn't as clear as I had originally thought. This makes the question of whether to merge the items more complicated. DraconicDark (talk) 16:10, 28 April 2020 (UTC)

Taxonomic classifications change alarmingly frequently, and the subfamily Glyptodontinae is currently classified under the family Chlamyphoridae, but that doesn't mean there didn't used to be a family named "Glyptodontidae". Whether that family was subsumed into Chlamyphoridae and reclassified as a subfamily, or whether Chlamyphoridae was formerly called Glyptodontidae and just renamed, I couldn't say, both are possible and there are numerous examples of each happening, as well as other permutations (Glyptodontidae containing several subfamilies, including Glyptodontinae as well as several others, and then it was just deleted and its subfamilies added to Chlamyphoridae for example). You'd need an actual taxonomist to be able to make a decision here; I'm just a well-read chemist myself and don't have a lot of training here beyond what I find in Wikipedia articles. --Jayron32 16:15, 28 April 2020 (UTC)

Explained at Glyptodon. Jayron32's suggestion of a former family subsumed into another is correct.--Khajidha (talk) 17:37, 28 April 2020 (UTC)

@Khajidha: So should the Wikidata items be merged? DraconicDark (talk) 18:25, 28 April 2020 (UTC)

Not being clear as to what Wikidata is or what it does or how it's used, I do not have any opinion on this. --Khajidha (talk) 21:08, 28 April 2020 (UTC)

Higher/lower voltages with respect to fire safety

In the United States, the standard voltage used is 110V. In many other countries, the voltage is much higher. Is there any consensus in the engineering community as to which one is safer? I would think higher voltages are safer since circuits with a higher voltage do not require as high of a current as lower voltage circuits, but obviously electrical engineering is a bit more complex than that, so I might be wrong. --Puzzledvegetable^{Is it teatime already?} 17:42, 28 April 2020 (UTC)

• The standard in the US (and Canada) is actually 120 V: see Mains electricity by country. However, that's just the nominal voltage; in practice it will vary from one place or time to another, and most of the time a variation of that amount is insignificant. --76.71.6.31 (talk) 20:01, 28 April 2020 (UTC)

Aren't both of you wrong? The standard in the U.S. is 117 rms. 67.175.224.138 (talk) 22:11, 28 April 2020 (UTC).

I cited a reference, but it's only Wikipedia. What's yours? --76.71.6.31 (talk) 22:21, 28 April 2020 (UTC)

Either one can kill you. What kills you is the electric current going through your body; see our article on Electrocution. The higher the current, the sooner it is fatal. The strength of the current is determined by Ohm's Law and is a function of voltage and resistance. Given the resistance of your body, if it is not isolated from ground, the current is directly proportional to the voltage. So at 220V the current will be twice that at 110V, and will be deadly sooner.  --Lambiam 18:48, 28 April 2020 (UTC)

Actually at 220 V you need twice less current to draw the same power, so the fuses typically have less amps. 93.136.9.236 (talk) 23:33, 28 April 2020 (UTC)

What you need is less, but that doesn't have to do with how the current manifests. Current is determined by Ohm's Law, and your resistance does not differ because you are standing in Europe and not the US. If your body resistance isn't changing, then the current resulting from 220 V traveling through you will be double. --OuroborosCobra (talk) 23:40, 28 April 2020 (UTC)

Do fuses have different resistances in Europe? 67.175.224.138 (talk) 23:44, 28 April 2020 (UTC).

For fire safety, the issue is the heat generated; see Joule heating. The heat is proportional to the product of current and voltage. So a voltage that is twice as high but a current that is only half as high produce the same amount of heat in the wiring. The amount of useful work is also proportional to that product; see AC power. So in the end the fire hazard, given the same amount of work, should be equal.  --Lambiam 19:02, 28 April 2020 (UTC)

The rate of heat generation in a resistance is indeed given by the product of the current through it and the voltage between its ends. But the voltage drop across a poor connection in domestic wiring is not equal to the supply voltage, since the connection is in series with some domestic appliance such as an oven or kettle.

Unless the connection is very bad indeed, its resistance will be far lower than that of the appliance. In that case the current in the circuit is approximately the power rating of the appliance divided by the supply voltage (so double the current for half the supply voltage). The voltage dropped by the poor connection is the current times its resistance (so doubled by doubling the current). That means for an appliance of given power, and a connection of given resistance, the heating at the connection is quadrupled by halving the supply voltage, consequently increasing the fire risk.

This is an over-estimate, as we have neglected the fact that the poor connection will reduce the current slightly (though it is easy to correct for that). catslash (talk) 00:50, 29 April 2020 (UTC)

My current [pun not intended] calculations suggest even a factor of 16, assuming that the resistance of the poor connection is small compared to that of the appliances hooked up. The reason is that (to achieve the same amount of useful work or heating) the appliance resistance increases proportionally to the square of the voltage. The relative contribution of the poor connection to the voltage drop over the serial circuit of poor connection plus appliances then decreases proportionally to the inverse of that square, giving another factor of four.  --Lambiam 05:26, 29 April 2020 (UTC)

I'm not following you. If the source voltage doubles and the load impedance is quadrupled to maintain the same power, then the ratio of load_impedance : I^2R_loss_of_poor_connection does indeed increase by 16X, but isn't that a meaningless ratio to even consider? -- ToE 13:13, 29 April 2020 (UTC)

I have now redone the algebraic calculations using a computer algebra system. I must have made some error in the calculations done by hand. To a first approximation, the heat generated in a bad (high-resistance) wire is indeed inversely proportional to the square of the source voltage, when in a serial circuit with an appliance of a given power consumption.  --Lambiam 12:17, 30 April 2020 (UTC)

@Lambiam: You have written "... produce the same amount of heat in the wiring." When you use the word "wiring" I will assume you mean the two conductors that join the power source to the appliance; in which case the amount of heat produced in the wiring is not the same regardless of the voltage at the source. As you know, the heat produced in a conductor is the square of the current multiplied by the resistance of the conductor. If an appliance draws a given power P, and we consider first one supply voltage, and then consider doubling that supply voltage but still supplying power P to the appliance; doubling the voltage will cause the current in the wiring to halve. The heat produced with half the current will be only one-quarter of the heat produced in the first case, simply because joule heating is directly proportional to the square of the current providing the resistance is unchanged. So the risk of fire caused by joule heating in the wiring decreases dramatically as supply voltage is increased to an appliance drawing a given power P. Dolphin (t) 15:20, 29 April 2020 (UTC)

By "wiring" I was thinking of the wires in the walls, which in older buildings are often not in good shape, and by "voltage that is twice as high" I meant the voltage drop across the wiring. However, the latter is not simply proportional to the voltage supplied by the supplier.  --Lambiam 12:17, 30 April 2020 (UTC)

Australian switched dual 3-pin socket-outlet. See AS/NZS 3112#Switch requirements. -- ToE 17:03, 29 April 2020 (UTC)

Here in Australia, with 240 volts, power points/wall sockets almost always have switches, meaning that devices can remain plugged in but not be electrically alive. My limited observation is that switches don't generally exist on US wall sockets with 110 volts. While not a direct consequence of the different voltages, perhaps the switches are there because of the higher voltage involved, and definitely make things safer. HiLo48 (talk) 02:39, 29 April 2020 (UTC)

A frequent feature in American homes is that some of the wall outlets will be controlled by a switch, though not necessarily a direct part of the outlet. The purpose of that setup is to allow floor lamps and such to be turned off by the switch. Outlets that are near water sources, such as sinks, may have a different type of switch, kind of a "breaker" which trips if there's a short. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:07, 29 April 2020 (UTC)

I always called them GFIs but wiki calls them RCDs. hmm. Pelirojopajaro (talk) 13:54, 29 April 2020 (UTC)

Just to make sure that I'm understanding this, the answer to my question is that the heat in a circuit is proportional to the product of the current and the voltage, and not just the current, as I assumed in my question. Therefore, higher voltages are not necessarily less likely to cause a fire. Correct? --Puzzledvegetable^{Is it teatime already?} 14:12, 30 April 2020 (UTC)

That is correct. Electric power is voltage times current, and heat generated from electricity is directly proportional to power. This type of heating, called Joule heating, is often written as "I²/R", but that's just a rewriting of the variables in terms of current and resistance, voltage is just I/R, so voltage is just as important here. It's important to note that V and I are not independent variables, they are linked via Ohm's law, and changes to one (given a constant resistance) will cause changes in the other. Since Electrical resistance is temperature dependent, the voltage-current relationship is significantly non-linear over a wide enough temperature range. When it is said that "it's the current that kills you, not the voltage", that's also true, but misleading. Current is dependent on the number of electrons whereas voltage is not. A few electrons at an extremely high voltage will produce only a small current. However, if we're dealing with the same system (say a specific wire) then increasing the voltage through that wire will have a direct increase in the current. Ignoring the temperature dependence I mentioned earlier, if you double the voltage across a wire, you'll also double the current through that same wire, meaning that circuit is now twice as dangerous. On the flipside, by changing the material, you can change the current delivered without affecting the voltage. You can get astronomical voltages across good insulators, and get minimal to no current from them (that's what makes them an insulator). The point with this tangent is to note that the voltage and current need to be dealt with separately, but not entirely so, because each affects the other, but not in ways where we can say they are identical. There's a reason why electrical engineers need a good amount of calculus to do their jobs, the real math here gets messy. --Jayron32 14:28, 30 April 2020 (UTC)

Being the question 'Higher/lower voltages with respect to fire safety', the answer is 'yes, higher voltage increases considerably fire safety'. That is by the way why Europe moved from 110 V to 240 V at the beginning of the 20th century: this way they had not to replace miles and miles of wire no more able to manage the rapidly growing loads. The point is that doubling the voltage you can double the impedance of your appliance to do the same work, thus halving the necessary current. And as Dolphin pointed above, by halving the current you divide by four the resistive heating of the conductors.

So the reasoning that by doubling the voltage you also double the current if the circuit doesn't change is true but here irrelevant as the circuit must and will change: a toaster is part of the circuit, and a toaster drawing 800 W by 110 V must more than double its impedance to draw 800 W by 240 V, thus more than halving the necessary current.

Generally speaking, by doubling the voltage trasmission losses are reduced by a factor of four, that is why very long overhead power lines are operated with tensions up to 1 MV. 2003:F5:6F0A:C00:5CC0:1EE6:8417:F1F (talk) 19:38, 30 April 2020 (UTC) Marco PB

Yes, but I understood from Lambiam that the heat is based on the product of the voltage and the current, so the lower current should be irrelevant since it requires a higher voltage, and the heat remains the same. Is this incorrect? --Puzzledvegetable^{Is it teatime already?} 20:31, 30 April 2020 (UTC)

This is exactly the point, because the heat does not remain the same: heat doesn't depend from the current but instead from the current squared, see Joule_heating and voltage doesn't appear at all in the equation. You halve the current, you divide the heat by four. Suppose the current is 2, then heat is 2² = 4. Half the current = 1 and 1² is 1.

Lambiam is mistaken: heat is not based on the product of voltage and current, but on the product of current squared and resistance (current itself depends on voltage divided by resistance.) 2003:F5:6F0A:C00:5CC0:1EE6:8417:F1F (talk) 21:54, 30 April 2020 (UTC) Marco PB

They are the same. By Ohm's Law, I = V/R, so V = IR. Then we find for the product of voltage and current: VI = IV = I×IR = I²R, the product of current squared and resistance. No mistake there. My initial mistake was very different. It was not to realize that the relative voltage drop across the wiring is affected by the voltage-dependent resistance of the appliances (hooked together with the wiring resistance in a serial circuit).  --Lambiam 07:32, 1 May 2020 (UTC)

Okay, if you looked at Lambiam's table of 110 vs. 220 V. We saw the R got quadrupled. So should we look at P as in temperature of wire in terms of VI or I²R? The problem with looking in terms of VI is you don't see the change in R when V is doubled and I is halved. This makes I²R more convenient to use? 67.175.224.138 (talk) 14:18, 3 May 2020 (UTC).

Okay going back to what Jayron said as voltage = I/R, when UK goes from 120 V to 210 V or so, part of the confusion is, is both I and R both changing? Voltage can go up if R is getting smaller with I being the same. Voltage can also go up if I is getting bigger with R being the same. And so and so forth. 67.175.224.138 (talk) 05:27, 1 May 2020 (UTC).

@Jayron32: and the IP address immediately above have written that voltage is I/R and power is I²/R. Both are incorrect. Ohm’s law tells us that voltage drop across a conductor is IR; and heat is generated by a conductor at the rate of I²R. Dolphin (t) 06:03, 1 May 2020 (UTC)

Okay, then as IR goes up, the safety is therefore the same? Stronger amps, but stronger resistance, if going up equally. 67.175.224.138 (talk) 06:39, 1 May 2020 (UTC).

An example with numbers may clear up some things. The assumption is that we have a series circuit consisting of an ideal voltage source and two resistance components. One component is an appliance or combination of appliances, for example an electric heater and a washing machine, consuming together nominally 3kW. The other component is formed by the wiring, rated at 0.02Ω.

For 110V

source voltage: 110V; nominal appliance wattage: 3000W; wiring resistance: 0.02Ω

appliance resistance: 4.03Ω; total resistance: 4.05Ω; current: 27.1A

appliance voltage drop: 109.5V; wiring voltage drop: 0.5V

actual appliance power: 2970W; wiring power: 14.7W

---

For 220V

source voltage: 220V; nominal appliance wattage: 3000W; wiring resistance: 0.02Ω

appliance resistance: 16.13Ω; total resistance: 16.15Ω; current: 13.6A

appliance voltage drop: 219.7V; wiring voltage drop: 0.3V

actual appliance power: 2990W; wiring power: 3.7W

For a voltage source that is twice as high, the heat generated in the wiring drops to 25%. If the wiring resistance goes up, that heating also goes up to a maximal value of 750W, after which it decreases. The maximal value is reached sooner for the lower source voltage; when the (faulty) wiring resistance reaches 8.1Ω, the still rising 220V wiring power exceeds the already dropping 110V wiring power.  --Lambiam 10:51, 1 May 2020 (UTC)

So notice that R is quadrupled when we went from 110 V to 220 V. Needless to say, I think it was pointless to say heat/temperature/fire safety is determined by the product of I * V. Just as it could be said I²R. In the latter, we saw the R quadrupled. And since R = V/I, a stronger R is a bigger V and smaller I. So I think to answer the original poster's question, we shoulda said this heat/fire safety in terms of R (or V/I), and not as VI or I²R. @Puzzledvegetable:. 67.175.224.138 (talk) 14:38, 3 May 2020 (UTC).

Weird, it seems like the discussion ended already. I'm curious to know why voltage and amps can't be independent of each other. Can you not increase the voltage of something with the amps staying the same? But as a separate, and and somewhat political question, if 220 V is safer than 120 V from a fire-safety point of view, then why hasn't U.S. attempted to change into that? 67.175.224.138 (talk) 05:29, 2 May 2020 (UTC).

For any electrical load that is resistive in nature, Ohm's law is applicable and says that the current is directly proportional to the voltage. Fire-safety is only one aspect of an electrical supply system; another is safety with respect to electrocution. The thing that protects us all from electrocution is the insulation surrounding all live conductors; the higher the voltage, the greater the amount of insulation that is required, but there is always a risk that the insulation surrounding an appliance will be damaged, allowing a person to come in contact with a live conductor. Keeping the voltage low is an alternative to using increasing amounts of insulation so it can be argued that people who live in a country where the domestic supply voltage is 110V are safer with respect to electrocution than those who live in a country where 220V is supplied. Dolphin (t) 12:30, 2 May 2020 (UTC)

The lower domestic supply voltage in the US, and consequent higher current, necessitates thicker wires. Thick wire use a lot of copper, which is expensive. This motivated the use of aluminum wiring in domestic premises in the US in the 1960s and 70s. Achieving and maintaining a good electrical connection to aluminum can be problematic, and poor connections a fire hazard (especially those carrying high currents). In this way the lower voltage did, somewhat indirectly, lead to a fire risk. catslash (talk) 16:24, 2 May 2020 (UTC)

"The higher the voltage, the greater the amount of insulation is required." So that's like saying the lower the amps/current, the more insulation needed. Insulation is synonymous with resistance right? Weird, I think of current as the amount of electricity, and voltage as the force that pushes the electricity. So UK is like the stronger force that pushes the same amount of current, but will actually be less current, which I visualize as faster electricity, whereas US has weaker force for more electricity. So has anyone said whether resistors/fuses are different in UK and US? 67.175.224.138 (talk) 22:27, 2 May 2020 (UTC).

The insulation being discussed is what surrounds the wire (rubber/plastic sleeve, etc.), not a poorly-conductive segment of the conductors themselves. DMacks (talk) 06:59, 3 May 2020 (UTC)

With regards to fire safety, people earlier were saying that fire safety is proportional to IV. But as V goes up, so does R. So that make UK safer for fire safety, but US safer for electrocution? 67.175.224.138 (talk) 22:38, 2 May 2020 (UTC).

Okay, it looks like with all the said, it can be summarized into a simple paragraph: "Higher voltage is safer for fire hazards, but lower voltage is safer for electrocution." Higher voltage being with lower amps and lower voltage being higher amps. And although this answers with power (IV) it doesn't answer with the 3rd variable R. But among R, there's 2 factors to consider: the wire resistance R, and the insulation around the wire. Does anyone want to take a dab at the economic feasibility, between US and UK, of the wire and insulation around the wire? 67.175.224.138 (talk) 13:26, 3 May 2020 (UTC).

April 29

Planets

In the Planet article, it says that in order to be considered a planet, an astronomical body needs to be massive enough to be rounded by its own gravity. About how much mass is required for this? Is there some kind of "minimum mass" for a planet? 193.210.228.37 (talk) 11:43, 29 April 2020 (UTC)

This will depend on how strong the material is. If it is the very weak solid nitrogen, the smallest size would be much smaller than for an iron nickel asteroid. Graeme Bartlett (talk) 11:45, 29 April 2020 (UTC)

Here is an assessment in terms of radius. The radius is more useful for astronomers because it's easier to measure than mass (you need e.g. a satellite to measure mass). As expected the necessary radius depends a lot on the body's composition. 93.136.9.236 (talk) 15:30, 29 April 2020 (UTC)

Pluto is certainly round, but Tyson decided it wasn't a planet anymore, so it isn't. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:49, 29 April 2020 (UTC)

Because being round isn't the only qualifying factor to make something a planet. Tyson didn't decide anything, either. The International Astronomical Union decided on a definition. Tyson has been a proponent of said definition, but it wasn't his call. --OuroborosCobra (talk) 22:49, 29 April 2020 (UTC)

He championed it, so he gets the blame. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:57, 30 April 2020 (UTC)

That's not how this works. That's not how any this works. --OuroborosCobra (talk) 16:03, 30 April 2020 (UTC)

"Roundness" is not the only criterion. Pluto fails on the other tests for planetness. Or, to be more exact, if Pluto is a planet, then so are umpty-dozen other things in and around its orbit and in and around the asteroid belt. Which would make planet a less useful term. --Khajidha (talk) 17:36, 30 April 2020 (UTC)

How I Killed Pluto and Why It Had It Coming by Michael E. Brown is well worth reading. Mikenorton (talk) 17:53, 30 April 2020 (UTC)

Ceres with some more than 900 km and a mass of 9.5 × 1020 kg was the smallest celestial body thought to have been rounded by its own gravity. I think I read somewhere that another body with about 400 km could be also rather round and this would be probaly an example of minimum mass. But yes, mass alone is not all: a stony body covered with water ice will possibly get round faster than some irregular chunk of iron-nickel. 2003:F5:6F0A:C00:5CC0:1EE6:8417:F1F (talk) 20:17, 30 April 2020 (UTC) Marco PB

The self-styled experts changed the definitions. Pluto never changed. And Pluto will be around long after those who dissed it are gone. ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:14, 30 April 2020 (UTC)

That's a very unscientific way of thinking. You do understand that the category "planet" is just a creation of the experts that is used to classify things so that we can understand them better? Also, experts had been debating Pluto's "planetness" basically since it was discovered.--Khajidha (talk) 22:31, 30 April 2020 (UTC)

And speaking of Ceres, when it was discovered in 1801, it was universally hailed as a newly-found planet. All decent science books from then on included Ceres along with Jupiter, Mars etc and all the other then-known planets. This situation obtained for almost 60 years (almost as long as Pluto was regarded as a planet), until it was reclassified as an asteroid after many similar-sized objects were found, and it dropped out of general consciousness so totally that most people these days have never even heard of it. That's unlikely to happen with Pluto, mainly because of all the hullabaloo after it was reclassified as a dwarf planet. I actually accuse most scientists of being closet astrologers whose equanimity was disturbed when they had to go back and recalculate all their horoscopes with Pluto left out. That's what this is really all about, mark my words. -- Jack of Oz ^{[pleasantries]} 23:47, 30 April 2020 (UTC)

I'm a supporter of Alan Stern's idea to call them "überplanets" ("main planets" sounds better to me). Pluto, Eris and Sedna are basically planet-sized and planet-shaped bodies that happen to be very far away from the Sun. Could've gone with that but that would mean avoiding a huge publicity grab... 89.172.8.118 (talk) 01:09, 1 May 2020 (UTC)

Baseball Bugs If you're interested, I strongly recommend this casual YouTube video from CGP Grey. Link here. --Puzzledvegetable^{Is it teatime already?} 00:00, 1 May 2020 (UTC)

What exactly does it mean for a photon to have an amplitude?

Given the extent to which photons cannot be described as classical waves, what exactly does it entail to measure the amplitude of one? As in, what are you physically measuring and called the 'amplitude', if the photon itself does not truly possess a 'height' the way a classical wave might. Apologies for any misconceptions inherent to the question itself. — Preceding unsigned comment added by Opossum421 (talk • contribs) 11:50, 29 April 2020 (UTC)

Amplitude in quantum mechanics refers to the level of probability of making that measurement at that value. See Probability amplitude (though that article like many at Wikipedia is probably too technical for someone unfamiliar with the field. Le sigh.) The basic way to think of the wave function is that it isn't a wave in space-space, it's a wave in probability space. Only the position wave function includes space-space in it, and even that isn't telling you where the particle is, it is telling you where the particle is likely to be if you were to try to interact with it. The particle isn't anywhere in particular until you interact with it, then it's wave function changes to a more narrow shape showing the increased probability that the particle is located where you interacted with it. There are lots of other wave functions, however, that do not have space or position in them, such as momentum or spin orientation or the like. If there was one thing about QM that most people get wrong (often because teachers either themselves don't get this, or don't make it explicit enough) is that quantum mechanics abandons the notion that a fundamental particle is an object in any sense we understand the term "object" to mean. Some properties of quantum particles are definite and well-defined, like electric charge and spin magnitude and things like that, but other properties (including position and momentum) are not well defined, meaning those values can only be described by giving the probability of finding the particle with that value for the measurement. If we graph all of those probabilities on a coordinate system, where the Y axis is defined as the probability of finding the particle with that value, and the amplitude is just the maximum probability of finding the particle with that value. The mathematics of the graph we draw obeys the mathematical rules of wave mechanics, which is why it is called a wave function. But again, it isn't a wave in the real world. It's a wave in probability space, which is a mathematical construct where the property being measured is plotted on some of the axes, and one of the axes is a probability. (This is a massive oversimplification, and ignores the difference between, for example Ψ and Ψ² and a WHOLE lot of messy math, but it captures the spirit, which is all we need here). I highly recommend a few channels on YouTube that explain this even better than I can through text are Science Asylum (Nick Lucid) and 3Blue1Brown (Grant Sanderson). Science Asylum does a fantastic job of explaining complex physics phenomena in a short, informative, and easy to digest way (pretty much the OPPOSITE of Wikipedia articles on the subject) while 3Blue1Brown covers complex mathematics in a slow, methodical, but still fantastically clear manner that also makes these things easy to understand. This video that Science Asylum did on the Wave Function is probably most germane to our discussion, but his entire series on QM would be useful. I don't have a specific 3Blue1Brown video to recommend on this direct topic, but he's done lot of videos the mathematics useful to understanding QM (such as this video on the uncertainty principle [2] which starts off his whole series on the Fourier transform, which is a mathematical tool with many wide-ranging uses, but which is SUPER important to understanding the mathematics of QM). I hope that helps some. --Jayron32 13:19, 29 April 2020 (UTC)

Like many simple questions in physics, there is a long and convoluted answer! To summarize shortly, whomever spoke of the photon's amplitude was being sloppy and imprecisely using terminology - and this sort of nonsense is very common amongst practicing physicists, because as a community, they've generally proven to themselves that they can be precise when they need to be, and they know that they don't usually need to apply that precision, because being excruciatingly correct all the time becomes very tiresome.

So - to the question: what is the amplitude of the photon?

To recite the rote-memorized answer, that would be a quite trivial question! Why, the amplitude of "the wave function", of course.

Now, the thing that's funny about quantum physics is that the books all speak in grandiose terms about the wave function; they have a special equation for the function; and they even have a special-purpose symbol, $\Psi$, which is exclusively reserved to represent the function. And if you read any book on the topic, the amplitude of $\Psi$ is a scalar; and it is an abstract quantity that can be used to predict the probability and the time-evolution of measuring some other physical observable. But what is it?

If you read a little deeper, and learn what the function is, you'll discover that it's actually unique to every single physical system. There are actually many different wave functions; the part that is common to all of them is that they must satisfy certain mathematical and physical requirements.

In the case of the quantum-mechanically-correct description of the photon, the actual equation that describes the wave function is not so surprising: it is derived from the very same relationships that describe the classical electromagnetic wave; and that wave is easily represented using the classical wave equation of conventional electrodynamics. That wave is a pair of coupled vector-quantities; the two components - a magnetic field and an electric field - have position and direction, but they have no extent.

By convention, we often simply work with the amplitude of the electric field and we ignore the amplitude of the magnetic field; but it doesn't really matter, because we can equally convert to calculate the amplitude of the magnetic field. Because the two fields are related by a very well-behaved governing-equation, we can easily switch between electric- and magnetic- field amplitude; this is simply a change of units, with a little extra-special mathematical care in the case the wave is propagating inside an imperfect material.

To make the equations for these amplitudes quantum-mechanically-correct, we just need to make sure that the time-evolution of the system abides by any quantized interactions; but in the case of the electromagnetic wave in a vacuum, there's no change from the classical case: the photon is the quantum particle, and its quantum mechanical wave function is the classical electromagnetic wave function.

The extra work to manipulate the equation so that it looks like the Schrödinger equation is really just a rearrangement of variables - and when you succeed in doing that, you've simply rearranged the same thing into a format that is particularly inconvenient for expressing macroscopic interactions between the photon and the outside world. But that form is useful for describing special cases in physics - things like photon-material interactions; or making statistical predictions about the probabilistic behaviors of individual photons; and so on. For those cases, the amplitude of $\Psi$ is used to predict the amplitude of E and B (the electric field and the magnetic field).

And of course, because the actual values of those field amplitudes are themselves a probabilistic attestation deduced from the amplitude of $\Psi$, we expect that $\Psi$ should evolve predictions that, over time, change in a manner that is statistically indistinguishable from the time-evolution of the amplitudes of E and B, in other words, the classical wave equation.

Nimur (talk) 13:27, 29 April 2020 (UTC)

Nimur's answer is also really good here, as it draws comparison to "classical" wave mechanics and quantum mechanics. I just want to point out that his analogy between the the classically defined EM wave equation and the quantum wave function is really useful here, but maybe not in the direction one thinks; for example, when one asks "what is waving" in the EM wave, we say "the electromagnetic field". This is similar to saying that the thing that is waving in an ocean wave is the ocean, except that the ocean is a real physical stuff we can pick up and touch and is an object. The electromagnetic field is not a stuff. Like all fields (classical or quantum) it is a mathematical construct. A field is just a set of numbers (which can be scalar, vector, or tensor) that we attach to space itself, describing a specific property that space has. The "manipulation of variables" and the other things Nimur talks about relating the quantum wave function to the EM wave itself are just a set of mathematical transformations. Just as you can do math to convert one function to another (for example, the way that the exponential function e^ix can be used to convert between circular motion and wave motion, there are also trivial mathematical tools we can use to convert between EM waves in the EM field to quantum wave functions in probability space. Such mathematical tools change our perspective on a phenomenon, but don't fundamentally change the phenomenon itself). So, when we say "what is waving" in an EM wave, it's the values we assign to the various points in space that we call the EM field, and when we say "what is waving" in the quantum wave function, it's the probability of the particle having the values in question. And because those functions are describing the same phenomenon from different perspectives (both describe light), there must be some mathematical tool to relate them to each other. --Jayron32 13:55, 29 April 2020 (UTC)

Thank you, Jayron. Your post has helped me to realize something: I am just assuming that everybody already knows what electric fields and magnetic fields are. Of course, if our readers are not intimately familiar, they can review those articles.

If I may coopt Jayron's terminology - and simplify by glossing some of the details - the "thing" that is "waving" is the Electric field (and the B-field, per our earlier discussion). The E-field is not "stuff." It is a set of numbers that tell us what force would be felt by a charged particle if it were at a specific place. Those numbers are what are waving up and down. If we had a way to measure those numbers, we could see them fluctuating in a more-or-less perfectly sinusoidal fashion. But the only way we can measure those numbers is to place a test-particle at some specific position, and watch how it moves.

The really tricky part is that if we actually put a charged particle there, the waving numbers that tell us what force the particle should feel would change because they would interact with the charged particle; it would have its own fields; and it would also be moving as the wave interacts with it. So these wave-equation representations are mathematical idealizations - they help us make useful predictions, and they only make helpfully-valid predictions when we consider statistically-large numbers of individual particles, so long as we are willing to ignore the effects of any individual particle.

It isn't necessarily obvious - great scientists became famous for finally realizing this very subtle detail! - but the E- and B- fields are deeply deeply related to the photon. Early physicists described electricity, and magnetism, and light; and for a lot of centuries, we thought of them as three separate things; but ... we now understand that all of them are actually the exact same thing under different conditions. Photons interact with E- and B- fields - but even more importantly, photons are actually made of E- and B- fields. There's nothing else inside of a photon except those fields - and those fields are not made of "stuff"! And if you're a real mathematical hot-shot with access to a good book, you can play some advanced vector-calculus-trickery to demonstrate that the E-field and the B-field are the exact same thing, too: they're not merely related or coupled fields - they're the same exact darned field. And this is the equally important bit, that's less-often shouted at students: if photons are made of E- and B- fields, then .... every E- or B- field you can conceive can be expressed as a photon. It might be a pathologically poorly-behaved photon - it might have a ludicrously useless wavelength; but ... there you have it. Photons-the-size-of-mountains, emitted by electric-wires-strung-between-peaks.

So if you're trying to figure out what "stuff" is "waving" when an electromagnetic wave ... "waves"... well, we can quickly find ourselves traveling down the rabbit-hole of physical definitions: what is "stuff"? Which physics-ese techno-jargon word corresponds to the plain English word "stuff"? (Matter? Mass? Momentum? Quantum state exclusivity?) And I am pretty sure our only conclusion will be that natural human languages are not really expressive enough to describe some of the physical realities unless we throw the weight of a lot of words at it.

Nimur (talk) 17:22, 29 April 2020 (UTC)

All of that being said, I really do recommend the OP watches the videos I recommended. Seeing someone explain something in a video format, with pictures and animations, can be a lot more enlightening than reading the same information. --Jayron32 18:15, 29 April 2020 (UTC)

The energy and momentum of a Photon which is the elementary particle of the electromagnetic field depend only on its frequency or inversely, its wavelength. Increasing the amplitude of a beam of light or radio wave doesn't create "stronger" or higher-amplitude photons, it just creates more of them. DroneB (talk) 21:39, 29 April 2020 (UTC)

Thank you tremendously folks, I hadn't anticipated such a thorough set of responses. To Jayron32's suggestion - I'm quite familiar with 3Blue1Brown and will surely venture down that particular rabbit hole along these lines, and have not heard of Science Asylum but will give it a proper go as well. I appreciate everyone's helpful explanations here. Opossum421 (talk) 11:12, 30 April 2020 (UTC)

Rooibus

Where can I buy a rooibus plant to grow in my garden?99.145.194.98 (talk) 19:42, 29 April 2020 (UTC)

Possibly you mean the Rooibus tea plant cultivated in South Africa. The American Herbal Products Association has a website that may help you locate sources and there is an Australian supplier. DroneB (talk) 21:22, 29 April 2020 (UTC)

Postage to and from Australia is very delayed at the moment due to limited international flights, so if you are in the USA, buy from USA. Graeme Bartlett (talk) 23:56, 29 April 2020 (UTC)

To import plants or seeds from abroad you need a phytosanitary certificate or a small lots permit, to make sure you are not bringing non-native pests or pathogens into the US. Please check USDA APHIS website for what documentation is necessary in your case, and how to apply for it properly. Best regards, Dr Dima (talk) 06:19, 30 April 2020 (UTC)

Googling [buy “Aspalathus linearis”|"Rooibos" seeds] produces several outlets.  --Lambiam 14:04, 30 April 2020 (UTC)

Try at your local plant nursery. 89.172.8.118 (talk) 19:00, 30 April 2020 (UTC)

April 30

Weider weight bench

Are there Weider olympic weight benches available anywhere?99.145.194.98 (talk) 05:28, 30 April 2020 (UTC)

Googling ""Weider Pro 395" brings up several companies offering these for sale.  --Lambiam 13:57, 30 April 2020 (UTC)

This isn't really a science question. --OuroborosCobra (talk) 16:04, 30 April 2020 (UTC)

Thanks for the help and kinesiology is science!99.145.194.98 (talk) 22:09, 30 April 2020 (UTC)

Your question didn't ask anything about kinesiology. You asked where to buy an olympic weight bench. That'd be like saying asking about where to purchase a tickle-me-elmo was a developmental psychology question. --OuroborosCobra (talk) 15:25, 1 May 2020 (UTC)

Light sail

In episode 2 of Cosmos: Possible Worlds they talk about using a light sail to send interstellar probes. At first they are powered by lasers from Earth, then they carry lasers to fire photons into the sail to keep it accelerating. But wouldn't the backward force of the photons leaving the laser be at least as large as the force on the sail? Bubba73 ^{You talkin' to me?} 20:42, 30 April 2020 (UTC)

Yes, it's the equivalent of blowing in your own sails on a regular sailing vessel. However, efficient light sails must be made of a reflective material, so if the sails reflect the light from the probe's lasers it has the same effect as just aiming the lasers the other way, so it would still work (but the sails aren't really needed at that stage). - Lindert (talk) 21:21, 30 April 2020 (UTC)

By the way, Mythbusters tested "blowing in your own sails on a regular sailing vessel". It works. I was surprised when I saw the episode (they didn't explain it very well) until I realized that it was equivalent to the thrust reversers on a jet plane: directing the air backward is, again, similar to aiming the fan the other way. --76.71.6.31 (talk) 21:51, 30 April 2020 (UTC)

Thanks, that is what I thought about the lasers and light sail. A light sail is supposed to have a very low mass. Is it feasible to have lasers and (presumably) batteries to provide enough energy to accelerate to relativistic speeds? Bubba73 ^{You talkin' to me?} 22:25, 30 April 2020 (UTC)

To accelerate to relativistic speeds on battery power (taking the batteries with you), you need batteries with an energy content that's large compared to the rest mass of the batteries. That won't work with chemical or even nuclear batteries. PiusImpavidus (talk) 08:30, 1 May 2020 (UTC)

Ah, yes, that is right. Bubba73 ^{You talkin' to me?} 22:20, 2 May 2020 (UTC)

May 1

Human psychology during pandemics

From a human psychology point of view, why are so many aspects of the current situation divisive amongst people across the world regardless of their background? Have any studies been done in the past to look at this? Clover345 (talk) 14:46, 1 May 2020 (UTC)

Examples? I'm guessing you mean religious vs. atheistic viewpoints? Like, "punishment from God" to nothing peculiar at all? But what other answer could you be looking for besides genetics. 67.175.224.138 (talk) 14:55, 1 May 2020 (UTC).

Genetics is not at all likely to be the answer or even what they are asking for if they are asking a psychology question. --OuroborosCobra (talk) 15:23, 1 May 2020 (UTC)

I'm not sure our original question is specific enough to receive a proper response; but when I find myself wondering about the softer sciences, I start by browsing the collection at JSTOR, a free digital library collection that provides a lot of scholarly research.

Via JSTOR: The Crowd in History: Some Problems of Theory and Method (Social History, 1978). "What might be called crowd history as a field of study in its own right has, rather surprisingly, received very little critical comment." Perhaps the interested reader will be able to take a deeper look into further materials.

For regular readers of AAAS's journal Science - rather more likely to overlap with our regular contributors on the Science Reference Desk - you might find this article very exciting: The Lessons of the Pandemic (Science, 1919). Like many other Science articles in the older archives at JSTOR, it is available at zero cost.

If I may editorialize, a bit: well written scientific analysis, from any century, provides an astonishing clarity that - to me - proves beyond reasonable doubt that despite the progress of a few scientific minds in a very few select areas of specific understandings of our natural world; and in spite of amazing the proliferation of advanced technology in daily life - the overwhelming majority of the knowledge that we have about our natural world has truthfully not changed very much at all since the so-called Dark Ages - because the overwhelming majority of knowledge is, and always shall be, embodied in the minds of very scientifically-uneducated individuals who constitute the so-called crowd.

Nimur (talk) 16:22, 1 May 2020 (UTC)

One thing that can be observed is that some people who are materially better off do not appear to fully grasp the economic hardship on others resulting from the closures. Economic relief may be denied or only given grudgingly and in insufficient amounts. The question when to re-open in what phases depends on risk assessments; these will be based on the information one receives, which may be rather different in different bubbles. The issue also depends on the relative values assigned to human life and health versus material wealth, which also may be based on one's ideology.  --Lambiam 16:39, 1 May 2020 (UTC)

That's common to pretty much every economic downturn, including the Great Depression. 89.172.65.59 (talk) 04:25, 2 May 2020 (UTC)

Source of names for 4 mild human coronaviruses?

• Human coronavirus OC43 ?
• Human coronavirus HKU1 Hong Kong ?
• Human coronavirus 229E ?
• Human coronavirus NL63 Netherlands? -- Jeandré, 2020-05-01t20:17z

You should read discovery papers. Ruslik_Zero 21:00, 1 May 2020 (UTC)

In most cases these are just the labels used in the labs that isolated and described the viruses to label the samples from which the strains were isolated. Each research group had its own labelling method. The "OC" in "OC43" stands for "organ culture",^[3] while "43" is probably a meaningless sequence number. Described in: McIntosh K, Dees JH, Becker WB, Kapikian AZ, Chanock RM. "Recovery in tracheal organ cultures of novel viruses from patients with respiratory disease". Proc Natl Acad Sci USA 1967;57:933–40. "HKU" is the abbreviation for the University of Hong Kong where the strain was isolated.^[4] Strain 229E was isolated by Dorothy Hamre and John Procknow and described in: Hamre, D., and J. J. Procknow, Am. J. Epidemiol. 83:238 (1966). Perhaps their publication explains the label, but I suspect it was simply the label of one lab sample among many, labelled ..., 228A, 228B, ..., 229D, 229E, 229F, ... . And "NL" is obviously the Netherlands.^[5]

Does pure ammonia expand when it freezes like water does?

Water, H2O, expands when it freezes. It has a different structure than NH3, because it only has 2 Hs, but isn't it frequently like H3O+, which might resemble NH3 structure?144.35.20.92 (talk) 20:44, 1 May 2020 (UTC)

No as far as I know. Ruslik_Zero 20:55, 1 May 2020 (UTC)

Our article on Ammonia gives among its properties densities of 681.9 kg/m³ at −33.3 °C (liquid) and 817 kg/m3 at −80 °C (transparent solid). As the mass of a fixed number of ammonia molecules will not change as the stuff freezes, this implies that a liquid cubic meter of ammonia will take up a volume of 681.9/817 m³ = 0.834 m³ when frozen.  --Lambiam 22:41, 1 May 2020 (UTC)

I had asked this question before about hydrogen peroxide, which is probably more similar to water than ammonia. And our article doesn't have its density as a solid, or talk much about it. People say this is a result of hydrogen bonding, but hydrogen bonding alone doesn't answer the question. As H2O2 and NH3 are both examples of hydrogen bonding. On a separate topic, I did have a list of other compounds that are like water with its density, I'ma look for it. 67.175.224.138 (talk) 00:38, 2 May 2020 (UTC).

The phase diagrams in this article [6] (pdf) imply that solid ammonia is denser than liquid ammonia. Higher pressure favors the denser phase, and solid ammonia is above the adjacent liquid ammonia on the pressure-temperature phase diagrams.--Wikimedes (talk) 04:31, 2 May 2020 (UTC)

In addition to the above, it is worth noting that water is not "frequently like H₃O⁺. At room temperature, for example, the concentration of H₃O⁺ = 10^-7 M. For perspective, water itself has a "molarity" of about 55.5 M. That means that in a given volume of water, only about 1.8*10^-7% of the molecules will be H₃O⁺. -OuroborosCobra (talk) 19:34, 2 May 2020 (UTC)

There is a one-sentence section at Freezing#Expansion that lists just two substances that expand when they freeze. The other one is bismuth. But it doesn't say how rare or common the phenomenon is (I've formed the impression that it's pretty rare, but I don't remember where I picked that up from, so don't believe me), or what sort of chemical properties cause it, or anything. I've just flagged the section, rather appropriately, as needing expansion. --76.71.5.208 (talk) 23:18, 2 May 2020 (UTC)

Plutonium does as well. Plutonium#Physical_properties last paragraph. But it is a rare phenomenon.--Wikimedes (talk) 03:46, 4 May 2020 (UTC)

How are inorganic archaeological artifacts dated?

Of course, any object containing the appropriate elements can be dated via isotope dating, but how do archaeologists date inorganic objects with any specificity? If a stone tool or fragment of pottery is found, is it simply dated according to other, more easily determinable nearby objects? Is it as simple as assuming a constant rate of wear-and-tear? Are there other convenient isotopes which decay at a useful time scale? Surely there are sophisticated techniques I am unaware of, and would be interested in hearing about. Opossum421 (talk) 23:27, 1 May 2020 (UTC)

When objects are found in an archaeological excavation, their context is recorded, particularly where they lay in the site's stratigraphy. Other datable objects are used to establish the age of the different layers and thereby the objects found within them. Mikenorton (talk) 00:14, 2 May 2020 (UTC)

Radiometric dating discusses several isotopes that are used in dating.--Wikimedes (talk) 04:17, 2 May 2020 (UTC)

Chronological dating contains many dating methods. Luminescence dating is an interesting one.--Wikimedes (talk) 04:20, 2 May 2020 (UTC)

As others have basically said, there are a lot of dating methods, and (ideally) more than one is used so as to validate results. When no method for a specific artifact is available, they may have to make use of referencing something else. So, for example, you find an arrowhead on its own made of a particular material and particular design, but for some reason you can't specifically date that arrowhead. You then look to see if other arrowheads of the same material, same design, and in roughly the same region have been found. Perhaps one was found together with other items, including a material that could be dated (hair? wood? there are tons of materials that we can date). Those materials then form a reference to date the arrowhead they were found with, and the arrowhead you found off on its own can be assigned a similar date based on referencing. It's not perfect, but it is better than nothing. This type of reference dating is done with extreme care and, given other evidence, can be revised. --OuroborosCobra (talk) 17:09, 3 May 2020 (UTC)

May 2

Is there a minimum practical size (thermochemically) for a blast furnace?

I know there are minimum economic sizes, with considerations for fuel and maintenance costs. I'm not asking about that.

From a thermal/chemical standpoint, there can be other considerations. For instance, you can't scale a bloomery furnace much below 30cm in inside diameter for reasons of fuel element size, air injection, and thermal loss through the (~10cm) clay walls. That means that in practice, the minimal size creates maximum bloom sizes between 2 and 20kg

I'm wondering if there are similar low-end constraints on a traditional (tall, brick, cast-iron-producing) coal+iron ore(+ lime) blast furnace. Anyone know?

Riventree (talk) 01:18, 2 May 2020 (UTC)

Making a miniature version of a full production-capable Blast furnace would mean sacrificing all Economies of scale and require much effort to assemble special tiny fire bricks, flues, bellows, etc., the ingredients would need to be unusually finely powdered and metered, and it would not be easy to keep a very small quantity of coal burning. However there is no lower size limit on performing the essential Smelting operation on Iron ore in a laboratory Crucible. DroneB (talk) 18:57, 2 May 2020 (UTC)

Maths question on pandemic graphs

In the pandemic graph, with time (t) on the x axis and infected people (c) on the y axis, I understand that the area under the curve remains the same if you reduce R(t) on the exponential curve. So in other words, it’s modelling the spreading out of the same number of infections over a longer period of time, in order to lower and delay the peak which I’ll call C-max. So how do you model this for countries which have eradicated this virus? The countries which locked down early seem to have reduced cases to 0 after a short period of lockdown but unless I’ve misunderstood, number of infections would have continued to rise if lockdown had not been imposed and so I don’t see how infected number of people would be the same by the end. So when you model this, which variable do you change? Or is it that this model becomes less accurate at the lower numbers of c since it will just tend towards infinity? 90.196.238.188 (talk) 13:15, 2 May 2020 (UTC)

Is it simply that the model assumes that in both scenarios (restrictions vs no restrictions) the same number of infections will occur which in reality wouldn’t be true? Or that R(t) gets closer and closer to 0 as the virus finds less people to infect? If this was the case, would this be so for every country? That R(t) will start tending towards 0 if the same measures are kept in place? 90.196.238.188 (talk) 13:30, 2 May 2020 (UTC)

We do not know enough to create a mathematical model with any certainty that it reflects the situation of the current pandemic, but an extremely simplistic model with a stationary population (ignoring demographics – so no births and no deaths from other causes) may help to clarify some things. Assumption 1 is that everyone is susceptible and may get infected if in contact with an infected person. Assumption 2 is that even vigorous contact tracing and isolation of contacts cannot entirely eliminate spreading (since they may have been infectious before they were identified and isolated). Assumption 3: no one (think people at high risk) can be kept indefinitely in isolation with zero chance of being infected. Under these assumptions, it is possible that the disease may get completely eradicated, like smallpox, before everyone is immune or deceased. But if we add assumption 4 that the disease will remain endemic with a low but non-zero lower bound on its incidence until no one is left to be infected, then eventually everyone will get infected, if not sooner then later, possibly much later. The total area under the curve of new cases against time (with linear scales and 0 at the bottom of the y-axis) is proportional to the total fraction of the population that has been infected. If that is the whole population, then all you can do is alter its shape. To explain the apparent paradox: the curve may have a thin but very long tail; its thinness is canceled out by its length. But if the disease is totally eradicated, that area may be genuinely smaller.  --Lambiam 15:36, 2 May 2020 (UTC)

Thanks, appreciate this is a very simplified model. So what would be the best way to represent that 4th assumption in this very simplified mathematical model? In my mind, if R(t) is less than 1, the number of infections will always tend towards 0? Do you basically make the assumption that R(t) will be just below 1? I know in reality, there’s a lot more to take account of but keeping with the simplified model. 90.196.238.188 (talk) 16:05, 2 May 2020 (UTC)

I guess representing what you’ve described mathematically would mean that the curve would tend towards infinity? So how do you vary R(t) to model this? — Preceding unsigned comment added by 90.196.238.188 (talk) 16:08, 2 May 2020 (UTC)

If we are working with proportions of the total population, so that the total solution corresponds to 1 (100%), the area under the curve is bounded between 0 and 1, inclusive. The effective reproduction number $R_{\mathrm {e} }$ at time $t$ can be found by multiplying the basic reproduction number $R_{0}$ by the fraction $S_{t}$ of the population that is susceptible at time $t$. This is described at Herd immunity#Mechanics; it ought to be mentioned also in the article on Mathematical modelling of infectious disease.

We can partition the total population into three compartments: $S$ for susceptible, $I$ for infectious, and $H$ for the rest, hopefully healed, but in any case the cumulative fraction that has at some time been infectious. At all times they sum up to the whole population: $S_{t}+I_{t}+H_{t}=1$. Individuals can transition from $S$ to $I$, and from $I$ to $H$. The transition rate from from $S$ to $I$ is given by $\mu R_{0}S_{t}I_{t}.$ The factor $\mu$ is needed because an individual's infectiousness lasts several days, and $R_{0}$ is the cumulative effect over that whole period. If we take one day as the unit of time, and we assume that an infected individual remains on the average infectious for 10 days, a reasonable value is $\mu =0.1$, the reciprocal of the infectious period. The transition rate from from $I$ to $H$ should depend on how long the members of compartment $I$ have been there, but for simplicity let us assume the continuous equivalent of a Poisson process and set the rate at $\mu I_{t}.$ Together this gives the coupled differential equations

${\frac {d}{dt}}S_{t}=-\mu R_{0}S_{t}I_{t};$

${\frac {d}{dt}}I_{t}=\mu R_{0}S_{t}I_{t}-\mu I_{t};$

We can eliminate $I$ by solving the first equation for $I$ and substituting the solution into the second:

${\frac {d}{dt}}\left({\frac {1}{S_{t}}}{\frac {d}{dt}}S_{t}\right)=-\mu \left({\frac {1}{S_{t}}}{\frac {d}{dt}}S_{t}-R_{0}{\frac {d}{dt}}S_{t}\right).$

I don't know if this can be solved analytically. I have simulated the process by a simple difference method: repeatedly replacing $t$ by $t+\Delta t$, $S_{t}$ by $S_{t}+\Delta S_{t}$, and $I_{t}$ by $I_{t}+\Delta I_{t}$, where

$\Delta S_{t}=-\mu R_{0}S_{t}I_{t}\Delta t,$

$\Delta I_{t}=~~(\mu R_{0}S_{t}I_{t}-\mu I_{t})\Delta t.$

I set $\Delta t=0.01$; smaller values give negligeably different results. To start the epidemic, we need a value for $I_{0}$. I used $I_{0}=0.0002$; the final outcome is not particularly sensitive to this, except for very small values of $R_{0}$. $H_{0}=0$, so $S_{0}=1-I_{0}$. In the end, $I_{t}$ tends to 0. I stopped the process 1000 steps (10 virtual days) after its value dropped below $5{\cdot }10^{-9}$.

Other than I had expected, $H_{\infty }$ was substantially lower than $1$ for moderate values of $R_{0}$; for example, $R_{0}=1.2$ resulted in $H_{\infty }=0.31448$, reached in 1201 days. For large values of $R_{0}$, such as $R_{0}>4$, I saw that $H_{\infty }\approx 1-\exp(-R_{0})$. For very small $R_{0}$, on the other hand, $H_{\infty }\approx I_{0}(1+R_{0})$. I have not attempted to see if these observations have a theoretical explanation.  --Lambiam 18:06, 3 May 2020 (UTC)

Thanks for the detailed answer. I see for simplicity you used differential equations. I’m guessing in reality, such complex modelling would require the use of matrices. Do you know if any of the studies have released the source code? 90.196.238.188 (talk) 19:07, 3 May 2020 (UTC)

Difference in assumptions in mathematical pandemic modelling?

Slightly related to my previous question, am I correct in understanding that Sweden’s model and the UK’s model simply makes different assumptions? So on the R(0) value, infection fatality ratio and whether there will be an intervention which can stop spread before it passes through the population (this last one seems to be the key one which is driving Sweden’s response)? 90.196.238.188 (talk) 21:54, 2 May 2020 (UTC)

The Swedish approach is based on a different strategy. Unlike what most other countries are doing, they are not aiming to stop the epidemic running its course. Rather, they are focusing on isolating the elderly from the rest of the population so that the epidemic can run its course in the non-elderly population and allowing the population to reach herd immunity. They have only taken rather mild social distancing guidelines to make sure the epidemic doesn't rip through the country too fast. They have argued that they don't need to take strong measures because their hospitals have more than enough capacity.

So, this is not really about a different judgement about parameters such as R0 and the infection fatality ratio, rather a judgment that a vaccine is going to come too late to stop the epidemic from running its course. This means that whatever the infection fatality risk is and whatever R0 is, all the deaths due to infecting the 1-1/R0 fraction of the population to reach herd immunity will happen anyway. There is then only the choice of limiting the infection fatality ratio by making sure the people who are at the greatest risk of dying don't get ill. And one has to make sure that hospitals should not face capacity problems. Count Iblis (talk) 00:45, 3 May 2020 (UTC)

While the Swedes are busy patting themselves on the back, their total death rate per million people peaked at more than 10 deaths/million/day averaged over 7 days, putting it in the bad boys club, Belgium(28) Spain(18) France (17) Britain(14) Ireland (14) Italy(14) Sweden (11) . In terms of total deaths per million they are on the same trajectory as the Netherlands and Ireland. Other countries have done far better than that, and some of course are worse. Greglocock (talk) 02:56, 3 May 2020 (UTC)

The Swedes have chosen to go through the epidemic faster than other countries, therefore their death rate per unit time should be higher. What matters is whether when it's all over, the number of deaths per head of the population will be lower. Count Iblis (talk) 03:07, 3 May 2020 (UTC)

and as I said, they are on the same trajectory as several other countries for overall deaths per million. At this stage we don't know if flattening the curve, or not, results in lower overall deaths, but Sweden doesn't stand out for either. I see no sign that their strategy is an outlier in death-related results. Greglocock (talk) 05:35, 3 May 2020 (UTC)

I see. What about Japan? I note they have a softer lockdown too although they seem to be somewhere in between Sweden and the rest of Europe? Are they making similar assumptions? 90.196.238.188 (talk) 07:30, 3 May 2020 (UTC)

Japan is the big outlier, way down in terms of confirmed cases (a problematical measure) and deaths (a slightly less problematical measure). I suggest you spend a bit of time exploring https://ourworldindata.org/coronavirus Greglocock (talk) 21:39, 3 May 2020 (UTC)

An added difficulty with making these comparisons is that, by looking just at policies, one may not be accounting for variation in compliance with policies in the different populations. So, population A with a light lockdown, but near 100% compliance, may have better numbers than a population with a harder lockdown, but only 25% compliance. --OuroborosCobra (talk) 22:07, 3 May 2020 (UTC)

May 3

How close can you be to a fighter jet passing you at mach 1 without sustaining injuries?

Some closes passes are shown here, but could the people have been injured if these planes had passed the public at, say, 10 meters distance? Count Iblis (talk) 01:22, 3 May 2020 (UTC)

Does getting your eardrums blown out count? ←Baseball Bugs ^{What's up, Doc?} carrots→ 02:03, 3 May 2020 (UTC)

Wouldn't earplugs prevent that? Count Iblis (talk) 03:08, 3 May 2020 (UTC)

Maybe, if you knew it was coming and had them to put in. ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:55, 3 May 2020 (UTC)

See Tourist killed by jet blast at notorious Caribbean airport.

If you get VERY close, This Sailor Got Sucked Inside a Fighter Jet Engine (and survived with minor injuries). Alansplodge (talk) 11:11, 3 May 2020 (UTC)

I mean, the scenario is slightly contrived. At some point you're very likely to be injured, though it's always possible to escape unscathed in freak circumstances. A jet traveling at Mach 1 is a large object moving very quickly and doing so using one or more turbofan engine(s). Air is a thing that has mass, and the jet compresses and heats the air in front of it as it travels. The engine also sucks in air and propels air+exhaust out the back. Close to the ground, this will kick up any debris lying around; foreign object damage is a major concern for jet aircraft. If the jet blast hits you directly enough, it will knock you over at least, per Newton's Third Law. The article linked by Alansplodge describes a worst-case scenario for this, and you can find plenty of videos of jet blasts knocking things over. In part this is probably standard human intuition misfiring. We kind of innately think of air as "insubstantial" since it doesn't seem to affect us much, but go stand in a hurricane and you'll see the power of lots of it moving at high speeds. --47.146.63.87 (talk) 19:37, 3 May 2020 (UTC)

DSLR

People say smartphone cameras are getting closer to DSLR quality but what are the main advantages of DSLR over smartphone? Is it night shots and more control over what’s in focus? 90.196.238.188 (talk) 14:35, 3 May 2020 (UTC)

DSLRs have much large aperture than smartphone cameras. Ruslik_Zero 17:21, 3 May 2020 (UTC)

Link: digital single-lens reflex camera. Also optical zoom and interchangeable lenses. There have been moves towards bringing these to phone cameras, and you can always jury-rig a lens attachment to a phone, but dedicated cameras have an advantage in being designed exclusively for photography and thus not needing to compromise for other things like form factor. Some cameras probably have an advantage in battery life, though this will vary widely, and some use swappable batteries while most smartphones these days have internal batteries. (You can still carry a power bank but this adds inconvenience.) --47.146.63.87 (talk) 19:45, 3 May 2020 (UTC)

Give it time. I recall 20 or 25 years ago when this same question was being asked about digital vs. film cameras. Digital has since become so good that film is kind of passé. And phone cameras are way much better than they were 10 or 15 years ago. So they're getting there. ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:35, 3 May 2020 (UTC)