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November 18

Owl Call Identifictaion Requested

Crossing Central Park (NYC) I heard an owl calling, something I have not heard for decades. The call was as follows, with length indicated in caps, since I can't write music:

LOW-mid.mid.midlow-midlow-midlow

The call lasted about three seconds, the first note could be imagined as a long A, with a short pause, two short, back-to-back C notes with another C that lower back into an A, with the C lowering back to A repeated twice after short pauses. (I have the relative contours right as to length and relative pitch, but cannot give an absolute difference in pitch, as I am only a music listener.)

One might imagine this written for the lyrics "_DON'T take the sub_way sub_way sub_way." The owl was about 50 feet into the woods off the road to my right, and after I paused to listen I heard another, more highly pitched owl responding, with the same call, about 100 feet to my left. This continued as long as I paused, about a minute or two, hearing maybe 20 calls.

Thanks. μηδείς (talk) 05:49, 18 November 2016 (UTC)[reply]

After some hunting around the best match I can find (not very close) is this which is a Great Horned Owl, reported here as being a rare visitor. This site also names the other owls to be found in Central Park. Richard Avery (talk) 07:46, 18 November 2016 (UTC)[reply]

The call to which you linked starts out the same, with the first four "notes" sounding the same, but the ending in the clip (and others posted at youtube) didn't end with the falling-off "sub_way sub_way sub_way". I suspect it was indeed a Great Horned Owl, though. What I heard was not a lone bird's territorial call at sunset but at least two birds alternating calls just before midnight. μηδείς (talk) 18:47, 18 November 2016 (UTC)[reply]

There is a rather long and varied duet between two Great Horned Owls here: [1]. They seem to lengthen their calls with less vocal hoots about midway through the recording and especially at the end. --Modocc (talk) 20:02, 18 November 2016 (UTC)[reply]

Yes, the "duet" link is, for all intensive purposes, the same one that I heard, just with a slight variation in the final hoots with the downward steps I have mentioned. Thanks, we can consider this resolved, unless others have further comments. μηδείς (talk) 22:14, 18 November 2016 (UTC)[reply]

So you don't mind that it might not fit for extensive purposes? —Tamfang (talk) 18:33, 19 November 2016 (UTC)[reply]

November 19

Bacterium or Virus that actively infects both a Plant and an Animal?

I am aware that certain fungi can infect both plants and animals, as in the case of aspergillosis. Is there either a bacteria or a virus that actively infects a host from both the Plantae and the Animalia? I am not talking about something like food poisoning caused by vegetation handled under unsanitary conditions, but an agent which actually causes disease symptoms in bot a plant and an animal, especially in humans. μηδείς (talk) 01:05, 19 November 2016 (UTC)[reply]

The bacterium Pseudomonas aeruginosa does. From the article, "In higher plants, P. aeruginosa induces symptoms of soft rot, for example in Arabidopsis thaliana (Thale cress) and Lactuca sativa (lettuce). It is also pathogenic to invertebrate animals, including the nematode Caenorhabditis elegans, the fruit fly Drosophila and the moth Galleria mellonella. The associations of virulence factors are the same for plant and animal infections". In humans it is "An opportunistic, nosocomial pathogen of immunocompromised individuals, P. aeruginosa typically infects the airway, urinary tract, burns, and wounds, and also causes other blood infections." --Modocc (talk) 02:19, 19 November 2016 (UTC)[reply]

Very interesting. I knew of Pseudomonas as an opportunistic infectious agent, but thought it was a fungus. Its mode of action seems analogous to types of fungi which are generalized omnivores.

I am curious if there is a more specialized pathogen, that attacks, for examples, cacti and birds which nest in them, or marine iguanas and the algae they eat? μηδείς (talk) 04:02, 19 November 2016 (UTC)[reply]

When Worlds Collide

In When Worlds Collide (1951 film), as the rogue planet approaches on its collision course with Earth, but before it hits, Earth experiences earthquakes, volcanic eruptions, and tsunamis. Is there a reference that discusses whether these things would really occur, and if so to what extent based on the mass of the rogue planet? Loraof (talk) 03:51, 19 November 2016 (UTC)[reply]

Such events would be caused by tidal forces. In extreme cases, tidal effects can do a lot more than just move water around a few meters. It is thought that tidal effects between some planets and their moons cause quakes and volcanoes, and in theory tidal effects could get strong enough to tear a planet to pieces. See also roche limit, which includes formulas used to approximate how close two objects of known composition can get before one is torn apart. Someguy1221 (talk) 04:15, 19 November 2016 (UTC)[reply]

The question comes up if the planets would be close enough, for long enough, to show substantial tidal effects. If they were headed straight towards each other at high speed, maybe not. Of course, this is quite unlikely. A near-miss is far more likely, and there tidal effects could be the main problem (although a change in Earth's orbit wouldn't be good, either). StuRat (talk) 04:24, 19 November 2016 (UTC)[reply]

The tidal effect is proportional to the other body's mass divided by the cube of its distance. —Tamfang (talk) 18:35, 19 November 2016 (UTC)[reply]

To show the importance of time, tidal heating may generate significant heating of the magma within the Earth, but it would take time for that magma to make it's way to the surface and erupt as volcanoes. StuRat (talk) 19:19, 20 November 2016 (UTC)[reply]

Induced coma and general anaesthesia

What's the difference between an induced coma and general anaesthesia?Llaanngg (talk) 13:23, 19 November 2016 (UTC)[reply]

Barbituates. --Jayron 32 16:30, 19 November 2016 (UTC)[reply]

Yes, but is the state of the brain in each case somehow different? Or are they the same, but induced through different means (barbiturate vs. opioid)? The induced coma article claims that it is a "a deep state of unconsciousness" and the General anaesthesia one claims it's a "state of unconsciousness." How much weight should we put on the 'deep' here? Is this just an careless way of expressing it? Or does it imply that there are different levels of unconsciousness, and induced coma is a deeper one. Both the articles about induced coma and Unconsciousness are quite short.--Llaanngg (talk) 00:53, 20 November 2016 (UTC)[reply]

Judging just from a look at the articles, it looks to me like in an induced coma the object is to achieve flatline EEG, while regular anesthesia just aims for a satisfactory level of unconsciousness. The articles could benefit from some expert attention. On a side note, opioids produce analgesia, not anaesthesia. If you are administered enough opioids to cause a loss of consciousness you'll probably die of an overdose without treatment. The anaesthetics are other drugs, such as propofol and the inhaled anaesthetics. --02:43, 20 November 2016 (UTC) — Preceding unsigned comment added by 47.138.163.230 (talk)

I think it's helpful to look at the roots of words. Anesthesia means literally "loss of sensation". It's also a given that pain-killing will also occur (as pain is a sensation) as well as unconsciousness (with general anesthesia). "Coma" means literally "deep sleep". So if a procedure is to cause loss of sensation (e.g. an op) then it's referred to as anesthesia. If its merely to induce deep unconsciousness for other reasons it can be referred to as an induced coma. --Jules (Mrjulesd) 17:53, 20 November 2016 (UTC)[reply]

The article at General anaesthesia (note subsection) implies that a coma would refer to a deeper state of unconsciousness, but it is not referenced. I also haven't found anything solid out there that spells it out, despite there being several other people asking the same question. Matt Deres (talk) 02:05, 21 November 2016 (UTC)[reply]

Naming of human settlements in the U.S.

How is a human settlement on United States territory officially called? In my opinion, it is "name of setlement, state". Example: Ramona, California. In human settlements of the Insular areas of the United States another Sufixes are in use. Example: Charlotte Amalie, U.S. Virgin Islands. --Mattes (talk) 20:36, 19 November 2016 (UTC)[reply]

Charlotte Amalie, U.S. Virgin Islands is correct. Loraof (talk) 01:24, 20 November 2016 (UTC)[reply]

That's not another suffix, the two forms are perfectly analogous. —Tamfang (talk) 08:13, 21 November 2016 (UTC)[reply]

November 20

Does the vas deferens contain stem cells?

Also, for the record, my question here is geared towards the human vas deferens. Futurist110 (talk) 01:10, 20 November 2016 (UTC)[reply]

I think not, at least not in any significant number. But human vas deferens fibroblasts can be cultured, and those fibroblasts can be used to generate induced pluripotent stem cells. [2] - Nunh-huh 01:17, 20 November 2016 (UTC)[reply]

OK. Futurist110 (talk) 05:14, 22 November 2016 (UTC)[reply]

Artillery question

Would an artillery shell which is filled mostly with high explosive but also contains buckshot (or ball bearings) to enhance the fragmentation effect still be considered a HE shell, or would it be classified as a shrapnel shell? Or would it be considered a hybrid between these types? 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 05:47, 20 November 2016 (UTC)[reply]

I can't find any evidence of such a shell existing, at least as a standardized piece of equipment. Since the name would be arbitrary to begin with, there's no answer to give unless an example can be found. Someguy1221 (talk) 06:25, 20 November 2016 (UTC)[reply]

During the Israeli-Lebanon war of 2006 (or 2007, I forget), the Lebanese used rockets with HE/buckshot warheads similar to what I described here -- I wonder how come nobody mentions this? 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 06:37, 20 November 2016 (UTC)[reply]

Lebanese army equipment was all manufactured in the US, France and Russia, none of which (as far as I can figure) make any rockets or artillery shells like the one you describe. If the Lebanese used one, I would thus assume it's not a standardized piece of ammunition, which may be the reason for the lack of information. Someguy1221 (talk) 06:48, 20 November 2016 (UTC)[reply]

I'm assuming you're aware of the Beehive anti-personnel round? Not exactly what you're asking, as it seems it doesn't actually have an HE charge, but perhaps interesting nonetheless. Fgf10 (talk) 16:35, 20 November 2016 (UTC)[reply]

Well, the beehive round is more like the "normal" shrapnel round in that it relies on the shell's velocity to scatter the flechettes -- I was asking about a round which uses (mainly) the explosive force of the filler to achieve this. 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 07:29, 21 November 2016 (UTC)[reply]

Some information at Shrapnel shell#Modern era, but it seems that most modern HE shells depend on the fragmentation of the external body of the shell to throw out damaging pieces of metal. The original shrapnel shells, not much used after World War I, had a small high explosive "bursting charge" to scatter the solid shot inside, which were generally a bit larger than a normal ball bearing and much larger than buckshot. Alansplodge (talk) 22:41, 20 November 2016 (UTC)[reply]

Yes, I know this -- what I'm asking about is a hybrid of the two, in which most of the wounding effects come from the ball bearings (rather than pieces of the external body, although they also contribute) as in a shrapnel shell, but the force which propels these comes mainly from the explosive charge as in a HE shell. (I'm sure the Lebanese used these during the 2006 war with Israel, I saw the reports on both Fox and CNN -- I wonder why there's no info about it?) 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 07:29, 21 November 2016 (UTC)[reply]

So you're saying that there are in fact artillery shells more-or-less like I described, and that these are called "all-round shrapnel shells" -- is that correct? 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 23:31, 21 November 2016 (UTC)[reply]

Note that terrorists, by definition, are NOT civilians but are classified as illegal combatants -- but other than that, this is useful info! 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 01:26, 22 November 2016 (UTC)[reply]

Firestorm

What weather conditions are optimal for the development of a firestorm (such as after the WW2 carpet-bombing of Hamburg and Dresden)? How often do these weather conditions occur in the Arabian Peninsula? (Or, in lay terms, is it possible to Dresdenize a city like Mecca?) 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 06:33, 20 November 2016 (UTC)[reply]

According to our article on Firestorm, the quantity of combustibles per square meter necessary for a firestorm to form is (40 kg/m²), which Mecca almost certainly lacks. --Guy Macon (talk) 07:03, 20 November 2016 (UTC)[reply]

Do gas mains (or oil pipelines) count toward this figure? Just askin' ;-) 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 07:37, 21 November 2016 (UTC)[reply]

You would want a lack of current rain, at the very least, and dry conditions would be ideal. Also, if some form of visual targeting is used, then a lack of cloud cover is required, and perhaps daylight, or at least strong moonlight, if there aren't any lights to guide the bombers to their target. Of course, modern bombers have many methods of finding a target that don't rely on visual cues. StuRat (talk) 18:43, 20 November 2016 (UTC)[reply]

And these are often seen in desert areas (although in the case of Saudi cities, their construction mitigates against it) -- what about upper atmospheric conditions? The WP article mentions a jet stream enhancing firestorm effects, and another source claims that a thermal inversion is important for the development of a firestorm -- is this true (and are there any further fine points like these), and if so, are these conditions often seen on the Arabian peninsula? (BTW, what if the attacking force uses not just bombers but also artillery and rocket launchers?) 2601:646:8E01:7E0B:F88D:DE34:7772:8E5B (talk) 07:37, 21 November 2016 (UTC)[reply]

Feynman Lectures. Lecture 38. Ch. 38–1 Probability wave amplitudes [3]

This means that the idea of a particle is limited. The idea of a particle—its location, its momentum, etc.—which we use so much, is in certain ways unsatisfactory. For instance, if an amplitude to find a particle at different places is given by e^{i(ωt−k⋅r)}, whose absolute square is a constant, that would mean that the probability of finding a particle is the same at all points. That means we do not know where it is—it can be anywhere—there is a great uncertainty in its location.
— Feynman • Leighton • Sands, The Feynman Lectures on Physics, Volume I

Also I have found in [4] that $|\psi |^{2}=\mathrm {const}$ .

Besides we know that length of complex number is real coefficient at e^iθ. In this case coefficient = 1.

But I don't understand are t and r constant? Can you show that if momentum is known ~~e^{i(ωt−k⋅r)} = const~~ |e^{i(ωt−k⋅r)}|² = const? And when then ~~e^{i(ωt−k⋅r)} ≠ const~~ |e^{i(ωt−k⋅r)}|²≠ const and we can determine where the particle is? Username160611000000 (talk) 08:57, 20 November 2016 (UTC)[reply]

The variables t and r are not constants. They are coordinates - time and position. The probability wave for the position of a particle with a precisely known momentum is e^{i(ωt−k⋅r)}, the probability of finding the particle at any position is proportional to the square of the absolute value of that probability wave and that is 1 - so it has the same probability of being anywhere. It's position is completely unknown. The value of that probability wave is a complex number, it is not a constant, only its absolute value is constant - and that is only because the momentum is known precisely. Dmcq (talk) 11:52, 20 November 2016 (UTC)[reply]

Yes, must be |e^{i(ωt−k⋅r)}|² = const.
But then it is always true, because |e^{i(ωt−k⋅r)}|² = 1. Even if k (and so momentum) is not constant, equation is true. To find the particle we must have |e^{i(ωt−k⋅r)}|² = 0 for some r. How is it possible? Username160611000000 (talk) 12:11, 20 November 2016 (UTC)[reply]

That being zero would mean there was no chance of finding the particle there. Dmcq (talk) 18:14, 20 November 2016 (UTC)[reply]

Zero square of amplitude means zero chance, it's clear. But zero square of amplitude is impossible because |e^{i(ωt−k⋅r)}|² = 1 , always. Or please show how non-constant momentum makes square of amplitude zero in some places and nonzero in other places.Username160611000000 (talk) 18:26, 20 November 2016 (UTC)[reply]

Ψ(r,t) = e^{i(ωt−k⋅r)} isn't a physically sensible wave function (because it can't be normalized). It's best to think of it as an unphysical limit of physically sensible functions, much like the Dirac delta function (which is closely related, as it's the Fourier transform of this thing). Physicists like it because it's mathematically simpler than more realistic wave functions, but when you work with it you have to be careful to remember that it's only a limit, not a meaningful object in its own right.

A wave function describing a real particle would square-integrate to 1 and would not have the same value everywhere in R³.

It isn't necessary for the wave function to be zero anywhere for the particle to be "findable". A Gaussian distribution is nowhere zero, but it still makes sense to sample a random value from a Gaussian distribution. It can be arbitrarily far from the center, but usually is pretty close to it. The squared norm of the Dirac delta function is the limit of normalized Gaussians as the standard deviation goes to 0, and the squared norm of the function you're asking about is the limit of normalized Gaussians as the standard deviation goes to ∞. -- BenRG (talk) 22:19, 20 November 2016 (UTC)[reply]

Thank you, but I know nothing about the wave functions so far, just the amplitudes in experiment with electrons and two holes. Username160611000000 (talk) 07:47, 21 November 2016 (UTC)[reply]

Culmination of constellation's visibility

I noticed there's some mess about when Coma Berenices (and possibly other constellations) is best visible. In case of Coma Berenices and apparently for Northern Hemisphere this book gives "midnight, 2 April", this website says "early May" and this book even gives three months: early March, 2 a.m., early April at midnight and early May at 10 p.m. I suspect it's because of different latitudes, but none of the sources makes such an adjustment which is strange. Which one to choose? Brandmeister^talk 11:26, 20 November 2016 (UTC)[reply]

It's because culmination time constantly gets 2 hours earlier every month. Sometimes it's in daytime so no one would pick those months. Among the nighttime culminations midnight (apparent solar time) might win tiebreakers like "highest average nighttime height" edit: and high latitude twilight avoidance (thanks 90.) but usually ~~people~~ American publications that use 40° latitude charts give the ~~month~~ season and maybe culmination date of a constellation as when it culminates in the dark part of the evening. Sagittarian Milky Way (talk) 17:46, 20 November 2016 (UTC)[reply]

i.e. In 2016 the center of the northern winter constellations culminate at midnight around the December solstice but the center of winter is after the solstice so in the coldest part of the year that point culminates in the evening. Sagittarian Milky Way (talk) 19:03, 20 November 2016 (UTC)[reply]

[Edit Conflict] The latitude makes no difference, as the constellation will still culminate (at which point it will be highest in the sky) at the same time on the same date regardless of its elevation.

Note that the three answers are not contradictory, the first gives the date (April 2nd) for culmination at true midnight (minimising any residual twilight from the set Sun, which can be significant at higher latitudes) the second for culmination at daylight-saving 11:00pm, and the third gives dates for culminations at three different (true) times depending on how late you want to be awake (later times may reduce interference from artificial lighting). {The poster formerly known as 87.81.230.195} 90.208.174.251 (talk) 17:57, 20 November 2016 (UTC)[reply]

This would not matter much in Brooklyn or Las Vegas. The astronomical magnitude scale shows the eye isn't sensitive to smallish brightness changes and cars and peoples' windows are not a big part of the evening skyglow when there are that many streetlights and signs that never turn off. There's parts of the suburbs without streetlights so how many headlights and rooms in the area are lit is more likely to be significant there. Sagittarian Milky Way (talk) 19:12, 20 November 2016 (UTC)[reply]

Uncoupling agents as a hypothermia countermeasure?

Given that hypothermia results from heat loss that outstrips the body's rate of thermogenesis, and that the main effect of uncoupling agents like 2,4-Dinitrophenol is to greatly increase the rate of thermogenesis to far above its normal range, could 2,4-DNP and\or other uncouplers be useful in counteracting severe hypothermia? It seems to me, at first glance, that this would be a very useful tool for first responders in areas with cold winters, as well as for people with occupations and\or hobbies placing them at a high risk of getting severe hypothermia (the examples I had in mind were cold-water fishermen and high-altitude mountaineers, specifically, but there'd also be lots of others)... so, then, how come this isn't (so far as I know) a thing? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 14:16, 20 November 2016 (UTC)[reply]

As an aside: A retired Jewish Doctor told me some years ago that she was disgusted that a French pharmaceutical company was promoting Chlorpromazine as an aid to hypothermic surgery without mentioning nor acknowledging they were using data which was gained by the Nazi's during their experiments in concentration camps. Apparently, downed aircrew where often retrieve alive from the cold seas only to die soon after from the latent effects of hypothermia. So the Nazi doctors came to focused on an insecticide used for intestinal parasites . Its action was to make the parasites lose grip on the gut wall, enabling the normal peristaltic motion of the gut to flush them out. Ie. It did not kill them it just paralysed them. It was observed that concentration camp inmates so treated, showed more indifference to the cold and so were experimented upon further to aid the survival of downed airman. Many inmates died during those experiments from immersion in ice cold water. Having done voluntary work in a psychiatric hospital I have witnessed this indifference to the cold for myself and thus was given the harrowing history (which was far more detailed and graphic to the little I have mentioned) . Yet the pharmaceutical company declared that it was its own discovery. Think about it. Without this data, who would have thought that hypothermic surgery could be performed by injecting the patent with a insecticide? Yet they never made mention of the source. Given her personal and family experiences and the pharmaceutical company's desire to distance itself from negative connotations I can sympathize with her grievance. All expression left her face as she recounted these thing. It was as if there was no way that she could speak and feel emotion at the same time. It was just voiced cold and matter of factually.--Aspro (talk) 17:52, 20 November 2016 (UTC)[reply]

Sounds like Nazi human experimentation#Freezing experiments should mention this rewarming method (with cite obviously). DMacks (talk) 03:14, 22 November 2016 (UTC)[reply]

Oh wait, sounds like they were just using the hypothermia data (which is noted at Nazi human experimentation#Modern ethical issues), not chlorpromazine as a treatment. That would be a chronological problem, and also I can't this or other drugs listed as http://remember.org/educate/medexp or similar resources. DMacks (talk) 03:27, 22 November 2016 (UTC)[reply]

The article you yourself linked to says under Health Effects:

"DNP is considered to have high acute toxicity. Acute oral exposure to DNP has resulted in increased basal metabolic rate, nausea, vomiting, sweating, dizziness, headache, and loss of weight. Chronic oral exposure to DNP can lead to the formation of cataracts and skin lesions and has caused effects on the bone marrow, central nervous system, and cardiovascular system."

Elsewhere it mentions the frequent fatalities associated with its illicit use. So, long story short, if you used it to treat hypothermia you'd likely kill the patient with it, and if you used it prophylactically for persons routinely exposed to cold you'd likely injure or kill them with longer-term side effects. There are much less dangerous answers to both problems.

That's not to say that further research might reveal ways to use it more safely for the purposes you suggest, but good luck with designing ethically acceptable means of testing such uses and getting them funded. {The poster formerly known as 87.81.230.195} 90.208.174.251 (talk) 18:13, 20 November 2016 (UTC)[reply]

The article also says that the main cause of the acute effects is the increase in thermogenesis:

"The factor that limits ever-increasing doses of DNP is not a lack of ATP energy production, but rather an excessive rise in body temperature due to the heat produced during uncoupling. Accordingly, DNP overdose will cause fatal hyperthermia, with body temperature rising to as high as 44 °C (111 °F) shortly before death. In light of this, when it was used clinically, the dose was slowly titrated according to personal tolerance, which varies greatly."

Which would be much less harmful if the person was starting out from a moderately to severely subnormal core temperature. If you're at normal (~37C) body temperature, then raising your core temperature by 7C is likely to kill you. If, on the other hand, you're moderately hypothermic (let's say 30C CBT), then raising your core temperature by 7C will merely bring you up to normal temperature.

Secondly, I never suggested using it prophylactically—merely having an uncoupler on hand to use in case one gets severely hypothermic. It's (or would be) kind of like an Epi-Pen in that regard. One doesn't prophylactically use one's Epi-Pen on oneself beforehand—one merely keeps it (relatively) handy in case they get exposed to the allergen their reaction to which necessitates the Epi-Pen.

Thirdly, if one is currently in the process of freezing to death, wouldn't worrying about possibly getting cataracts and whatnot in the longer term be rather low on one's list of priorities relative to not freezing to death right now?

And, fourthly, I don't believe I suggested that 2,4-DNP was necessarily the uncoupler of choice anyway—I was more using it as an example of an uncoupler. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 21:37, 20 November 2016 (UTC)[reply]

No drugs needed. Count Iblis (talk) 03:52, 21 November 2016 (UTC)[reply]

High voltages, low currents, and Ohm's law, oh my!

One often hears people talking about how some electrical discharges and circuits (such as lightning, electrostatic discharge in computer components, electric fences, etc.,) can have incredibly high voltages but nevertheless don't (usually) kill people, and this being supposedly a result of these power sources producing really high voltages but really low currents. My question is this: how can this be possible, given that Ohm's law states that, in all cases, $V=IR$ , and, therefore, for a given voltage (holding $V$ constant) applied across a given object (meaning, at least at first, that the resistance— $R$ —is being held constant), one should always get the same current flowing through the object, no matter its source? According to Ohm, shouldn't a (supposedly) 500-volt electrostatic discharge be considerably more lethal than 120- or 240-volt household wiring, and almost as deadly as stepping on a live third rail (generally 600 or 625 volts in the US)? So why, then, is it that no-one ever hears about unwary ITs being fried by errant ESD—are the huge-voltage-tiny-current claims false, or is it Ohm's law that's wrong? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 14:54, 20 November 2016 (UTC)[reply]

The very high voltage only exists for a very tiny moment. Real voltage sources are not ideal voltage sources. For most static electricity situations, the total capacity is low, so as soon as current is flowing, the charges equalise and the voltage (and with it the current) breaks down. The total energy (

\int I(t)U(t)dt

) of such a discharge is therefore quite low. --Stephan Schulz (talk) 15:14, 20 November 2016 (UTC)[reply]

Ah, thanx—that makes sense. (That still leaves the electric fences, but I'm guessing that the source in that case [of the information, not the voltage! X-P] was probably just wrong\misinformed.) Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:30, 20 November 2016 (UTC)[reply]

Very similar question: one often hears that a step-up transformer is able to increase the voltage in its output loop beyond that in its feed loop because it decreases the current in the output loop to compensate (and the reverse for a step-down transformer), and that this lower current means less loss en route to distant customers, which is supposedly why we use step-up transformers and humongous transmission voltages (our very own article [!] on Electric power transmission makes this claim, as does Railway electrification system)—but, according to Ohm's law, voltage and current are directly proportional, not inversely proportional! So, since $V=IR$ , shouldn't increasing the voltage in the output loop also increase the current in step with the increase in voltage, not decrease it? Is the resistance in the output loop somehow massively greater than in the feed loop, so as to satisfy $V=IR$ and allow a low current even with an enormous voltage? Did Ohm have his head screwed on wrong? Or is it all our modern electrical engineers who have it back to front? Or something else entirely?

Utterly perplexed... Whoop whoop pull up ^{Bitching Betty | Averted crashes} 14:54, 20 November 2016 (UTC)[reply]

The thing to realize is that power is a constant, i.e.,

P=VI

= constant. (Otherwise we could magically create energy by putting our power source through one transformer after another.) So for a given P, if V goes up I has to go down and vice versa. Shock Brigade Harvester Boris (talk) 17:21, 20 November 2016 (UTC)[reply]

I think the key point you're missing is that Ohm's law is not a universal law of electricity. It only applies to (ideal) resistors. Transformers are nothing like ideal resistors, and don't obey Ohm's law. (Transformers have internal resistance, but a transformer as a whole is not like a single resistor with a single value of R.) -- BenRG (talk) 21:41, 20 November 2016 (UTC)[reply]

A good way of looking at it is internal resistance. It's the reason why you can get a nasty shock from a 12v car battery; but at the same time get unhurt from many thousands of volts, like a high voltage generator we had at my old physics lab. Sources with high internal resistance can be harmless even at many thousands of volts, while sources with low internal resistance can give you a nasty shock at a few volts. --Jules (Mrjulesd) 18:07, 20 November 2016 (UTC)[reply]

It is neither the voltage nor current but the total energy which is of prime importance. We have an article Earth leakage circuit breaker but even that does not explain it in my view. Even Electrocution doesn't seem to point out that it is the complete depolarization of the cells that stops them from functioning. These articles need expanding. Edison tried to use this as his USP in so much that his DC system didn’t depolarized the same as AC did. --Aspro (talk) 18:13, 20 November 2016 (UTC)[reply]

Power really rather than energy. However the concept of internal resistance is useful in explaining why voltage sources differ from a pragmatic angle.--Jules (Mrjulesd) 18:22, 20 November 2016 (UTC)[reply]

Power is the 'rate' of doing work which is the wrong unit. I'm talking about energy. --Aspro (talk) 20:05, 20 November 2016 (UTC)[reply]

Others mentioned transitory effects because of the way it was formulated, but I suspect Whoop whoop pull up's original confusion was much simpler than that.

Consider a spark plug powered by a car's battery, which we can model as an ideal DC voltage generator of 24V. During a spark, current is fairly low, because it has to go through air which has a huge electric resistance. However, if you wire yourself to the battery (need I say not to do this?), you have quite a lower electric resistance. Consequently, assuming the battery generates the same voltage, the current is higher, hence the expended power is higher as well (and, as explained above, power is what really matters). Even though

V=IR

could reasonably be applied in both cases,

R

is not the same for the various "resistances" (air, human) considered. Tigraan^{Click here to contact me} 22:36, 20 November 2016 (UTC)[reply]

There's one point we should clarify here - the voltage across a spark plug is around 25 kV. It's generated by the ignition coil, not directly from the battery - see ignition system. Car batteries are dangerous in many respects, but don't pose a risk of electric shock. You should indeed avoid touching the HT leads in a car, but the battery terminals won't hurt you. Tevildo (talk) 07:23, 21 November 2016 (UTC)[reply]

OK, so my example was a terrible one on multiple accounts. Sorry. Tigraan^{Click here to contact me} 17:40, 21 November 2016 (UTC)[reply]

Ohm's law states that V and I are proportional (or equivalently that V/I is constant) in a conductor to which it applies, such as an ideal Resistor component. But it is not a conservation law for transformers! The quantity that an ideal power transformer conserves is Electric power

P

given by

P=VI\,

A utility power transformer does not alone decide the power that it delivers to customers; that depends on the sum of the currents that they individually draw. The power generator "sees" a load whose effective resistance is (resistances of all customer equipments in parallel ^{see ref.}) x (overall step-down voltage ratio from generator to customers).

Electric fences usually emit only short pulses of high voltage from a source that has high internal resistance and low capacity. This ensures a low-energy non-lethal discharge, extends battery life and avoids starting fires in dry foliage. DaDoRonRon (talk) 00:40, 21 November 2016 (UTC)[reply]

I suppose that part of it is that nature doesn't dish out voltages that often; more often it dishes out currents: for example, the transfer of surface charges in the case of static electricity requires a certain amount of charge to move to balance the equation. But the path for that charge to move via spark is very high in resistance, so the voltage that accumulates before the spark starts is very high. Part of it also is that capacitance in many unplanned situations is very low; a foot with a sock on it is not really designed to power a TV set. :) This means that very high voltage can build up before much charge is accumulated. Wnt (talk) 03:33, 21 November 2016 (UTC)[reply]

One significant correction to the original question: lightning very frequently does kill people. --76.71.5.45 (talk) 09:29, 21 November 2016 (UTC)[reply]

... because the large capacitance means that currents can be high as well as voltages. Dbfirs 14:18, 21 November 2016 (UTC)[reply]

November 21

About Transformers

Hi, We know that the number turns in primary coil, in comparison to the otherone, play the role of determining input/output in terms of voltages. My question is that can the thickness of wire on either side cause difference in output of power coming out, in terms of voltage or in any other terms ?124.253.244.210 (talk) 10:16, 21 November 2016 (UTC)[reply]

The thickness of the wire determines the resistance of each winding. The resistance of the winding affects the current that flows in that winding. The current flowing in one winding affects the current flowing in the other winding. But apart from that important consideration, the thickness of the wires is not a primary cause of difference. Dolphin (t) 11:31, 21 November 2016 (UTC)[reply]

The article Transformer explains that the windings in a real transformer have finite non-zero resistances. Winding resistance is inversely proportional to the cross-sectional area of the wire used. Consider the winding resistance

R

as a resistor in series with an ideal winding. Any current

I

in the winding causes a voltage drop according to

V=IR

and loss of power as heat according to

P=IV

.

Power transformers are usually wound with sufficiently thick wire to make these losses negligible. If the wire diameters of the primary and secondary windings are inversely proportional to the voltage (turns) step-down ratio, the joule losses in the windings will be approximately equal. However in transformers for high frequency currents the effective wire resistance is increased by the Skin effect. At high frequencies a braided bundle of thin wires or Litz wire can achieve lower loss than a single round wire. DaDoRonRon (talk) 11:39, 21 November 2016 (UTC)[reply]

DaDoRonRon, it is not true that winding resistance is inversely proportional to the cross-sectional area of the wire (except for direct current). For alternating current, which is the kind you make go through transformers, skin effect applies.

To answer the question, in the ideal-transformer model (zero resistance), thickness does not matter, but for the real thing Joule effect is of course taken into consideration in the design. Tigraan^{Click here to contact me} 11:49, 21 November 2016 (UTC)[reply]

Our article states that "An object of uniform cross section has a resistance proportional to its resistivity and length and inversely proportional to its cross-sectional area." which is consistent with resistance normally being specified at DC. DaDoRonRon (talk) 12:31, 21 November 2016 (UTC)[reply]

...and when it isn't DC, we can calculate how different it will be at whatever frequency we are using. See Skin effect. --Guy Macon (talk) 15:22, 21 November 2016 (UTC)[reply]

DaDoRonRon does have a point that there is no textbook definition of "wire resistance" that depends on frequency. I think their original answer did not insist enough on that crucial point, but there is no need to argue when everyone seems to agree on the physics. Tigraan^{Click here to contact me} 15:38, 21 November 2016 (UTC)[reply]

DaDoRonRon can have points all day, but I don't think he'll be replying anymore to this (or any other thread). It's an old friend coming back for a visit. He's been shown the door. --Jayron 32 16:29, 21 November 2016 (UTC)[reply]

Also, there most certainly is a textbook definition of "wire resistance" that depends on frequency. See Impedance. --Guy Macon (talk) 16:55, 21 November 2016 (UTC)[reply]

Oh well. WP:DENY, I guess. Tigraan^{Click here to contact me} 17:33, 21 November 2016 (UTC)[reply]

High wattage circuit breakers

Orders of magnitude (current) confirms 15 or 20 Amps is the typical North American circuit breaker rating. At a nominal 120 Volts that's about 1.8 or 2.4 kilowatts. It says 10 and 20 Amps for Europe which is about 2.3 and 4.6 kilowatts since 230 Volts is the nominal standard. Why is Europe so awesome? Do any parts of the world exceed 2300/4600W? What part of the 230+ Volt world has the lowest percentage of sub-20 Amp circuits? Do these high wattage countries have more powerful microwaves, hotplates, vacuum cleaners, blenders etc? Are some electrics like blankets, lawnmowers or chainsaws more common there? Does a higher wattage common socket enable uses/inventions that North America doesn't even use? Sagittarian Milky Way (talk) 19:23, 21 November 2016 (UTC)[reply]

I'm on 240V. The only appliance I have that is rated over 1.8 kW is my immersion heater for hot water at 3.2 kW, and that is on its own circuit breaker, hard wired in. The most powerful plug in motors I have on a standard 10A plug are 1.8 kW on my electric chainsaw,rotary saw table and log splitter. Judging by the current load the vacuum cleaner must be about 1.8 kW as well. In Australia we also have 15A plugs, but they don't tend to be used in houses. So, it would appear that the USA's inferior electrical system (designed by politicians) sets the standard for the world. Funny dat. UK fuses (in the plugs) used to be 13A, at 240V. Greglocock (talk) 20:18, 21 November 2016 (UTC)[reply]

Hot water heaters here have two circuit breakers connected together with one handle. That's the only way we can get 240V in the home (connect a +120V wire and -120V wire to each other instead of to neutral) Sagittarian Milky Way (talk) 21:36, 21 November 2016 (UTC)[reply]

Wait, an immersion heater is a rod you stick in something instead of putting the something on the stove? American water heaters are tanks in the basement fed by water mains. See, you have appliances I've never heard of. Sagittarian Milky Way (talk) 21:57, 21 November 2016 (UTC)[reply]

In the UK, high-power individual plug-in appliances are limited by a 13A fuse (just over 3 kilowatts) to protect the socket and flex from overheating, but a typical ring circuit is protected by a 30A circuit breaker (7.2 kilowatts or 7.5 at the 250v that I usually get). 3 kilowatt kettles and portable heaters are common here. Dbfirs 20:25, 21 November 2016 (UTC)[reply]

It is correct that the typical circuit in North America is 15 A or 20 A at 120 V, but it is standard to provide for higher power levels for appliances that require them such as electric stoves, central air conditioners, etc. My house has four such circuits, as large as 40 A at 240 V. We just don't have that much power at ordinary outlets. --76.71.5.45 (talk) 21:23, 21 November 2016 (UTC)[reply]

Yes, we have higher power for hard-wired appliances here in the UK too, including power showers up to 10.8 kilowatts with a 50 amp circuit breaker. (I don't have anything so decadent!) Dbfirs 00:18, 22 November 2016 (UTC)[reply]

Do you have plug-in railguns and hobby woodcarving lasers? Sagittarian Milky Way (talk) 01:01, 22 November 2016 (UTC)[reply]

Not personally. Are they common in your part of the galaxy? Dbfirs 08:48, 22 November 2016 (UTC)[reply]

November 22

How much medical personnel is needed to harvest and transplant organs?

How much medical personnel is needed to harvest and transplant organs? How difficult would it be for a mafia to kill people and steal their organs to sell them in a black market? Llaanngg (talk) 01:28, 22 November 2016 (UTC)[reply]

Basically, it's high-tech and has to be done fast, also compatibility has to be assessed (not any organ is fit to be transplanted into a specific person). There are countries wherein it is legal to sell your own kidney. Tgeorgescu (talk) 01:36, 22 November 2016 (UTC)[reply]

Servo horns

Servo horns are mechanical connectors for servo (radio control)s. They come in a great variety of sizes, shapes and materials. http://hitecrcd.com/uploads/hornssm.jpg As shown in this picture, a cross-shaped horn having a number of holes in the arms can be used as a lever to pull or push a segment of piano wire connected to another mechanical part. With a little modification, this horn can be connected to a larger mechanical part (e.g., a robotic arm) to rotate it directly.

Many servo horns have a raised circular rim in the center. The above-shown horn has exactly this feature. https://www.rcplanet.com/23957-tm_thickbox_default/Futaba_Servo_Horn_F_Spline_Small_X.jpg

Some other servo horns do not have this feature. http://gogo-rc.com/store/image/cache/data/Motor-Servo-ESC-Gyro/Servo/IMG_0385-01_go_x550-500x500.jpg

If you use the servo as an airplane actuator, the raised rim is useless. If you use it as a robot arm motor, the raised rim actually gets in the way between the servo and the movable part. Sometimes, I just use a knife to remove it.

What's the purpose for this useless(?) feature? -- Toytoy (talk) 02:44, 22 November 2016 (UTC)[reply]

I suspect that a long time ago Mr Futaba got his radio antenna wire wrapped up between the screw and the horn. It's tempting to think it might be a stiffener, but the webs are all wrong. It could be something to do with making sure the plastic flows nicely in the mould, but again, it doesn't look quite right for that. Greglocock (talk) 08:26, 22 November 2016 (UTC)[reply]

I can show you a round aluminum server horn made with this useless(?) feature. http://www.nihonbashimokei.net/data/rc-nihonbashi/product/ars-3216htg-hv.jpg and http://alturn-usa.com/products/CAD/Robot.jpg Mine comes with two plastic horns and two aluminum horns of the same designs and nearly identical dimensions (plastics shrink!). Anyway, I suspect it has anything to do with plastic moulding. I tried to attach this server horn to a wooden part. It's painful to create a 10 mm diameter and 2 mm deep recess into the wood because I don't have a tool for this job. -- Toytoy (talk) 09:00, 22 November 2016 (UTC)[reply]

Feynman Lectures. Lecture 38. Ch. 38–2. Fig. 38–3 Determination of momentum by using a diffraction grating. [5]

That is, the waves which form the diffraction pattern are waves which come from different parts of the grating. The first ones that arrive come from the bottom end of the grating, from the beginning of the wave train, and the rest of them come from later parts of the wave train, coming from different parts of the grating, until the last one finally arrives, and that involves a point in the wave train a distance L behind the first point. So in order that we shall have a sharp line in our spectrum corresponding to a definite momentum, with an uncertainty given by (38.4), we have to have a wave train of at least length L
— Feynman • Leighton • Sands, The Feynman Lectures on Physics, Volume I

Suggest please what is meant: does the wave train look like this or like this?Username160611000000 (talk) 06:40, 22 November 2016 (UTC)[reply]