The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
I heard on the news that the Detroit Police Department is wearing expired vests. So, what makes them lose effectiveness over time, or is it just a case of the manufacturers only having tested them after so long, so only guaranteeing them for that period ? StuRat (talk) 00:33, 27 July 2013 (UTC)[reply]
High temperatures and UV rays from the sun can slowly degrade polymers such as Kevlar; however, the expiration date is set conservatively, with a large safety margin (i.e. the vests expire well before degradation reaches a serious level). 24.23.196.85 (talk) 01:10, 27 July 2013 (UTC)[reply]
This paper discusses some of the aging effects (due to use and environmental exposure) of one kind of balistic fabric material. Over its lifetime a vest in real daily police use will be exposed to a range of environmental attacks (sunlight, heat, sweat, gasoline, spent firearm propellant, tear and pepper spray and their solvents, cleaning materials, damp and mildew, and general rough handling). 146.90.36.204 (talk) 01:13, 27 July 2013 (UTC)[reply]
Part of the protection that Kevlar provides is the weave of the fabric. A spinning bullet interrupts the weave, causing the force to be spread over a larger area. Over a period of years the fibers in woven Kevlar fabric start to separate, reducing the ballistic protection provided by the vest.OsmanRF34 (talk) 01:21, 27 July 2013 (UTC)[reply]
Very true; in fact, the specs say that if the vest is hit by a bullet, even once, it's no longer usable and must be sent back to the manufacturer for repair. 24.23.196.85 (talk) 01:35, 27 July 2013 (UTC)[reply]
Really? I'm surprised. Safety harnesses (see below) are single use only, at least in Ontario. Once the fabric wears out or the harness experiences a drop, the harness is trashed. If I was in a situation where I needed to wear kevlar, I don't think I would get a warm fuzzy feeling about seeing a "reconditioned" tag on it... Matt Deres (talk) 01:42, 27 July 2013 (UTC)[reply]
A similar instance occurs with safety harnesses, which actually come stamped with an expiration date. Besides UV light and the other factors mentioned above, marking the material with permanent ink or marker also degrades the fabric. Matt Deres (talk) 01:39, 27 July 2013 (UTC)[reply]
Before we veer into some crazy side discussion about safety harnesses... I think Stu's question makes a lot of sense. Obviously after a vest is shot you're not going to use it again. Similar to how once a bike helmet is used it won't be used again. The real question Stu's asking is whether they really age. I suspect the answer is largely no (or not very quickly). That some reporter dug up the expiration date on bullet proof vests and then tied that into the Detroit bankruptcy doesn't surprise me either. If you really wanted to get into it, I say you email the author of that piece with what you're really getting at Stu: "Does this actually matter or is this just some neat rhetorical device?" Shadowjams (talk) 08:12, 27 July 2013 (UTC)[reply]
I'm not sure what's crazy about it; they're both safety devices made out of fabric that have to withstand tremendous forces in order to be useful. They're also very rarely put to the test (as it were), which makes them prime candidates to get put on the agenda when it comes to cutting costs. I bring them up not simply due to the similarities, but also because they're much more widely used and because I've had experience with them that might be relevant. For example, the individual strands of fabric wear out fairly easily considering how sturdily made the harnesses are; if they can wear out to the point of being unsafe without experiencing a drop or even getting up to the stamped code date, then you really do have to wonder about how good the vests are if they've been held onto past the code date. Matt Deres (talk) 13:13, 27 July 2013 (UTC)[reply]
Comparing a harness that has to support a distributed weight over long periods to a material that has to absorb a smaller force over a very short period is vastly different. I don't have the materials-engineering vocabulary to articulate it exactly, but the differences between the two applications is dramatic; "both safety devices made out of fabric" is about as far as the similarity goes. Shadowjams (talk) 17:27, 27 July 2013 (UTC)[reply]
Yeah, nothing crazy about Matt's post. The question can dwell into expiration of fabric. I'd even add that regarding climbing ropes, I seem to recall from an old Chouinard Equipment catalog that five year old unused, properly stored ropes, when subjected to the standard UIAA test (a factor two fall) broke on the first attempt! I wouldn't trust my life to them. The same applies to bullet proof vests. BTW, a harness doesn't have to support a distributed weight over a period of time. It is a security device that mostly doesn't get stressed.OsmanRF34 (talk) 17:37, 27 July 2013 (UTC)[reply]
Please don't make me try and find some numbers.... but you can't possibly compare the tensile strength of a 3" strip of fabric taking a 300 lbs weight falling, let's say 15 feet, to a few centimeters of fabric taking a 700 J force? Imagine shooting a handgun at a seatbelt, or through one of the climbing harnesses you're talking about. That's the point I'm trying to make. [you're probably right though about the aging factor... but it'd be nice to have some refs on it]. Shadowjams (talk) 19:50, 27 July 2013 (UTC)[reply]
Ok, thanks all. It sounds like they really do expire. That makes them even more expensive, if they have to be replaced every few years. I wonder if it's possible to make a more robust version. StuRat (talk) 06:06, 30 July 2013 (UTC)[reply]
Hatchet wasp "attacking" spider?!
Earlier today as I was walking through the garden I saw something very unusual: a small spider being chased around by a (much smaller still) hatchet wasp. For about a minute or so I watched the spider jump about frantically as the wasp followed in hot pursuit (it seemed to land, or at least come very close to landing on the spider several times). I haven't been able to find anything detailing this sort of behavior, but then again the research on these things seems pretty scant. Can anyone with insect knowledge shed some light on this?
70.112.97.77 (talk) 00:58, 27 July 2013 (UTC)[reply]
I can think of three possibilities:
1) It's trying to kill and eat the spider.
2) It's trying to kill or drive off the spider to protect itself or other hatchet wasps from the spider.
In WPs Hero shrew article, it says "The hero shrew's unique anatomy allows it to bear the weight of a 160 lb (73 kg) human without harm". Does this mean that a human can stand on the thing's back, or ride it around (the article doesn't say how big this critter is), or what? I'm struggling to visualize... --Kurt Shaped Box (talk) 10:18, 27 July 2013 (UTC)[reply]
From the article on shrews:"All shrews are comparatively small, most no larger than a mouse. The largest species is the Asian house shrew (Suncus murinus) of tropical Asia, which is about 15 cm long and weighs around 100 grams." Thus I would think it means one can step on it and not kill it like one would other creatures that size or smaller. -Modocc (talk) 10:44, 27 July 2013 (UTC)[reply]
If you read the cited source, you'll see exactly what they're talking about - a ritual dance involving jumping on the shrew. I was skeptical at first, so I independently verified it with the 1917 Bulletin of the American Museum of Natural History, Volume 28, which is available for free from their website.
“
The natives of these regions, especially the Mangbetu, who are well acquainted with this shrew, first called our attention to its abnormally strengthened back-bone by their performances upon captive specimens... Whenever they have a chance they take great delight in showing to the easily fascinated crowd its extraordinary resistance to weight and pressure. After the usual hubbub of various invocations, a full-grown man weighing some 160 pounds steps barefooted upon the shrew. Steadily trying to balance himself upon one leg, he continues to vociferate several minutes. The poor creature seems certainly doomed. But as soon as his tormentor jumps off, the shrew after a few shivering movements tries to escape, none the worse for this mad experience and apparently in no need of the wild applause and exhortations of the throng.
”
— J.A. Allen, American Museum of Natural History, 1917
It turns out that there were a series of expeditions to the Congo between 1917 and 1919 by the museum. More reports are also available on the same website. I have found them to be fascinating reading. They are providing excellent context for my next major literary foray, The Crime of the Congo, a non-fictional work that I had already queued up from Project Gutenberg as I have completed reading most of that author's fiction.
For the record, after reading of the shrew on Wikipedia's main page on Friday morning, I researched this material extensively before I left to work; and the Armored Shrew became the topic of our lunch-hour conversation. Like many of my conversations, it was received with much incredulity. Nimur (talk) 15:08, 27 July 2013 (UTC)[reply]
Thanks for the info. If someone just told me about that 'shrew dance', I'd probably just assume that they were testing my gullibility. It does sound fairly Boratian, doesn't it? But then again, these strange local customs do actually exist - e.g. the one where they throw a goat from a bell tower and then get drunk - or the one where they run around on a field, fighting each other over a sheep carcass and then get drunk - or the one where they chase a rolling cheese down a v.steep hill and then get drunk. I'd assume that they all get drunk after the shrew thing too? --Kurt Shaped Box (talk) 15:55, 28 July 2013 (UTC)[reply]
Having read quite a bit about Africa, particularly material written in the latter half of the nineteenth and early twentieth century, I would not entirely rule out that the publication is embellishing details. Much historical writing about Africa is exaggerated, or fictitious; or partakes in scientific racism, or other cultural insensitivity; many encounters with local people portray them as "savage" (whether for political ends or for the purposes of entertainment). One hopes that these unfortunate portrayals were extinguished by the enlightened time of this 1917 expedition. One hopes that respectable naturalists refrain from exaggeration or cultural insensitivity, and sticks to strict observationalism. Yet, when one studies literature and science written by Europeans and Americans from that era on the topic of sub-Saharan Africa, one is left with the distasteful impression that perhaps not every writer attains these enlightened goals. (We know, from even a cursory study of history, that racial equality and respect was not a universal reality in America or Congo in 1917).
I have never been to Congo, nor met a Mangbetu who danced on a shrew. All I can truthfully say is, the American Museum of Natural History archived the writings of its member, one J.A. Allen, who wrote in 1917 that during travels to the Congo, he met more than one person from the Mangbetu group, and from other areas, who stepped on the shrew. He drew very detailed diagrams of the shrew, and took a few photographs; yet no photograph depicts anyone stepping on a shrew.
Thanks. I thought this might be a bit difficult. It turns out the bee was dying and when I came back to work it was still there and is now dead. So I've taken some more pictures of the bee but now upside down. They are at the above link. I've also kept the bee in a plastic bottle in case there is something else I can do to id it. I wouldn't bother but there does not seem to be too many pictures of bees from up here. I noticed that there are something like 12 - 15 species of bee in the Arctic. Also based on the underside pictures and second paragraph it's probably not a carpenter bee. CambridgeBayWeather (talk) 12:09, 28 July 2013 (UTC)[reply]
No clue what it's called, but I see them frequently enough (including today) in my southern Northern Ontario fields. So it's not strictly Arctic. Or maybe I'm seeing a slightly different bee, but to my layman eyes, it looks the same as your pictures. Not the most helpful answer, but might help you narrow it down. RIP bee. InedibleHulk(talk)23:40, 28 July 2013 (UTC)[reply]
flare
why dont/where /how do we use flare coming out of an oil refinary.Why do we waste such an enermous amount of energy for free.We must think of uitilixzing it somehow.Has anyone thought of using it to generate energy in steam power plant.Please suggest me some reaing on this topic.
175.101.60.14 (talk) —Preceding undated comment added 15:51, 27 July 2013 (UTC)[reply]
Gas flare and Associated petroleum gas are the relevant articles. The answer to your question is "because it's cheaper just to burn the gas than to transport it anywhere useful", but your point about wastage is valid, and discussed in the articles. Tevildo (talk) 16:28, 27 July 2013 (UTC)[reply]
I have to think that if they properly valued both the cost of the pollution they are generating/global warming and the fuel value lost to future generations, then this would be seen as the waste it truly is. Unfortunately, the company doing that doesn't have to pay those costs, and doesn't care if others do. This is one argument for the proposed cap and trade system. StuRat (talk) 20:02, 27 July 2013 (UTC)[reply]
[Dons ex-refinery employee hard-hat]. As Rmhermen correctly said, refinery flare flames are safety devices: specifically, they're pilot lights.
Oil refineries produce from crude oil a number of commercially saleable liquids and gases. These processes leave a relatively small residue comprising a mixture of inflammable gases which would cost more to purify than they could be sold for.
This mixed residue is called refinery gas [what, no Wikilink?], and most of it is used to produce power and heat for operations in the refinery itself. However, the complex and sensitive nature of various refining processes means that the volume of refinery and other inflammable gases being produced is subject to occasional surges which the normal procedures and equipment cannot absorb.
These excess gases have to be vented to the atmosphere (otherwise, kablooie!) and since they may in sufficient concentration be toxic to the refinery workers and residents in surrounding areas (who may predominantly be refinery workers' families) they have to be burnt. They are directed up the flare stacks, ignited by the pilot flame, and make an alarming but harmless roar. In the refinery I used to work at and still work within earshot of, this happens less that half-a-dozen times a year at most.
Refinery flares are therefore in general no different to the millions of household gas water heaters and central heating boilers which also have a permanent pilot light lit.
If someone else would care to crunch some numbers, I suspect it could be shown that the flares' contribution to pollution is utterly insignificant compared to that caused by the utilisation of oil and gas that we all directly or indirectly benefit from. I don't say that this is a good thing, and I'd be the first to welcome better viable alternatives. {The poster formerly known as 87.81.230.195} 90.213.246.168 (talk) 21:09, 28 July 2013 (UTC)[reply]
Long reverberation time in stone-walled cathedrals
Are thick-stone-walled cathedrals intentionally built with a long reverberation time, or is the echo-like feel of entering the cathedral a product of its big and sturdy architecture? Also, where can a person find a stone-walled cathedral in the Midwestern region? Sneazy (talk) 16:51, 27 July 2013 (UTC)[reply]
To calculate the 60dB reverberation time of a room, you use Sabine's formula, t=0.161 V/A with V=volume in m3 (or 0.049 V/A if measuring in cubic and square feet), A is equivalent absorption surface = S1.a1 + S2.a2 +... for all the surfaces, S = surface area in m2 and a is the absorption coefficient of the material.
Carpet, foam, curtains will absorb higher frequencies better, window glass will mainly absorb low frequencies.
The formula explains the long reverberation: low absorption materials and a large space. If you double the length, width and height then the time will also double. Since the main material is so low absorbing, changes to small areas can make a big difference, For example, if all surfaces have a=0.02, covering just 3% with carpet having an absorption coeff of 0.6 will halve the reverberation time.
Long reverberation was preferred in Catholic churches. It fitted the music, specially the chants. After Vatican II, that changed, because the liturgies were no longer conducted in Latin. Before, it didn't matter whether the people could hear clearly what was being said from the altar, since most didn't speak latin. But after Vatican 2, people were supposed to listen to what was said, and the reverberation became a problem. (http://www.catholica.com.au/misc/ChurchAcousticDesign.pdf) Ssscienccce (talk) 22:53, 27 July 2013 (UTC)[reply]
Interesting... Is this also true for Eastern Orthodox churches (where the liturgies are conducted in Old Church Slavic, and the chants are much like the Catholic ones in the musical sense)? 24.23.196.85 (talk) 07:59, 28 July 2013 (UTC)[reply]
Hagia Sophia featured an eleven second reverb, the Alexander Nevsky Memorial Cathedral in Sofia a bit less with nine seconds, according to this CD of Slavonic Orthodox Lithurgy.
"The reverberation time RT60 is very high in the largest TST. The volume of the Church of the Ascension at Kolomenskoje is equal to V=1700 m3. That is about 2 times less than in 5-dome Trinity Cathedral in Kostroma (V=3200m3 [1]). And the values of RT60 in AK are about 2 times larger" (the text is hard to interpret at places, I think some Russian expressions were translated verbatim, like "two times less" when "half" is meant)
Such values of RT60 depend greatly on the inner attire of TST. For example, IC has the volume of V=1670 m3 that is very close to the volume of AK. But the values of RT60 in IC are much smaller than in AK. This is caused mainly by the large wooden icon-stand that was installed in IC. During the measurements in AK in this church there was no icon-stand and the room was completely empty with only several wooden benches on the floor. Still shorter reverberation time was estimated in more lower TST with exedrae on the sides of the tetrahedron. For example, in IF RT60 is equal to 1.4-1.7 s at middle frequencies. In this case the huge icon-stand with 8 rows of icons and various wooden decorative structures as well influenced greatly on the sound absorption in the church.
It should be noted that there is no large increase of RT60 at the low frequencies in IC and IF that is typical for 5-dome churches and cathedrals. This fact deals with the large surface of windows in IC and IF that absorb sound energy at the low frequencies. The surface of windows in AK is much smaller in comparison with IC and IF.
The problem of speech intelligibility is of the great importance in Russian Orthodox churches. In general the speech intelligibility in TST is better than in 5-dome churches.
Thanks! You've confirmed my hypothesis, as well as my own personal observations re. the acoustics in the Church of the Assumption in the Kremlin. 24.23.196.85 (talk) 07:33, 29 July 2013 (UTC)[reply]
Oocyte activation by the first sperm prevents most zygotes with two fathers, and in any case polyspermy that results in aneuploidy is certainly lethal in humans. Therefore, you get a random race. I am not aware of any systematic research into whether sperm from any particular race might have a competitive advantage - there's no great reason to suspect it, and while you could measure times in vitro in an artificial environment, there's no way to conduct the experiment with any real validity without actually creating zygotes in a human woman, which would never get approval (by the woman, let alone the IRB). If the sperm were delivered naturally, rather than as an artificial mixture, there is some possibility that orgasm would cause the woman's sexual preference among the men to translate to a higher chance of conception by his sperm (essentially, by muscular movements that give his sperm an advantage) but I'll admit that idea is not widely accepted. Wnt (talk) 21:20, 27 July 2013 (UTC)[reply]
See [2], etc. The rape - no pregnancy claim was obviously not literally true per se, but it did have some percentage of truth to it, massively amplified by wishful thinking among those with a certain point of view. Wnt (talk) 23:46, 27 July 2013 (UTC)[reply]
Indeed you can't get a child with two fathers. Though there is a rare chance of superfecundation where twins can be born of two different fathers. This is rare in humans, though more common in fellow mammals such as dogs and cats - members of a litter of street cats or dogs can have obviously different characteristics. 88.112.41.6 (talk) 21:39, 27 July 2013 (UTC)[reply]
I guess you guys are just more important than the three of us answering here, so I'll defer to your "consensus", but at least leave me the opportunity to answer the request for source? Wnt (talk) 00:26, 28 July 2013 (UTC)[reply]
Why do people feel tired after eating ?
I can understand a huge meal doing this, as most of the body's energy must be used for digestion, but I've observed this for small meals, too. Does the body release insulin in anticipation and cause blood sugar to drop early in the digestion process ? (Not a medical question, just asking about the normal digestion process.) StuRat (talk) 22:22, 27 July 2013 (UTC)[reply]
I think I know one reason. If you are eating way past your bedtime, regardless of the meal size, then you might still feel tired due to your circadian rhythm reminding you that it is bedtime. Sneazy (talk) 22:53, 27 July 2013 (UTC)[reply]
If blood sugar levels were to drop, you would feel hungry after the meal. I know how this feels, I eat about 4000 Kcal per day but when travelling and eating in restaurants it's difficult for me to eat this amount. It can then happen that when I start to eat I will feel more hungry than before; when I am done eating and completely full, I can be extremely hungry. I then need to wait a few hours and eat again. But I won't feel tired (unless I would force myself to eat a normal diet). Count Iblis (talk) 23:47, 27 July 2013 (UTC)[reply]
I weigh 58 kg, I do 50 minutes of fast running per day, but I have had a fast metabolism for most of my life also before I was exercising at this level. I can tell you that low blood sugar after a meal is a strange feeling that most people will never experience. On the one hand your belly is completely full, on the other hand it feels like you are starving. Count Iblis (talk) 00:02, 28 July 2013 (UTC)[reply]
Can you plant and grow bing cherries or "sweet cherries" in the temperate region of the Northern hemisphere in your backyard, assuming that you own a property of a suburban land division? How long does it take to grow the cherry tree to its first fruit? Will it ever produce fruit by itself, or do you need two trees to produce offspring? How do you pollinate each flower? If it produces fruit through parthenocarpy, then would the offspring fruit be seedless like navel oranges? Sneazy (talk) 00:50, 28 July 2013 (UTC)[reply]
I typed "How to grow bing cherries" into Google, and got a multitude of sites which seem to answer most of your questions. --Jayron3201:14, 28 July 2013 (UTC)[reply]
The multitude of questions are really linked or are supposed to fall under one big inquiry - the parthenocarpy of cherry trees. Sneazy (talk) 01:41, 28 July 2013 (UTC)[reply]
No, Parthenocarpy is something entirely different. What you seem to be asking about is Self-pollination. Cherries are self-pollinating, so you can grow a single tree and get fruit. Although parthenocarpy does occasionally occur in cherries, and you might find a seedless fruit or two on a tree, there are no cherry varieties that consistently produce only seedless fruit. A truly parthenocarpic variety like navel oranges doesn't have "offspring". They can only be propagated by cloning. The rest of your questions are specific to your geographical area and can be answered by Googling, or contacting your local extension agent. Dominus Vobisdu (talk) 07:30, 28 July 2013 (UTC)[reply]
I have read that two vectors cannot be divided. Why is it so? Here is something contradictory - according to the Newton's second law of motion, force (a vector) divided by acceleration (a vector) gives mass (a scalar). What is the reason behind this? Publisher54321 (talk) 03:23, 28 July 2013 (UTC)[reply]
In the case of Newton's 2nd law, you divide the vectors' magnitudes. However, given that vectors can be adequately represented as 1-row or 1-column matrices, which can't be divided, there really isn't such a thing as vector division. For example, I know of no such operation that would return a "quotient" in a way that reverses the vector cross-products or dot products. It seems like you can only do this to reverse multiplication of a scalar by a vector.--Jasper Deng(talk)03:48, 28 July 2013 (UTC)[reply]
In general, vectors don't have a form of division. However, there are vector spaces that do admit division, like the complex numbers. Other examples of vector spaces that may have some form of division are: polynomials over a ring, complex valued functions from a topological space, formal power series, differentiable functions, analytic functions, many others. You may want to read our articles Vector_Space#Algebras_over_fields and Algebra over a field; granted, you aren't guaranteed division. If you are talking, strictly, about the geometric concept of vectors as arrows, then there is no natural, or obvious, division. It's important to remember that "multiplication" is a certain general type of combining objects and that "division" is the inverting of this, so there is no reason to expect that things called a "product" are going to have some nature of inverse operation.Phoenixia1177 (talk) 04:19, 28 July 2013 (UTC)[reply]
With my vague understanding of set theory, the layman's answer is basically that much rests on the definition of "multiplication" used, and the more complex your data type gets, the harder it is to make a definition of multiplication and have a compatible definition of division.--Jasper Deng(talk)04:29, 28 July 2013 (UTC)[reply]
Yes and no. Division isn't really a thing; there may be multiplicative inverses, or multiplication may be cancellable, but there isn't a clear "division operation" in most cases. As for commonality, it depends on how far you cast your net. Every field is a vector space over itself and has division. More abstractly, if we can consider elements of a Module (mathematics) to be vectors, then every abelian group is a Z-module, so every field is a Z-algebra with multiplicative inverses. On the other hand, if we mean real vector in euclidean space considered as an arrow, then there isn't any obvious geometry based division. Finally, the obvious examples of vector structures that have some form of division, and aren't really abstract at all, would be the complex numbers; and slightly more abstract, the Quaternions and the space of continuous functions from the reals into the reals. Ultimately, the answer depends on just what you allow "vectors" to be; I would say the lay answer would be that vectors don't divide unless you provide additional structure.Phoenixia1177 (talk) 06:09, 28 July 2013 (UTC)[reply]
Geometric algebra is one framework that systematically allows a vector to be divided by another vector, to produce a multivector that contains a scalar and a bivector component. One application is to define a geometric calculus that combines the vector-calculus operators div and curl into a single object, that works in arbitrary numbers of dimensions, and that can be inverted in ways that relate to Stokes' theorem and the Cauchy integral theorem. This derivative operator also allows Maxwell's equations of electromagnetism to be written in a particularly concise form. Complex-number division and quaternion division fall out as the specialisation of the system to 2 real dimensions and 3 real dimensions respectively. Jheald (talk) 14:46, 28 July 2013 (UTC)[reply]
There are some good-looking answers at [3] - apparently you can define a vector C = "A / B" such that B × C = A, but the problem is that there is no "1" such that "1" x A = A, which somehow degrades the system. I'm sure there are better things written about this! Wnt (talk) 14:50, 28 July 2013 (UTC)[reply]
As indicated by Jheald above and Muphrid in Wnt's math.stackexchange link here, there is a natural manner in which vector division is defined. In effect, one must be a little more flexible about what our "numbers" are allowed to be (in addition to the already allowed scalars and vectors). The dot and cross products each discard necessary information, but together they constitute a complete product that can be inverted; in particular, left- and right-division by vectors are not only possible, but are standard operations. — Quondum15:34, 28 July 2013 (UTC)[reply]
When an Alzheimer's disease patient dies, how is it decided whether the disease caused the death? Or in other words, how is it decided whether someone died of Alzheimer's? Of course I understand that it's sometimes easy, e.g. the patient is killed in an automobile accident, but I'm addressing a situation in which the proximate cause of death is something physiological. Bringing this up because of Category:Deaths from Alzheimer's disease. Neither Douglas Engelbart nor Auguste Deter are in the category, even though Engelbart's "death came after a long battle with Alzheimer's disease", and Dr Alzheimer's description of the disease was based largely on his observations of its effects on Deter. Moreover, the category says "People who died with Alzheimer's disease" — with, not from, so it doesn't help me understand the definition of "dying from Alzheimer's". Nyttend (talk) 11:42, 28 July 2013 (UTC)[reply]
Alzheimer's mostly kills by infections due to loss of basic motor and reasoning skills. Patients fall and suffer fractures, in a later stage they won't be able to walk, become incontinent, have difficulty eating and swallowing; these lead to bedsores, bladder infections, pneumonia caused by inhaled food. Treatment becomes more difficult because the patient can't cooperate. (alzheimers.co.ukslate.com: how does alzheimers kill) Ssscienccce (talk) 12:15, 28 July 2013 (UTC)[reply]
Can we get to the original question? I'm only asking about the medical classification of the deaths of Alzheimer's patients as being "caused by Alzheimer's" or "not caused by Alzheimer's", and seeking to apply it to our categorisation of people who had the disease before their deaths. Nyttend (talk) 18:59, 28 July 2013 (UTC)[reply]
To clarify, although it's only anecdotal, the cause of death in my grandfather's case was indeed listed as pneumonia (he died after a week with a with a fever) although he had been declining for a decade. (I remember being quite surprised at the time that this, and not Alzheimer's was listed as the cause). In the case of my grandmother it was listed as stroke, even though in her case the immediate cause was aspiration pneumonia caused by an inattentive attendant. My suspicion is that just looking at death certificates or press reports you may not get a very accurate picture. If, as I understand, your concern is for wikipedia purposes, it might be better to classify as dying suffering the disease, rather than from it. Or perhaps look at AIDS and give death from complications of Alzheimer's. μηδείς (talk) 01:27, 29 July 2013 (UTC)[reply]
"Cause of death" is, really, a legal definition, rather than a medical one. The purpose of that box on the death certificate is to make sure that the appropriate legal steps have been taken - if a homicide is suspected, that the police have been involved; if an accident took place, that the appropriate investigators have examined the circumstances; and if a disease was involved, that a report was made that to the public health authorities (who are interested in epidemic diseases like cholera and chronic ones like Alzheimer's). I apologise if that sounds like pedantry, but when you see "cause of death" statistics, for different jurisdictions and for different times, it's important to realise that many are happy to simply record "homicide / accident(misadventure) / disease / old-age" and leave it at that. For old age, and diseases related inextricably to it (of which various neuro-degenerative ailments like Alzheimer's are prime examples) there's a tricky legal and statistical problem - an 80 year old dies with a panoply of disease, a number of which may contribute to their death, but putting the blame on one is difficult. A typical 80 year old might have high blood pressure, cardiovascular and pulmonary dysfunctions and diseases, one or several latent carcinomas, and various organic or endocrine diseases or dysfunctions. If that person has Alhzeimer's too, and they die, how should the death certificate be written, and how should their death be recorded in the country's death statistics database? The UK's Alzheimer's Society says "Although dementia is a life-shortening illness, another condition or illness (such as pneumonia...) may actually cause a person's death. This other condition or illness will most likely be listed as the cause on the person's death certificate. Pneumonia is listed as the ultimate cause of death in up to two-thirds of people with dementia." The UK has undertaken a process to improve the determination listed on death certificates, as described in this document for England (similar changes were undertaken in Scotland and I think in Wales and Northern Ireland). Those changes try to capture, on the death certificate, more of the underlying cause, and not just to have an unhelpful determination like "old age" or "organ failure". They want doctors to list both the proximal cause (pneumonia, organ failure, heart failure, infection) and the underlying cause - and for someone elderly, that underlying cause might well be several serious complaints, occurring together, of which Alzheimer's is but one. For almost all elderly people in poor health, no detailed post-mortem pathological examination is performed - there's no legal basis for doing so (legally it doesn't matter which disease killed them) and medically little reason (because they've followed the downward path most people do). From a scientific perspective, if you were to do the most thorough examination of each person, it's still not clear what you should really record. Looking at the body of the deceased person, and running every scientific test available, you'd end up with a laundry list of contributory factors: the observable neural effects of Alzheimer's; the consequences of poor diet and hydration; general diseases often associated with age; and infections. Just as asking why the Roman Empire failed, it's a facile TL;DR to say "Visigoths" just as the death certificate has the doctor to write "pneumonia subsequent to Alzheimer's disease" - from a practical perspective, that's a sufficient cause of death. -- Finlay McWalterჷTalk22:33, 28 July 2013 (UTC)[reply]
"The most frequent cause of death for people with AD is aspiration pneumonia. This type of pneumonia develops when a person is not able to swallow properly and takes food or liquids into the lungs instead of air." [4] --Anthonyhcole (talk · contribs · email) 00:05, 29 July 2013 (UTC)[reply]
100% percent of pneumonia deaths are caused, most immediately, by brains rotting due to lack of oxygen. Granted, 100% of all deaths are caused by brain damage, so that kind of specificity actually makes the matter vaguer. But, very basically, even those without Alzheimer's die from the same effects (though maybe not the same causes, and certainly quicker).
As Finlay McWalter said, it's not always clear where to draw the line between proximate and immediate causes, and no wide standard. Depending how you look at it, the most significant contributor to death is staying alive. InedibleHulk(talk)03:33, 31 July 2013 (UTC)[reply]
(... assuming it was made of stronger stuff ...) The moon would have an even more irregular orbit that is does with the current oblate spheroid shape. Dbfirs19:46, 28 July 2013 (UTC)[reply]
Because the separation between the Earth and the Moon is not sufficent, compared to their radii, for them to be considered as point masses, so tidal forces become significant. These depend on the geometry of the orbiting bodies. Tevildo (talk) 20:39, 28 July 2013 (UTC)[reply]
All the water and air would flow to the centre of the faces, so there would be circular shaped oceans surrounded by a ring of atmosphere, but away from there the elevation would be so high and atmosphere too thin. It would be impossible for living things to travel between the faces through the cold vacuum. Graeme Bartlett (talk) 20:52, 28 July 2013 (UTC)[reply]
Well, lets assume the Earth as a tetrahedron (it's got to be Platonic) with the same mass (or volume or surface area) as the real Earth, made from Unobtainium of uniform density and braced with polarity-reversed tachyon shielding to keep it from collapsing. What would the gravity field look like? If I start in the middle of one of the sides, and walk towards one of the edges, do I walk up the potential well, or down? Intuitively, the gravity field should be weakest at the 4 corners, or do I have that wrong? --Stephan Schulz (talk) 09:36, 29 July 2013 (UTC)[reply]
Stephan, to determine the gravitational force exerted by a tetrahedron shape of uniform density (or of any known density function), you have to solve Newton's law of gravitation in its integral form. This is a straightforward and classical mathematics problem; there is no "trick" or "gotcha"; it's just a volume integral of a vector function with somewhat ugly limits of integration to constrain the mass to a tetrahedron-shape. Depending how familiar you are with integral calculus, you should be able to solve (or make a computer solve this for you). Such integral field equation solution problems are typical "homework" questions in Marion & Thornton's classical physics text (they have a chapter explaining how to perform these types of calculations for gravity). In my younger years I could probably crunch the numbers on such a problem in ten or twenty minutes, or maybe faster; but now that I'm out of practice, it'd probably take a bit longer.
If I were forced to analyze the problem "intuitively" rather than analytically, I would try to reason about the magnitude of forces based on Gauss's law for gravity, and regarding the direction of forces, I would consider the symmetry of the tetrahedron. The force of gravity falls off as r2, but the mass increases as r increases (following the integral of the intersection of a sphere and the tetrahedron - which is less than, but upper-bounded by, r3 - and therefore this is not trivial!) At the corners, your test-point is quite a bit farther away, but the gaussian pillbox it defines is only enclosing slightly more mass; and because a sphere is a convex hull around the tetrahedron, I intuitively suspect the increase in enclosed mass is small and negligible. So, the center-of-each-face ought to have a stronger field. I would defer to a complete analytic integral solution for the field and its potential-function, though; I'm having trouble "intuiting" the gradient of the magnitude of the integral of a three-dimensional geometric intersection. (I suspect most people also have trouble with that "intuition"). Nimur (talk) 16:15, 29 July 2013 (UTC)[reply]
To put it briefly, that the force of gravity would be less at the vertices would matter less than the fact the direction the gravity would be pulling you would not be "down" (normal to the face) but sideways, toward the center of the face. μηδείς (talk) 16:26, 29 July 2013 (UTC)[reply]
Not towards the centre of the face, but at an angle to the face (maybe between 45 and 60 degrees in the "habitable zone", assuming the sea doesn't spread too far towards a vertex). Houses and roads would have to be built at an angle between 30 and 45 degrees to the face, but that's no worse than building on a steep hillside. (I haven't tried Nimur's integral, and I'm only guessing how the water would behave, so I'm happy to have my estimates corrected by a rigorous mathematical analysis) I'm struggling to estimate the 3-D shape of the four seas. Each would form some sort of dome. Dbfirs16:59, 29 July 2013 (UTC)[reply]
The angle would differ depending on where you were, but the net drag would be toward the center of the face, in the same way that when falling down a mountain, one falls toward the foot of the mountain, although "straight" down would be into the mountain. One might as well view the tetrahedron as four mountains. The center of each face would be the nadir of a three-branched valley formed between any three of the mountains. μηδείς (talk) 19:20, 29 July 2013 (UTC)[reply]
Yes, a component of gravity towards the centre of the face, but we wouldn't need to "fall down". I've climbed and traversed mountains steeper than that. I think the "mountains" would be too big to perceive them in the same way as we do Earth mountains, and the four oceans would hide the "valleys" from being noticed. Life would be lived on a steep slope, and the (imaginary) inhabitants would adapt. Dbfirs07:50, 31 July 2013 (UTC)[reply]
In deference to better physicists than myself, that's not "Nimur's integral," insofar as anyone may "own" an integral, it's probably either Green's or Gauss's integral (or maybe Laplace... I'm struggling to recall my history of famous formulations of classical physics). Suffice to say, every student of physics and mathematics ought to have performed this integral, for at least the last hundred and fifty years of physics pedagogy. I do not recollect that Isaac Newton ever performed the integral to solve for the potential for an arbitrary mass-density distribution; he was far too practical to consider such nonsense.
Regarding calculating the height of the seas: I would first assume that the mass of the ocean is negligible compared to the mass of the planet; and then I would assume that sea level follows the gravitational equipotential; the solution for which we would obtain by solving the same integral mentioned above. Were we to get serious about things, we would have to assume the mass of the ocean is non-negligible and iterate the solution numerically. Nimur (talk) 17:12, 29 July 2013 (UTC)[reply]
Back to the "What would be different..." part, there would be more differences than just a crazy gravity field.
Total Lunar eclipses would be slightly more rare because of the shape of Earth's shadow, and solar eclipses would be slightly less rare. The additional eclipses would, however, only shadow the tips where there is no atmosphere anyway: a purely theoretic result.
Scratch the above. Both, lunar and solar eclipses would be significantly more common, because...
The Moon would spiral out from Earth even more slowly than it does now (because the tetrahedron (terrahedron?) is assumed to be rigid), and thus Earth would spin faster, and the Moon would be closer. On top of that, the Moon would have a shorter orbit, so there would be more chances at eclipses.
The seas would be so deep that the water near the bottom would be solid (a kind of ice different from the one we have on Earth).
Planet Earth would have ceased to be a planet in 2006, because planets must have assumed a hydrostatic equilibrium, and we have assumed that the terrahedron is strong enough to withstand gravity. Defeat by definition.
Forgive me for my understanding of these concepts which I have picked up from popular science books. For the photons of light to travel at the speed of light, they have to have zero mass right? How then is a black hole able to bend light? Surely gravity can only pull or affect something that has mass and therefore weight? Sandman1142 (talk) 22:06, 28 July 2013 (UTC)[reply]
If you try to calculate the deflection of light due to a gravitating body by plugging the photon's relativistic mass into Newton's law of gravitation, you get an answer that's off by a factor of two. Newtonian gravity just becomes too inaccurate of a model to use when dealing with objects moving at or near the speed of light. To get the right answer when dealing with very fast particles, you need to use general relativity. And from the perspective of general relativity, whether a particle has a mass or not is irrelevant to the question of whether the particle is affected by gravity. A hypothetical particle that has even a zero relativistic mass would still have the same world line as a photon, as long as the hypothetical particle also travels at the speed of light of course, and as long as neither the photon nor the hypothetical particle undergo any proper acceleration (e.g., the photon doesn't go through a lens or something). What happens with either particle is the gravitating body curves spacetime, and the particle travels in a straight line along that curved spacetime. The particle's world line is determined entirely by the geometry of spacetime as affected by the gravitating body, and the particle's initial conditions of direction and speed. The particle's mass never enters into consideration in the least.
A particle that isn't affected by gravity wouldn't even be a meaningful concept in general relativity. In Newtonian gravity, it would mean a particle that moves in a straight line relative to a global inertial frame of reference. But a global inertial frame of reference is just a bit of fiction that works well as an approximation under appropriate circumstances. In reality, no such global inertial frame of reference exists, because spacetime isn't flat, due to the curvature of spacetime due to gravitating bodies.
It may help toward understanding this to point out that in the view of general relativity, gravity isn't a proper force. When gravity appears to act as a force, it's really just acting as a fictitious force due to using an accelerating frame of reference. E.g. when an apple falls to the earth, the apple isn't really accelerating toward the Earth, in that in the apple's proper frame, which is an inertial frame of reference, the apple is just standing still. It's really the frame of reference that's attached to the surface of the Earth that's accelerating upward, which is making the apple appear to be accelerating downward. All small objects appear to accelerate toward the earth at the same rate in a vacuum regardless of what the object's mass is or what it's made of, because the acceleration involved is really the acceleration of the reference frame attached to the Earth, not anything having to do with the small object. If you want, you can multiply that acceleration by the small object's mass to give a "force", but in determining the dynamics of the object you're just going to turn around and divide that "force" by the same mass of the object in order to determine the acceleration anyway, so you might as well not bother multiplying the acceleration by the small object's mass in the first place. The acceleration is really the more fundamental thing that's actually going on, not the fictitious "force". This is basically the equivalence principle of general relativity. Red Act (talk) 06:57, 29 July 2013 (UTC)[reply]
Deriving relativistic kinetic energy for curvilinear motion
With Newton's second law and the work-energy principle, derivation for motion along a straight line is straightforward. But I ran into problems when trying to derive it more generally by taking the line integral (over the particle's path C):
Now there's no question about the second term, which is purely a scalar function. But it seems like the correct formula is only obtained when and are in the same direction, which is not necessarily true. I know I must be doing something wrong here, or otherwise I misunderstand the meaning of that dot product (which seems to be treated as the same as the scalar product in the article on kinetic energy). What is it?--Jasper Deng(talk)23:35, 28 July 2013 (UTC)[reply]
IIRC, "v squared" is a scalar value. One can not take a square root of a vector AFAIK. To get a particle to move in other than a straight line requires a force of some sort - such as gravity. Collect (talk) 00:22, 29 July 2013 (UTC)[reply]
But that's not very helpful. When v is written without being boldened, we mean its magnitude only, without regarding its direction (hence why the integral cannot be evaluated by separating components of v, since the magnitude of the whole vector is in the denominator). In addition, when you take the dot product of a vector with itself, you get the same as its magnitude squared. Also, a derivation of kinetic energy has to be valid for any net force and path - Newton's second law and the definition of work make no distinction between different kinds of forces.--Jasper Deng(talk)00:24, 29 July 2013 (UTC)[reply]
Although it's counter-intuitive, the following might be a partial answer:
Well, I omitted the step where I differentiated the momentum expression with respect to time, using a product rule (differentiating the vector and scalar components separately). In the first term, I differentiated the vector v in the numerator with respect to time. In the second term, I differentiated the denominator with respect to time, so in that term the vector v in the numerator is unaltered. Taking its dot product with the vector v in the integrand produces a purely scalar-valued expression:
Now that I think about it, , so your original integral is the same as and you must have made a mistake in evaluating it. -- BenRG (talk) 05:13, 29 July 2013 (UTC)[reply]
No. It's not that. I tried to evaluate it last night and found that the correct formula comes if I can use your assumption. But I'm not sure if you can assume that. In an extreme case, what if the change in velocity were almost orthogonal to the velocity at that instant? Remember, in magnitude form, the dot product of two vectors is the product of their magnitudes and the cosine of the angle between them - the assumption seems to be only valid if that cosine is 1, which would essentially be the two vectors having the same direction.--Jasper Deng(talk)05:22, 29 July 2013 (UTC)[reply]
But BenRG isn't just making an assumption which may or may not be valid depending on the path. It's always true in general that , because . Red Act (talk) 21:48, 29 July 2013 (UTC)[reply]
It can be proven much simpler if one notes that by definition . Differentiation of both side of this equality leads to . Ruslik_Zero09:51, 30 July 2013 (UTC)[reply]
That's right and matches the headline. But what about the 'ascender' part? Chimps and Orangutan are not our ascenders. If we move through the phylogeny tree, we see the chain of ascenders of modern man: H. erectus < H. habilis < Forest ape. What is older that these forest apes? (if they are extinct). OsmanRF34 (talk) 04:20, 29 July 2013 (UTC)[reply]
I've never heard the term "ascender" used in a biology context either, though I've heard the word many times in a mountaineering context. And yes, for the species immediately preceding any given species in the timeline of evolution, the correct term is "ancestor". 24.23.196.85 (talk) 07:38, 29 July 2013 (UTC)[reply]
So, your question is which is the closest species that has some present day descendants that are humans and others that haven't changed enough to be considered a different species? 202.155.85.18 (talk) 07:29, 29 July 2013 (UTC)[reply]
The answer to that question is; there are no currently known extant species from which humans could have evolved. The oldest known extant species are the tuatara (200 million years old, diverged from synapsids 315 million years ago), horseshoe crab (440 million years old, diverged from deuterostomia 555 million years ago), the coelacanth (400 million years old, diverged from mammalia 555 million years ago), platypus (110 million years old, diverged from primates 220 million years ago) and the lamprey (360 million years old, diverged from mammalia 416 million years ago. Since they all appeared after their phylogeny had diverged from homo sapien's they cannot be an ancestor. 202.155.85.18 (talk) 09:17, 29 July 2013 (UTC)[reply]
"Inheritance" of the notion of female beauty
The female beauty standards vary across cultures, but I wonder how such criteria as wide hips, thin waist and long legs are transferred to many people seemingly at subconscious level? I know such traits are related to biological advantages, but many people share such notions without even knowing it, so what's the mechanism of transfer? Is it at gene level or something else?--93.174.25.12 (talk) 07:56, 29 July 2013 (UTC)[reply]
Jealousy more or less blindly driven by the urge of competition is certainly one strong influence. You can for example find that same strange behavior of "jealousy about food" among all higher Organisms where one entity tries to get the food of the other no matter an even bigger portion may actually be right in front of its nose. So in consequence whatever features the most dominant or most active male entity makes his choice on all his competitors will adapt to subconsciously. --Kharon (talk) 14:26, 29 July 2013 (UTC)[reply]
When I see areas being defined as having high Endemicism, it is in places far away from where I live (near Washington, DC). Even the few areas in the United States that I've heard of are either on the Hawaii, the West Coast (Spotted Owl/California condor) and Florida. On the other hand, it seems like most of the eastern half of North America is just one large ecological region. So, for the Eastern half of North America north of Florida, are there any mammals, bird, reptile or amphibians whose entire habitat is smaller than a circle one hundred miles across? (This was sparked by a comment that an acquaintance made that "you could pave the entire state of Maryland over and no animal would go extinct")Naraht (talk) 19:35, 29 July 2013 (UTC)[reply]
Travel from Toledo, OH to College Park, MD, and you will cross several ecoregions.Endemism doesn't really say anything about the scale. For instance, we have a list of Endemic birds of eastern North America, which lists several endemic species. But they have decently large ranges, and it seems that you are interested in endemic species that also have very small ranges. For those, in eastern USA, I'd start looking at the New_Jersey_Pine_Barrens, [7], and also the Barrier islands along the coast. There are definitely some endemic and endangered plant spp there. Finally, there are a number of ecoregions in the eastern USA, each with their own plant and animal communities. These regions can have remarkably distinct flora and fauna, though the differences are not always visually striking. SemanticMantis (talk) 20:16, 29 July 2013 (UTC)[reply]
While both the Maryland Darter, the Franklinia (sort of) and the Venus Flytrap qualify, I limited my initial request to Mammals, Birds, Reptiles and Amphibians. I agree with the restatement "it seems that you are interested in endemic species that also have very small ranges", perhaps I should have phrased in in that way. It seems like even the extinct Mammals/Birds of the US East Coast like the Red Wolf and the Carolina Parakeet had relatively *wide* ranges more than 500 years ago. Naraht (talk) 20:43, 29 July 2013 (UTC)[reply]
Birds and mammals don't tend to have such small ranges in areas like the Eastern US with large contiguous ecological regions. It's not like there are deep mountain valleys or flightless birds on offshore islands. If I remember correctly, the completeion of Route 18 in New Jersey was delayed by some sort of herp. But I cannot recall the exact species. μηδείς (talk) 21:02, 29 July 2013 (UTC)[reply]
An interesting perspective on this very problem can be found in 1493: Uncovering the New World Columbus Created, which discusses the way in which human migration (via the so-called Columbian Exchange) has created ecosystems which are much larger and more homogeneous across wider areas of the Earth's surface. The author even coins a term "Homogenocene" to represent the current ecological age as a break from the Holocene epoch, considering this trend. I can't say one way or another how "correct" this view is, but it is an interesting perspective. --Jayron3223:39, 29 July 2013 (UTC)[reply]
Can mathematical models cause errors in astrophysics
I know when astronomers and planetary scientist tries to predict future events of solar system and stars they typically use mathematical models and physical laws. Is mathematical models and astrophysical law equations always a reliable predictions, or mathematical models can often be wrong? I saw [8] when calculating remaining planets around white dwarfs, they use mathematical models and physical laws. I thought mathematical models and physical law/equations has to make reliable predictions, or they are nothing more than a basketball shot.--69.226.33.213 (talk) 20:32, 29 July 2013 (UTC)[reply]
Well models are not reality. They will be making huge simplifications. There is also the possibility that there is an error in the interpretation of what is happening. So the model does not apply. Lastly, errors can increase over time so making future predictions less and less accurate. For example this happens in weather forecasting. Predicting the future positions of the planets gets less and less accurate over long periods of time too. Graeme Bartlett (talk) 21:33, 29 July 2013 (UTC)[reply]
If you want to "predict" there is already some unknown factor or even an chaos element in it and its not the mathematical method causing an error instead your expectation of the capability of that method is wrong. You can be certain that a position of an known Planet can be calculated very exactly thousands of years into the future but you cant calculate how strong the solar wind will be around Earth next week. --Kharon (talk) 00:10, 30 July 2013 (UTC)[reply]
So these math problems only gives us the estimates but not the exact variable? So we won't know where the habitable zone in 7 billion years future until somebody can actually try to find the habitable zone around giant star? Have anybody really tried to find habitable moon/planets around giant star. So the mathematical formulas will only give me the guesses and is rather waste of time until better studies are done? --69.226.33.213 (talk) 01:43, 30 July 2013 (UTC)[reply]
Read the section you referred to. It applies to "an idealized object which is perfectly opaque and non-reflecting". The sun is not a perfect black body, so any results you obtain with the equation will not perfectly describe the behavior of the sun. In physics teaching, this is often joking referred to as "Assume a perfectly spherical cow of uniform composition existing in a perfect vacuum", which, believe it or not, we even have an article about: Spherical cow. No, models are not useless, and they are much better than mere guesses, but they do have their limitations. Dominus Vobisdu (talk) 06:01, 30 July 2013 (UTC)[reply]
One of the most noticeable differences between an amateur and a professional scientist is that the professional knows when to place stock in a particular model. This is a skill that develops with experience, and perhaps is augmented by some innate aptitude. Amateurs blindly "plug and chug," and will spit out any result of any equation, without any understanding of why they are using that equation.
Specifically regarding error in equations: entire branches of applied mathematics - like sensitivity analysis - are dedicated to the study of how much error is introduced if a particular input is wrong. Statisticians use confidence intervals and other probabilistic measurements to describe reliability of data. Engineers and experimental scientists use signal to noise ratios to express how imperfect any part of their measurements are. So, error can be quantifiable. If we assert with high confidence that all our errors are small, we can be very accurate. Again, a distinction is that a professional scientist should be better at ensuring that they accounted for everything. Occassionally, they screw up; eventually, if the screw-up is big enough, somebody else catches it and fixes it. Nimur (talk) 14:25, 30 July 2013 (UTC)[reply]
Also, when I was a student, a Prof. told us in the very first astrophysics lecture that in many cases it is good enough to be within a factor ten of the correct answer. This has to do with the type of questions one wants to be answered in astrophysics as opposed to theoretical physics. In recent years things have changed in this field, but there is still some truth in it.Count Iblis (talk) 14:32, 30 July 2013 (UTC)[reply]
The reliability of mathematical models critically depends on the nature of what is being modeled. In the case of a planet in it's orbit, there are hardly any unexpected events that can cause significant deviation from idealized behavior - and small errors in our measurement of the initial position and velocity don't get wildly magnified as the model runs off into the future. Contrast that with systems like weather and solar wind - which are chaotic systems (in the mathematical sense of "chaos theory") where the smallest error in initial conditions or the smallest error in the model will rapidly blow up into huge errors. That's "the butterfly effect" - and explains why weather forecasts for more than a few days into the future fare no better than chance but the position of a large moon or planet can be known to high precision over a period of hundreds of years. SteveBaker (talk) 14:35, 30 July 2013 (UTC)[reply]
And even something as seemingly stable as the solar system is not necessarily stable in the very long term - see n-body problem and stability of the Solar System ... "an error as small as 15 metres in measuring the position of the Earth today would make it impossible to predict where the Earth would be in its orbit in just over 100 million years' time". Gandalf61 (talk) 15:53, 30 July 2013 (UTC)[reply]
Yes - but over 100 million years, unknown factors such as the fate of the sun and the consequent variation in the pressure of the solar wind and such mean that we don't know the position that the earth will be at, even if we had it's current position accurate to a nanometer. SteveBaker (talk) 20:27, 30 July 2013 (UTC)[reply]
The question of the stability of the solar system can be investigated. Recent studies have come up with surprising results, see e.g. here:
"These results also answer to the question raised more than 300 years ago by Newton, by showing that collisions among planets or ejections are actually possible within the life expectancy of the Sun, that is, in less than 5 Gyr. The main surprise that comes from the numerical simulations of the recent years is that the probability for this catastrophic events to occur is relatively high, of the order of 1%, and thus not just a mathematical curiosity with extremely low probability values. At the same time, 99% of the trajectories will behave in a similar way as in the recent past millions of years, which is coherent with our common understanding that the Solar System has not much evolved in the past 4 Gyr. What is more surprising is that if we consider a pure Newtonian world, the probability of collisions within 5 Gyr grows to 60 %, which can thus be considered as an additional indirect confirmation of general relativity." Count Iblis (talk) 21:07, 30 July 2013 (UTC)[reply]
This IP editor has been asking essentially the same question about mathematical models over and over again for months, and getting essentially the same answers each time. I don't get it. Looie496 (talk) 15:31, 30 July 2013 (UTC)[reply]
Or maybe I can think about comparing astrophysics and professional scientist to predict traffic light evolutions in Orange County in the future. Ever since I was tens of years old, I have tried to predict what will the traffic lights in Irvine Base Areas will happen 6-10 years in the downtime. I noticed the specifications of red, yellow, and green bulbs changes overtime, and the traffic inspection paints happens on traffic light enamel. I could have tried to do models I just didn't have that time to do that and I am kind of lazy to do that. First, models have to be consistent and has to work on many conditions. Few years ago, most South Orange County cities still didn't have incandescent yellow light bulb replaced, only the reds and greens were replaced around 2001-2002 areas [9] I was expecting the yellow light specifications will be the amber LED colorings, but some of the cities post 2010 actually came out to egg-yellow LED lighting the new LED standards, and between 2011-2012 City of Irvine went ahead, replaced some obsolete red, yellow, green bulbs with newer systems scarlet-red, lemon-yellow, cyan-green according to the metric table listing. I can learn, when I try to create model to predict what will the traffic light bulbs will happen 5-10 years in the future, each time things actually presents and performs, my older speculations becomes more and more inaccurate. Without calling the city department to ask them questions, I can try to create a model templates according to what I have studied in the past if I have time to do that, but the problem is the models have to be consistent, and has to work to all-related cases it is hard to do because I have to organize informations, predictions intersections undergoing constructions usually happens to cause replacement of signal poles, (same as what happened on Culver and Walnut Avenues in Irvine two years ago) requires depth understanding and enough studies to be done to imitate like-intersections undergoing construction, but each time newer inventories comes up without even knowing it.--69.226.33.213 (talk) 00:58, 31 July 2013 (UTC)[reply]
My $0.02, nothing is infinitely correlated or infinitely predictive, and this has nothing to do with the correctness of a model. If I take a baseball and throw it in the air, I can use projectile motion to calculate its position and velocity 15 or so seconds later, and assuming that there are no confounding variables the result should be highly predictive. Now if I try and rely on the same model to tell me where that baseball will be several minutes from now than it becomes a bit more difficult, taking it to the extreme if we try to figure out where it will be a million years from when I first tossed it in the air, the best you could do would be to say that it's most likely somewhere in the general vicinity of earth. This doesn't mean that there's no such thing as projectile motion. See also correlation function. (+)H3N-Protein\Chemist-CO2(-)11:25, 31 July 2013 (UTC)[reply]
July 30
Domesticated Hedgehog
Is there a scientific difference between a hedgehog and a domesticated hedgehog as there is between wolves and dogs or lions and cats? — Preceding unsigned comment added by 68.115.240.26 (talk) 01:49, 30 July 2013 (UTC)[reply]
As for your other examples, wolves and dogs now appear to be the same species, so that's a reasonable comparison, but lions and (house)cats are not the same species at all and cats certainly are not domesticated lions, but rather cats are domesticated African wildcats. StuRat (talk) 06:14, 30 July 2013 (UTC)[reply]
Wierdly enough though the dog and the wolf are (as of 1993) considered the same species and cats and the African wildcat are not even though dogs were domesticated much earlier. But I don't think that represents a higher amount of genetic change. And sorting out the genus and species of the smaller members of Felinae (like the wildcats) gets ugly.Naraht (talk) 13:33, 30 July 2013 (UTC)[reply]
I had never heard of a domesticated hedgehog until a few seconds ago. My only comment is that according to our article, it is a hybrid of two separate species and therefore different to any wild hedgehog. Is a mule different to a horse? Yes. Alansplodge (talk) 22:58, 30 July 2013 (UTC)[reply]
Feeling pain in a dream
I just woke up from a long nightmare. The last part of the dream was someone digging their hands into my chest, as if they were trying to murder me. The odd part was that I ran away from the person in my dream, but then I saw myself laying down where I was in real life right after, in my dream. The person was right there, which is when they dug their hands into my chest. An even more odd thing is that I felt the pain and woke up from it. I am very shaken and I don't think that I can get back to sleep which is understandable, considering I felt like I was being murdered. Has there been any reports of people feeling pain in their dreams? Please don't tell me that it can't happen, I felt the intense pain in my chest, in all of the correct spots. SL93 (talk) 08:38, 30 July 2013 (UTC)[reply]
I wasn't asking for my dream to be interpreted. I have a habit of saying exactly what I want, rather than implying it. SL93 (talk) 09:34, 30 July 2013 (UTC)[reply]
You didn't help me. I know of lucid dreaming, but nothing that you linked mentions pain being involved. Someone even asked for a reliable source on the talk page for REM sleep. No wonder it isn't in the article. SL93 (talk) 09:37, 30 July 2013 (UTC)[reply]
Yes, there have been reports (other than yours) of people feeling pain in their dreams. It isn't common. Just search for pain+dreams in Google Scholar. You will then be in a position to update the REM sleep article. Sean.hoyland - talk10:05, 30 July 2013 (UTC)[reply]
I feel pain in dreams fairly often. I was surprised a few years ago when I saw a thread on here about c-fiber nerves and people claiming that it is impossible to feel pain in dreams. I challenged it and nobody offered up a counter argument or any sources, and I haven't been able to find anything concrete one way or the other myself. But I assure you, when someone in a dream is cutting my fingers off or stabbing me, I feel it. 82.44.76.14 (talk) 10:31, 30 July 2013 (UTC)[reply]
It appears than dreams can contain just about anything that real life can contain - including hearing, taste, feeling etc. Brains can be weird <g>. Collect (talk) 15:57, 30 July 2013 (UTC)[reply]
The OP asked if there had been reports of people feeling pain in dreams. Certainly I have had dreams in which pain played a part, after surgery, when there was a sore back, when there was tonsilitis, when there was a belly ache. The dream incorporated and explained the pain in some way. The dreaming process sometimes explains away stimuli in such a way that you can just go on sleeping. A Google book search will reveal countless books discussing pain in dreams, though many of them are "new age" self=help books or religious books. Freud, in "The interpretation of dreams" p190 mentions "headache dreams" and "toothache dreams." Edison (talk) 16:11, 30 July 2013 (UTC)[reply]
On the occasions I have felt pain in my dreams it was always because of some real stimulus outside of my dream, such as thinking my leg is in a bear trap to wake up and find I am in a twisted position and it has fallen asleep, or feeling that my mouth is full of ashes and my teeth are falling out to wake up and find I am gagging because I have been mouth-breathing and my mouth is entirely dried out. This seems to be the brain rationalizing the real stimulus into the dream. But I cannot say I have ever felt a pain in a dream without some external reason. The exception seems to be sex dreams, which can be very pleasantly authentic feeling without any real external stimulus. μηδείς (talk) 16:21, 30 July 2013 (UTC)[reply]
Original research: I've personally had at least four dreams involving physical pain: on one occasion, I dreamed that I was shot to death right through the heart by a crooked cop; on another, that my mouth and throat were stuffed full of sharp plastic shards (this was when I was down with strep throat and bacterial sinusitis on top of that); on the third occasion, that I was bayonetted to death (also right through the heart) by a Red Army soldier (of the four, this was the dream I remember best -- in that dream, I was a White Army squad leader during the Russian Civil War, and me and my squad had just ambushed a Red Army column but bit off more than we could chew, so we all died fighting); and on the final occasion, I dreamed that I was shot in the chest by a gangster, but survived and had enough strength to strangle my attacker. So yeah, it can totally happen -- and also, you CAN dream that you were killed without actually dying (contrary to a popular misconception). 24.23.196.85 (talk) 02:44, 31 July 2013 (UTC)[reply]
Well, the difference here is, I don't have casual sex with every Tina, Dolly and Helen that I meet (unlike the characters in that movie), which means I'm safe on that end. 24.23.196.85 (talk) 08:40, 31 July 2013 (UTC)[reply]
So the possibility exists that you suffered real chest pains which were then incorporated into you nightmare. For chest pains, you would want to consult a doctor, but these are possibly dreamt chest pains. I'm not sure what the medical guidelines are for reporting those. Also, your sleeping position could possibly cause cramps in the chest muscles, which your brain interpreted as the heart. But, do you get chest pains during the day ? If so, you should definitely see a doctor. StuRat (talk) 09:05, 31 July 2013 (UTC)[reply]
Deaths Cap in homeopathy and well poisoning
I have read that death's cap is responsible for most mushroom poisonings; including the deaths of a few roman emperors; which makes me wonder how often it was used in medieval well poisonings. I know that is a rather sensational subject from hundreds of years ago and the numbers were inflated for propaganda purposes; but there were at least some recorded instances. I recently tried adding an article I found on PubMed Central about the use of Amanita phalloides homeopathically to treat B-cell chronic lymphocytic leukemia. I'm aware that a single study needs to be repeated; which is why I'm not contesting the removal. Now I don't believe in water memory; but what is the difference between homeopathy and diluting something in water; wouldn't diluting poison with water allow for smaller amounts of it to be consumed? They do say that the poison is in the dosage. I'm also confused as to whether alternative medicine journals are legitimate primary medical sources when they are from the NIH or other government health organizations. Here is the link; http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3151460/. Thanks for answering these questions. — Preceding unsigned comment added by CensoredScribe (talk • contribs) 19:04, 30 July 2013 (UTC)[reply]
Homeopathy is tricky. The homeopaths have their own special backdoor into the medical racket, allowing them to dispense "treatment" and "medicine" according to their own peculiar standards. In practice, there are two different kinds of homeopathy encountered: one uses astronomically high dilutions to sell tap water to unsuspecting customers, given sometimes even by mainstream physicians who have applied medical ethics (i.e. profit) to determine that the act of patients giving them money for bogus medicine will make them feel better. The other, more interesting variety, uses absolute minimum homeopathic dilutions to deliver genuinely therapeutic treatments - the catch being that they are not medically evaluated and they have the potential to harm as readily as heal. For example Zicam causing loss of smell, [10], and others. In this case, the amanitin is diluted by "D4" (1:10000) and 5-40 drops are given. Sometimes homeopathy includes a hidden dilution factor (a drop flicked into a bottle of sugar pills) but it is not guaranteed and I don't think we should count on it here. So there is possibility of a beneficial effect, but the utmost skepticism is required! Wnt (talk) 20:14, 30 July 2013 (UTC)[reply]
Right, be careful about identifying homeopathy with homeopathic dilution. Homeopathic dilution is the aspect of homeopathy that probably gets the most attention, but it isn't the whole discipline, just one practice. Homeopathy is an entire framework of (mostly if not entirely bogus) theory and (possibly occasionally effective and/or dangerous) practice. --Trovatore (talk) 20:23, 30 July 2013 (UTC)[reply]
Symbol/syntax for denoting an (intentional) error?
Hi, in the Orders of magnitude (speed) article, I wanted to highlight the fact that the speed measured in the Faster-than-light neutrino anomaly is "incorrect". Is there a symbol or syntax I could use to denote an intentionally incorrect value or item (in science)? I thought I could use italic for the values, or maybe an asterisk or something similar, but I frankly have no idea. Thanks in advance. --CesarFelipe (talk) 21:44, 30 July 2013 (UTC)[reply]
Just asterisk/footnote or other text annotation would be good. It's a disproven or contradicted statement, not a typo that would merit "sic". DMacks (talk) 22:00, 30 July 2013 (UTC)[reply]
MPN / CFU relationship
Hi Guys,
Is there a direct relationship or conversion factor for comparing fecal coliform results by enzyme reaction reported as MPN and results by membrane filtration reported as CFU? Thanks! — Preceding unsigned comment added by 148.66.156.178 (talk) 22:02, 30 July 2013 (UTC)[reply]
No simple conversion seems possible but models have been proposed to compare results of the two methods. Also results may depend on circumstances, for example a Korean study found MPN for e. coli larger than CFU, except in winter. On the other hand, enterococci were lower in MPN than in CFU.
Maybe this helps: "Modeling the relationship between most probable number (MPN) and colony-forming unit (CFU) estimates of fecal coliform concentration." Gronewold AD, Wolpert RL. pdf download. Ssscienccce (talk) 10:51, 31 July 2013 (UTC)[reply]
July 31
Visibility of the New Moon
The Wikipedia entry for New Moon states 'The Moon is not normally visible at this time except when it is seen in silhouette during a solar eclipse'. While driving from California to New York on Interstate 10 I pulled over early in the morning in the desert to stretch my legs. I look up at a perfectly clear night sky and saw the entire surface of the moon un-illuminated. What phase of the moon was I viewing? — Preceding unsigned comment added by 66.65.43.46 (talk) 00:56, 31 July 2013 (UTC)[reply]
The Muslims often take good pictures of very young crescent moons for Ramadan
Yep, that's the only possibility. In addition to what you've quoted regarding the new moon, it's also only above the horizon during daylight hours (layman's definition, not astronomical). It's impossible to see a new moon, or see a dark disc where a new moon would be, at night. — Lomn01:38, 31 July 2013 (UTC)[reply]
No! No! A new moon can be above the horizon shortly after dusk or before dawn, and it is illuminated by earthlight, which is easily bright enough to make it visible in good conditions. (Earthlight is over ten times as bright as moonlight.) Looie496 (talk) 02:32, 31 July 2013 (UTC)[reply]
In an area with high levels of nighttime illumination (such as in large cities), it's also possible for a new moon to be darker than the night sky, and therefore visible as a dark circle. 24.23.196.85 (talk) 02:54, 31 July 2013 (UTC)[reply]
Skyglow over cities is caused by light that originates on Earth, and is then scattered back down from within the Earth's atmosphere. The Moon cannot obstruct this light, because the light goes nowhere near the Moon. So, no dark spot in your city sky. TenOfAllTrades(talk) 03:26, 31 July 2013 (UTC)[reply]
I have seen a full lunar eclipse (as well as a full solar eclipse) and it was not a eclipse. I was in the high New Mexico desert at the time, late November perhaps 4 or 5 AM, and it seemed to me that it was being illuminated by the Earth's albedo. I have posed the question because there seems to be differing views as to what I saw is in fact possible. — Preceding unsigned comment added by 66.65.43.46 (talk) 02:54, 31 July 2013 (UTC)[reply]
That's not a bad hypothesis, and it's one I've thought about too. Unfortunately, the sky at twilight – after the Sun has set, but before the sky is fully dark – is bright enough to completely wash out the Earth-lit new Moon. The Moon can't get much further than about 7 degrees from the Sun before it starts to show a visible crescent, which means that even under ideal conditions you're just at the edge of civil twilight. (Go ahead and search—you'll find that there just aren't any images of a new, non-crescent moon at twilight. I will also note, for the record, that I have another reason to hate Stephenie Meyers—her silly vampire books and movies hopelessly contaminate Google image searches for new moon and twilight.) TenOfAllTrades(talk) 03:34, 31 July 2013 (UTC)[reply]
A quick question or two: (1) it was in the early morning - was it just before sunrise? (2) Was the Moon in the same direction as the rising Sun? A new Moon would be close to where the Sun would rise. In a lunar eclipse, the Moon would be opposite the Sun. Bubba73You talkin' to me?04:01, 31 July 2013 (UTC)[reply]
List of lunar eclipses here. In 2012 there was a penumbral on nov 28, mainly visible between 14:00 and 15:00 UT, that would be between 7:00 and 8:00 local time? Ssscienccce (talk) 09:50, 31 July 2013 (UTC)[reply]
I don't believe that one can see a perfectly unsunlit moon for the reasons given above, but it's certainly possible to see a very thin crescent moon with the rest of its visible disc faintly illuminated by earthshine. This has traditionally been referred to as "seeing the new moon in the old moon's arms", as in the ballad of Sir Patrick Spens. There's a good image here. Deor (talk) 12:46, 31 July 2013 (UTC)[reply]
To somehow detect Everett's "other worlds": some new "telescope?"
Hi, I'm certainly a layman, but i was thinking that if Hugh Everett's Many worlds interpretation does turn out to be true, then physics needs a new "telescope", that is, an instrument as radical as the telescope(and microscope too) were in the 1600s, in order to "see", that is, somehow detect the other worlds. I'm wondering if the new "telescopes" could be experiments at temperatures much colder than they've done yet. I mean, I have heard of physicists getting down to .1 degree kelvin, but maybe the clues could be there at .00001 kelvin, or even 10^(-40) kelvin? And what are the hopes right now for getting these much lower temps? Thanks76.218.104.120 (talk) 09:29, 31 July 2013 (UTC)[reply]
Please forgive me for not checking, wikipedia says they are down to a few hundred nanokelvins now. I got up late at night with an idea that was exciting to me and i didn't read up. But the general direction of my first and second questions still stand, but they are much closer to 10^-10 than I thought. So, my second question can be revised to what would be the outlook for even lower temps, for example 10^-10000 k? Thanks again.76.218.104.120 (talk) 09:50, 31 July 2013 (UTC)[reply]
The problem is that the different interpretations of quantum mechanics yields the same predictions for the outcome fo experiments. The only way to test the Many Worlds Interpretation (MWI) is to think of some very unusual experimental set-up that would yield a different prediction. As long as you have the usual type of quantum experiments where the observer is treated classically, you will not see any difference. The only way to get a difference is to think of an experiment where the observer is itself going to participate in a non-trivial way in the epxeriment where a de-facto calssical description of the observer (due to quantum decoherence) is not a valid assumption anymore.
David Deutsch came up with such a thought experiment back in 1985. This involves an observer who can be perfectly isolated from the environment. He prepares an electron with its spin polarized in the x-direction. He then measures the z-component of the spin. This measurement can yield two outcomes with equal probability, spin up or spin down. In the MWI both measurement outcomes really exist, but you'll find yourself in one of them. So, how would you know that the other outcome also exists? Well, since you are perfectly isolated, you could in principle reverse all the steps made in the measurement process, ending up with the electron spin polarized in the x-direction. However, you would not know anymore that you had previously measured the spin of the electron, as your memories would also have to be reset in that reversal process.
What you can do is keep in your memory is the information that you did perform the measurement, the transformation back to the initial state is then still a unitary transform (an easy exercise to check) and thus relaizable in principle. You cannot keep the result of the measurement; because this is different in the two branches where the measurement result is different, you won't get the electron back in its original spin state polarized in the x-direction. This verifies the physical existence of the two different measurement outcomes, because both are needed to get back to the initial state.
Of course, this experiment as originally proposed by Deutsch is impossible to perform in practice, however a few years later people came up with the concept of a quantum computer. If very large scale quantum computing ever becomes a reality, you can contemplate simulating an artificially intelligent entity inside a quantum computer and that entity could then perform Deutsch' experiment. Count Iblis (talk) 13:45, 31 July 2013 (UTC)[reply]
Is the moon in the sky exactly as often during the day as during the night?
That is, if I timed how long the moon was in the sky during the day, and how long it was in the sky during the night, over a long enough period, would I get identical answers?
Be as picky as you want about details, and do please try to quantify your answers: I'm interested in whether the various skewed orbits and other factors make a difference. 86.164.26.17 (talk) 10:02, 31 July 2013 (UTC)[reply]
You should to a first approximation. Now, you see more of it at night because the full moon is at it's zenith at local midnight, while the new moon is at zenith at local noon. But averaged out over an arbitrarily long time, there are no stretches of any 24 hour cycle when the moon is in "the sky" more than at other times. --Jayron3211:16, 31 July 2013 (UTC)[reply]
You never get things like that exactly the same if you measure closely enough. Lets just consider the moon to be a point to simplify things. The sun isn't a point so that makes days longer than nights. Refraction makes the light go further so that also makes days longer. Discounting those two factors the earth has a width so with a point sun the day is a little bit shorter than the night if it rotates evenly. The earth goes in an ellipse round the sun which I believe means the northern hemisphere gets slightly more daylight as the earth is farther away from the sun and goes slower when it is tilted towards the sun. Discounting the tilt as well we have the problem of the speed of the moon as it orbits the earth, I think it should be going slightly faster when it is on the sun side so that would mean it is more often out at night. And I haven't started on earthquakes or the other planets. Anyway the most important factor discounting refraction is the width of the sun I believe so you would have it more often in the day than the night. Dmcq (talk) 13:27, 31 July 2013 (UTC)[reply]
Seems that Dmcq has already given the reasons for the two not being equal: sunrise and sunset are defined as the moment the upper edge of the sun is at the horizon, and atmospheric refraction. When considering twilight, the difference becomes larger, see definitions here. I calculated the total time the moon would be visible at 0°W 20°N during 2013, using the tables you can generate here; results: moon visible during the day for 134937 minutes (2248.95 hours or 93.706 days), and during the night for 128363 minutes (2139.38 hours or 89.14 days). Some other numbers: total daytime 184.68 days, nighttime 180.31 days; moon for 182.85 days, no moon for 182.15 days Ssscienccce (talk) 13:40, 31 July 2013 (UTC)[reply]
Same calculation for 0°W 60°S gave: moon during the day: 132561 minutes, during the night: 125251 minutes. (92.06 vs 86.98 days) Total daytime 185.18, nighttime 179.82 days; moon for 179.04 days, no moon 185.96 days total.
Are you counting the new moon? Because although the moon is in the sky during the day, a human eye will not be able to see it. Also the thin crescent moons are not that obvious during the day, so are you counting that as being in the sky, as you have to look carefully to notice it. So the moon is certainly less noticeable in the day than at night. guess you knew that! Graeme Bartlett (talk) 21:51, 31 July 2013 (UTC)[reply]
I am somewhat puzzled by the advent of solar-power generators capable of producing electricity from the infrared part of the spectrum. For starters, I've heard that they can generate power in all kinds of weather—even at night! Also, I've heard that the only reason we (end users) don't have them is because they produce electricity at about 10 THz, making them totally useless with which to run our appliances, which typically run at approximately 50 to 60 Hz.
I have two quick questions; namely, how exactly can a solar cell generate power at night? Does it utilize infrared light, from the Sun, absorbed by Earth's atmosphere during the daylight hours, or perhaps, infrared radiation arriving here from more-distant stars? Also, (and I apologize in advance, if this is stupid question) just how impractical and unwieldy would it be to just nix all of 50-60 Hz electronics and replace them with 10 THz ones—the way we did with regular gasoline, and are currently doing with analog television?
Would be the cost greatly outweigh the benefit of living free of power plants, dams, coal mines, transmission lines, et al? Thank you.Pine (talk) 20:57, 31 July 2013 (UTC)[reply]
Sounds unlikely. Do you have a source? Maybe it got a bit distorted in transmission or in recall. The frequency you say it produces electricity is only a little lower than for infrared, and certainly it could not be transmitted over power lines. See Terahertz radiation, Infrared. Photovoltaic Solar cells produce direct current. Mirror arrays can concentrate sunlight on a boiler and probably could produce power from infrared if there were enough of it hitting the mirrors, but residual heat in the atmosphere, or moonlight, or starlight would not be enough. And if the mirror array did boil water, then the generator would produce plain old electricity unless you somehow had a microwave generator or like which ran on heat, or the concentrated black electromagnetic energy just heated something up so it radiated black body radiation. Edison (talk) 21:34, 31 July 2013 (UTC)[reply]
Solar cells can respond to near infrared with wavelengths shorter than about 1 micron. You don't get much of this at night at all, as it comes from the sun like normal visible light. When you get the longer wavelength thermal infrared, solar cells do not turn this into electricity in any significant amount. THey are just as likely to do the reverse process and radiate thermal infrared. The issue here is entropy. Note there are other techniques such as solar thermal power. Graeme Bartlett (talk) 21:42, 31 July 2013 (UTC)[reply]
I found an IEEE article from 2012 which discusses Terahertz power generation. It generally has less than 1% conversion efficiency in generating the Terahertz output, and such EM radiation is reduced severely as it travels through the atmosphere. Edison (talk) 22:05, 31 July 2013 (UTC)[reply]
How do you determine if a bird is fat?
I was just looking at this photo and it made me think. How do you determine the difference between a really fat bird and a bird that's just very fluffy? Is there a reliable thing to look for? --87.114.229.248 (talk) 22:06, 31 July 2013 (UTC)--87.114.229.248 (talk) 22:06, 31 July 2013 (UTC)[reply]
Resolved
