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June 8

who edit Wikipedia in a sensitive subject... who is right ??

Hi: According to Wikipedia in English, in the page about "omega 3" they said that are really controversial the good effects that brings omega 3 in any mammal... if fact they cite research that say that is not a really big help for the human body.

In Wikipedia in Spanish, they say exactly the opposite, omega 3 is really a big help for the human body. Who is right ?? who edit this articles ?? what about if I am a mediocre doctor who just write something that I learned 100 years ago ?? to whom I have to believe is a kind of articles that involve a live risk ?? many thanks in advance. chau and sorry for my funny English — Preceding unsigned comment added by 186.2.50.237 (talk • contribs)

Because Wikipedia can be edited by anybody, you should always have doubt about what you see in a Wikipedia article. If you want to check, you should look at the sources that the article cites. If there are no sources, you should have a lot of doubt. Anyway, my understanding is that the English article is correct. (For convenience, our article is Omega-3 fatty acid. Looie496 (talk) 03:02, 8 June 2012 (UTC)

It's difficult to evaluate the claims in the Spanish article as instead of citing sources, they simply provided links, and those links are now broken. Someguy1221 (talk) 04:10, 8 June 2012 (UTC)

... and I hope that even mediocre doctors don't use Wikipedia as their medical text! Dbfirs 07:28, 8 June 2012 (UTC)

Spanish wikipedia has horrible administrators, some of them, revert all Ip's edits without reading them at all. Many requests are ignored. They take community decisions by votes instead of arguments.. though I have to admit that there is a lot more vandalism there than here. 65.49.68.173 (talk) 16:28, 8 June 2012 (UTC)

Thanks to all of you. chau — Preceding unsigned comment added by 186.2.50.237 (talk) 04:41, 9 June 2012 (UTC)

I think the English Wikipedia is not so reliable on these sorts of medicine related issues, because it gives far to much weight on the Institute of Medicine (IoM) reports; these reports are extremely conservative when it comes to accepting claims of benefit, while the burden of proof needed to include possible negative health effecs is extremely low. While this may be a good thing for compiling reports meant as advice to health care workers, what we need on Wikipedia is a balanced approached, one that gives equal weight to equally reliable evidence. Count Iblis (talk) 17:53, 8 June 2012 (UTC)

Helium dihydride cation

Why can't this exist? I'm talking about the species HeH₂²⁺, isoelectronic with the trihydrogen cation.--Jasper Deng (talk) 05:38, 8 June 2012 (UTC)

It's too unstable. HeH⁺ is already the strongest known acid, i.e. it is more likely to dump a hydrogen ion than any other compound ever discovered. There are a handful of funky looking sources claiming you can get HeH⁺ to accept a hydrogen atom, but not a hydride ion. My guess is that even if you could, the second hydride would be dumped almost immediately. Someguy1221 (talk) 05:47, 8 June 2012 (UTC)

Careful. The relevant unit would be a simple proton (H⁺), not "hydride" (H^–), which would instead give the neutral helium dihydride result. But the massive instability is certainly the key. Our helium hydride ion discusses the species mentioned by Someguy, and also evidence for and stability of HeH₂⁺ and others in this monocationic series. One could certainly so some ab initio calculations on HeH₂²⁺ to see what would be happening there. DMacks (talk) 14:22, 8 June 2012 (UTC)

Whoops, my bad. Thanks DMacks. Someguy1221 (talk) 18:16, 8 June 2012 (UTC)

And just to prove I'm not just making stuff up, HeH₂²⁺ (CAS #12519-50-5) has been studied in this manner, and is unstable in normal situations but is stable in the presence of high magnetic fields (like "surface of a neutron star")--see doi:10.1103/PhysRevA.81.042503. Note that all refs I found consider it as a chain--not sure if exactly linear in all cases, but anyway more structurally related to beryllium hydride than the trigonal trihydrogen cation you mention. DMacks (talk) 14:46, 8 June 2012 (UTC)

Water on the line

I work 50m away from a railway line in SE England. Yesterday, after a day of rain, when a train passed along the line, there was a spectacular noise followed by plumes of steam that stretched for a couple of hundred metres and reached well above the tree line. I've lived and worked next to this stretch of track for 17 years and have never seen anything like this before. I have not heard on the news that a train-load of people have been fried on the London to Dover line yesterday afternoon, so was the train acting like a Faraday cage and were the passengers in any danger? 83.104.128.107 (talk) 15:51, 8 June 2012 (UTC)

There's some missing info in your question. Is this an electric train, powered by an electrical rail ? If so, the electricity would want to go towards earth/ground, and standing water by the non-charged rail, in conjunction with electrical connections between the rails through the train's undercarriage, might have allowed that. There would be little "motivation" for the electricity to go into the passenger compartments, so a Faraday cage isn't necessary. StuRat (talk) 17:20, 8 June 2012 (UTC)

(@StuRat - Most of South East England is a 'third-rail' electrical pick-up system.)

At OP - Can you clarify more closely where you are in SE England and where this incident occured? Your description sounds like some kind of flashover effect, but I hadn't heard anything about such through my railfan contacts, and most modern trains have a specific trip so they don't produce the kind of effect you saw. Sfan00 IMG (talk) 15:49, 10 June 2012 (UTC)

Hello Sfan00, It was on the stretch of track between Faversham and Selling at around 16.40 on 14th June. The line crosses a farm where I saw it, I checked the train times and no journey in either direction seems to tie in, but there are often freight trains passing. I would love to know what the passengers/driver felt or saw! A friend did suggest that I might have seen a steam train, but there was no 'chugging' and there was no majestic view of it in the distance when I dashed up to the bridge. I am hoping that with more rain to come and the ground still saturated it might happen again this week. 80.176.84.184 (talk) 17:28, 10 June 2012 (UTC)

As I said, I hadn't heard anything, you could try asking on the uk.railway newsgroup :) Sfan00 IMG (talk) 18:25, 10 June 2012 (UTC)

@80.176, you said that this happened on June 14, which, unless you are talking about last year, will not occur for another four days. What did you intend? Falconus^{p t c} 21:03, 10 June 2012 (UTC)

D'oh - I meant the 7th! Looked at the wrong Thursday on the calendar! Having been a long-term stalker of the ref desks I am absolutely appalled at the lack of clarity of my original question and subsequent blunders (let alone somehow first posting the question on the Language Desk!). I think I'll stick to reading and not asking... but I would like to say what an amazing resource the Ref Desks are and it is incredible that we have such informed and interesting people to probe with our musings! 83.104.128.107 (talk) 08:34, 11 June 2012 (UTC)

Yes, the most difficult part in asking a question here is generalizing it so everyone around the world understands it. To you, a "train" is a third rail electrical train, while to me in Detroit, it's a diesel-electric train, with the third-rail system reserved for subways, which you call "the underground" (which to me means a covert organization) or "the tube" (which to me means television), etc. But learning to explicitly state all of your assumptions is a valuable lesson, in any event. StuRat (talk) 16:32, 11 June 2012 (UTC)

What cause brownian movement?

What makes gas molecules extert brownian motion ..? is it related to the electron shell?, proton/neutron core? quarks? the energy at least intermediately has to be stored somewhere. Electron9 (talk) 16:15, 8 June 2012 (UTC)

The energy isn't stored anywhere. It's a manifestation of the thermal kinetic energy of the individual molecules impacting the larger object.

In simpler language, you should know that temperature is really a measure of the average kinetic energy of the individual molecules in a substance. For solids this is just molecules vibrating, but in a fluid like water or air the molecules are free to move around each other. So when you put a small enough particle in a fluid and look at it under a microscope, only a few molecules are going to be hitting the (relatively) larger object at any one time, and the odds are good that they won't be hitting symmetrically. Therefore they impart a small amount of net kinetic energy to the larger object, and it moves slightly. It is a random process, so averaged over infinite time there will be no net motion, but it can still jiggle the object around a great deal over short time scales. -RunningOnBrains^(talk) 16:39, 8 June 2012 (UTC)

Short version of the above - Heat. Roger (talk) 16:51, 8 June 2012 (UTC)

No, it's not heat. Temperature and heat are very different concepts. 203.27.72.5 (talk) 07:22, 9 June 2012 (UTC)

If brownian motion is plainly a manifestation of thermal vibration. What parts of the atom is vibrating? electron shell? nuclei? quarks? Electron9 (talk) 17:44, 8 June 2012 (UTC)

The entire atom or molecule. StuRat (talk) 17:47, 8 June 2012 (UTC)

Further: Any macroscopic analogy is going to be at least partly incorrect, because we are dealing with individual molecules and atoms here, which are subject to quantum mechanics. However, I will do my best.

In a gas or liquid, molecules are never still. They are constantly moving, at a velocity that can be predicted by the temperature of the gas. As a 2-dimensional analogy, imagine a whole bunch of billiard balls flying around a giant pool table, with no friction to slow them down. They will stay at a constant velocity until they hit either the walls of the table or another billiard ball, and then they will go off at a different velocity in a different direction. It is very nicely illustrated by the image I posted at right: You have to remember that there is no friction at this scale, so unless the material is cooled or warmed, the average thermal velocity of the molecules is going to stay the same. Now if you introduced a larger object to the table, say a bowling ball, it's going to be jostled by the constant collisions, and so if you were standing far enough away that you could only see the bowling ball, it would appear to be vibrating randomly, just like in Brownian motion. -RunningOnBrains^(talk) 17:59, 8 June 2012 (UTC)

Thanks! (maybe should be added to the article), does thermal vibration cause the nuclei and electron shell to vary their distance to each other? ie will the atom deform in some way like air does for sound? Electron9 (talk) 19:43, 8 June 2012 (UTC)

Not really, but here we're really getting to the point where macroscopic analogies break down, because electrons really aren't in one place at any time, and electron shells aren't really a physical object: see electron cloud. I am also probably extremely unqualified to speculate on the exact quantum mechanical processes which take place when two atoms collide; it probably depends strongly on which atoms we're talking about. But it can be safely described as a purely elastic collision for the point of describing Brownian motion.-RunningOnBrains^(talk) 20:09, 8 June 2012 (UTC)

Actually, I disagree with Runningonbrains here. Absolutely, thermal vibration causes variations in the position of the different atomic particles relative to one another. As Runningonbrains rightly pointed out the (average) velocity of the particles that make up an object can be predicted from the object's temperature. The velocities of the individual particles vary according to the Maxwell-Boltzmann distribution. The deformation of the atoms or molecules leads to a higher energy state i.e. electrons repel one another altering the shape of their orbitals. These high energy states are relatively unstable and if a more stable configuration can be assumed, then it will be. Many molecules that decompose do so more quickly at higher temperatures. This can be modelled by saying that those molecules that have high energies are deformed by the motion of their constituent particles and assume a more stable state by breaking chemical bonds. The number of molecules with high energies is a function of temperature as predicted by the boltzmann distribution. Electrons can even be heated so much that they leave the atom all together, as in a thermal plasma. 203.27.72.5 (talk) 07:49, 9 June 2012 (UTC)

That would mean that the atomic nuclei (protons-neutrons) will deform in a plasma just like electrons does at a lower temperature? Electron9 (talk) 08:27, 9 June 2012 (UTC)

Somewhat more important in practice is that some thermal energy is stored by the rotation of molecules at temperatures above about 600 -700 K, (ie angular momentum, as well is the linear momentum mentioned above) and by the lengthening of inter-atomic bonds with increasing temperature, which further increases the fraction of heat energy stored in rotation. This is evidenced by the fact that noble gasses show specific heat independent of temperature, but large molecules have considerable variation in specific heat throughout the measureable temperature range. Only the fraction of heat energy stored in linear momentum drives brownian motion. Wickwack124.178.139.104 (talk) 11:42, 9 June 2012 (UTC)

A side question, would a tube between two vessels with a diameter just slightly larger than a single atom (or molecule) and a length significantly less than the average brownian motion distance. And funnel on one side make more atoms to move to one side than the other? especially when the mol/m³ is low. Electron9 (talk) 19:43, 8 June 2012 (UTC)

No. You're proposing a variation of Maxwell's demon, which violates the second law of thermodynamics (in this case, by creating a pressure and temperature gradient. The Brownian ratchet may also be of interest. — Lomn 20:01, 8 June 2012 (UTC)

Radiation

Do objects which absorb radiation reemit it? How does this work? — Preceding unsigned comment added by 176.250.228.38 (talk)

It depends a lot on what type of radiation you're talking about. All kinds of matter absorb and emit all kinds of radiation all the time. It is a continuous process. --Jayron32 18:54, 8 June 2012 (UTC)

Most things which absorb radiation with then give off radiation as well, although it may very well be a different form of radiation. For example, if you shine visible light (one form of radiation) onto a black object, it will radiate the energy back out, not as visible light, but as infrared light/heat. StuRat (talk) 19:02, 8 June 2012 (UTC)

There are chemical processes in which the electron shells surroundng atoms absorb radiation at one wavelength and them re-eimit it at another wavelength - this is called fluorescence. And there are processes in which radiation is absorbed and its energy is converted into a different form - see pair production, photoelectrochemical processes, photosynthesis, photoelectric effect, photovoltaic effect, concentrated solar power. Gandalf61 (talk) 10:18, 9 June 2012 (UTC)

Rules of science

Often in Biology, everything doesn't follow the known rule we've established. Can the same thing be said for physics & chemistry? 176.250.228.38 (talk) 20:00, 8 June 2012 (UTC)

Yes. I know in chemistry, there are lots of experiments that don't do what a hypothesis (based on literature precedent for similar experiments) says they "should"--exceptionally low yield, different geometric form, different part of a complex molecule reacts, nothing happens at all, or a totally different reaction occurs instead. There are probably a near-infinite number of combinations of experiments that would follow some not-yet-detected pattern, or an observed pattern that does not have any known underlying cause, where the data is all "out there" but nobody has even looked yet (i.e., to explain apparently random variations in yield, etc.). DMacks (talk) 20:06, 8 June 2012 (UTC)

This answer isn't exactly right. The answer to your question really depends on what question you're asking.

If you're asking if there are exceptions to the established laws of physics and chemistry, the answer is no. You're not going to get gravity to be different from one experiment to the next, and you're not going to get sodium and chlorine to react and form anything other than sodium chloride. They are called "Laws" for a reason.

However, if you're asking whether or not experiments can produce unexpected results, the answer is most definitely yes. You can never (probably) have an experiment that is completely controlled, where you know every bit of information about the initial conditions. You can get a different yield than you were expecting from a reaction, but this would be due to some contaminant you didn't know about, or some environmental factor that was different like temperature or moisture in the air. Maybe the yield is highly sensitive to the initial ratio of reactants, and the expiriment Maybe even some unforeseen quantum mechanical effect could change the expected results, if it's an especially complicated chemical reaction. Or, incredibly rarely, maybe your physics experiment has discovered a whole new particle or effect.

However, if you do exactly the same experiment every time, you will get exactly the same result. The reason that biology is such a messy science with many unexpected results is that there are just too many unknown factors to take them all into account; an organism is unimaginably complex, certainly not as easy to describe with simple laws as E=mc² and F=ma.-RunningOnBrains^(talk) 20:22, 8 June 2012 (UTC)

Another important point is that science is neither a set of laws, nor a primarily deductive enterprise. Science is a method for inductive reasoning. --Stephan Schulz (talk) 20:31, 8 June 2012 (UTC)

Running, your second third last paragraph sounds like determinism, which is not a necessary component of science, and indeed is contrary to some widely held interpretations of quantum mechanics. --Trovatore (talk) 20:36, 8 June 2012 (UTC)

Deterministic probabilities ? :-) Electron9 (talk) 21:52, 8 June 2012 (UTC)

I guess I interpreted the OP's question as a macroscopic sort of thing. You're right in that there is always some QM-related uncertainty (even if miniscule at macroscopic levels); but those would not be unexpected to an experimenter, and certainly follows the "known rule" as the OP put it. I would say that my final sentence above really sums up my point; maybe I shouldve just stuck to that :) -RunningOnBrains^(talk) 22:46, 8 June 2012 (UTC)

It depends a lot on what the OP means by "rules". Most scientific "rules" are actually models of some sort, and all models are approximations of reality, so there will always be real examples that lie outside of the predictions of the model. --Jayron32 03:04, 9 June 2012 (UTC)

As a former analogue electronics engineer, now involved in certain aspects of chemistry, I must say there is an immense difference between electronics and chemistry at an engineering/design level. In electronics, everything is ultimatey based on a limitted number of component parts - resistors, capacitors, inductors, conductors, and active devices (transistors etc). The behavior of these devices behave according to well established simple laws - so simple that a 12 year old can, if sufficiently interested, design a stereo (I did when I was 12). Real parts don't exactly follow these laws, but they are close enough. Understand those laws properly, and you can understand anything in analogue ectronics.

Chemistry is very different. The "almost fundamental" component parts of chemistry are the atoms. The behaviour of atoms in any situation can (in theory) be predicted by quantum mechanics. In practice, that's just too hard, so the "laws" of chemical engineering are fortutious theories like the kinetic theory of gasses, and the theories of chemical kinetics with regard to reaction rates. These theories have so many gaps, exceptions, an approximations, that to former electronic engineer, it is very frustrating. To calculate current in a circuit, I can alway do that to at least 3 figures accuracy - 6 figures, if I need it, is not hard. To calculate the rate of a chemical reaction, chemists are doing well if they get within the correct order of magnitude. As a further example, the kinetic theory of gasses pupports to give an understanding of specific heat (thermal capacity) and how it varies with temperature. It accuately gives the specific heat for noble gasses (but who cares), and is roughly right for low valency atoms, but seems to be very inaccurate otherwise.

In electrical enginering, if say, a power company wants a $100M EHV transmission line and distribution system, the engineers do some caculations, order the materials, get it built, and it will work just fine. In chemical engineering, if a company wants a new $100M processing plant, the engineers do some calculations, scour the world for somebody who has done something like it, tweak the calculations, then build a pilot plant and muck about with it untill it works. Then with that experience, do more calcs, scale it up, then order all the materials etc and build the BIG ONE. Then sometimes find out it doesn't work at all well, and $100M has been wasted.

In short, chemistry must conform to valid scientific laws, but those laws are too difficult, so in practice, rough semi-empirical approximations are used. And things don't always go according to plan.

If you understand "basics" like chemical kinetics (and that is not at all easy), you still may not be able to understand real world applications. By "understanding" I mean able to calculate and preduct accurately what will happen.

Ratbone124.182.45.112 (talk) 03:34, 9 June 2012 (UTC)

It's a subtle and important point - every science has different objectives to deal with different topic matter. Physics sets hard-and-fast absolute rules of what is impossible. Chemistry is more about figuring out what is practical. And biology is a science of the possible, where every "rule" has an exception. The continuum continues further in disciplines like psychology, where there is doubt if it is even a science, and perhaps, even to the tropes of fictional writing. Wnt (talk) 15:26, 9 June 2012 (UTC)

cephalosporin

are there any once a day oral cephalosporin antibiotics?--Wrk678 (talk) 22:56, 8 June 2012 (UTC)

It depends what it's for. Apparently some of the cephalosporins taken PO are administered qd, such as cefpodoxime (Vantin) for otitis media. DRosenbach ^{(Talk | Contribs)} 13:04, 12 June 2012 (UTC)

Mercuroketones

Are any of these possible?

Can any mercuroketones (compounds containing a C=Hg double bond), such as those in the image, exist? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 22:58, 8 June 2012 (UTC)

I don't really think so, because good luck getting mercury to hybridize its s and/or d orbitals to allow a covalent bond.--Jasper Deng (talk) 03:18, 9 June 2012 (UTC)

Then explain organomercury compounds. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 04:38, 10 June 2012 (UTC)

These either all have single bonds to Hg or Hg is part of a cation.--Jasper Deng (talk) 04:54, 10 June 2012 (UTC)

In your earlier post, you implied that the problem was with trying to get mercury to form covalent bonds. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 23:48, 12 June 2012 (UTC)

Look at Oxymercuration reaction in which a cyclic mercurinium ion is formed, Hg has lost an electron and has three bonds, two with adjacent carbons. Still no double bond though, and I could find no evidence of it on google either. Graeme Bartlett (talk) 07:28, 9 June 2012 (UTC)

Mercury can definitely form double bonds with oxygen, I don't know about carbon though. Plasmic Physics (talk) 12:02, 9 June 2012 (UTC)

"Mercuroketone" is probably not a good term for it, since mercury not very electronegative. More likely "methylene mercury" (or other coordination/description of the carbon part, as usual for ligands on metals) or a "mercury carbene" (161 google hits) complex or something like that. Hg⁺=CH₂ (apparent Hg(III) species), CAS#1234574-43-6, has been studied theoretically. But I'm also seeing lots of examples where what your type of connectivity is written as a Hg(II) ylide, for example, Hg⁺–C^–H₂, rather than a Hg=C double bond. As for some of the coordination examples, you have to be careful not to exceed an electron-count of 18 (the transition-metal analog to the octet rule used in main-group elements). DMacks (talk) 14:42, 9 June 2012 (UTC)

Mmm, can't comment on the specific molecules here, but I remember from my chemistry undergraduate that transition metals can form some pretty funky organometallics from time to time. In principle these molecules look plausible. LukeSurl ^{t c} 09:34, 10 June 2012 (UTC)

Have you yet attempted to model the proposed molecules? Nevermind the mercurylidenemethylidene group, the angle strain of tetragonal carbons alone, is enough to destablise most of these molecules. Plasmic Physics (talk) 10:13, 10 June 2012 (UTC)

Oooh yes, that triangular arrangement in 4 looks especially painful. I'm assuming the question is mostly about the idea of the C=Hg bond rather than the carbon skeletons I'm guessing the OP has just made up. 10:30, 10 June 2012 (UTC)

What about 1, 2, 6, and 7? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:28, 12 June 2012 (UTC)

How should the carbon be modified to form a kinetically stable stable bond? Perhaps, a persistent carbene? Plasmic Physics (talk) 13:58, 10 June 2012 (UTC)

June 9

"35dB-90dB[μV]" equal to 56 - 31000 mV ?

On the page "How to use the booster." it is said that "35dB-90dB[μV]" is the necessary voltage level for a 75 Ω antenna signal. Is that equalient to 56 - 31000 mV ..? Electron9 (talk) 01:54, 9 June 2012 (UTC)

No, you are 3 orders of magnitude out. To calculate db[μV] take the Log of the voltage in microvolats and multiply by 20, thus 56 mV corresponds to 95 dB[μV] and 31000 mV (31V) corresponds to 150 dB[μV]. 35 dB[μV] corresponds to 56 μV. 90 dB[μV] corresponds to 31.6 mV. However, 35 dB is rather high for the required input at the teminals of a TV set. A modern analogue TV should get a good picture with 20 dB[μV] or even less. 35 db[μV] would be good at the input to the antenna distribution cable system in a high rise building, where there is significant loss in the cable runs and in the splitters. A digital TV should in theory do rather better but in practice you need to allow a good margin to avoid dropouts and friezes. Keit120.145.6.122 (talk) 06:36, 9 June 2012 (UTC)

Like I suspected then, at least the multiplier 20 was correct. Btw, do you have any sources regarding the 20 dB[μV] level? maybe there's even a general difference between analog (CVBS) and digital (DVB) in regard to minimal signal strength? Electron9 (talk) 06:49, 9 June 2012 (UTC)

I answered from memory. However, a quick web search turned up this paper (as well as a lot of useless chat rooms about TV!), which seems to cover things quite well: http://www.eecs.berkeley.edu/~sahai/Presentations/Dyspan_2005_tutorial_part_I.pdf. On page 82 it gives the good picture minimum level for a digital TV as -85 dBm (dBm is an impedance-independent measure referenced to 1 mW). This corresponds to 15.4 μV across 75 Ω, i.e., 23 dBμV. Keit120.145.6.122 (talk) 11:03, 9 June 2012 (UTC)

Why is the oldest person always 114 years old?

For some years now, every time the allegedly oldest person in the world dies, the person's age has been reported to be 114. Just today we saw this item, saying this woman was the oldest person in Europe, and she died today at the age of 114. Why always that same age and never 113 or 115? Michael Hardy (talk) 02:38, 9 June 2012 (UTC)

It's often 115, that number you saw was just for Europe. I would argue it's simply a statistical issue. If you look at the US Social Security Administration's most recent actuarial table, they calculate the probability of making it from 115 to 116 to be only 25%. And making it from 114 to 115 is only a measly four percentage points better. So I would look at that and say that you start with a fixed population maximum of people born in 1897, and have that population experience greater-than-exponential decay from age 10 onward (it is less-than-exponential prior to that). The reason the "oldest person at the moment" is almost always 115 is that the decay function, although a bit noisy at those ages, would predict less than one survivor for all ages past 114. You'll see that creep up in the future as the starting population for each given year is increasing, as is post-adulthood life expectancy. Someguy1221 (talk) 02:53, 9 June 2012 (UTC)

I have seen data on that the maximum life expectancy is 130-150 years for a human, so 114 years is getting close. And thus the deterioration of the body is likely becoming exponential. Electron9 (talk) 03:02, 9 June 2012 (UTC)

Now, modifying the human might allow indefinite lifespan, like genetically rebuilding it and adding repair nanobots. So I'm assuming they're talking about without genetic engineering, cyborgization, or reanimation. (perfect preservation already existing, in the form of liquid nitrogen). So how do they propose living 27.5 years over the record? Calorie restriction? Sleeping through most of it?(/coma/hibernation/near death) Hysterectomy? Sagittarian Milky Way (talk) 18:54, 9 June 2012 (UTC)

You must be a young whipper snapper if you don't remember la chẻre Jeanne Calment. See list of the oldest verified people. μηδείς (talk) 03:15, 9 June 2012 (UTC)

She was a pistol. Her comments about Van Gogh are especially funny. ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:25, 9 June 2012 (UTC)

I guess the disagreement is probably due to low sample size, but this study suggests that mortality per year (i.e. chance of dying in any given year of life if you live that long) is 50% from age 110-115, and they speculate that that number may even increase beyond this age. -RunningOnBrains^(talk) 04:41, 9 June 2012 (UTC)

The small sample sizes are, indeed, a problem. The mortality tables I use professionally are only based on actual data up to age 95. After that, there just isn't enough data to get robust results, so they arbitrarily extrapolate from age 95 up to age 120, which they set as having a mortality rate of 100%. A detailed explanation of the process can be found here (be warned, it is quite technical). --Tango (talk) 16:31, 9 June 2012 (UTC)

So we will soon lose the last people who have lived in three centuries.Hayttom (talk) 04:25, 12 June 2012 (UTC)

The last ones in the lifetimes of anyone old enough to read this, at any rate. Come 2101, there'll be a whole pile of people born in the 1980s and 1990s who'll still be alive. -- ♬ Jack of Oz ♬ ^{[your turn]} 23:56, 12 June 2012 (UTC)

Baeyer's Reagent

Isn't Baeyer's reagent an alkaline solution of potassium permanganate?? The article on it states it to be neutral. — Preceding unsigned comment added by Roshan220195 (talk • contribs) 10:21, 9 June 2012 (UTC)

I guess that it is a matter of time. A fresh solution of potassium permanganate should be neutral. As time passed the permanganate decomposes slowly, as it does, the solution becomes more and more alkaline. Plasmic Physics (talk) 10:57, 9 June 2012 (UTC)

Plasmic Physics is right. To add the numbers to show why: A pure solution of potassium permanganate should be neutral, because potassium hydroxide is a strong base and permanganic acid is a strong acid, with a pKa of -2.5[1]. The salt of a strong base and a strong acid always forms a neutral solution. However, as the permanganate ion decomposes to the manganate ion, manganic acid has a pKa of about 7.4, making it a weak acid. --Jayron32 12:40, 9 June 2012 (UTC)

What is the decomposition mechanism for permanganate? O=[Mn-](=O)(=O)=O.O=[Mn-](=O)(=O)=O → O=[Mn-](=O)(=O)OO[Mn-](=O)(=O)=O → O=O.O=[Mn-](=O)=O.O=[Mn-](=O)=O ? Plasmic Physics (talk) 13:17, 9 June 2012 (UTC)

It isn't really decomposition, it's oxidation: The permanganate will oxidize just about anything, producing manganate and some sort of oxide, or elemental oxygen. The manganate will spontaneously disproportionate to permanganate and managanese dioxide, so given any trace reductant, there should develop an equilibrium between manganate, permanganate, and manganese dioxide which will account for the rising pH. --Jayron32 15:41, 9 June 2012 (UTC)

In that case, I don't think that it is in an equilibrium - it is not a closed system, oxygen escapes from the solution as singlet oxygen, which then decays into triplet oxygen. According to your process, I think that the reaction would be:

MnO⁻
₄ + 2 H
₂O + 3 e^- ↔ MnO
₂ + 4 HO⁻

4 MnO⁻
₄ + 4 HO⁻
→ 4 MnO²⁻
₄ + O
₂ + 2 H
₂O

3 MnO²⁻
₄ + 2 H
₂O ↔ 2 MnO⁻
₄ + MnO
₂ + 4 HO⁻

Plasmic Physics (talk) 06:47, 10 June 2012 (UTC)

Electronic eavesdropping

How does one find a bug that has been put in a house or car? Kittybrewster ☎ 11:40, 9 June 2012 (UTC)

Make a simple visual search of places where things can be quickly hidden first. Then, search using a radio scanner. Set the scanner sensitivity fairly low so you don't waste time on legit radio transmissions. See http://en.wikipedia.org/wiki/Scanner_(radio). Scanners are very good at picking simple radio transmitter bugs because when they get to the right frequency while auto scanning, you'll hear your own voice(s), and or the scanner will howl. Sometimes bugs placed by professional outfits or govt angencies are placed that display great inguinuity and cannot be found with scanners. Books have been written about this. I'm not trying to imply anything about your goodself, but be aware that there is a common pschological condition, often occuring in people who are otherwise normal, and sometimes brought on by stress, where people believe that they are being spied on, when they are not. Sometimes businessmen think they are just so darn good that the opposition must surely be spying on them. Very very few actually do so with bugs. Most industrial intelligence is obtained quite legally and simply by employing specialist researchers scanning documents in the public domain. I've learnt what I needed to know about competition by sharing a beer in a pub combined with monitoring employment adverts and press releases. Did you check our wiki article http://en.wikipedia.org/wiki/Covert_listening_device ? Also be aware that it is possible to write computer malware that activates the microphone in a laptop and monitor your voice, as well as your keystrokes, without you knowing, over the internet. Always have reputable computer security installed, and make sure your software firewall is set up with optimised port restrictions. If your PC has Vista, that's good - make sure that installing software requires an administrator password. Wickwack124.178.139.104 (talk) 11:55, 9 June 2012 (UTC)

Radio scanners will miss spread spectrum digital burst transmissions. Use a digital spectrum analyzer to find bugs. As for computers, disable recording devices like microphone and webcams physically and audit software. If you use any Microsoft software your computer is f-cked by design. Electron9 (talk) 18:53, 9 June 2012 (UTC)

You are correct, however a spectrum analyser requires a trained operator or electronic technician to operate it and interpret the display. Anybody with at least average intelligence can operate a radio scanner, and if there is an ordinary FM bug, the scanner operated in the same room as the bug will make the presence of the bug obvious. Wickwack124.178.60.220 (talk) 03:00, 10 June 2012 (UTC)

Note that radio scanners only work for devices which transmit radio signals. Other approaches are a hardwired bug, with the wires going outside the house to where somebody taps into them, a device that records and is retrieved later, or one that uses existing communication channels, like the phones lines, cable, wireless internet, cell phones, etc., to transmit signals. StuRat (talk) 18:52, 9 June 2012 (UTC)

Measuring vibrating glass is not that uncommon technique. Electron9 (talk) 18:55, 9 June 2012 (UTC)

Yes, that's another one, where you bounce a laser off the window at a shallow angle, and the refracted laser beam vibrates as the window vibrates, due to sounds in the room, allowing someone at the receiver to listen in. StuRat (talk) 23:45, 9 June 2012 (UTC)

I think that might be an urban myth, Stu. Being a bit of a nerd, I once made a laser communication system. I tried it out bouncing the beam off a window. I did detect sound from within the room to degree, and noise from inpinging wind to a greater degree, but as for understanding converstions, it was a dead loss. That doesn't mean someone else could not have succeeded, but it seems unlikely. The laser I used was a visible wavelength. A covert device would have to be infrared, which window glass doesn't reflect as well. Wickwack124.178.60.220 (talk) 03:10, 10 June 2012 (UTC)

Indeed, this technique is easy and cheap, as it can be done using a laser pointer, photodiode, and any old device with an audio-in jack. My physics advisor proposed that we make one as our senior project, even had all the materials ready, but I (stupidly) chose a different one. -RunningOnBrains^(talk) 03:14, 10 June 2012 (UTC)

Was your physics advisor talking about a voice link (ie from laser pointer to photodiode, nothing in between except air) or a bounce-off-glass eavsdropper system? I just tried my ear against the window (with my other ear plugged) while my lady was inside talking on her phone. All I could hear was the wind. Wickwack124.178.60.220 (talk) 03:25, 10 June 2012 (UTC)

Correct, you need a window. We were going to have a laser pointer, and a photodiode hooked up to the audio-in jack of a computer set up on one end of the hallway, with the laser pointer pointing at the glass on the other end of the hallway, reflecting back to the photodiode. Theoretically you could use other surfaces too, but windows are best because they are relatively rigid, often single-layer, and not weight-bearing so they are free to vibrate. In retrospect it probably would have been a bad idea for privacy reasons, since that glass was to my advisor's office :D -RunningOnBrains^(talk) 05:52, 10 June 2012 (UTC)

You would have been lucky to get that to work. The window vibration will not impart amplitude modulation, it will only impart phase modulation (http://en.wikipedia.org/wiki/Phase_modulation) via doppler effect. A photodiode connected to the audio input of something will not inherently recover the audio, unless you position it just right so that the beam is centred not quite on the diode. That would be extremely critical. In the system I tried, I used circuitry to pulse the laser diode on/off at 40 MHz. That allowed me to use a 40 + a bit MHz oscillator at the recive end to mix with the incomming beam signal and recover the doppler modulation imposed by the vibrating glass. Wickwack120.145.7.109 (talk) 06:32, 10 June 2012 (UTC)

Such eavesdropping would certainly be easier on a day with no wind. However, even on a windy day, it might be possible, if the wind causes different frequency vibrations than the sounds inside. The choice of window also might be critical, as some are more free to vibrate than others. I suspect a large, single pane would vibrate more, and a wooden or vinyl frame would tend to absorb vibrations more than a metal frame. StuRat (talk) 14:46, 11 June 2012 (UTC)

Wind noise is essentially white noise, covering all audible frequencies, and cannot be filtered out. Wickwack121.215.63.236 (talk) 00:50, 12 June 2012 (UTC)

Stalin's little thing worked. DriveByWire (talk) 00:05, 13 June 2012 (UTC)

The Nonlinear junction detector (NLJD) is a device that detects semiconductor devices (diodes/transistors), as used in the majority of listening devices. However, the heavy use of electronics in cars and homes nowdays might render the NLJD much less useful in 'bug' detection than before the widespread use of electronics in almost every electrical appliance. It might not be able to sort the bug from the many other electronic devices in most houses/cars. It would also, most likely, be unable to detect the Thing (listening device) mentioned by DriveByWire as the 'Thing' uses no semiconductors. See Also Technical surveillance counter-measures, which has a link to "Listening In: Electronic Eavesdropping in the Cold War Era" (PDF). (4.1 Mb). "Bugging" from Popular Science, Aug. 1987 may also bbe of interest, though well out of date. - 220 of ^Borg 16:48, 14 June 2012 (UTC)

Crepuscular rays

I saw some interesting atmospheric phenomena the other day (gallery is here; I've included all the shots I took, but the first, fourth, and fifth are the best). So, I'm assuming these are some kind of crepuscular rays, but the shape of them is what I'm curious about. I've seen crepuscular rays on countless occasions, of course, but the shortness of these is something new to me. The sky looked like someone had gotten crazy with a clone brush. What exactly is at work here? Are the rays only showing up in places where there's a certain amount of humidity/water vapour and then disappearing in the drier air below? These pictures were taken around 8am, facing (roughly) east; the sun is off-frame to the left. As you can see, there was a variety of clouds out that morning. I'm afraid all I had on me was my iPhone, so the quality is less than ideal. The images have not be manipulated in any way (other than the standard jpeg compression). Matt Deres (talk) 13:26, 9 June 2012 (UTC)

The weird thing is, sun rays (crepuscular rays) are lighter than the background, while yours appear darker. I might say they were smoke in the upper atmosphere blown into lines, but that doesn't explain why they would appear to radiate from the Sun. StuRat (talk) 18:47, 9 June 2012 (UTC)

What you are seeing is the shadow of the cloud. I suspect that you are right that there is a layer of smoke that is causing the short length of the rays. The normal cloud free condition of the sky would be for it to be as bright as the crepuscular rays, and the cloud shadow makes it darker. It would be easier to see a bright ray against a dark background than a slightly darkened sky against the bright sky, but the difference in this situation would be the smoke making higher scattering of light in that part of the sky, bringing up the contrast. Graeme Bartlett (talk) 23:15, 9 June 2012 (UTC)

I'm not sure what smoke has to do with it. I didn't see any smoke, nor smell any (not that I necessarily would, if it was that high) and we'd just had a few days of gentle rain, so there's no particular reason to think there'd be a forest fire or something. Also, none of the pictures appear smokey. I'm wondering if the cloud on the right is involved; its left edge seems to indicate that it was gradually thinning out towards the phenomenon. Matt Deres (talk) 15:42, 10 June 2012 (UTC)

Yes they are, although faint. As a (formerly obsessive) sky watcher and optics enthusiast, I have no problem seeing them. 129.2.171.55 (talk) 22:11, 10 June 2012 (UTC)Nightvid

These are indeed a type crepuscular rays, although you are seeing them from an atypical angle, which is part of the reason for their strange appearance. Were you near a body of water perchance? My theory is that the rays seem to end mid-sky because the sun is reflecting up off a body of water nearby, interrupting the cloud's shadow nearer the ground. Pretty much speculation on my part though, tough to tell just from a few photos. -RunningOnBrains^(talk) 00:30, 11 June 2012 (UTC)

Nope, no sizable water nearby. Matt Deres (talk) 01:02, 11 June 2012 (UTC)

How many times did sex evolve independently?

How many times did sex evolve independently? 82.31.133.165 (talk) 17:17, 9 June 2012 (UTC)

Are you sure that it did ? (As opposed to only evolving once and being passed down to all species which evolved from that one.) Our evolution of sexual reproduction article says, in the 2nd sentence, that "All sexually reproducing organisms derive from a common ancestor which was a single celled eukaryotic species.", and provides source(s) to back up that claim. StuRat (talk) 17:40, 9 June 2012 (UTC)

That sentence by itself wouldn't mean much more than that all sexually reproducing species are eukaryotes. More strongly, it seems very unlikely that meiosis -- the special type of cell division involved in sexual reproduction -- evolved more than once, since it requires a large number of special mechanisms in order to happen, and as far as I know those mechanisms are always implemented in essentially the same way. Looie496 (talk) 19:06, 9 June 2012 (UTC)

It appears that isogamy was the first stage of sexual reproduction that involves gametes that look alike and so cannot be classified as "male" or "female." In several lineages (plants, animals), this form of reproduction independently evolved to anisogamous species with gametes of male and female types to oogamous species in which the female gamete is very much larger than the male and has no ability to move. There is a good argument that this pattern was driven by the physical constraints on the mechanisms by which two gametes get together as required for sexual reproduction, see Isogamy#Evolution. DriveByWire (talk) 19:21, 9 June 2012 (UTC)

If you mean meiotic reproduction, just once, per above. But the transfer of genes between bacteria and the recombination of genes in viruses are separate phenomena. μηδείς (talk) 18:04, 10 June 2012 (UTC)

What is: "colonic sorting" in biology?

thanks.

As I understand it, it's a mechanism by which the colon of certain animals such as rabbits separates small particles and fluids from larger, less digestible particle. Looie496 (talk) 18:56, 9 June 2012 (UTC)

a simple table which sums up all types of reproduction?

1.monoecious: types, dioecious: male-female, male or female with intersex. thanks.

This list of modes of reproduction would be very long indeed. Here are a few that you might not be familiar with: Apomixis#Types_of_apomixis_in_flowering_plants, and Fungus#Reproduction. Combined, these links present a few dozen different modes, and only cover a small fraction of what plants and fungi actually do. Some fungi have thousands of mating types. Animals have slightly less variety in reproductive modes, but there is still parthenogenesis, which occurs in things like aphids and even some lizards and fish. So- good question, but I've never seen a comprehensive list that covers all forms of life :) SemanticMantis (talk) 13:15, 10 June 2012 (UTC)

I don't understand how to mount the camera on a barn door tracker

Doesn't the angle between the camera body and the tracker matter? What are the guidelines for tilting and aligning the camera, once the polar finder has been aligned with the poles? 76.104.28.221 (talk) 19:10, 9 June 2012 (UTC)

I'm not sure I understand your confusion. Where else would the camera be pointed, if not towards the object you're trying to photograph? See this photo: http://www.astropix.com/BGDA/SAMPLE2/207A.JPG

The barn door tracker acts like a simple equatorial mount with only a right ascension adjustment knob. If you imagine moving the angled board around a full circle, the direction in which the camera points will also move in a full circle. The size of the circle depends on how far the camera lens' direction is from the axis. If it's pointing along the axis, the circle has 0 size, and the camera only rotates. If it's pointing 90 degrees away, the camera traces a great circle around the sky. This is exactly how a star behaves in relation to the celestial pole. --140.180.5.169 (talk) 20:14, 9 June 2012 (UTC)

i.e. does it matter how the camera is oriented, in order for the barn door tracker to work? Can the camera point towards any object in the celestial sphere and still track the object? 207.114.92.194 (talk) 17:45, 11 June 2012 (UTC)

Yes, if your polar alignment is accurate, you can point the camera at any part of the sky. The earth only rotates around one axis, if the "mount" counter-rotates around the same axis, it doesn't matter where the camera is pointing; in relation to the "sky", it will remain stationary. Maybe have a look at Equatorial mount. Vespine (talk) 22:54, 11 June 2012 (UTC)

Jamming the GPS system on a global scale

Apparently, it's quite easy to jam GPS signals, GPS jammers are even available commercially. But this raises the question of how reliable this system is. Could e.g. China shut the GPS system down globally using satellites in case of war with Taiwan? Count Iblis (talk) 23:31, 9 June 2012 (UTC)

Yes, along with GLONASS and GALILEO. But most major militaries have frequency hopping positioning satellites ready to be switched on in an emergency. 71.212.248.104 (talk) 04:47, 10 June 2012 (UTC)

Do you have a source on the frequency hopping GPS satellites? Even if this technology actually exists it's not available in any of the portable GPS receivers that the army uses. Anonymous.translator (talk) 05:30, 10 June 2012 (UTC)

I'd also like to see a source on China being able to jam the GPS system. Seems like it's possible on a local scale, but on a global scale? You'd need a hell of a transmitter. -RunningOnBrains^(talk) 05:46, 10 June 2012 (UTC)

Or a weak transmitter on board an airplane, or near the up-link which probably is the weakest system point. With a huge (state) budget one could employ satellites for the job. Electron9 (talk) 11:32, 10 June 2012 (UTC)

June 10

Liquid air energy storage

Thermal energy storage doesn't mention condensing and cooling air to a liquid, storing it, and then boiling it off to recover stored energy. How efficient is that for power storage? I know pumped hydro is about 80%, but it doesn't have a very high capacity on flat terrain. What I really want to know is whether building water towers for pumped hydroelectric is more cost efficient per kilowatt hour than building giant thermos tanks for condensing liquid air. 71.212.248.104 (talk) 04:45, 10 June 2012 (UTC)

Insulating and maintaining liquid air isn't easy or efficient. In order to keep it cold enough and under enough pressure to be useful will make the system vastly less efficient than pumped hydroelectric. Besides, you don't need to build tanks for pumped hydroelectric. You just need to dig a big hole at high elevation. --Jayron32 04:53, 10 June 2012 (UTC)

Ah-ha! Clearly you're right, because for pumped hydro on flat terrain you only need a big hole in the ground, since the gravity pressure is only necessary at a generator turbine which could be at the bottom of the hole. (Unlike everyone's plumbing which is why water towers need to be in the air, or tanks on hills, etc.) 71.212.248.104 (talk) 05:01, 10 June 2012 (UTC)

You may be interested in Pumped-storage hydroelectricity. Of course, like other forms of hydroelectric, it only works in areas of the proper relief. You can't build an efficient hydroelectric plant in a wide open plain for any purpose. --Jayron32 05:07, 10 June 2012 (UTC)

... but if you build it on the coast you can get "more than 100% efficiency" with careful tidal timing (really gleaning energy from the earth-moon orbital system). Dbfirs 08:42, June 10, 2012 (UTC)

Um, what I'm saying is, you just need two (sets of) water tanks, one far underground, and one at the surface, which is fairly inexpensive and would work in Kansas. That is much less expensive than using a water tower, which is the traditional way municipal water pressure is powered in flat areas. You can't predict water usage and you can't predict electricity usage, but you can shape the load with pumped storage tanks in both cases, for only 20% overhead lost to heat at the turbine, pump, and piping in the case of electricity. (Modern engineers would probably want to try to recover some of that lost heat, since it's fairly localized.) 71.212.248.104 (talk) 10:52, 10 June 2012 (UTC)

If you happen to have a convenient deep underground cavern with easy access, then that idea sounds workable, but you are limited by the size of the underground tank. Dbfirs 11:29, 10 June 2012 (UTC)

Subjective experience and age

How does subjective experience--by which I mean consciousness--change with age? I'll give a few examples to indicate what I'm looking for. Children are easily excited, and their emotions are more extreme. Children are happier, sadder, and angrier than adults in happy, sad, and frustrating situations, respectively. Teenagers and people at mid-life tend to be more depressed. The young perceive time as passing much more slowly older people; one year seems like an eternity to a 6 year old, but passes quickly for a 70 year old.

Does anyone know of a more complete list? My examples above are mostly based on my experience, which is limited because I'm quite young and represent only 1 sample point. --140.180.5.169 (talk) 09:26, 10 June 2012 (UTC)

I don't know of a more complete list. Except for a couple of things which I'll note, I think you are wrong anyway. I have passed retirement age. I dissagree that children have stronger emotions - what changes is perception about what is sad, happy etc. To a child, not getting your favorite food today can be something to cry over. To an adult, it's unimportant. This is not about strength of emotion - its about intelligence and experience. Children know that by over-acting, they can persuade adults. Acting angry or sad is not being angry or sad, even if you convince yourself. Psychologists have written learned papers about depression in teenagers. Never the less, I think it's nonsense. Having been through it, depression when a teenager is not deep and is quickly overcome. What's different for teenagers and adults in recent years, is that doctors will more readily prescibe for it, and that is a bad thing. On the other hand, depression is recognised as real these days, and that's a good thing. You won't know what sadness is until you loose a loved one after years of being together.

One thing that DOES change is perception of size and distance. When I was about age 3 to 5, my mother and I used to go and spend a week now and then with grandfather at his house, in another city. When there, Grandfather would walk me to a shop and buy me an icecream. It seemed a long walk. We also used to go to the beach - another long walk. Grandfather died when I was 5. When I was 18, the company I worked for sent me to that city. I decided to retrace old steps. I was greatly surprised that the distance to the shop was 1 block! The distance to the beach was 1 block! Recently, my old primary school was opened as a tourist office. I had a look. I was suprised how small the rooms were.

You are correct that time passes more slowly for children.

Another thing that changes is your view of the general competence of adults. When you are a child, you think adults are 100% capable. When you are a late teenager, adults frustrate you whith their slow decision making and inability to learn new things quickly. When you get to retirement age, your perception changes - older people now seem competent, but teenagers seem rash and impatient. But, thinking about it objectively, those pesky teenagers are right!

When people get really old and start to loose their faculties, then the more deeper and more evolutionary early parts of the brain take over. Then, emotions can get rediculously strong. I had a relative who lived in an old-folk's home - the sort where nurses were on duty 24 hours a day. I was there once when the kitchen was 30 minutes late in serving lunch. A number of inmates staged a riot (in their wheelchairs, in slow motion)! You'd never get that at any other age, and you'd never get that in people who are mentally fit - they would just accept that sometimes things go wrong.

What does change as people get older is the rate at which emotions can change. In management training, I was taught that when given bad news, people go thru 5 stages of grief: disbelief, why me/anger, sadness, resignation/acceptance (sometimes bargaining), and finally, seeing an advantage. See http://en.wikipedia.org/wiki/5_stages_of_loss. Some psychologists don't accept this, but you get that. In my experience it is right - it has helped me a lot in counselling employees and family. Now, children can go thru all 5 stages in seconds. Teenagers can take longer. Middle aged folks can take months. Some folks remain stuck at one stage - that can happen with young adults, but not very often. I don't think children ever get stuck.

Wickwack124.182.162.50 (talk) 10:37, 10 June 2012 (UTC)

"One year seems like an eternity to a 6 year old, but passes quickly for a 70 year old" is something that everyone says, but personally, I don't think it's true. What is true as we age is that events we remember can be longer ago, and that any given length of time is a smaller fraction of our total experience. When people say "time goes by so fast these days!", they're always reminiscing about the past, not looking to the future; "1 year ago" doesn't seem like such a big deal now as it did when you were 6, but on the other hand "next week" will take just as long to get here now as it did then. FiggyBee (talk) 14:36, 10 June 2012 (UTC)

I would disagree with that point, the time until next week does grow shorter with age, the "are we there yet" question shows a slower movement of future time as well. The OP should look for information at Gerontology or in these journals. 65.95.22.197 (talk) 17:02, 10 June 2012 (UTC)

I have to agree with IP 65. Knowing you had to wait A WHOLE HOUR!!! for something as a kid was torture that took forever. Nowadays that's hardly even enough time to get ready for anything. μηδείς (talk) 18:00, 10 June 2012 (UTC)

For a 5-year-old a year is 1/5th of their entire lifetime, for a 70 year old it's only 1/70th. Roger (talk) 20:27, 10 June 2012 (UTC)

Reaction times do slow down with age ( A Literature Review on Reaction Time by Robert J. Kosinski

Clemson University , http://biology.clemson.edu/bpc/bp/Lab/110/reaction.htm ) and working memory capacity decreases as well. So I think it is fair to say that subjective experience indeed changes with age. 129.2.171.55 (talk) 22:26, 10 June 2012 (UTC)Nightvid

2012 event

will the 2012 event eventually kill us all or not??????????!!!!!!!! — Preceding unsigned comment added by 77.35.15.169 (talk) 12:24, 10 June 2012 (UTC)

I changed your heading from the non-informative "!!!!!!!!!!!". I guess you refer to something from 2012 phenomenon which says: "Scholars from various disciplines have dismissed the idea of such cataclysmic events occurring in 2012." PrimeHunter (talk) 12:31, 10 June 2012 (UTC)

Not only that, but recent archaeological discoveries have demolished the notion that the Mayans considered 2012 to be the "last" year. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:34, 10 June 2012 (UTC)

There never was any evidence that the Mayans considered 2012 the last year. They computed numbers up to a point far into the future, far more than they would ever need. They stopped computing at the end of a particular mathematical cycle because it made rational and aesthetic sense to end there rather than some point chosen at random. You can find many places where there are ordinary Gregorian calendars, or the precise dates of Easter, computed for a number of years into the future, but they all stop at a certain point, too. Nobody has ever interpreted this to mean a prediction of the end of the world (!) at that point, and they should never have done so for the Mayan calendar either. -- ♬ Jack of Oz ♬ ^{[your turn]} 23:51, 12 June 2012 (UTC)

Microsoft Excel can only show times up to 31/12/9999 23:59:59 and after that every time is show as ################. Surely this means that some brilliant soothsayer/programmer foresaw that the world will be reduced to a series of hashes at that point. Or maybe there will be free hash browns. 203.27.72.5 (talk) 01:48, 13 June 2012 (UTC)

From a scientific perspective, nothing unusual, besides an astronomical observation, should happen on 21 December. Don't believe every thing you hear. Plasmic Physics (talk) 13:38, 10 June 2012 (UTC)

Most scientists believe that every single human alive on the 21st December 2012 will die, either on that day or at some point in the following months or years. LukeSurl ^{t c} 14:56, 10 June 2012 (UTC)

But not as a result, direct or indirect, of "the 2012 event". FiggyBee (talk) 14:58, 10 June 2012 (UTC)

Quite. --LukeSurl ^{t c} 15:06, 10 June 2012 (UTC)

See section 3 page 12 and further of this article. The universe is making a transition to a state of exact supersymmetry, we are now in the false vacuum state, after the transition we'll be in the real vacuum. A long time ago, a bubble containing the exact supersymmetric vacuum nucleated and started to expand at the speed of light. We're now just a quarter of a light year from the edge of that bubble. When we cross the boundary of that bubble, a huge amount of energy will be released killing all of us in a fraction of a second.

However, since we all have an infinite number of identical copies, the closest one located about about 10^{10^29} meters from here see here, and they may be further away from the edge of the expanding bubble that exist there, they may survive into next year. Because these copies are identical to you, you will actually survive as your copy without having any knowledge about being killed here. So, it will look like this was all a doomsday myth. Count Iblis (talk) 16:29, 10 June 2012 (UTC)

I can personally affirm that as the absolute truth, Count Iblis. because this has all happened befor .I know because I was there.190.148.132.194 (talk) 17:15, 10 June 2012 (UTC)

Citation required. 84.209.89.214 (talk) 18:04, 10 June 2012 (UTC)

LOL (at both the two most recent posts immediately above) HiLo48 (talk) 00:03, 11 June 2012 (UTC)

Could we escape this fate if the bubble nucleated at such a distance from us that the universe's expansion would keep it from ever reaching us? 203.27.72.5 (talk) 01:42, 11 June 2012 (UTC)

Yes, making certain assumptions about the expansion of the universe. See Observable Universe. Of course, we couldn't see the described bubble coming if it was moving at the speed of light - the image wouldn't reach us until the bubble hit us. Schild's Ladder is a book involving such a bubble expanding at half light speeed. 209.131.76.183 (talk) 13:03, 11 June 2012 (UTC)

Hold on. Tomorrow will be there after it all blows away. (video) DriveByWire (talk) 14:27, 11 June 2012 (UTC)

How would Uranus have been pronounced in the 1780s / 1790s

Johann Elert Bode suggested the name Uranus. However, the Wikipedia (and many other places) always refer to this as the 'Latinized name'. I assume this is intended to distinguish it from the Greek pronunciation of the name. (Something like 'oo-RA-nos') My question is, how would an english speaker in the 1780s/1790s have pronounced the latinized version of the name Uranus? --CGPGrey (talk) 15:22, 10 June 2012 (UTC)

Not a definitive answer, but Greek οὐρανός would have been pronounced, in English, with the stress on the first syllable (OO-rah-nohs), since the vowels in the last two syllables are short. I assume that the Latinized version would have been pronounced something like YOO-ruh-nuhs if the normal rules for such things were being followed. Deor (talk) 17:41, 10 June 2012 (UTC)

We don't have a speech recording from the 18th century. Those who spoke of the planet would likely have known enough Greek to argue about how the ancient Greeks pronounced Ouranos. Since both the English words "anus" and "your" were current at the time I'm sure someone could have thought of a joke that would have been for once original. Anyone embarassed by that hilarity might persist in calling the planet Georgium Sidus (King George III's star), as did British Victorian nautical almanacs as late as 1850. 84.209.89.214 (talk) 17:55, 10 June 2012 (UTC)

New question to add on to this: what about georgium sidus? I assumed it would be with a 'g' as in George sound, but I've come across a few sources saying it with a 'y' sound at the start. --CGPGrey (talk) 13:05, 11 June 2012 (UTC)

Old newtonian problem solved?

Resolved

Is this accurate: http://www.dailymail.co.uk/news/article-2150225/Shouryya-Ray-solves-puzzles-posed-Sir-Isaac-Newton-baffled-mathematicians-350-years.html

and has it been verified by others? Bubba73 ^{You talkin' to me?} 15:41, 10 June 2012 (UTC)

From the linked report one can gather only that someone (who?) has hailed a schoolboy a genius for "cracking puzzles (what?) that have baffled the world of maths (who?) for 350 years. I see no more to support this claim than an article title translated as "Analytical solution of two unsolved problems in fundamental particle dynamics" on a PC screen, and the utterly non-newsworthy information that "scientists can now calculate the flight path of a thrown ball and then predict how it will hit and bounce off a wall". Poorly educated journalists should keep away from subjects that they are not equipped to understand. 84.209.89.214 (talk) 17:38, 10 June 2012 (UTC)

Actual experts weight in here. Someguy1221 (talk) 19:22, 10 June 2012 (UTC)

Thank you, that helps a lot. It seems like just about every time there is something like this, it doesn't quite pan out. Bubba73 ^{You talkin' to me?} 21:50, 10 June 2012 (UTC)

physics

what distance of gravitional field? prove that — Preceding unsigned comment added by Krishan chodhary (talk • contribs) 15:47, 10 June 2012 (UTC)

Isaac Newton obtained his law of gravitation in which gravity works over unlimited distance, by observation and experiments that are readily demonstrated on laboratory scale. The law also allows planet movements to be explained. We cannot explain fully how gravity works and in extreme cases, such as over great distances and velocities aproaching the speed of light, Newton's classical description must be modified by Einstein's geometric theory of curved spacetime. We can provide these references but cannot give you proof of what appears to be a universal Physical constant. 84.209.89.214 (talk) 17:15, 10 June 2012 (UTC)

Physics numerical

uniform electric and magnetic fields with strength E and B are directed along the y axis 

a particle with specific charge q/m leaves the origin in direction of x- axis with an initial velocity Vo

Find a) the co ordinate of the particle when it crosses the y- axis for nth time b) the angle alpha between the particle velocity vector and the y-axis at that moment — Preceding unsigned comment added by 101.214.105.47 (talk) 17:19, 10 June 2012 (UTC)

Please do your own homework.

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know.

Shelled Pistachios lose their flavor?

On Amazon.com there's two types of pistachios they sell, shelled and unshelled.

Shelled http://www.amazon.com/gp/product/B0013NBM5A/ref=s9_qpp_gw_p325_d1_g325_ir06?pf_rd_m=ATVPDKIKX0DER&pf_rd_s=center-7&pf_rd_r=1BA24ZE30ANCCZK60VPV&pf_rd_t=101&pf_rd_p=470938451&pf_rd_i=507846

Unshelled http://www.amazon.com/Keenan-Farms-In-Shell-Pistachio-Naturally/dp/B001IZIEGS/ref=sr_1_1?ie=UTF8&qid=1339350634&sr=8-1

But the shelled pistachios have significantly worse reviews, with most people saying it tastes worse, and even rancid. They are both made by the same company, so I'm wondering if shelling a pistachio would have a real impact on its flavor, and if so, why? ScienceApe (talk) 17:53, 10 June 2012 (UTC)

Most foods keep longer if they're kept intact; I don't see why pistachios would be any different. Matt Deres (talk) 18:38, 10 June 2012 (UTC)

Perhaps the company sorts out the better pistachios to be shelled, since the edible part's appearance could be directly judged by the consumer. -RunningOnBrains^(talk) 00:36, 11 June 2012 (UTC)

Denver heart attacks

Does Denver have a higher than average rate of heart attacks? What about heart attacks in Denver marathon? Thanks.--Rich Peterson76.218.104.120 (talk) 18:39, 10 June 2012 (UTC)

Compared to the average rate of heart attacks in which population? Falconus^{p t c} 20:50, 10 June 2012 (UTC)

Exceeding the RDA of vitamins

I vaguely remember from college that taking excessive amounts of some vitamins and minerals isn't a problem as they can be easily broken down and excreted by the body while others build up and cause damage. If I was to take double/triple the RDA of all the main vitamins and minerals what would be the first symptoms (or should I say which would be most toxic)? Thanks 87.115.195.33 (talk) 19:20, 10 June 2012 (UTC)

Vitamin A toxicity, see Hypervitaminosis A. The dose that can cause toxicity is the lowest above the RDA for all vitamins and minerals. This may also be due to the RDA being too low for other fat soluble vitamins e.g. vitamin D. The minimum dose of vitamin D that is thought to be capable of producing toxicity is about 40,000 IU/day for many weeks (although the lowest dose that has been observed to actually produce toxicity is 77,000 IU/day). The RDA is just 800 IU/day. However, it may be that the natural dose of vitamin D is 10,000 IU/day as that's what we would get from the Sun were we to spend a lot of time in the Sun at Noon time in the tropics. Count Iblis (talk) 19:43, 10 June 2012 (UTC)

And some minerals are toxic when taken to excess. Although selenium is required in the human diet, it is toxic in excess. The saem would apply to copper. Graeme Bartlett (talk) 12:28, 11 June 2012 (UTC)

That's great, thanks for the answers :-) 87.115.195.33 (talk) 21:07, 11 June 2012 (UTC)

is a twelve minute mile impressive

I know there was a time that the four-minute mile was record-setting, so I'm wondering if a twelve-minute mile is considered impressive IF the person weighs the same as three average-sized people (or if the average weight is obese today, then three averaged-sized people from 50 year ago) for that height and age? This is not a trollish question, if yout think it's bad just answer it matter-of-factly okay. --80.99.254.208 (talk) 20:07, 10 June 2012 (UTC)

The world record for a mile has dropped a fair bit below 4 minutes now, see Mile run world record progression. For info an average human Walking speed is about 3 miles per hour (a mile every 20 minutes) so a 12 minute mile (5mph) is not quite twice as fast as normal walking pace. Read into that what you will. Over a short distance twice walking pace is nothing, over a mile it does become a question of endurance so I suppose it depends on who you ask...I'm 'relatively' sporty and i'd like to think I can do a faster than 12 minute mile but i've honestly never tried...Maybe I will tomorrow and then reply with my time...might end up embarrassing myself though! ny156uk (talk) 20:43, 10 June 2012 (UTC)

Yes, in my younger days, I used to walk a mile in 12 minutes (occasionally overtaking slow joggers), but for someone who is three times normal weight, I'd say "well done" and keep up the exercise! Dbfirs 21:25, 10 June 2012 (UTC)

Yes, a 5 mile-per-hour pace for a full mile would be pretty impressive for a 500 pound man. Looie496 (talk) 22:53, 10 June 2012 (UTC)

I agree with the other responders. A 12 minute mile is a very brisk walk. Maintaining a very brisk walk for a mile when you are very unfit and overweight is impressive. The important thing is your heart rate - if whatever exercise you are doing is enough to get your heart rate up, then it is worthwhile exercise for you. I would expect maintaining a very brisk walk for a mile would get your heart rate up a lot if you are three times your healthy weight. --Tango (talk) 23:03, 10 June 2012 (UTC)

I echo everything above. The fitness standard required for those joining the British Army in non-combat roles is a 1.5 mile run in 14 minutes, which equals 9 mins 20 secs for a mile, so you're really not that far away. Remember to put rest days in your programme to avoid joint and muscle injuries. Alansplodge (talk) 23:17, 10 June 2012 (UTC)

Top racewalkers do a 20km (12½ mile) walk in around 1 hour 17 minutes. This is about a 6.2 minute mile. Now that's impressive. Tonywalton ^Talk 23:35, 12 June 2012 (UTC)

Is this possible at all? Someone who weighs that much, yet who doesn't collapse during such an effort would have to be very fit, and therefore he would have to exercise vigorously on a regular basis. But then, how could he be so extremely obese? People who exercise a lot don't put on a lot of weight if they eat a lot. Count Iblis (talk) 23:23, 10 June 2012 (UTC)

By some measures (e.g. BMI) some sportsmen are considered obese. I'm thinking especially of Sir Steven Redgrave and Matthew Pinsent - and nobody would consider those Olympic athletes unfit! A lot depends on what the weight consists of. Muscle weighs more than fat. --TammyMoet (talk) 12:24, 11 June 2012 (UTC)

That's because the BMI is stupid. It was dreamed up by a dietitian who didn't undertand basic math, and is promoted by psuedo-medical types who also don't understand math. It assumes weight rises as to the square of length. It doesn't. For an object of any shape, if its size goes up in all linear dimensions proportionately, mass will rise according to the cube of any dimension. I am 1.67 m tall and weigh 80 kg. According to the BMI, I am borderline obese. Well, I do look pretty fat, though when I was younger and better muscled from exercise, I didn't look fat but weighed about the same. But I have a cousin who is 2.01 m tall and weighs 120 kg. According to the BMI formula he has the much the same BMI and he too is obese. He's the skinniest person I know. Also, judging by the manual work he does, he's one of the fittest. Also, some people have wide frames. I am 420 mm from shoulder to shoulder, and 350 mm from hip to hip. I used to do weight lifting at the YMCA. The trainer there was the same height as me, but nearly 700 mm across the shoulders. If he tried to get his weight down to 80 kg, he'd collapse from starvation. Wickwack124.182.46.250 (talk) 14:47, 11 June 2012 (UTC)

Does the Plastic to Oil Machine really work?

I am curious about Japanese inventor Akinori Ito and his invention that turns plastic bags into oil.

What can you do with the oil product once produced?

http://earth911.com/news/2011/02/21/japanese-inventor-turns-plastic-bags-into-oil/

http://www.mnn.com/green-tech/gadgets-electronics/stories/from-waste-to-fuel-invention-turns-plastic-bags-back-into-oil

All my best,

Mark B. Strauss, M.S.

its not real oil — Preceding unsigned comment added by 64.38.226.77 (talk) 20:36, 10 June 2012 (UTC)

Certainly possible just from the chemistry aspect (not turning it into real crude oil of course, but something similarly combustible) but I can't imagine it's economically feasible, and it certainly wouldn't result in a net gain of energy. -RunningOnBrains^(talk) 00:39, 11 June 2012 (UTC)

Most plastics can theoretically used as an exothermic fuel source. Although of course, there may still be no net gain of energy after including the cost of converting the plastic to something more efficient/safe/usable. In any event, I have no idea if this machine actually works as advertised. The rate at which newfangled green power generation technologies fade into the aether after appearing in the news has made me extremely skeptical of all such claims. Someguy1221 (talk) 09:51, 11 June 2012 (UTC)

Note that it would take a huge number of plastic bags to pay off the cost of the device, so it's not something each home should have. StuRat (talk) 23:05, 12 June 2012 (UTC)

How can I remove moles on my face?

Questions and responses removed because we cannot provide medical advice as a matter of policy. Please consult a qualified medical professional; best of luck. Falconus^{p t c} 00:08, 11 June 2012 (UTC)

Remove them by going to a dermatologist and seeing what they recommend. StuRat (talk) 15:22, 11 June 2012 (UTC)

Photoelectric threshold frequency

I'm trying to design sunglasses that convert UV light into solar energy. Is there a cheap material with a photoelectric threshold frequency in the blue-violet (or near-ultraviolet) range? 68.173.113.106 (talk) 20:57, 10 June 2012 (UTC)

I think you are confusing the photovoltaic effect with the photoelectric effect. To answer you directly, yes, the alkali metals do, however, they would not work for your purpose since they are very reactive with air and water, are not even partially transparent, and do not have an appreciable photovoltaic yield, among other reasons. A suitable substrate for your purpose might be a thin film solar cell. 129.2.171.55 (talk) 21:54, 10 June 2012 (UTC)Nightvid

Another phenomenon of interest is fluorescence. But if you want your sunglasses to see UV you are going to have to preserve the information of the direction of UV rays. Fluorescence will radiate it in all directions Graeme Bartlett (talk) 01:16, 11 June 2012 (UTC)

An Image intensifier is a vacuum tube device by which one may obtain an image in IR (infrared) light. DriveByWire (talk) 13:35, 11 June 2012 (UTC)

If you want to "see" UV you can get a CCD or other optical detector camera, remove the hot filter and replace the glass lens with a fused silica or fluorite lens. Then attach a 350nm UV pass filter. A normal camera glass lens is a bit opaque to UV. And air is opaque at the shorter wavelengths. Graeme Bartlett (talk) 11:08, 12 June 2012 (UTC)

Primary cortices of "other" senses?

If each sense has its own cortex (primary visual, primary auditory, etc.) what about the "others" such as equilibrioception, feeling the fullness of one's bladder, etc.? Is there such thing as the "primary equilibrioceptive cortex (EQ1)" ?

129.2.171.55 (talk) 21:40, 10 June 2012 (UTC)Nightvid

The vestibular cortex is the equilibrioceptive part -- not much is known about it, but it does seem to exist. I would expect that bladder fullness activates the somatosensory cortex, although I don't specifically know. Some other nonstandard senses, such as the systems that detect blood sugar levels or blood salt levels, activate parts of the insula. Looie496 (talk) 22:50, 10 June 2012 (UTC)

June 11

testosterone

Questions and responses removed because we cannot provide medical advice as a matter of policy. Please consult a qualified medical professional; best of luck. Anonymous.translator (talk) 02:16, 11 June 2012 (UTC)

paranormal effects of Ghungroo anklet

Context: in my editing during Ghungroo in the wikipedia encyclopaedia, the editing reached in a question and to get answer i had added my e-mail id; which resulted in automated warning. then i learnt, said question can be asked here also. thus i type entire "edited portion" by me, here, for further help: If some one can clarify, this point is written. In the web page: www.travelblog.org/Asia/blog-455847.html, the following can be read: “We had a camp fire inside, in the verandah. We were able to hear the insects noise loud. It appeared as if somebody with ghungroo was walking just beside the window mesh and Doppler's effect was also there (The Ghungroo sound used to get loud and again disappear as if somebody is running towards the window and again going far.) If we had heard the ghungroo aawaz inside the bungalow don't know what we would have done? The sound still murmurs in my ears. I do not know who or which animal was roaming around with ghungroo and whom it was trying to romance among the three?? :):)”. Even outside the forest, such sound is heard by the person editing it, in residential colonies. In his enquiries, there are no animals, birds or insects which create such sound. It is told that, female ghosts, a particular classification in Kerala “Rakhta/ blood Yakshi” do create this sound. Can some one clarify this phenomena ? This sound is heard not by every one, but only a few selected ones in the group, at that particular sound. for reply's sake e mail is added, eventhough no need while editing wikipedia: <email redacted> — Preceding unsigned comment added by Ramanathan2108 (talk • contribs) 09:35, 11 June 2012 (UTC)

It's not that an email address is unnecessary, it's that it's not permitted. As it says at the top of the page: Do not provide your contact information. E-mail or home addresses, or telephone numbers, will be removed. You must return to this page to get your answer. Karenjc 12:06, 11 June 2012 (UTC)

A Ghungroo is a musical anklet tied to the feet of classical Indian dancers. DriveByWire (talk) 13:27, 11 June 2012 (UTC)

There is no such thing as a ghost. Look for the rational explanation. 217.158.236.14 (talk) 13:33, 11 June 2012 (UTC)

A Ghungroo in Bollywood. (video) DriveByWire (talk) 14:50, 11 June 2012 (UTC)

Boiling point for Phosphorus

The WP article on Phosphorus, http://en.wikipedia.org/wiki/Phosphorus, gives the boiling point (presumably for standard pressure, 1.01325 Bar) as 280.5 C (553.7 K). This ought to be for the standard state, P₂, to be consistent with WP articles on other elements. However, the NIST-JANUF tables (4th Ed) give the standard state as P₁ and the boiling point as 1180 K at 1 Bar. The Ihsan-Barin tables (2nd Ed 1993) give the standard state as P₂ and the boiling point as 1180 K at 1.01325 Bar. Googling "boiling point phosporus" returns sites similarly disaggreeing. It is of course quite unlikley that P₁ would have the same boiling point as P₂, and unlikely that P₁ would boil higher than P₂. What is the most common standard state? What is the correct boiling point for P₁ and P₂? It would be nice if the WP articles gave the allotrope/catenate for which the triple, melting, boiling, and critical points are quoted - it would save some confusion. Which one of the two standard pressures (1 Bar as per IUPAC, 1 ATP as per USA) the data is applicable to would be good too. Keit120.145.195.11 (talk) 10:17, 11 June 2012 (UTC)

That is because the standard state is P₄. Plasmic Physics (talk) 10:39, 11 June 2012 (UTC)

Gee, Plasmic, you've just made things more difficult. Do you have any references for P₄ being the standard state, noting that NIST-JANUF bible gives (on page 1817) the standard state as P₁? The What are the temperature points for P₄? I thought that phosphorus in the gas state can only be P₁ or P₂. Therefore the standard state in the gas phase at least must be P₁ or P₂. If that is so, and the minimum energy state when liquid is P₄, then shouldn't there be a different temperature for boiling and for condensation? I haven't come across two temperatures, as in one listed for boiling, one for condensation. Keit120.145.195.11 (talk) 11:09, 11 June 2012 (UTC)

Sorry, I don't have the data on hand. I just did some reading of the article myself, apparently the allotropy of phophorus is a messy business, much like sulfur. It changes from one crystallomorph to another as the solid heats up, while at the same time starting to decompose into different molecular structures. The diatomic and monoatomic molecules tend to exist only in the gas phase, mixed with other molecular varients. Plasmic Physics (talk) 23:23, 11 June 2012 (UTC)

If you want some insight, try octasulfur. Plasmic Physics (talk) 23:54, 11 June 2012 (UTC)

Does anyone know which more kinetically stable: a phosphorus icosaherdron, or a phosphorus tetrahedron? Plasmic Physics (talk) 23:54, 11 June 2012 (UTC)

Does anyone know (from a reliable reference, or from logical process) the boiling points for P₁ and/or P₂? Keit58.170.163.132 (talk) 01:06, 12 June 2012 (UTC)

From the same phosphorus article, I can tell you that it is next to impossible to experimentally find the boiling and melting points. They are just way too reactive, they polymerise faster than what they condense at. If you're lucky, you can maybe find theoretical calculated values. Plasmic Physics (talk) 01:20, 12 June 2012 (UTC)

Since all references I can find give the boiling point as either 554 K or 1180 K, theoretical values will do nicely, as even if not accurate, they would identify which of these two values is wrong, and which value should be used in other calculations. If only I knew where to look. Keit124.182.173.247 (talk) 01:42, 12 June 2012 (UTC)

ChemSpider gives 741.33 K, 769.65 K, and 796.85 K for P, P₂, and P₄ respectively, as their boiling points. (According to the EPIsuite predictions.) Plasmic Physics (talk) 02:47, 12 June 2012 (UTC)

P₄ thermolyses into P₂ at ~1180 K, P₂, thermolyses above an even higher temperature. Plasmic Physics (talk) 02:53, 12 June 2012 (UTC)

ChemSpider gives 435.17 K, 450.94 K, and 480.72 K for P, P₂, and P₄ respectively, as their melting points. (According to the EPIsuite predictions.)

I was in the university library today, and had a look at a few references. The OP is correct in stating that the NIST-JANUF tables (4th Ed) on page 1817 gives the reference state at all temperatures as P₁, and it gives the boiling point as 1180.008 K at 1 bar. Further, the Phsophorus data has been repeated unchanged (except for a minor fidle to change from 1 ATM to 1 Bar) in each edition since 1961. So an error is unlikely, it would have been corrected by now. Also, the 1180 K boiling point is supported by P's position in the Periodic Table, 554 K would be unlikely. I also looked in Thermodynamic Properties of The Elements, published by the American Chemical Society. It gives the standard state as red phosphorus solid up to 704 K, the sublimation point, and P2 gas above. Note that http://en.wikipedia.org/wiki/Phosphorus says red phosphorus sublimes at approx 416 C ie 689 K. I can't say why Plasmic Physics thinks the standard state is P4, however one might expect the standard state to be the lowest energy state, and you do find different authors choose different standard states. I assume that WolframAlpha, which gives the boiling point as 553.7 K is in error, and this error has been copied into the Wikipedia article. I found ChemSpider to be hopelessly inaccurate. Try looking up the boiling points for O₂ for example. Ratbone58.170.167.209 (talk) 12:46, 12 June 2012 (UTC)

The reason P₄ is standard is because that is the most common molecule in the gas when white phosphorous boils or red phosphorous sublimes. You are going to get a similar issue with sulfur and the S₈ molecule. The gas contains S₄ also and at higher temperatures S₃ and S₂ and only at extreme temperatures will you find significant S₁. There is no point insisting on P₁ as you will not find it around at transitions between liquid and gaseous phosphorous. Graeme Bartlett (talk) 12:57, 12 June 2012 (UTC)

Further to my post above, perhaps in a sense Wolfram Alpha is not in error - the ACS publication gives the boiling point of white phosphorus (not a standard state) as 554 K, exactly as Wikepedia does. Regarding Graeme's view, it's not me that is doing the insisting, it's NIST (USA standards body) insiting on P₁, and the American Chem Society insisting on P₂. Regarding sulphur, NIST-JANUF 4th Ed gives the standard state as a mixture of S₁ and S₂ with traces of other forms. See page 1859. With carbon, NIST-JANUF give the standard state as graphite up to submlimation (~3900 K), then the monatomic form. But the most dominant form in gas appears to be C₃. If heated, C₃ splits into more and more C₂ and C₁. So where something transitions, and what it transitions to, is not necessarily the key. What reference gives P₄ as the standard state for phosphorus? Ratbone58.170.167.209 (talk) 13:28, 12 June 2012 (UTC)

Acid (not the trippy kind)

I'm looking for some scientific assistance on aspects of a story I'm working on; I am doing my own research as well but I haven't done science since I was at school, and I'm struggling to find specifics in terms I can understand and work things out from. Acid; probably sulfuric or nitric.

1) If it's strong enough to burn skin through ordinary clothes (let's say ordinary outdoor clothes, so through a jacket and shirt), what effect would it have on buildings and other structures?

2) How long would it take to significantly damage the structure of an average building?

2a) Would that depend on the kind of brick/stone it was made of?

3) What would it do to a car and to road surfaces etc.?

4) What effect would extra layers of clothes have - would a drop of acid keep dissolving subsequent layers, or does it stop after a while?

5) Are there any widely available materials that are completely resistant to these acids?

6) What's it like to be burned by acid?

6a) Does it sizzle?

6b) Does it smell like burning?

Any assistance greatly appreciated :) Morgana Fiolett 15:21, 11 June 2012 (UTC)

I numbered your questions for you. StuRat (talk) 15:27, 11 June 2012 (UTC)

1) You would need a lot more acid to substantially damage the structure of a building (versus just cosmetic damage), unless you applied it specifically to critical points, like rivet heads in a steel frame building. StuRat (talk) 15:30, 11 June 2012 (UTC)

2a) Yes. Limestone is particularly subject to acid erosion, since it's alkaline and an alkaline material plus an acid react to create a salt and water. Note that Egyptian granite artifacts brought to places with acid rain, such as Cleopatra's Needle, have also been damaged because of it. StuRat (talk) 15:35, 11 June 2012 (UTC)

Cement is also very susceptible to dissolution by acid as is mortar. Destroying the mortar between the bricks will do significant damage. 203.27.72.5 (talk) 21:19, 11 June 2012 (UTC)

3) Cars would tend to rust after having the paint and protective cladding dissolved off the metal. StuRat (talk) 15:42, 11 June 2012 (UTC)

4) Acids might go right through the gaps between the threads in clothes, just like water. Battery acid (sulfuric acid) can also dissolve some clothes. StuRat (talk) 15:44, 11 June 2012 (UTC)

5) Platinum is often used in labs with strong acids, to contain them without reacting. See Reactivity series. StuRat (talk) 15:47, 11 June 2012 (UTC)

Both nitric and sulfuric acid can be contained in glass, which is much more widely available than platinum. Also many plastics do not react with those acids. 203.27.72.5 (talk) 21:23, 11 June 2012 (UTC)

6) A minor acid exposure can cause irritation, itching, and drying of the skin. A major acid burn likely feels like other types of burns, such as a thermal burn, sunburn, or radiation burn. StuRat (talk) 15:51, 11 June 2012 (UTC)

6a) With a strong enough acid, which produces a gas as product of it's reaction, yes, it would sizzle. Incidentally, you can get this same effect with a base, like 3% hydrogen peroxide, when applied to a cut.

H₂O₂ fizzing when it's applied to a cut isn't a simple acid/base reaction; the breakdown is a catalysis, performed by the enzyme catalase. Tonywalton ^Talk 17:13, 11 June 2012 (UTC)

Yes, by "this same effect", I meant "fizzing when a chemical is applied to flesh". StuRat (talk) 17:28, 11 June 2012 (UTC)

6b) No, it would smell like a chemistry lab. StuRat (talk) 15:54, 11 June 2012 (UTC)

Probably, but that really depends on what the acid's acting on. Concentrated sulphuric acid on sugar, for instance, dehydrates the sugar, leaving a solid foam of carbon and a smell of, well, caramel. Morgana, you might try researching into the activities of John George Haigh regarding dissolving flesh in concentrated sulphuric acid. Tonywalton ^Talk 17:19, 11 June 2012 (UTC)

When sulfuric acid reacts with hair it also smells just like burning hair. I found that out the hard way. 203.27.72.5 (talk) 21:23, 11 June 2012 (UTC)

Incidentally, if you're looking for a fictional substance which can dissolve through anything and not be "used up", you want a catalyst. Those cause a chemical reaction, but remain unchanged as a result of the reaction, so they are free to go on and cause more reactions. They speed up reactions which would have occurred anyway, eventually. Chlorofluorocarbons, for example, are a class of catalysts which destroys the ozone layer. Table salt is also a catalyst for the rusting of iron and steel. Perhaps your fictional "catalyst from hell" could react with atmospheric oxygen, water vapor, and carbon dioxide to create carbonic acid, or with atmospheric nitrogen to create nitric acid, using sunlight to power the reaction, and dissolve nearly everything in it's path ? StuRat (talk) 16:05, 11 June 2012 (UTC)

(1, 2, 4) In general, chemicals are consumed when they react. As an acid dissolves or burns or whatever, it becomes neutralized or more dilute, reducing its power to do...whatever. It takes a lot of a corrosive chemical to do widespread damage by corrosion (vs substantial but locallized damage, as StuRat notes).

For any acid or other chemical, there are generally many materials that are resistant to it. Concentrated sulfuric acid and nitric acid are sold commercially in glass containers. Concentrated hydrochloric acid is sold commercially for cement cleaning in glass or plastic bottles...it dissolves a bit of the cement but only "a bit" (neutralizing itself in the process). For any acid, there's something it will destroy and something else that will be inert, and sometimes you can even find complementary cases. Acid "strength" is only part of the picture, since some of the specific chemical components (not just "what makes it acid") can be separately reactive in other ways. HF is a weak acid that eats glass and even small amounts on skin are incredibly painful and possibly fatal but it does not react at all with many plastics, and other much stronger acids don't react with glass at all and small amounts on skin have no or only much less drastic effects. DMacks (talk) 15:37, 11 June 2012 (UTC)

You might go for a combination of acids; see for example Aqua regia, which will dissolve platinum. Or go completely fictional and use Alkahest. Tonywalton ^Talk 15:58, 11 June 2012 (UTC)

Re your question on the effects of a strong acid on a building, the effects are slow and undramatic, especially for a one-time event. Acid will cosmetically damage limestone, but it would take a long time to have any structural effect, and then only if it is reapplied, as the reaction with the stone has the effect of neutralizing the acid. Most other masonry material would be reasonably resistant in periods of less than a decade. Concrete would gradually become powdery and weakened, but it would take a lot of acid. Much the same thing would happen for steel: it would corrode, but for any steel element other than sheet metal that'd be about it unless it was applied over a long term, as in years. As StuRat notes, targeted application would have more effect, but riveted and bolted connections are still quite substantial and tend to be redundant. Even with sheet metal it will take a long time. I had a broken battery case in my VW Beetle in college (the battery in a Beetle is under the back seat on the floor pan): I mopped it up with baking soda, but gave up after a while when I ran out of soda. I sold the car four years later and the battery hadn't fallen out into the road yet.Acroterion (talk) 15:38, 11 June 2012 (UTC)

By the way, I replaced the battery (the victim of a friend who tried to jump-start his car from mine and cross-connected the cables, blowing out a couple of cells) and used up a box of baking soda when I changed batteries, but it was still fizzy when I ran out of soda. Acroterion (talk) 21:32, 11 June 2012 (UTC)

For question number 2, not that this should ever be tried EVER (not only for the obvious reasons but because it involves handling mercury) but for certain metals amalgamation is the surest way to "dissolve" a structural member. -RunningOnBrains^(talk) 18:09, 11 June 2012 (UTC)

What should never be tried ? Leaving a leaky battery in a car ? Where's the mercury ? StuRat (talk) 18:14, 11 June 2012 (UTC)

I was replying to the OP's question, I see that my indentation was a little ambiguous. I assumed that part of his story involved damaging a building with acid, so I offered a possible alternative material. The obvious part was not to try it on an actual building. I've edited it to make clearer what I was answering. Also, why are we whispering? :) -RunningOnBrains^(talk) 19:40, 11 June 2012 (UTC)

I fixed your indentation to show you were replying to the OP, not Acroterion.

I whisper when saying anything that's not an answer to the Q.

Now, what mercury are we talking about ? StuRat (talk) 20:01, 11 June 2012 (UTC)

Click the link... amalgamation is when you mix mercury to another metal. Really cool video here; but again, mercury is dangerous so DON'T TRY THIS AT HOME KIDS! -RunningOnBrains^(talk) 22:09, 11 June 2012 (UTC)

OK, I was not familiar with that definition of "amalgamation", but was going with the more common "any combination of two or more things". StuRat (talk) 23:27, 11 June 2012 (UTC)

I thought the definition of amalgam as a mixture of things was derived from the term for mixture of metals with Hg. And also, haven't you ever heard of a dental amalgam? Maybe I'm just biased because my family have bad teeth, but I thought they were a pretty common thing. 203.27.72.5 (talk) 02:33, 12 June 2012 (UTC)

Based on the lack of "mercury" or "hydrargyrum" in the word, I doubt that origin. Looking up the etymology, I found this ref (page 274): [2]. It states "AMALGAM is an old alchemical word, probably a perversion of the Latin malagma, a mollifying poultice, traceable to the Greek malassein, to soften. Other early writers associate it with ana, together, and gamos, marriage. Bacon (1626) writes the word 'amalagma'." So, it seems that the general meaning came first, and then the more specific application to amalgams of mercury. And yes, I've heard of dental amalgams. StuRat (talk) 21:51, 12 June 2012 (UTC)

I have a gruesome story about acid which seems relevant. My dad worked in the petrochemical industry for a while, and witnessed a fatal industrial accident. A worker was on a catwalk above a large, open vat of highly concentrated sulfuric acid. He had a full "chem suit" on. But, apparently forgetting where he was, he removed his head gear/mask to scratch an itch, accidentally inhaled, and passed out immediately from the fumes. He then fell between the bars, into the vat, and fully dissolved in a rush of bubbles. When they emptied the tank, nothing was left of him and his clothes, except his gold wedding ring. If I remember correctly, I believe this was a Union Carbide plant, near Whiting, Indiana (a suburb of Chicago). So, their lax safety practices (like even having a catwalk above an open tank of acid), might have been a warning about their later Bhopal disaster, had anyone cared to listen. StuRat (talk) 20:15, 11 June 2012 (UTC)

Nothing left except the gold ring, eh?. This is the same as an ancient movie I saw once. Give me some time and I'll remember the name of the movie or who starred in it. Bit sus, Stu. Wickwack121.221.229.129 (talk) 01:26, 12 June 2012 (UTC)

Since everyone so far has seemed to assume that the OP is only asking about a splash or a smear of acid, I will consider what would result if a bulk container of concentrated sulphuric or nitric acid ruptured in a builing. Having a large amount of acid pour into a brick room and flow out it's windows and doors would expose the brick work to a great deal of acid. If the bricks themselves do react with the acid then the acid will more or less uniformly degrade the walls it contacts in proportion to it's flow. If the bricks do not react quickly with the acid, then the effects will be worse, with the acid being consumed only in degrading the mortar and thus quickly destroying the structural integrity. Assuming a cement floor it will also react with the floor helping to neutralize the acid. It will also react mainly with the lower parts of the wall as that is where it will make most contact. This would severly weaken the structure leading to the walls probably falling inward. If the masonry is reinforced with rebar or similar this will also be attacked by the acid once it has penetrated a significant distance into the wall. Also of interest is that sulfuric acid releases a lot of heat when it makes contact with water. A few mls added to a cup of water will make it boil briefly. In a building with a water supply it could make a big mess. 203.27.72.5 (talk) 21:48, 11 June 2012 (UTC)

Its a question of scale and basic stoichiometry. The amount of material the acid will dissolve cannot exceed the amount of acid there is (as measured by the actual number of acid molecules). So, if you had, say, 1000 kg of concentrated sulfuric acid, that's 980 kg of actual sulfuric acid (conc. H2SO4 is 98% ), and since sulfuric acid has a molar mass of 98 g/mole thats 10,000 moles of sulfuric acid. Thus, if you know what you are dissolving, you can calculate the maximum mass of the substance the acid would dissolve. Say something like limestone (basically calcium carbonate.) 1,000 moles of calcium carbonate (100 g/mole) would weigh 1000 kilograms, meaning that 1 metric ton of concentrated sulfuric acid would dissolve 1 metric ton of limestone. Since bricks, concrete, and other building materials probably contain some material which may be unreactive to sulfuric acid, you would dissolve even less. So basically, to dissolve a brick and mortar building you would need a minimum of an equivalent mass of sulfuric acid, and likely a substantial amount more. Highly impractical. --Jayron32 22:16, 11 June 2012 (UTC)

You wouldn't need more acid because of parts that don't dissolve, as the acid would not be "used up" by those, but would go around them and dissolve the next bit. StuRat (talk) 22:26, 11 June 2012 (UTC)

"Since bricks, concrete, and other building materials probably contain some material which may be unreactive to sulfuric acid, you would dissolve even less." Acutally, this would cause more damage to the structure. You will always dissolve the same amount of lime (as you pointed out). If there are unreactive materials (like sand and aggregate in concrete) then they will not contribute to neutralizing the acid but once the lime around them is dissolved they will also no longer contribute to the structural integrity. Regular concrete is about 1/6 portland cement which is in turn only 2/3 lime at most (so regular concrete is about 11% lime).

It is also important to consider the kinetics in addition to the scale and stoichiometry. If the acid's bulk flow out of the door is fast enough you might only have say the equivalent of half of it actually react to near completion with the lime and metal in the structure.

If we imagine a bulky box of 1000L(1840kg) of sulphuric acid suddenly released into a square room of say 25 square meters with a closed door, the acid will make contact with the walls up to a height of 4cm. The acid will be in contact with 25 square meters of floor and 0.8 square meters of wall. If the floor is concrete it will react with the acid and neutralize it. If we assume that the proportion of reaction with the floor and walls is determined only by the surface area of contact then ~3%(57kg) of the acid is used up in reacting with the walls. For lime (molecular mass 56g/mol) only 560kg is dissolved by 1t of sulfuric acid. If the walls are made out of regular concrete, then the lime portion of 513kg of concrete will be dissolved. That's 25.7kg per meter of wall. If we take the density of concrete as 2300kg/cubic meter and assume the acid dissolves away a perfect prism extending into the foot of the wall, the hole is 27.9cm deep (though it still contains undissolved aggregate). That's more than enough to topple most concrete walls. 203.27.72.5 (talk) 23:10, 11 June 2012 (UTC)

When bricklayers have finished a wall, the wall usually looks a bit messy becausee of odd bits of dirt and mortar adhering to the sides of the bricks. They use "brick cleaner" to get this off, usually by brushing the brick cleaner on to the bricks with a small paint brush. Brick cleaner is a mild acid, HCl I think. It does not attack brick at all, but does loosen sub-millimeter thicknesses of mortar. So, if any common acid has any effect on concrete or brick, it will be too slow and feeble to have any significance re Morgana's story. Wood is somewhat resistant too. But metal window and door frames - different story. Wickwack121.221.229.129 (talk) 01:26, 12 June 2012 (UTC)

The chemistry of cement is complex, but it probably would have been more useful if I consider the lime component of cement to become slaked lime upon setting. It is also important to note that since

$Ca(OH)_{2}+H_{2}SO_{4}\rightarrow CaSO_{4}+2H_{2}O\Delta H_{r}=206kJ/mol$
and a total of 1840kg(18776 moles) of sulfuric are reacting, the total energy liberated is 3,800,000kJ, which in and of itself would probably damage the structure.

Wickwack, I don't see how your argugment leads to your conclusion. Dilute HCl is used to dissolve messy edges on brickwork therefore if any common acid has any effect on concrete or brick, it will be too slow and feeble to have any significance? Dilute HCl is a long way away from concentrated sulfuric acid. You can't even dip a paint brush into sulfuric without the whole thing just about dissolving. Also it turns wood to charcoal by dehydration, so wood doesn't have much resistance to it at all. 203.27.72.5 (talk) 01:46, 12 June 2012 (UTC)

Wow, that's a lot of answers! Thanks Stu for numbering the questions, I was having a bit of a train of thought and it all came out in a bit of a jumble. The story's going to involve repeated application of acid; say if there was acid falling from the sky on a fairly regular basis, how long would it take for a building to be damaged beyond the point that it offers a reasonable amount of shelter- months, or years? Morgana Fiolett 07:55, 12 June 2012 (UTC)

As can be gathered from the various answers above, that would depend on a lot - type and concentration of acid, whther the building has a clay tile roof, galvanised steel roof, colour bond roof, etc, and whether the structure is Georgina style (ie no overhang to keep the acid rain off windows, doors, and walls), or has a roof overhang. It could range from minutes to years (or longer - acid rain occurs is some areas due to air polution - damge to buildings due to this takes decades and more). Perhaps you could tell us what you want plot wise, and we can then say what in the way of acid type and quantity is needed to do that. Wickwack124.178.61.192 (talk) 10:19, 12 June 2012 (UTC)

Basically, my premise is a post-apocalyptic scenario involving effectively actual acid rain; my survivors have to move from building to building in clear spells in search of ever-rarer safe shelters as the buildings gradually disintegrate. I want them to be able to survive for fifteen-twenty years though; I don't know whether to have the rain getting stronger at times than others, or more frequent, or whether to have it shifting areas so that they can find some areas that have less damage than others? (There's a trope about an isolated valley in there somewhere, I'm sure.) If there was normal rain falling in between acid rainfalls, would that alleviate the effects? The only character with any scientific knowledge is a weatherman, who can forecast the acid storms with some accuracy, but they won't know how it actually came about. At the beginning of that 15-20 year time scale there are plenty of buildings with adequate shelter so they can choose the ones with the most suitable facilities, but at the end of it they should be racing against the storms in hope of getting to the one building in reach that's more solid than the one that just practically fell down around them... As it's fiction, I just need something vaguely plausible, but if this sounds scientifically impossible, I'll have to reassess my plot and go for something over a more condensed time scale. Morgana Fiolett 14:12, 12 June 2012 (UTC)

For what it's worth, almost any roof under normal circumstances has a life expectancy of between 15 and 50 years from the time it's installed, the low end being a standard asphalt shingle or mopped-asphalt roof, and the upper end being copper, lead or tile roof. Some could make it to 100 years between major repairs. That's with normal climatic conditions. In this case, I mean "roof" to be the water-resistant membrane, not the structure, since a building's of little use even if structurally sound if the roof leaks in a general way, particularly if the leakage is toxic. I think your 15 or 20 year time frame would work for the purposes of your story as leaks develop in more cheaply-built materials relatively quickly, which would have a deleterious effect on the structure over a period of years. It's worth mentioning that even in usual circumstances, abandoned buildings deteriorate rapidly. Here's what Chernobyl looks like after 25 years [3]. A search for photographs of abandoned buildings will give you an idea of what happens: Detroit is the poster child for this sort of thing. Plain old water's pretty insidious. Acroterion (talk) 15:38, 12 June 2012 (UTC)

Now that I know what you're writing about I have a story that's much more relevant. My lab has some very badly designed fume cupboards. The scrubbers are woefully undersized for the volumes of acid vapor that get sucked up when we're digesting material for analysis. As a result, a lot of acid vapor makes it up into the flues of the roof and condenses either on the pipework or in the air, and settles back down on the roof in a fine mist.

Up until recently the roof was made of corrugated galvanised iron. This was eaten away by the acid at such a rate that it needed to be replaced once a year. The corrosion was heavily dependent on the seasons; in the wet season, the roof was fine as it was constantly being rinsed by the rain. In the dry season it corroded very fast, as the vapors would condense instantly upon hitting the cold air outside the flue and you could see the mist form and settle back down. We got tired of replacing the roof so often, so we started using a specially designed, corrugated plastic sheet to cover the iron. So far this has lasted well with no noticeable corrosion.

Before that we tried coating the roof in a marine grade sealant, which didn't really do anything to stop the acid. Another building here affected by acid damage is a shed that stores suldides. The sulfides decompose and form massive clouds of sulfur dioxide. Acid rain is basically caused by sulfur dioxide, so having clouds of it form in the shed and escape through the doors and louvers destroys the structure. It's just been redone recently with the same plasic sheeting as the lab, but a few months ago the panels on the roof and walls were so corroded that many were only held onto the frame at one or two points and they just flapped in the breeze. The whirlybird roof ventilators were rusted into solid clumps of metal and the concrete floor was severely damaged all over.

If there was a literal storm of relatively concentrated acid I'm sure it would destroy many structures in one hit and people would be severely injured. The agitation would ensure a very complete reaction with metals and concrete, and aftwards, the acid on the ground would fume as it all dries up which would gas people breathing it in, even if they were inside (unless there were adequate seals on the door, windows, etc. Agriculture would also be entirely shot. On the other hand, concentrated acid storms are really not likely to happen, and even if they did, the conditions required for them to be a long term feature of global weather patterns would be...improbable. 203.27.72.5 (talk) 22:00, 12 June 2012 (UTC)

Nasty sulphides stored in an obviously inadequate & unsecured manner, replacing roof iron once per year? Must be a government lab. It would have much been cheaper to install proper fume cupboards & chimneys etc rather than replace a building roof each year. Wickwack60.228.240.65 (talk) 01:56, 13 June 2012 (UTC)

It's not a government lab. Yes, it would have been better to install fume cupboards that were up to the task, but the supplier advised that these were adequate, and the previous Chief Chemist took them at their word. The shed has been rebuilt with acid resistant materials, but recently we've found that just storing the sulfides in piles outside is better anyway, because both the heat and fumes dissipate much more quickly. The heat is a problem because it speeds up the reaction so you get way more fumes and the fumes are a hazard to the workers here as well as the structures. We're talking about the bulk storage of thousands of tonnes of sulfide ore awaiting transport. I don't know what you mean by saying it's unsecured. Would someone want to steal it? And the storage requirements are met, so whether or not it's adequate depends on who you're asking. 203.27.72.5 (talk) 03:35, 13 June 2012 (UTC)

Hi energy protons direct conversion into electricity?

The article Helium-3#Power generation mentions direct conversion of protons into electric energy with efficiencies at 70%, how would such conversion device be built?, a simple metal plate or more complex?

A second-generation approach to controlled fusion power involves combining helium-3 (³₂He) and deuterium (²₁H). This reaction produces a helium-4 ion (⁴₂He) (like an alpha particle, but of different origin) and a high-energy proton (positively charged hydrogen ion) (¹₁p). The most important potential advantage of this fusion reaction for power production as well as other applications lies in its compatibility with the use of electrostatic fields to control fuel ions and the fusion protons. Protons, as positively charged particles, can be converted directly into electricity, through use of solid-state conversion materials as well as other techniques. Potential conversion efficiencies of 70% may be possible, as there is no need to convert proton energy to heat in order to drive a turbine-powered electrical generator.

Electron9 (talk) 20:31, 11 June 2012 (UTC)

I don't really have any idea if this would work, but I imagine a coil of wire with the proton beam path as a tangent to the coil. As the protons move past they drag electrons with them and induce a current through the wire. 203.27.72.5 (talk) 03:39, 12 June 2012 (UTC)

One way is to have the protons leave one electrode and head for another one and then impact it. The other electrode will be at a high positive voltage, so that the proton loses velocity and impacts at low speed. However this is unlikely to be practical for voltages much above 1 million volts, and you really have to get the protons moving in one direction, and any fast electrons going in another direction. Having a variety of energies mean that some higher energy protons will have energy wasted, and low energy protons may not reach the electrode. Graeme Bartlett (talk) 10:58, 12 June 2012 (UTC)

air bag safety

I am trying to find the minimum weight and height for a child to safely sit in the front seat of a car with an airbag. All I can find so far is Yahoo! Answers or else "check with the vehicle's manufacturer" and I want to know what either the Canadian or U.S. government safety recommendations are. Many thanks! I have a reference question (talk) 20:36, 11 June 2012 (UTC)

The danger is that the airbag will only hit their head, and not their chest, and thus break their neck. So, how high the child's head is above the seat is important, and this would be affected by the use of a booster seat, whether the car seat is adjusted all the way up (assuming it has an up-down adjustment), and the model of the car. So, it probably doesn't depend on their weight as much as their height, and those other factors I mentioned. Also, some airbags have a slower inflation setting for children, which allows them to be used for younger kids. So, unfortunately, the model of the car is a critical part of the decision. StuRat (talk) 22:20, 11 June 2012 (UTC)

According to the U.S. Centers for Disease Control and Prevention website section "Child Passenger Safety": "All children younger than 13 years should ride in the back seat. Never place a child in the front seat facing an airbag." See also safercar.gov at "Air Bag Safety".

• Yale School of Medicine says more specifically:"In fact, no child younger than 13 or under 65 pounds should sit in the front seat of a car equipped with passenger-side air bags, according to both the Department of Transportation (DOT) and the National Transportation Safety Board (NTSB)" here "Air Bags: Not for Children". Nb. found by 'Googling' "air bag children" - 220 of ^Borg 16:26, 13 June 2012 (UTC)

How will Venus to survive tip of RGB

From the youtube I saw [4] two guys from UC Berkeley demonstrate the reasons Earth will escape enough to avoid engulfment over RGB, is it still possible enough from Venus to expand its orbit enough so it misses over tip of RGB's engulfment. Since this site said Venus is most likely swallowed up, is it still currently possible for Venus to escape enough to avoid destruction? Is tip of RGB definitely 1.2 AU, and loss sun's mass definitely 33%, or is it just part of Dr Smith's calculation. Do other guys agree with the same calculation or people come up with different calculations every year? Is it also possible for sun's tip of RGB to miss Earth's orbit by degrees of fractions, because at 1.2 AU solar radius, it is unlikely for Venus to avoid engulfment. Can tip of RGB make sun lose to more than 42% of the solar mass? When you look at the similar stars going through the similar fate, how will you only get the estimates, is the estimates more direct or the information from foreign stars is still chaotic?--69.226.45.43 (talk) 20:38, 11 June 2012 (UTC)

I'm not sure we have all that reliable predictions as to what will happen. As you have noted, some estimates say Earth is toast; others say it will survive. I think the best we can say is that, being closer, Venus stands less of a chance of surviving, but there is no perfect prediction one way or the other. --Jayron32 20:43, 11 June 2012 (UTC)

Not that I don't enjoy these kind of questions, but any reason you're so interested in this topic? As I've mentioned before this is an event extremely far in the future that has an incredible amount of uncertainty in it, so there's not much that we can say definitively. I wouldn't even go as far as to say that Mercury won't escape; it's certainly possible! Again, not criticizing, just curious as to why you're so curious.-RunningOnBrains^(talk) 22:14, 11 June 2012 (UTC)

Me and my stupid summarizing mind, I failed to read your whole question. I'm not sure the exact uncertainty involved in these calculations, but I suspect they are quite large (possibly as large as you allude to above). I'll see if I can dig up some exact figures later today. -RunningOnBrains^(talk) 22:19, 11 June 2012 (UTC)

I wasn't really paying attention for first. I just talking too much about why earth will be swallowed up just because Dr. Smith's variable coming up later. I didn't know this site still exist. Is there any difference between Second Red Giant Phase and Asymptotic Giant. The Pogge's calculation calls it Second Red Giant Phase but Smith calls it Asymptotic Giant, I am not sure which way should I identify it. I thought Asymptotic Giant is not a Red Giant.--69.226.45.43 (talk) 22:25, 11 June 2012 (UTC)

There are probably hundreds of scientific papers on this topic and they all make different assumptions and therefore reach different conclusions. The only definitive answer anyone can give you is the same answer you've been given every other time you've asked these questions: nobody knows. --Tango (talk) 01:14, 12 June 2012 (UTC)

No offense, but unless you are an astronomer this question is pointless for you. If you are an astronomer, then you are perfectly capable of answering the question yourself. Regardless of whether the Earth gets engulfed or not, its surface temperature is certain to rise over 500°C, making it uninhabitable. Anonymous.translator (talk) 01:45, 12 June 2012 (UTC)

Okay I got it clear now. Thank you all for your help and the time in clarifications.--69.226.45.43 (talk) 02:09, 12 June 2012 (UTC)

When the sun swells up there will be a lot of tidal drag on the planets that will slow them down and let them fall in, and once they are in the solar envelope, there will be a bow shock wave that also causes more dragging and spiralling in. Once in the sun's atmosphere the planets will not be vapourised immediately but will take thousands of years to do so, when the planet gets deep inside the evaporation will be even faster, and it is very unlikely that the star will shrink back on down and leave a planet corpse behind. If it was left it may just be an iron core. Graeme Bartlett (talk) 10:50, 12 June 2012 (UTC)

It's not certain Venus will exist by that time, see here:

A long-term numerical integration of the classical Newtonian approximation to the planetary orbital motions of the full Solar System (sun + 8 planets), spanning 20 Gyr, was performed. The results showed no severe instability arising over this time interval. Subsequently, utilizing a bifurcation method described by Jacques Laskar, two numerical experiments were performed with the goal of determining dynamically allowed evolutions for the Solar System in which the planetary orbits become unstable. The experiments yielded one evolution in which Mercury falls onto the Sun at ~1.261Gyr from now, and another in which Mercury and Venus collide in ~862Myr. In the latter solution, as a result of Mercury's unstable behavior, Mars was ejected from the Solar System at ~822Myr. We have performed a number of numerical tests that confirm these results, and indicate that they are not numerical artifacts. Using synthetic secular perturbation theory, we find that Mercury is destabilized via an entrance into a linear secular resonance with Jupiter in which their corresponding eigenfrequencies experience extended periods of commensurability. The effects of general relativity on the dynamical stability are discussed. An application of the bifurcation method to the outer Solar System (Jupiter, Saturn, Uranus, and Neptune) showed no sign of instability during the course of 24Gyr of integrations, in keeping with an expected Uranian dynamical lifetime of 10^(18) years.

Count Iblis (talk) 19:16, 12 June 2012 (UTC)

June 12

NASA Venus transit video artifacts

In the NASA SDO Venus transit video, [5] (APOD) and [6] (YouTube), there are a couple of points, such as at 0:33 and 0:43, where Venus appears to be fractionally translucent. As Venus crosses into the sun's field, the edge of the sun appears to be partially visible through the planet's disk. Is this just my imagination, an optical illusion, an artifact of the video encoding/decoding, or is this evidence that the video was manipulated and possibly fabricated? I'm not questioning the transit itself. I just think the video seems odd. -- Tom N (tcncv) talk/contrib 05:45, 12 June 2012 (UTC)

I had noticed this too; I instantly knew it would be a hit in the flat earth community. Ever take a digital picture of a really bright object (like the sun)? The over-exposed area will tend to bleed into nearby under-exposed areas. This is called Sensor blooming, and I suspect a similar effect is at play here. The image at right is an extreme example; not only does light bleed over the silhouette of the tower, but into the landscape itself. It can be mitigated by high-quality sensors, but I wouldn't be surprised if there was a slight effect left over, and the folks at NASA would probably find it scientifically dishonest to digitally remove it for aesthetic purposes.

Another related effect is called halation, but I think it's solely a phenomenon related to photographic film. That link doesn't show any examples, but this is a good one. -RunningOnBrains^(talk) 06:06, 12 June 2012 (UTC)

Have you seen the entire video? Sensor blooming cannot explain why Venus often appears transparent. There should be absolutely no way for the solar surface behind Venus to be visible, because it's blocked by Venus. --140.180.5.169 (talk) 06:45, 12 June 2012 (UTC)

It may appear transparent (really, translucent), but it's not. If you watch closely (I recommend full-screen), you'll see that there is never any anomaly over Venus's disk when it passes completely over a bright part of the Sun, but it does show an anomaly when passing near a bright spot. This is especially evident from a close watching of the sequence starting at 0:58. In response to the below points, I doubt any filtering in the traditional sense is to blame. Obviously some digital filtering had to be done to get a single wavelength from the raw signal, but this should be pixel-by-pixel, and not result in any translational anomalies. I am still convinced that it is a purely optical effect. -RunningOnBrains^(talk) 08:01, 12 June 2012 (UTC)

You don't isolate a wavelength with a digital filter. (Or an analog one for that matter.)--Srleffler (talk) 17:15, 12 June 2012 (UTC)

Hmm, I can't find any papers on the exact mechanism by which satellites isolate different frequency channels; I always just assumed it was a digital filter of some kind. Do you have any insight on exactly how this is done? -RunningOnBrains^(talk) 17:35, 12 June 2012 (UTC)

I would say that the video is "manipulated" for those sequences, and I doubt that NASA would deny it. There would need to be some pretty intense filtering to remove the background brilliance of the sun, and it's very difficult to do intense filtering without producing artifacts. Looie496 (talk) 06:38, 12 June 2012 (UTC)

Why does filtering need to produce artifacts? Have you ever looked through a neutral density filter or any other type of solar filter? Any solar filter that isn't total junk introduces absolutely no visible distortions or artifacts. --140.180.5.169 (talk) 06:45, 12 June 2012 (UTC)

The more I think about it, the more I think this might be the result of lossy video encoding and compression. If the encoder attempted to generate the darkened disk by defining some kind of transform to previously rendered data or , it could yield "close", but imperfect results. It would be interesting to see the set of original still images. -- Tom N (tcncv) talk/contrib 07:11, 12 June 2012 (UTC)

I've now found some still images at http://svs.gsfc.nasa.gov/vis/a010000/a010900/a010996/index.html, including a couple that match the video at about 0:45. The still images show a much more distinct black disk that lacks the haze which gives the appearance of translucency in the video. -- Tom N (tcncv) talk/contrib 07:42, 12 June 2012 (UTC)

I'm not sure which images you are referencing, but the still images look exactly the same to me. The bleed-through is especially evident in the 193 Angstrom image, just as in the video. -RunningOnBrains^(talk) 07:46, 12 June 2012 (UTC)

Here are some excerpts of both the video and the equivalent still image.

Video screen capture and crop

Equivalent original still image (also cropped)

I think the difference in quality is fairly obvious, and supports the assertion that any apparent translucency in the video is an artifact of the video creation process. -- Tom N (tcncv) talk/contrib 08:27, 12 June 2012 (UTC)

It's unfortunate that the pictures aren't larger. But my thought as a non-expert is that the video is of lower resolution, which lends itself to being "fuzzier". Note that the texture of the sun is also less sharply defined in the video frame than in the still picture. ←Baseball Bugs ^{What's up, Doc?} carrots→ 09:18, 12 June 2012 (UTC)

Hmm, I see what you mean now, but if you look at the still composite I linked above the translucency is still apparent. This single image also features it. -RunningOnBrains^(talk) 15:32, 12 June 2012 (UTC)

Three explanations of light inside Venus' disk might apply:

1. Long decay in display. The effect is seen in an oscilloscope that continues to show weakly a trace after it has been switched off, and I have seen a full image persist many seconds after a TV is switched off in a darkened room. It would give a persistent ghost image of the edge of the Sun after it was in reality obscured by Venus
2. Long decay in image sensor.
3. Reflection by Venus. The surface of Venus' outer atrmosphere is neither a mirror nor totally black but something in between, and it may receive some light by backscatter within its atmosphere or by reflection from our own Earth, Moon or even interplanetary dust.

Case 1. would vary depending on the many different kinds of display that we each have. Case 2. probably involves a number of different sensors used by NASA etc. Cases 1. and 2. would give identical effects on video but 1. would not be seen on a still frame. The edge crossing at 0:43 in the NASA video comes closest to identifying them because there is visible activity on the Sun's edge. Is there edge activity visible "through" Venus or is the edge seen there just a static hangover from earlier frames? I can't decide. I do see low level light patterns on Venus at 0.42 before the transit even begins which support explanation 3. Added a small 2nd thought.DriveByWire (talk) 14:00, 12 June 2012 (UTC)

That last possibility is a really interesting one; I hadn't even considered it. I really want to get a NASA scientist on the phone. -RunningOnBrains^(talk) 15:32, 12 June 2012 (UTC)

Anybody who is looking for scientifically-valid data in the individual pixels of a compressed image or video frame is bound to be surprised. Here are some of the (many) practical engineering details that will derail your analysis, presented in roughly decreasing order of interference, ranging from "a standard engineering procedure that catastrophically destroys the scientific validity of individual pixels" and ending with "a standard design principle that only destroys the validity slightly, but in a recoverable way."

1. Start by reading about image compression.
2. Next, read about the optical design of practical lenses and telescopes, in our article on photographic lens design.
3. Read about digital signal processing, CMOS imagers, and digital signal processing in general.

Finally, if you're looking at SDO data, you can access Solar Dynamics Observatory raw data free of cost - so there's really no excuse for anyone to still be looking at summary preview images, and drawing conclusions about Venus' optical properties. JPEG and video-files are also available, because they "look neat" and help citizens and scientists appreciate the sort of "big picture" objectives; but if you actually want to analyze the observations, you have got to use the scientific data. Its error margins are well-known and well-documented, and all that information is available free-of-charge. Nimur (talk) 16:39, 12 June 2012 (UTC)

Thank you for the links - particularly the image download site. Unfortunately, the downloadable images appear to only be available as .jpg files, which may have already gone through some degree of lossy compression. I accessed a few leading transition images for AIA 171 gold (here), AIA 193 bronze (here), and AIA 304 red (here), and they all still show "translucent-like" artifacts. Additional higher resolution images may be retrieved from this page by selecting a date/time range of 2012-06-05 22:00:00 to 2012-06-05 22:30:00, but they show the same. Perhaps lossless versions of the images may exist but are just not publicly accessible. The download link seems to have sufficient bandwidth for such data. Anybody out there with insider connections? -- Tom N (tcncv) talk/contrib 20:55, 12 June 2012 (UTC)

To clarify the Q, I see two odd things:

1) The slight reddish color of Venus on the video. The is absent in the still.

2) Small red tendrils extending into the black disk of Venus on the still. This seems like the harder to explain issue, to me. StuRat (talk) 21:24, 12 June 2012 (UTC)

Assuming I'm seeing your "tendrils", they're ordinary imaging artifacts, with about the same level of reality as the Martian canals. --Carnildo (talk) 00:05, 13 June 2012 (UTC)

Fascinating video. I had originally considered the idea that the effect may have been a reflection by Venus' atmosphere, but I don't believe that's the case. Being somewhat familiar with the mechanics of digital video, I'd say those are primarily compression artifacts. The aforementioned halation may be playing somewhat of a role as well, though I doubt it is to any significant degree, given the still/video comparison posted by Tom. Evanh2008 ^{(talk|contribs)} 00:57, 13 June 2012 (UTC)

I tend to agree with Evanh and others regarding the likelihood that compression artifacts are the main culprit. Given that the image sampling rate is on the order one frame per minute and that NASA would be expected to be using high quality image sensors designed to handle stark contrasts, I would expect a better quality from the raw data. The Hinode image File:Hinode Views the 2012 Venus Transit.jpg appears much more crisp - with a very nice atmosphere glow. I'm hoping to find similar quality in the SDO images. I found some potentially better quality FITS and .jp2 images here and here, but I have not been able to view them. I tried GIMP, but it was unable too open either format. Does anyone have a recomendation for a free FITS viewer that can handle these filess and runs on a WIN7 machine? -- Tom N (tcncv) talk/contrib 03:43, 13 June 2012 (UTC)

Petrol engine horsepower rating?

The power rating of the petrol engine in an ordinary car. That is usually specified as "100 hp". Does that mean the chemical power of the fuel that goes in, or the power that comes out as mechanical energy? (ignoring heat) Electron9 (talk) 13:11, 12 June 2012 (UTC)

Horsepower of an engine indicates the power output. If it were the input, all engines that use the same fuel would be have the same rating, no? See Horsepower#Measurement and dynamometer. There are several different points where output can be measured, and where it is measured will depend on circumstances. I don't know where the standard place of measurement for cars is, but I'd guess the crankshaft or flywheel. SemanticMantis (talk) 13:45, 12 June 2012 (UTC)

You can generally get the power rated for a custom car either "at the fly wheel" or "at the road" (i.e. at the tyre using a dynamo). 203.27.72.5 (talk) 22:13, 12 June 2012 (UTC)

The power of petrol engines is given slightly different ways by different organisations, but never as the "chemical power" of the fuel consumed, as that has no practical value. The energy value of fuel is given as kilojoules per kilogram (metric countries) or calories per pound (USA). The most common way of rating an engine is what is known as the brake power, measured at the flywheel, and given in kilowatts (metric) or horsepower (USA). This is the actual mechanical power output of the engine in usuable or complete form. The term "brake" comes from the use of a brake in a dynamometer test to load up the engine. For accuracy, it is important to understand what is meant by usuable & complete. For example, for a car engine, apart from friction and thermodynamic losses inside to the engine, power is absorbed by the water pump, alternator, and radiator fan, thus reducing the power available at the flywheel to move the car along. Power is also aborbed by the power steering pump and airconditioner (if fitted), but these are not installed for a power output test or calculation. Power output varies slightly according to the energy value of the fuel, the ambient air temperature and humidity, and the altitude. Therefore, there are agreed standards in the USA and Europe on the fuel to be used when testing, and engines are rated at 25 C (Europe), and 64 F (USA, if I remember correctly) at sea level, 50% humidity. For a marine engine, as supplied it will not have a radiator fan (usually) and often not an alternator either. In such cases the power rating quoted will not allow for losses in the raditor fan and alternator. The SAE in USA, and the ISO in Europe publish standards on exactly how to do it, and how to apply corrections for unstandard condtions (eg high altitude) Ratbone58.170.167.209 (talk) 13:49, 12 June 2012 (UTC)

Car testers are most familiar with the chassis dynamometer or "rolling road" installed in many workshops. This measures rear wheel brake horsepower which is generally 15-20 percent less than the brake horsepower measured at the crankshaft or flywheel on an engine dynamometer. This video shows workshop measurement of a car's power. The measured power curve in kW is shown at 3:39 and comments below the video include one about the factory spec power by cuddlyable3, a former valued Wikipedia editor. DriveByWire (talk) 14:21, 12 June 2012 (UTC)

If one were to use a diesel engine for driving a electric generator then how these power ratings are done have significant impact. It was like I suspected, but I rather be sure. I added this information to the Petrol engine#Power measurement article. Electron9 (talk) 15:00, 12 June 2012 (UTC)

Motorcycle on an incline

A motorcycle mass $m$ drives on an incline $\theta$ its wheels separated by a distance $L$. Its center of mass is between the wheels and height $h$. What is the normal force at the wheel in the back? Apparently the answer is $mg({\frac {h}{L}}\sin \theta +{\frac {1}{2}}\cos \theta )$; how do you derive this? --150.203.114.37 (talk) 14:53, 12 June 2012 (UTC)

Draw a diagram of the motorcycle seen from the side that shows its center of mass and the two points where the wheels touch the incline; these 3 points when joined form an isosceles triangle. The base of the triangle is the length L and is at angle $\theta$ to the horizontal. The gravitational force on the motorcycle is $mg$ and its direction is vertical (down). You have to split this force into the following two components which do different things to the back wheel. 1) The component normal to the slope. Each wheel bears half of this force. (As a check: if $\theta$ is zero, the motorcycle is on the level and this component is just the whole weight $mg$. Remember cos(0) = 1.) 2) The component parallel to the incline. Think of the front wheel contact point as a fulcrum and this force component tries to rotate the rear of the motorcycle into the road, resisted of course by additional force at the back (lower) wheel. Look for the lever relations that involve $h$ and $L$ to calculate this additional force. DriveByWire (talk) 19:59, 12 June 2012 (UTC)

We have to assume the motorcycle is stationary and not being driven because we lack data on its weight, speed and acceleration. DriveByWire (talk) 21:15, 12 June 2012 (UTC)

Mass is given by $m$. As long as the acceleration is zero I think the speed is irrelevant; the motorcycle need not be stationary. --150.203.114.37 (talk) 21:29, 12 June 2012 (UTC)

That is true for the normal force at the wheel. However for any driving movement except free-rolling downhill there must be additional tangential traction force at the wheel(s) for accelerating and braking. DriveByWire (talk) 23:55, 12 June 2012 (UTC)

Sleeping

question removed by author.

Waking by alarm clock, spouse, children, neighbor workshop etc.. or by yourself ? Electron9 (talk) 15:32, 12 June 2012 (UTC)

A classic question. Getting more exercise and more sleep might help, but for me going to sleep at night is rarely as enticing as staying in bed in the morning :) SemanticMantis (talk) 15:37, 12 June 2012 (UTC)

See Circadian rhythm. You may also be interested in Delayed sleep phase disorder, not that the phenomenon you describe sounds like a disorder.--Srleffler (talk) 17:19, 12 June 2012 (UTC)

Discuss melatonin with your doctor. Subjectively I find it induces yawning and that tingly "time to sleep" feeling if taken on an empty stomach in a darkened room. μηδείς (talk) 20:39, 12 June 2012 (UTC)

The warm part is likely due to body heat that built up overnight under the blanket. StuRat (talk) 21:15, 12 June 2012 (UTC)

No, the circadian clock cranks up metabolic activity shortly before dawn, when it functions normally. And it probably isn't the answer the OP is looking for, but the most straightforward way to get that warm and comfortable feeling at night is sex. Looie496 (talk) 22:43, 12 June 2012 (UTC)

...followed quickly by a wet and sticky feeling. StuRat (talk) 04:20, 13 June 2012 (UTC)

I do not want your stupid comment on my question but every attempt to either hide it or remove it has been rebuffed by people who apparently think comments about your semen have a place on the reference desk. So I have removed my OWN question. Thank you to the people who tried to help before this person decided to post their jokes here and others felt it a good use of their time to keep adding back the crude sex joke when the question author clearly does not like it. Sleeper4545454545 (talk) 19:12, 13 June 2012 (UTC)

Conker tree (Aesculus hippocastanum)

I have a small Conker tree (Aesculus hippocastanum) growing in a medium-sized pot. The tree is ~10 years old and 6 foot tall. It has not flowered yet. It is doing wonderfully and continues to grow at a rate of about 1.5 feet per year. This however is becoming a slight problem as it is now dominating my tiny garden and starting to block access. As the constrains of the pot have not deterred its growth, I need to reduce its height or at least stop it from growing any taller somehow. Will it be able to take some pruning and cutting back, or will doing so likely kill it? I do not wish to plant it in the ground at this time as I would like to be able to take it with me when I move. Aesculus hippocastanum (talk) 15:15, 12 June 2012 (UTC)

I suggest you look into bonsai techniques, specifically pruning of both branches and roots. That is probably the simplest method for controlling size given your situation. 65.95.22.197 (talk) 15:35, 12 June 2012 (UTC)

Conkers recently came up on the ref desks, here: [7], and the consensus was that they can take a lot of pruning. Bonsai would be fun for this specimen, I also encourage you to look into it. You could knock it back to 3 feet tall, and fit it into a ~2 gallon pot (better if it's wider than deep). After a year or so, you can train it back to a nice shape. Note that it will still continue to grow, the idea is that you prune it back as necessary. You could try downsizing the pot only, and leaving the top intact, but that may just force the tree to self-thin its branches, i.e. cause some to die off. SemanticMantis (talk) 15:43, 12 June 2012 (UTC)

Do you know any free sources for beginners on bonsai? Itsmejudith (talk) 22:43, 12 June 2012 (UTC)

The Royal Horticultural Society has a page here, if that helps. Tonywalton ^Talk 23:52, 12 June 2012 (UTC)

Uses of gravitational acceleration

How much of today's technology depends on the discovery that gravitational acceleration is similar for all objects on different places on earth, neglecting friction? Did we have to discover this fact in order to build spaceships? in order to build airplanes? If so, how does it help us? Jobnikon (talk) 16:34, 12 June 2012 (UTC)

Well, if it differed, then a plane designed for one part of the world would be very difficult to control in another, with a much greater or less amount of relative lift.--Gilderien Chat|List of good deeds 16:48, 12 June 2012 (UTC)

Not much. Scales come to mind. Note that gravitational acceleration is not exactly the same everywhere on Earth. See Gravity of Earth and Gravimetry.--Srleffler (talk) 17:24, 12 June 2012 (UTC)

Are you talking about the fact that bowling balls and feather pillows accelerate at the same speed when they fall? I mean, it is pretty fundamental, in the sense that anything which uses the gravitational acceleration equations somewhere is going to be based on that understanding. But figuring out exactly where that would matter is a tough thing; if gravitational acceleration was correlated with mass, for example, then a whole lot of things would have to be wired up differently to take that into account. It would be easier (in a sense) to answer that question than to ask what technologies would be capable if people were simply ignorant of the acceleration equation. --Mr.98 (talk) 22:01, 12 June 2012 (UTC)

You couldn't understand orbits, and you couldn't predict the flight paths of spacecraft. That means no space ships, no weather satellites, no GPS, no satellite phones. You also couldn't predict all of the motion of celestial bodies, for example the motion of those colliding. As Mr.98 said, it's pretty fundamental though. If NASA blew up one rocket because they didn't know that gravity acts the same on all ojects, they'd have enough data from that mission to deduce that relationship. On earth, gravity is generally opposed by air resistance, which depends on the objects shape and composition. This effectively masks the fact that all objects fall at the same rate, so no realising the details and thinking that objects fall at different rates due to gravity can still lead to accurate predicitons.

I suppose the reason this is hard to answer definitively, is that if you imagine a world where we didn't realise that the earth's gravity provides the same acceleration to all objects, the people in that world also can't have yet understood air resistance. They possibly don't realise the air is there at all. Their level of understanding of the natural world is so far backward it's hard to believe that they could have much modern technology at all. 203.27.72.5 (talk) 22:28, 12 June 2012 (UTC)

None of today's technology depends on that discovery. All of our technology would work even if that discovery had never been made. People might not understand why it worked, but it would still work. Looie496 (talk) 22:38, 12 June 2012 (UTC)

Actually, some fairly fundamental technology does. If gravitational acceleration varied from place to place, or if it was not known to be constant, it would not be possible to sell goods measured by weight: you couldn't order a ton of wheat from Chicago, another ton from Denver, and count on having two equal-sized loads of wheat arrive at your bakery in New York. --Carnildo (talk) 00:12, 13 June 2012 (UTC)

I think the OP really wanted to ask, "what technologies required an understanding of the way gravity works in order for them to be developed?". 203.27.72.5 (talk) 00:56, 13 June 2012 (UTC)

Disagree, you'd either calibrate spring type scales centrally, or use 'comparison of mass' type scales. Greglocock (talk) 01:05, 13 June 2012 (UTC)

Calibrating the spring type ones centrally wouldn't work. They would need to be calibrated with a standard weight before use in any given location (many extemely accurate pan balances do this in real life automatically which accounts for the variation in acceleration due to gravity that exists). 203.27.72.5 (talk) 01:20, 13 June 2012 (UTC)

As a side note; my understanding is that the force of gravity is slightly different on different objects. If you think of a massive object, e.g. the moon, the gravitational acceleration between it and the earth should be greater than something small e.g me at the same distance from earth. Maybe this just depends upon your reference frame though; is the moon accelerating toward the earth at the same rate as a small object but it hits it earlier because it also accelerates the earth in its direction? 203.27.72.5 (talk) 01:03, 13 June 2012 (UTC)

Prediction reliability threshold

Is there some scientific threshold, which separates mere coincidences in the alleged prediction capabilities from actual prediction power, that is, a threshold that says that the number of correct prediction is unusually high for coincidences? How many correct prediction it equals? — Preceding unsigned comment added by 176.241.247.17 (talk) 17:10, 12 June 2012 (UTC)

Yes, but it's not a fixed number of predictions. How many predictions you need depends on how distinct the predictions of the competing hypotheses are and how precise your data is. If you have two hypotheses with very different predictions, and your data is very precise, one experiment may be enough. If the predictions of the theories are very similar and the data is not very precise, one may need to do an enormous number of experiments to decide which hypothesis to reject.

You may be interested in Statistical hypothesis testing and Confidence interval.--Srleffler (talk) 17:32, 12 June 2012 (UTC)

Or in statistical significance, which is the precise name for the concept being discussed here. Looie496 (talk) 17:46, 12 June 2012 (UTC)

If you're having trouble working out what the "other" hypothesis should be, see null hypothesis. 203.27.72.5 (talk) 22:38, 12 June 2012 (UTC)

Closed bottle at "room" temperature equilibrium?

If bottle is filled to half with a liquid like water and then closed with a cap. The pressure in the gas section will initially be the same as the surroundings. But when time goes to infinity the pressure in the gas portion will have an elevated pressure by the vapor pressure over the surroundings, ie 2.3 kPa in this case? Electron9 (talk) 17:20, 12 June 2012 (UTC)

That's right, the vapor pressure will reach its equilibrium value, and that means that the total pressure will have become larger. The fact that you go from equal pressure to unequal pressure with the environment means that you could extract work from this system, and that in turn derives from the fact that you started out with a non-equilibrium situation. In equilibrium, all the water should have been evaporated, because ambient relative huidity is typically a lot less than 100%. So, it's not completely nonsense to say that a car can in theory run on pure water... Count Iblis (talk) 17:39, 12 June 2012 (UTC)

...until the water completely evaporates, at which point you have to do more work to reliquefy it. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 18:39, 12 June 2012 (UTC)

At which point you open the valve for more tap water.. Electron9 (talk) 18:41, 12 June 2012 (UTC)

Just as a point of order, a back-of-the-envelope calculation (which may be wrong, I made a bunch of questionable late-afternoon assumptions), you could get about 2kJ of energy from a cubic meter tank using this method, assuming you could perfectly extract all the available energy (for reference, gasoline provides about 350 kJ per liter, or 1320 kJ per gallon). I'm loathe to believe that running a car on this puny amount of energy will ever be possible, and if you could, I think you'd be better off using solar power so that the car would never have to stop to refuel. Also, our article on Compressed air cars suggests that the efficiency of energy extraction would be less than half that of an internal combustion engine. -RunningOnBrains^(talk) 21:48, 12 June 2012 (UTC)

You're going to be limited by carnot efficiency. This is a heat engine because the water draws heat from its surroundings in order to evapourate and increase the pressure. The difference between $T_{H}$ and $T_{C}$ is going to be minute, so your efficiency will be terrible. To increase the efficiency, you need to heat the tank of water (and now you have a steam engine). 203.27.72.5 (talk) 02:45, 13 June 2012 (UTC)

The Carnot efficiency is irrelevant here, because the water will be evaporating at ambient temperature leading to entropy increase. If the evoporated water is to be used in a close circuit, then since the internal state will return to the starting point, and then the total entropy change is due to the heat that goes into the system and out of the system, in which case you recover the Carnot formula for the maximum work that can be done.

Now, our case is different, the water vapor is allowed to escape and you can refill the water tank. The maximum amount of work that is available for the car can be computed as follows. Take the initial state to be the car with an empty water tank, the final state is the same car again with an empty water tank but where a water reservoir contains a little less water and there is a bit more water vapor in the atmosphere. Also some work has been performed by the car engine. We assume that both the initial and final states are states of thermal equilibrium with a well defined teperatures and pressure equal to that of the environment.

Then since the car's state can be taken to be the same in the initial and final state, we only need to focus on the water reservoir and the water vapor in the atmosphere. The question is then simply how much work you can extract by bringing some amount of water from the water reservoir to the atmosphere. Water has a latent heat of L = 2450 kj/kg at room temperature, so if we evaporate 1 kg of water and also let it perform an amount of W of work, we would have extraced a amount of heat of Q = L + W. This leads to an entropy change of - (L+W)/T, where we are not yet taking into account the entropy increase due to the added evaporated water.

The increase in entropy due to the phase change of water from liquid to vapor would be L/T if the relative humidity would be 100%. Because the water will be added to the atmosphere at some lower relative humidity r, it will be expanding by a factor 1/r more and thus the entropy increase will be larger. If we treat the vapor as an ideal gas, then the entropy increases by an amount -N k Log(r). The total entropy increase due to entropy drop from extracting the heat Q from the environment and the evaporation of the water is thus -N k Log(r) - W/T. At best this is zero, so the maximum amount of work we can extract is

-N k T Log(r), which is about 69 kJ/kg at 20°C and relative humidity of r = 0.6. Count Iblis (talk) 18:25, 13 June 2012 (UTC)

With the vapor pressure of Acetone at 14.7 kPa and Ammonia at 800 kPa, I think they are quite interesting candidates for an experiment. The catch being that they are quite corrosive and toxic. Electron9 (talk) 18:33, 12 June 2012 (UTC)

And a bit more expensive than water. But with the both, you could extract energy in that way, then take the gas in the head space as a fuel for an ICE then wait for more to volatalise. 203.27.72.5 (talk) 22:43, 12 June 2012 (UTC)

Surface Charges on an open circuit

I have a circuit diagram with two 1.5 volt batteries in series, an 8 ohm resistor and a 32 ohm resistor. These would all be connected in a single loop, however there is a gap between the battery and the 32 ohm resistor. What do the surface charges look like on this open circuit? Would there be any net charge at any point? Widener (talk) 18:09, 12 June 2012 (UTC)

Yes. Let us consider the two 1.5 V cells in series to be a battery and ignore the internal node where the two cells connect to each other. There would be a net positive charge on the components and wires connected to the positive terminal of the battery and a net negative charge on the components and wires connected to the negative terminal of the battery. Jc3s5h (talk) 18:29, 12 June 2012 (UTC)

Wouldn't that cause a current to flow? Widener (talk) 18:34, 12 June 2012 (UTC)

No, because the battery's own polarity opposes any resulting current. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 18:38, 12 June 2012 (UTC)

I don't think that explanation clearly describes this situation. Polarity is the wrong word here; and voltage does not "oppose" current. Nimur (talk) 19:31, 12 June 2012 (UTC)

Okay. How do the surface charges differ in this circuit than if the circuit was closed in a loop? You would have charges on the nodes of the battery, but nowhere else? Whereas you would have surface charges at every point in the wire if the circuit were closed? Widener (talk) 18:40, 12 June 2012 (UTC)

The battery "pushes" electrons in one direction. This builds up a concentration gradient of electrons which eventually creates a high enough potential difference that it equals the battery's electron-pushing power and stops the current. In a closed circuit, the electrons would be pushed all the way around the circuit to the other side of the battery, which sucks them up. In an open circuit, this cannot happen, and the growing potential difference that the electrons must be pushed up shuts down the current flow. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 18:57, 12 June 2012 (UTC)

You get a surface charge gradient in a closed circuit don't you? I thought that was what made electrons flow and caused the current. Widener (talk) 19:03, 12 June 2012 (UTC)

Electric current is the movement of electric charge. It is perfectly valid to have a non-uniform charge distribution, and no current (no motion of charge) in the steady state. Of course, to get to steady-state, charge needed to move (if the initial distribution was uniform). That temporary flow of current is what we call a transient response and requires the tools of electrodynamics for full analysis. In basic physics, we usually learn about electrostatics (steady-state) first, because the math is simpler; but you are correctly intuiting that there's something "unphysical" about it. You cannot move from one static condition to another without a dynamic transient response in-between; but typically, you just ignore that bit until you develop the tools to fully study it. Nimur (talk) 19:26, 12 June 2012 (UTC)

So there's no difference in the surface charges between an open and closed circuit? Widener (talk) 19:40, 12 June 2012 (UTC)

An electric capacitance exists between any two physical objects - even between the two leads of a resistor, or between one portion of a conducting wire and another portion of the same wire. However, in your scenario, that capacitance is very small. If you needed to estimate the actual net charge that builds up, you would need to measure or model the parasitic capacitance of your system; you could add this to your schematic. In practice, you can directly measure this very small capacitance using an oscilloscope or certain multimeters; in principle, these work by measuring an R-C time-constant). Then, you can trivially calculate the net charge, using the capacitor equation, charge Q = C V (capacitance times voltage). Nimur (talk) 18:45, 12 June 2012 (UTC)

The initial current flow that charges the capacitance of the gap in the circuit, which Nimur calls the transient response, is what Maxwell called Displacement current. Good luck measuring what is probably an extremely small (brief) time constant. DriveByWire (talk) 21:07, 12 June 2012 (UTC)

No, that is not correct. In fact, displacement current is a different other physical parameter with a different definition. A system that is not in equilibrium may undergo a transient response, which may include both a displacement-current and a true current. Nimur (talk) 00:37, 13 June 2012 (UTC)

So just to clarify - In the steady state, the surface charges between the open and closed circuits are exactly the same? And what do they look like around the resistors? Widener (talk) 23:09, 12 June 2012 (UTC)

The short answer is no. To clarify the question, we have (a) a closed circuit comprising two 1.5 V batteries plus a 32 ohm resistor and an 8 ohm resistor (total 40 ohms) all in a loop, and (b) the same thing but a gap between the 32 ohm resistor and the battery. In cases like this it is best to define a reference point, considered to be zero voltage and zero charge (after all the whole world has a voltage and charge with respect to some point in space). Let the ref point be the junction of the battery and the 8 ohm resistor. In case (a) we have 3 x 8/40 ie 0.6V at the junction of the two resistors, and of course 3V at the opposite end of the 32 ohm resistor (2.4 V across the 32 ohms). This means there is a certain charge at the junction of the two resistors, and a larger charge at the opposite end of the 32 ohm resistor. Just what the charge is numerically (in coulombs) we can't say - as Nimur has implied, you need to know the (stray) capacitance across each part. In case (b) no current is flowing due to the break, so there is no voltage across either resistor. Therefore, with respect to the reference point, there is no charge at any terminal of either resistor, but there is still a charge across the battery. The 2 situations are different, becasue in (a) we have 3 V across the 2 resistors, and in case (b) with have 3 V across the gap instead. Keit60.228.241.58 (talk) 03:03, 13 June 2012 (UTC)

Polarization?

A large thin plastic disk with radius R = 1.5 m carries a uniformly distributed charge of -Q = -3 x 10^-5 C. A neutral circular piece of aluminum foil is placed d = 3.0 m from the disk, parallel to the disk. The foil has a radius of r = 2.0 cm and a thickness of t = 1.0 mm.
(i) Show the charge distribution on the foil
(ii) Calculate the magnitude and direction of the electric field at the center of the aluminum foil, inside the foil
(iii) Calculate the magnitude q of the charge on face of the foil facing the charged disk.
For the first part, due to polarizability, I assume we have negative charges on the side facing away from the disk and positive charges on the side facing the disk. What about the other two parts? I would have thought you would need to know the dielectric constant of the foil. Widener (talk) 18:33, 12 June 2012 (UTC)

Is this homework? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 18:44, 12 June 2012 (UTC)

No, it comes from here: [8] Widener (talk) 18:52, 12 June 2012 (UTC)

Part (iii) is easy once you have the answer to part (ii), of course. --130.56.77.205 (talk) 23:10, 12 June 2012 (UTC)

"herpautochory" -- seed dispersal

At Seed_dispersal#Types_of_dispersal, we have "herpautochory (the seed crawls by means of trichomes and moist changes)." Being interested, I tried to find more information -- with no avail. Googling "herpautochory" returns an echo chamber sourced from our page, there are zero hits in google scholar as well as google books. I suspect there is a nugget of truth here, so what is it? Perhaps a simple typo? Any help is appreciated. SemanticMantis (talk) 18:42, 12 June 2012 (UTC)

I agree with you that the term cannot be sourced, and neither can the term "boleochory" that appeared in the same sentence. I have removed that sentence. It was added on 10 January of this year by an IP editor, 79.158.109.74 (talk · contribs), without any explanation. Looie496 (talk) 19:06, 12 June 2012 (UTC)

What a lesson on the influence of WP! ~250 pages have pulled that content since then... Still, you've helped me quite a bit, because you read "boleoautochory" and wrote "boleochory"-- effectively undoing the IP's mistake ;) Searching boleochory led me to this book: [9], it is a fairly authoritative textbook, and defines boleochory, as well as herpochory, so it was more of a typo/idiosyncratic terminology situation after all-- "herpautochory" should have been "herpochory". I'll put the article back together with the right words, and cited the book. SemanticMantis (talk) 19:40, 12 June 2012 (UTC)

Hah, that's pretty funny. I swear I've never seen any of this terminology before. Looie496 (talk) 22:13, 12 June 2012 (UTC)

Resolved

Speed of Response of Interconnected Ganglia

Hello. Why can interconnected ganglia in the autonomic nervous system respond faster than ganglia not interconnected? Thanks in advance. --Mayfare (talk) 19:13, 12 June 2012 (UTC)

Might be the diff between serial and parallel computing. That is, if there are a certain number of calculations to perform, then performing them all at once is quicker than one at a time. StuRat (talk) 19:45, 12 June 2012 (UTC)

I don't understand the question. All of the ganglia of the autonomic nervous system are interconnected -- at least, they are all connected to the central nervous system in some way or other. How did this question arise? Looie496 (talk) 22:11, 12 June 2012 (UTC)

I assume they are talking about a variable degree of interconnectivity, as some ganglia have many more connections than others. StuRat (talk) 00:20, 13 June 2012 (UTC)

It is to my understanding that: Sympathetic ganglia connect to each other to form the sympathetic trunk. Whereas parasympathetic ganglia run from the brain stem or the sacral region and synapse in or near the affected organ but do not connect to each other. --Mayfare (talk) 01:53, 13 June 2012 (UTC)

Okay, that makes sense. But why do you think the connection pattern determines the speed of response? It is natural that the "fight or flight" responses of the sympathetic system take place on a faster time scale than the "rest and digest" responses of the parasympathetic system, but there are lots of differences between the systems beyond their levels of connectedness. Looie496 (talk) 03:57, 13 June 2012 (UTC)

June 13

What's in a Galileo thermometer?

I've checked the relevant article, but what I see is: A Galileo thermometer … is a thermometer made of a sealed glass cylinder containing a clear liquid…. I'd like to know what the "clear liquid" might be.

Why do I ask? I had one of these thermometers on top of a cabinet. Something (a CO monitor, not that that's relevant) inexplicably fell off the wall and knocked the Galileo thermometer onto the floor, where it smashed, disgorging the contents onto the carpet. I mopped it up as best I could, expecting it was water, but a) though odourless it felt more like a very light oil and b) the carpet is now "lifting" as if the liquid has soaked into the backing and has caused some damage. At the worst I'll just put up with it (or enjoy the "claim on my home contents insurance" game) but I'd be interested to know in more detail what's inside one of these things. Tonywalton ^Talk 00:05, 13 June 2012 (UTC)

The article to which you linked says "The clear liquid in which the bulbs are submerged is not water, but some inert hydrocarbon[clarification needed] whose density varies with temperature more than water's density does. DriveByWire (talk) 00:14, 13 June 2012 (UTC)

Oops, I missed that. Thanks for pointing it out. I'd appreciate some clarification as tagged in the article, though. The stuff was completely odourless. No "oily" smell at all, and nothing like carbon tet, ethanol or anything ketone-y. Tonywalton ^Talk 00:22, 13 June 2012 (UTC)

In short, the most likely reason for it feeling like a very light oil is that it is a very light oil, and consequently, it dissolved some oil-soluble component in the carpet backing. --Carnildo (talk) 00:23, 13 June 2012 (UTC)

I assumed this novel thing would be patented (and from that would have at least an example of what sort of thing could be used) but am unable to find it on the uspto web-search system. A statement about its origin and link to patent would be a nice bit of history for our article if anyone can find it. DMacks (talk) 00:26, 13 June 2012 (UTC)

Google Patents turns up a few — e.g. this one for a liquid in them. It seems like there are a lot of candidates for what the liquid could be, though, and any patent worth its money wouldn't be exclusive about which one they'd use. You'd be better off tracking down the manufacturer and asking them than looking into patents. --Mr.98 (talk) 16:20, 13 June 2012 (UTC)

If the glass cylinder is sealed, and there's little or no headspace, how does the volume (and density) of the liquid change at all (Beyond the small change in the volume of the glass cylinder itself)? 203.27.72.5 (talk) 01:11, 13 June 2012 (UTC)

I have three of them and they all have "headspace" in the cylinders. I used to have four, but one fell onto hardwood flooring and smashed. Unlike Tony Walton's experience, the liquid in mine smelled awful -more like a rotting meat than anything else. I am really glad no carpets were involved. Bielle (talk) 01:45, 13 June 2012 (UTC)

non colinearity in solar cells or solar panels

With early solar panels, if a cell was shaded, it turned into a resistor and could burn out as all the power of the panel was pushed through it. Diodes were added which protected the cells of the panels but still, the loss of power from a shaded cell was more than expected from the percent of the panel that was shaded. At present solar panels have dropped to around a dollar a nominal watt. We are getting to the sort of cost that would make it economically worthwhile to put solar panels on East and West rooves as well as Northern roves (I am in New Zealand). Does the technology exist to get the full power from panels that are not colinear without them interfering with each other. Then we come to electric cars. Already the Prius has a roof retrofit that gives about 10km of power for a day in the sun. A nice little bonus. Soon it will be possible to clad the entire car, if desired, with solar cells. No two cells will be pointed in the same direction. Does the technology exist to get the full power that each cell is absorbing to the in-car battery. William

William Hughes-Games Waipara, N. Cant, NZ — Preceding unsigned comment added by 125.239.168.182 (talk) 02:45, 13 June 2012 (UTC)

To understand this, you need to know 2 things: a) a given panel under certain illumination supplies a certain voltage up to a certain load, and above this load, voltage falls with increasing load. b) for any given panel under a certain illumination, there is an optimum working voltage that gives maximum power. The optimum voltage increases with increasing illumination. Optimum Power Point Tracking regulators allow the panel voltage to be independent of the load voltage and thus can optimise the panel voltage to extract maximum power.

There's 2 ways of handling the situation:-

(1) use diodes in series with each set of panels and parallel the panels + diodes. In this case the panel set that has the strongest illumination will tend to supply all the power, and the worst angled panel set will tend to contribute less than it can, as it will produce insufficient voltage to bias it's diode on. Where the load is suficient to pull the strong panel's voltage down so that the weaker panel can contribute, so it isn't as bad as it may seem.

(2) use separate electronic optimum power point tracking regulators for each panel set and parallel (with diodes if not already provided) the regulator outputs, not the panels. You get more total power this way, becasue the separate regulators allow each panel to operate at its own optimum voltage. A further major advantage is that each panel set can be for a different area and thus design nominal voltage. NOTE: this is the way to go with vehicles, which must have at least some panels not optimumaly angled wrt the sun and with various dissimilar areas in each position, but for a building, you will always get the maxiumum possible energy if you install ALL panels at the optimum angle and direction. A lot of houses where I live have installed panels, becasue of the government subsidy. But they always have them installed to conform to their existing roof pitch and direction. If they used a frame to install all the panels at the optimum angle and direction, it might not look so neat, but they would get a lot more energy. Keit60.228.241.58 (talk) 03:26, 13 June 2012 (UTC)

Linearity of Lorentz Transformations

I asked this question in the math ref desk but I didn't get an answer, so I figured I'd have better luck here.

The article on the Lorentz transformation says that "[i]f space [and time are] homogeneous, then the Lorentz transformation must be a linear transformation." It's intuitively clear that this is true (if there are terms like x^2, it would be difficult to see how homogeneity would be preserved), but despite going through a dozen pages of Google I could not find a (reasonably) rigorous proof of this statement (that didn't introduce any new assumptions), except for one that was far too involved for something this simple. Can anyone here help? Thanks. 65.92.7.168 (talk) 04:14, 13 June 2012 (UTC)

The Math desk thread got several replies. I think Quondum's was pretty good. "If space and time are homogeneous, then the Lorentz transformation must be a linear transformation" makes no sense in a vacuum. If you adopt the usual definition of "Lorentz transformation" then it's true simply because the Lorentz transformations are, in fact, linear. The interesting part of these arguments is usually the assumptions they make, not the reasoning from there to the conclusion. -- BenRG (talk) 05:37, 13 June 2012 (UTC)

I was short on time when I wrote that and I still am, but a quick follow up: the assumption that space(time) is homogeneous doesn't really help, because it doesn't say anything about the Lorentz transformations. Presumably the idea is that the Lorentz transformations should preserve that property, but I'm not sure I know what it means to preserve homogeneity. One could say that it means to preserve the spacetime metric, but it hardly seems fair to assume the metric up front. The trick in these arguments is to come up with a set of assumptions that uniquely characterize the Lorentz transformations without appearing to do so—and ideally that sound as obvious as possible. There are certainly arguments of this kind, one of them in Einstein's original paper, but I don't think I've ever understood what their assumptions are, counting all the hidden ones. Somebody somewhere has done it (probably in print, not on the web). -- BenRG (talk) 15:00, 13 June 2012 (UTC)

Extreme peaks in electricity demand

As I understand it, a lot of money is sometimes spent on electricity generation capacity which will only ever be called upon under extreme peaks (less than 40 hours per year year was the figure quoted for Melbourne, Australia where I live).

My two questions are:

1. Besides for intense heat (where most people are using their air-conditioning), which other circumstances can place an electricity network under this sort of upper-peak capacity strain?

2. Besides for clothes dryers (which use a LOT of electricity per minute), which industrial or household processes can be shut down or put in standby under these circumstances, without causing massive disruptions? Can an aluminium smelter, for example, be put in standby when a heatwave hits (or is due to hit)? Which other industries can / should be made to wind down for these 40 hours or so per year?

(Don't bother suggesting people be made to switch off their air-conditioning under extreme heat. It's not realistic). 203.45.95.236 (talk) 07:04, 13 June 2012 (UTC)

Turning off aluminium smelters (and other smelters) is not going to solve your problem. They chose the location so it has a cheap, reliable energy source and the power plant is built to cater for it. If you make the power plant smaller and tell the smelter operator that they will have to shut down when people turn their air conditioners on, they will build it somewhere else. Quite often they're the main driver of developement in the region (see Gladstone), and the inhabitants running their air conditioners are incidental to the smelter existing. They can (and do) shut down for maintenance, etc. but it's a major drama starting them back-up especially if the shutdown was not planned.

Electrical appliances like TVs and stereo equipment use a lot of power just sitting on standby. Turn those off and you'd save a fair chunk. Individually, they don't use as much as a dryer but together they do add up. Refridgerators use a fair bit, but they can shouldn't be shut down lightly.

All of that extra capacity isn't wasted money anyhow. It allows for future expansion without needing to install new generators.203.27.72.5 (talk) 07:46, 13 June 2012 (UTC)

Regarding question 1 although it's not a peak load in the way you mean, as it's more of a daily rather than yearly thing, I remember when I visted Dinorwig Power Station, a pumped-storage hydroelectricity plant in Wales, that one of the main daily surges in demand in the UK is just after a popular television show ends when millions of people simultaneously switch on their electric kettles to make a cup of tea (this recollection is confirmed in the article). In anticipation Dinorwig, which operates as a so-called "short term operating reserve" spins up some of it's turbines just as the credits start to roll. This sort of energy storage approach is an another approach to reduce the need for traditional generating reserve, by providing a storage reserve instead. Equisetum (talk | contributions) 09:35, 13 June 2012 (UTC)

It is intertesting that the OP hails from Melbourne Australia, which has had for many decades an off-peak electicity scheme. This is where the consumer has two sets of meters and distribution within the building. Off-peak electricity is charged at a lower tarrif - in return the power company/authority turns the off-peak supply on and off each day at times to suit themselves. When I worked in a Melbourne high-rise office, each tenant on each floor had electric storage heaters - these had a large thermal storage media inside. In winter they stored heat energy when the power company turned the off-peak electricity on, and continued to deliver heat to the offices from storage while the off-peak electricity was off.

Most (if not all) power companies charge their major comercial customers not a flat rate of so many cents/kw.hr, but a complex scheme whereby there is a base rate, plus a surcharge for electricity consumed during defined peak periods, and an additional surcharge for electricity consumed at a rate above a certain level or levels. This does encourage major consumers to even out their electricity usage. This is not to be confused with the off-peak system mentioned above, because it lets the consumer draw whatever power he wants whenever he wants - but it may cost him.

Trying to reduce peak load by messing about with small dwelling / domestic airconditioning is pretty futile, but commercial and high-rise building aircon is a different story. Such installations can have signicant thermal storage, and nearly all will have so-called economy cycles fitted. By adjusting the temperature setting up a degree or two, and sometimes increasing airflow to compensate, the time each day the building stays in economy mode can be lengthened without affecting tenants too much.

How much a power company can reduce its worst case (as in a few really bad days in a year) depends somewhat on their attitude and their relationship with their major customers. Many telephone company sites, computer server farms, hospitals, and the like may draw megawatts in each site, and together may be a significnat fraction of a power company's total load. Such sites have diesel generator backup in case of power company supply failure, plus battery backup to cover the diesel startup time. To assit a power company on days of extreme load, these customers can run the diesels. Often they don't mind doing so, becasue their fuel cost is often less than or comparable to the power compnay electricity charge, and they need to run the diesels at least once a month anyway to keep the diesels in prime condition. The downside for the power company is that they MUST accept the immediate full load if someone's diesel fails for some reason.

At least with aluminium smelters I know about, they have their own power station. If a deal can be worked out between them and the local power company, it can be the case that they can help with peak demand, with or without shutting down or reducing smelting. This is the case with Alcoa in Western Australia. Large chemical and fertiliser plants can aslo assist by reducing or shutting down. Sometimes deals are reached with power companies, as it may not cost that much to shut down a large automated plant for short period, and the power company may be prepared to pay some compensation.

Many governments have Emergency Supply Laws, where, in a declared emergency, such as unprecedented peak demand, or a power station unplanned outage, they can direct industrial consumers to shut down, in order to keep consumers deemed to be more important (hospitals, shopping centres, electric trains systems, etc), with power.

Ratbone120.145.53.76 (talk) 13:18, 13 June 2012 (UTC)

Our article on the UK's National Grid Reserve Service says that automatic relays cut off the power to steel works if too much load is placed on the Grid, with a standing reserve of industrial diesel generators kicking in at the same time to alleviate the strain. These generators are supposedly contracted to provide power for up to two hours. Note that in the UK, this is probably enough to negotiate some extra juice being pumped in through the HVDC Cross-Channel lines. Brammers (talk/c) 14:49, 13 June 2012 (UTC)

You might find the articles Control of the National Grid#Short term and instantaneous load and generation response mechanisms and Load management interesting reads, oh and Demand response Also this website http://www.dynamicdemand.co.uk/grid.htm which shows the balance of the grid at any time (UK grid this is). I find all this stuff fascinating but have struggled to find an 'accessible' (i.e. not insanely technical) book that provides an interesting read on how the grid works. ny156uk (talk) 17:50, 13 June 2012 (UTC)

Soluble component of starch

Which component of strach is more soluble in water --- Amylose or Amylopectin?

It says amylose is insoluble in the link

But it says amylose is soluble in the links Bulto95 (talk) 16:17, 13 June 2012 (UTC)

References, --Gilderien Chat|List of good deeds 16:23, 13 June 2012 (UTC)

Essjay - grandiose delusions?

Are Essjay's alledged PhDs, etc, classifiable as grandiose delusions? Zaminamina (talk) 18:51, 13 June 2012 (UTC)

No one here is qualified to diagnose someone's through the computer. As far as I know, his behavior would be classified as deceit, but as to what the cause of that deceit was, it would be anyone's guess. --Jayron32 18:57, 13 June 2012 (UTC)

Please identify the plack spots on my garden furniture

Hi, My garden plastic furniture, as well as the marble tiles in my garden are all spotted with thousands of black spots, depicted in the photo. What are these, and how can they be cleaned? Gil_mo (talk) 19:19, 13 June 2012 (UTC)

Plastic garden table stained with black spots

Pertinent details missing: Where in the world is this photo taken? What is the proximity to flower beds, mulch, or wooded areas? Are the spots dry, or sticky/ moist? Do they have any odor? Have you tried cleaning with basic soap and water? Supplying this info will help us give you better answers. SemanticMantis (talk) 19:48, 13 June 2012 (UTC)

Looks like mold, but we'd need a much closer up picture. Suggest you try something that says removes mold and mildew. μηδείς (talk) 20:22, 13 June 2012 (UTC)

where can I buy lecithin for home cooking and baking?

I don't want it as a nutritional supplement. However I've contacted several suppliers and they all seem unwilling to do business with the average homemaker. 72.229.155.79 (talk) 20:58, 13 June 2012 (UTC)