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May 17

What is the formal definition of a chromosome, or is there one?

I cannot seem to find a definition anywhere, Wikipedia contains conflicting information across different pages, as do Google results. Hoping someone here with a genetics background can enlighten me. Chromosome says that it is DNA complexed with protein, but I've read elsewhere that it is DNA complexed specifically with chromatin. The latter definition would exclude prokaryote circular DNA from being classed as chromosomes then.

Also, in regards to karyotype, I've heard at various times both a chromatid and a pair of homologous chromosomes being referred to as a chromosome in the singular. Humans have 23 pairs of chromosomes, right? So 46 different (heterozygous?) chromosomes in a normal interphase cell? Sorry if I have overcomplicated this, I just can't get my head round it at the moment. Thanks for any clarification in advance -Zynwyx (talk) 09:30, 17 May 2012 (UTC)[reply]

The first part is is a very good question, which urges me to do some remedial reading on prokaryote biology - thanks for asking it. As explained in chromosome, the term is used "loosely" to refer to the bacterial chromosome, which is technically a genophore. But when the literature is full of people speaking of the bacterial chromosome, that attempted redefinition seems like a failure.

So what is the criterion to distinguish a natural chromosome from a mere plasmid?

First, let's consider what it's not. A plasmid has an origin of replication, or it couldn't replicate; and the vast majority of the time it contains a partitioning locus (see segrosome), or otherwise strains would be spontaneously cured of it with fairly high frequency. (See [1] for some pretty data) The plasmid apparently is associated with the nucleoid at least the majority of the time (see previous and [2], though I'm a bit tentative about this). The distinction I find in one book is "Probably the best distinguishing characteristic of a plasmid is that it has a more typically plasmid origin of replication with an adjacent gene for a Rep protein rather than a typical chromosome origin with an oriC, along with a dnaA gene and other genes typical of the chromosomal origin of replication".[3]

Problem is, bacterial artificial chromosomes and phage artificial chromosomes don't meet this definition! As explained here, they're based on the F episome (F plasmid), but as our article explains, they do have the Rep gene. Also, lest we have any confusion, these are closed circular constructs which often undergo homologous recombination after injection to a mammalian host to form linear transgene arrays [4] - they are not eukaryotic chromosomes with telomeres and such like yeast artificial chromosomes. So they're called "artificial chromosomes" in the bacterial sense.

So what distinguishes a plasmid from an artificial chromosome? I'm going to guess that much of the distinction is practical: how much stuff you can cram into it. A high copy number plasmid is going to recombine with itself, sometimes lose pieces of the gene, and the ones with the least stuff in them will replicate best. So when you want to shove a lot of stuff into bacteria, you want one of these more well behaved low copy number vectors with more natural-ish performance. And so, in the absence of a very clear definition, we have this sort of continuum from natural chromosome to artificial chromosome to plasmid to the odd random piece of lost circular DNA, without as clear a defining line as we might like. But I might still be just not thinking about something important... Wnt (talk) 15:24, 18 May 2012 (UTC)[reply]

-----Thanks for the detailed response! I think genetics has become a victim of the speed of its own success in this respect, as there aren't concrete definitions for loosely thrown about terms. So at least there are genetic differences between a plasmid and a bacterial genophore (in terms of origin of replication, which is good to hear. A chromosome in eukaryotes though, as I have now learnt, can refer to:

a single strand of DNA, what might commonly be referred to as a chromosome when you look at a karyotype. E.g one member of the homologous pair of chromosomes we inherit from each parent.

in mitosis, two strands of identical DNA joined at the centromere (the X-shaped things, to put it in very non-technical terms). Each strand is in itself an individual 'chromosome' as by the first definition, and indeed one of the strands is a replica of the other, but in this situation each strand is called a chromatid and the two identical 'chromatids' joined together are known as sister chromatids. When the sister chromatids separate, each one then becomes known as a (daughter) chromosome again.

So this was confusing me when reading up on mitosis/meiosis as I found it hard to keep track of what chromosome/chromatid was going where and so on. I don't personally think it is a good use of a word, much like bringing two identical twins together and calling them one person, but I guess that is personal opinion and its something that is ingrained and unlikely to change. By the wider definition though of simply a large 'section' of genomic DNA (chromatin or not), both situations above are fine (albeit still slightly muddling to me). Another semantic muddle-up I was having was homozygous and homologous. Homozygous refers to when the two 'sister chromatids' are identical in gene sequence and allelic content (barring random mutation). Homologous refers to when two separate chromosomes or two 'chromatids' are the same in terms of what genes they encode for and the sequence of the gene loci, but differing in alleles. So in chromosomal crossover, two 'chromosomes' (which are composed each of two homologous and homozygous 'sister chromatids') come together to form what is known either as a tetrad or bivalent (more confusion...) and then homologous but heterozygous 'chromatids' (ie one chromatid from each X-shaped 'chromosome') recombine to form in total four homologous (but heterozygous to one another) chromatids, which then separate to become four homologous chromosomes. Phew! I'll definitely be thinking more carefully in future in using these terms!! -Zynwyx (talk) 13:25, 19 May 2012 (UTC)[reply]

Or maybe not! According to this page the two sister chromatids together can also be called a dyad. Is it correct to call the two sister chromatids (the "X") a chromosome then? -Zynwyx (talk) 13:42, 19 May 2012 (UTC)[reply]

I'm not following your reasoning with the last step. If you have 46 pairs of chromosomes, they first replicate their DNA to form 92 chromatids. They're still 46 chromosomes because each has 1 centromere. (That isn't as odd as it sounds because, remember, there's a period when the DNA isn't fully replicated, and you only definitively know that's over after they are starting to be pulled apart) Now the chromosomes pair up to form 23 bivalents, with exchanges between sister chromatids. I'm unaware of people calling them anything but 46 chromosomes and 92 chromatids at this point. After the exchanges, which involve crossing over and thus are incomplete, the bivalents get pulled apart, so you have 23 chromosomes and 46 chromatids in each. Then the chromatids are pulled apart and you have 23 chromosomes and 23 chromatids in each gamete (1n). So 1 chromosome = 1 to 2 chromatids, 1 bivalent = 2 chromosomes, 1 tetrad = 4 chromatids, 1 dyad = 2 chromatids. Unless I fouled up ;) Wnt (talk) 14:38, 19 May 2012 (UTC)[reply]

Thanks, that makes a lot more sense. I have been thinking this whole time of a chromosome as being a helix of DNA which holds one allele of a gene, and thus couldn't get round in my mind how the X things are called chromosomes as well. Obviously each X derives from one chromosome that is being replicated, and the replicas are joined at the centre, so it makes a bit more sense in my mind now. So 23 pairs of homologous chromosomes > 46 individual chromosomes > 92 chromatids during mitosis (but still 46 chromosomes). That clears it up in my mind a bit, thanks! -Zynwyx (talk) 08:30, 20 May 2012 (UTC)[reply]

Metal loses its magnetic properties when in a liquid state?

I was just reading this, http://en.wikipedia.org/wiki/Electromagnetic_projectile_devices_%28fiction%29#Literature

And it stated that "Later it was shown that molten metal cannot be accelerated by a magnetic field as metal loses its magnetic properties in a molten state, and Clarke admitted his error gracefully." Is that true? In a railgun however, a piece of metal need only be conductive, not magnetic. Would molten lead or an eutectic of lead and bismuth still remain electrically conductive? ScienceApe (talk) 15:28, 17 May 2012 (UTC)[reply]

Mercury (element) is a metal in it's liquid state at room temperature, and, I believe, retains electrical conductivity. Perhaps what they mean is that, if significantly perturbed, the liquid will then break into droplets, making it more difficult for an electric charge to pass between them. StuRat (talk) 16:25, 17 May 2012 (UTC)[reply]

Mercury is absolutely conductive in its liquid state, see mercury switch for a common electrical use of mercury. --Jayron32 17:23, 17 May 2012 (UTC)[reply]

It must be possible because you can buy one. This link is to the Permanent Magnetic Pump (PMP), but the data sheet also mentions an Electromagnetic Pump (EMP). --Heron (talk) 18:35, 17 May 2012 (UTC)[reply]

A railgun uses an electric current to generate a magnetic field in the projectile, accelerating it through simple magnetic repulsion, and can use any projectile material that remains conductive during acceleration. A coilgun is essentially a linear motor that accelerates a magnetized projectile, and so requires that the projectile be able to remain magnetic during acceleration. You can (theoretically) fire a liquid metal from a railgun, but not from a coilgun. --Carnildo (talk) 23:04, 17 May 2012 (UTC)[reply]

If a nail is heated red hot, as with a propane torch, it reaches its Curie temperature and is no longer attracted by a magnet. Other ferromagnetic metals have varying Curie temperatures. Iron would still be electrically conductive when heated red hot, though its resistance would change.Molten iron and steel would still be conductive, since an electric arc furnace relies of current flow from an electrode to molten metal in the crucible. Metals typically have an increase of resistivity of 1.5 to 2.5 in the liquid form compared to the solid form at the melting point per [5]. Edison (talk) 23:54, 17 May 2012 (UTC)[reply]

I would agree with Heron, induced eddy currents would make it possible. An eddy current separator accelerates nonmagnetic metals in the same way. Ssscienccce (talk) 08:14, 18 May 2012 (UTC)[reply]

Could the molten metal imagined by Clarke have been above its Curie temperature but contain a suspension of a powdered metal with a higher Curie temperature? 84.209.89.214 (talk) 14:40, 18 May 2012 (UTC)[reply]

Yes! A ferrofluid! You can even make one yourself fairly easily. That said, it's useful to understand why something is no longer magnetic in liquid phase. It's essentially the definition of a liquid - the material no longer has any defined ordering of its atoms (or constituent molecules), and ferromagnetism in particular depends on electrons being able to align in some fashion and remain that way. It's a useless suggestion if the atoms are turning and tumbling over each other as happens constantly in a liquid. That said, one could have something putty or gel-like which flows like a liquid but has defined crystal structure at a small scale. This is kinda a cheat, but it might work for some sci fi - a very-much-zoomed-in example would be like a giant tank of those mini-magnetic spheres. They will flow. SamuelRiv (talk) 02:11, 21 May 2012 (UTC)[reply]

Space radiation

Lets say you are in outerspace and you are protected from everything (pressure, you have oxygen, temperature, etc) except radiation. About how long would it take before you die from radiation? ScienceApe (talk) 15:46, 17 May 2012 (UTC)[reply]

- which part of outer space? Some parts are full of radiation, and you'd die immediately. Some parts have very little, and you'd die of old age. 91.125.207.125 (talk) —Preceding undated comment added 16:03, 17 May 2012 (UTC).[reply]

I don't think there's enough radiation, near the Earth, to kill from radiation sickness. There is enough, however, to cause genetic damage and cancer. But these things aren't always fatal, so there would be a decreased life expectancy, not certain death. If you were closer to the Sun, the radiation damage might be more severe.

Note that I'm assuming that you are excluding UV light. If that is included, I'd expect bare skin in space (if somehow protected from the cold and vacuum) to quickly burn, crack, and bleed. Death might occur within hours or days, from dehydration and infection. StuRat (talk) 16:04, 17 May 2012 (UTC)[reply]

The article Health threat from cosmic rays may be of interest to you. LukeSurl^{t c} 18:46, 17 May 2012 (UTC)[reply]

Define "radiation". What portions of the electromagnetic spectrum are you including, and what portions are you excluding? --Carnildo (talk) 23:05, 17 May 2012 (UTC)[reply]

Quoting from a NASA report (Shielding Strategies for Human Space Exploration dec 1997):

In prior manned space missions, the GCR have been considered negligible since the mission times were relatively short and the main radiation concern was the very intense SEP events which can rise unexpectedly to high levels, delivering a potentially lethal dose in a few to several hours which could cause death or serious radiation illness over the following few days to few weeks if precautions are not taken [4]. The most intense such event known occurred on August 4–5, 1972 between the Apollo 16 and Apollo 17 missions [5].

GCR = galactic cosmic rays, SEP = Solar Energetic Particles Ssscienccce (talk) 15:21, 18 May 2012 (UTC)[reply]

Mario Rabinowitz

I was reading through the Ball lightning article, and it has some stuff about black holes by someone named Mario Rabinowitz. At a first look the Mario Rabinowitz article makes him out to be someone very impressive, but I can only see links to ArXive papers, and I can't see any affiliation with physicists I've actually heard of, or a position at a university or laboratory. Looking at the article's history, it looks like almost all of it was added to wikipedia by people who haven't done other things. So I'm a bit suspicious. I can't find anything worthwhile about this person by searching Google (I find facebook and patents and whitepages and stuff, and copies of the papers). So I'm concerned that a] this person doesn't really exist at all (that the article is a hoax) or b] that this person does exist, but isn't a physicist anyone has heard about (and so maybe should't be on wikipedia). Or is he really a famous physics guy who I've just failed to hear anything about? 91.125.207.125 (talk) —Preceding undated comment added 15:59, 17 May 2012 (UTC).[reply]

He's a real person,[6] but it sure looks to me like he's a non-notable person who wrote an article about himself as an autobiography, and the article should be deleted. But WP:AFD would be the place to bring that up, not here. Red Act (talk) 19:29, 17 May 2012 (UTC)[reply]

Cooperative lightning? Intelligent grass?

I'm sure this is a very dumb question, but assuming that what this website says is true: When lightning begins to travel downward from a cloud, many objects that have built up a charge emit streamers. This could come from anything such as a blade of grass or a power pole. The first streamer to make contact with the bolt defines the final path the lightning will take. - then how do the ground objects "know" when it's time to start emitting streamers? A simple answer please, for this admittedly ignorant non-scientist. Textorus (talk) 16:03, 17 May 2012 (UTC)[reply]

See our article on lightning. Fundamentally, though, positive streamers from the ground emerge for the same reason that negative streamers from the cloud (i.e. the formative lightning strike) emerge -- there's a large electric charge differential present. — Lomn 16:11, 17 May 2012 (UTC)[reply]

You might also want to know how gravity works. That is, how does the Earth "know" there is a star 93 million miles away which it should orbit ? Some rather non-intuitive explanations emerge, such as space being curved, or the even weirder gauge boson theory. StuRat (talk) 16:20, 17 May 2012 (UTC)[reply]

Now that you mention it, that is a fascinating question, but I'll save that one for another time. (The aether gets knotted up into a rope, maybe, like a yo-yo string?) Textorus (talk) 18:01, 17 May 2012 (UTC)[reply]

Lightning is a very complicated electrodynamic phenomenon. In addition to the visible incandescent stream of hot gas that you see, there are also wide-band electromagnetic waves (radio waves), preceding the lightning strike, occurring in tandem with the lightning strike, and coupling with the complicated streams of moving ions and electrons. The radio signals from a lightning strike are often called "sferics." Like all other radio waves, they travel approximately at the speed of light. The actual event of a lightning "striking" may be preceded by a very quick burst of radio-energy; and then as the streamer forms, all sorts of electromagnetic effects start happening and interacting with each other chaotically. The gas gets hot and incandesces, releasing visible light (incandescent light); but the gas is also ionizing and forming an electrically conductive stream, providing a current path, releasing more radio-wave emissions; and of course, the radio-waves emitted will affect air surrounding the lightning streamer. Here's a fairly advanced science web-site: Lightning Modeling, that reviews some of the physics necessary to accurately describe what's happening during a lightning strike. Nimur (talk) 17:10, 17 May 2012 (UTC)[reply]

The website is above my pay grade, but it makes sense that there must be some connecting force linking earth and sky. So when conditions are right for a storm, you're saying there's already a lot of ions and electrons moving between the two, invisible to our eyes? (I'm sure I must have learned that in Physics 101 but that was a l-o-n-g time ago.) Textorus (talk) 18:01, 17 May 2012 (UTC)[reply]

I think this is actually a very interesting question. The electricity clearly finds a quite specific path. My suspicion is that even in the absence of thunderstorms we are surrounded by an amazing display of static electricity which is simply, most often, too weak for us to see. Certainly I know that during a thunderstorm I can feel frequent little shocks from a mattress if it contains metal springs - sort of the sensation of being first bitten by a mosquito, but of course without the mosquito or subsequent irritation. Sometimes I've ever observed sparks from a window screen though lightning was not nearby. Has anyone ever sought to visualize the wider web of static electricity, or is it simply impossible, or indeed, am I deluded? Wnt (talk) 18:04, 17 May 2012 (UTC)[reply]

HAIL Project sought to visualize the wider web of electrodynamics in the atmosphere on massive geographic scales. By monitoring perturbations in the continuous background of electromagnetic signals (specifically, several LORAN transmitters), data was collected to drive a complete realtime model of the ionization and the electromagnetic environment for the continental US. Nimur (talk) 20:58, 17 May 2012 (UTC)[reply]

A former colleague of mine attempted to measuring electric fields near clouds by photographing polarization changes through an optical telescope. This is called the Kerr effect, and refers to the change in optical properties of certain materials (like atmospheric air) when exposed to very strong electrostatic fields. I recall thinking the idea was crazy (the signal should be well buried in the noise); but that's why it's research... Nimur (talk) 21:02, 17 May 2012 (UTC)[reply]

I can't search videos online at work, but I encourage you to find the ultra slow motion videos of lightning, it's frikken awesome! The "streamers" they talk about are a LOT slower then the speed of light and there are videos of them propegating though the sky, once they "contact" eachother, the lightning bolt actually fires like a flash. It's one of the most incredible natural phenomena on earth I think. Vespine (talk) 22:56, 17 May 2012 (UTC)[reply]

And of course, the fine folks in Gainesville, shoot off rockets trailing metal wires. Nimur (talk) 00:33, 18 May 2012 (UTC)[reply]

The streamers start due to electrostatic forces: attraction between positive and negative charges. To use an analogy: Imagine a cardboard box with a bunch of nails, screws, iron filings and stuff. If you hold a magnet above it, the objects start moving, pointing in the direction of the magnet. Get the magnet closer and eventually one of the objects will jump up and attach to the magnet, usually pulling others along. Electric charges act in a similar way. Ssscienccce (talk) 08:24, 18 May 2012 (UTC)[reply]

Thank you for that plain-English answer. I thought it must be something like that. Textorus (talk) 18:50, 18 May 2012 (UTC)[reply]

Of course, there's still the question of exactly why forces can act at a distance, whether electromagnetism, gravity, etc. StuRat (talk) 18:54, 18 May 2012 (UTC)[reply]

Where can drugs get in your body? What barriers are there?

I'm aware of the blood-brain barrier but presumably there are plenty of other barriers. Where can orally or intra-venously taken drugs get in your body? Presumably anywhere blood can get, but where is that? Can drugs get inside cells? What about bones? What about your eye lenses? What effects whether they DO get there?

I think that's enough questions for now... although I've got a lot more.

What would be a good place to start looking for the basics of this stuff?

Egg Centri c 19:50, 17 May 2012 (UTC)[reply]

Hair analysis shows that detectable levels of many drugs can be found in your hair. Long-haired drug users effectively carry a timeline of their drug consumption imprinted in every hair of their head. --Jayron32 20:31, 17 May 2012 (UTC)[reply]

Plus, if a man has long hair, can't we just assume he's a drug addict ? :-) StuRat (talk) 18:56, 18 May 2012 (UTC) [reply]

An oral drug should get into a cell at least once, in the intestinal epithelium, in order to enter the body. These and many other drugs usually have their effects inside a cell. Sort of an exception are antibiotics, which act inside a cell, but not one of yours! ;) There are exceptions, though - any drug which blocks a cellular receptor, for example. So far as I know any injected drug (well, "biologic") with a name ending in "-mab" (monoclonal antibody) will not get into a cell for meaningful purposes (it might get endocytosed with a receptor and have a trip to the lysosome, but that hardly counts). Vaccines don't get into cells, at least not the old fashioned kind available on the market. There's nothing quite like the blood-brain barrier and even that allows some things to pass. Certainly bones are visited by Fosamax and its ilk. The lens of the eye is a curious case, as it receives sustenance from the aqueous humor from the ciliary body; thus drugs must go by this indirect route; nonetheless they can arrive. For example, acetaminophen overdose can form cataracts in experimental animals receiving the drug systemically;[7] however, this occurs after it is first processed by the liver to form a more toxic metabolite. [8] Wnt (talk) 21:28, 17 May 2012 (UTC)[reply]

The place to look for basic information is any introductory pharmacology textbook. Generally speaking in order to get into cells a chemical needs to be lipophilic, meaning capable of dissolving in fats or oils. That's basically the same thing required for a drug to cross the blood-brain barrier. The exception is substances that are transported by active uptake mechanisms. Looie496 (talk) 23:47, 17 May 2012 (UTC)[reply]

Thanks all

Egg Centri c 17:57, 18 May 2012 (UTC)[reply]

Note that the skin acts as a barrier to some, but not all, drugs. Hence transdermal patches. StuRat (talk) 18:58, 18 May 2012 (UTC)[reply]

May 18

sun become a red giant 5 or 7.5 billion years?

Wait, I am a little confused here. said sun enter red giant about 5 billion years from now. Dr. Schroeder and Smith's website said tip of RGB is 7.59 billion years from now. Does it take 2.69 billion years for the sun to become a red giant, or sun last of red giant for 2.69 billion years. So when sun leaves main sequence in 5 billion years, it becomes a yellow subgiant first, and it slowly work the way to red giant by gradual increase of size/luminosity, or once it branches off main sequence it goes directly to red giant by large increase of size/luminosity. Is the end of red giant alot larger in diameter and luminosity then the beginning of red giant?--69.233.254.22 (talk) 01:32, 18 May 2012 (UTC)[reply]

According to Formation_and_evolution_of_the_Solar_System#The_Sun_and_planetary_environments, the process that leads to the formation of a Red Giant starts at around 5.4 billion years, and the process of growing into that phase will last until 7.5 billion years, so yes, it seems to take roughly about 2-3 billion years from starting on the red giant path and becoming a full-fledged red giant. --Jayron32 01:39, 18 May 2012 (UTC)[reply]

(edit conflict)Unfortunately I can't find the original website (I'd appreciate if someone could link to it) but 7.59 billion years seems like way too much precision to tell when the Earth will be consumed by the sun, especially since it is up for debate whether that will even happen.

The timeline you have sketched is about right though. See the text and charts at Sun#Life_cycle. The sun will start fusing helium at its core in around 5 billion years, which would technically be the start of its red giant phase. From there its luminosity and size will continue to increase over the course of 2-3 billion years, while the surface temperature will decrease. At some point in this expansion phase the earth may be destroyed, likely towards the end if it does happen. -RunningOnBrains^(talk) 01:49, 18 May 2012 (UTC)[reply]

The diagram sketches [9].--69.233.254.22 (talk) 02:35, 18 May 2012 (UTC)[reply]

Eclipse question

I've heard that there will be a partial solar eclipse in my area on Sunday, and I plan on watching it. What type of eye protection do I need? Thanks! 24.23.196.85 (talk) 05:59, 18 May 2012 (UTC)[reply]

Regular sunglasses aren't enough. I believe the usual precaution is not to look at the Sun at all, but rather look at a projection of the Sun. If you have no equipment for this, you can create a makeshift pinhole lens by poking a pin through a sheet of paper (preferably dark colored) and letting it shine on a white sheet of paper. StuRat (talk) 06:06, 18 May 2012 (UTC)[reply]

Welding goggles are sufficient protection if you can get your hands on them. There are a good number of eclipse-viewing goggles available online (I tried to link to amazon, but apparently even admins are blocked by the spam filter. How rude!), though I'd be sure I bought from a reputable vendor if I was going to risk my vision. Alternatively, it is safe to look at the sun for brief periods close to sunset ([10]); depending on where you are it may still be visible then. Just don't stare, and I'd still recommend sunglasses.-RunningOnBrains^(talk) 06:50, 18 May 2012 (UTC)[reply]

And it should be number 14 welding glass, not any other number. Sagittarian Milky Way (talk) 17:06, 18 May 2012 (UTC)[reply]

This construction or variants of it should give good results compared to other designs as it will (a) prevent stray light from hitting the screen, (b) reduce the loss of light from transmission through the screen. An adjustable back board(made by using one box inside another) will help you make the image-size brightness tradeoff. Use a binocular instead of the pinhole for better imagesStaticd (talk) 09:11, 18 May 2012 (UTC)[reply]

You can also get good projections simply by walking under some trees - pinholes in the leaves will project images of the sun (of varying focus) for a unique effect. Wnt (talk) 14:09, 18 May 2012 (UTC)[reply]

I was at Watford Junction train station at the time of the 1999 total eclipse, and was fortunate enough to see multiple images of the eclipse reflected off one of the dark glass windows of the waiting room! --TammyMoet (talk) 15:36, 18 May 2012 (UTC)[reply]

And in 2004 I didn't have a real filter, I saturated a piece of paper with butter, started looking through it for the shortest instant needed to form an image. Constantly blinked or waited 1 or 2 seconds depending on brightness. Was this bad? I stopped doing it when it when it seemed too bright (how high I don't remember but no more than about 12°). Another time I was holding eclipse glasses over 7x35 mm binoculars (the front, never the back), (I made a shield cause the glasses frames were too small), the sun was ~26° high, and either I put them up too late or probably forgot to cap the other lens despite planning to do that but holy crap, I saw sun! I shut my eyes as quick as I can. Thank goodness at least an eye doctor later didn't say anything (and I didn't ask). So don't do anything stupid/risky like this. They got so hot from the method above that a piece of glass broke off inside. Sagittarian Milky Way (talk) 17:57, 18 May 2012 (UTC)[reply]

What if you pointed a video camera at the eclipsed sun for 15 seconds or so (without looking through the view scope of course) and then watched it later on the TV? Obviously, the harmful rays would not be recorded and conveyed through the TV screen when you watched it, but the question is if 15 seconds pointed at the sun would seriously damage a video camera's CCD. 20.137.18.53 (talk) 15:43, 18 May 2012 (UTC)[reply]

Yes, I would expect that to damage the camera. Specifically, I'd expect the area of sensors which had the Sun focused on them to lose sensitivity. StuRat (talk) 18:13, 18 May 2012 (UTC)[reply]

That sounds a lot like electronics advice that should be given by a trained professional. ;) It seems worth saccing a videocamera for a total eclipse, but not an annular. ;) Wnt (talk) 16:54, 18 May 2012 (UTC)[reply]

Or get one of these, only $100,000 for the 95mm x 95mm CCD alone :) 20.137.18.53 (talk) 17:15, 18 May 2012 (UTC)[reply]

I suppose you could design a system where a camera has something like a "finder scope" that measures the max light level and puts the appropriate filter in front of the main lens. However, sudden changes in brightness, like the Sun coming out from behind the Moon, might still damage it. StuRat (talk) 19:23, 18 May 2012 (UTC)[reply]

Maybe a suitable photochromic lens material could be developed, like the one in those eyeglasses that become sunglasses outdoors? Which works too slow and too mild for this purpose. Even photographic solar filters transmit 0.01% of light, 10 times that of filters for eyes. Sagittarian Milky Way (talk) 21:49, 18 May 2012 (UTC)[reply]

An important thing to remember when picking a filter is that direct sunlight is a mix of infrared, visible, and ultraviolet light, any of which can cause eye damage. A lot of filters (eg. filters designed for cameras) only block visible light, leaving the infrared and ultraviolet components to blind you. --Carnildo (talk) 23:20, 18 May 2012 (UTC)[reply]

Thanks everyone, I used a #10 welding filter glass (couldn't find a #14) and didn't have any problems (but I did take the precaution of turning away whenever I felt my eyes start to get tired). 24.23.196.85 (talk) 05:00, 21 May 2012 (UTC)[reply]

Phd stipend life sciences

Please don't post the same question to multiple places. Your question will be answered at Wikipedia:Reference desk/Miscellaneous, section "phd salary". Nyttend (talk) 12:44, 18 May 2012 (UTC)[reply]

What's the most likely problem with my tire inflator/spotlight?

I have a tire inflator with spotlight from the same manufacturer but not the exact same model as this one. I can charge it with the AC adapter all night until the battery charge status indicators (as seen on the fourth page of the PDF) show full charge with two red LEDs and one green one. But after I unplug the AC adapter, within a few minutes, if I push the battery charge status pushbutton, only two red LEDs come on, indicating a supposed medium battery charge level. I opened the thing up after having fully charged it and seen the indication of only medium charge (when the adapter was unplugged). The thing has two 6V sealed lead acid batteries. I tested both with my multimeter every couple of hours for the past two days. One has maintained a level of 6.57V, and the other has maintained a level of 6.43V. What is the most likely reason I get only two red LEDs and not the green one when the adapter's not plugged in? 1) Something's wrong with the charge indicator circuitry, 2) Something's wrong with the batteries (are sealed lead acid batteries labeled 6V supposed to be getting up to 6.4-6.5V? do bad things happen when they are overcharged by this much?), or 3)something else (ideas appreciated)? Thanks. 20.137.18.53 (talk) 12:59, 18 May 2012 (UTC)[reply]

The article Lead-acid battery indicates that 2.1V per cell i.e. 6.3V from your battery is a normal open-circuit voltage at full charge. The slightly higher voltage that you measure is not likely any fault and a gelled electrolyte battery would happly accept continuous 6.7V float charging. There could be a calibration error in the green charge indicator or you may be using the unit at an unexpected temperature. 84.209.89.214 (talk) 14:23, 18 May 2012 (UTC)[reply]

about sodium silicate

Hi, I want to remove or de-active sodium silicate property from west water of textile unit. — Preceding unsigned comment added by 115.248.240.58 (talk) 14:37, 18 May 2012 (UTC)[reply]

What is a textile unit? Sodium silicate is soluble in a highly alkaline solution. Plasmic Physics (talk) 14:40, 18 May 2012 (UTC)[reply]

Sodium silicate is often shipped with clothes, to absorb humidity and prevent mildew. It's normally grains in a small packet. Are you saying the packet ripped open and the grains are on the clothes ? If so, just brush them off, they aren't toxic to the touch (but wash your hands after, as they can be alkaline, and you wouldn't want them to get in your eye). If they leave a residue, run it through the washing machine. StuRat (talk) 18:19, 18 May 2012 (UTC)[reply]

I think the OP is not a native English speaker, possibly third world, and meant to ask: "I want to remove or de-activate sodium silicate properly from waste water produced by a textile (manufacturing) unit. How can I do this?" Wickwack60.230.203.253 (talk) 04:03, 20 May 2012 (UTC)[reply]

Neutralize with acid and filter the residue. 24.23.196.85 (talk) 05:02, 21 May 2012 (UTC)[reply]

Gibbs free energy

Can someone explain what Gibbs energy is simply and give a biological example of Gibbs free energy. This article [11] is far too complicated. 176.250.232.230 (talk) 15:25, 18 May 2012 (UTC)[reply]

Simple explanations will invariably gloss over subtle details: this is tricky, because Gibbs energy is distinct from enthalpy and Helmholtz free energy, only by a small variation in definition. Roughly, Gibbs energy refers to the available energy from a chemical reaction, accounting for the pressure-volume work that must accompany that reaction. For example, if fermentation will release gaseous CO2, we use the Gibbs energy to quantify the work done, minus the "useless" work in expanding the CO2 as a gas. A worked example: ethanol metabolism thermodynamics. Nimur (talk) 15:47, 18 May 2012 (UTC)[reply]

thanks but people refer to Gibbs energy with things like linking amino acids to build proteins, converting ATP to ADP and the Krebs cycle and I just don't see the link with Gibbs which seems to be a thermodynamics concept rather than anything biological. — Preceding unsigned comment added by 176.250.232.230 (talk) 15:54, 18 May 2012 (UTC)[reply]

Lets try this really simply, by explaining "free energy" first. Free energy is energy availible to do work. Period. It just means that it is energy which could do something useful. There are forms of energy which are not free, that is there is energy which will cost you more energy to get to use. Roughly speaking, this is what entropy is. Basically, all of the energy in the universe is constant (First Law of Thermodynamics) but the amount of free energy is decreasing as the amount of entropy is increasing (Second Law of Thermodynamics). The different types of "free energy" and related measurements (like Gibbs Free Energy, Helmholtz Free Energy, Enthalpy) are just different variations on that theme; they are mathematical ways of expressing free energy in terms of highly constrained experimental set ups. One way to look at energy is to divide it into thermal energy and mechanical energy; that is energy which changed the temperature of a system, and energy which moves something around. So, free energy has a thermal component (changes in temperature) and a mechanical component (moving stuff). When you deal with gases, the mechanical aspect of their free energy deals with changes in volume (think, heating a balloon) and pressure (think, heating a steel tank). The difference between the various types of free energy is in how they treat that mechanical component of gases. In Gibbs free energy, your calculations assume that the system is at a constant pressure. For any system which is exposed to the earth's atmosphere, this works well, because any production or consumption of a gas will have a negligible effect on the entire earth's atmosphere, so any open container is a good Gibbs system. In Helmholtz free energy, your calculations assume a constant volume, which happens when you have a closed system, say a sealed tank, where the pressure will tend to vary a lot, but the volume remains constant. Gibbs free energy is also very important, because it is (via the Second Law of Thermodynamics) a mathematical way to calculate spontanaity. That is, any process which itself has a decrease in free energy associated with it will be spontaneous, because the universe spontaneously loses free energy, so any process that does that is likewise spontaneous. If you have a process which has an increase in free energy, TANSTAAFL, so there has to be a connected process which lowers the free energy by a greater amount, so the net change in free energy is always decreasing. So, to put it in simplest terms:

Free energy is energy which the universe has availible that you can tap into to do something useful
Gibbs free energy is a specific way of measuring that energy which works well in systems that are "open" to the environment
Gibbs free energy is important in calculating how "spontaneous" a process is; processes which cause a decrease in free energy occur spontaneously
(to your last question) Thermodynamics is inescapable. It's not like the laws of thermodynamics stop working in biological systems. That's why they are "laws" of the universe. They always work and continue to work, so when looking at, say, the assembly of a protein, if it has a negative Gibbs value, we know that it occurs spontaneously. This is kinda important info to know as a biochemist or molecular biologist who is concerned with how biological processes can occur.

Does this all help? --Jayron32 16:53, 18 May 2012 (UTC)[reply]

Thanks alot. That's a really good explanation. Most books or articles I look at just get too mathematical. — Preceding unsigned comment added by 176.250.232.230 (talk) 17:56, 18 May 2012 (UTC)[reply]

Zeta potential

I have a basic understanding of zeta potential (a property of colloidal systems). My question is, do hydrocolloids such as protein have a zeta potential? What about dissolved ions; do they have zeta potentials? ike9898 (talk) 18:48, 18 May 2012 (UTC)[reply]

Well, just trying the first example that came to mind, I searched "sickle cell" "zeta potential" and got actual numbers for red blood cells.[12] (also mentioned in Erythrocyte sedimentation rate). Trying the same for amyloid got what looked like a weaker set of results but nonetheless [13] [14]. My thought is that the measurement for proteins should be sort of weird because their aggregation depends so much on interactions that vary widely across the surface, but I really don't know. Wnt (talk) 19:00, 18 May 2012 (UTC)[reply]

Well, blood cells really don't fit into the category of things I am asking about. I'm taking about aqueous dispersions of hydrophilic polymers such as protein, starch, or polyacrylamide. Amyloid doesn't really fit into this category well, either. ike9898 (talk) 19:18, 18 May 2012 (UTC)[reply]

Apparently starch granules have a zeta potential. (Google dumps lots of results; apparently it's important for paper making) Polyacrylamide delivers more random results ... if it's a true gel, I don't know how you define a zeta potential, but that doesn't mean it can't be done. ;) Here's zeta potential for BSA [15]. Wnt (talk) 23:21, 18 May 2012 (UTC)[reply]

Thanks. The last sentence, especially, is a good lead. ike9898 (talk) 01:00, 19 May 2012 (UTC)[reply]

Flat universe have zero total energy?

I read from some dude on the internet (not exactly reliable) that the WMAP and Boomerang experiments (no idea what those are) indicate that the universe is flat and there is no net warpage of space time. Then he claimed that a flat universe can have zero total energy and thus come from nothing. What on earth is he talking about? Is this complete bs? ScienceApe (talk) 19:11, 18 May 2012 (UTC)[reply]

Those experiments (WMAP and BOOMERanG experiment) don't indicate that space is flat, they are simply consistent with it being flat. There is a margin of error in the results of any experiment, so all we can say is that zero curvature is within the range the experiments give. To conclude that it is absolutely flat, you need to use theoretical arguments, rather than experimental ones. --Tango (talk) 19:50, 18 May 2012 (UTC)[reply]

Check out Lawrence M. Krauss's 2009 lecture A Universe from Nothing. SkyMachine ⁽⁺⁺⁾ 20:52, 18 May 2012 (UTC)[reply]

Even theory doesn't imply that it's exactly flat, though I think the theoretical constraint is a lot stronger than the experimental constraint. -- BenRG (talk) 21:25, 18 May 2012 (UTC)[reply]

See Zero-energy universe. It's an idea that doesn't make much sense to me (because you have to break general covariance in order to talk about the energy of the universe), but plenty of legitimate physicists believe in it.

(Incidentally, the universe is spatially flat (approximately). Spacetime isn't flat.) -- BenRG (talk) 21:25, 18 May 2012 (UTC)[reply]

Can you explain what that means exactly? That the universe is spatially flat. ScienceApe (talk) 22:00, 18 May 2012 (UTC)[reply]

It means that if you look at the distances to all of the galaxies at a particular cosmological time (a particular era in their evolution), the number of galaxies within a distance R of you is proportional to R³. The number of galaxies would increase more slowly as R increased if space was positively curved, or more quickly if it was negatively curved. This is like measuring the amount of Earth's surface that's within a given distance of you—for small distances it grows like R², but the increase slows down for distances in the thousands of kilometers. The difference between spacetime and space in this context is like the difference between Earth and the surface of Earth. Earth is flat (it's a ball in 3D Euclidean space, approximately) but Earth's surface is positively curved on average. -- BenRG (talk) 23:08, 18 May 2012 (UTC)[reply]

Why isn't spacetime flat? If space is approximately Euclidean, isn't spacetime approximately Minkowski? I would call Minkowski spacetime flat - would I be wrong? --Tango (talk) 23:09, 19 May 2012 (UTC)[reply]

Yes, Minkowski spacetime is flat spacetime. However, even if the spatial portion

\mathrm {d} \mathbf {\Sigma } ^{2}

of the FLRW metric is Euclidean, the FLRW metric does not describe a flat spacetime, i.e. the FLRW metric is not equivalent to the Minkowski metric, unless the scale factor is constant. Red Act (talk) 14:59, 20 May 2012 (UTC)[reply]

Mootractor

For lack of a better name. I made this image in the hopes that I can make a collage of images for different views of the moon.

Can I assume that a view from the north pole would be 180 deg different from that of the south pole and 90 deg different from the equator? I am hoping that people from around the world can hold their monitors up to the moon and let me know which angle shows on the top of the moon, or is there a way to do this with math once I get one from the equator?--Canoe1967 (talk) 22:26, 18 May 2012 (UTC)[reply]

You might find Commons:Category:Moonrises and Commons:Category:Moonset to be useful (hmmm, the first two photos I looked at from each, both from Germany, had about the same angle, though one was rising and one setting. Thinking about the moon is a good way to strain your spatial comprehension. ;) File:Gaisberg_and_rising_full_moon.jpg File:Monduntergang_2011-04-17_002.JPG) Wnt (talk) 23:13, 18 May 2012 (UTC)[reply]

http://wms.lroc.asu.edu/lroc_browse/view/WAC_GL000 may be the same as a view from the equator. I may have to query further on an astronomy site.Canoe1967 (talk) 00:11, 19 May 2012 (UTC)[reply]

Resolved

My original plan won't work. The wise people on the astronomy site tell me that there are variations in rotation of up to 180 deg of view from any one point on earth from moonrise to moonset. Our view in the northern hemisphere would be close to 180 degrees sideways from that of the southern hemisphere. Now I just need a practical use for the mootractor image I made.--Canoe1967 (talk) 02:51, 19 May 2012 (UTC)[reply]

May 19

Eye movement limit

Can you move your eye so that your iris and pupil are not visible to reasonably sizeably open eyelids? The reason I ask is that you quite often see in tv/movies etc folk with just the white of their eyes visible, but I just tried to do this both up and down in my webcam and it didn't work, and then I asked a girl friend of mine with a very different build to try it on webcam and it didn't work, so it isn't a male/female thing and probably isn't a build thing. Doesn't work with left/rights either with me at least, I just realised that when typing this. Indeed, I can't even move my eyes to the extent that I can't see anything (although she says she can). If it isn't possible and we're not freaks, is this a physical or a mental limitation? Egg Centri c 02:49, 19 May 2012 (UTC)[reply]

It would be limited by the flexibility and elasticity of the optic nerve and of the muscles moving the eyeball. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 02:52, 19 May 2012 (UTC)[reply]

That's the physical limit. Is there a mental one? (i.e. brain preventing demands to the muscles that "don't make sense") Certainly when I move my eyes it isn't really a conscious process of trying to move the muscles as such, if you see what I mean. I just move the eyes to what I want to look at. Egg Centri c 03:00, 19 May 2012 (UTC)[reply]

I would guess not, seeing as it is possible to roll back one's eyes until it hurts, implying to me that that is where the physical limit kicks in. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:03, 19 May 2012 (UTC)[reply]

Sure, but for me at least that hurt is very similar to the sensation of looking at the sun. How do you perceive it out of interest? (And where does looking at the sun 'pain' come from now I think to add it, I was just assuming it was psychological) Egg Centri c 03:19, 19 May 2012 (UTC)[reply]

You know this is the thread that is actually going to get WMF sued... ;) Wnt (talk) 14:19, 19 May 2012 (UTC)[reply]

Are you saying I should not suggest trying to do eye exercises to gain stretchiness? Unique Ubiquitous (talk) 16:58, 19 May 2012 (UTC)[reply]

Clomethiazole

Just watching trainspotting at the moment (which is actually where my eye question inspriation came from!), and Renton is going through all the drugs he and his fellow junkies enjoy aquiring illicitly. As an opioid addict myself, which I've mentioned before (although never one who has ever stolen things or manipulated people or in any way whatsoever behaved like a twat to get drugs, and in fact I resent that stereotype for a million reasons - but primarily cause it means that a lot of so called medical "professionals" treat me like shit - oops rant detected... in my view it comes about because twats are far more likely to use drugs, rather than drugs make you a twat, I've even met highly functional crack and meth addicts, I know a head of a desk at a bulge bracket bank who's one... but I digress...) I've never heard of this drug, unlike the others, and looking it up... ok, it's a sedative and a hypnotic. That usually means it may have some recreational potential. But is that the only reason they would want it, and does anyone know of it actually being abused? Is it a potentiater (sp?) for something? One suspects there was another reason for including it in the list. Egg Centri c 03:19, 19 May 2012 (UTC)[reply]

Clomethiazole ... hmmm... I see it affects GABA signalling; but GABA-A rather than the GABA-B of Gamma-Hydroxybutyric acid ... still, apparently both have been attempted as withdrawal-medications for alcohol according to the worst translation of a medical article I've seen in some time. It appears on a long list of abusable substances, but this seems to be mostly a weird practice of alcohol addicts. [16] [17] My suspicion is that all this GABA crap, including GHB, is a poor substitute for ethanol which is a straight up GABA agonist and much better-known to partiers. Wnt (talk) 03:56, 19 May 2012 (UTC)[reply]

I wouldn't say GHB is a poor substitute for ethanol, speaking from personal experience (admittedly with GBL but AIUI that's for all intents and purposes the same thing) - indeed I'd say it's probably better due to the lack of calories! They are very similar admittedly. Lots of thingies effect GABA signalling though, right? Egg Centri c 04:36, 19 May 2012 (UTC)[reply]

Well, some Russians I've known said they used to get high on isopropanol in Gorbachev's time. All alcohols seem to have certain things in common - they taste bad, they're toxic, and they get you drunk - some are just worse than others. The nice thing about beer is you can usually figure out the right dosage even after you're drunk. ;) Wnt (talk) 14:42, 19 May 2012 (UTC)[reply]

The right dosage is until you pass out, crash the car, or run out, right? Egg Centri c 18:28, 20 May 2012 (UTC)[reply]

Proton-Proton Chain Reaction

In the artile (http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction), the third paragraph reads:

“In the Sun, deuterium-producing events are so rare (diprotons being the much more common result of nuclear reactions within the star) that a complete conversion of the star's hydrogen would take more than [ten billion] years at the prevailing conditions of its core.”

I am confused by this, as well as the next sentence, which states

“The fact that the Sun is still shining is due to the slow nature of this reaction; if it went more quickly, the Sun would have exhausted its hydrogen long ago.”

Regarding the first sentence; I find several problems for me to conclude that the article is accurate. First of all it claims that ‘deuterium-producing events are so rare in our sun, that it would take 10 billion years to convert all the sun’s hydrogen into helium’ - yet this IS what the sun’s lifetime is predicted to be.

Secondly; the article goes on to explain this very deuterium process (one proton, one neutron) in the subtitle: The proton-proton chain reaction - opposed to the “much more common” nuclear reactions the author proposes is taking place in our sun.

Thirdly; the sentence, in parenthesis, infers that the sun’s main fusions product is “diproton” - or, helium-2. The article for helium-2 (http://en.wikipedia.org/wiki/Diproton#Helium-2_.28diproton.29) explains that diproton is only a “HYPOTHETICAL” helium isotope. It appears to me that helium-2 not only defies the Pauli exclusion principle, if even possible, it would convert the sun’s hydrogen into helium exponentially faster than 10 billion years!

As for the second sentence; it could be accurate IF it unambiguously defined WHICH “reaction” (deuterium or diproton) is responsible for the “slow nature” of the sun’s actual proton-proton fusion process.

To me, it seems the contributor may have cut-and-pasted information from another stellar nucleosynthesis web-page, without making the necessary changes to the wording to fit the context of the actual article. — Preceding unsigned comment added by 64.180.243.107 (talk) 03:50, 19 May 2012 (UTC)[reply]

The diproton, once formed, immediately decays back into two protons. Therefore, the only reactions that produce anything are those extremely rare ones that produce deuterium. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 14:09, 19 May 2012 (UTC)[reply]

See the present state of the article. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 15:51, 19 May 2012 (UTC)[reply]

Equilibrium in analytical mechanics

My textbook asked me to prove that a system with scleronomic constraints is in equilibrium if and only if ${\frac {\partial V}{\partial q_{j}}}=0$ for all generalized coordinates q_j, assuming that all non-constraint forces are conservative. (V is the system's potential energy).

I was able to prove this, but as far as I can tell this proof holds for all holonomic systems, not just scleronomic ones. But because it was stressed that the constraints are scleronomic, I suspect that I'm making a mistake somewhere.

Proof of above statement: System is in equilibrium iff ${\vec {F}}_{i}=0$ , where ${\vec {F}}_{i}$ is the total force on the ith particle. $Q_{j}=\sum _{i}{\vec {F}}_{i}\cdot {\frac {\partial {\vec {r}}_{i}}{\partial q_{j}}}$ , where Q_j is the generalized force associated with the jth generalized coordinate. So, if ${\vec {F}}_{i}=0$ , then Q_j = 0. But $Q_{j}=-{\frac {\partial V}{\partial q_{j}}}$ , so ${\frac {\partial V}{\partial q_{j}}}=0$ is a necessary condition for equilibrium.

Now we prove that it is a sufficient condition. To do this, we find the ${\vec {F}}_{i}$ 's as a function of the Q_j's by making virtual displacements $\delta q_{j}$ to the generalized coordinates. The the virtual work is $\delta W=\sum _{j}Q_{j}\delta q_{j}=\sum _{i}{\vec {F}}_{i}\cdot \delta {\vec {r}}_{i}$ . Writing $\delta q_{j}=\sum _{i}\nabla _{i}q_{j}\cdot \delta {\vec {r}}_{i}$ (we've tacitly expressed the generalized coordinates as functions of the r_i's; $\nabla _{i}q_{j}$ stands for ${\hat {x}}_{i}{\frac {\partial q_{j}}{\partial x_{i}}}+{\hat {y}}_{i}{\frac {\partial q_{j}}{\partial y_{i}}}+{\hat {z}}_{i}{\frac {\partial q_{j}}{\partial z_{i}}}$ ).

From this, it follows that $\sum _{i}{\vec {F}}_{i}\cdot \delta {\vec {r}}_{i}=\sum _{i}(\sum _{j}Q_{j}\nabla _{i}q_{j})\cdot \delta {\vec {r}}_{i}$ , implying that ${\vec {F}}_{i}=\sum _{j}Q_{j}\nabla _{i}q_{j}$ . Therefore, if Q_j = 0, system is in equilibrium. QED.

Now, as far as I can tell I haven't used the assumption that the constraints are scleronomic, but maybe the assumption sneaked in there somewhere.

This is a pretty pedantic question, but it's been bugging me for a little while now, and I'd appreciate anyone's help in alleviating this confusion. 65.92.6.118 (talk) 06:26, 19 May 2012 (UTC)[reply]

Plastic cities

http://rt.com/usa/news/donna-summer-cancer-singer-627/

Donna Summer believed that it was the September 11 deadly smokes that gave her cancer.

During the WW2, the U.S. and Nazi Germany destroyed many cities by carpet bombing: Dresden, Coventry just to name a few. These WW2 cities certainly did not have much polymers and advanced man-made materials. People could live in the ruins. The cities were rebuilt within years after the end of the wars.

What will happen if a today's city is destroyed by a major earthquake or a meteor impact? Will they become too toxic for anyone after the fire? -- Toytoy (talk) 07:02, 19 May 2012 (UTC)[reply]

Despite Donna Summer's beliefs, most people don't worry too much about a comparatively low probability of harm from pollutants in city air (except possibly in Tokyo). A major earthquake or meteor impact might temporarily increase these pollutants, but it is likely that any single city so afflicted would be rebuilt within a few years, by which time nearly all air-borne pollutants would have been blown away (possibly to harm life on the rest of the planet, but the risk to any one individual is usually considered to be really small). Dbfirs 07:37, 19 May 2012 (UTC)[reply]

Plain old wood ashes and fumes are plenty toxic, when inhaled. People just don't think of it in those terms, when they say people died from "smoke inhalation". StuRat (talk) 22:11, 19 May 2012 (UTC)[reply]

While I know that Donna Summer didn't smoke, and died of lung cancer, she did work most of her life in an industry historically riddled with smokers. For her to blame the September 11 attacks is pretty non-scientific, but of course she wasn't a scientist. However, we haven't conducted many experiments on the longer term health impact of destroying several large, modern city buildings, and hopefully won't, so some speculation is valid. HiLo48 (talk) 22:40, 19 May 2012 (UTC)[reply]

I saw a clip a couple of days ago, of her in a recording studio, smoking a cigarette. So she definitely smoked. How much, I couldn't say. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:27, 20 May 2012 (UTC)[reply]

TMZ confirms she was a smoker.[18] This pic[19] looks like what I saw on TV. My guess is she was in denial. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:29, 20 May 2012 (UTC)[reply]

Hmmmmm. Our article on her says "...despite being a non-smoker, and the cancer was unrelated to smoking". Is this bullshit? HiLo48 (talk) 00:38, 20 May 2012 (UTC)[reply]

She was certainly a smoker at some point. I didn't record the coverage the other day, but I had the impression she had quit relatively recently. ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:22, 20 May 2012 (UTC)[reply]

Actually, the Royal Air Force destroyed Dresden before the yanks even got there. 1.124.213.85 (talk) 18:12, 20 May 2012 (UTC)[reply]

Species identification

Could I have help identifying this flower species (1, 2) and this monkey species? — Crisco 1492 (talk) 09:13, 19 May 2012 (UTC)[reply]

No. 2 looks like Bouganvillea spectabilis which illustrates our article. Richard Avery (talk) 19:45, 19 May 2012 (UTC)[reply]

I believe the monkey is a macaque, possibly a crab-eating macaque but there are lots of macaque species that all look generally similar. Looie496 (talk) 22:02, 19 May 2012 (UTC)[reply]

And number 1 is clearly a lily, and looks a lot like Lilium candidum, better known as the madonna lily, although that isn't native to Indonesia. Looie496 (talk) 22:09, 19 May 2012 (UTC)[reply]

I think you should be cautious with lilium candidum. The flower in your picture seems too shallow and does not show the characteristic textured 'line/s' down the length of the petal normally evident in lilium candidum, see here. Richard Avery (talk) 14:53, 20 May 2012 (UTC)[reply]

Unidentified houseplant

Could anyone help identify this houseplant please? I suspect it of being some sort of geranium. DuncanHill (talk) 09:21, 19 May 2012 (UTC)[reply]

No it's a Begonia tiger. --TammyMoet (talk) 10:38, 19 May 2012 (UTC)[reply]

Of course it is, I should have known it's a Begonia! Many thanks :) DuncanHill (talk) 10:50, 19 May 2012 (UTC)[reply]

Geiger tube

Find the potential difference between the tube and the wire in a Geiger tube. Variables are $r,L,R,Q$ ; you can guess what they stand for. Doing these integrals is prone to mistakes all the time. This is what I have tried so far. We must find the potential difference between a ring segment of the tube and the corresponding segment of the wire, each of width $dL$ . This makes the ring segment carry charge $-{\frac {QdL}{L}}$ and the little wire segment carry charge ${\frac {QdL}{L}}$ . To find this, we must find the potential difference between a point on this ring segment and the little wire segment. By Gauss's law, the electric field due to the little wire segment a displacement ${\vec {x}}$ away from the wire is $k{\frac {QdL}{x^{2}L}}{\hat {x}}$ if $x>r$ and zero otherwise. This potential difference is therefore $-\int _{r}^{R}k{\frac {QdL}{x^{2}L}}{\hat {x}}\cdot d{\vec {x}}=-\int _{r}^{R}k{\frac {QdL}{x^{2}L}}dx=k\left({\frac {Q}{RL}}-{\frac {Q}{rL}}\right)dL$ . I am wondering if you have to do lots more integrals; is there an easier way perhaps? The version of Gauss's law I have been given is: the electric field in the tube is solely contributed by the wire, and the field outside the wire is the same as though the wire were infinitely thin; the outer tube does not contribute as long as we are not near the ends of the tube. --150.203.114.37 (talk) 10:59, 19 May 2012 (UTC)[reply]

This approach is completely wrong, and the OP has apparently realised it. See similar question from the same OP below. Keit60.230.198.136 (talk) 11:48, 20 May 2012 (UTC)[reply]

Do microwave ovens use less energy than stoves?

For comparable tasks, I mean. Suppose a 1400kw microwave versus an ordinary GE electric stove (I have no idea what it's kw rating is, but it plugs into a standard 220v outlet). I wonder if anyone can make a reasonable calculation of energy required in two cases:

1. Raising a 16 oz. mug of instant coffee + tapwater to drinking temperature, which takes 1 minute in my microwave, versus perhaps 5 minutes on the stovetop. (Actually I haven't heated coffee on the stovetop in the last twenty years, so I'm guessing at the 5-minute idea.)
2. Heating a small frozen dinner, 5 minutes in the microwave versus 30 minutes in a 350-degree oven, which takes 5 minutes to preheat.

Am I truly saving energy by using the microwave for these tasks, or not? A plain-English answer, please. Textorus (talk) 15:05, 19 May 2012 (UTC)[reply]

In general, yes, the microwave oven is more efficient, because you heat less air. However:

A) Excess heat isn't a bad thing in winter. That means that much less heat your home heating system needs to generate. If you have electrical home heating, this should cost about the same (unless you pay a different rate for that electricity). Of course, the reverse is true in winter, if you find yourself running the A/C more because of the excess heat generated by the oven and stove.

B) Old microwave ovens tend to lose efficiency.

C) You compared with an electric stove and oven. The calculations are different with gas, since it's usually about 1/3 the cost.

D) Certain things (large frozen items) are not well suited to microwaving, and you may need to nuke them for a very long time on the lowest setting, to allow the heat to conduct into the center. This lowers the microwave's efficiency, as the heat generated near the surface is also radiating out.

E) The efficiency of a stove is highly variable. If you have the heating element extending beyond the edges of the pot, and the pot is uncovered and boiling away, you are wasting lots of energy.

To find the answer in your specific case, you might want to buy an outlet monitor, which will show exactly how much electricity is used by an outlet. Of course, in your case, you'd need one that can handle 110V/220V, so you can compare the stove and microwave. And the oven/stove may be hardwired in or have an inaccessible outlet.

If so, you can try watching your house electricity meter. Note the rate of electricity usage before you turn each device on, and after. Then multiply this by the time needed. This isn't perfect, though, as a fridge or something might just happen to kick on or off at the same time. So, you might want to do this a few times.

Incidentally, I bet the cost of the electricity to heat the coffee will be trivial in either case, when compared to the cost of the coffee. This is because you are heating such a small amount of water. Compare this, say, with the amount of water heated for a bath or shower. StuRat (talk) 15:11, 19 May 2012 (UTC)[reply]

Thanks for these observations, Stu, but I'm not concerned with the cost (convenience is paramount in these situations), and I'm not going to buy an outlet monitor that I would never use again for anything. It's an academic question - I was thinking that someone who knows what must surely be a very simple equation or two could quickly calculate the amount of energy used in each case (e.g., 1 min. @ 1400kw/hr. = 23.33 kw, no? But how much for five minutes on a stove burner?). Textorus (talk) 15:39, 19 May 2012 (UTC)[reply]

Unfortunately, you won't have the inputs needed for such equations, like the insulation factor on the oven, amount of heat that escapes around the edge of your pot, etc. So, the best we can do, short of measuring those, or measuring electricity use by the devices directly, is to make generalizations. StuRat (talk) 15:45, 19 May 2012 (UTC)[reply]

Stu, if we were building an ultra-precise set-up for a rocket launch or a nuclear pile, we might want to consider those other variables. But it should be a very simple, straightforward calculation to determine how much energy is drawn from the household electrical system during the time the stove is turned on. That's what I'm asking, and I've given specific values for those times. Textorus (talk) 15:53, 19 May 2012 (UTC)[reply]

The answer is, obviously, to check the power consumption of your stove. It's probably printed on the stove's UL sticker. A typical GE electric range or cooktop, like this one, is rated around five kilowatts, or about 1500 watts per burner at full power. Total energy is almost perfectly approximated by "power times duration" because the coil is almost a perfect resistive load. Nimur (talk) 16:04, 19 May 2012 (UTC)[reply]

1500W×4 = 6kW. I take it the smaller burners are more like 1000W ? StuRat (talk) 16:29, 19 May 2012 (UTC)[reply]

I'm in the habit of avoiding false precision; as the exact stove model is unknown, the wattage is unknown, the number of burners is unknown, the altitude above sea-level is unknown, and the total time required to boil a "cup" of unknown volume (between 8 and 16 fluid ounces?) is also unknown, I'd avoid even using as much precision as that, StuRat. One significant figure and an approximate order of magnitude should suffice; I trust that our OP knows how to multiply, as long as they can find the relevant parameters. In the words of a pedigreed physicist, "approximately four burners times approximately 1500 watts each is approximately 5 kilowatts, which is approximately the safe level for a typical household kitchen circuit breaker in the USA." Incidentally, this fast-and-loose definition of "equality" is the difference between analysis and arithmetic, which is why they pay physicists more than accountants. (Or less. I can't recall; there may be a lost minus-sign in my figuring; that sort of arithmetic nuance is left as an exercise for the reader). Nimur (talk) 16:40, 19 May 2012 (UTC)[reply]

How much energy the oven and stove uses, is, of course, variable. So, your best bet is to look at the house electricity meter, as I suggested, to see how much is actually used when heating the given items. StuRat (talk) 16:00, 19 May 2012 (UTC)[reply]

There is a label of some kind inside the oven door, but it's long since become illegible from heat and cooking stains. I don't need a precise, down-to-the-last ion answer, just a general comparison between the two heating methods. Using standard ratings, such as in the link Nimur provided, would be sufficient. And no, I'm not going to go stand in the yard and guestimate how fast the little wheel in the electric meter is turning; that would be highly inexact for anyone but the meter reader, perhaps, and not worth the trouble. Textorus (talk) 17:03, 19 May 2012 (UTC)[reply]

OK, then, without having any idea how much power your stove or oven uses, the best we can do is go with the general trend and say that it uses more than the microwave. We could even guesstimate it at twice as much. Good enough ? StuRat (talk) 17:17, 19 May 2012 (UTC)[reply]

Yes, possibly rather more than twice. "1 min. @ 1400kw/hr. = 23.33 kw" should read "1 min. @ 1400watts. ≈ 23 watt-hours" (or 84000 joules) and a 1.5Kw stove ring for 5 minutes would use 5 min. @ 1500 watts ≈ 125 watt-hours (or 450000 joules) so perhaps up to six times as much (though with the benefit of room heating mentioned above). Dbfirs 07:09, 20 May 2012 (UTC)[reply]

But that's a large burner on full, which is probably more than is needed to heat that small amount of water in 5 minutes. StuRat (talk) 15:16, 20 May 2012 (UTC)[reply]

Agreed. I was taking figures suggested above where there would certainly be considerable wastage of heat. I could certainly boil a large cupful in 5 minutes or less with a small pan on a 1000w radiant ring. Dbfirs 19:27, 20 May 2012 (UTC)[reply]

I made some experiments on a Electrolux cf 502(stove), and Melissa AG820CQM (microwave).

In both cases I thrid to heat 2.1 dl (aprox. 7 oz.) from around 22 C to boiling.

To heat the water 2.1*100*4.2*(100-22)/3600/1000 kWh=0.019 kWh is needed.

The stove used about 1640 W for 3 minutes, approx. 0.08 kWh, 23% efficency.

The microwave used about 1220 W for 2 min, approx. 0.04 kWh, 46% efficensy.

If the claimed 800 W microwave power are correct 0.027 kWh, 66% become microwaves.

The losses in the stove are mostly due to the heat remaining in the stove when the water starts to boil. The stove could be used more effeciently by turning it of before the water strts boiling and use some of the after heat.

The electriciy consumption was mesured by counting impulses on the meeter and clocking them and substracting background consumption.(1000 imp/kWh)

The stove will be more efficient when boiling larger volumes. I expect the microwave oven to be much more effecient than a conventional oven but i have not tested. Gr8xoz (talk) 18:50, 20 May 2012 (UTC)[reply]

Thanks for doing the experiment. That 0.08 kWh for the stove versus 0.04 kWh for the microwave is exactly twice as much energy, just as I guessed. Do I get a prize ? Incidentally, the cost of using the stove is about 1 cent and the microwave half a cent, at my electricity rates (12.5 cents per kWh). StuRat (talk) 19:42, 20 May 2012 (UTC)[reply]

Specific entropy

Is it possible to express specific entropy in terms of the gas constants cv, cp and pressures p1 and p2 only? I can't derive anything close to this without using v1 and v2. The closest I've got is s2-s1=cvln(p2/p1)+cpln(v2/v1) but i don't want the vs in there. This is an is isentropic process so were saying PV^n is constant. 94.116.0.41 (talk) 15:17, 19 May 2012 (UTC)[reply]

I haven't done chemistry in years. I remember PV/nT(1) = PV/nT(2). Pressure, Volume, molecule count (not moles) and temperature. I think you can remove any variable from each side, so P/nt(1) = P/nT(2). I don't know if this helps.--Canoe1967 (talk) 19:07, 19 May 2012 (UTC)[reply]

Why isn't dinitrogen toxic?

Since dinitrogen, the acetylide dianion, and the nitrosonium cation are isoelectronic with carbon monoxide and the cyanide ion, why aren't they as deadly poisonous as carbon monoxide and the cyanide anion? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 16:24, 19 May 2012 (UTC)[reply]

Different different elements have different electronegativity, formal charge, HSAB interactions, bond strength, intrinsic stability/transportability to the biochemical target, etc. I think even your own hypothesis flawed, assuming that CO and CN- are "deadly poisons" by the same biochemical pathway. DMacks (talk) 17:18, 19 May 2012 (UTC)[reply]

They are deadly poisonous in the same way—they bind to iron in hemes to the exclusion of oxygen. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:03, 20 May 2012 (UTC)[reply]

Cyanide poisoning is not caused by reaction on bloodborn hemoglobin or exclusion of oxygen transfer. Carbon monoxide is not instantly deadly down to the cellular level even at low concentrations. DMacks (talk) 02:49, 20 May 2012 (UTC)[reply]

Is this similar to iron, cobalt and nickel being the only three normally magnetic elements but stainless steel made from iron, cobalt and nickel is not normally magnetic?--Canoe1967 (talk) 19:12, 19 May 2012 (UTC)[reply]

One other thing to think about: Why would life have evolved to be incompatable with dinitrogen, which has always been the predominant gas in the atmosphere? If you reach a possible conclusion that is incompatable with easily observable facts, then your conclusion is obviously wrong, in other words, since all life has been bathed in dinitrogen for billions of years, there's no need to assume it would be toxic in any way. Any line of thinking that leads you down that road isn't very productive, so your assumptions must be horribly flawed. --Jayron32 20:28, 19 May 2012 (UTC)[reply]

Because the lone pair on the carbon of CO is not as tightly bound as those on dinitrogen, CO is the stronger Lewis base. The same goes for the cyanide anion. Nitrosonium's lone pairs are even more tightly bound than dinitrogen's. Therefore, CO and cyanide coordinate much more easily to the iron than dinitrogen or nitrosonium. This is my best explanation of the toxicity of these isoelectronic species.--Jasper Deng (talk) 02:58, 20 May 2012 (UTC)[reply]

But then why isn't the acetylide dianion even more toxic than either CO or cyanide? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:09, 20 May 2012 (UTC)[reply]

As I noted before, how were you planning to get that dianion to the heme–Fe target? DMacks (talk) 03:12, 20 May 2012 (UTC)[reply]

How do you think? It's obvious, isn't it? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:16, 20 May 2012 (UTC)[reply]

It's obvious that you did not actually think about that. DMacks (talk) 03:28, 20 May 2012 (UTC)[reply]

Two reasons:

Acetylide cannot exist in solution. It would immediately hydrolize to acytelene and hydroxide ion: C₂^2-(aq) + 2H₂O(l)→ 2 OH^-(aq) + C₂H₂(g).
Even if it could exist in solution, the lone pair is more tightly bound on acetylide than CO because in CO, the bond is polarized toward O, lessening C's grip on its electrons, which is not the case with acetylide.--Jasper Deng (talk) 03:31, 20 May 2012 (UTC)[reply]

That depends on whether the iron atom is a stronger Lewis acid than the proton or vice versa.
Non sequitur from what you said earlier about that. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 03:39, 20 May 2012 (UTC)[reply]

The proton is stronger by a lot. I know this for a fact for that particular anion; it would hydrolyze before coming anywhere near the iron.--Jasper Deng (talk) 03:42, 20 May 2012 (UTC)[reply]

Dasyuromorphia and carnivores

Since Dasyuromorphia are carnivores, why there are ranked as separate order and not suborder of Carnivora? Also, what hinders thylacine from being placed in Canidae or Caniformia?--176.241.247.17 (talk) 22:27, 19 May 2012 (UTC)[reply]

For the first part, Carnivora are placental carnivores, and the Dasyuromorphia are marsupials. Now, as to why they don't have an order of marsupial carnivores, I do not know. StuRat (talk) 22:31, 19 May 2012 (UTC)[reply]

The simple answer is that the order Carnivora is not inclusive of all carnivores. Being carnivorous does not put one, taxonomically, in the order Carnivora. As for your second question, again, marsupials are quite separate from the other carnivorous orders and quite different, no matter what type of taxonomical scheme you are using. Their apparent visual and behavioral similarity is an example of parallel evolution and does not indicate that they are closely related. --Mr.98 (talk) 23:05, 19 May 2012 (UTC)[reply]

Problems with xenon tetroxide xenate formulas

Didn't the xenic acid article say that there is no such thing as a completely de-protonated xenate salt?--Jasper Deng (talk) 23:22, 19 May 2012 (UTC)[reply]

Xenate. Not perxenate. Completely deprotonated perxenates are known—and xenon tetroxide is perxenic anhydride, not xenic anhydride. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:06, 20 May 2012 (UTC)[reply]

Yeah, but the xenon tetroxide article makes mention of XeO₄^{2-, which is a completely deprotonated xenate. See the synthesis section.--Jasper Deng (talk) 01:16, 20 May 2012 (UTC)[reply]
That has been fixed. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:33, 20 May 2012 (UTC)[reply]
My only concern is that I'm not too certain of the products of that reaction.--Jasper Deng (talk) 01:36, 20 May 2012 (UTC)[reply]}

May 20

Locomotives

There are 0-4-0s, 2-2-0s, and 0-2-2s. Why aren't there 0-2-0s? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 01:57, 20 May 2012 (UTC)[reply]

In France an 0-4-0 would be an 020. Sticking to American notation, an 0-2-0 would have a single axle, which seems unsatisfactory, rather like a car with one axle. Acroterion (talk) 02:03, 20 May 2012 (UTC)[reply]

Although, strangely, there have been 0-3-0s. Gandalf61 (talk) 08:51, 20 May 2012 (UTC)[reply]

And 1-1-2-1-1's. Tevildo (talk) 12:40, 20 May 2012 (UTC)[reply]

The answer is, therefore, simple statics. It would be possible to have a 0-2-0 or 0-1-0 locomotive on a Lartigue-style monorail, with the centre of gravity of the loco below the rail. However, the minimum stable configuration for a loco with all its wheels on the ground would be a tricycle 1-2-0 or 0-2-1 arrangement, but this would require three rails rather than two. Tevildo (talk) 12:51, 20 May 2012 (UTC)[reply]

There is an important confusion to be resolved here. In Europe, an 0-4-0 engine has 8 wheels, 4 on each side - the number '4' refers to the number of axles. In the US, the numbers count wheels instead so an 0-4-0 has only 4 wheels, 2 on each side. Hence, in US notation, an 0-2-0 would be silly - an engine with just one wheel on each side! Obviously such a thing cannot exist on a conventional railroad. In Europe, an 0-2-0 is a simple 4-wheeled, 2 axle vehicle with all four wheels being driven...like a 4WD car. I strongly suspect that our questioner is using European notation and asking why there aren't any engines with just four driven wheels on two axles. The answer probably relates to stability on the track or something...but I don't know details. SteveBaker (talk) 15:03, 20 May 2012 (UTC)[reply]

There were plenty of two-axled locomotives (0-4-0 in American (Whyte) notation, 0-2-0 in French) during the age of steam after the initial era of development: they were most typically small shunting engines, though they might occasionally be used for pulling small, local freight or passenger trains. In the subsequent diesel and electric era, 0-6-0s tended to be the minimum, probably because railway vehicles in general became larger and heavier, and train lengths on average greater: the locos themselves therefore also had to become more powerful and heavier, and a heavy 0-4-0 would exert too great an axle load on the track, whereas the same loco in 0-6-0 format would have a maximum axle load 33% less.

(Early in locomotive evolution, of course, 0-2-2s and 2-2-0s also existed – Stephenson's Rocket was an 0-2-2.)

To further avoid confusion, it should be mentioned that the UK (and hence the Empire/Commonwealth countries) also employed the Whyte notation system rather than the Continental.

By the way – Welcome back, Steve! {The poster formerly known as 87.81.230.195} 90.197.66.211 (talk) 22:49, 20 May 2012 (UTC)[reply]

Geiger tube continuation

I calculated the potential difference between the wire and the tube of a Geiger tube to be ${\frac {1}{2\pi \epsilon _{0}}}{\frac {Q}{L}}\ln {\frac {R}{r}}$ approximately. The power-supply voltage is slowly increased until you see a glow in the air very near the inner wire. This means that the air breakdown potential has been reached (and they provide a value for this and for all the other variables). Calculate this power-supply voltage. How do you do this calculation? Do you just substitute the numbers into the above equation or is it more involved than that? --150.203.114.37 (talk) 04:54, 20 May 2012 (UTC

It would help if you told us what L, Q, R, & r signify. Operating a geiger tube at a voltage high enough to cause a gas discharge (as indicated by a glow) is harmful to it. Normally, you just set the voltage to the value stipulated by the gieger tube manufacturer. This is at a "plateau" of maximum sensitivity that occures just below the glow point. Ratbone60.230.203.253 (talk) 07:23, 20 May 2012 (UTC)[reply]

Maybe; I didn't actually observe this; it's just a theoretical problem. Q = charge on the wire, R = radius of the tube, r = radius of the wire, L = length of the wire/tube. --150.203.114.37 (talk) 07:30, 20 May 2012 (UTC)[reply]

Then the formula given is the relation between the potential difference and charge as for any coaxial pair of conductors separated by a perfect insulator of permitivity e₀. It is a concept useful in calculating the capacitance http://en.wikipedia.org/wiki/Capacitance and has no relavence whatsowver to the operating voltage of a gieger tube. Under operating conditions,gieger tubes are operated with DC voltage, and no current flows in the tube capacitance, except for the very brief recharging current after each detected particle. Note that the term e_o, strictly speaking, being the symbol for the permitivity of free space, is in any case incorrect. It should be the permitivity of the gas used (e₀.k, which will however be sensibly close to e₀). Keit120.145.31.247 (talk) 09:22, 20 May 2012 (UTC)[reply]

Ah. What is the correct furmula then, and how do you derive it? --150.203.114.37 (talk) 10:44, 20 May 2012 (UTC)[reply]

As Ratbone said, this is not something you would normally calculate. The manufacturer of the geiger tube will tell you what voltage to use. The manufacturer determines the optimum voltage by testing. However, an approximate calculation of the optimum operating voltage can be found by using the formula for breakdown voltage of a gas under partial vacuum, given in http://en.wikipedia.org/wiki/Breakdown_voltage. As Ratbone also said, the optimum voltage is just a bit below the breakdown voltage. If you are intending to make your own geiger tube, say so, we can then give you information of more practical benefit. Keit60.230.198.136 (talk) 11:41, 20 May 2012 (UTC)[reply]

This is just a homework question; I am not trying to make one of these. I have never seen this formula before. Presumably you are required to use some of the parameters given. It's possible you also need some other well-known constants, but I wouldn't know. This is the exact text of the question: The power-supply voltage is slowly increased until you see a glow in the air very near the inner wire. This means that the air breakdown potential of

3\times 10^{6}V/m

has been reached. Calculate this power-supply voltage (give a numerical value) and explain your calculation. The length L = 80 cm, the inner radius r = 0.7 mm, and the outer radius R = 3 cm.--150.203.114.37 (talk) 13:20, 20 May 2012 (UTC)[reply]

In which case you have attempted a $10 answer to a 10 cent question, unless the orginal question gave you the gas pressure, and you can ignore the answers given previously by Ratbone and myself. Whoever wrote this question has a very simplified view of geiger tubes. All you need to answer the question is given in the question, plus a bit of the most elementary algebra. You can ignore the length, as a long tube has the gas under the same voltage gradient stress as a short one. The answer, fitting the data given, is 88 kV. As it is a homework question, I won't say any more than that. Real geiger tubes have an operating voltage of around 500V, because radius of the tube is smaller, and because the gas inside, which isn't air, is under partial vacuum, but this is no concern to you - you need to write what the teacher wants. The moral of this posting is that if you have a homework question, gives us the exact question, and what you have attempted in order to solve it. Then you may get an answer from us appropriate to your needs. Keit60.230.198.136 (talk) 15:14, 20 May 2012 (UTC)60.230.198.136 (talk) 14:57, 20 May 2012 (UTC)[reply]

The problem is I don't know where to start. Is there an equation which relates all these variables that I have to manipulate? --150.203.114.37 (talk) 15:49, 20 May 2012 (UTC)[reply]

Are you pulling my leg? This is very simple for a chap prepared to get involved with integrals etc as in your first version of this posted question. You don't need no fancy formula - its just proportion. As stated in your assignemnt question, the breakdown voltage of air at standard atmostpheric pressure is approx 3 x 10⁶ V / m. So if you have a breakdown distance of 2 m, you'd need 2 x 3 x 10⁶ V (ie 6 MV); if you have 0.1 m, you'd need 0.1 x 3 x 10⁶ V (300 kV). And the breakdown is between the 0.7 mm radius wire and the 3 cm radius cylinder.... Go for it tiger. Keit58.167.225.195 (talk) 16:10, 20 May 2012 (UTC)[reply]

I see. But how do you know that that is the breakdown distance? There being a glow "very near" the wire may imply that the breakdown distance is less than that. Or is the breakdown distance necessarily the distance between the wire and the tube in a Geiger tube? --150.203.114.37 (talk) 17:45, 20 May 2012 (UTC)[reply]

(edit conflict) I think that is missleading, the field is not homogenius, the glow was just closest to the inner conductor. You need to integrate the field along a radius or use a ready formula for this case. Due to rules regarding homework i do not want to give it here but you could calculate the linear charge that gives the desired field at the surface of the inner conductor and then use the formula you gave. Gr8xoz (talk) 17:49, 20 May 2012 (UTC)[reply]

Does this matter? Is the power-supply voltage not still 87900 V? What calculation am I supposed to do? I don't understand. --150.203.114.37 (talk) 21:38, 20 May 2012 (UTC)[reply]

When the glow is seen, breakdown has already occurred. Any non-linearity in electric field strength as alluded to by Gr8xoz has no relavence to your assignment question, because you want the breakdown initiating voltage, which is under conditions existing just before breakdown actually occurs. In practice, as you slowly increase the voltage, breakdown occurs suddenly with only a very small increase. So this additional small voltage, between no breakdown and breakdown started, can be neglected. Yes, obviously, the electric field stress on the gas/air (ie a field which accelerates the odd free electron fast enough to strike a gas molecule with enough kinetic energy to knock off another electron) is between the wire and the cylinder. As it is homework I don't want to put it any plainer than that - what you learn and remember is proportional to the mental effort YOU put in, not what I put in. If you want to understand the breakdown phenomena, and thus why the breakdown voltage is proportional to the breakdown distance, FOR THE DIMENSIONS AS GIVEN, start with our article on Paschen's Law, http://en.wikipedia.org/wiki/Paschen%27s_law. You could have found this yourself if you checked the Wiki link I gave you previously. The relationship between distance and voltage becomes very non-linear at small distances, but this is at distances very tiny compared to the dimensions given in your question. Given that whoever wrote the question has a very simpilfied view of geiger tubes (the dimensions are very innappropriate, and geiger tubes use particular gases under partial vacuum, not air at standard pressure, tubes are operated just below breakdown, not at breakdown), I suspect only a very simple answer is required. Is this from a high school math assignment? What year? Or a university physics course? Keit124.182.148.14 (talk) 01:32, 21 May 2012 (UTC)[reply]

Domestic Heat Pump connections

I am planning to install an air to water domestic heat pump for heating and am looking at the manuals. The 12 kW heat pump works best when there is a 35 litres/minute flow through it (it then heats the water by about 5C each sweep through the exchanger). If the flow rate falls below this, the manual says it becomes inefficient and cuts in and out. The circuit I wish to heat is already partly structural to the house and would not take that flow rate (indeed I am guessing even approaching that flow rate would be rather noisy), but the heat loss on that circuit is more than 5C. So is there any reason I cannot just "short" part of the outflow from the heat pump directly into the inflow for the heat pump mixing with the return? It seems a little too obvious as a solution...obviously the later part of the heating circuit will be a little cooler from having a high temp drop across them (thats ok, the later parts of the circuit will be the larger surface area ones), but if the heat pump is happy I know the actual heat is being transferred into the house a reasonable efficiency, right? --BozMo talk 07:22, 20 May 2012 (UTC)[reply]

Heat pumps are more efficient when the heat exchanger is at a lower geometric mean temperature throughout the exchanger, and more efficient if the hottest part of the exchanger is lower in temperature. So, if you reduce the water flow, the hottest point, and the geometric mean temperature, must increase, lowering efficiency. If you try and get around this by providing a bypass flow, what you are doing is increasing (by mixing cold with warm) the inlet water temperature, again increasing the mean exchange temperature. So efficiency is still lost, but not necessarily at quite the same degree. Ratbone58.167.225.195 (talk) 16:36, 20 May 2012 (UTC)[reply]

I'd think just lowering the flow rate through the entire system would lower efficiency less than the bypass. I don't quite understand what they mean when they say it will "cut in and out", though. StuRat (talk) 16:45, 20 May 2012 (UTC)[reply]

A heat pump water heater is essentially the same technology as a heat pump (ie compressor type) airconditioner. A temperature sensor controlls when the compressor runs - if the water is too hot, the sensor turns the compressor off. Thus the system cycles just like an airconditioner does. If frequent cylcing occurs, effieciency drops for 2 reasons: a) a high exchanger temperature itself reduces efficiency, b) the system spends a higher fraction of time in running up and stabilising. Ratbone124.182.19.67 (talk) 01:05, 21 May 2012 (UTC)[reply]

Right, but none of that explains why a lower flow rate would cause it to cut off more often. On the contrary, a lower flow rate should allow the water inside the cooling part of the loop to get cooler, since it stays in there longer, and the house should remain hotter than at a higher flow rate. Both of these should make it stay on longer. StuRat (talk) 02:59, 21 May 2012 (UTC)[reply]

Stu, it seems like you are talking about an airconditioner as in some domestic airconditioners that use a chilled water system, like large commercial buildings - the commpressor (termed a chiller) at a convenient location chills water, which is then piped to one or more fan-coil units within the dwelling. I assumed that the OP was asking about the more common heat pump water heater, ie a compressor unit, which "pumps" heat from outside air to heat water for your kitchen, bathroom, etc. On reviewing the OP's wording, it does seem that he's talking about an airconditioner. In which case, my explanation was the wrong one. I'm not familiar with domestic grade chilled water systems, but assuming they are much the same as commercial systems, the on-exchanger water temp in cooling mode will be about 12 to 14 C, and the off-exchanger water temperature will be a design temperature of around 6 to 8 C. If you halve the flow rate, the compressor (somewhat simplifying) in moving the same amount of heat, will cool the water to 0 C. The water must not be allowed to freeze - that will stop functionality and damage the system. So a sensor on the off-exchanger temperature will prevent it by shutting down the compressor until the water warms up again. If the OP installs a water bypass, the water will still get too cold, because some of it doesn't go thru the fan-coil unit and get warmed up, and will cause cycling. This problem sometimes occurs in commercial buildings, where (say) a 10-storey building has been constructed, but so far there are tenents on only one or 2 floors, so the fan-coil units on the empty floors are not functioning. A common solution is to temporily use electric heating, or pump in outside air thru some fan-coils, to make the chiller work cycle more slowly. This of course wastes energy. In heating mode, there shouldn't be much of a problem, though the chiller will cycle on and off within a short time frame for the same reason a standard aircon will, if way too large for the space conditioned. When its running, it will heat the room(s) too quickly, and the thermsotat will turn it off. Installing a water bypass won't fix that either, except that the higher exchanger temperature will make the system a little less efficient. Ratbone120.145.172.175 (talk) 03:33, 21 May 2012 (UTC)[reply]

I did consider the idea that water inside the cooling loop might freeze, but thought that they could just add anti-freeze to fix that problem. Or is the issue that frost will build up around the cooling loop ? I can see why that would reduce efficiency, but would it damage the system ? StuRat (talk) 03:52, 21 May 2012 (UTC)[reply]

I have never heard of adding antifreeze to chiller water, though that does not say it's not done. I note that common antifreeze (glycol) has a lot less specific heat than water, so adding it will promote short cycling and reduce heat & cooling capacity. You are right about frost on the exchanger though - I should have remembered that. Frost is, compared to moving air, a thermal insulator, so once frost forms, the exchanger temerature drops, forming more frost, ending up with bulk ice and no efficeincy. When I mentioned damage, I was refering to the fact, that at freezing, water expands. Ratbone120.145.172.175 (talk) 04:26, 21 May 2012 (UTC)[reply]

Hexahydrogen sulfide

There is hydrogen sulfide (H₂S). There is sulfur hexafluoride (SF₆). Why isn't there hexahydrogen sulfide (H₆S)? Whoop whoop pull up ^{Bitching Betty | Averted crashes} 12:34, 20 May 2012 (UTC)[reply]

Hydrogen does not hybridize its orbitals. So many hydrogens can't quite overlap with all six of sulfur's hybridized orbitals.--Jasper Deng (talk) 16:35, 20 May 2012 (UTC)[reply]

I fail to see how hydrogen orbitals come into it. It's probably because hydrogen isn't electronegative enough to stabilise sulfur in the +6 oxidation state, whereas fluorine is electronegative enough. However, a few molecules are known with sulfur bonded to six carbons (Organosulfur compounds#Sulfuranes and persulfuranes) which casts doubt on the electronegativity argument. --Ben (talk) 16:40, 20 May 2012 (UTC)[reply]

Well, hypervalency tends to occur when you have enough stretched-out orbitals that can link many atoms.--Jasper Deng (talk) 18:26, 20 May 2012 (UTC)[reply]

See Hypervalent molecule; one of the models of hypervalency is the Three-center four-electron bond, which can be predicted and explained using molecular orbital theory. To put things simply, a 3c-4e bond requires p-p-p mixing between the two colinear ligands and central atom. Hydrogen doesn't have any electrons in p orbitals availible to create the 3c-4e bond, so it won't bond like SF₆ will. There's even a nice picture of the type of bonding that happens in SF₆ in the hypervalent molecule article which shows the p-p-p interactions in forming the 3-center bond. Hydrogen doesn't do this. --Jayron32 18:36, 20 May 2012 (UTC)[reply]

If hydrogen couldn't form 3c-4e bonds, then the bifluoride anion wouldn't exist either—but it does. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 19:32, 20 May 2012 (UTC)[reply]

It does so only when it is the central atom, not when it is the ligand. Hydrogen can form such multi-center bonds as the central atom, and does so in the 2-electron 3-center bond in diborane as well. The difference is that hydrogen isn't the central atom in the hypothetical SH₆, sulfur is. --Jayron32 19:44, 20 May 2012 (UTC)[reply]

Thanks. Whoop whoop pull up ^{Bitching Betty | Averted crashes} 05:02, 21 May 2012 (UTC)[reply]

Resolved

Ingredients of paint primer?

Two questions:

I need to know what ingredients are used in paint primers (the kind used for wood and plaster - not metals) - our article Primer (paint) has no indication of ingredients and Paint discusses paint ingredients in general but doesn't explain what's different about primers.
Some interior paints say that they include a primer. From the description in our two articles of what a primer does, this seems unlikely to be true. Can anyone explain what's really going on here? (And again, what are the ingredients in primer+paint products that aren't in the paint-only stuff?)

Background: I'm using a 100 watt CO2 laser-cutter to cut 5mm plywood - when the plywood is painted with some kinds of paint (I'm using and off-white glydden interior eggshell), the laser has a hard time cutting it - with other kinds of paint (acrylic craft paint), it's no problem. Shinier paints seem to cut worse than flat paint and our paint article suggests that shinier paints have silica, glass and metal flakes in them...glass and metal can't be cut with such a low power CO2 laser, so this explains a lot. However, I still want to use primer on my plywood...and I'd like to find a way to put a shiney/waterproof surface on my plywood without impeding the laser too much. (No, I can't paint the plywood after it's cut because I'm also etching a design into the paint using the laser.) SteveBaker (talk) 14:48, 20 May 2012 (UTC)[reply]

Have you seen the Wikipedia article specifically on paint primers? The main purpose of a primer for wood is to seal it, so that the colour coat is not absorbed into the wood, leaving it pale & uneven and permeable to moisture. Paints intended for wood that incorporate a primer are quite common. Primer ingredients commonly are linseed or synthetic oils, or synthetic latex (such as PVA). In at least some cases, the only difference between (for wood) a primer and a topcoat is the proportions of the ingredients, the primer having a higher percentage of oil or latex. See http://www.buildings.com/tabid/3334/ArticleID/2846/Default.aspx. Why not use a matt paint compatible with the laser, then spray on a clear varnish to seal and make shiny. Ratbone58.167.225.195 (talk) 15:53, 20 May 2012 (UTC)[reply]

One thing to keep in mind: paints are not subject to ingredients disclosure laws like, say, foods are. Many paints and pigments are proprietary formulas whose exact compositiona and processes for making them are kept highly secretive as trade secrets. Actually getting the specific composition of a specific primer may be impossible. I had a friend that was a research chemist worked in the industry, and he had a ten-year non-competition agreement with the company he worked for; that is if he ever left the company he worked for, he couldn't work for any company in the pigments industry for ten years. They take that secrecy pretty seriously. --Jayron32 17:43, 20 May 2012 (UTC)[reply]

Electricity of x cm³ of water at y m

Can you calculate how much electricity can be generated by x cm³ of water at y m.88.9.109.2 (talk) 15:04, 20 May 2012 (UTC)[reply]

Yes, by determining the potential energy of that mass at that height, then applying an efficiency factor to determine the percentage of that energy actually converted to electricity by your dynamo. Do you want the formula ? StuRat (talk) 15:10, 20 May 2012 (UTC)[reply]

Looks like Stu thinks you want a hydro-electric (water-fall) answer, which is most likley the case. But you might want an answer appropriate to ocean thermal power (OTEC). Which is it? Is this homework? Keit60.230.198.136 (talk) 15:31, 20 May 2012 (UTC)[reply]

No homework. Just curiosity. Can you give me two formulas: 1. water is flowing through a dynamo/turbine. 2. water is enclosed, hanging in a cable attached to pulley connected to a dynamo. 88.9.109.2 (talk) 15:36, 20 May 2012 (UTC)[reply]

Water has a density of 1 g/cm³ (depending on temperature, impurities, pressure, etc.). So, we have X grams, at Y meters. The formula for gravitational potential energy is U = mgh, where m = Xg, g (on Earth) = 9.8m/s², h = Ym. This gives us U = (9.8)XY gm²/s². That's 1/1000th of joule. We also need to apply the efficiency, let's say 0.0 < E < 1.0. So, the amount of electricity generated, in joules, is (9.8)EXY/1000. The formula would be the same for the water in a container, although there the mass of the container must also be considered, and the cable is more complicated, because each portion has a different gravitational potential energy. StuRat (talk) 17:13, 20 May 2012 (UTC)[reply]

State of radon related to radioactivity?

Would radon be a gas if its radioactivity were neglected? Astatine and francium both would be vaporized by their own decay.--Jasper Deng (talk) 18:33, 20 May 2012 (UTC)[reply]

Quite likely it would, since the boiling point trends displayed by the noble gases don't have an obvious discontinuity between Xenon and Radon. I don't think radioactivity has anything to do with phase of matter at room temperature, which is a function of an atom's electronic properties, not its nuclear properties. That is, a substance is a gas because it has weak intermolecular forces, the strength of which can be preicted solely by the structure of the electron cloud around the molecules/atoms that make up the substance. --Jayron32 18:42, 20 May 2012 (UTC)[reply]

Well, radioactivity does tend to change temperature, so there's an indirect effect there. Francium's article said it would be liquid especially because it's radioactive.--Jasper Deng (talk) 18:46, 20 May 2012 (UTC)[reply]

Yes, but it doesn't change the boiling point or melting point. You're confusing the two temperatures. The boiling point of a substance is the expected temperature for it to boil, NOT the temperature it exists at. That is, Radon will boil at 211 K regardless of where the source of the heat making it 211 K comes from. If it gets to 211, it boils, and that has nothing to do with radioactivity. Likewise with Francium, the sentence in the article is misleading. What it is saying isn't that the 27 Celsius MP quoted in the article is effected by its radioactivity, it is saying that the extra heat generated by its radioactivity means that in room temperature air (nominally 25 degrees Celsius), the excess heat from the radiation is more than sufficient to raise Francium's temperature the extra 2 or so degrees to melt it. That is, sitting in the average indoor room, Francium will be a liquid because it will always be somewhate warmer than that room. Again, the radiactivity of the Francium doesn't affect the temperature it should melt at, it affects the temperature it is. Two different ideas. --Jayron32 19:01, 20 May 2012 (UTC)[reply]

Radon's boiling point is −61.85 °C so any radiative self-heating wouldn't be significant re: the state in a cozy room (or lab). Maybe if you had some radon at Vostok Station where the air temp was -62 °C radioactivity might push the temperature of your radon just above the boiling point. 19:19, 20 May 2012 (UTC)

Ultra High power mechanical HVDC-breakers?

I am doing some calculations on the feasibility of a ultra high voltage ultra high current world wide HVDC-grid. I speculate about 2000 kV to ground and 25 kA nominal current, 100 GW per bi-pole. The fault current after a few ms short circuit would be in the range of 100 kA. One of the big problems in constructing such a system are of course the selective removal of faulty components while maintaining operations of the rest of the system.

My question is if the needed size and cost of DC-breakers capable of breaking 2 000 kV and 100 kA can be estimated?

The challenge are of course to contain the blast and to cool the arc (generating >200 GW?) to temperatures below the temperature were significant ionization occurs while maintaining the isolation.

Today no HVDC-breakers are in operation, HVAC-breakers in the range of 800 kV and 50 kA exists, as I understand it they can only break a few kV so they will only interrupt the current near a zero crossing in the AC-current. ABB have developed a "Hybrid" HVDC-breaker, using semiconductors. Proactive Hybrid HVDC Breakers. It seems that it would be extremely expensive to use semiconductors in all the breakers at this level. Could large fuses in parallel with a smaller mechanical breaker be used? Fuses normally break the current before the zero crossing so it should work equally well with DC. Smaller fuses at for example 36 kV and 20 kA breaking current has a mass of about 2 kg, giving about 3 kg per gigawatt short circuit power, would it scale so that a fuse for 2000 kV and 100 kA would have a mass around 0.003*200 000 kg= 600 kg? (It seems small) Could it be done with explosives for fast and synchronized operation as in some MV Is-limiters? How can the breaking capacity of a given breaker chamber be approximated without complicated fluid dynamics and FEM-simulations? Gr8xoz (talk) 20:32, 20 May 2012 (UTC)[reply]

By "of curse", do you mean "of course" ? StuRat (talk) 20:43, 20 May 2012 (UTC) [reply]

Yes, corrected that. Gr8xoz (talk) 20:54, 20 May 2012 (UTC) [reply]

ABB Group, Siemens and Alstom likely have the information you seek, though they may consider it proprietary. Do you really want to be replacing extremely expensivbe fuses whenever lightning hits a line, as opposed to a breaker opening and reclosing? There are fuses with explosive charges to open the circuit and programmable sensors and actuators to provide any desired operating curve, for a price. If you want to de-energize a HVDC line, could you do it via the "valve hall" or electronics which are transforming AC to DC at the sending end? Limiting fault current is inherently easier with HVDC than with HVAC. Edison (talk) 03:10, 21 May 2012 (UTC)[reply]

Yes someone in the suggested companies has probably looked in to this. It is well outside their current product lines. The most powerful HVDC-liks built today are around 7 GW, this are 100 GW, almost twice the maximum consumption in the UK. For point to point HVDC-links the converter station ("valve hall") or HVAC-breaker on the AC side can be used to interrupt the fault current. The need for HVDC-breakers comes when building large redundant HVDC-grids with many converter stations. You do not want to shutdown the whole world wide HVDC-grid in this example as soon as there is a fault anywhere. Why do you think it is easier to limit the fault current in HVDC-systems? That is not the impression I got.

If a fully reusable low maintenance breakers can be made at a reasonable price, that is of course better but I get the impression that some consider it almost impossible. That is the reason ABB introduces semiconductor based solutions. I think one time controlled fuses could be competitive if they can be made at 10 % of the price of a reusable breaker. I think it would be possible to repair used fuses.

Lighting is not a likely source of faults as the system I think of are built with Gas Insulated lines. Gr8xoz (talk) 06:52, 21 May 2012 (UTC)[reply]

Planetary Resources, Inc.

This may belong on the Language help desk but I think I get better answers here. Planetary Resources, Inc. are a company that plans to mine asteroids. (NEOs) Are there any logic behind the naming? To my it sounds like they are one of very few mining companies that plan to use non planetary resources. Almost all other use resources on the planet Earth while the NEOs are not planets. The IAU states: "the term 'minor planet' may still be used, but generally the term 'Small Solar System body' will be preferred." Minor planet. So are they basing their name on the obsolete term "minor planet" or are there any other rationale. Gr8xoz (talk) 20:51, 20 May 2012 (UTC)[reply]

Planetoid is another name for a minor planet. But perhaps they mean they will provide resources to a planet (Earth), rather than from one. StuRat (talk) 20:56, 20 May 2012 (UTC)[reply]

(ec) I doubt if anybody can give you an authoritative answer, but my take on it is that "planetary" is a word that to most people has nothing to do with the Earth, but just conjures up "space". --ColinFine (talk) 20:58, 20 May 2012 (UTC)[reply]

electric car batteries

If there were a way to charge a battery while driving without hydrocarbons, how many kwh would be needed to supply the electric motor with the needed energy? — Preceding unsigned comment added by 70.162.248.221 (talk) 21:12, 20 May 2012 (UTC)[reply]

Well, there have been solar powered cars, but that requires an extremely light car (dangerous in an accident) completely covered with solar cells, and a sunny day with the Sun high in the sky. A more practical solar-powered car might be able to charge up while parked in the Sun (assuming it spends at least 90% of the time parked), with a range slightly increased by driving in sunlight. I assume the reason no manufacturer offers such a car is that a solar-cell covered car is rather ugly. StuRat (talk) 21:17, 20 May 2012 (UTC)[reply]

Then, of course, there's regenerative braking. This only charges the car battery while braking, so the net effect is to reduce the overall rate at which it discharges. StuRat (talk) 21:25, 20 May 2012 (UTC)[reply]

The Chevy Volt has a 111 kW motor [20]. So, if you want to be able to drive it continuously, you would need to provide that much. If you wanted to be able to drive it for 10 hours using the batteries, they would need to hold 1110 kWh. Of course, these both assume maximum power consumption. If you are driving more conservatively, you can get by with less. StuRat (talk) 21:37, 20 May 2012 (UTC)[reply]

Depending on a loot of factors but generaly around 20 kW in 100 km/h :Electric_car#Running_costs_and_maintenance

Are these "proper" moobs?

On the gynecomastia article, the representative and first image is File:Gynecomastia_001.jpg.

Without trying to show false sensitivity - are we sure that's not just a fatty? Is there proper diagnosis that that image is actually Gynecomastia? Obviously a doctor or medical specialist cannot diagnose over the internet but does it "look right"? I've actually started a very similar discussion on the representative talk page but makes sense to come here too. Egg Centri c 21:47, 20 May 2012 (UTC)[reply]

It seems to me that there is no sharp line. All men have breasts, and those on the overweight are larger, as are those with men suffering from certain problems, like hormone imbalances. I wouldn't expect them to look much different. I suppose a thin man with large breasts might be a better example, though. StuRat (talk) 22:00, 20 May 2012 (UTC)[reply]

Is a breast that is large purely cause the guy is fat, and which will shrink to normal if he loses weight, actually gynecomastia though? Indeed, I don't think any of the images on that page are very great as displaying tits - something like this is more like it...Egg Centri c 22:10, 20 May 2012 (UTC)[reply]

On a mere fatty (pseudogynecomastia), I'd expect more continuity between the upper slope of the "breast" and the surface of the flesh above it. —Tamfang (talk) 22:50, 20 May 2012 (UTC)[reply]

Wood as an electric conductor

In Jurassic Park, the Dr. Alan Grant Sam Neill is walking with John Hammond's Richard Attenborough grandkids and comes upon an electric fence that appears to have been deactivated. To check it, Grant throws a stick at the fence. If wood doesn't conduct electricity, what good would this do? DRosenbach ^{(Talk | Contribs)} 22:52, 20 May 2012 (UTC)[reply]

I've forgotten where this is in the narrative but wasn't it raining just before this? A wet stick would possibly spark if the voltage were high enough. Dismas|^(talk) 23:06, 20 May 2012 (UTC)[reply]

It makes for good cinematography. It was just a film. Would an american audience comprehend SIDE?--Aspro (talk) 23:10, 20 May 2012 (UTC)[reply]

(edit conflict) x2 Wet wood is a good conductor of electricity, and even dry wood conducts to a certain extent, especially at the high voltage (but low current) generated by most electric fences. Allowing the wood to rest against a modern animal electric fence would short the current to earth. I'm not sure whether the fence in Jurassic Park was designed to kill. If so, then it might have been both high voltage and high current, producing an electric arc when shorted (or perhaps that's what the director wanted viewers to think.) Dbfirs 23:14, 20 May 2012 (UTC)[reply]

Think I might have to explain that. SIDE =Switch off, Isolate, Dump & Earth. Until then then an electric fence is not to be trusted. It was just a film for the Hoi polloi -not a science lesson... Dinosaurs... I ask you! It was on par with the utter hogwash that came out of seaQuest DSV.--Aspro (talk) 23:30, 20 May 2012 (UTC)[reply]

Also, Some EF's pulse -such as cow fences. So the stick might might just hit between energising and thus show no observable effect.--Aspro (talk) 23:38, 20 May 2012 (UTC)[reply]

A random piece of wood cannot be counted on as either a good conductor or a good insulator. Oven dried wood has resistivity of about 10E14 to 10E16 ohm-meter , while damp wood has resistivity of 10E3 to 10E4 ohm-meter. The conductivity increases with temperature. If an electric fence in Jurassic Park were built to kill dinosaurs who touched it, it might well be energized at 480 or even 2400 volts, with a high current, low impedance source, in contrast to the high voltage, low current pulses of a typical farm electric fence. A piece of tree limb wood connected from the high voltage J park fence to ground (such as the nonenergized metal of the fence) might blacken, then catch fire. It might or might not pop the breaker. When a tree limb touches 2 utility conductors at 4kv or higher, the current may be sufficient to cause a fuse or breaker to open the circuit. A carefully dried wooden stick might not conduct as much current. Edison (talk) 02:54, 21 May 2012 (UTC)[reply]

"SIDE =Switch off, Isolate, Dump & Earth." Thanks for the explanation of the acronym. I think I understand S, I, and E. But what is "Dump"? Wanderer57 (talk) 04:32, 21 May 2012 (UTC)[reply]

D = Dump = Discharge to earth. 84.209.89.214 (talk) 23:32, 21 May 2012 (UTC)[reply]

Asthma meds for otherwise healthy people

What would puffing on an inhaler do for someone who does not have asthma? Would it basically be the list of adverse side effects listed at the Salbutamol page?

Note: I don't have asthma, don't have anyone in my immediate circle of friends/family with it and therefore don't have easy access to an inhaler. I don't plan on finding one to experiment with no matter what the answers I get here are. I have not died and come back to see if maybe an inhaler caused my death. I'm not asking for medical advice. I'm simply asking out of curiosity. Dismas|^(talk) 23:27, 20 May 2012 (UTC)[reply]

The first effect of overdoses might be hands trembling. However, puffing only once won't have any noticeable effect for sure. It's quite difficult to overdose on the inhaler (which is different from taking a pill or taking salbutamol intravenously). OsmanRF34 (talk) 00:10, 21 May 2012 (UTC)[reply]

I've had a few puffs of friend's inhalers in the past, just for a laugh and to see what would happen, I noticed no effect. Vespine (talk) 04:44, 21 May 2012 (UTC)[reply]

Albuterol is often prescribed to people who do not have asthma. (I have had it prescribed to me a number of times to just clear up bronchial weirdness and phlegm that occasionally trails behind long after the cold has gone.) In my experience (just anecdotal), a single deep pull of albuterol (expel all air from lungs, inhale a puff and suck it as deep into lungs as possible) usually makes one feel a little jittery and makes the lungs feels a little "funny." (But it does seem to work wonders at "clearing them out".) Which are some but definitely not all of the adverse effects listed there. I would expect those effects to be there whether you are "otherwise healthy" or not; they are the effect of inhaling that kind of drug into your lungs, not an interaction produced between the drug and your injury/disease. I know, not super helpful, but I thought I'd offer it up. Not medical advice, etc. etc., all of what I am describing was prescribed to me personally by a doctor, please do not replicate without a doctor, etc. --Mr.98 (talk) 13:12, 21 May 2012 (UTC)[reply]

Some asthma meds are widely used as quasi-legal (with a prescription) doping for endurance athletes. I think 40% of the Tour de France participants are certified asthmatics, and up to 80% of UCI professional cyclists share this fate [21]. --Stephan Schulz (talk) 21:24, 21 May 2012 (UTC)[reply]

May 21

Oxidation state trend

Why is it that the vertical trend of stable oxidation states varies so much horizontally across the periodic table? Why is that carbon(IV) is more stable than lead(IV), but osmium(VI) is more stable than iron(VI)? Is there an underlying trend, or is it completely unpredictable? Plasmic Physics (talk) 00:50, 21 May 2012 (UTC)[reply]

These particular cases are due to relativistic effects. Normally, oxidation states increase going down a group because of the progressively looser hold by the atoms on their valence electrons. However, relativistic effects may counter-act that in the cases of Tl, Pb, Bi, and 7th-period elements. They reduce the effect of lower effective nuclear charge by stabilizing particular electron configurations more than in lighter members of the same group.--Jasper Deng (talk) 01:10, 21 May 2012 (UTC)[reply]

I would expect that the stability for high oxidation states should increase down the periods for every group. What is so special about the electron configurations for those cases? Plasmic Physics (talk) 02:28, 21 May 2012 (UTC)[reply]

It seems that the s orbitals for Tl, Pb, Bi are mostly inert, and removing their electrons has less of a stabilizing effect. I don't understand the core reason - since I don't understand relativistic effects. But what is clear is that as the electrons are further and further from the nucleus, the nucleus cares less and less about a full octet than about keeping all those electrons, it appears.--Jasper Deng (talk) 02:53, 21 May 2012 (UTC)[reply]

Could obesity be caused by a lack of magnesium in the diet?

Let me explain why I think magnesium could be a factor in obesity. First, the obvious explanation, i.e. too much calorie intake is, I think, not so plausible, because obese people actually don't eat that much calorie-wise, at least not the obese people who I know. Also, most people will be at some constant weight, so they are in dynamical equilibrium between calorie intake and calorie use. Then that dynamical equilibrium could in theory be reached at any weight, there is no good reason why at an intake of say, 3000 Kcal/day you have to weigh 100 kg, and not 70 kg or 150 kg. Most of the calories are burned by muscles, the fat tissue doesn't use a lot of energy, so you could just as well be at the same equilibrium of 3000 Kcal energy intake and energy use, but at a much lower body weight.

In fact, I weigh only 60 kg, yet I eat on average 3600 Kcal per day. Some of my obese family members eat way less than I do, but they are also eating a lot less healthy foods. Now, some time ago I posted here about magnesium, when I checked my diet I found that my magnesium intake was way too high (I get about 1 gram of magnesium per day from eating whole grain bread, potatoes, brown rice, whole grain pasta, and bananas). However, since then I've read that my magnesium intake may be normal, and that most people are actually magnesium deficient. Now, magnesium plays an important role in metabolism, so I thought that perhaps one can explain why some people are obese as follows.

If I eat X calories a day then I could gain or lose weight until I reach dynamical equilibrium where I burn X calories per day. If my food does not contain enough magnesium, then the metabolism becomes less efficient, it would take a cell longer to burn energy, so for my body to burn X calories per day would take a larger store of energy in my body, therefore I would be a lot fatter. Count Iblis (talk) 03:10, 21 May 2012 (UTC)[reply]

It's always possible, but magnesium is a rather basic nutrient, and if a deficiency caused obesity, I rather think they'd have noticed it by now. I suspect it's a different aspect of their unhealthy diets that are to blame. Also note that it's not always easy to tell how much someone eats by casual observation. I knew an obese woman who seemed to eat modestly at each meal. However, if I went back later for, say, a 2nd piece of pie as a snack, I'd discover that the entire pie had somehow disappeared. :-) StuRat (talk) 03:19, 21 May 2012 (UTC)[reply]

A confounding factor: Most obese people I know may not eat that much more, but what they do that's different, is do a lot less physical work. For example, for quite a while I worked in a company that occupied two floors of a building, and the way the company laid out their operations, and the type of work done, meant that most empoyees had to frequently walk from one floor to the other. Most of us took the stairs as that was quicker. But, with no exceptions, obese folks used the lift every time. Also you could use free public carparks about 6 block away, or you could park in the basement at $20 per day. Guess which folks paid the $20/day for their own cars! We ran construction projects. I've never seen an obese labouror, and seldom seen a fat one. Ratbone120.145.172.175 (talk) 03:44, 21 May 2012 (UTC)[reply]

You don't live in the UK do you! I've seen plenty of obese brickies here in my time... --TammyMoet (talk) 09:50, 21 May 2012 (UTC)[reply]

No, I live in Australia. Brickies are paid per number of bricks laid and use labourors/assistants to keep them constantly supplied with bricks and fresh mortar. So the faster they go, the more money they make. Are they paid by the hour, or at a flat rate, in the UK? Bricklaying is damm hard work. Ratbone60.230.230.72 (talk) 10:57, 21 May 2012 (UTC)[reply]

Is it even possible to avoid ingesting magnesium to the extent that one becomes deficient? Roger (talk) 09:55, 21 May 2012 (UTC)[reply]

Yes, actually. 57% of the US populace has inadequate magnesium intake according to the USDA. Also, certain medications affect magnesium absorption: transplant patients are often put on supplements because ciclosporin (an important anti-rejection drug) impedes magnesium absorption. --NellieBlyMobile (talk) 22:41, 21 May 2012 (UTC)[reply]

Cartography - resolution?

Is there a concept in cartography that is analogous to resolution? By this I mean, will a particular map have a property that, say, features smaller that 10m2 are ignored? I just want to know what this is called so that I can read about it. ike9898 (talk) 14:04, 21 May 2012 (UTC)[reply]

You may find the Coastline paradox article interesting. hydnjo (talk) 14:22, 21 May 2012 (UTC)[reply]

steam engine plans

I want to make a working 1/12 scale model of this: http://upload.wikimedia.org/wikipedia/commons/f/f3/TrevithicksEngine.jpg or something much like it, any ideas where I can find plans of how it's made?

Kitutal (talk) 16:12, 21 May 2012 (UTC)[reply]

10 seconds of google revealed this thread http://www.model-engineer.co.uk/forums/postings.asp?th=47132 which might be of use. There are also companies that sell live steam models. --TrogWoolley (talk) 16:27, 21 May 2012 (UTC)[reply]

Purple lightning

I was watching this video, http://www.youtube.com/watch?v=RDDfkKEa2ls&feature=related

And some of the lightning looks quite purple. What causes that? Is it just the camera, or does it really look purple? ScienceApe (talk) 16:27, 21 May 2012 (UTC)[reply]

If the camera is using a whitebalance for fluorescent lighting, which is greenish, then other real white stuff like lightning can look purplish. Graeme Bartlett (talk) 21:47, 21 May 2012 (UTC)[reply]

Comparing climate between places in Australia & North America

Is there somewhere in North America that would have a very similar climate to that of Sydney, Australia for example? Or Perth, Australia?

More generally, are there tables that answer such questions?

Eg, is the climate of Portugal very similar to that of Northern California (both being on the west coast of a continent and at the same latitude)? Thanks, CBHA (talk) 19:43, 21 May 2012 (UTC)[reply]

Well, there's the Köppen climate classification system. Per that map, Portugal and coastal California are both Mediterranean climates (Csa or Csb on the Koppen map). Sydney appears to be in an oceanic climate region, of which there is little in North America (perhaps around Seattle or Vancouver) but loads in Europe. It may also be useful to pull up single-factor maps such as average temperature or rainfall. — Lomn 21:03, 21 May 2012 (UTC)[reply]

Diversity of sexuality

I've sometimes heard people claim that the diverse sexual tastes of humans is a feature that distinguishes us from all the lower animals. However, I'm heavily skeptical of any claim that separates humans from the natural world and declares "we are unique amongst all species". So, do other animals have unusual sexual interests? Are there chimps, for example, who enjoy sadism, necrophilia, pedophilia, or foot fetishes (assuming these aren't typical amongst the species)? --140.180.5.169 (talk) 19:57, 21 May 2012 (UTC)[reply]

The concept of sexuality is something which, in itself, is a human-created idea. To assign sexuality to an animal is to disrepect it by anthropomorphizing it. Animal behavior is to be understood on its own terms, not by analogy to human behavior. Animals are not incomplete people, and we cannot hope to understand them properly by starting with the premise that they are. Assuming that animal behavior has human analogues, or that the classifications of animal behavior fit in the same schema as we have created for human behavior is a wrong-headed tack. --Jayron32 23:01, 21 May 2012 (UTC)[reply]

(ec) I don't accept the validity of the artificial wall you're putting up between human and animal sexual behavior. While the word "sexuality" was invented by humans, sex itself predates the first multicellular organism and is several orders of magnitude older than all of human history. To say that human sexuality is unique in a profound way, instead of being just a variant of one of the oldest and most widespread biological phenomena, seems very anthropocentric and anti-Copernician. — Preceding unsigned comment added by 140.180.5.169 (talk) 23:31, 21 May 2012 (UTC)[reply]

Bonobos, dolphins, and chimpanzees are known to engage in sexual intercourse even when the female is not in estrus, and to engage in sex acts with same-sex partners. Like humans engaging in sex primarily for pleasure, this behaviour in the above mentioned animals is also presumed to be for pleasure, and a contributing factor to strengthening their social bonds. 84.209.89.214 (talk) 23:23, 21 May 2012 (UTC)[reply]

I don't think one can properly distinguish the motives for mating in animals. They mate because they feel the urge to, without realizing why. That they get pleasure from it does not mean its purpose is not reproduction. Rather, animals (and humans too) derive pleasure from sex precisely because sex leads to reproduction. Animals having the urge to mate even when there are no fertile opposite-sex partners available takes nothing away from the fact that the urges they feel have the purpose of reproducing. -Lindert (talk) 23:28, 21 May 2012 (UTC)[reply]

Yes, but they don't have complex language with a distinction between signified, signifier and referent. Chimpanzees enjoy injuring other chimpanzees, they don't experience sadism. Animals copulate with animals that lack reproductive maturity. Animals don't have "children" as a unique category of social being, and so are incapable of "pedophilia" as sex with a child. Animals do not have sexuality, they have sex and sexual behaviours. Animals lack culture, because they lack language. Animals lack sexual taste, because they lack language. Fifelfoo (talk) 23:33, 21 May 2012 (UTC)[reply]

Dogs hump legs HiLo48 (talk) 23:34, 21 May 2012 (UTC)[reply]

Odd notation in genetics (greater than symbol?)

In this paper (on royal jelly), specifically in the section "Royalactin changes Drosophila phenotypes via Egfr," there is some form of gene notation using greater than (>) symbols that I am not familiar with and cannot find a non-dense reference on google, or indeed really any useful information at all. I think it may be describing something to do with gene silencing but I'm not sure. Could anyone shed some light on this for me? Examples are:

P0206>dPI3K
P0206>dEpgrRNAi
ppl>dPI3KDN (I know the DN part means dominant-negative as thats noted in the text, but thats it)

Specifically, I'm wondering what the > and d stand for, and what relationship the > symbol denotes for the genes either side of it. Hoping someone who knows about genetics will recognise it. Thanks!! -Zynwyx (talk) 20:37, 21 May 2012 (UTC)[reply]

May 22

Rotating in space

If an electric motor was allowed to run while floating free in space, then presumably the armature would rotate one way and the casing the other, but at what relative speeds? I'm guessing there must be an equation relating the two rotational speeds, that somehow also involves the masses of the components. For example if the casing was very heavy compared to the armature then it it seems intuitive that it would rotate more slowly. But what exactly would that equation be? 86.177.105.243 (talk) 00:05, 22 May 2012 (UTC)[reply]

@@ Line 509: / Line 509: @@
 = May 22 =
+== Rotating in space ==
+If an electric motor was allowed to run while floating free in space, then presumably the armature would rotate one way and the casing the other, but at what relative speeds? I'm guessing there must be an equation relating the two rotational speeds, that somehow also involves the masses of the components. For example if the casing was very heavy compared to the armature then it it seems intuitive that it would rotate more slowly. But what exactly would that equation be? [[Special:Contributions/86.177.105.243|86.177.105.243]] ([[User talk:86.177.105.243|talk]]) 00:05, 22 May 2012 (UTC)