Wikipedia:Reference desk/Science

WP:RD/S C #eee #f5f5f5 #eee #aaa #aaa #aaa #00f #00f #000 #00f science Wikipedia:Reference desk/Science Science WP:RD/S

Welcome to the science reference desk. Want a faster answer? Main page: Help searching Wikipedia     How can I get my question answered? Select the section of the desk that best fits the general topic of your question (see the navigation column to the right). Post your question to only one section, providing a short header that gives the topic of your question. Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when. Don't post personal contact information – it will be removed. Any answers will be provided here. Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context. Note: We don't answer (and may remove) questions that require medical diagnosis or legal advice. We don't answer requests for opinions, predictions or debate. We don't do your homework for you, though we'll help you past the stuck point. We don't conduct original research or provide a free source of ideas, but we'll help you find information you need. Ready? Ask a new question! After reading the above, you may ask a new question by clicking here. How do I answer a question? Main page: Wikipedia:Reference desk/Guidelines The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

Choose a topic: Computing desk Science desk Mathematics desk Humanities desk Language desk Entertainment desk Miscellaneous desk Archives   See also: Help desk Village pump Help Manual   skip to bottom

November 12

Seriously hurt in the news

We frequently hear in the news that somebody has been "seriously hurt", or even worse "herido muy grave" or the worst "en estado crítico" (I don't know how you would say that in English, normally somebody who is in "estado crítico" dies soon). What do those expressions mean in medicine? --Taraborn 00:35, 12 November 2007 (UTC)

See Medical conditions. These labels are used by hospitals when speaking to the press and are not used among doctors. They vary in meaning from hospital to hospital. --Milkbreath 00:49, 12 November 2007 (UTC)

Proving once again that WHAAOE. (But, dang, Milkbreath, ya beat me to it. *I* was just about to post that link!) —Steve Summit (talk) 00:56, 12 November 2007 (UTC)

Thank you, guys. --Taraborn 01:07, 12 November 2007 (UTC)

I think what you call "en estado critico" is, in english, "in critical condition". Just an FYI. Kuronue | Talk 21:06, 13 November 2007 (UTC)

Jupiter's moon Io

Is there a certain reason Io has volcanoes? —Preceding unsigned comment added by 72.38.227.206 (talk) 00:35, 12 November 2007 (UTC)

Well, I guess the answer is only really stated clearly in the very first paragraph of Io...you couldn't be expected to find something buried that deeply! SteveBaker 01:36, 12 November 2007 (UTC)

tidal stress--( Mulligan's Wake)(t) 01:37, 12 November 2007 (UTC)

I think that prior to IO being surveyed by the Vikings during the 70s, it was just assumed that small remote planets and satellites would all be frozen solid. The incredible phenomenon of IO suggests there could be life-friendly worlds far away from their host stars. Not IO, however. It is estimated that there the entire surface is continually subducting from the heat of tidal friction and is completely renewed every hundred years or so. Myles325a 03:49, 14 November 2007 (UTC)

Jeez! Those Vikings discovered everything! -SandyJax 16:49, 14 November 2007 (UTC)

stpm trial exam question

can ketone react with chlorine?thank you —Preceding unsigned comment added by 218.208.91.249 (talk) 02:04, 12 November 2007 (UTC)

Perhaps haloform reaction may be of some help. (EhJJ) 03:11, 12 November 2007 (UTC)

"brush harder, not smarter"

my dentist told me this. what does he mean? —Preceding unsigned comment added by 172.130.26.74 (talk) 02:08, 12 November 2007 (UTC)

Why don't you ask him?

The usual exhortation is to do something smarter, not harder -- that is, to do it with more care and deliberation, and less brute force. —Steve Summit (talk) 02:16, 12 November 2007 (UTC)

As Steve has pointed out, the usual phrase is "do it smarter, not harder". In this case, your dentist may be using a play on words to imply that you are not brushing hard enough. If you're not sure, you may want to ask him. (EhJJ) 03:14, 12 November 2007 (UTC)

You might be remembering it backwards, or your dentist might have said it backwards accidentally, or your dentist might have said it backwards intentionally. There’s no way we can know which is the case; you’ll really need to ask your dentist. MrRedact 07:50, 12 November 2007 (UTC)

Your dentist wants you to brush gently (not harder), and with a technique (smarter) that gently cleans effectively. If you use a hard brush and scrub, you'll damage your teeth and irritate your gums. If you gently massage your gums with a soft-bristle brush, and hold the brush at the right angles, you will prevent gum disease. --Mdwyer 20:40, 13 November 2007 (UTC)

It means that either you or your dentist are dyslectic. Myles325a 03:51, 14 November 2007 (UTC)

Bamboo nomenclature

72.50.180.145 03:54, 12 November 2007 (UTC)After reviewing your article on bamboo, I'm not sure what you call stalks of bamboo. Are they called stems? I've heard them called that before. Are they called branches? Are they called culms? which according to wikipedia- "originally referred to a stem of any type of plant". What is the appropriate scientific way to name them, because I'm confused.

A bamboo's shoot is like a tree's trunk, whereas a bamboo's stem is like a tree's branch. A culm is a stem of a monocotyledon, which bamboo is an example of. Someguy1221 05:06, 12 November 2007 (UTC)

Laxative agent in apple juice that's not in pear juice?

Here's the strange thing: I get diarrhea when drinking apple juice. Even a single glass causes discomfort, and multiple are certain to cause diarrhea (which passes quickly, however). I thought this could be oversensitivity to sorbitol, but I have no problems drinking even large amounts of pear juice, which likewise contains sorbitol. Ditto for any other fruit juice you'd care to name (though I haven't tried prune juice). I have no problem with raw apples either. It could be an additive of concentrated apple juice (I haven't tried "natural" apple juice either), but I wouldn't know which one.

Though apple juice really is the only thing giving me problems (so it's easy enough to avoid), I'm curious if anyone could guess what could cause diarrhea and is just about only found in (concentrated) apple juice -- a rare chemical or ratio of something? 82.95.254.249 07:57, 12 November 2007 (UTC)

Apples have traditionally "gone through" some people, especially if the apples are the first of the season. I suspect it may be malic acid that causes this, but I stress this is pure speculation. DuncanHill 14:24, 12 November 2007 (UTC)

Well, any hint is welcome, but I don't think it's malic acid. Tart apples are my favorite, actually, and I have no problem with other foodstuffs that contain malic acid (according to the page). 82.95.254.249 20:10, 12 November 2007 (UTC)

I'd take another look at Sorbitol. It has known laxitive effects (many sugars do). As for the difference between pear and apple juice, I can only speculate. Apples and pears are VERY similar. It is possible that the pear juice has been modified in some way? Perhaps it was pasteurized or cooked in some way that would have converted its sugars? --Mdwyer 20:37, 13 November 2007 (UTC)

I get a little stomacheache from eating an apple core, and also from fresh apple juice made from whole apples. I deduce that the core or seeds has something in it that disagrees with me but not the majority of people. I am vaguely aware that apple leaves are supposed to contain some poison also.Polypipe Wrangler 22:10, 13 November 2007 (UTC)

Apple seeds do contain a small amount of a cyanide compound. Snopes has the dirt on it. --Mdwyer 03:44, 14 November 2007 (UTC)

Pseudo-stable isotopes

For isotopes with half-lives longer than the known lifespan of the universe, such as Calcium-48, how do they determine that the isotopes aren't stable? --67.185.172.158 08:06, 12 November 2007 (UTC)

Simple: get a lot if atoms of that isotope and wait for them to decay. Remember that half-life is a statistical property: a half-life of X doesn't mean that no atoms will decay until X has passed. An unstable atom can decay at any time (which is exactly why we talk of "half-life" and not the full life).

The initial suspicion that an isotope is unstable comes from theoretical considerations, and obviously it's hard to determine the half-life for an extremely long-lived isotope with great accuracy. 212.178.108.2 10:55, 12 November 2007 (UTC)

In that case, the article cited in footnote 2 explains it. They thought it was a likely candidate (via theory) for a very rare form of decay, and then spent a lot of time watching it in a test chamber (via experiment) to see if it actually did decay in that way. And it did. Hooray. --24.147.86.187 17:00, 12 November 2007 (UTC)

A radiation counter can pick up a single atomic decay (so long as you shield it from background radiation) - if you have a mole of whatever isotope you have then you have 6x10²³ atoms. A mole is the atomic weight expressed in grams - so a mole of a solid or liquid is a pretty reasonable quantity to measure. The age of the universe is something like 4x10¹⁷ seconds - and a mole is 6x10²³ atoms - so in half the age of the universe, you'd expect 3x10²³ radioactive decay events - so you'd expect about half a million decay events per second. Easily detectable! More to the point of course, if this is a synthetic isotope (made in a nuclear reactor for example) so that it's "fresh", it'll be radiating a lot more than it will be when it's reached it's half-life. SteveBaker 18:51, 12 November 2007 (UTC)

For "a lot more" read "twice as much".

Calcium 48 makes up only 0.187% (1/535) of the atoms in natural calcium, so making a whole mole of the stuff (about 48 grams) would be rather expensive. And then, its half life is not a mere 4x10^17 seconds, but about 100 times that. But all this just means that the number of decay events you'd have to detect is smaller, not that there are too few to detect at all. No doubt the paper by Balysh et al., cited in the article, would have details of how large a sample was used and how long it was observed. --Anon, 23:26 UTC, November 13, 2007.

Penguins in the Arctic? Polar bears in the Antarctic?

Just checked the articles on Penguins and Polar Bears. Just wanted some confirmation, are there any penguins that live in the Arctic? Also are there any Polar bears in the Antarctic? 64.236.121.129 14:20, 12 November 2007 (UTC)

No, and no. DuncanHill 14:22, 12 November 2007 (UTC)

But they could live there, couldn't they? I mean, if you kidnapped some and dumped them there, they'd manage ok, wouldn't they (not that I advocate kidnapping well dressed animals or anything)? Jeffpw 14:44, 12 November 2007 (UTC)

No - Penguins are almost 100% confined to the southern hemisphere (they do creep just above the equator - but not by much) - but not all species live in snow and ice - some are very happy in the tropics. Polar bears are strictly Arctic. Whether you could transplant them depends on an awful lot of things. Bears live on the mainland as well as in the arctic ocean - if you dumped them in the antarctic, they'd have no way to make it do warmer areas in the winter - also the species they are used to hunting would not be present. Some penguins have these very complicated behaviors where one sex gathers together far inland to look after the eggs while the other partner goes off for food. It's not clear how this would work in a continuously moving sea of ice floes - do penguins navigate somehow? I'd be quite surprised if relocating either species would actually work out well. SteveBaker 15:03, 12 November 2007 (UTC)

• The polar bears would probably eat the penguins. Seriously. The kind of penguins that live in Antarctica generally are unable to recognize any land animal as a predator, because the only land animals that have ever attacked them in the last million years or so have been humans or escaped domesticated animals like dogs. If you took a polar bear to the Antarctic, the polar bears would very quickly gravitate towards this huge, meaty, docile food source. If you were to transplant some penguins to the Arctic, then even if they could solve the problem of figuring out which landmass to lay their eggs on, eventually the polar bears (which will eat any kind of animal if they're hungry enough) will figure out that they're easier to catch than their usual prey, and kill them off in short order. --M@rēino 18:54, 12 November 2007 (UTC)

Well, the penguins are prey to Leopard seals and killer whales - so they have seen large predators - just not big, furry white ones. It's hard to know what their reaction might be. Certainly they are generally unafraid of humans. (True story: When my kid was about 5 years old, we went to SeaWorld, saw the Shamu show and then visited the Penguin enclosure for a behind-the-scenes tour. My son (who happened to know a lot about penguins) picked the moment with the most other people around to proudly (and loudly!) announce that "Shamu is a killer whale - he likes to eat penguins!" - the penguin-keepers were torn between their roles as "teaching kids about nature" and "being good corporate employees" - so simultaneously, one of them said "Yes, that's right!" and the other one said "No, Shamu is friendly with EVERYONE!" - then each of them realised their horrible blunder and changed their story - it took them quite a while to decide what story line was the best and lamely trot it out. It was hilarious!) SteveBaker 21:06, 12 November 2007 (UTC)

I will gladly concede that penguins face aquatic predators (and avian predators) -- just not land-based predators. --M@rēino 21:25, 12 November 2007 (UTC)

Wrinkled face

Is there an illness that makes the skin wrinkle excessively? Keria 17:04, 12 November 2007 (UTC)

There is evidence that smoking can do that to you. SteveBaker 18:39, 12 November 2007 (UTC)

I'll point out that our article on Wrinkles sucks, so this question might require some real digging. --M@rēino 18:57, 12 November 2007 (UTC)

As far as I know, wrinkling of the skin is caused by old age, however certain skin disorders may contribute to wrinkles too. Topical Corticosteroids as often used by Eczema sufferers are notorious for thinning the skin and this can definitely cause premature wrinkles. GaryReggae 22:41, 12 November 2007 (UTC)

As the piccie in Wrinkle article demonstrates, masturbation makes wrinkles a lot, lot worse. Myles325a 22:51, 12 November 2007 (UTC)

We must have a few doctors on the science desk, a dermatologist who would like to add a "disease" section on the article wrinkle? Keria 02:55, 13 November 2007 (UTC)

There is a wrinkly skin syndrome, but the description seems to indicate that the wrinkly skin is not on the face. --JWSchmidt 03:55, 13 November 2007 (UTC)

medical ebooks

i am a 2nd year mbbs student of osmania medical college.please give me information about sites to download medical ebooks,so that i can clear evrything. —Preceding unsigned comment added by 117.97.30.240 (talk) 17:13, 12 November 2007 (UTC)

You could try our sister site [Wikibooks] although I don't know if there is anything medical there. Chances are if it's anything too specialist, the best option (apart from spending extortionate amounts of money on textbooks) is a visit to your local library. GaryReggae 22:39, 12 November 2007 (UTC)

Try www.freebooks4doctors.com. - Cybergoth 03:16, 16 November 2007 (UTC)

how does our organism recognize foreign tissue?

How can our organism distinguish between our own tissue and foreign ones? Of course, the DNA is different, but how our immune system can know it? —Preceding unsigned comment added by 80.58.205.37 (talk) 18:28, 12 November 2007 (UTC)

Antigens on blood cells.

Mainly due to Antigens. These are molecules on the outside of cells, and they are distinctive to each type of cell. For example, there are 2 two types of blood antigen (A and B), so someone with blood type A has "A" antigens on their blood cells. If you give these to someone else whose blood is also A (or AB), the body accepts it, while if you give it to someone of blood type B, the body can't recognise the blood antigens, so it attacks the blood, which causes all kinds of nasty effects inside the body. Laïka 18:50, 12 November 2007 (UTC)

See also, here: "To be able to destroy invaders, the immune system must first recognize them. That is, the immune system must be able to distinguish what is nonself (foreign) from what is self. The immune system can make this distinction because all cells have identification molecules on their surface. Microorganisms are recognized because they have unique, foreign identification molecules on their surface. In people, identification molecules are called human leukocyte antigens (HLA), or the major histocompatibility complex (MHC). HLA molecules are called antigens because they can provoke an immune response in another person (normally, they do not provoke an immune response in the person who has them). Each person has unique human leukocyte antigens. A cell with molecules on its surface that are not identical to those on the body's own cells is identified as being foreign. The immune system then attacks that cell. Such a cell may be a microorganism, a cell from transplanted tissue, or one of the body's cells that has been infected by an invading microorganism." Rockpocket 18:52, 12 November 2007 (UTC)

Potassium chloride and seeds

Is potassium chloride a fertilizer like potash, or a toxin like chlorine? How it will affect the growth of seeds? 199.89.180.65 18:37, 12 November 2007 (UTC)

• The majority of the potassium chloride produced is used for making fertilizer. Just remember that it might kill your livestock if they eat a bag of it. --M@rēino 19:00, 12 November 2007 (UTC)

Mobile phone buzzing

What's the buzzing/beeping that speakers make just before a mobile phone rings? I can't imagine the phone is making an electric/magnetic field strong enough to affect speakers from 3+ metres away. And if it is, is it safe to keep it near my computer? Laïka 18:45, 12 November 2007 (UTC)

Don't worry about it. My cellphone does that to the earbuds on my MP3 player and my FM clock-radio - it's simply the phone acknowledging the cell tower prior to ringing the ringer. The radio signal from the phone is plenty strong to induce a current in a wire of a suitable length to act as an antenna. It should be OK to keep it by your computer - the voltage is enough to hear on a speaker because they are analog devices - also loudspeakers and headphones have long wires leading to them that make great antennas. But no, it won't affect your computer (except that you might hear it on the computer's speakers). SteveBaker 20:55, 12 November 2007 (UTC)

Yes, this is very common although some speakers appear to be sensitive to this and others not - depending on whether the cable is shielded - otherwise the cable will act as an aerial and pick up the electromagnetic waves! It won't do any harm though, other than the obvious interuptions to your audio. GaryReggae 22:37, 12 November 2007 (UTC)

It occurs to me that there is an easier way to explain this convincingly. I should point out that inside the phone is a small computer - you know it survives the most intense radio signals the phone can put out without screwing up. What's more, radio signals decrease in strength as the square of the range. The computer in the phone is at most maybe 2cms from the radio transmitter. So if you put your phone maybe 20cms from your desktop computer, the signal strength will be 100 times less. If the computer in the phone can take the full strength, then for sure your desktop machine will be OK with just 1% as much radio interference. SteveBaker 23:42, 12 November 2007 (UTC)

Also, a desktop computer usually has a metal case (with few holes), so the electromagnetic waves most probably can't get in it. This doesn't apply to notebook computers of course. – b_jonas 11:21, 13 November 2007 (UTC)

Note that this only happens with TDMA (especially current-generation GSM) phones. You aren't hearing the signal itself, so much as the signal being turned on and off something like 50 times per second. This induces currents in some analog circuits -- especially inexpensive amplifiers that lack correct shielding. --Mdwyer 20:34, 13 November 2007 (UTC)

A designer wishes to calculate the thickness of a steel cable for suspending a passenger lift

The total length of cable supporting the lift when it is at the ground floor is 8 m. The mass of the lift when full of passengers is 950 kg. The designer has decided to incorporate a safety factor of 10 into the lift cable: i.e., the cable must be able to withstand 10 times the load it will actually be exposed to in service, before it fails. The steel selected for the cable has a failure stress of 900 MN m– 2 . (a) Using this information, calculate the required diameter of the cable. Show all your working. Assume that the cable is a single piece of steel, with a circular cross-section. (Ignore any effect of the weight of the cable in your calculation.) The downward force F on the cable is calculated by multiplying the total mass m (expressed in kg) by g, the acceleration due to gravity: F = m × g Take the value of g to be 10 m s–2. (In reality there is extra force needed to accelerate the lift upwards, but as this is relatively small, there is no need to consider it here.) Hint: the safety factor means that the cable will fail – the stress will reach its failure stress – when it is loaded to 10 times the intended design load. Use this to calculate the cross-sectional area of the cable, and so its diameter. I found the diameter to be between 11 and 12 mm —Preceding unsigned comment added by Flano1 (talk • contribs) 19:09, 12 November 2007 (UTC)

So do I. 169.230.94.28 19:23, 12 November 2007 (UTC)

This does sound a bit like a homework question to me ("Show your working" is always a dead giveaway!) - please look at the top of this page where you'll see it's our policy not to help people with homework. Sorry. If there is any specific part of the math/physics you can't understand, then feel free to ask specifically about that problem and we'll be more than happy to assist. SteveBaker 20:50, 12 November 2007 (UTC)

It looks like you have all the information you need here to calculate the answer. The most obvious thing to do is ignore the factor of ten for the moment, work out the diameter of cable needed just to support the normal load then multiply it by ten. Our page on Elevators may be of use. GaryReggae 22:34, 12 November 2007 (UTC)

Erm...cough...no, no, NO! Cable strength is proportional to cross-sectional area - not diameter. To get a factor of 10 safety factor, you only need to make the cable sqrt(10) times thicker. About 3.16 times thicker should be fine. SteveBaker 00:01, 13 November 2007 (UTC)

Using the fracture strength of the cable, instead of the yield strength, sounds like a bad idea as well, particularly for applications involving lifting humans... Titoxd^{(?!? - cool stuff)} 00:13, 13 November 2007 (UTC)

Percent solutions

While interviewing undergraduates to be a lab helper, our lab manager was bemoaning how every single candidate could not explain how to make a molar solution. She then asked one how to make a 1% percent agarose solution [weight/volume] and the hapless student suggested filling a measuring cylinder to 1ml with agarose then filling it up to 100ml with water. After we were done laughing at his expense, I noted that doesn't bode well for the future of American science, as it should be simple to work that one out by logic. But this got me thinking that might not actually be the case: a percentage solution [w/v] is considered to be 1g in 100ml. But logically, it should probably be 1mg in 100ml. Are the w/v units different simply by convention, or is there a logical explanation? Rockpocket 19:16, 12 November 2007 (UTC)

In CGS (centimeter, gram, second) units, the base unit of mass is the gram, while the 'natural' unit of volume is the cubic centimeter (equal, of course, to one milliliter). So you shouldn't be deceived by the 'milli' prefix; what you should be thinking is 1% (w/v) is 1 gram in 100 cubic centimeters, which just happens to be 1 gram in 100 mL. TenOfAllTrades(talk) 19:42, 12 November 2007 (UTC)

It’s ultimately only a guess, but I don’t think the name "percentage solution" was originally perceived to be a good name for what it means due to how it works out in CGS units, but rather due to the density of water, which is by far the most common solvent. 1 ml of water weighs almost exactly 1g. (Indeed, the gram was originally defined to be the mass of 1ml of water.) So 1g of solute in a water solution that totals 100ml means that the solute comprises very close to 1% of the solution by weight (w/w). MrRedact 20:15, 12 November 2007 (UTC)

Both suggestions are perfectly logical explanations. Thanks. Rockpocket 23:52, 12 November 2007 (UTC)

Running cool water over frozen drink defrosts. Why?

Sometimes I put canned soft drinks in the freezer, but I occasionally forget about them. When I finally take them out, they're frozen solid. I found out that I can run cool water (regular temperature) from the kitchen sink over the cans, while holding them at a 45 degree angle and rotating. After a few seconds, they're completely thawed out. If I take too long, they might even be too warm to drink! How does cool running water thaw out a frozen drink in only seconds? Steohawk

The drink you remove from the freezer isn't actually frozen solid — not even close. In fact, the ice that forms is neither contiguous nor very dense, because the sugar and carbonation bubbles in the soda pop interfere with the water molecules forming a contiguous lattice. It feels as though it's frozen solid because it only takes a fairly large chunk of fairly firm slush to provide enough support for the metal can to make it feel solid. Furthermore, water that feels "cool" (or "room temperature") relative to our normal body temperature actually contains a great deal of heat. Figure the soft drink is at 32°F. If you measure the "regular temperature" water coming from your tap, you'll probably find it is between 60°F and 70°F, for a ∂T of roughly 30 to 40 Fahrenheit degrees. The metal drink tin conducts heat very efficiently and uniformly to the contents, so your not-really-very-frozen drink "thaws" in a hurry.

Also, please sign your talk page and reference desk posts properly, not by manually typing in your user name, but rather by ending your post with four tildes ~ ~ ~ ~ (but without the spaces between them). --Scheinwerfermann 20:45, 12 November 2007 (UTC)

The water is cool - but it's above freezing - hence it's warmer than the ice and can melt it. I doubt your drink was frozen solid anyway because the pressure from the liquid expansion would likely split the can. Since the freezer will tend to freeze the liquid towards the outside of the can before the inside, there may not be all that much ice to melt and it's right at the surface where the warmer (but still cool) tap water can melt it most easily. I don't see anything weird going on here. SteveBaker 20:46, 12 November 2007 (UTC)

(edit conflict - GMTA) Who knows? Are we talking about soda (pop) like Coke? A few guesses: The can isn't really frozen solid, it just has a coat of ice all over the inner surface that makes it feel solid. When the water in the drink starts to freeze, it will freeze from the outside in because the heat is being lost through the can, and from the bottom up because of convection. As ice forms, the remaining solution will get more and more concentrated, depressing its freezing point more and more and helping it stay liquid. Running water is terrific at transferring heat, and if it's at all warmer than the can it will transfer heat to it. Do this: next time, take one of your cans outside and saw it in half with a big serrated knife (or get your mom or dad to do it). See if it is really frozen solid. --Milkbreath 20:58, 12 November 2007 (UTC)

Make sure to set up a video camera on a tripod and post the video to YouTube when you're done. It could easily be right up there with Diet Coke and Mentos!  :-) SteveBaker 21:38, 12 November 2007 (UTC)

Thanks for the answers.

BTW, I did sign my last entry with four tildes (no spaces), but it got "converted" when I saved the page. Let me try it again. Steohawk 22:08, 12 November 2007 (UTC)

Yup, that time your signature worked. I'll add that if you are eager to see inside an aluminum pop can, the easiest and safest way to remove the lid is to turn an ordinary can opener (preferably a gear-drive Swing-a-way type) 90° so its wheels are horizontal rather than vertical, clamp the rim of the pop can between the cutter wheels, and turn the can opener's handle to remove the lid and part of the rim. Be careful, for this will create a sharp edge. However, it will not create metallic dust or other contaminants (as the hacksaw method would), so you won't spoil the contents of the can. --Scheinwerfermann 23:02, 12 November 2007 (UTC)

Melted ice contains white particles, even purified water.

Whenever I freeze water and melt it, white particles can be seen floating around in the water. They're never visible prior to freezing. I thought that this might have been caused by some impurity in the water from my faucet, so I froze water from my neighbor's purifier. After melting it, the same white particles were visible. They're even visible in hot water, so I know that they aren't tiny bits of left-over ice. What are these particles and why do they only show up after melting ice? Steohawk 22:14, 12 November 2007 (UTC)

Make sure that the container in which you make your ice is clean and not shedding dust or other bits of material and that you cover it with something similarly clean when you're freezing/melting it. Maybe put the water in a zipper bag during the freeze and thaw? My older plastic ice-cube trays used to flake off over time, and also the paint and other coatings in the freezer itself did. DMacks 22:29, 12 November 2007 (UTC)

Possibilities. 1. Your neighbour’s purifier is full of gunk 2. Your freezer is full of gunk 3. The container you are using is full of gunk 4. The back of your retinas are full of gunk. Myles325a 22:30, 12 November 2007 (UTC)

Is it possible that you live in a 'hard water' area? I could maybe imagine there being a lot of dissolved minerals in the water that could maybe come out of solution when you freeze the water - making these little particles. What we need here is an experiment. (Woohoo! Actual science on the science desk!) The way to prove whether I'm right is to filter the water you get by melting the ice cubes and then re-freeze it in the same freezer/ice-tray that you used the first time. If you melt it again and it has particles in it the second time around then the bits must have come from the freezer/ice-tray. If there are no particles the second time around then DMacks & Myles are wrong and I guessed right. As an alternative experiment, you could try washing your hands with a little soap and the melt-water (wait for it to get to room temperature first) and see if you get lots more lather than you do with tap-water. Soft water produces more bubbles from soap than hard water and if I'm right and you somehow removed minerals by freezing and filtering - then you'd expect to get more lather with the melt-water. SteveBaker 23:57, 12 November 2007 (UTC)

You might enjoy our zone refining article.

Atlant 17:50, 13 November 2007 (UTC)

E = mc2 - how does c relate to the weight and energy units?

Everyone can see why there is so much energy in a given mass, because c, the speed of light is so large, and c2 is enormous. But why couldn’t we just define c as a single unit? How does Einstein define a measure of speed with weights and energy values?

After all, c = sqr(E/m). I have often thought about this, tho obviously I am a layperson in physics, and my brain hurts. Myles325a 22:25, 12 November 2007 (UTC)

We actually have a page about the E=mc2 formula, with some info. One hand-waving reason that "c² is more reasonable than c" here is that we need the units of measurement (dimensional analysis) to fit. A conversion factor for an energy/mass equivalence must be expressed in terms of distance²/time²). Once you have the relationship though, you can do algebra and other manipulations to express t he equation in different forms. Saying "c = sqr(E/m)" is not ideal because it fails if m=0. DMacks 22:37, 12 November 2007 (UTC)

In SI units, mass is in kg, lightspeed is in meters per second and energy is in Joules - which is SI shorthand for: "kilogram meters-squared per second per second" ...which is a bit of a mouthful.

You can work in any scale units you like so long as you are consistent. If you want to define your unit of length as (say) the distance light travels in a second (a "lightsecond") instead of the more usual 'meter' that's OK. The E=mc² equation still works - except that now we've defined c to be 1.0, instead of the result 'E' being in Joules (which is shorthand for kg.m².s^-2) the answer would be in some new unit of energy because we defined our unit of length to be a lightsecond instead of the meter. Let's name it after it's creator and call it the 'Myle'. The Myle is shorthand for kg.lightsecond².s^-2. One kilogram of matter would contain exactly one Myle of energy...very convenient! But the Myle would be a truly phenomenal amount of energy because it's in units that contain lightseconds²!! The Myle simply isn't a very useful energy unit. A 'AA' battery stores about one kJ - but and inconvenient 0.09 attoMyles! On the other hand, you could instead define your time unit as the time light takes to travel in one meter - c is still equal to 1.0 but your unit of time is microscopic - so your unit of energy is still enormous. You can shuffle the unit definitions around however you like - but the result is the same. Dimensional analysis is always very instructive in these cases. SteveBaker 23:31, 12 November 2007 (UTC)

It's also worth noting that when working with problems in relativity, velocities are generally expressed as fractions of c rather than using more common measurements. Donald Hosek 02:00, 13 November 2007 (UTC)

You might want to read up on Maxwell's equations. That's where Einstein got almost all of the theoretical support for the conclusion that E=mcc. --M@rēino 14:27, 14 November 2007 (UTC)

November 13

Technology and Depression

Does the more engrossing technology becomes translate to the more depression society feels as a whole? (Forgive the grammar.)

Living in our parent's house, no post-secondary education or real job experience between us, my brother and I were absolutely entranced with World of Warcraft. "Real" decisions with all their complications were so far removed from the crisply cut and dry work/rewards system in game. I alone spent 7 months of 16-hour grueling days rising to the rank of Grand Marshal of the Alliance and I loved it. We lived every waking hour in WOW for 2 years. Our parents eventually gave us an eye-opening ultimatum and he killed himself with a plastic bag and duct tape about a year ago. I was in a psych unit for about 2 months after. I know now what horror and true self-loathing feel like.

In my heart I know this is a sign of things to come for society as a whole. I'd like to know if depression and technology are correlated. Sappysap 02:13, 13 November 2007 (UTC)

There are arguments that they are, but I don't believe so. Depression is fairly clearly genetic, and has existed in some form for all of history. Suicide, of course, being the worst-case fate of depression, has also been in existence and I don't believe suicide rates have increased or decreased significantly with time, though certainly environmental factors have played a role. I have no significant sources with which to back this, though, other than the fact that "melancholy" was a diagnosis since at least the early 17th century. SamuelRiv 03:14, 13 November 2007 (UTC)

First you need to clear up what exactly do you mean by depression. There are many subtypes of clinical depression, some of which include atypical depression, and dysthymia. Each clinical type will have certain characteristic features; so people with dysthymia may be less prone to committing suicide than people in a major depressive episode.

Correct me if I am wrong, but your hypothesis is that technology allows people who are lonely and/or need human contact but lack it to withdraw into artificial worlds where they get only superficial human interaction. This leads to increasing loneliness and eventually to suicide or total withdrawal from society?

Also when you use the word technology, do you mean only modern technology? If not, then we would need to analyze the impact of earlier technological advances like electricity and the printing press (among many others) to see if they provide support to your hypothesis.

What would be key to supporting your argument is that technology heightens the impact of someone who would be prone to withdrawal from human interact, because I would imagine there is and always has been a certain subset of people who are lonely but still take actions to further isolate themselves.

As to SamuelRiv's claim that depression is "fairly clearly genetic", I do not agree. First it is unclear what that claim even means. Minimally, it would mean there is a genetic component, which does seem to be the case. But the more natural interpretation is that he is claiming it is almost entirely due to genetic factors, which would need some serious evidence to justify. This issue also points to the need to clear up what exactly you are talking about when you use the word "depression", since each clinical subtype is likely to have different roles a genetic component would play.--152.2.62.27 13:00, 13 November 2007 (UTC)

I also disagree that depression is a mainly genetic phenomonen, I have always believed that it is mainly due to the way one has been brought up combined with the life experiences they have been through, ie somebody that was emotionally neglected as a child and/or bullied will take a major battering to their self esteem which is definitely a factor in causing depression. These psychological scars can take years to heal, that is if they ever heal completely. On the flip side, somebody with a more fortunate upbringing is less likely to suffer from these issues BUT there are many exceptions to this and I think there may be a genetic element too...HOWEVER this may be psychological - if a parent suffered from depressive conditions then they may 'rub off' on the child.

Going back to technology, I think it CAN cause depression but it can also be a godsend. For example a depressive person could spend hours "wasting their life" (in the eyes of some people) playing video games but then again this is usually a distraction and definitely doesn't mean that without this activity, the person would be a raving extrovert. On the flip side, I think the internet can be very helpful as people with mental illnesses including depression can find out information about their condition and discuss it with other suffers which is often very cathartic. People may be willing to engage in others online when they wouldn't otherwise.

A third argument is that it is not technology itself that is causing depression but the way society has become. People nowadays are generally more insular than they were say thirty years ago, many people don't know their next door neighbours any more and the general perception is that things are unsafe and there are a lot of people around who cannot be trusted. This could (arguably) be caused partyly by technology, for example people not bothering to talk to each other because they are more interested in watching the TV or gaming while before the advent of these solitary pleasures, most pastimes were more social, take the fact that a lot of families no longer sit around a table for meals, instead choosing to eat in front of the TV. Take also the fact there are a lot more single person households than ever before, this causes isolation but I cannot see any attribution to technology except for perhaps the learning of antisocial trends from television. GaryReggae 15:51, 13 November 2007 (UTC)

Though it is out of date you might fight Oswald Spengler an interesting read—it's not about technology, per se, so much as it is about modernity. But in any case, people have wondered for years (since the 19th century, anyway) if the creation of new technological means of representation and life have "sucked the soul" out of modern human interactions, encouraging depression, vice, etc. I myself am doubtful of a straightforward relationship—first, I don't believe there probably were "good old days" where everybody was chummy and warm and happy (most people are sons-of-bitches no matter what time or place you live—but that's the misanthrope in me talking), and second, I'm unconvinced that technology itself is responsible for the sorts of changes people are talking about. When a family eats its dinners in front of the television rather than together at a table, it isn't the television that has compelled them to do so; it's just as easy not to have the TV on during dinner (even easier as time goes on, with the ability to time-shift shows). As for depression—that's a complicated issue, but I don't see any obvious technological links there. If it wasn't the game, who isn't to say it wouldn't have been some other form of escapism? --24.147.86.187 01:03, 14 November 2007 (UTC)

• Sappysap, I could point you to a ton of Wikipedia articles, such as Uncanny Valley and Escapism. But ultimately, there's still a lot of research to be done before we can say for sure how artificial worlds affect psychology. In a situation like yours, I doubt any answer from us volunteers at Wikipedia will be completely satisfying (especially ones like me who aren't even pre-med students!). You may want to consider the possibility that this is your life calling, enroll in your local college, and get yourself a degree in pyschology. If that's not an option, at least ask your psychologist (or someone you like from when you were at the ward) to refer you to one of their colleagues who specializes in technology, so that you can do some meaningful research. --M@rēino 14:42, 14 November 2007 (UTC)

I don't think it's the technology. If you'd spent two years doing nothing but...I dunno...reading medieval poetry or learning to ride a unicycle - the end result would be kinda similar. MMPORG's are fun - they're fine for a couple of hours of entertainment a few days a week - they help you unwind - but in the end when you finally chuck it in - you haven't achieved anything. Heck you said it was "grueling" - so you weren't even getting the relaxational benefits. But it's pretty clear that making your entire life (16 hours a day, 7 days a week for 2 years) happen in a place some dumb game designer (like me - I'm a game designer) dreamed up is not a healthy thing. I mean, 11,000 hours at minimum wage would buy you a small house or a pretty decent sports car! The problem here isn't the technology - it's the sheer waste of human effort in doing it in the first place. I know where you're coming from - I'm closing in on 10,000 edits over 3 years to Wikipedia. That's an addiction that I recognise - but it's a lot less than 10,000 hours - probably more like 1,000 hours - but still, it's a serious chunk out of my life. The difference is that it doesn't feel like time wasted - there is a positive result - the accumulation of all human knowledge where almost anyone can read it...that's something worth spending your time on. The technology of Wikipedia and that of an MMPORG are not all that dissimilar!

So it's not the technology...it's what you use it for and to what extent you let it take over your life.

I've come across horror stories of MMPORG addiction too (although nothing like on this scale) - we had a neighbour who's smart "straight-A's" teenage son spent 50+ hours straight (no sleep, no food) wrapped up in Everquest. When his parents found out, they took away his computer promising to return it after one week. He then came over to my house, smashed a window, broke in and stole my wife's laptop. Sadly, being a sleep-deprived teenager, he forgot to steal the battery charger and phoned us up the following day to see if we happened to have a laptop charger he could borrow...which certainly raised some suspicions! When we got the laptop back, yep - Everquest had been mysteriously installed on it. But that incident blew away any trust his parents had. He wound up in a downward spiral that put him in military school then downtown Bagdad when he should have been in college.

SteveBaker 19:46, 14 November 2007 (UTC)

Mareino: I desperately wanted to write and publish an article about the surrealism currently going on in games like WOW. There are sweat shops in China, with families on the line, farming WOW gold (I know for a fact b/c I've actually learned a litte phonetic Mandarin from them, even here in hick Virginia, US). They are in so much pain. There are so many "heroes" in-game that have gained their status by giving up on real life completely, much like myself and my brother. Both of our best friends, the guild, dissuaded me from publishing anything after his suicide, as it would dishoner a fallen comrade and the guild as a whole. I know this is irrational, but I guess I agree with them. It's insane to even state, but: you had to have been there with us; through all the guild wars, and the insurmountable dungeons. Of course, none of them really existed but they were as real to me as any war imaginable. That is why I tremble for the future of technology and virtual media. Sappysap 04:16, 15 November 2007 (UTC)

Reduction of Fe(III) in FeCl₃(aq) to Fe(II)

How would I reduce the iron in FeCl₃(aq) to Fe(II) in order to use the indicator ferrozine, which only detects Fe(II), not Fe(III)? Can I somehow use ascorbic acid to do so? What would the chemical reaction be? Could this be undergone by using titration? Chickenflicker-♣ 02:48, 13 November 2007 (UTC)

Edit: What would the reaction between Hydrazine (N₂H₂(l)) and FeCl₃(aq) look like? Chickenflicker-♣ 03:03, 13 November 2007 (UTC)

I've seen hydrazine used as a 2-electron reductant (standard way to prepare Pd(0) complexes is from Pd(II) salts, and also the Wolff-Kishner reduction). Not sure about a good 1-electron reduction mechanism. DMacks 14:39, 13 November 2007 (UTC)

Thinking more technically here, hydrazine is N₂H₄. N₂H₂ (diazene) also exists and is a good reducing agent (via hydrogen transfer, such as in hydrogenation of unsaturated compounds), but it's mighty unstable (the pure material decomposes even around –180°C). DMacks 16:50, 13 November 2007 (UTC)

horizontal component of speed

This is a mix between a maths and physics questions...anyway here goes:

An object is being pulled up a slope by a force at a constant speed of say 3m/s. At the top of the slope is a horizontal plane. What is the object's speed at the beginning of its journey along the horizontal plane? Is it 3m/s, or is it just the horizontal component of velocity (ie. 3cos(angle))? Is the vertical component of velocity wiped off or does it become a part of the horizontal velocity?

    ->>______________________
   -  /
  -  /
    /
   /
  /
 /
/

What about when a ball rolls down a slope and then travels along flat ground? At the beginning of its journey along the flat ground, is its speed just the horizontal component of its previous velocity or is the magnitude of both components of its previous velocity?

                     /
                    /
                   /
                  /
                 /
                /
____________<<-/  

Thanks! D3av 02:53, 13 November 2007 (UTC)

In these types of problems you usually approximate the sharp corner as a tiny rounded corner. Then assuming ideal conditions the ball rolling down the hill will conserve kinetic energy when it hits flat ground. The ball being pulled up a hill will also conserve kinetic energy, but the end result depends on the precise wording of the problem. If you are constrained to be on the surface of the slope, then you move forward with kinetic energy conserved. However, if unconstrained, you fly upwards as a projectile with some initial velocity at some angle. SamuelRiv 03:09, 13 November 2007 (UTC)

So say the one moving down the hill is unconstrained (just freely rolling down). Would you use the horizontal component of velocity or the whole thing?

When you say "with kinetic energy conserved", does that mean that the vertical component of velocity becomes part of the horizontal component?

We had one question in which the object seemed to be constrained as it was being dragged up the slope. The instruction was to use only the horizontal component of velocity when it reached the horizontal plane at the top. Was this a correct instruction? thanks. D3av 03:26, 13 November 2007 (UTC)

As SamuelRiv said, it depends on the precise wording of the problem, in particular how the force is being applied. For example, if there's something pushing on the object from the left (slowly, so that gravity keeps the object from ever leaving the surface), moving at a constant rate to the right,

       ______________________
      /
->|  /
->|*/
->|/
  /
 /
/

then the horizontal component of the velocity will remain constant, and the vertical component will go to zero at the corner. If there's a rope attached to the object that goes up and over the corner (so the corner acts somewhat like a pulley), and the rope is being pulled at a constant rate,

       ______________________->
      /______________________
     //
    //
   */
   /
  /
 /
/

then the component of the velocity parallel to the rope will remain constant, so the horizontal and vertical components will change as the rope's direction changes. If there's something pushing the object in the direction parallel to the slope,

       ______________________
\     /
 \   /
  \*/
   /
  /
 /
/

(...well, I don't know how to make it look like that thing is moving diagonally upwards with ASCII, but anyway) it will actually move faster when it reaches the horizontal surface because the component of the velocity in the direction of the slope will remain constant, but it gets a new perpendicular component when it passes the corner. So it all depends which of these different problems you're trying to answer. —Keenan Pepper 06:27, 13 November 2007 (UTC)

The vertical component won't magically go to zero, regardless of whether the object is being pushed or pulled. Unless the object is held on some sort of track, it will rise into the air above the horizontal plane until gravity brings it back down. Clarityfiend 08:00, 13 November 2007 (UTC)

Unless, as I said, it goes "slowly, so that gravity keeps the object from ever leaving the surface". This is a quasi-static approximation, which is actually very good for everyday objects. If you pull a rubber ball up a ramp with a string, slowly, it doesn't fly into the air when it reaches the top. —Keenan Pepper 23:04, 13 November 2007 (UTC)

Thankyou very much for those answers. They are very comprehensive and clear. I'm still unsure though (as I wrote above) about what will happen to a ball that is rolling down a slope and then continues along a flat surface. (The only forces acting on it are the force of gravity and the normal). Thanks D3av 09:45, 13 November 2007 (UTC)

It depends on the nature of the ball. When the ball hits the bottom of the slope, there is a collision. What happens next depends on what the ball and the ground are made out of. A softer ball might bounce a few times - eventually rolling off along the new direction at roughly the speed it was moving down the slope - much conserving kinetic energy (assuming we ignore losses). What happened was that the kinetic energy temporarily turned into elastic energy from the compression of the ball's material - then that elastic energy was turned back into kinetic energy directed in a different direction. That little energy interchange permitted the velocity vector to change but the kinetic energy to be conserved. A harder ball might simply give up the vertical component of its energy in the collision (so the kinetic energy turns to heat) - continuing to move along the horizontal region at a speed equal to the horizontal component of it's former motion. Without more information, you can't say which it will be.

When the ball is travelling up the slope, the same thing happens. At the top of the slope the ball will continue upwards (because of conservation of momentum) - travel on a free parabolic trajectory and eventually collide with the top of the slope. Once again, the ball will either bounce - causing it's new velocity to be horizontal - with the same speed as it was travelling up the slope - or it'll impact without a bounce - causing it to lose the vertical component of it's former velocity.

With real balls, the behavior is somewhere between the two cases - energy is lost in the bouncing and the actual horizontal speed will be a blend between the old horizontal speed and the old net speed.

SteveBaker 12:26, 13 November 2007 (UTC)

To original poster, remember kinetic energy is NOT a vector. It has no horizontal and vertical components. When kinetic energy is conserved, all velocities must be taken into account. SamuelRiv 14:56, 13 November 2007 (UTC)

But also remember that kinetic energy is NOT necessarily conserved -- it's only conserved if the collision between the ball and the horizontal surface, described by Steve, is elastic. To make this clear, imagine rolling a cannonball down a steeply sloped, almost vertical wall into a lawn. It'll probably stop dead at the bottom and embed itself in the grass. However, the same ball coming down a gentler slope, one that's almost horizontal, will probably keep rolling and may well do it without much loss of kinetic energy. --Anon, 23:32 UTC, Nov. 13.

Thanks for all your help. I see that my maths course has oversimplified the issue and ignores bouncing/heat/other forces etc. I'll just try to follow the guidance of the question! D3av 00:03, 14 November 2007 (UTC)

That's pretty typical of a math course - they aren't really interested in the physics - because it's a math course! However, in this case, it's not the same thing as simply ignoring friction and air resistance for the sake of getting an answer. In this case the outcome will be quite utterly different for an elastic versus an inelastic collision. The issue of whether kinetic energy is conserved or whether conservation of momentum applies is not an insignificant matter. SteveBaker 18:46, 14 November 2007 (UTC)

Well, momentum (unlike kinetic energy) is always conserved, but that's not useful to consider in this problem. Whether or not the ball's speed changes, its velocity certainly does, since it's moving in a different direction; therefore there is a transfer of momentum between the ball and the Earth. --Anonymous, 23:01 UTC, November 14, 2007.

stars come down

hi my question: there is a place in space that stars come down(and it is a holy place...god says in quran)Have you heard anything about it? —Preceding unsigned comment added by 213.207.252.64 (talk) 10:20, 13 November 2007 (UTC)

Could you provide a reference to a specific passage in the Quran? Scripture often contains sections that need to be read very carefully to determine what they refer to exactly. My first guess would be a black hole. (EhJJ) 11:41, 13 November 2007 (UTC)

It's possible that the book refers to meteorites as stars. -- JSBillings 11:43, 13 November 2007 (UTC)

If "come down" means "move deeper into a gravity well", then I suppose you could say that "stars come down" as they fall into a supermassive black hole. Those are found at the centers of many (and perhaps all) galaxies. But there is no one, single place where this happens in the universe - it's likely that there are at least as many of these objects as there are galaxies - and there are billions of those. The nearest one of these things is at the center of our own Milky Way galaxy.

I find it amusing that a black hole might be considered a holy place. Good and useful things (like stars and planets) go in - nothing (including light or information) comes out! But really, any sufficiently vague and ancient statement (with likely translation errors and a wide laxity of interpretation) can be made to fit some real thing. I don't think I'd read anything into this statement beyond that whoever wrote it didn't know what a star really was and thought that the idea of the seemingly immutable stars 'falling' was a dramatic image. Spotting an occasional meteor and (incorrectly) assuming that was a 'falling star' might serve to reinforce that. Sadly, with what we now know, it really doesn't make any sense.

SteveBaker 12:07, 13 November 2007 (UTC)

Stars falling can refer to the yearly Leonid meteor shower which offered a particularly vivid display on November 13, 1833 bringing quite a few people to thinking it was the end of the world. Not that this particular event would be mentioned in the Qur'an but maybe a very early occurence of this same phenomenon. Keria 14:26, 13 November 2007 (UTC)

Meteor showers such as the Leonids and Perseids are not localised to one place though. Every place on the earth and the moon is peppered by them at one time or another. So this couldn't be a specific holy place - except by the very broadest interpretation. SteveBaker 14:56, 13 November 2007 (UTC)

The representations of the meteor shower often show a sort of fountain of stars as if there was an origin to these moving lights. I'm not sure what the questioner means by "that stars come down" but if it's from I could definitely see a 7th century scholar/mystic (or Muhhamad!) seeing a holy fixed source of stars in space in the apparent origin of these showers. Keria 15:55, 13 November 2007 (UTC)

Yes - both the Leonids and the Perseids appear to come from one place in the sky - in fact their names come from the constellations that they appear to radiate from (that point is called "the radiant"). The reason is that these showers each come from tiny bits of an ancient broken-up comet that have spread out into a gigantic loop around the sun. The earth moves through each of those streams once a year - which is why the debris appears to come from a fixed spot in the sky year after year - over the exact same couple of days each year. But the problem with this theory is that this 'radiant' effect means that the 'shooting stars' radiate outwards - they are all heading off to different places - not falling to the same place. SteveBaker 23:51, 13 November 2007 (UTC)

• The Qur'an also says that Mohammed flew from Mecca to Jerusalem on a magical flying horse, so you should probably take the unrealistic parts as metaphor. --Sean 19:06, 13 November 2007 (UTC)

The authors of the quran can't possibly be intending to refer to black holes, since its authors lived long before anyone knew anything about black holes!

I did a search on "stars" within English translations of the quran.[1] There are only about 15 hits, although that number varies by translation. There are a couple passages that look like what you might be referring to:

Al-Waqi’a 56:75 refers to what is translated in the Yusuf Ali translation as "the setting of the stars". The Muhammad Asad translation, however, uses the phrase "coming-down" instead of "setting". (The Muhammad Asad translation also uses "parts" instead of "stars", so the meaning of that passage is particularly unclear.) If this is the passage you are referring to, then no, there is no place where stars literally come down. They merely appear to come down, to the west of wherever you happen to be located, due to the rotation of the Earth.

Al-Hajj 22:18 talks about the stars "bowing down" before god, or "prostrating" themselves, depending on the translation. And Al-Hajj is about the pilgrimage to Mecca, which is certainly considered a very holy place among Muslims. If this is the passage you are referring to, then no, no such bowing motion is observed in stars, even when they appear to be over Mecca. MrRedact 19:30, 13 November 2007 (UTC)

Thanks! Excellent work! That's pretty much what I thought - vague translations of an uncertain source...but it's nice to see it confirmed. SteveBaker 23:51, 13 November 2007 (UTC)

This is mona who asked the question,at the first place I want to thank you because of your answers then I want to excuse you because of the vague question(my own language is not EN so it is expected,I'v studied the persian translation and it was wrong)I want to correct it: The exact passage:I swear to sites of stars(75) if you know it's a big sworn(76) my question:where are sites of stars and why are they so important that GOD swear them

Alabama? --Trovatore 00:02, 14 November 2007 (UTC)

What snake a specific book refered to

I'm looking for a king of snake who may havg from a tree over the Amazon river or one of its tributaries, would look like a vine to someone in a canoo going under that tree, and can swim away if grabbed off the tree. I looked in Category:Reptiles of South America, but couldn't find it - the Emerald tree boa looked promissing, but I couldn't find evidence of it being able to swim. Od Mishehu 11:29, 13 November 2007 (UTC)

Perhaps an anaconda? (EhJJ) 11:38, 13 November 2007 (UTC)

I don't think it hangs from trees. In the book in question (Magic Tree House book called Afternoon on the Amazon), Jack tries to get to the river bank by pulling on a "vine" that turns out to be a snake. Od Mishehu 11:47, 13 November 2007 (UTC)

It doesn't look like the author actually used real animals in her books, but just made up a convenient fictional snake as a plot device. -- JSBillings 13:27, 13 November 2007 (UTC)

I'm afraid I don't understand the question. It was a fictional snake. --Milkbreath 12:01, 13 November 2007 (UTC)

Indeed, it's difficult to see why a snake would hang above a river unless it could swim. But why would it do that? It's not as though tree snakes normally go fishing.--Shantavira|^{feed me} 13:52, 13 November 2007 (UTC)

I thought all snakes could swim, no? I know hedgehogs can. Lanfear's Bane | t 16:46, 13 November 2007 (UTC)

Condensation and Deposition

I need to measure the rate at which Condensation and Deposition (physics) occurs at standard temperature and pressure from a particular inorganic vapor with a melting point in the 1000s K produced by combustion. I have had no luck so far. How do physicists and chemists measure condensation and deposition? As a biologist without a whole lot of physical chemistry, I haven't even been able to figure out where to begin. I am hoping that there is a formula but I fear that these rates are determined only empirically. Thanks for any help. Spc303 12:59, 13 November 2007 (UTC)

You'll want to look up the phase diagram for your substance and then calculate the vapor pressure, I believe (not an engineer). Deposition is a much harder problem, and I'm afraid I don't know the physics of that phase transition, but for condensation you may also need to look up the constants a and b in the Van der Waals equation, but not for these formulas. Here's the relevant equations: for the change in the number of particles going from liquid to gas phase, we have

$\Delta N={\frac {16\pi \gamma ^{3}}{3n_{l}^{2}}}(k_{B}T\log(p/p_{0}))^{3}$

where $\gamma$ is the surface tension (in erg/cm³), $n_{l}$ is the number of molecules per volume of the liquid, and $p$ is the vapor pressure defined by

$p=p_{0}\exp(-L_{0}/RT)$

and $L_{0}$ is the last thing to look up is the latent heat of vaporization per molecule. Note T is temperature, kB is Boltzmann's constant, and R is the gas constant. SamuelRiv 14:53, 13 November 2007 (UTC)

Biofuels

Hi there, I have a question about Biofuels which I can't find a meaningful definitive answer to online. Where I work, we have a well used old 2002 reg Citroen Berlingo van (diesel) and I have been toying with the idea of using waste cooking oil as fuel as we have tons of the stuff available. According to some sources I have seen (for example [2]), engines need to have conversion kits fitted, such as adding an extra fuel tank or heating the fuel, however I have read many cases of people just putting used cooking oil (once it has been filtered) straight into their fuel tank - not neat but making sure they have half a tank of standard mineral diesel as well as half a tank of veg oil, ie 1 50:50 mix. Would this work or damage the engine? It's located in the South of the UK so unlikely to be susceptible to really cold temperatures. It's probably too much hassle to get the engine modified but if we can use it as it is then great! The van is a bit of a 'rattler' so it's not exactly the end of the world if it needs repairing after any 'experiments'.

Disclaimer: I know we need to pay fuel duty tax if we use more than 2,500 litres a year and obviously we would monitor this and ensure that we do not go over this amount so this is NOT a legal question, just a technical one. GaryReggae 15:36, 13 November 2007 (UTC)

There are two problems you need to deal with:

• Filtering
• Cold weather

Typical cooking oil has lots of bits of food in it - and sometimes some dissolved water - and it absolutely must be carefully filtered before you put it into the tank of your car. I'm told that coffee filters work - but they fall apart and clog before you get enough oil through them. You need something more robust if you are going to be doing this for any length of time.

Cold weather is the more serious matter. Veggy oil is not very runny at cold temperatures - that can strain your fuel pump and clog the system up pretty badly. A proper conversion kit uses a secondary tank of straight diesel to start the car - then picks up waste heat from either the exhaust or the hot water entering the radiator to heat up the fuel lines coming from the veggy oil tank so you can switch over to veggy oil a few minutes after you've started the car when everything is nice and toasty. If you lived here in Texas, you could probably not bother and just stick filtered oil straight into your car, switching to straight diesel for a few weeks in the depths of winter - but in the South of England - no, you're going to need a heating system of some kind at least 6 months of the year. Certainly you can experiment. I don't see the harm in mixing diesel and oil - although I suppose they might settle out if not mixed leaving the oil being the first thing that enters the system on a cold weather start. If I were you, I'd start in the summer when things should work OK even if you're not really getting it right - and as the weather gets colder, and the thing gets harder to start - or just won't run at all - then you need to switch to diesel (at least in some mixture). But before you do any of those things, you need to figure out a good way to filter your oil.

SteveBaker 16:37, 13 November 2007 (UTC)

This is something you need to ask a sharp mechanic (like the Car Talk guys) if your mind is really set on it. A car has internal computer regulators of the octane level of your fuel, among everything else including the integrity of the fuel-air mixture that is injected into the engine. It may be that this regulation will cause serious consequences to your exhaust or fuel efficiency, and will certainly damage the engine if you're using just any old cooking oil. The engine is designed to self-clean and lubricate, and it only can do this if the nature of the exhaust of the fuel you put into it is predictable. Cooking oil almost certainly burns different and probably is dirtier than gasoline, causing premature wear in your pistons. I would definitely recommend against it for all these reasons. But please, if you feel you must do it, ask an actual mechanic. SamuelRiv 20:37, 13 November 2007 (UTC)

Whaaat??? Your answer is almost 100% wrong! We're not talking about gasoline. This is a diesel engined vehicle. The whole business of octane levels are relevent to gasoline powered cars because you need to avoid knocking - but diesel engines are quite different in that regard because they rely on the fuel detonating without a spark. The idea of running diesel cars on left-over cooking oil is extremely well understood (see Vegetable oil used as fuel for example). Thousands of people have done it - it works (although without engine modifications - it only works well in warm climates and with older vehicles). The issues are the ones I've described. The issues you bring up are not a problem.

The onboard computer (the ECU) doesn't "regulate the octane level" - how the heck could it possibly do that? There isn't some separate 'octane supply'! What it generally consists of is a small microphone that detects the onset of 'knocking' in the engine due to insufficient octane in the fuel. The computer then alters the fuel/air mixture to avoid the damage that might result. But that's all irrelevent here because this isn't a gasoline engine!

SteveBaker 23:45, 13 November 2007 (UTC)

A computer that can arbitrarily alter hydrocarbon composition of a mixture would be impressive, indeed... -- 128.61.20.199 13:35, 14 November 2007 (UTC)

Thanks for all your ideas. I will do some more research on conversion kits and suchlike. It seems another option is to add methanol to the veg oil to thin it down but I don't know how well that works. GaryReggae 12:36, 14 November 2007 (UTC)

A big problem with methanol is that it is an organic solvent, so you'd want to talk to someone who knows organic chemistry before subjecting the various fuel lines and seals in your engine to that. It may also react with or give the fuel undesirable properties, but I don't know enough chemistry to say. Anyway, running diesels on veg oil works beautifully from my experience, just make sure you take the precautions that Steve mentioned. -- 128.61.20.199 13:32, 14 November 2007 (UTC)

Yeah - absolutely. Even the 10% ethanol that they are putting in gasoline these days can seriously mess up the engine of an older car. I don't know about methanol - but the results could be similar. Ethanol dissolves rubber and some kinds of plastics. On modern cars they've been aware that ethanol was likely to be added to gasoline for many years - so modern cars have seals made of plastics that don't dissolve - but I've noticed all sorts of seal failures in my 1963 Mini since we switched over to E10 a year or two ago here in Texas. There are other considerations too - ethanol produces acidic residues from combustion that can destroy the viscosity of non-synthetic motor oils, it also conducts electricity so E25 ethanol (coming soon to a gas station near you!) can short out the fuel gauge sending unit or your fuel pump on some older cars.

I don't know which (if any) of these things also applies to methanol - but unless you know for sure that methanol has been sucessfully tested by a lot of people - I wouldn't use it. Mix your veggy oil with regular diesel fuel by all means - but getting creative is not advised! SteveBaker 18:39, 14 November 2007 (UTC)

"From the Fryer to the Fuel Tank", by Joshua Ticknell is the standard book I read on the subject. (try to get the second or any later edition). Also, ReNew magazine published in Aust has articles occasionally. Lots of considerations. The price of old chip oil in rural victoria has gone up in the last 5 years as a result of conversions.Polypipe Wrangler 20:14, 14 November 2007 (UTC)

PMTs, APDs, DUV, and other exciting acronyms

Anybody happen to know roughly what the window of time for detection of a Cherenkov radiation event in a particle detector is? This statement in the first paragraph of the photomultiplier tube article struck me as possibly outdated and unsubstantiated:

"Semiconductor devices like avalanche photodiodes have replaced photomultipliers in some applications, but photomultipliers are still used in most cases."

Modern APDs (like the ones my group produces for UV wavelengths) approach the gain afforded by PMTs and enable single-photon detection with enormously lower cost, power requirements and physical size, but higher reliability, bandwidth, speed, quantum efficiency, etc. The advantages of APDs for this application seem to be somewhat confirmed by a presentation given by a KEK scientist about a next-generation Cherenkov radiation based neutrino detector, though his avalanche gain figure is outdated (gain over 10⁴ is quite doable with APDs these days).

The only area I can think of where PMTs might still hold an advantage regards the window of event detection. I have no idea how much time you have to detect the photons emitted from a Cherenkov event, nor exactly how PMTs are operated (are they run continuously or sampled?). Typically high-gain APDs are operated in Geiger mode--they are biased below breakdown and then pulsed into breakdown wherein they operate at very high gain for the duration of the pulse. While this pulsed/sampled mode of operation can be run reasonably fast (several MHz with pulses on the order of tens of nanoseconds with a good device), I don't believe it's possible to operate very high gain APDs in breakdown continuously for risk of various destructive breakdown mechanisms. The higher quantum efficiency of APDs may make up for the possible deficiency of Geiger mode measurement, but I have no figures for PMTs to compare with.

So enough rambling, here are my burning questions: How long do you have to detect Cherenkov radiation? Presumably the data coming from these neutrino detectors is sampled for digital processing, but how quickly? Are the photomultiplier tubes operated continuously? Does the whole system operate in a sort of "triggered" mode, operating the photomultipliers continuously and waiting for a detection? What percentage of "missed" events is considered acceptable? Does anyone happen to know what the quantum efficiency of a PMT typically is? (I'd guess it's mostly limited by the front-end optics and the quantum efficiency of the photoelectric effect in the photocathode) Have I lost more than 90% of the readers of this page by now? Would I be better off learning Japanese and calling up the KEK to ask my questions? :)

Thanks for any insight you can provide.

Bonus: Impressive picture of the innards of the Super-Kamiokande detector. Notice the two technicians and the size of each photomultiplier tube (enhanced by the big lens assembly). Now consider the infamous Super-K PMT implosion accident which took out over half of these in a chain-reaction, and the motivations for using micron-sized APDs rather than enormous evacuated tubes become apparent!

-- mattb 16:42, 13 November 2007 (UTC)

IceCube operates continuously with triggered events. Trigger timestamping is at <5 ns, and the resulting current pulses are digitized at 250 MHz. The size of pulses allows one to discriminate events running from 1 to ~200 photons per 15 ns. Quantum efficiency for those PMTs is typically 20%, with practical efficiencies with a single photon closer to 10%. Many events involve more than 1 photon. For IceCube like environments, you also need to operate at -40 ° C. Dragons flight 17:44, 13 November 2007 (UTC)

Ah ha. So the necessary detection window is around 15 ns, and the detector needs to be able to distinguish ~1-200 photons in that time? That's a lot shorter window than I thought. Continuous linear operation with high gain is indeed a big boon.. You can operate APDs continuously in their linear (low gain) region and outpace the PMTs, but high-gain Geiger mode operation is necessary for single-photon detection and isn't nearly as fast... With a 15 nanosecond window for counting photons, I don't think current APDs alone could cut it, since the shortest Geiger pulse I've seen an APD used with is a bit under 1 ns. Even with good active quenching, there's a necessary 1 or 2 ns delay before the APD can be reset for another count, so if you're right there's no way they could distinguish between 200 photons in 15 ns using Geiger mode... Realistically, I don't think Geiger mode is even practical for periods in the tens-of-picoseconds range; generating timing pulses that short is difficult, and quenching fast enough is even more difficult. That explains the interest in hybrid detection schemes using an APD in linear mode... Thanks for the info. -- mattb 18:51, 13 November 2007 (UTC)

This page indicates a maximum 30% quantum efficiency and overall gain up to about 10⁸. DMacks 17:47, 13 November 2007 (UTC)

Free-falling

When we fall, we fall at a rate of approximately 33 feet per second, per second.

Is this an 'exponential' rate? Do we fall at a speed that is increasing 'exponentially?'

Peter Lamont —Preceding unsigned comment added by 72.39.249.249 (talk) 17:09, 13 November 2007 (UTC)

Free-fall due to gravity on earth is at an acceleration of about 32 feet per second, per second. Note it's "per second per second", so it's not a simple speed of falling. What it is, is how fast the speed increases: every second of falling, one is falling at a speed 32 feet/second faster. It's a linear increase in speed because gravity is an approximately constant force in this situation. DMacks 17:15, 13 November 2007 (UTC)

Neglecting air resistance and the changes in gravity due to distance and (eventually) relativistic effects - then over time:

• Your accelleration is constant (at roughly 9.8 ms^-2 - which is about 32 feet per second per second).
• The speed with which you are falling is increasing linearly.
• The distance from your starting point is increasing exponentially (but not in the mathematical sense of e^x).

SteveBaker 17:38, 13 November 2007 (UTC)

Actually, that's not an exponential increase. For the speed to increase, then we need to have $v(t)=ab^{t}+c$. The velocity, is, in fact, given by $v(t)=ct+v_{0}$ where c is a gravitational constant (32 ft/s² or 9.8 m/s²) and v₀ is the starting velocity. Donald Hosek 17:43, 13 November 2007 (UTC)

The technical term for this type of increase is "increasng polynomially." That's faster than "increasing linearly" but slower than "increasing exponentially." in this case the ploynomial is of order 2 (the square,) so we can also say is is O(n²) -Arch dude 17:51, 13 November 2007 (UTC)

• And therefore we can alternatively say "increasing quadratically". --Anon, 23:35 UTC, Nov. 13.

(ec) Let's run successive integrals: constant acceleration, linear speed, quadratic position. DMacks 17:52, 13 November 2007 (UTC)

see also Big O notation. -Arch dude 17:53, 13 November 2007 (UTC)

Also note that terminal velocity will tend to put a cap on acceleration if you give it enough time (and aren't in a vacuum) GeeJo ^(t)⁄_(c) • 18:18, 13 November 2007 (UTC)

Length of US State Coastlines

I saw a VERY cool entry called "List of Countries by Length of Coastlines"

And naturally wanted to extend that to our 50 States and was hoping to find that already computed somewhere .....

Has it been?

Many thanks! Drew

(Email address removed to prevent every spammer sending you 'special offers'... Skittle 18:45, 13 November 2007 (UTC))—Preceding unsigned comment added by 71.160.50.240 (talk) 17:58, 13 November 2007 (UTC)

I did some digging, but I have yet to find such a list. In fact, I'm having trouble finding the info state-by-state. We could compile a list and create a Wikipedia page with it if we ever find the facts. --Milkbreath 20:12, 13 November 2007 (UTC)

You could start here - just the Atlantic and Gulf coasts. Cheers Geologyguy 20:17, 13 November 2007 (UTC)

Here are the rest. Cheers Geologyguy 20:18, 13 November 2007 (UTC)

It's also traditional to bring up How Long Is the Coast of Britain? with these questions, emphasizing that states like Maryland can see a particularly large coastline versus what might be expected. — Lomn 20:19, 13 November 2007 (UTC)

One source (see the section "Coastline and Shoreline" and note the disclaimer about being "generalized" and made from "small scale maps" -- meaning the figures probably err on the side of being too small)

NOAA has been creating high-quality shoreline datasets recently. These would probably provide the most accurate info on coastline length currently available. I'm not sure if they have finished building the datasets yet, or whether one would need to use something like GIS to get a list of coastline lengths by state. The main webpage for the project is here (NOAA's Coastal Geospatial Data Project). The project includes far more than just coastline data. A page specifically on shoreline data is here (NOAA Medium Resolution Shoreline, GIS Data). More info here. There's tons of pages and info on these NOAA pages, so it might take some effort to figure it out. Still, the best coastline length data would be from there. I looked at the two PDFs linked above and it looks like they use NOAA data, but not quite the most recent data, athough I could be wrong. Pfly 21:10, 13 November 2007 (UTC)

This is a famous introductory problem in fractals. Basically, there is no such thing as an "accurate" estimate of coastline length, as the coastline is self-similar to multiple scales. Any measure of coastline must state what the length of the "meter stick" used to measure is. Otherwise, you get a length that approaches infinity with increasing precision. Two separate books will quote two wildly different figures because of this. SamuelRiv 23:47, 13 November 2007 (UTC)

You could give the length as a function of how close you looked, couldn't you? Also, what if you took the difference in area covered in high tide and low tide, and divided that by the average distance between the high- and low-tide shores? — Daniel 00:34, 14 November 2007 (UTC)

Coastlines are fractal-like and self-similar among some scales. But when measuring at a scale fine enough to be affected by tides, for example, the analogy of coastlines as fractals breaks down. The problem then is one of defining what to measure -- low tide line, high tide? Also, rivers often widen into estuaries and merge with the sea without an obvious demarcation. At the kind of measuring scales common to things like the NOAA projects, these are the problems with coastline measurement, not fractals. Plus, when measuring at this fine scale, differences of coastline definition (where in the tide zone to measure, where to delineate estuary vs. sea, etc) are not going to result in the kind of wildly divergent figures one finds at the larger scales where fractal issues arise. In short, the idea that coastlines are fractals is only true under certain conditions and scales -- coastlines are not fractals in the strict, mathematical, absolute at all scales notion. Like most real world phenomena, mathematical analogies break down at some point, once you look close enough. I suspect the fractal coastline thing has been a little too much of a good introductory example for explaining fractals. But remember the difference between pure mathematics and the real world. Coastline lengths do not "approach infinity" as measurement precision increases. Just go to the beach with a centimeter-long ruler and take a stab at measuring the coastline and you'll find the problem isn't so much that you are measuring something fractal but more that you keep getting your feet wet as the waves wash up; and what time is low tide today anyway? (sorry for being overly critical, the fractal coastline thing is a minor pet peeve of mine!) Pfly 07:35, 14 November 2007 (UTC)

It is fractal to many orders of magnitude. Obviously it is not a perfect fractal. But let's say you were on a rocky coastline, not a sandy one, so tides wouldn't affect things too much. How do you measure into the crevasses of each rock where water gets in with a centimeter ruler? The point is that there's practically no such thing as an "true" coastline measurement. SamuelRiv 13:14, 14 November 2007 (UTC)

I absolutely agree - there is no correct answer to this question - it depends ENTIRELY on how precise your ruler is. The difference between measuring with a 10m ruler and a 10cm ruler is huge - it's not going to be a 1% error or even a 10% error - the length you end up with could easily double when you measure around every little inlet and rock rather than taking 'short cuts' across them. Truly, we should not be encouraging people to look at the results of this or that survey - the correct answer to this question is that the length of a coastline is unknowable. SteveBaker 18:30, 14 November 2007 (UTC)

I understand the inherent problems, but I still don't see the harm in making a list of US states ranked by coastline length based on a measurement system standardized for the nation. Just make sure to point out the uncertainty of the figures and the vagueness of what a coastline is in the first place. That's what is done with lists of rivers by length (or ought to be done!), even though measuring river lengths has the same problems as measuring coastlines. Pfly 22:44, 14 November 2007 (UTC)

User:Milkbreath made the page yesterday List of U.S. states by coastline. Perhaps the participants in this thread may want to add the appropriate caveats and links to fractals. Cheers Geologyguy 22:48, 14 November 2007 (UTC)

Squinty eyes focus better

I notice my vision has changed since I originally got my glasses (I'm near sighted). However, I notice that if I slightly squint my eyes while wearing glasses, my view is much clearer. What's the science behind this? Am I slightly reshaping my cornea by squiting or is less light passing through causing my iris to have a higher "f stop" or something entirely different? --24.249.108.133 18:13, 13 November 2007 (UTC)

First, we have to get this out of the way... This is not a medical diagnosis. Some Wikipedia users feel that any description of the human body is a diagnosis and delete both the question and answer out of fear of someone suing Wikipedia.

It is very common to have extremely clear vision when you first get a new pair of glasses (or contacts) and then it blurs. Being nearsighted, your eye is basically too long for the lens. When you get new glasses, the muscles learn to relax and make it just a tad longer. By squinting, you do two things - you put pressure on the lens and you shorten the length of the eye. Combined, this corrects the nearsightedness problem. Of course, you can't squint hard enough to correct any significant level of nearsightedness - so walking around squinting all the time is not an option to wearing proper corrective lenses. -- kainaw™ 18:19, 13 November 2007 (UTC)

Kainaw, could you please avoid dragging meta-discussions onto the desks? Particularly argumentative misrepresentations of people's motivations.

On another note, that wearing glasses causes vision to worsen is certainly not proved, according to my (actually trained) optician. It could be, but the evidence isn't there yet. (Also, I'd always heard that squinting was more to do with creating a smaller aperture than reshaping the eye, but I'm not so sure about that one). Skittle 18:44, 13 November 2007 (UTC)

I haven't considered it a "meta-discussion". There is no discussion. Some users delete questions and answers without any discussion at all. If you know of a better way to stop the problem, feel free to make a suggestion. I should note that this isn't an issue with one user and one question. I've seen many questions deleted by many users and I've seen no means of controlling the problem. -- kainaw™ 20:26, 13 November 2007 (UTC)

The relevant discussions have been had on the talk page many, many times. When people consider a removal contentious, they tend to discuss it on the talk page and determine consensus. The reasons for deleting medical questions and advice are explained at some length all over the talk page archives, as well as at a user subpage (I forget which one). If you have a problem with a particular deletion, please bring it up on the talk page and explain why you feel that particular deletion was inappropriate. If necessary, you could also place a message on the talkpage of the person who carried out the deletion. If you have further to say on this topic, please take it to the talkpage. Skittle 15:23, 14 November 2007 (UTC)

Squinting also has the effect of making your depth of focus deeper, similar to how a pinhole camera doesn't need a lens. --Mdwyer 20:28, 13 November 2007 (UTC)

I think that's a big part of it, and another part may be that "stopping down" the lens (using only the central part of it) reduces aberrations. Photographic lenses certainly benefit from being stopped down (until diffraction becomes important), and I would expect the relatively simple optical design of the eye to likewise benefit. -- Coneslayer 20:39, 13 November 2007 (UTC)

So in short, the answer is yes, squinting is like increasing your eyes' f-stop (i.e. narrowing their aperture). --Anonymous, 23:37 UTC, November 13, 2007.

AND the change in eye shape, as kainaw mentioned. I think they are both significant. --Mdwyer 03:42, 14 November 2007 (UTC)

So it's a bit of both, eh? Interesting. (BTW, I wasn't seeking medical advice. I was just curious about this phenomenon.) --24.249.108.133 18:14, 14 November 2007 (UTC)

Doot dooo do doo doo. Before the scourge of corrective lenses were foisted upon me, I found that rubbing my eyes would cause the blackboard to swim into focus for a few seconds. This was primarily the deformation of my eye. The amount of deformation needed to change (correct?) focus is incredibly small, as I discovered later when eye surgeons started poking at my eye with ultrasonic probes -- the focus would change in a little donut-shaped zone around the probe. It was a truly strange experience. As a child, I didn't know about the pinhole effect, yet, though, so I didn't get to try that one. It must be a common occurance, though, since teachers are told to watch for their kids rubbing their eyes, as it might be a sign of poor vision. --Mdwyer 01:45, 15 November 2007 (UTC)

Hard Coupled Forces Problem

Imagine a train consisting of N frictionless cars following a locomotive that is accelerating the whole train forward with acceleration of magnitude a. Assume that the first car behind the locomotive has mass M. If the tension in the coupling at the rear of each car is 10% smaller than the tension in the coupling in the front of the car, what is the mass of each sucessive car as a fraction of M? What is the total mass of the cars in terms of M? [Hint: You might find it helpful to know that 1+x^2+x^3+...+x^n=(1-x^n+1)/(1-x).] —Preceding unsigned comment added by 164.107.244.238 (talk) 18:59, 13 November 2007 (UTC)

How is it that you're in a position to give us a hint, yet unable to solve the problem yourself? -- Coneslayer 19:45, 13 November 2007 (UTC)

There's a good chance the hint was provided with the original homework question. — Lomn 20:14, 13 November 2007 (UTC)

It was in the original homework problem and you don't need to be a jerk coneslayer

Nor do you, and Coneslayer posed a fair question. Anyway, as noted above, the Reference Desk will not do your homework for you. If a particular part of the problem is causing difficulty, then perhaps we can assist or advise. — Lomn 20:29, 13 November 2007 (UTC)

(To anyone else who has solved this: it's quite neat, isn't it?) To the OP:We could certainly help you better if you told us what you have attempted and what you are having trouble with. No-one's likely to just post a full solution, and we can't give specific advice without knowing the problems you're having. For example, are you familiar with Newton's second and third laws of motion? Algebraist 20:42, 13 November 2007 (UTC)

When setting the accelerations equal

A1=A2

you can say

F1/M1=F2/M2

(T1-T2)/M=(T2-T3)/(M*x)

T2=.9T1

T3=.9T2=.81T1

so

(1-.9)=(.9-.81)/x

so x=.9

thus each car is ninety percent the mass of the car in front of it.

The total mass

Mt = M(1+x+x^2+...+x^(N-1)) = 10M*(1-.9^N) —Preceding unsigned comment added by 164.107.244.238 (talk) 21:44, 13 November 2007 (UTC)

Well done! Unfortunately, you have made a slight error: your arguments do not apply for the Nth (ie rearmost) car, which is not being pulled from behind. You need to consider this case separately. Algebraist 21:49, 13 November 2007 (UTC)

The condition in the question(T(K+1)=.9T(K)) will never apply to the last car. If it did, every tension would have to be zero. —Preceding unsigned comment added by 164.107.244.238 (talk) 22:33, 13 November 2007 (UTC)

Indeed, that condition must fail at the bacl. Thus the last car experiences a forward pull of TN=0.9^(N-1)T1, and no rearward pull at all. Algebraist 22:37, 13 November 2007 (UTC)

Is there a landmass under Antartica and the Arctic?

If the ice and snow melted, would you see dirt/rock? 64.236.121.129 20:09, 13 November 2007 (UTC)

Antarctica is a continent and thus has a landmass. The Arctic has no such continent. — Lomn 20:14, 13 November 2007 (UTC)

Antartica, yes. Arctic, no. —Preceding unsigned comment added by 134.84.156.80 (talk) 20:15, 13 November 2007 (UTC)

By the way, not even all of Antarctica is "under" any ice; there's a certain amount of bare dirt. Not a lot of it, but some. See Dry Valleys. --Trovatore 20:18, 13 November 2007 (UTC)

So is the Arctic just a big slab(s) of ice floating on water? 64.236.121.129 20:20, 13 November 2007 (UTC)

Generally speaking yes. See Arctic. And just for fun, you might want to read up on USS Nautilus (SSN-571). --LarryMac | Talk 20:27, 13 November 2007 (UTC)

If you look on Google Earth, there is actually nothing at all shown at the North Pole (the Arctic) while there is a large icy landmass at the South Pole, the overall earth imagery must have been taken during the Northern hemisphere summer months as the Arctic virtually all melts away...of course while it's summer at the North Pole, it's winter at the South, hence why the Arctic is not there but there is loads of ice at the Antarctic. GaryReggae 20:42, 13 November 2007 (UTC)

Antarctica is actually (at least) two landmasses. --Carnildo 23:42, 13 November 2007 (UTC)

Geography of Antarctica doesn't say that. On the other hand, Geography of Greenland does say that Greenland, the largest Arctic island, may really be three islands. The thing is that the ice sheet (in both places) is so thick that it's hard to measure what's under it. --Anon, 23:49 UTC, November 13, 2007.

The sub-ice configuration of both Antarctica and Greenland is fairly well known, and has been published in National Geographic atlases for years. One version for Antarctica is here. Cheers Geologyguy 23:58, 13 November 2007 (UTC)

The problem is that the land is pushed down under that mass of ice. If the ice ever melted, the land would tend to 'spring back' - increasing the height of all of the terrain. On the other hand, if that amount of ice were ever to melt, the sea level rise would inundate more of the antarctic continent. Between those two counteracting effects and our relatively poor view through all of that ice, I think it's hard to guess the precise result. SteveBaker 01:45, 14 November 2007 (UTC)

Uplift totally wins. The isostatic rebound at equilibrium would be roughly 35% of the ice thickness, which at places means more than 1 km of uplift. The sea level rise due to melting is only ~70 m. Discounting small islands, both Antarctica and Greendand would be a single land mass after accounting for the uplift. Dragons flight 16:06, 14 November 2007 (UTC)

Wow! That's a lot! I'd have guessed the rebound was only tens of meters...but a whole kilometer of uplift! Amazing! SteveBaker 18:24, 14 November 2007 (UTC)

Values around 300-350 meters of rebound are well documented in the Hudson Bay and Scandinavian Shield areas [3] - I couldn't find a ref quickly but I agree with Dragons Flight that values more than 1000 m are known. Cheers Geologyguy 21:08, 14 November 2007 (UTC)

But that uplift takes place over thousands of years (the Hudson Bay / Scandinavian Shield uplift mentioned is still ongoing since the last ice age), whereas the sea level rise could be much faster. Short term, the sea level rise might be the much more important effect. --169.230.94.28 21:57, 14 November 2007 (UTC)

Well, that's got to depend on how fast the ice vanishes. If it all disappeared overnight, I'd agree - but if it melts slowly, then presumably there would be time for the land to start to 'relax' before the flooding gets bad. SteveBaker 02:06, 15 November 2007 (UTC)

Anyone have a 'sub ice' map of Greenland? I'd be interested in seeing that... --Kurt Shaped Box 02:10, 14 November 2007 (UTC)

Guess what! asked and answered right here! To a degree, at least. Cheers Geologyguy 15:41, 14 November 2007 (UTC)

Very cool. Thanks very much. :) --Kurt Shaped Box 23:05, 14 November 2007 (UTC)

I've created a map of what a fully deglaciated Antarctica would look like after accounting for isostatic rebound and sea level rise: [4] Dragons flight (talk) 04:08, 19 November 2007 (UTC)

Hard disk

Is there an associated change of mass, even at the atomic level, when writing data to a hard disk? In other words, does a full disk drive weight more than an empty one? Rockpocket 20:37, 13 November 2007 (UTC)

I doubt it, an HDD is a sealed unit so nothing can 'get in' to increase the weight. All the read/write heads do is rearrange magnetic particles but those particles have to be there in the first place. GaryReggae 20:39, 13 November 2007 (UTC)

(Off on a tangent...) Hard drives aren't actually sealed; there's always some path that allows a slight flow of air in and out of the head/disk assembly, if only to easily accommodate changing barometric pressure.

Atlant 02:33, 14 November 2007 (UTC)

No, you aren't adding or removing particles. You are flipping existing particles upside down. --Mdwyer 20:48, 13 November 2007 (UTC)

Doubtful, even on a subatomic level. Writing to a hard drive does not involve adding or removing matter, and as far as I'm aware, the magnetic potential energy of a full hard drive does not differ from that of a blank one. --Carnildo 23:45, 13 November 2007 (UTC)

Well (blatantly armchairy speculation ahead, beware), when you write to a disk, you're changing its entropy. A disk with all 0's or all 1's on it has less entropy, and is therefore (I think) somewhat less energetically favorable, than a disk with random patterns of 1's and 0's. A disk with highly ordered patterns of 1's and 0's obviously has less entropy, too. It takes energy to create order, and (if what little I know about thermodynamics is accurate) there's a certain energy cost in creating order / decreasing entropy above and beyond (or perhaps I should say below and beneath) the energy required to actually (in the case of a magnetic disk) induce the tightly-polarized little magnetic fields and stuff.

Now, as we all know, devices with potential energy stored in them weigh slightly more, due to e=mc². So does a device that contains an infinitesimal amount of anti-entropic energy in it therefore weigh an even more infinitesimal (less tesimal?) amount more? (I have no idea, and actually I kind of doubt it, because I think I'm mixing apples and apples here, because I'm not at all certain that entropic randomization of an unstably ordered state is an exothermic reaction; rather, it's an inevitable result of other, exothermic reactions. But anyway. Perhaps someone who knows more about this stuff than I do can set me straight.) —Steve Summit (talk) 03:28, 14 November 2007 (UTC)

I think Information entropy and Thermodynamic entropy are different. --antilived^{T | C | G} 05:07, 14 November 2007 (UTC)

I'm surprised that nobody has mentioned the actual effect utilized to store data in most modern hard disks, giant magnetoresistance. GMR does not involve rearranging particle position per se, and I'd be hesitant to foster weird notions about physics by comparing particle spin with macroscopic orientation. Basically GMR enables a large resistance change in material stacks with differing magnetic orientations because it takes more energy to cause electrons to flow through a magnetic domain if their spin is aligned differently than the magnetic moment ordering of the domain. (see ferromagnetism) Back to the original question, any mass gained or lost is totally negligible; macroscopic object masses fluxuate many orders of magnitude more when a dust particle settles on them. (if that's unsatisfying, then the random accumulation of electrical charge may provide compelling reason not to care much about these infinitesimal mass changes) -- 128.61.20.199 13:19, 14 November 2007 (UTC)

Entropy is not a form of energy. Sure, you need non-thermal energy to reduce entropy, but this energy is converted to heat (which can easily leave the hard drive).

The energy, and thus the mass, of the hard drive is probably higher when there are long runs of all 0's or all 1's, compared to alternating 0's and 1's, due to magnetic repulsion between domains of the same orientation. Icek 18:07, 14 November 2007 (UTC)

As far as I'm aware, hard drives use data encodings designed to eliminate such long runs, because long runs tend to result in read errors. --Carnildo 00:18, 15 November 2007 (UTC)

That's interesting, could you point me to some source of further information? Icek 14:51, 15 November 2007 (UTC)

Check out Run length limited encoding. -- Coneslayer 18:43, 15 November 2007 (UTC)

Antenna length

The radio station WUMB has a page here that suggests a way to make a dipole antenna. They say each branch of the antenna should be 30.5" long to optimize your reception of their 91.9 MHz signal. I've read Dipole antenna, which shows a formula of 468/f(MHz) feet, which leads to an antenna length of 0.5223 inches. Can anybody explain how WUMB arrived at their figure, and how this might be generalized to any radio frequency? Thanks, jeffjon 20:39, 13 November 2007 (UTC)

I think you've made a math error; 468/91.9 = 5.09 feet, which is 61 inches (or 30.5 inches for each half). -- Coneslayer 20:43, 13 November 2007 (UTC)

Hmm. The only thing that's better than a math error, is a math error that's so obvious I can't fathom how I might have made it. Thanks. jeffjon 21:36, 13 November 2007 (UTC)

If you're wondering where this seemingly arbitrary number comes from, it corresponds to a little less than half the wavelength (in air). Exactly half the wavelength gives two quarter wave rods, which provides the best resonant properties. However, the antenna input impedance is slightly reactive if its length is exactly a half wavelength, and adjusting the length down a bit causes the reactive part to vanish. Purely real impedances are very easy to match, and conveniently the real impedance of a slightly-less-than-half-wave dipole is fairly close to 75 Ohms. You can, of course, match an antenna with complex-valued input impedance, but it's additional trouble, could involve this beauty, and easier for hobbyists to just cut the antenna to appropriate length. The main reason you want the antenna impedance matched to the line is so the maximum amount of power can be delivered to whatever is at the other end (a LNA, perhaps) instead of being reflected back. As I hinted at above, the dipole configuration mentioned has a real impedance that results in a very low amount of reflected signal due to antenna mismatch when connected to a 75 Ohm line, less than 2%, actually. -- 68.158.1.192 06:53, 14 November 2007 (UTC)

Simplified Electromagnet Math

I'm doing the grade-school experiement of making electromagnets in my kitchen. I'd like to maximize the strength of the magnet while minimizing the copper. It seems like I should be able to plug in all the variables and get an answer. The math is pretty daunting, however! Is there a simplified way to handle this math:

$H\approx 787\ {\mbox{ampere.turns/meter}}$

For sxample, say I've got a .25" thick nail, and I'm going to run 12VDC through it. Is there a way I can draw a graph of turns vs ohms and figure out where the sweet-spot is? In the above math, is 'meter' actually the square-meter cross-section of the nail?? --Mdwyer 21:08, 13 November 2007 (UTC)

Yeah, this is a pretty gnarly optimization problem. It wouldn't be so difficult if you assumed an ideal voltage source (capable of supplying unlimited current) and a linear magnetic material (that reacts the same to an arbitrarily strong magnetic field and never saturates), but that's totally unrealistic. In practice, you're going to be limited by the internal resistance of the power supply (which limits the amount of current you can draw from it) and the magnetic saturation of the ferromagnetic iron core (which limits the magnetic field, at about the amount you quote). Frankly, I think trial and error would be easier to do than a thorough theoretical analysis, and guaranteed to give realistic results. —Keenan Pepper 23:18, 13 November 2007 (UTC)

The theory is straightforward: magnetic inductance is linearly proportional to the number of turns. Increase turns, and you increase inductance. The resistance encountered per length of wire depends on the wire's gauge (look it up), but will be negligible until you approach many hundreds of turns. Basically, from what I recall, we got about 8 ohms per meter of very thin copper wire in one experiment. Magnetic inductance is also directly proportional to current, so at constant voltage, doubling the number of turns doubles the length but reduces current by half, so your net gain is zero. Basically, in the end resistance is not going to be your problem. SamuelRiv 23:54, 13 November 2007 (UTC)

On this page there's a chart showing that, given a fixed voltage, they got a stronger magnet by using fewer turns of thicker wire, but the magnet had heat problems. That just seems backwards to me, for some reason. I always see magnets use lots of turns of really fine wire, but the page seems to suggest that, given a big enough power supply and a way to handle the heat, you should actually be using fewer turns of thicker wire. Am I still missing something? --Mdwyer 03:55, 14 November 2007 (UTC)

Yes. The magnetic field is proportional to the current flowing through the wire, and the current is limited by the internal resistance of the power supply and the resistance of the wire itself. The way to make a strong magnet is not only to have many turns, but also thick, low-resistance wire, and something to pump a heck of a lot of current through it. The resistive cores of the most powerful magnets in the world look like this: Florida-Bitter coils. See how much metal there is? The only holes are for cooling, because they have to pump dionized water through the magnet incredibly fast to keep it from overheating. —Keenan Pepper 06:13, 14 November 2007 (UTC)

Geez... Given the insane power and cooling requirements for those 35 T magnets, I'd think a high-temperature superconducting coil would be more effective. -- 128.61.20.199 12:47, 14 November 2007 (UTC)

They do use superconducting magnets as well, but the problem with that is that superconductors tend to expel magnetic fields (the Meissner effect), and if the field becomes too strong, the superconductor changes phase into a normal, resistive state. So there is a fundamental limit to the strength of a superconducting magnet, and the strongest magnets are in fact hybrids, with a resistive core inside a larger superconducting magnet. —Keenan Pepper 15:50, 14 November 2007 (UTC)

If I were faced with the design problem you stated, I would want to know the characteristics of the 12 volt DC supply. Is it like a car battery, which could supply hundreds of amps for a short while, or is it like a lab power supply, which might be current limited to a few amps or a fraction of an amp? As for heat problems, how long does the magnet have to stay energized?Many electromagnetic devices are designed for temporary or intermittent operation. As for the wire, magnet wire is made with a thin coating of insulation, so that more turns can be placed closer to the magnetic core than would be the case for, say doorbell wire or wire intended for 120v or 240 v AC service. Joseph Henry in the 1830's created the first truly powerful electromagnets (able to lift thousands of pounds) by using many turns of insulated wire. One thing he did was to place several windings on the same core, in parallel, so that each winding had a low resistance and therefore could carry relatively high current, but the multiple windings added their magnetizing effects. A nail makes a poor electromagnet compared to a u shaped or horseshoe shaped core, where the pole pieces are in proximity. Edison 20:27, 14 November 2007 (UTC)

BINDEEZ

Hi, I need a FULL ingredient list for Bindeez/ aqua dots —Preceding unsigned comment added by 68.193.21.179 (talk) 21:30, 13 November 2007 (UTC)

Since they're not considered food, there's little chance of a publicly-available breakdown of what goes into the product. — Lomn 21:33, 13 November 2007 (UTC)

Pretty sure it would be a trade secret. bibliomaniac15 A straw poll on straw polls 23:21, 13 November 2007 (UTC)

Unless it was patented. But I don't see any obvious patent claims, so maybe it's just a trade secret. --24.147.86.187 01:13, 14 November 2007 (UTC)

If you're interested in the "date rape drug" aspect, see our article on it Bindeez, which describes what chemical was substituted for the one that was supposed to be there. --24.147.86.187 01:13, 14 November 2007 (UTC)

Two identical humans

Two identical human both "weights" 70 kilograms.

They both jump off a high place (simultaneously). One of them is attached to a frictionless pulley and to a 65 kilograms counterweight. Assuming no air resistance, will both of them hit the ground at the same time? Will the one with the counterweight survive the fall if the height is deadly for the other human? 202.168.50.40 23:21, 13 November 2007 (UTC)

More homework, is it? We don't do homework. --Anon, 23:51 UTC, Nov. 13, 2007.

I'd be a bit surprised if none of us do homework. :P —Tamfang 23:59, 13 November 2007 (UTC)

Some of us don't even have homework :-p Someguy1221 00:20, 14 November 2007 (UTC)

But for the OP, this is a very straightforward problem; draw out your force diagrams, and calculate the acceleration (including direction) of everything involved. Someguy1221 00:22, 14 November 2007 (UTC)

And they 'll both die since no air resistance implies no air so they'll asphyxiate. Donald Hosek 00:24, 14 November 2007 (UTC)

When I was designing problem sets (for the one class I taught that required them) I really wanted to kill off a few of our hypothetical astronauts and things like that, just to see if the kids were paying attention. --24.147.86.187 01:16, 14 November 2007 (UTC)

I think they both hit the ground at the same time because for the human with counterweight, his/her mass is effectively only 5 kilograms (70kg - 65kg = 5 kg). But Gallieo says that both masses will fall at the same rate in Vacuum, so they must both hit the ground at the same time regardless of their different mass. 202.168.50.40 03:46, 14 November 2007 (UTC)

Hmm, interesting no one has picked up on the important principle here. Try a reductio ad absurdum thought experiment. Suppose that the second man weighs 65 TONS instead of kilograms. Now would they both hit the ground at the same time, because Galileo said that in a vacuum objects will fall at the same speed? Hmmm…ponder ponder….

The answer is that the second man will fall much more slowly because of the inertia of the 65 kg weight he has to move. Inertia of massive objects is very important in these considerations. For example people think that because objects are weightless in space, you can just pick up a 1 ton block and throw it around like a beach ball. Wrong. The block doesn’t want to move. You would still have to exert a lot of effort to get it going. Try this next time you are down at an ice rink. Get your skates on and put some castor type skates on a grand piano. Now try moving the piano around the rink (get permission from the manager first). You will find that it’s very difficult to start and stop the piano, and to steer it. It will feel like the piano is trying to waltz YOU around the rink. Similarly, while heavy objects will fall at the same speed in a gravitational field (let’s just forget about air resistance, it’s negligible for massive objects), they don’t behave the same way if something is trying to stop them. That’s why a large asteroid will make a huge hole, while a little meteorite won’t. The second man can be said to weigh only 5 kg, and he would feel like he only weighs 5 kgs when he is wearing his counterbalance on the ground. But does that mean he will fall with the same speed as the 70 kg man, because different weights fall at the same speed (a la Galileo)? No, because the second man is, to use the old-fashioned terminology, expending most of his “potential energy” doing the work of lifting the 65 kg weight upwards, while the first man can covert ALL of that potential energy into kinetic energy. And I eat my homework. Myles325a 05:28, 14 November 2007 (UTC)

Geez guys! Can't someone figure this out properly?!?! It's easy!

Once the rope goes tight - there are two forces on the guy - gravity pulling downwards - the tension on the rope pulling upwards. Gravity wins but only just so there is an almighty jerk and the pull on the rope is such as to let the guy fall and the weight start rising - we can all imagine that.

But what's going on? The appeal to Galileo's (fictitious) dropping weights off of the tower of Pisa is utterly incorrect here. Galileo's result is something of a coincidence: The force due to gravity is proportional to the mass of the object (F=mg) - but the accelleration of an object due to gravity is given by F=ma. Because F=ma and F=mg then for an object dropped off of a leaning ancient monument, we're talking about the same force so we can equate those two force formulae: mg=ma and therefore g=a - which is what Galileo noticed. But if the mass you are accellerating isn't the same as the one that gravity is yanking on, then all bets are off...and that's what's happening here.

However, when there are ropes and weights and all sorts of other stuff going on, things are a bit more complicated. Gravity is pulling on both the man and the weight - but in opposite directions (from the point of view of the man) so the total force due to gravity is F=(mass_man - mass_weight).g (proportional to the difference of their masses)- but the accelleration of the pair requires both of them to be accellerated so F=(mass_man + mass_weight).a - the sum of their masses. In this case, we are accellerating 70+65=135kg with the gravitational force due to 70-65=5kg. So, now we're accellerating 135/5 = 27 times slower than the guy with no rope. SteveBaker 02:02, 15 November 2007 (UTC)

Note that only the acceleration is what has been reduced, given that we have no limits on the height of the original fall, save that it is "deadly for the other human" you can still end up with a velocity which would be fatal. If the hight of the fall began at the minimum fatal distance than we would expect our the tethered adventurer to survive, however from a height of X time that minimum deadly height (and here the OP should do his homework, the means to work out a solution were given in some of these replies) the final velocity would be deadly.

Blade metallurgy chart

I've been searching for a chart that shows the properties of differant metals found in blades including iron and titanium alloys. I can't seem to get anything showing many metals or numarical properties that I'm searching for. I would like to know strength hardness and efficiency of hardening on the alloys, not just this can hold an edge better then that. —Preceding unsigned comment added by 71.102.37.149 (talk) 23:41, 13 November 2007 (UTC)

How about this one? http://www.crkt.com/steelfct.html It's missing some- for instance, one of my favorites is 154 CM, and it isn't included because CRKT doesn't use it. DeepSkyFrontier (talk) 23:27, 18 November 2007 (UTC)

November 14

What is a Dye Front?

I'm working on a PCR and I need to divide the distance of the bands that I have by the Dye Front, but our teacher never explained what this looks like. Can someone give an explanation or a diagram? Thanks. M@$+@ Ju ~ ♠ 00:29, 14 November 2007 (UTC)

It is the front edge of the dye as it travels across the gel. this page has a good picture with notation. If you can't see the dye front, it may have left the gel - which means you can't do analysis on it. -- kainaw™ 03:32, 14 November 2007 (UTC)

Thanks very much! M@$+@ Ju ~ ♠ 03:58, 14 November 2007 (UTC)

labrador coat

Is it normal for a labrador puppy to have a thickest fur along its "back"? (My dog-related English speaking is not the best, I hope you understand it anyway). Thanks in advance. 217.129.241.186 00:30, 14 November 2007 (UTC)

My girlfriend worked a couple years in the kennels of a large guide dog school that uses labradors almost exclusively, so she’s something of an expert on labs. According to her, yes, it’s normal. She also says that a lab’s coat changes substantially during its first year (it becomes coarser, sometimes wavier, and is two layers), so what a puppy’s fur looks like is a lot different from what it will look like as an adult. MrRedact 16:41, 14 November 2007 (UTC)

Our Lab/Retriever has much coarser hair down her back than elsewhere - I think it's typical of the breed. I somehow got the idea that it was an adaptation for swimming since the soft fur is what ends up getting wet and the coarse fur stays pretty much above water...but I can't imagine why that's an advantage! I have a book about labs at home - I'll see what it says. SteveBaker 18:19, 14 November 2007 (UTC)

My girlfriend confirms this idea. The soft, downy, oily "undercoat" supposedly keeps the dog warm and dry while swimming. The course, hard, dense "guard coat" is what's mainly on their back. MrRedact 03:42, 16 November 2007 (UTC)

Chemtrail Crazies, why is the bandwagon so big?

In my opinion the whole chemtrail phenomenon is a load of bollox. From what I've seen from online research is that those who believe in chemtrails piece together unrelated elements onto a big illogical ball of *cough*. I don't know of anyone on MCAGCC 29 Palms running around with black lung desise and their hedaches are from working with the Marine Corps or PTSD. What I want to know is how people hear these illogical reffrences to contrails being poison and then they fall for it the second they hear about it? Is it that people in general are just that gullible or the bandwagon effect at it's best, because thats all I can think of. This may seem like a nonsensical rant but I have a friend who claims the government is out to get us all and everyone seems to believe him when he mentions chemtrails. —Preceding unsigned comment added by 71.102.37.149 (talk) 01:52, 14 November 2007 (UTC)

Perhaps the chemtrails page can help you. Delmlsfan 02:27, 14 November 2007 (UTC)

(ec) This comic may provide some insight. I agree that the chemtrails stuff is one of the dumbest conspiracy theories around (on par with the "we never landed on the moon" thing). I can't give you any kind of definite explanation, but I do have some ideas about this kind of behavior. I think the whole thing has to do with Occam's razor. This tells us that a good strategy for finding the best explanation for some observation is to pick the simplest explanation that fits the facts. While this usually works, it does leave the person using this strategy very exploitable (and the human mind is shaped by eons of lying and exploitation). For instance, if my best friend is sleeping with my wife, all he has to do to stop me from figuring it out, is make sure that all clues I might find have a simpler explanation than him sleeping with my wife. Humans seem to have evolved an ability to second guess Occam's razor (which may actually be what we call paranoia) in order to save ourselves from being exploited this way. Part of this happens when an explanation that is more complex involves someone benefiting from us using Occam's razor. Once I realize that there is an alternative hypothesis where my best friend wants me to accept the simplest explanation, that hypothesis gets a lot of attention, and I reject Occam's razor.

I think this is the mechanism that generates conspiracy theories. Al-Qaeda explains 9-11 without leaving anything open. The Apollo program is a perfect, flawless explanation for the moonlanding footage. All the simplest possible explanations work perfectly. Yet the suggestion that somebody wants us to believe the easiest explanation seems impossible to reject. We are so tempted by these theories, because if we weren't we'd be easy to take advantage of. And then there's the confirmation bias. Once the paranoia kicks in, subtlety seems to go out the window. All we're concerned with is finding confirmation, rather than verifying the hypothesis. I'm not sure how to explain that, but I think it has something to do with defense mechanisms lacking the subtlety of everyday mechanisms, since they're only activated situations that warrant defense. risk 02:52, 14 November 2007 (UTC)

Well, there there are a number of reasons why people believe weird things like "chemtrails":

• People don't like to believe that they are easily fooled. So, when they think that they've seen through a "trick" or "conspiracy" it makes them feel smart. Or, if they're being told that they're being tricked or fooled, they may prefer to accept a belief that makes them feel more clever than the people they believe are attempting to trick them. The problem is, they can fool themselves this way, and after that admitting that they have made a mistake by believing foolish claims is even worse than saying that they used to be tricked by someone quite clever.
• People like to feel that they are special. This is related to the previous item. Part of the reason why people believe some weird things is that it makes them feel like one of the "brave few", or some such thing, in that they have some special knowledge, power, purpose, or experience that most others do not have.
• People don't like things they don't understand. Things like vapor trails and epidemiology are difficult because they require an understanding of science, however "planes spread sick juice" is simple. This leads me to the next point.
• It's easy to make claims that are hard to refute, even when those claims are entirely false. Some people tend to accept any claim that is not easily disproved if they like it, and thus fall into the trap of believing things simply because they don't understand why it's probably wrong. For example, "you're an adulterer" is an easy claim to make, but can be impossible to refute even if 100% false.
• People prefer beliefs that reinforce their worldview. If they are the kind of people that want to believe that the government is evil, then "chemtrails" or "9/11 was an inside job" is an appealing idea to them. This appeal becomes more important to them than evidence for or against those ideas. You see this in debates where you show evidence against the conspiracy you're discussing, and rather than show evidence for their conspiracy, they start talking about other conspiracies as though they are evidence that their conspiracy is true.
• People like the idea that someone powerful controls things. Thinking that things occur due to intelligent intervention allows people to believe that things don't just happen randomly, but that there is someone who has control of things, and thus they might also be able to control things. Random is scary, thus it's comforting to think that someone is in charge because then you can feel that things happen for a reason, and perhaps you can understand that reason.
• We tend to see patterns, even in entirely random events. This is related to the previous item, but it also means that people may see all sorts of data, and assume that they're related. They often see the "hits" that fit the "pattern" they're looking for, while ignoring most or all of the misses. Even in entirely random data, patterns will appear if you have enough data. The problem is that some people assume that "there are no coincidences" or "it's too coincidental to be random", therefore believe that any patterns must have some cause.

These reasons, among others, are why people tend to believe weird things for very poor reasons. You can take a look at most weird beliefs and see some combination of those factors at play. This is why we need to be skeptical of claims that are not supported by good science.

That being said, there really aren't that many people on the "chemtrail bandwagon", but the vocal few make it seem like it's a more popular belief than it really is. If you go to a place where people promote the idea, you'll see a disproportional percentage of the population that believe the idea. But start asking around randomly and you'll see how few people actually believe it. Hope that helps! -- HiEv 11:33, 14 November 2007 (UTC)

I would add:

• The truth is neither as interesting nor exciting as the fiction.
• People would love to find an 'easy way out' of whatever bothers them.

This is why things like water fuelled cars continue to be bandied about. It's interesting to hear that some guy with almost zero science training (they always have "almost zero science training") was able to figure this out when all of those smarty-pants scientists couldn't. People like to think that there is a quick fix for global warming just around the corner - and the only thing that's stopping it is 'big oil' suppressing the technology - that makes the whole problem "somebody elses fault".

It's really the same with this one: You're feeling tired and stressed out? Well no wonder! The government are spraying us with all of those tranquilizers! It's much easier to blame "them" than to start taking exercise and eating right. The truth, by comparison, is very boring. Water fuelled cars simply cannot work because of the first law of thermodynamics...jeez - what a boring answer! You scientists are just too dumb to figure out how to get around it like that guy in the garage did...Or: Contrails look different at different times because of weather conditions in the upper atmosphere - and besides, there are contrails all over the USA - how would the politicians doing this manage to avoid breathing them in themselves - and how would the enormous chemical industry needed to produce these chemicals manage to keep it so secret? All of those explanations are just tedious compared to a good old conspiracy theory.

SteveBaker 18:16, 14 November 2007 (UTC)

And I would add:

• People who believe there is something out there to get them live longer than those initially sceptical about any danger.

This said, it might not fill our lungs full of harmful chemicals but it might still cook us (or freeze us depending on the versions): "In 1999 the Intergovernmental Panel on Climate Change (IPCC) calculated that contrails from the world fleet of 12,000 civil airliners contribute as much to global warming as the CO2 their engines pour out as they burn jet fuel." from newscientist Oct 2002 and then "They may look benign, beautiful even, but by reflecting heat rising from the ground, contrails have a small but significant environmental impact." from Nature 2001. Keria 09:08, 15 November 2007 (UTC)

Er - isn't that the wrong way around? There was a story on TV about a guy who collected temperature readings across the whole of the USA on the few days after 9/11 when all air traffic in the USA was grounded. He found that temperatures were HIGHER in the absence of contrails because they are white and increased the earth's albedo - thereby reflecting more sunlight out into space. Certainly they'd also work to reflect heat back towards ground too - but that wasn't the finding. Temperatures were significantly cooler in the few days following 9/11. That's not to say that contrails are good for the environment - they are full of greenhouse gasses - so in the longer term, they are harmful. SteveBaker 13:50, 15 November 2007 (UTC)

Actually, as contradictory as it may seem, you're both kind of right. The fact is, contrails may either warm or cool the planet depending on various conditions including time of year, time of day, contrail optical depth, contrail coverage, and other things. Overall they currently appear to have a small net warming effect over time, but as air traffic increases so will the amount of warming they cause. See Contrail#Contrails and climate for starters and then "Contrails, Cirrus Trends, and Climate" for more detailed information. -- HiEv 15:26, 15 November 2007 (UTC)

Also, I disagree with the claim that "People who believe there is something out there to get them live longer than those initially sceptical about any danger." That is only the case for real threats, if you worry about nonexistent threats you will cause yourself stress and may waste energy or pass over opportunities in an attempt to avoid that "something dangerous" that doesn't exist. People should find a good balance in their skepticism, and rely on scientific consensus for complex issues they are unfamiliar with. Science may be wrong sometimes, but it's right far more often than uninformed opinion is, and science corrects itself over time. -- HiEv 22:03, 15 November 2007 (UTC)

Polystyrene cups

What are the effects from drinking hot beverages out of Polystyrene cups? Zain Ebrahim 09:13, 14 November 2007 (UTC)

Could you please explain what you mean by the question? Do you mean the effects on the person drinking from the cups or the environmental impact from the practice of using these cups?

In the first case, I would have thought there was no difference to the person drinking whether they were using a polystyrene cup or a ceramic one. Polystyrene is generally inert at the temperatures it would be subjected to by this use (if it melted due to the heat or reacted with the fluid then it would not be a suitable material for that use).

The environmental effect on the other hand is more serious, recycling of plastic cups is very limited at best and due to the fact that they are so inert, they sit in landfill sites for hundreds of years and biodegrade incredibly slowly (if at all). Producing plastic cups utilises precious hydrocarbons which could (arguably) be put to better uses than disposable cups and once they are landfilled or incinerated, the hydrocarbons are gone.

There are also several different types of polystyrene cups; the rigid plastic ones you commonly get by water coolers and the foamy (expanded polystyrene) ones hot drinks come in. I assume you mean the latter?:GaryReggae 12:47, 14 November 2007 (UTC)

On the other hand, polystyrene undergoes depolymerization pretty easily. (Note - we need a separate page on depolymerization). So it can be recycled into styrene monomer and reused. Options such as wax-covered paper cups are not nearly so recyclable. Delmlsfan 13:10, 14 November 2007 (UTC)

The critical thing with plastics is not whether they are "recyclable" - it's whether they are "recycled". It's no use having a substance that's easy to recycle only to find that people toss them into landfill anyway. That's the problem with plastic cups of all kinds - they are very rarely collected up and sent to the special recycling center that can turn them back into something useful. Paper cups are not recyclable - but at least when you toss them away, they behave much like a pile of leaves would - and decay down to normal soil nutrients in a matter of months.

Thanks a lot. I was referring to the effects on the person drinking from the cups. And yes, I meant the expanded polystyrene ones. Zain Ebrahim 13:38, 14 November 2007 (UTC)

The polystyrene is pretty inert - it's not going to harm you - the only thing that might be a problem is if the manufacturers use some kind of mold-release agent to make it easier to get the cups out of the injection molding machine. But I seriously doubt they'd use anything problematic. SteveBaker 16:57, 14 November 2007 (UTC)

chemicel contents of ferro titanium

i would like to know the chemical contents of ferro titanium.if the compound has silica 3.64%,iron (fe2o3) 35.25%,titanium(tio2) 48%,mangansese 2.55%,tungsten(w)6.09%,molybdenum 2%,carbon 0.79%,alumina 1.93% and its bulk density is 2.476% what is this material called. —Preceding unsigned comment added by Sammy 203 (talk • contribs) 10:03, 14 November 2007 (UTC)

Help on the Antlion

Please help?

I'd like to know how to go about answering the below mentioned challenge copy 1 vs copy 2 presented to 3-12 year olds

Copy 1.I also challenge you to find a sticky substance in an antlions hole. The antlion is the larvae. The struggles of the ant to get out of the hole cause the dry sand to slide down trapping the ant. The antlion flicks sand up from below with it's pincers until the ant is within reach and it can grab the ant. Anything sticky would cause the sand granules to stick together making the antlions hole useless.

Copy 2. The Antlion Antlions think ants are so delicious, they spend their whole lives building sneaky traps in dry sunny soil for ants to fall into. They even live in these traps - and can dig one in just 15 minutes! They then prepare a gooey, sticky substance called Larvae, and wait for some unlucky ant to fall in! When the gooey Larvae traps the ants in the hole the antlion get very happy and say "yummy, a fine breakfast is served!"

Any help in this regard would be much appreciated

Confused!÷ —Preceding unsigned comment added by 196.36.13.113 (talk) 13:51, 14 November 2007 (UTC)

I can't make heads or tails of this. The whole thing is in non-standard English and makes little sense. Where did it come from? There is nothing sticky in the hole, especially not the larva (larvae is plural) which is the ant lion itself. They don't spend their whole lives catching ants; they only do that as larvae. See Antlion. --Milkbreath 14:55, 14 November 2007 (UTC)

Plasma rockets

If the rocket shoots out plasma instead of fire, will it go faster? 64.236.121.129 18:04, 14 November 2007 (UTC)

Thrust is not dependent on "fire" versus "plasma", but rather the mass and velocity of the expelled material. — Lomn 18:45, 14 November 2007 (UTC)

Plasma would come out faster I think since it's hotter and it's trying to escape to an area of lower pressure. 64.236.121.129 19:59, 14 November 2007 (UTC)

Fire is plasma, you will find many ions in a flame. Graeme Bartlett 20:12, 14 November 2007 (UTC)

I seem to recall a physics teacher I had say that fire was NOT plasma. He could be wrong, but I trust him over a random internet guy. But confirmation on this would be appreciated. 64.236.121.129 15:06, 15 November 2007 (UTC)

Neither criteria you mention (heat, pressure) is meaningful. Heat does not affect mass or velocity, and high-to-low pressure is a function of all rockets. A more useful consideration is the combustion characteristics of your fuels: what provides the most thrust per fuel mass? What byproducts are acceptably non-toxic? What fuels are feasible to store and handle? These are the sort of concerns rocket engineers have. — Lomn 20:38, 14 November 2007 (UTC)

You don't know that heat creates high pressure? You don't know that? The hotter the medium, the greater the pressure. The greater the pressure, the faster the medium will move between the area of high pressure and low pressure. Btw, you contradicted yourself. You said pressure is not meaningful, then you said pressure is a function of all rockets, which means that pressure is certainly meaningful. You are contradicting yourself. 64.236.121.129 15:06, 15 November 2007 (UTC)

I was unclear. I intended to make it clear that "pressure", while essential to the operation of rocketry, has no relevance to a comparison of "fire" versus "plasma", as it is not dependent on those terms. Consequently, it's not meaningful to your original question.

It should also be noted that, in any event, peak thrust isn't terribly relevant to rocketry once you can actually lift off. Rather, it's far better to sustain acceleration for as long as possible, due to acceleration's second-order relationship with position. — Lomn 19:10, 15 November 2007 (UTC)

Just a guess: plasma is an electrically neutral collection of charged particles. You would have to separate the charges and accelerate them out of the rocket. IT seems that since the energy of the escaping particle depends on the variable electrical field and not the fixed chemical energy of combustion, the velocity could be faster. You would have to alternate charges being expelled. —Preceding unsigned comment added by DHeyward (talk • contribs) 06:15, 15 November 2007 (UTC)

Plasma rocket engines are in the works. I can't explain all of the physics behind it, but I saw the someone talking about it on tv so I think that qualifies me to answer your question. From what I understand this particular engine can't generate the thrust needed to get into orbit, but once it is in space its the most fuel efficient way to travel. So I think it can go faster over the long run, with a lower acceleration over a much longer time frame. Man It's So Loud In Here 16:59, 15 November 2007 (UTC)

Gold Detecting

Would gold show up on a metal detector? What kinds of metals show up? —Preceding unsigned comment added by 209.203.103.2 (talk) 18:24, 14 November 2007 (UTC)

Yes. See prospecting and metal detector.--Shantavira|^{feed me} 18:31, 14 November 2007 (UTC)

Our article about eddy currents may interest you.

Atlant 13:29, 15 November 2007 (UTC)

Is it possible for gold to come grow back (I know its not like a tree) but be created again from the place it was taken. If the same elements that created it the first time could create it again, or once it is taken and the place picked dry gold remains gone from that place forever?

Gold is an element, so it can't be created on Earth in useful amounts. Only stars can create gold in large amounts. So once the gold in the Earth is used up, there will never be any more. However, it is possible that there is lots more of it buried deeper inside the Earth, and this could slowly rise to the surface mixed with molten rock (magma), and be deposited on the surface by volcanoes. So, the gold that you collect could be replaced over many thousands or millions of years, but not likely in your lifetime. --Heron (talk) 18:19, 17 November 2007 (UTC)

I had never heard of stars creating gold before, that is very interesting. Just curious, how does that work, where does the gold go?

See stellar nucleosynthesis and Nucleosynthesis#Explosive_nucleosynthesis. The big bang left us with a lot of hydrogen, some helium, and very little else. Essentially all the heavier elements are created either by nuclear fusion (for the lighter ones) or supernova explosions (for the heavier ones, including gold). These explosions conveniently distribute the elements throughout the universe. That's why only younger stars and star systems contain significant amounts of heavier elements - see Metallicity (keep in mind that for astronomers all elements but hydrogen and helium are "metals"). --Stephan Schulz (talk) 16:39, 19 November 2007 (UTC)

Dried Banana Chips

Do dried banana chips still contain high level of potassium similar to fresh bananas?84.66.175.242 19:23, 14 November 2007 (UTC)

Speaking out of pure speculation, I would expect the majority of the potassium to go with the water (i.e, out of the bananas) during the drying process, being that potassium ions, K⁺, dissolve very well in water. shoy (^words _words) 19:59, 14 November 2007 (UTC)

When banana is dried evaporation is used. Only water and volatile substances leave it. Potassium is not volatile at the low temperatures used, so it stays in the banana. But be aware some dried banana has sugar and oil added, so it may not be so healthy as you think. Graeme Bartlett 20:05, 14 November 2007 (UTC)

Why don't they make bullets out of steel instead of lead/tungstun?

What advantages does lead have over steel bullets? 64.236.121.129 20:01, 14 November 2007 (UTC)

Off the top of my head, bullets are swaged when fired. This means softer metal is better. If the bullets were as hard as the barrel, I imagine the barrels wouldn't last very long. Friday (talk) 20:03, 14 November 2007 (UTC)

Yea but, if they were made of steel, you could shoot into water with them. I saw on Myth Busters, they fired bullets into water to see if water was "bullet proof", and the bullets fired from high powered rifles had a tendency to shatter when they hit the water. If they were made out of steel, they wouldn't shatter. 64.236.121.129 20:08, 14 November 2007 (UTC)

Tungsten is not soft, however like lead it is dense, so it will carry more kinetic energy for a given velocity and size. This means it will travel further through air before slowing down. Graeme Bartlett 20:11, 14 November 2007 (UTC)

Some bullets are meant to expand on impact, some are not. However it's possible when hitting something like water than even a full metal jacket bullet might come apart. I imagine this would happen more with smaller ones than bigger ones, but I'm just speculating. Bullets are designed for traveling through air rather than water. You might want something less bullet-like if you're intending to shoot through water. Friday (talk) 20:13, 14 November 2007 (UTC)

Yea, full metal jackets were used, and they shattered on the water. I think steel bullets would have greater armor penetration than lead. The tungsten sounds promising, but I read on the tungsten page that steel with small amounts of tungsten greatly increases its strength. So I think tungsten/steel bullets would be best. Yea bullets are designed at shooting through the air, but if you are trying to kill someone under water, it would be best if your gun's bullets can penetrate that water. 64.236.121.129 20:19, 14 November 2007 (UTC)

Even if the bullet stays together, I can't imagine you'd get much penetration. Water is dense. It's probably not as bad as shooting into sand, but it's not going to be easy to get through it. Friday (talk) 20:24, 14 November 2007 (UTC)

Yea, water is dense, but if the bullet can stay together, it means that at higher velocities, you will indeed get greater penetration. This contrasts with the fact that at higher velocities, a lead bullet will simply be more likely to shatter, which ironically means, a handgun has more penetration into water than a machine gun. 64.236.121.129 20:29, 14 November 2007 (UTC)

Even the handgun bullet doesn't penetrate very well through water. Are you imagining some military application? I can't think of why you'd want your gun to be able to shoot someone underwater. Wait til they come up, and shoot them then. Or, if they're in a submarine, they have torpedoes for that. Friday (talk) 20:33, 14 November 2007 (UTC)

I didn't say handgun bullets penetrate very well through water did I? I said handguns ironically penetrate through water better than high powered rifles because the bullet doesn't shatter. You are arguing strawmen, please don't do that, it's annoying. Yes, I am imagining a military application. I'm talking about guns and bullets, DUH! What application would we be using this for? Lol. You shoot at someone in the water to KILL them! DUH!!! A gun that can shoot at targets in the water AND the air is superior to a gun that is limited to only targets in the air. You can't figure that out? Lol. 64.236.121.129 21:41, 14 November 2007 (UTC)

Please remain civil and do not snap at people trying to answer your questions. Using terms like "duh" and "lol" are not helpful, nor are ad hominem attacks. This is a discussion, not a pure Q&A, so deviations from your original question, which mentioned NOTHING about water, are no more "strawmen" than your sidetracking to the topic of shooting into water is. Thank you.

Not exactly bullets, but they use steel shot in shotguns for waterfowl. The lead shot was poisoning the wetlands. The shot has a sleeve around it to keep it from wrecking the barrel. It doesn't hit as hard or go as far, but who wants lead in their duck dinner? --Milkbreath 20:24, 14 November 2007 (UTC)

Wait that makes no sense dude. Steel should hit harder than lead. 64.236.121.129 20:29, 14 November 2007 (UTC)

No, lead is denser. Steel is harder but when you're looking at retained velocity of the shot, density helps. Friday (talk) 20:31, 14 November 2007 (UTC)

No, mass helps. Newton's 2nd law. Hmm, I do agree that lead would retain velocity better than steel. But I think a steel/tungsten alloy would be best, as is indicated in the tungsten article. 64.236.121.129 21:41, 14 November 2007 (UTC)

For a bullet of a given size, density is effectively mass. A lead bullet masses more than a steel bullet because lead is denser than steel. — Lomn 22:07, 14 November 2007 (UTC)

You are stating information I am already aware of. Please do not state information I am already aware of. State information that I am not aware of. 64.236.121.129 15:07, 15 November 2007 (UTC)

Please remain civil and do not snap at people trying to answer your questions. Thank you.

Supposing steel is superior to lead for firing into water -- why do you want to fire into water? That's a really really really specific consideration with no practical relevance that I can see. Even if you want to shoot a guy who's underwater, doesn't he have to surface? Waiting to shoot him then (with a standard bullet) not only solves the problem of bullet composition but also energy dissipation, aiming despite refraction, and a host of other things. Ultimately, the answer to this "why don't we change..." is the same as several others -- it's a solution in search of a problem.

(ec) I see Friday has raised the same question above. — Lomn 20:34, 14 November 2007 (UTC)

You can't figure out that a gun that can shoot at targets in both the water and the air is better than a gun that can shoot at targets only in the air? You can't figure that out? 64.236.121.129 21:41, 14 November 2007 (UTC)

Please remain civil and do not snap at people trying to answer your questions. Thank you.

Again, it's a solution in search of a problem. BTW the people here are trying to be helpful and answer your questions. There's no need to get testy. It's less obvious that a shooting-through-water gun is useful than a shooting-through-air gun. You still need to be able to see your target and aim at it, for example. Water makes this harder. Friday (talk) 21:45, 14 November 2007 (UTC)

No, I already stated the problem, shooting into the water. I didn't ask you to debate if shooting into the water is a serious problem or not, or if you can't see your enemy in the water, or if your enemy can breath underwater or not. I didn't ask any of that. I originally asked why we don't use steel bullets, and now I'm wondering why we don't use steel/tungsten alloy bullets. I'm not interested in your speculation regarding shooting people in the water. Please stay on topic. Thanks in advance. 64.236.121.129 15:12, 15 November 2007 (UTC)

Please remain civil and do not snap at people trying to answer your questions. Thank you.

BTW, the article sucks but there are Spearguns meant to shoot through water. But, it's quite different than a bullet. Friday (talk) 20:35, 14 November 2007 (UTC)

Yea but a speargun isn't that good at killing targets in the air! Versatility is always better than a gun with limitations. 64.236.121.129 21:41, 14 November 2007 (UTC)

Shooting through walls, for example, is arguably even more useful than shooting through water, depending on the application (in some applications, excessive penetration is specifically not wanted.) Mass of the projectile (as well as hardness) generally help penetration- but it means less ammo that an individual is able to carry. Also note that copper-jacketed lead bullets are cheap to produce. Do you want ammo that penetrates more, if it's significantly more expensive? Like most practical questions, it's about making trade-offs. Friday (talk) 21:52, 14 November 2007 (UTC)

(ec) And so by your argument everyone should be driving SUV's, or even better, tanks around, just in case it needs to go off road into the mud or cross trenches, even though all you do is dropping the kids to and from school and grocery runs? Can you think of actual military or civilian needs to be able to shoot guns in water? Why sacrifice capabilities for its usual purpose for some obscure, not commonly used purpose? --antilived^{T | C | G} 21:56, 14 November 2007 (UTC)

That's a poor analogy because you are comparing military to civilian use. I'm comparing military use with military use. A civilian doesn't need to cross trenches, or drive off road. Military units do. It's a silly analogy.

I can think of many reasons to shoot into water, but we aren't debating the usefulness of that. We are talking about why guns don't shoot steel/tungsten alloy bullets. Sacrifice what capabilities? Be specific. What capabilities will be sacrificed if we use steel/tungsten bullets? 64.236.121.129 15:19, 15 November 2007 (UTC)

Y'know, thinking this through, I realized that at no point has the OP submitted evidence that steel bullets would be preferable for shooting into water. With no evidence of my own, I would surmise that steel's hardness (relative to lead) would in fact make catastrophic fracturing even more likely than that of lead bullets, since shattering is a characteristic of hard things. Anyway, I think it's quite clear that Friday and I aren't idiots. If a gun/bullet system existed that fired at submerged targets just as well as at normal ones, with no meaningful disadvantages, it would be... nice? But let's face it -- there are disadvantages. Given that the last time soldiers were forced to fire rifles at submerged targets as a best-case option was... let's see... never, I'm quite confident in my opinion that this is a dead-end. Defending against frogmen is a possibility, but the concussive force of a small depth charge (that is, a hand grenade) works just as well and is already widely deployed. Our opinions are most certainly not the result of failing to imagine that a magic gun would be cool. — Lomn 22:04, 14 November 2007 (UTC)

True, I didn't submit evidence that steel bullets or even better, steel/tungsten bullets are good at shooting into water. So you can take that as my question. Would they? You obviously don't know, so you can abstain from answering that question. Also, questioning the reasoning behind shooting into the water is irrelevant. We aren't talking about whether it's important or not. I think it is, you don't. Who cares? That's not what we are talking about. That's not even a science question, it would be a military tactics question. So you are addressing points that don't even have to do with science. 64.236.121.129 15:24, 15 November 2007 (UTC)

Well, force comes from mass times acceleration (F=ma), and steel has about 70% the density of lead,[5] so for two bullets that are identical except the lead is replaced with steel, the steel bullet will pack about 70% the punch of the lead bullet (though the fact that it weighs less will cause it to have a bit more acceleration.) This is why depleted uranium, which is extremely dense, is used in some military ammunition. -- HiEv 04:47, 15 November 2007 (UTC)

You also might be interested in supercavitation. In the article it talks about a navy project to use supercavitaing bullets to destroy under water mines. The article also links at least two guns specifically designed to be used underwater. -- dcole 14:06, 15 November 2007 (UTC)

You have the right idea in theory, but there are several reasons why a 100% steel bullet is not a good idea. First, remember that the barrel of your weapon is likely steel as well, so the rifling grooves are not going to "bite" into the bullet as well as they would a lead or copper-jacketed one, since the grooves and the projectile are of about equal hardness. The bullet will be spun up somewhat, but not as much as is usually desirable. Along similar lines, the bullet will also wear on the grooves alot more, thus reducing their ability to bite. This is the reason you use a bronze brush when cleaning a weapon, and not a steel one... the steel brush will erode the inside of the barrel.

Now, that said, note that a common NATO bullet, the M995 Ball, uses a hard core (steel originally, tungsten now), surrounded by lead, and copper jacketed. This is the best of both worlds: soft outer parts for maximum effectiveness in the barrel with minimal erosion, followed by a very hard core which will penetrate body armor up to certain levels. Arakunem^Talk 15:25, 15 November 2007 (UTC)

What about steel/tungsten alloy bullets? 64.236.121.129 16:22, 15 November 2007 (UTC)

It's difficult for me to believe you're doing anything but trolling, when you ask questions and then apparently don't read the answers. As pointed out above, several times: all steel bullets would be hard on the barrel. Guns as commonly exist today wouldn't work well like that. You might want to do something other than forcing a bullet down an undersized rifled barrel if you want an all steel projectile. See swaging and rifling for relevant articles. And please, if you're honestly looking for answers, be a little more productive in how you interact with the people trying to provide you with useful information. Friday (talk) 16:26, 15 November 2007 (UTC)

What part of steel/TUNGSTEN ALLOY bullets was not clear? He addressed 100% steel bullets. I'm asking about steel/tungsten alloy bullets. 64.236.121.129 16:30, 15 November 2007 (UTC)

Steel is an alloy and frequently contains tungsten. So your question is not very meaningful. If you want to know why materials much harder than lead aren't good for bullets, we've told you. Do you have another question? Friday (talk) 16:34, 15 November 2007 (UTC)

Haha, you're mad. You are just fillibustering. Yes, I know steel is an alloy, but not all steel contains tungsten, and he was addressing 100% steel bullets, not steel alloyed with tungsten. My question is about steel/tungsten bullets. You're mad obviously, so why not just abstain from the discussion? You don't have to participate, or follow me around. It can be considered harrassment. 64.236.121.129 16:44, 15 November 2007 (UTC)

I will however answer, as I do see that I was not quite clear on something which (if someone didn't know this) would indicate that 64's followup was perfectly valid. Which is: a steel/tungsten alloy would be better than steel at the penetrating side of things, as tungsten is harder than normal steel, but the flipside would be that the projectile would be much harder on the barrel as a result, including getting less spin out of the rifling. This extra hardness is likely why the M995 ball now uses tungsten as the core over steel. You still need to jacket/coat it with a nice soft lead/copper coating though, so it doesnt rip your barrel up, and so it does get nice bite from the rifling. Arakunem^Talk 16:37, 15 November 2007 (UTC)

Ok what if you have a thin coating of copper or lead, surrounding steel/tungsten alloy? Wouldn't that solve the problem? 64.236.121.129 16:44, 15 November 2007 (UTC)

If it's thicker than the depth of the rifling grooves, then yes. This is essentially the M995 ball we're at now. I wish that article had a good cross-section.... *ponders adding one* :) Arakunem^Talk 16:47, 15 November 2007 (UTC)

Similar to current ammo that's already being used? Sure, I think that would work. Also, barrel wear concerns could be addressed by using a sabot- but this is probably only practical for larger-diameter rounds. Also, sabot ammo is complicated and presumably much more expensive to produce than standard ball ammo. Friday (talk) 16:49, 15 November 2007 (UTC)

Yep, the copper and lead basically act like the sabot, sans the discarding. They're just there to get the round out of the barrel. Soon as they hit the target though, the core goes through, leaving the copper/lead behind. So I guess it does eventually discard! :) —Preceding unsigned comment added by Arakunem (talk • contribs) 17:02, 15 November 2007 (UTC)

Mystery of the freezing water

A few years ago in a trip to Norway early one morning, my friends and I were about to get on the road with our car. The temperature was about 0 degrees celsius or maybe below, and we found that we'd left two glass bottles with some water in the car overnight. Picking up the (sealed) bottles for taking a sip, we saw the water freeze and turn into ice in just an instant. Can anyone please explain this? Gil_mo 21:25, 14 November 2007 (UTC)

Your water was supercooled. Picking it up jarred it and started nucleation. Friday (talk) 21:30, 14 November 2007 (UTC)

Prior to taking a sip, you may also have opened the bottle. This release of built-up pressure allowed the supercooled liquid to form some ordinary Ice-1 whereas, prior to being de-capped, the liquid hadn't yet cooled enough to form one of the more-exotic phases of water ice.

Levitating at the center of the world

There's a well in my backyard. It's lined with a material that makes it almost impervious to heat and pressure. (Pretty amazing stuff!) The well is so deep it actually runs straight through the center of the world and connects up both Antipodes. Yesterday my poor cat Mr. Cuddlypants accidently fell into the well. Am I safe to assume Mr. Cuddlypants is now levitating at the center of the world? Sappysap 22:04, 14 November 2007 (UTC)

Eventually, yes, due to atmospheric drag. Negate that and he'll cycle back and forth from endpoint to endpoint. — Lomn 22:14, 14 November 2007 (UTC)

Should I continue to throw food down the well for him? —Preceding unsigned comment added by Sappysap (talk • contribs) 22:43, 14 November 2007 (UTC)

If you throw food down and hit him, it will kill him. Of course, if there's any water levitating at the center of the well, your cat would have died when he hit it. If there isn't any water, what was the point of the well? — Daniel 22:45, 14 November 2007 (UTC)

What if there's water at the other end - i.e. the antipodal point is somewhere in the ocean? Man... --Kurt Shaped Box 22:53, 14 November 2007 (UTC)

The air at the center of the well (I’m assuming there’s just air in the well) is very dense. That means that the terminal velocity of food thrown down the well would be plenty slow enough to keep the food from killing Mr. Cuddlypants. But there’s no point in throwing food down the well, because he’s already dead from the compression. That’s assuming he didn’t get shredded from friction bouncing against the well liner here and there before the compression got to him. Or die of internal injuries due to all the bouncing around. One way or another, I’m afraid Mr. Cuddlypants is now frolicking with the Invisible Pink Unicorn (blessed be her holy hooves). MrRedact 00:03, 15 November 2007 (UTC)

It's true that there is no gravity at the center of the earth - so with air resistance, kitty will wind up there pretty soon (what with terminal velocity, etc) - but so will everything else that ever fell into the well. Rainwater, for example - so I'd expect there to be an nice section of the well that's full of water and leaves and all sorts of other crud. But maybe we can neglect that too - perhaps there is one of those cute little rustic roofs over each end of the well with a few thousand miles of rope coiled up and a bucket on the end. But I'm also wondering whether there actually would be air down there in the first place? I mean, think about this. With no gravity, any air that's not PRECISELY at the center of the well will be flung gently outwards by centrifugal force due to rotation of the earth. At some distance from the center, gravity and centrifugal force would cancel out - and we'd have normal air pressure from there on out. But I wonder if, in fact, there might be a fairly decent vacuum in the middle of the well? I've been working too long today - so I can't be bothered to figure it out - but I'm sure someone here will. SteveBaker 01:35, 15 November 2007 (UTC)

No, the density of the air in the center would be much greater than at the surface, not less. Just like with water in the ocean, the deeper you go, the greater the pressure, due to the weight of all the air or water above you. With air, the greater pressure (at thermal equilibrium – and remember that the sides of the well are perfectly insulating) corresponds to a greater density. The rate at which the density increases would decrease as you approach the center of the earth, due to the decreasing gravity, but the density itself would be monotonic increasing to the center. Both the rate of density increase and the centripetal force approach zero as you approach the center of the Earth, so I’m not sure off the top of my head if the peak air density would be precisely at the center or slightly above it, but whatever less density there would be at the precise center than at the peak would be very small. MrRedact 02:35, 15 November 2007 (UTC)

A few comments: (1) Except at the ends, the well is in fact full of liquid air. (2) That may be a good thing for your cat, considering that the Coriolis force will make him drag along the well wall as he falls (unless you live on the North or South Pole). He'll appreciate being slowed down. (3) Then again, maybe not: liquid air, being 20% oxygen, is a strong oxidant - explosive feline combustion could become an issue. --mglg_(talk) 03:53, 15 November 2007 (UTC)

There's no such thing as a centrifugal force, right?128.163.224.198 (talk) 23:19, 17 November 2007 (UTC)

With a name like "Mr. Cuddlypants", I suspect it isn't so much that he fell into the well, so much as he jumped. --Mdwyer 01:37, 15 November 2007 (UTC)

Perhaps he became suicidal when the court denied his petition to change his nmame to Mr. Bigglesworth.

Atlant 13:41, 15 November 2007 (UTC)

Thats a great question but considering the mass of the earth will be around any object placed at its center would it really float there in the midle (assuming no debris or water) or would it be attracted by one of the walls? Keria 08:41, 15 November 2007 (UTC)

No - actually there would be a teeny-tiny gravitational force pulling you into the exact center. It's only at the precise center of the earth that the gravity goes to zero - so away from the center there is a tiny amount of gravity. Imagine if you have your back to one side of the wall of the well which (let's say) is two meters in diameter. Behind you is almost a complete hemisphere of the earth - minus one meter. In front of you is a little more than a hemisphere of earth - plus one meter. So the gravitational forces of the two hemispheres ALMOST cancel out - but a one meter wide slice of the earth in front of you is pulling you gently forwards - away from the wall and towards the center of the well. In truth though, that would be an amazingly gentle force - I doubt you'd notice it. SteveBaker 13:41, 15 November 2007 (UTC)

He probably hasn't got there yet; the Earth is about 6000 km in radius. His average terminal velocity is going be pretty low, as the atmosphere would behave like a liquid; it could be lower than 20 m/s. As such, the fall may take him 3 or 4 days. Taking his overshoot into account and hence the harmonic oscillations, it could be up to a month before you could consider him to be stationary at the centre of the Earth. Laïka 14:21, 15 November 2007 (UTC)

Maybe it’s time to play with some actual numbers:

The barometric formula says that the density of the air in the well will equal the density of liquid air at a depth of only about 55km. So the air in the well is all liquid except for about the top 55km of gaseous air! That’s only a rough approximation though, mainly due to the ideal gas law becoming increasingly inaccurate as the density approaches that of liquid air.

I’ll assume Mr. Cuddlypants’ terminal velocity in air at the surface is roughly that of a skydiver in free fall, which is about 54 m/s. Given the equation for terminal velocity, and given that the density of liquid air is about 710 times greater than air at the surface, that means that Mr. Cuddlypants’ terminal velocity at about 55km down is already down to only about 2 m/s. It’s safe to say that Mr. Cuddlypants is nowhere near the center of the earth yet.

Assuming that the earth’s density is uniform (not a great assumption), the gravitational force inside the earth is proportional to the distance from the center of the earth. The radius of the earth is about 6371km. Combining that information with the equation for terminal velocity says that Mr. Cuddlypants' terminal velocity will be down to only 2 cm/s when he's still about 637 m from the center. Mr. Cuddlypants isn’t going to experience any oscillation around the center of the earth, or even overshoot the center of the earth. Rather, his position from the center of the earth will decrease exponentially with time. That is, his motion is overdamped.

Calculating the pressure of the air at the center of the earth would involve doing an integration that I’d have to think about. But the pressure is proportional to the density of the material above it. So given that the earth’s mean density is only about 6.3 times that of liquid air, and given that the pressure at the bottom of the earth’s mantle is about 1.4 Matm, we can estimate that the pressure of the air at the center of the earth is getting up close to 1 million atmospheres. So on the way down, Mr. Cuddlypants’ soft tissues will turn to pulp, his bones will be crushed, and his molecules will undergo all sorts of chemical reactions that wouldn’t normally occur if it weren’t for the extreme pressure. So it really doesn’t even make sense to talk about where Mr. Cuddlypants is, but only to talk about the statistical distribution of the locations of the atoms that used to be a part of Mr. Cuddlypants. Throwing cat food down the well is really a complete waste of cat food. MrRedact 21:14, 15 November 2007 (UTC)

Wonderful, simply wonderful! Oh well, then I guess a rescue mission is out of the question. RIP Cuddlypants. Thanks for a comprehensive answer! SteveBaker 04:47, 16 November 2007 (UTC)

Err, the problem is, if the well really is almost resistant to heat and pressure, the temperature and heat will not be much different from the surface. Thus, the liquid air may only be at the core. Oh, and since more than 70% of the surface of the world is overed with water, the ocean will probably start filling the center of the well, and global sea levels will plummet, but as the slightly increased heat in the well heats the water, you may have yourselves a geyser. Oh, and if the gravitational forces are strong enough or if the currents in the core or magma are strong enough, it may bend or even fracture the well. Oh "well". Hope this helps. Thanks. ~AH1^(TCU) 15:37, 17 November 2007 (UTC)

Err, the air in the well, which indeed is assumed to be at the same temperature throughout as at the surface, is liquid due to the pressure acting on it by the air above it. It’s not liquid due to heat – if you want to liquefy air by varying its temperature, you cool it down, not heat it up.

Oh, and since the volume of a 1m diameter well going through the earth is only 1.0x10⁷ m³, if it got filled with water it would cause the 3.6x10¹⁴ m² of the world that’s covered by water to lower by all of 28 nm. I would hardly call that a "plummet" of global sea levels, as it'd be vastly too small to measure.

If the well's antipodal point happened to be in the middle of an ocean, you could just increase the length of the well by only 0.001% to make the well stick up above the ocean's surface enough to prevent even the largest of waves from spilling into the well. It's a completely negligible increase in the extreme difficulty of building the well. MrRedact (talk) 17:40, 17 November 2007 (UTC)

Why do raw carrots taste like soap?

I've recently began eating raw carrots, peeled, they seem to leave a soapy aftertaste. I've tried fresh carrots from other store and they have the same soapy taste. Is this because of chemicals used by farmers, should I be worried? Or perhaps certain chemicals, fatty acids... gives the taste buds a similar sensation as when one smells a bar of soap? Thanks for any info! --Taktser 22:08, 14 November 2007 (UTC)

I don't get that taste, myself. This may be a stupid question but have you washed them? Friday (talk) 22:23, 14 November 2007 (UTC)

Well, if he/she peeled them him/herself, then washing isn't really a major issue. I don't taste anything soapy though I do think they have an odd texture and smell. It might be one of those things that some people are more sensitive to than others (i.e. like those people who really can't enjoy raw tomatos because they taste more acidic to them than to others). --24.147.86.187 23:02, 14 November 2007 (UTC)

I'm guessing the point of Friday's question was "did you wash them with soap? because then that's your problem, right there...". --Trovatore 01:54, 15 November 2007 (UTC)

I cannot answer your question, as raw carrots taste fine to me. But McDonalds fries taste, to me, distinctly of soap. Yughhh! I cannot eat them. I've never heard this from anyone else, so maybe things do taste different to different people.--Shantavira|^{feed me} 08:25, 15 November 2007 (UTC)

Are you thoroughly rinsing your peeler? 130.88.140.8 11:08, 15 November 2007 (UTC)

Yes, I washed them, then peeled them (washed the peeler as well), then washed them again (and my tap water tastes fine). It's only with raw carrots, if they're cooked/heated in any way they loose the "soapy" taste and become quite sweet. I guess it must be some sensitivity on my tongue then... --Taktser 13:11, 15 November 2007 (UTC)

I believe some people feel that cilantro tastes soapy, and it is in the same family as carrots, Apiaceae. Maybe there's a link there? -- JSBillings 17:55, 15 November 2007 (UTC)

Well, the vexed question remaining: Maybe your soap tastes of carrot? Is it orange, at least? Pallida Mors 22:18, 16 November 2007 (UTC)

Potassium/Rubidium-filled bullets?

How effective would bullets containing small pieces of metallic potassium or rubidium be against human targets? Would it be possible to seal small fragments of the pure alkali metal inside lead bullets which were designed to shatter upon penetration - with the end result being that the target/victim takes a round to the chest, then almost immediately begins to be incinerated from the inside out as the alkali metals react with his own bodily fluids...

I'm mainly thinking of the possible applications with regard to sniper rifles and near-certain one-shot kills. I'm aware that pure alkali metals are expensive. --Kurt Shaped Box 22:44, 14 November 2007 (UTC)

Sounds to me you'd get a similar result much more easily using a bullet containing an explosive charge, or one designed to explode on impact. Exxolon 23:34, 14 November 2007 (UTC)

As far as I know, the alkali metals have not been used in any significant way in warfare. What has been used is white phosphorus. On exposure to air it burns, brightly and with extreme heat. It can be used to generate screening smoke, and it is a potent incendiary agent (it can be used to start fires). More insidiously, it has also been used as an antipersonnel weapon. During both World Wars, white phosphorus shells, grenades, and rockets were used extensively. In more recent decades, the use of white phosphorus against personnel has been restricted by various chemical warfare treaties. (Incidentally, the same treaties would likely forbid the use of alkali metals in bullets intended for use against people.) TenOfAllTrades(talk) 01:10, 15 November 2007 (UTC)

Thanks for your answers guys. What I was really getting at, though was whether it would actually work from a purely technical standpoint - not so much the actual feasibility of using alkali metal rounds in standard warfare. I was thinking more along the lines of your archetypical Jackal-esque 'gentleman sniper' looking for a new, interesting, reputation-enhancing methods of carrying out a hit, rather than the soldier on the battlefield. --Kurt Shaped Box 01:52, 15 November 2007 (UTC)

Perhaps that's how Fatima Blush bought it? Between underwater bullets and alkali bullets, we all seem to be in a James Bond mood here.

Atlant 13:46, 15 November 2007 (UTC)

I recall that lead azide was at one point used in exploding bullets, although, it seems kind of unstable for that kind of application. As far as I know, exploding bullets are specifically banned by the Geneva Accords on Humane Weaponry (although we don't have a real article on that), although that wouldn't really matter your your gentleman sniper. -- dcole 14:17, 15 November 2007 (UTC)

Along the same lines, you might be interested in Georgi Markov, who was supposedly murdered by a ricin-filled pellet. (Delivered by an umbrella, though, not a gun.) -- 20:15, 16 November 2007 (UTC)

Sodium bicarb and ethanol turning black

What might react with baking soda to turn a solution of water and ethanol black? Solution probably also contains ethyl acetate, fusel oils and trace amounts of methanol. Lowerarchy 23:05, 14 November 2007 (UTC)

You don't seem to be getting anyone answering, so I'll take a wild ass guess at it. I seriously doubt the sodium bicarbonate, water or ethanol itself is turning black. However, sodium bicarbonate can act as a base (or an acid). Add to that "abnormal" effects on pH seen in mixtures of organic solvents and water, and it is possible you have a non-pH-neutral solution. There might be some acid/base catalyzed reactions occurring with trace contaminants in the solution. (The fusel oils, perhaps?) You only need a small amount of colored compound to give a very visible effect. That's my inexpert guess, anyway. It sounds like a bit of a witches brew. It can be hard for chemists to understand what is happening in simple reactions. When you have complex (and partially unspecified) mixtures, it becomes even more so. There are so many reactions and chains of reactions that can occur, it's hard to track them all down (especially if you're only seeing trace amounts of products - and you only need trace amounts to get strongly coloured solutions) . -- 20:09, 16 November 2007 (UTC)

November 15

Criticality weapon

Has anyone or any country (more likely) developed a criticality weapon, i.e. ray (radiation) gun versus a bomb (neutron bomb)? Adaptron 06:23, 15 November 2007 (UTC)

I doubt it, there's no easy way of controlling the trajectory of neutrons, so everyone (including yourself) will be radiated, which is not very useful as a gun. --antilived^{T | C | G} 10:28, 15 November 2007 (UTC)

Not true at all. Put the entire thing into a sphere of neutron-reflecting material. Open up one small hole in one end. Bang, a simple, aim-able neutron gun. They use stuff like that in physics all the time—even a reactor's neutrons can be directed in a single direction for use with experiments via this method. Not useful as a weapon, but it isn't about not being able to aim them. --24.147.86.187 04:37, 16 November 2007 (UTC)

Can you carry that around though? It's not very portable... --antilived^{T | C | G} 10:08, 17 November 2007 (UTC)

It could be. There are lots of small neutron sources. The problem is that they are usually one-shot sorts of things. The problem isn't the casing, though—you could basically have a sphere of beryllium surrounded by something else (since beryllium is itself toxic) and that would work (beryllium is an excellent neutron reflector). Again, it'd be a lousy weapon for reasons already stated, but aiming isn't the issue. --24.147.86.187 (talk) 17:08, 17 November 2007 (UTC)

See Particle beam weapon, Directed-energy weapon. The utility of such a gun would be limited; the kind of radiation required to do immediate damage to a person is vast; the largest recorded dose of ionising radiation (180,000 REM) still resulted in the victim surviving 36 hours. If you really want to kill someone with radiation, spiking their sushi with nuclear material or lobbing a dirty bomb at them would be far more effective. Remember that the most destructive elements of the atom/hydrogen/neutron bomb are not the initial radiation burst, but the explosive force and nuclear fallout which results. Laser guns do exist to some extent, but most, like the Active Denial System, are so big that they need to be mounted on the back of a Humvee and all they do is warm up the skin, although Israel shoots down enemy missiles with the giant Tactical High Energy Laser. Laïka 13:49, 15 November 2007 (UTC)

You could do it, though it would probably be heavy (imagine a pulsing TRIGA on wheels) and not be at all useful in the short-term since radiation takes a long time to kill by itself. (Imagine how un-useful it would be to have your enemies know that you had just killed them, but that they'd be alive another day or so. Think we have problems with suicide attackers as it is?) So nobody has tried or bothered to. --24.147.86.187 04:41, 16 November 2007 (UTC)

Emergen-C etc

Welcome to Wikipedia. About your question: please see Vitamin C and Multivitamin and the cited references on those pages. A useful source of information. --JWSchmidt 16:38, 15 November 2007 (UTC)

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis, prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page.

This question has been removed. Per the reference desk guidelines, the reference desk is not an appropriate place to request medical, legal or other professional advice, including any kind of medical diagnosis or prognosis, or treatment recommendations. For such advice, please see a qualified professional. If you don't believe this is such a request, please explain what you meant to ask, either here or on the Reference Desk's talk page. --~~~~

We're not really qualified to offer this sort of advice about your health here at the Reference Desk. If you're looking for guidance on the appropriateness of combining vitamin supplements (particularly in larger doses) you should speak to your pharmacist or physician. TenOfAllTrades(talk) 13:08, 15 November 2007 (UTC)

Discussion moved to talk page. --Milkbreath 14:26, 15 November 2007 (UTC)

There was no request for medical advice. --JWSchmidt 16:38, 15 November 2007 (UTC)

in vitro-fertilisation

what is the latest research or essay about the in vitro-fertilisation in mare? --84.56.215.32 14:02, 15 November 2007 (UTC)

here is one starting point for information on in vitro fertilisation. --JWSchmidt 16:17, 15 November 2007 (UTC)

Human Eating Habits

Which is the correct posture for human beings while they are eating? Sitting or standing and why?

Food or water taken in. How hot or cold must your food/liquids be for the body to efficiently digest it?

196.26.82.146 14:21, 15 November 2007 (UTC)

As temperature goes, everything will equalize to body temperature pretty quickly. As long as you're not causing injury while swallowing, you're fine. Similarly, posture is irrelevant so long as you're swallowing and not choking. — Lomn 14:53, 15 November 2007 (UTC)

You may want to check out peristalsis, which is the process by which food is transported from the mouth to the stomach (i.e. swallowed). Because the food travels down the esophagus via muscle contractions, it makes no difference what position you are in when you swallow. — dcole 17:44, 15 November 2007 (UTC)

I dispute this. One of the reasons waterboarding is so effective is that it is administered in the prone position where gravity can not pull food or liquid from the throat down into the esophagus. As a result it can accumulate and block the trachea or windpipe. 71.100.6.233 01:19, 16 November 2007 (UTC)

Principles of plasmid dna isolation

While isolating the plasmid dna from e coli, wy do we always take new lysis solution every time? —Preceding unsigned comment added by 59.180.126.4 (talk) 15:44, 15 November 2007 (UTC)

Do you have any enzymes in your lysis solution? Some people include lysozyme in the solution they use. --JWSchmidt 16:01, 15 November 2007 (UTC)

I'm not quite sure what you mean by "new" lysis solution. In the standard alkaline lysis procedure for plasmid purification, a strong solution of sodium hydroxide lyses the cells due to its basicity (this is the simple version of the explainations). The lysate is then neutralized by an acidic solution of sodium acetate. If you mean "new" as in "not used", this solution no longer has the pH to lyse cells (not to mention the contamination with the plasmid/other cell components). If you mean "new" as in "not old" (that is "freshly prepared"), the lysis solution needs to have a high pH in order to lyse the cells. If not tightly sealed, carbon dioxide in the atmosphere can dissolve in the solution, forming carbonic acid and dropping the pH. That said, I've successfully used plasmid-prep lysis buffer that was over a year old - you just need to keep it tightly sealed. -- 18:02, 15 November 2007 (UTC) —Preceding unsigned comment added by 128.104.112.105 (talk)

principles of plasmid dna isolation

Whatr is the difference in using isopropanol and ethanol as precipitating agent in plasmid isolation? —Preceding unsigned comment added by 59.180.126.4 (talk) 16:23, 15 November 2007 (UTC)

I may be wrong but, is the answer "do your own homework"? SGGH ^speak! 16:45, 15 November 2007 (UTC)

From the article "Ethanol precipitation": "Isopropanol can be used instead of ethanol; the precipitation efficiency of the isopropanol is higher. However, isopropanol is less volatile than ethanol and needs more time to air-dry in the final step (see below)." - I'll also note that, due to it's lower polarity (higher hydrophobicity), it takes a smaller volume of isopropanol than ethanol to precipitate the (hydrophilic) DNA. -- 18:08, 15 November 2007 (UTC) —Preceding unsigned comment added by 128.104.112.105 (talk)

My protocols guide includes steps for isopropanol precipitation and ethanol precipitation in the DNA/RNA exctraction methods it details. I haven't figured out why yet, but I've always been sure they had a reason for using both. Someguy1221 19:19, 15 November 2007 (UTC)

Because ethanol is more volatile than isopropanol, most isopropanol precipitation steps include a 70% ethanol wash immediately afterwards. This removes residual traces of isopropanol. The residual ethanol than can be easily removed by evaporation (say by leaving the tubes open on the bench). If it's at two different steps in the protocol (i.e. not immediately afterwards), it's probably a volume issue. You only need ~0.7 volumes of Isopropanol, whereas you need ~2.5 volumes of Ethanol. If you have 25 mL of sample in a 50 mL tube, it's simple to do an isopropanol precipitation, whereas to do an ethanol precipitation, you would need to split the sample into multiple tubes first. -- 20:41, 15 November 2007 (UTC) —Preceding unsigned comment added by 128.104.112.105 (talk)

Most probable method of attaining pure fusion?

There are many types of attaining pure fusion, laser fusion, bubble fusion, etc... Which is the most probable to occur? 64.236.121.129 18:35, 15 November 2007 (UTC)

Do you mean in terms of generating electricity? The ITER project looks to be the first one likely to result in a net power generation, even if only experimentally. It uses magnetic confinement fusion as contrasted with inertial confinement fusion, of which laser and (allegedly) bubble fusion are subtypes. — Lomn 18:53, 15 November 2007 (UTC)

I guess for both electricity and as a pure fusion weapon. What would be the most probable for each. 64.236.121.129 18:55, 15 November 2007 (UTC)

I think most people would consider the Teller-Ulam design to be a "pure" fusion weapon. Energy to start the fusion chain reaction has to come from somewhere, and a fission bomb is the most compact and easily-engineered representation of that seed energy. Note in the "Soviet developments" section that the first (non-T-U) Soviet H-bomb's fusion component was 1/4 that of the fission component; their largest test (Tsar Bomba) had a fusion component 30 times the size of the fission seed. — Lomn 19:01, 15 November 2007 (UTC)

But that's not a pure fusion weapon. A pure fusion weapon is usually described as a weapon that doesn't use a fission reaction at all to avoid the nasty radiation fallout. The Tsar bomba was certainly a clean bomb, but not 100% clean. 64.236.121.129 19:04, 15 November 2007 (UTC)

In that case, I doubt anybody is working on a pure fusion weapon. There's simply no currently-plausible means of packing the energy of a fission bomb into a fusion bomb apart from the fission bomb itself. Were I to stray into the realm of science fiction, I'd say that an antimatter-matter reaction might serve the same purpose -- but as soon as it's feasible, I'd just make antimatter bombs instead of fusion bombs. Our article on pure fusion weapons notes that the US is not developing one and has found no credible design after approximately 40 years of research. — Lomn 19:18, 15 November 2007 (UTC)

At the moment the prospects for generating fusion power are pretty dim; it turns out to be just very, very hard. Personally I'd put my bet on laser fusion, because it doesn't suffer from the problem of trying to confine plasma (which is very, very hard) and instead just tries to implode fusion fuel (which—thanks to weapons design—they have a very good idea of how to do already and a lot of experience with, it's "just" a matter of adapting it to the different driver and "just" getting rid of the asymmetries, etc.). Personally I'd be surprised if ITER worked as well as they hope it will, but hopefully I will be wrong! In any case, even if the technical bits are in some way overcome, the economics of it don't work out very well, though if the price of enegy severely rises they could be competitive. But note that I am not at all a plasma physicist or someone with deep scientific knowledge on the subject: these opinions are ones that I have glommed from other scientists and from reading around the non-technical literature. There is a lot of pessism there (except, of course, for the boosters, whose incomes require them to be optimists) and with good reason: they've been chasing after this dream since the 1940s and every advance has been coupled with a greater understanding of the difficulties ahead.

As for a pure fusion weapon—seems doubtful to me. You need to be able to generate many gigabars of pressure on the fusion fuel to get adequate compression before heating it. That's hard to do—there aren't a lot of things that can do that to anything of any size. Fission bombs as drivers make it look easy, but they're already extraordinarily powerful at that. Though the comparison is not exactly perfect, compare how much fusion material is compressed in a hydrogen bomb with a fairly mid-sized (in terms of mass) fission weapon, versus how tiny of a bit of fusion material they can compress using something as massive as the National Ignition Facility (a laser the size of a football stadium). The question is: where are you gonna get the energy from to compress the fusion material? Unless there's some other compact energy source out there, it's not gonna happen. Which is essentially what the US government concluded—if the boys at Los Alamos and Livermore couldn't come up with a pure fusion weapon after working on it for five decades, it's unlikely that it can be done at all with existing technology.

Note that I don't know anything much about bubble fusion, though I don't exactly see any possibilities for weaponization even if it turns out to be valid. --24.147.86.187 (talk) 00:19, 17 November 2007 (UTC)

How do whales drink water?

The amount of salt in the ocean is too much for a human to drink... But can whales drink it? I'm assuming they do... But if that's true, what allows for them to tolerate that much salt, but not humans? 64.236.121.129 18:57, 15 November 2007 (UTC)

• Here's a good writeup: [6]. The short answer is they get most of their water from their prey, don't sweat, and have giant kidneys for excreting salt. --Sean 20:16, 15 November 2007 (UTC)

Interesting, thanks. But I guess this kinda turns into a catch-22. How does their prey (fish for example) get fresh water? 64.236.121.129 20:28, 15 November 2007 (UTC)

Fish, whales, and other marine vertebrates have a relatively simple problem: their kidneys just have to produce urine that is slightly more concentrated than sea water. This ability allows them to extract as much water as they need, since they have an unlimited supply of sea water. The real masters at the urine-concentrating business are desert dwellers like kangaroo rats, some of which don't drink at all. Their urine can be five times as concentrated as human urine, and 14–17 times as concentrated as the rats' body fluids. [7] --mglg_(talk) 20:44, 15 November 2007 (UTC)

(By the way, to help with Google searches, the general term for this topic is Osmoregulation.) Fish have complex salt regulation systems to keep water balance. From the "Gill" article: "Gills' large surface area tends to create a problem for fish seeking to regulate the [[osmolarity] of their internal fluids. Saltwater is less dilute than these internal fluids; as a consequence, saltwater fish lose large quantities of water osmotically through their gills. To regain the water, they drink large amounts of seawater and excrete the salt. Freshwater is more dilute than the internal fluids of fish, however, so freshwater fish gain water osmotically through their gills." (Basically, saltwater fish actively pump salt out of their body, and freshwater fish pee a lot.) External link: [8] -- 21:03, 15 November 2007 (UTC) —Preceding unsigned comment added by 128.104.112.105 (talk)

Zinc restoration

I have some steel items that were zinc plated to keep them from rusting. In some places the zinc has been penetrated and needs repair. Can I simply use a piece of zinc as the anode and the damaged piece as the cathode and replate the steel with a few volts of electricity in an electrolyte acid bath of some kind or do I have to throw the work piece away? 71.100.6.233 21:01, 15 November 2007 (UTC)

it may be plated, or hot dipped galvanized in molten zinc. Unless you know what you are doing, or the items are very cheap you can use a commercial plater to do the job for you. Even with a few holes the zinc should still stop rust. Graeme Bartlett 01:46, 16 November 2007 (UTC)

The cost of a commercial plater is more than triple the cost of the item. Extensive surface area was also damaged from being under rotting oak tree leaves that released tannic acid which leached the zinc. Oh well, I was hoping to proceed with enough information to do things right but in absence of any information I will be forced to experiment with use trial and error. 71.100.6.233 03:23, 16 November 2007 (UTC)

There's some misinformation here. You can plate something with zinc through a redox reaction or electrolysis, as you seem to already know. Basically, you don't want an acid bath, because then all products of the reaction go into making hydrogen gas. Instead, use a solution with zinc ions in it (you can get it from a science supply store. I'm not sure about OTC zinc solutions). Put your material in the solution and have a separate salt solution with an electrode (you can use an electrode with a higher oxidation potential than zinc, in which case you won't need to hook up a battery. Otherwise, use one with a lower oxidation potential), connect the two solutions with a cotton wad dipped in ammonia for a salt bridge, and then connect the two electrodes to a battery to plate your zinc. I may have some of the details wrong, but basically this is your setup. Just make sure you have a zinc solution to put the material to be plated in. SamuelRiv 04:52, 16 November 2007 (UTC)

You can get hold of zinc sulphate from a fertilizer shop, and zinc chloride can be available as a solder flux. The science supply store zinc compounds will probably be many times the cost of the others, you don't need huge purity, it does not matter if you have some cadmium in your plate. -- Graeme Bartlett (talk) 22:07, 16 November 2007 (UTC)

November 16

Theory of Everything Solved?

There's this beautiful pattern called an e8 polytope discovered a century ago that is "the highest finite semiregular figure" known to exist. Well, some surfer dude in Hawaii apparently resolved all 4 forces of nature into a matrix of an E8 design. In a 3-sentence paper entitled "An exceptionally simple Theory of Everything," he states:

"All fields of the standard model and gravity are unified as an E8 principal bundle connection. A non-compact real form of the E8 Lie algebra has G2 and F4 subalgebras which break down to strong su(3), electroweak su(2) x u(1), gravitational so(3,1), the frame-Higgs, and three generations of fermions related by triality. The interactions and dynamics of these 1-form and Grassmann valued parts of an E8 superconnection are described by the curvature and action over a four dimensional base manifold."

From the article I read, these 3 sentences beat out string theory hands down as a workable TOE. I'd like to know what the science desk thinks about surfer-dude as the next Einstein. Sappysap 00:52, 16 November 2007 (UTC)

If it is indeed right it has alot of implications. For example there could not possible by any more smaller particles which would put an end to debate about particles being brocken down into smaller ones for ever and ever ...--Dacium 01:05, 16 November 2007 (UTC)

There is really no shortage of theories - but they have to predict something that we can test experimentally or it's just messing around with math. String theory basically describes the universe fairly well - but it's not testable. Remember - relativity wasn't initially taken at all seriously. It's initial publication made scarcely a ripple in the world of science. Fortunately, it predicted the amount that the light from a distant star would be bent as it passed close to the sun - and an experiment (during a solar eclipse) several years later showed that the amount of deflection was exactly what the theory predicted. Within mere weeks of the experimental results being made public, Einstein and his theory were transformed from almost complete unknowns into superstardom. Einsteins genius lay partly in the fact that even as a purely theoretical physicist, he was none the less able to end each of his papers with suggestions for experiments that could be performed to test them. Without testability, you don't have falsifiability - and a theory without falsifiability is useless to us. That's why string theory is in so much trouble. But it's early days yet - lets see where this goes. SteveBaker 01:10, 16 November 2007 (UTC)

Steve, you're a great all-around science guy, but you're not a great historian! Relativity did make quite a ripple even before 1919 amongst the European (specifically) German physics community (and if you think that is limited, remember that at the time that was the theoretical community of preeminence in such things), and Einstein, despite his quasi-outsider status, was taken up as being extremely significant by none other than Max Planck (who is probably responsible for relativity having been taken seriously at all at the time). Now 1919 did change things, but don't let its overwhelming popular attention make you think it hadn't previously garnished specialist attention! (And yes, Einstein's primary physics work was "purely theoretical", but in his life he spent quite a lot of time dealing with the practical and the empirical. He wasn't a patent clerk specializing in electrical engineering for nothing! And it's not a coincidence that synchronization of clocks by electrical signals plays a major role as a "thought experiment" in his early papers—it was one of the hottest practical technical problems of the day, and dozens of patents on the subject passed his desk while he worked in Bern.) --24.147.86.187 04:34, 16 November 2007 (UTC)

Wikipedia's article on the theory is An Exceptionally Simple Theory of Everything. The theory predicts 20 new particles which have not been observed, so if the particles actually wind up being observed, that would be a major confirmation of the theory. Of course, if the particles are never observed after much effort, then the theory won't look so hot. So the theory is essentially falsifiable, which is a very good thing. MrRedact 01:21, 16 November 2007 (UTC)

Both string theory and this simple theory claim that something nearly magical will happen when the Large Hadron Collider goes into operation. The hope is that the tests in the collider will somehow prove or disprove some of these theories. Luckily, my theory that there is no such thing as gravity is in no way dependent on any tests in the collider. -- kainaw™ 02:20, 16 November 2007 (UTC)

It's a 30 page paper [9], you have only reproduced the 3 sentence abstract. Also, he has the benefit of a PhD in particle physics, even if he hasn't been using it lately, which goes against his Einstein-like talented outsider cred. More significantly than what is mentioned above, this theory should allow the masses of all particles to be calculated from first principles (provided you can handle the 200-dimensional complex calculus of variations which the present paper hasn't even attempted it). It is a highly non-trivial calculation, but it should be possible to determine directly whether this theory is consistent with the families of particles we observe. (The catch is that there may be alternative ways of embedding physics in E8 that would also provide a theory of everything but allow masses to be tuned.) Dragons flight 09:14, 16 November 2007 (UTC)

Einstein was not an outsider. He got an undergraduate physics degree in 1900 and completed a PhD in 1905. "Patent clerk" sounds like unskilled clerical work, but in fact he was a patent examiner, i.e. he evaluated the technical merit of patent applications.

As for the paper, the only sensible thing to do (for those of us unqualified to evaluate it) is to wait and see what the community makes of it. Lee Smolin's endorsement does carry some weight. I'm a bit suspicious of the paper because it's exactly what most physicists would expect a unified theory to look like. People have been looking for exactly this sort of thing for decades; they've even been paying special attention to E8 at least as far back as 1984. This isn't like special relativity, where a conceptual breakthrough was needed; if this is right, then people have known what to look for and where to look for it for decades, and I have to wonder why it took so long to find. But I hope it's right, of course. -- BenRG 15:30, 16 November 2007 (UTC)

The real difference between this and special relativity is that the world wasn't really looking for a theory - they had the Aether and they were quite happy with it...except for the annoying little detail that nobody had been able to find the stuff, even with quite some rather clever experiments. But they weren't generally looking for a new theory - just a clever experiment to prove what they already thought they knew. Einstein had to come up with something from cold, hard thinking. That was incredible. His theory explained why existing UNEXPLAINED experiments came out the way they did - and also made testable predictions.

But this is a lot different. In this case, most physicists firmly believe that there must be a grand unified field theory - we just haven't found it yet. But the universe does seem to work OK without one. Experiments seem to come out the way we expect them to - although it would be nice to fix up the rough spots between relativity and quantum mechanics. It's really just a messy leftover that there isn't just one field that explains the whole thing. So LOTS of people have been looking around for a theory that might fit (and there are many theories out there that sound reasonable). But there isn't really an experimental flaw in our separate field theories that this "fixes" - it doesn't explain any anomalous results we're getting. So we won't know whether this is a breakthrough or not until we have a clear set of predictions and a do-able experiment to test them. Still, the prediction of 20 new particles - some at least of which we should be able to see in equipment we'll someday have running - now that's something you can take to the bank!

-- SteveBaker (talk) 22:00, 16 November 2007 (UTC)

Well, the aether theory was the theory of everything of its day! It was not conservative, it was revolutionary! They had unified all physics through the idea of an electromagnetic aether! They were still looking to figure out exactly how it worked, but it was the TOE of its day! And not finding the stuff was not that huge a deal for most of the physicists; Lorentz came up with some very snappy equations to understand why Michelson-Morley would be null, and those equations turned out to be useful and true even without the aether! The Lorentz contraction was born out of work to preserve the aether, not relativity! Einstein's theory didn't explain Michelson-Morley in a rigorous way—he just thought the idea of an aether was superfluous, but SR doesn't explain why there can't be an aether. --24.147.86.187 (talk) 00:48, 17 November 2007 (UTC)

Special relativity says that there is no preferred frame of reference. Since an Aether would have it's own frame of reference that would certainly be pretty much 'preferred', relativity blew away the idea of the Aether. In fact, later, after General Relativity was dealt with, Einstein worried about the "Newton's bucket" thought experiment (does the water in a rotating bucket push up the sides if it is the only thing in the Universe). He alternated between the view that the presence of all the other matter in the universe would be the only thing to let the bucket know that it was rotating...and the view that in fact he might be prepared to admit the existence of a different kind of Aether that would provide an absolute frame of reference for rotating objects. SteveBaker (talk) 17:29, 17 November 2007 (UTC)

Vanadium

What would be the effects of injecting Vanadium into your bloodstream? --24.58.159.152 02:09, 16 November 2007 (UTC)

See Vanadium for currently known toxicity factors. As far as I know, there are no published studies on the toxicity of injecting it directly into the bloodstream of humans. Even with lab rats, the concern is contamination through ingesting and inhaling vanadium compounds and ions. -- kainaw™ 02:14, 16 November 2007 (UTC)

Vanadium metal is solid at room temperature, so you can't really inject it. The effect of injecting some compound of vanadium would depend on the specific compound. —Keenan Pepper 23:43, 16 November 2007 (UTC)

Burning track in the sky

I've just seen the trail of a burning object in the sky. I'm pretty sure that it was slow, close and scattered enough to exclude the possibility of being a shooting star. What was it? An old satellite entering the atmosphere? Along with that amazing vision I also saw a Renault Clio that my father sold 6 years ago 30 miles away from my hometown here in Portugal, it was a night full of weird happenings. I'm a bit drunk, but I swear that I've seen that. I also swear that I'm not on drugs. Cheers! I hope you help me with the burning object thing. 217.129.241.186 02:51, 16 November 2007 (UTC) (Portugal)

Meteors can travel pretty slowly - distances are hard to judge when you're looking straight up - meteors do break up and scatter. I think you were lucky enough to see a pretty spectacular meteor. "Shooting star" is a little unscientific for the science desk! The correct terminology for these things is messy. When you see them a long way outside the earth's atmosphere they are "Asteroids", when they enter the atmosphere, they are "Meteors" and if they hit the ground - they are "Meteorites"...all exactly the same thing though. SteveBaker 04:16, 16 November 2007 (UTC)

I thought they were called meteoroids before they got here. --Tardis (talk) 20:13, 16 November 2007 (UTC)

Perhap you saw Comet 17P/Holmes? I can't help with the Renault thing ... --LarryMac | Talk 17:04, 16 November 2007 (UTC)

Thanks! I don't really need help with the Renault thing. It was just an amazing coincidence. About the comet, I don't think it was a comet, it was more like a burning thing that lasted for 3 or 4 seconds. 84.90.44.231 (talk) 11:04, 17 November 2007 (UTC)

Hi. Was it by any chance an iridium flare? They look like a very bright star, lasting for a few seconds, moving slowly accross the sky. I've seen a few of those before. Inexperienced people may mistake one for a bright meteor, but I've seen a meteor before so I can tell the diffence. Use this website to help you find iridium flares. Some iridium flares can, in rare cases, be as bright as a cresent moon, but look much smaller. Perhaps that was what you saw? Hope this helps. Thanks. ~AH1^(TCU) 15:20, 17 November 2007 (UTC)

Arenn't the Leonids slow-moving? Delmlsfan (talk) 01:32, 18 November 2007 (UTC)

Can you tell if a movie was made in colour if you watch it on a b/w TV set?

I often hear people say, of black and white movies that have been colorized and then presented on TV: “If you don’t like the idea, then just turn down the colour”. When I was a kid, my family owned a b/w TV, and I always thought that I could tell the difference between a b/w movie and one that had been made in colour, even when I had not seen the movies earlier. The colour movie/TV series looked fuzzy and unclear on a b/w set, and the more saturated the colour was, the fuzzier and murkier it looked. Otoh, the images from b/w movies looked sharp, pure and distinct on the same TV. As this is the Science Desk, I will suggest a line of research. Take an original print of some b/w classic like “Casablanca” and the recent colourized version of said film. Now select a scene from the b/w version and show it on a colour TV set. On another IDENTICAL TV set, show the same scene from the colourized version, but with the colour turned off. Now, do these images look the same? Could an observer who doesn’t know which is which tell one from the other? Has anything like this ever been done, and does anybody else think they can tell the difference between b/w and colourized b/w with the colour turned down. I reckon I can, and I will not watch such shows for that reason. Myles325a 03:07, 16 November 2007 (UTC)

Experiments? Here on the science desk? Heresy!

But seriously. Yes, you can kinda tell. The transmission format for colour TV was intended to be backwards compatible with black and white TV - but it wasn't 100% perfectly correct. The easiest way to see this is in a set of colour bars - the ones that go through all 8 fully saturated colours (Red, Green, Blue, Cyan, Magenta, Yellow, Black, White). In most versions of that pattern, the Green and Magenta bars are adjacent. If you look closely, you'll see a line down between the two...which is utterly not there in the original colour signal - it's an artifact of the way colour is encoded.

No this is an artifact of the color electron guns and pixels spots not lining up perfectly in your TV. Nothing to do with the signal. On a digital display you won't see this at all.--Dacium 05:42, 16 November 2007 (UTC)

(For the technically minded: Colour is encoded as two colour difference signals called 'Y' and 'U' - which is encoded in the phase of the colour carrier signal - which is too high in frequency for a black and white TV to display. However, when the phase abruptly changes by 180 degrees, the instantaneous frequency briefly halves and then the black and white ('V') part of the signal catches a glimpse of the colour carrier and displays either a dark or a light line. A US-style NTSC TV shows either a dark or a light vertical line between some of the colour bars. A UK-style PAL TV shows this as a wobbly vertical line because PAL (which stands for Phase Alternate Line encoding or thereabouts) alternates the phase of the encoding on alternate lines - which accounts for the better colour quality on PAL TV's than NTSC ("Never Twice Same Colour"!).

So, to cut to the chase, the deal is that while a pure colour doesn't mess up the monochrome signal, an abrupt change in colour can. So, in theory, with the right kind of picture (such as a test-card), you could tell the difference between a true monochrome TV signal and a colour TV signal - even on a black and white TV. I guess turning down the colour saturation on a colour signal would produce a similar result.

SteveBaker 04:34, 16 November 2007 (UTC)

You are saying that a change in colour will cause a change in lumanance? Color is sent at completely independent point in time during what is called the color burst. It is a burst of high freqency data that black and white TV's do not expect and cannot read. After the colour burst comes the luminance for the line, which comes at a much slower pace than the colour bust. Which is why TV's are backwards compatible. If you take a colour TV and ignore the color (ie. the colour burst) the picture is EXACTLY the same as a black and white TV. This has nothing to do with what the original poster was asking however. Once a movie is re-made into colour, the luminance values will change compared to the original luminance values of the black and white movie. This makes it look different. Some movies keep the original lumanice in-tact so you cannot tell the difference. If you plug a test card into a B&W TV and I plug one into a colour TV and turn the colour burst off, you will not see any difference 'due to abrupt change in colour'--Dacium 05:34, 16 November 2007 (UTC)

You're completely wrong about the colour burst. All that does is to provide a zero degree phase sample of the colour subcarrier. The actual colour data is transmitted along with the monochrome video - but modulated with a carrier wave that's too high in frequency for monochrome TV's to detect. The remainder of your reply (being predicated on that wrong assumption) is also wrong. Remember that electronics were pretty primitive when colour TV's first appeared and there is no way that they could have delayed the decoded colour information from some packed-up form at the start of the scanline to resynch it with the video. At any rate, if you do as I suggest and take a close look at some colour bars on NTSC or PAL, you'll see the artifact I describe and you will thereby be convinced! SteveBaker (talk) 17:15, 17 November 2007 (UTC)

SteveBaker, from what I understand, NTSC color TV mainly uses YIQ, instead of YUV, with the "I" and "Q" in the color subcarrier, and the "Y" (luminance) is the only channel used by black and white TVs. Also, Myles325a, regarding the original research suggestion, that wouldn't work. In theory, a black and white program should only have changes in Y, with I and Q being neutral. However, colorized images will have a different Y than than un-colorized images, since the Y value affects and is affected by the I and Q values. You would see a difference between a black and white version and a colorized version of the same video because the luminance would be different. Now, a simpler explanation for the difference you saw in color vs. black and white video on your TV could be due to a problem in your TV where some of the I and/or Q signal slopped over into the Y signal, which wouldn't happen with black and white video, since there effectively isn't an I or Q signal. (Note: This is somewhat guesswork on my part. I have some knowledge of colorspaces, but I'm no expert on electronics.) -- HiEv 16:42, 16 November 2007 (UTC)

YIQ and YUV are substantially the same thing - they are both colour difference systems - but with slightly different formulations. You're right though - it's the 'Y' that's the luminance - not the 'V' - my bad! I was getting confused with the HSV colour system where 'V' is luminance. SteveBaker (talk) 17:15, 17 November 2007 (UTC)

Snooker on an old set? And for those of you watching in black and white, the pink is next to the green. Lanfear's Bane | t 11:10, 16 November 2007 (UTC)

In my experience, yes, one can tell the difference between a colour and a black & white film when watching on a b&w set. As the OP said, contrast is higher on a b&w original. I suspect partly that this is because of differences in lighting for filming in b&w and filming in colour - when filming in b&w the cameraman must "think" in terms of contrast, when filming in colour he "thinks" in terms of colour, and adjusts the lighting accordingly. DuncanHill 11:26, 16 November 2007 (UTC)

Solar Energy Radiation

How much energy does the Sun radiate per second? I looked on the Sun page, a few other likely pages, and Google and couldn't find any answer. There does seem to be some info on how much energy Earth picks up from the Sun, but this represents only a fraction of its output. Thanks! --pie4all88 03:30, 16 November 2007 (UTC)

It's 3.846x10²⁶ Joules per second, also known as 3.846x10²⁶ Watts. It's listed in the Sun article, under "Physical characteristics" in the box on the right. MrRedact 03:58, 16 November 2007 (UTC)

Oh! Thanks; my mistake. I didn't realize luminosity referred to this. Thanks again! --pie4all88 06:30, 16 November 2007 (UTC)

It's about 3% more if you also count neutrinos; see proton-proton chain reaction for details. And there's also the solar wind carrying away about 10⁹ kg/s (equivalent to about 10²⁶ W, about a quarter of the mass loss due to electromagnetic and neutrino radiation). Icek 07:58, 16 November 2007 (UTC)

Great! Thanks for the helpful information! ---- pie4all88 (talk) 20:43, 16 November 2007 (UTC)

Chemistry tool

what is the name of this tool used in chemistry Picture [[10]] —Preceding unsigned comment added by 71.98.111.198 (talk) 04:05, 16 November 2007 (UTC)

It's called a retort. —Steve Summit (talk) 04:08, 16 November 2007 (UTC)

...and what you use it for is to boil a liquid in the bottom of the speherical part - so that the vapor rises up to the top. The liquid then condenses onto the colder glass at the top - and runs down into the spout so you can collect it. This can be use for distilling purer liquids from impure. SteveBaker 04:39, 16 November 2007 (UTC)

Nice, errr, retorts, gentlemen. Rockpocket 07:43, 16 November 2007 (UTC)

Genetic Algorithms

Does anyone know how genetic algortihms can be used for multivariable function optimization? I searched on the internet but couldnt find much help. —Preceding unsigned comment added by 202.83.169.98 (talk) 06:43, 16 November 2007 (UTC)

Do you have any constraints on the form/representation of the function? If not then one has to know the optimization problem in order to know which representation might be useful. Icek 08:10, 16 November 2007 (UTC)

Genetic algorithms work in a variety of ways. For example, if you have a problem with a "hilly landscape", meaning that as you approach an optimal solution your results get better, then you can use genetic algorithms as a hill climbing mechanism. To implement this, create a "seed" population with random values for each of the variables. Those values will be each individual's "genes". Then, test all of the population to find the individuals that "score the best" (most closely resemble the solution you're searching for.) Then "breed" your population by mixing the genes of the top performers to create new individuals (replacing the poor performers), perhaps with slight "mutations" (minor changes to the value of a few variables) to some offspring. Then, rinse and repeat the testing and breeding steps, usually many thousands of times, until you either find a solution or have stopped seeing improvements for a number of generations. Please be aware, however, that you may get stuck in a local maximum if there are many hills and the population is too small. Because of that you may want to run the algorithm from scratch a couple of times to see if you get consistent results. Anyways, that's a simplified description of one way to do it, but there are many others, and the best method(s) will depend on your particular problem. -- HiEv 15:34, 16 November 2007 (UTC)

My experience in choosing genetic algorithms to solve problems is that the "fitness test" decides the algorithm. Obviously, you have to model your solution somehow. Then, you have to have a fitness test to decide how well the solution works. It can be very difficult to develop a fitness test that works. Once you like your fitness test, you have a good understanding of how to genetically develop the solutions. -- kainaw™ 15:43, 16 November 2007 (UTC)

Yeah, and you have to be careful with your fitness tests because genetic algorithms are often quite good at finding "creative" ways of taking advantage of the system to get good scores. It may, for example, take advantage of the fact that one solution is more common than others, and simply evolve to give that solution all of the time. -- HiEv 17:02, 16 November 2007 (UTC)

Heehee. For brevity, I left out the example I use each time I teach genetic algorithm class. When I took my first GA class many years ago, we had to choose something simple to solve with a GA. I decided to make a calculator that took two single digit numbers as input and added them together. My fitness test was the absolute distance from the proper sum of the two input numbers. After a mere 100 generations, the little guys all started answering 9. I tried it again - same result. I tried it again - same result. Then, I realized that 9 was the average answer, so answering 9 all the time was an easy way to score well on the fitness test. -- kainaw™ 17:06, 16 November 2007 (UTC)

Guys, guys.....dont mean to sound rude but I asked about optimzation of multivariable functions. Everything you have told me, I already knew. I used GA to optimize the following function, f(x) = 10*sinc(x-10)-3*sinc(x-40)+5*sinc(x-30)+sin(3*x). I had a few problems at first. Like HiEv pointed out, my program got locked into the local maximum. Infact since the initial population was random, the results of my program were random too. I had 1 in a 20 chance of reaching the global. But I managed to fix that, by introducing mutations. Before I was using uniform mutations, but with multi non-uniform mutations, I was able to better my odds. Now the chances of reaching the local maxima are 1 in 10.

But all this was a learning process. My ultimate task is to optimize multi-variable unseen functions. But before that I need to know how to optimize multi-variable functions. And thats where I am stuck. I cant find help anywhere. Would you know of any link or a book I could consult?

Again I really appreciate the time you guys took to write. Thx. — Preceding unsigned comment added by 202.83.169.98 (talk) 09:44, November 17, 2007 (UTC)}

It appears you are asking for a basic hill climbing algorithm. Did you check that article already? -- kainaw™ 14:00, 17 November 2007 (UTC)

Did you try using a larger population like I suggested above? If you explain more specifically how you're attempting to find a solution, what you've tried, and some of your results (number of variables? range of values? population size? how soon until evolution plateaus on average? etc.) then we might be better able to give you some more specific answers. -- HiEv 22:17, 17 November 2007 (UTC)

Heinz Tomato Ketchup

Old style bottles looked like this.

In the old style bottle of tomato ketchup if you did not shake the bottle before use you would get a small amount of runny 'tomato sauce water'. I had always assumed that it was some sort of colloid and that a little separation had occured at the top of the bottle between uses as it sat in my fridge. However, the new bottles are 'upside down' and dispense from the bottom. Normally I would shake the bottle as a matter of routine, however yesterday evening I forgot and was surprised to see some of the aforementioned 'sauce water'. The sauce is stored dispenser side down in my fridge and dispenses from the bottom, unlike the old bottle which had the dispenser on the top and was stored dispenser side up. Am I misunderstanding the process here or does separation occur at the top and bottom of the bottle? (I was adding it to the side of some Tesco bacon and leek quiche for the curious, hardly sophisticated but terribly tasty after a minute in the microwave). Lanfear's Bane | t 11:07, 16 November 2007 (UTC)

Ketchup is a "thixotropic" substance; it has a very high viscosity if you don't exert a stress on it - much higher than water (especially when cold, as it would be straight out of the fridge). In addition to this, ketchup is also much denser than water. As a result, when left to settle, you will get some separation at the top where the liquid floats on top of the sedimentary colloid, but also some at the bottom: because the ketchup is now acting almost like a solid, it can't fit down the the funnel-shaped part near the cap there is a small space between the lid and the ketchup. Excess water will then drip down into this cavity. The same effect wouldn't happen to the same extent with old bottles because the "bottom" of these bottles was practically cylindrical (although it probably would if you held them upside down). Laïka 17:45, 16 November 2007 (UTC)

Here are two articles about the Heinz "Trap Cap" which was introduced back 2000, supposedly to ameliorate the runny liquid problem. I don't know if it's still in use. --LarryMac | Talk 20:23, 16 November 2007 (UTC)

It is, but in my personal experience, it's not very effective (perhaps I'm not screwing the lid on tight enough) - it's a pain getting the slightly slimy water down all over your hand every time you use it! Laïka 01:03, 17 November 2007 (UTC)

A dog in motion

What is the physics law "a dog in motion stays in motion"? --124.254.77.148 12:04, 16 November 2007 (UTC)

Could be a garbled or intentionally tongue-in-cheek version of Newton's first law of motion, which states "An object will remain at rest, or continue to move at a constant velocity, unless a resultant force acts on it". Gandalf61 13:08, 16 November 2007 (UTC)

That would be Newfoundland's first law. --Milkbreath 13:31, 16 November 2007 (UTC)

:-) Of course it should be pointed out that dogs do tend to be acted on by all sorts of external forces. Dogs have their own grand unified field theory in which all forces may basically be resolved into one force that is precisely opposed by the tension in a collar and leash. Also, I have observed that the force of attraction of a cool tiled kitchen floor is generally enough to reduce my dog's velocity to near zero on most summer days! -- SteveBaker (talk) 21:36, 16 November 2007 (UTC)

Universe ending next May?

Somebody above mentioned the Large Hadron Collider scheduled to go into operation next May and my jaw dropped when I read the "Safety concerns and assurances" section. I can think of 3 holes in CERN's argument that the probability "the LHC might trigger one of several theoretical disasters capable of destroying the Earth or even the entire Universe" is extremely small.

1. Though the standard model may predict that "LHC energies are far too low to create black holes," the whole point of the collider is to fill in the gaping holes of the standard model. Not a sound argument.

2. Cosmic rays may be millions of times more energetic, but the plan isn't to send single protons to collide. The bundles sent at blistering speeds will contain thousands of protons.

3. One of the subcontractors for the LHC is called KEK. Which, translates to "LOL" in orcish. [11]

Are these all crackpot concerns? -- Sappysap (talk) 20:21, 16 November 2007 (UTC)

3 is regardless of the scientific issues the LHC will tackle.---- droptone (talk) 20:24, 16 November 2007 (UTC)

When a scientist says "chance is extremely small" - they probably mean 10²⁶:1 against or something. That's truly negligable. However, there is a long history of these concerns. The guys on the Manhatten project had a theoretical concern that the amount of neutrons released by the atom bomb might (at extremely small probability) cause a chain reaction throughout the entire atmosphere of the planet. When I first went to work from college, I worked at Philips Research Labs in the UK - they took me on a tour of the building on my first day - and somewhere at the back of the glass blowing shop, there was an enormous concrete slab set into the floor - a couple of feet thick - on a huge toothed track so that it could be winched back to reveal a stairwell going down under the building into a concrete bunker. The building had been around for a very long time - before the first world war - and it seems that it was being used to test some of Ernest Rutherford's ideas about the nature of the atom. They were doing various primitive atom-smashing experiments - and at they time, they literally didn't know what would come out when you smashed an atom - it could be nothing - or it could end the universe. So they thought carefully about it - built the biggest safety barrier they could reasonably afford and went on to do the experiments. The trouble with cutting edge physics is that there really isn't much point in doing experiments where you already know the answer. To specifically address your three concerns, I'd say that (1) might be of some very minor concern - but any black hole formed would be nano-sized - it would fall to the center of the earth and either evaporate due to Hawkins radiation or sit there eating an atom or two per year until the end of time. (2) and (3) are nonsense. The total energy from a thousand protons going at even a sizeable fraction of the speed of light is simply not going to produce enough cosmic rays to boil an egg - let alone destroy the universe! SteveBaker (talk)

Here's an undergrad physics major's opinion: most of the arguments that "nothing can possibly go wrong" are, as you say, pretty weak. They depend on predictions of certain theories, but the whole point of building the accelerator is to look for new phenomena that disprove the theories! However, one argument is rock solid: the events produced at the LHC will be no different from events already occurring in the atmosphere every day. The only difference is that at the LHC, they will happen in a predetermined place, next to a bunch of big expensive detectors. —Keenan Pepper 00:08, 17 November 2007 (UTC)

Nitpick: it wasn't the neutrons the Manhattan Project scientists were worried about. It was the temperature. Full report available here. --24.147.86.187 (talk) 00:54, 17 November 2007 (UTC)

Nitpick squared: Why is it so hard for everyone to spell Hawking? Over at talk:theory of everything, one contributor consistently spelled it "Hawkings", didn't notice that everyone else spelled it differently, didn't bother to look it up. --Trovatore (talk) 03:52, 17 November 2007 (UTC)

I apologize for the previous display of crackpottery. It was a bit tongue in cheek. I'm sure that reactions in the LHC will be perfectly harmless and will vastly help our understanding of the universe. The problem I have is that this thing is gargantuan in scale with a circumference of 16.5 miles and a pricetag of $3 billion US...the world has never known such a mechanism. We're going to turn it on and find out what happens in a few months. Sometime in the future an even more colossul experiment will come along where its safety (similarly) is predicated on theories trying to prove its own safety. We might not be so lucky then. I don't want to be reading the newspaper and sipping coffee some morning in May just before some physicist yelps "oops" and I get cut off in midthou Sappysap (talk) 03:04, 17 November 2007 (UTC)

TV satellite downlink antenna shape? Datasheet?

On looking at the various footprints for SES Astra TV satellites in their neato flash map here, it's clear that they've done some clever stuff with footprints (for example to include the Canary Islands in footprints otherwise concentrating on mainland western Europe; the footprints aren't the simple bendy-ellipse you'd get from a simple circular-section parabolic antenna. I guess each transponder (of which modern sats appear to have around 15 to 20) has its own retargetable parabolic antenna (retargetable because it seems pretty common to retask a whole satellite to serve a different market) and they build that footprint out of a bunch of simple antenna prints. But I've not managed to find a proper diagram of an actual modern TV comsat to confirm this or show what can (and can't) be done. Is there anywhere I can get a decent datasheet for a modern TV comsat (all I can find are simplistic "energy makes it go!" type stuff)? -- Finlay McWalter | Talk 20:54, 16 November 2007 (UTC)

I think they do steerable beams using Phased array antennas. You might be able to find out more by going directly to the manufacturers of the satellites, like Space Systems/Loral and Boeing. ---- Mdwyer (talk) 22:41, 16 November 2007 (UTC)

Horton GO229 German WW2 All wing fighter

Is there available any dvd or films or pictures on the horton go229 nazi world war 2 jet

Pictures, yes. Horten Ho 229. You can find more through the external links in that article. -- Someguy1221 (talk) 22:10, 16 November 2007 (UTC)

Do Variations in Earth's Gravity Field Cause Variations in Air Pressure?

After reading about Earth's gravity field, and how altitude, latitude, and geological density cause variations in the earth's gravity field (amounting to up-to 0.5% between the pole and the equator), I have a question about the effect of gravity on air pressure. If sea levels can vary by as much as 200 meters due to variations in the strength of gravity, wouldn't the weight of the air column also vary with the strength of gravity to some small but measurable degree? There is mention of how buoyancy causes a decrease in apparent weight, but that's something else altogether.

Okay, it just now occurred to me that maybe there's no air pressure effect because air is compressible and water isn't. In other words, it's not just weight of the air column directly overhead, but the weight of air masses perpendicular to that vertical line. I don't know. Please help. DeepSkyFrontier (talk) 23:30, 16 November 2007 (UTC)

I'm working on finding a proper reference, but I'm pretty sure that statement that topography and geology can produce 200-meter anomalies in sea surface height is wrong - I think it should be more like 2 meters, and that is on a very broad (many kilometers) scale - the bulges would not be noticed on the sea surface. Cheers Geologyguy (talk) 00:29, 17 November 2007 (UTC)

I agree. I've removed the absurd claim of 200 m, but am also having trouble finding a cited number. Given variations like +/- 50 mGal, the sea level anomaly probably should be closer to a few to several meters. Dragons flight (talk) 18:01, 17 November 2007 (UTC)

Hm. I'm really not sure that it's absolutely absurd. It sounds unreasonable, and is certainly more than I know how to expect. But, on the other hand, we know that variations amounting to several meters are caused by mild weather systems, and that the earth's rotation causes about 25 miles of bulging all by itself. If we're comparing the difference between polar sea levels and equatorial sea levels, and correcting for oblateness, why shouldn't a difference of ((9.780equator/9.832pole)-1=) 0.5% be able to cause a 0.5% difference in surface level all on its own? If we take only the water volume into account- about 3,800 m deep on average according to our Ocean article, then 0.5% would come out to about 18 m- and proportionally more for deeper areas of the ocean. But why wouldn't the effect extend into the earth's crust and upper mantle, causing a small upheaval wherever gravity is lower? Not talking about oblateness here, but difference in gravity due to oblateness and the small but complementary effect that is not attributable to the earth's rotation. As for the effect on the strength of gravity, I think variation due to oblateness is almost the only factor here. Earth's gravity map shows an almost insignificant variation according to the Gravity Recovery and Climate Experiment- for instance, high in the upper Mid-Atlantic and low in the Indian ocean. Variations of 80-to-100 milligals from highs to lows- which amounts to less than 1/100th of a percent of the strength of gravity- and thus would amount to less than half a meter of variance if only the ocean depth is taken into account. Again, I doubt that it's the only factor. 200 meters may sound high, but 2 meters sounds so low. Whoever wrote 200 meters wasn't doing us any favors by not explaining what it meant. DeepSkyFrontier (talk) 19:23, 17 November 2007 (UTC)

The question reminded me of the equatorial bulge. --JWSchmidt (talk) 02:52, 17 November 2007 (UTC)

Um, what about undular bores? They are atmospheric waves (thus differentiating in air pressure) shaped by the Earth's gravity. Does that count? They are, however, more related to changes in weather than gravity. They can also be very large. Hope this helps. Thanks. ~AH1^(TCU) 15:12, 17 November 2007 (UTC)

Are undular bores a tidal effect? Something more to ponder. Thanks for bringing it up! DeepSkyFrontier (talk) 19:23, 17 November 2007 (UTC)

There is no question that there are variations in earth's gravity from place to place - and that air pressure is certainly affected by that...that's just basic physics. Each air molecule is feeling a net gravitational force towards the denser regions of the earth's crust - the molecules therefore move in that direction - the increase in the number of molecules arriving at that location will increase the pressure of the air until such time as the force due to the excess air pressure pushing outwards equals the excess gravity pulling inwards. So the air pressure is clearly going to be higher over areas of denser rocks or in the vicinity of large mountains or something.

The degree to which that change is noticable is debatable. Typical gravitational variation is of the order of one ten-thousandth of the force due to gravity. If this reflected a change in air pressure of a similar percentage - it would be quite utterly negligable.

SteveBaker (talk) 16:53, 17 November 2007 (UTC)

For my purpose, "noticeable" should be replaced with "measurable and predictable." If the effect exists, then how strong could it be? I'm with you (SteveBaker, you're a hero of the reference desk. Thank you for weighing in). I'm guessing that it isn't measurable over the noise of the weather. Nonetheless, doesn't the weight of air mass lateral to the air molecules directly above a specific point act laterally to increase the pressure? Not instantaneously- especially with large distances involved- but eventually? What I mean to say is that if a low gravity location had lower air pressure, it would still be surrounded by air at normal pressure due to normal gravity- at least some distance away. Thus, being crowded by higher pressure air would, over time, cause air pressure to be higher in the gravitational sink. This would happen only because air is compressible. For gravity variation in the ocean, the same effect would express as a bulge in sea level since water is almost non-compressible. Anyway- speaking of air pressure- my instinct tells me that this would act to wash away almost all of the pressure difference caused by small local variation in the strength of gravity (again, I mean variation due to density difference, not due to oblateness). At the end of the day, "utterly negligible" would be be about right :) DeepSkyFrontier (talk) 19:23, 17 November 2007 (UTC)

Gravity anomalies due to geology or bathymetry are on the order of (typically) 20-80 mGal as Dragons Flight says. These anomalies can be computed -- not by the GRACE experiment cited above, but by using the average sea surface height as measured to centimeter accuracy from satellite radar altimeters. So the kinds of variations in sea-surface height due to geology and bathymetry are going to be on scales of tens of centimeters. There are definitely also very-long-wavelength (tens to hundreds of km) sea-surface height variations, but they are on the order of a few meters vertically (2-4). here is one link for info about satellite-derived gravity measurements, and here is another. Those measurements were initially used to create predicted bathymetry maps, but as more sea-based bathymetry measurements, and more years of satellite data for better averaging were acquired, these data are now used extensively as first-look interpretation tools for inferring sub-sea geology. Cheers Geologyguy (talk) 21:54, 17 November 2007 (UTC)

November 17

scenario - the best type of telescope?

Hi. Let's say a person wants to buy a telescope. I'll list the scenario. What type of scenario is best for said person? I'm asking for approximate diameter of mirror/lens, type of telescope (eg. Achromatic refractor, Apochromatic refractor, Newtonian reflector, Dobsonian reflector, Schmidt-Cassegrain, etc), approximate weight, approximate cost (CAD), approximate f/ratio, approximate limiting magnitude, etc. I'm not asking for brand name, location of where to buy, exact price, exact objective diametre, exact parameters, etc. Here's the scenario. Said person:

• is not buying from a department store
• lives in a light polluted reigon
• will be observing from said person's residence
• lives in an area of approximately 2000 people per sq km
• lives in an area where most winter, and a few autumn and spring nights drop below freezing
• is especially interested in planets, comets, and deep-sky objects

Said telescope:

• should not cost over $300 (CAD, excl. tax)
• should include eyepieces and accesories
• should not weigh more than 20 pounds
• should be able to be easily carried through a 1-m wide corridor
• should not take up more than 1 sq m or total area space
• should not break on the first use
• should perform better than a 2 in department store refractor
• should be easy to have finderscope accurately positioned
• should not take more than 30 mins to locate a mag. 5 object
• should not require constant maintenance
• should not completely fog up under 0C in less than 5 mins
• should not take more than 10 mins to set up or carry back to its storage area
• should have a warranty lasting more than 3 months
• should not have an owners' manual of more than 25 pg long
• should not completely go out of focus due to a very slight accidental tap
• should not have an extremely wobbly mount
• should have eyepieces capable of at least 70x and not go over its maximum theoretical magnifacation
• should have a limiting magnitude of better than mag. +8
• should not capture too much light pollution when observing within 60 degrees of a streetlight
• should have a maximum FOV of at least 10 arcmins
• should actually work.

Sorry if my parameters wer a little too exact, but I only want vauge answers. If nessecary, please indicate some good brands. Thanks. ~AH1^(TCU) 00:11, 17 November 2007 (UTC)

You've mixed in technical specifications with some rather arbitrary looking manufacturing specifications (length of manual? come on, is that really going to be your limiting factor between one and another? my washing machine has a longer manual than that and that doesn't mean very much—I only flip through it when something is not operating right). I really doubt anyone on here is going to have that much knowledge about specific telescopes to be helpful; you'd probably get a better response by just saying you're interested in a relatively cheap telescope that isn't too hard to use and will let you see some neat stuff in the sky. You also seem to have very exacting standards for a telescope that you don't want to cost very much at all—that's usually not how things work with precision instruments. Even if you came to me with questions like that about something I knew more about (e.g. buying a new laptop) it would take me AGES to check through the stats of any given option to see if it really lined up with your desire. Better to ask more generally and then check through the stats yourself on whatever people offer up. --24.147.86.187 (talk) 19:18, 17 November 2007 (UTC)

Here's my advice. Get the biggest brand name (not Walmart Christmas special) reflecting telescope you can afford and most of the other requirements on your list will be under control. I suggest you buy a Celestron AstroMaster 130EQ. Most of your requirements are within 10% or so. Also, download Celestia for free to provide some 3D context to the objects you're looking at. Good luck! DeepSkyFrontier (talk) 19:39, 17 November 2007 (UTC)

Hi. Um, the problem with a 130mm reflector is, does it actually cost less than $300 dollars, including mount and eyepieces? Also, the scenario takes place in a light-polluted reigon. It would also be rather heavy, and might be hard to fit in the back of a car (eg. from the store). What is the best aperture for this scenario? I'd think that approximately 100mm would be suitable, what do you think? How did you decide on Celestron, 130mm, and reflector? Is it because you own one yourself? Thanks. ~AH1^(TCU) 22:57, 17 November 2007 (UTC)

My advice is just advice. It costs slightly more than $300, but you may be able to find one for closer to $300 if you shop creatively. It weighs 24 lbs. You can use the internet to find more information on it if you have any other questions. If you are in a light polluted area you should get the strongest, widest aperture telescope possible. It's a matter of contrast. Dim stars seen through an orange haze is worse than bright stars seen through the same haze. I personally wouldn't buy this telescope, but I certainly wouldn't buy anything smaller either. Celestron makes good equipment. I hope you find what you're looking for. DeepSkyFrontier (talk) 08:12, 18 November 2007 (UTC)

CGPM news; possible redefinition of the kilogram

Does anyone know what's going on at the 23rd CGPM? It's supposed to be almost over, but I can't find any mention of news from it, or even something like a conference program. They could be redefining the kilogram as we speak! The BIPM web page is hardly helpful.

BTW, shouldn't this guy be wearing a mask or something? What if he sneezes on that thing? —Keenan Pepper 00:51, 17 November 2007 (UTC)

P.S. Which is a more perfect sphere, that thing, or one of the Gravity Probe B gyros? —Keenan Pepper 00:52, 17 November 2007 (UTC)

The kilogram is overdue for change. Firstly, it's the only remaining SI unit that's defined in terms of a single physical object - and that's not a good thing. Secondly, by comparing the standard kilogram to various copies, it's becoming obvious that the standard kilogram is slowly changing mass (by 30 micrograms already - that's A LOT!). The ideal definition should be in terms of some exact number of some particular kind of atom. Defining the kilogram this way is definitely something that needs to be done because if the mass of the standard kilogram is changing - and since that is the very definition of the kilogram - what it means is that all of our scientific measurements are slowly becoming less and less accurate - and that's a disaster! However, I hope that what the Australians are doing is building a new kilogram that WILL NOT be "the" standard - but merely an example of a mass containing the right number of atoms of whatever atom is finally chosen (probably silicon - and that photo looks a lot like a ball of silicon and it's about the right size to be a kilogram of silicon). This can then be compared to the mass of the present standard kilogram - simply to verify that the new and old standards agree to within some reasonable precision. SteveBaker (talk) 16:31, 17 November 2007 (UTC)

The problem with defining the kilogram as a number of atoms is that it's very hard to accurately measure the number of atoms in a solid. According to Wikipedia, the uncertainty in Avogadro's number is around 50 parts per billion, which is larger than the total drift in the standard kilogram (30 parts per billion). Even if the kilogram does get redefined in this way, I think it will be quite some time before synthesis of a kilogram from the new definition will be easier or more accurate than copying a physical artifact. -- BenRG (talk) 18:05, 17 November 2007 (UTC)

Defining X number of Y atoms to be exactly 1 kg is equivalent to creating an exact redefinition of Avogadro's number. And the point is not that it should be easy. Surely defining a second as a number of vibrations of a specific atom, or defining a meter as the distance light travels in specific fraction of a second are not easy definitions. Nearly all clocks and meter sticks are still based on comparisons to physical objects that were created based on those definitions. The point is to have a fixed definition that can be reproduced by indepedent experts at different places and times in order to ensure that the standards of measurement don't accidently change over time. In other words it is a way to ensure that a standard kilogram produced in 2010 means the same mass as one produced in 3010. Of course this is dependent on someone showing they can produce a kilogram standard today from a number of atoms definition (or similar) with an accuracy comparable to the present uncertainty in the standard. Dragons flight (talk) 18:54, 17 November 2007 (UTC)

But the point is that right now, the very definition of the kilogram is a moving target - this year, a kilogram is a different number of silicon atoms than it was last year. A definition of a firm number for Avagadro's number (and by implication defining the mass of the kilogram) creates a standard that is absolutely firm and unchanging for all time. Our ability to make a mass exactly equal to one kilogram becomes difficult - but we'll know exactly what the standard. I'd also argue with the 30 migrograms error - in truth, we have no idea what the error is. The only way to see if the kilogram standard is changing mass is to compare it to replica standard masses. But what if all of the kilogram standards are in error? Maybe the error in the replica masses is 100ug - but the official standard is 130ug off? How would we know? SteveBaker (talk) 20:23, 17 November 2007 (UTC)

Thanks, but none of your responses really answer my question. I already knew all that stuff. I just want to know what actually happened at the recent CGPM. —Keenan Pepper 20:12, 17 November 2007 (UTC)

Some very old technology

Heiss Patent Amalgamator

Does anyone know anything about this device: the "Heiss patent amalgamator for saving fine gold from sand, gravel and pulp." I can find nothing on line. Click through on picture if you want more context. - Jmabel | Talk 04:37, 17 November 2007 (UTC)

Also, if someone is expert on technology from this era (circa 1900), there may well be other pictures on Commons:Category:Seattle and the Orient you could help describe better: there are several hardware store interiors, etc., with many objects I could not identify. - Jmabel | Talk 04:42, 17 November 2007 (UTC)

Not an expert, but know a little about mining. The picture appears to be a machine used in the amalgam process of gold recovery. Finely crushed ore and mercury would be mixed together. The mercury forms an amalgam with any gold present, and the amalgam can be readily seperated from the tailings (waste material). Gold is then recovered from the amalgam by heating. DuncanHill (talk) 04:50, 17 November 2007 (UTC)

There were dozens and dozens of different designs for things like this in the early 20th century. I'm not sure that this one is anything special—it looks pretty standard. If you search for "amagamator gold sand" on Google patents you can find dozens of similar devices. --24.147.86.187 (talk) 05:29, 17 November 2007 (UTC)

physics

What happens to its Kinetic Energy when a freely falling object stops on reaching the ground ?

A small percentage is converted to sound energy, smaller percentage converted into heat energy, the rest is transffered into the earth.220.237.156.78 (talk) 09:45, 17 November 2007 (UTC)

Way wrong.

This is a completely inelastic collision, so the speed imparted to the Earth is given by the formula at Inelastic collision#Equations of Motion. To avoid subscripts say v is the object's initial speed and m is its mass; M is the Earth's mass and V is the speed imparted to the Earth by the collision, its initial speed being 0. Then we have mv+0 = (M+m)V, but since the object's mass is negligible compared to the Earth's, we can replace M+m by just M, and we have mv = MV. Therefore (mv)² = (MV)² and therefore MV² = (MV)²/M = (mv)²/M = (mv²)(m/M); you'll see in a moment why I did this.

Now, the initial kinetic energy of the object is mv²/2 and that of the Earth and object afterwards is (M+m)V²/2, which again we can simplify to MV²/2... but by the above calculation, this equals (mv²/2)(m/M), or m/M times the initial kinetic energy. In other words, only about m/M of the object's initial kinetic energy -- a tiny, tiny fraction -- becomes kinetic energy of the Earth.

In fact almost all of it goes into heat. --Anon, 10:10 UTC, November 17, 2007.

In addition, if the object breaks, then some of the kinetic energy is converted into surface energy, as a result of breaking chemical bonds within the object. Gandalf61 (talk) 10:58, 17 November 2007 (UTC)

Treating the Earth a single rigid body is not appropriate. Assuming nothing breaks at impact (i.e. you don't break the object, or leave a hole in the dirt), then the largest fraction will be acoustic energy in the Earth (e.g. pressure waves eminating from the point of impact). This eventually becomes heat, but not immediately. As the first anon said, most of the energy goes into the Earth. Dragons flight (talk) 17:41, 17 November 2007 (UTC)

The first anon contrasted "converted to sound energy" and "converted into heat energy" on the one hand with "transffered into the earth" on the other. Since the original question was about kinetic energy, it seems clear to me that "transferred into the Earth" meant that it becomes kinetic energy of the Earth. On that basis the original answer was wrong and that's how I was responding to it.

Also, since we're talking about an object that falls to the ground and stops, we are talking about either the ground or the object "breaking", or more generally deforming, on impact. I think most of the energy will go into that, which means it then goes directly into heat; but I must admit to not knowing how to calculate the amount that goes into sound waves within the Earth.

--Anon, 06:28 UTC, November 18, 2007.

Virtually all of the recoverable deformation of the ground goes into pressure waves. Normally you would not refer to something as "breaking" unless the deformation is permanent. The fact that something is briefly compressed is not the same as it being broken. Dragons flight (talk) 07:35, 18 November 2007 (UTC)

Bear in mind that all energy is conserved, thus the landing object, say a ball, will undergo all the opposite effects of a falling ball. For example, the falling ball will tend to

A)Speed Up, B)Cool down, C)Lose Pressure and D) Expand.

We know this because the ball, when it lands, will tend to

A)Slow Down and Stop, B)Warm Up, C) Compress and D) Compact - opposite of 'Expand'.

In this way, all energy is conserved. Interestingly, the Cosmos seems to be falling, judging solely from the effects it's undergoing. Peter Lamont (November 17)

This makes no sense to me. Why do you think a "landing" object will "undergo all the opposite effects" to a "falling" object? "Falling" and "landing" are not time-reversals of each other. Also, it's obvious that falling objects speed up because of gravity, but the other three effects are news to me. What makes falling objects cool down? The answer is, they don't usually cool down, and if they do it's caused by some other effect like evaporation. This response is incorrect and I urge the original asker to ignore it. —Keenan Pepper 20:25, 17 November 2007 (UTC)

enerhiya

ano poh ba ang enerhiya? —Preceding unsigned comment added by 124.106.204.177 (talk • contribs) 10:15, 17 November 2007

ano ba yun. hydnjo talk 15:34, 17 November 2007 (UTC)

What language is this? —Keenan Pepper 20:31, 17 November 2007 (UTC)

I'm pretty sure it's Tagalog. MrRedact (talk) 20:55, 17 November 2007 (UTC)

GAA triplet repeats

Is it correct to wikilink "GAA triplet repeats" (found in Friedreich's ataxia) to Trinucleotide repeat disorders? Lova Falk (talk) 15:03, 17 November 2007 (UTC)

Yes, see this. --JWSchmidt (talk) 19:59, 17 November 2007 (UTC)

Thank you! Lova Falk (talk) 08:47, 18 November 2007 (UTC)

Waterboarding

Could waterboarding be used as a medical procedure against alcoholism and drug addiction? If not, why not? 71.100.6.233 (talk) 15:04, 17 November 2007 (UTC)

How about - "Because it's a really horrible, painful and terrifying torture technique" ? SteveBaker (talk) 15:58, 17 November 2007 (UTC)

Also, because it is probably illegal, at least under US law. In the past we have prosecuted water boarding as a war crime, and as a "Violation of the Laws and Customs of War". -- dcole (talk) 21:17, 17 November 2007 (UTC)

Why hasn't the US government then required that this be acknowledged by persons in authority, namely the new Attorney General? Isn't the failure to require such acknowledgment evidence of a double standard or is waterboarding an acceptable and legitimate means of persuasion when people do not agree or see thing exactly your way? 71.100.6.233 (talk) 22:26, 17 November 2007 (UTC)

Friedreich's ataxia

In Friedreich's ataxia it says: "Median age of death is 35 years, while females have better prognosis with a 20-year survival of 100% as compared to 63% in men." I'm not a physician, but isn't that preposterous? Even healthy women don't have a 20-year survival of 100%. (I have googled, but I can't find information on the 20-year survival of women.) Lova Falk (talk) 15:50, 17 November 2007 (UTC)

I would assume that the intended meaning is that the disease doesn't kill any of the women in their first twenty years, not that none of them die. I agree that the wording is misleading. I'll see if I can find a source. TenOfAllTrades(talk) 15:55, 17 November 2007 (UTC)

Hmm. This article (PDF) (Klockgether et al. (1998) "The natural history of degenerative ataxia: a retrospective study in 466 patients" Brain 121(4):589-600) indicates that "...survival was not influenced by gender", though women tended to become reliant on walking aids or wheelchair-bound sooner than men. TenOfAllTrades(talk) 16:16, 17 November 2007 (UTC)

Nazi freeze/atomic bombs

I found an interesting old New York Times article the other day, dated January 14, 1945 (headline is Nazis talk less of new V weapons, for those of you with ProQuest access). An interesting excerpt:

Swedish scientists, who have a reputation of being extremely well informed and just about as wide awake as their German colleagues, are most skeptical of German claims and rumors.

Especially is this so in the cases of the so-called freeze and atomic bombs, the former numbered V-3 in Reich propaganda releases. About the freeze bomb Swedish scientists say its principle is well known.

The German method of freezing has been the subject of much detailed study by Swedish chemists. The Swedes have come to the conclusion that at present it is completely impossible to manufacture an effective freeze bomb weighing less than eighty tons. They admit the possibility, however, that the Germans have been able to combine extreme cold and explosive effects in a somewhat lighter projectile.

Regarding the atomic bomb, Profs. Theodor Svedberg and Georg von Hevesy, both Nobel prize winners in physics and atom experts, agree that such a bomb is very far from realization and that there are scientific reasons to believe that the engine will prove a complete "flop" if one day in the distant future it is produce.

No other discussion of these particular weapons is in the rest of the article. All jokes about their great estimate on atomic weapons aside (the US crash program to make a bomb was pretty incredible—it is not what you would necessarily expect a country to have invested in, and did require scaling up brand-new physics work to previously unprecedented industrial levels, so I give them quite a break in being wrong), my questions were:

1. Any guesses as to what the "freeze bomb" would actually be and how it would actually work? What were the "well known" principles? Googling "freeze bomb" and many variants turns up very little actually reality and a whole lot of bad sci fi.
2. Would such a weapon really have much combat value, or would it simply be a bizarre Nazi science terror weapon?

Pure speculation is invited and encouraged. My understanding, by the by, is that the actual V-3 weapon was a long-range cannon, so it's possible/likely that any other information leaked was just propaganda and/or misunderstanding by reporters, etc. --24.147.86.187 (talk) 19:40, 17 November 2007 (UTC)

Sounds like the science fiction effect at work (like the CSI effect, only describes how people think all things are possible back in the '40s and '50s because of all the pretty paintings of flying cars on the cover of "Popular Science" ;). All I can reason is that the bomb is actually a delivery vehicle for a large quantity of liquid nitrogen or other cryogenic fluid (which, if used in normal bomb-sized amounts, evaporates before it can cause anything but superficial damage to anyone but a few hatless and hapless bystanders). 80 tons all at once, however, would be enough to overcome the rate of evaporation and possibly overwhelm ventilation systems in underground bunkers and kill the inhabitants through severe frostbite and/or asphyxiation. I'm trying to imagine how you could manipulate the dew point using a pressure wave, but the worse that could happen would be, well, frost. Only problem is that when pressure rises, so does the dew point (which is the same as the frost point, assuming its cold enough). Making things colder is a lot harder than making them hotter in an unconfined space like a battlefield. So, all my meager reasoning takes me farther away from the feasibility of capitalizing on such a route. For creating a low pressure zone, you could use something like a fuel-air bomb, but that doesn't last very long. And it raises the temperature of the air. I think that a freeze bomb is a manifestly dumb idea, worthy of the best traditions of Nazi psuedoscience. DeepSkyFrontier (talk) 20:04, 17 November 2007 (UTC)

I can easily believe that Hitler would have directed his scientists to develop such a weapon. Some think (see Horrifying Utopias) that Hitler believed in a theory of fire and ice, attributable to the fiction writer Hanns Hoerbinger—that things could be expained in terms of a conflict between the two forces. Fire played a major role in Nazi ritual; torchlit ceremonies were standard fare. It is even supposed by some that this belief in the destiny of Aryan fire contributed to the Nazi defeat at Stalingrad where German soldiers froze to death for lack of winter clothing, unneeded by Hitler's Aryan fire-beings. Given all this, it would have been natural for him to want to counter the fire of his enemies with Nazi ice. Speculative enough fer ya? --Milkbreath (talk) 22:06, 17 November 2007 (UTC)

A couple of (very) sci-fi thoughts: 1) Magnetic refrigeration bombs. 2) Create a field capable of dampening all molecular motion (although I think the uncertainty principle would nix this one). 152.16.59.190 (talk) 06:18, 18 November 2007 (UTC)

Construction of a tower

Say I wanted to construct a 65 m tall hollow tower of steel that would be able to support its own weight plus a further 15,000 kilos, how would I calculate how thick the walls would theoretically need to be, assuming a density for the steel of 8g/cm³? Would 10cm be enough? --80.229.152.246 (talk) 21:19, 17 November 2007 (UTC)

What's the diameter of the tower, for one thing? If it's much less than 65m, then your first mode of failure will be buckling, for which the article lists the relevant equations. Otherwise, check out compressive strength. Basically, you need to know, in addition to density, the Young's modulus for steel. I might try working out the equations tonight, but try it yourself and see what you get. SamuelRiv (talk) 03:02, 18 November 2007 (UTC)

In addition to the above articles, the area moment of inertia article will give you the equation you’ll need for the area moment of inertia of a hollow cylindrical cross section, which you’ll need in the equation for the maximum axial load without buckling. And the structural steel article has a bunch of links to articles on various standard steel alloys, such as the popular A36 steel. Those articles will list the yield strength of the alloy, so you can ensure that there won’t be more compressive stress on the tower than it can handle. Finally, actually designing a real tower would also involve using a factor of safety. What value you’d use for the factor of safety would depend to some extent on what the tower’s being used for. MrRedact (talk) 03:39, 18 November 2007 (UTC)

The nature of the load really makes a difference. If that 15,000 kg is distributed vertically, like a bunch of platforms going up the side of the tower, then it won’t require steel as thick as if the load were entirely at the top. If the load has a lot of wind resistance, like if it’s a windmill or huge billboard or something, then it’s going to require thicker steel due to the force of the wind. Or if it’s a load that might be lopsided, like a room containing a bunch of people who might all stand on one side, then compressive strength at the top of the tower and the required resistance to buckling are both going to need to be greater than if the load is guaranteed to be well-balanced. MrRedact (talk) 04:35, 18 November 2007 (UTC)

Is there a difference between a sunrise and a sunset?

If one were given a stack of photos of sunrises and sunsets here on earth,would it be possible without any other information to tell the difference between the two?216.10.164.215 (talk) talking zero —Preceding comment was added at 22:54, 17 November 2007 (UTC)

If there are clouds like these in the pictures it might be possible. --JWSchmidt (talk) 00:35, 18 November 2007 (UTC)

If the date is known, a bright star may help, if available. Pallida  Mors 01:22, 18 November 2007 (UTC)

See the previous discussion at Wikipedia:Reference desk/Archives/Science/2007 September 8#Sunrise vs. Sunset. hydnjo talk 02:56, 18 November 2007 (UTC)

If the land features are recognizable, you can tell east from west; but perhaps that counts as "other information". —Tamfang (talk) 10:49, 18 November 2007 (UTC)

November 18

water, ethanol, and dehydration

Drinking ethanol is said to cause dehydration due to a diuretic effect, and drinking water is normally hydrating. What concentration of ethanol in water would have a roughly neutral effect on the body's level of hydration? --Allen (talk) 00:10, 18 November 2007 (UTC)

A hole in a rib?

I was checking out this x-ray (from our AOTD on lung cancer; this is not my x-ray and I'm not seeking medical advice) and it looks like there's a sizable hole through one of this person's ribs (left side of picture, but "right" side of body). It doesn't seem to be addressed in the description pages. Any ideas as to what it is? Matt Deres (talk) 04:45, 18 November 2007 (UTC)

It's a tumour visible on the film. Because the cancerous tissue is much denser than the surrounding lung tissue, it appears whiter on the film. -- Flyguy649 ^talk 05:25, 18 November 2007 (UTC)

Are you looking at the left side of the picture? The right side of the picture has a tumour in it. That's described in the caption. I'm talking about what looks like a hole drilled through one of the ribs on the left side of the picture. Matt Deres (talk) 05:34, 18 November 2007 (UTC)

Could just be the blood vessels in the lungs that happens to look like a hole. Notice that there are similar structures on opposite side. --Diletante (talk) 05:42, 18 November 2007 (UTC)

(EC) Oh, that left. Sorry, I looked at the photo, saw the box, and spewed forth. If you mean the little hole over the fifth rib (roughly opposite the lower third of the box and to the left of the great vessels), I believe that's and illusion. It's just a place where the shape of the lymphatic tissue/lymph nodes have somehow made it look like there's a hole there. I don't think that the hole is any less dense appearing than any other bony tissue. It just appears that way because of the shape of the surrounding dense tissue. Again, that's just a guess. -- Flyguy649 ^talk 05:51, 18 November 2007 (UTC)

TIME

WHAT CONCEPT ACTUALLY ALLOWS TO THINK OF GOING PAST OR IN FUTURE IN TIME,WHAT EXACTLY FORMS THE BASE OF IDEA,IS THERE ANY RELATION OF GRAVITY WITH TIME,PLEASE EXPLAIN59.95.46.193 (talk) 10:39, 18 November 2007 (UTC)