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: You could use the method of heating the oxide with charcoal, or coal as in steel manufacture. You could use electrolytic methods, as in zinc or aluminium. On the small scale you may reduce by using a powder of another more electronegative metal, such as aluminium. [[User:Graeme Bartlett|Graeme Bartlett]] 03:44, 5 November 2007 (UTC)
: You could use the method of heating the oxide with charcoal, or coal as in steel manufacture. You could use electrolytic methods, as in zinc or aluminium. On the small scale you may reduce by using a powder of another more electronegative metal, such as aluminium. [[User:Graeme Bartlett|Graeme Bartlett]] 03:44, 5 November 2007 (UTC)
::Of course, you might want to be careful abound that last one, lest you end up with a [[thermite]] reaction. <span class="sigShoyrudude555"><font color="royalblue">[[User:Shoy|shoy]]</font> [[User talk:Shoy|(<sup>words</sup> <sub>words</sub>)]]</span> 14:27, 5 November 2007 (UTC)
::Of course, you might want to be careful abound that last one, lest you end up with a [[thermite]] reaction. <span class="sigShoyrudude555"><font color="royalblue">[[User:Shoy|shoy]]</font> [[User talk:Shoy|(<sup>words</sup> <sub>words</sub>)]]</span> 14:27, 5 November 2007 (UTC)
:By the way, "deoxidizing" is usually referred to as [[Reduction (chemistry)|reduction]]. -- 21:08, 5 November 2007 (UTC)


== How do you control snails in an aquarium ? ==
== How do you control snails in an aquarium ? ==

Revision as of 21:08, 5 November 2007

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October 30

Protein domain

Two different protein domains can have a same function why or why not?

If I consider two amino acid sequences of a protein with same domains but changes in 2 or 4 amino acids in the chains. Could it have the same function?

Respected teacher

My problem is that I considered 2 chains of the same protein

Chain A ---> n l i i l a n n s l s s " n g n v" t e s g c k e c e e l e e k n i k e f l q s f v h i v q m f i n " t s "


Chain B ---> n l i i l a n n s l s s " n g n " v t e s g c k e c e e l e e k n i k e f l q s f v h i v q m f i n t s

Those in double Inverted commas are excluded from the domain chain but they are considered in the same domain why? —Preceding unsigned comment added by Biomedicalpersonal (talkcontribs) 05:12, 30 October 2007 (UTC)[reply]

I do not understand what you are trying to ask in the second part of your question but I may be able to help you with the first part. A few amino acid changes in non-critical regions or mutations resulting in similar amino acids (see Neutral mutation) in critical regions may not cause measurable effect on the functioning or the interactions of the protein domain. It would help if you clarify the second part of your question about protein chains. By the way, these appear to be homework questions. The users on the reference desk are not supposed answer your homework questions, but if you have do not understand a concept or need help getting pointed in the right direction by all means ask away. 71.226.56.79 04:22, 31 October 2007 (UTC)[reply]

food poisoning ???

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. --~~~~
This question was removed for a reason. We cannot offer medical advice on either poisoning or allergies. If you are worried about the state of your friends health you should consult a medical professional immediately, especially if you believe she has been poisoned. Discussing it on the internet first should not be your immediate priority. Lanfear's Bane | t 21:41, 30 October 2007 (UTC)[reply]

production of particles by pair production method .

in the wikipedia article "pair production" only example of electron-positrion pair production is given . is it possible to produce particles like proton , neutron and neutrino by pair production method ? —Preceding unsigned comment added by Shamiul (talkcontribs) 12:30, 30 October 2007 (UTC)[reply]

Neutrinos and antineutrinos can be produced from a photon.[1] Like electron-positron pair production needing other particles involved in order to conserve energy and momentum, neutrino-antineutrino production from a photon needs other particles involved in order to conserve spin, which is a form of angular momentum.
Protons and neutrons aren’t elementary particles, but instead bound states of quarks. So in quantum chromodynamics, low-order particle interactions would include pair production of quarks, not of protons or neutrons. A quark and its antiquark can be produced from a gluon.[2] MrRedact 21:02, 30 October 2007 (UTC)[reply]
While these predictions from theory are very likely correct, we are currently far from being able to experimentally confirm pair production of neutrinos (because our neutrino detectors are not very sensitive). Icek 01:08, 31 October 2007 (UTC)[reply]
Somehow my previous answer to this question has been removed? However there is no neutrino pair production. They are electrically neutral and as such do not couple to photons. The nearest equivalent would be production of a neutrino/anti-neutrino and corresponding anti-lepton/lepton from a w particle. MrRedact is right about quarks being pairproduced, but note they can also be produced from photons as well as gluons (they are charged). Pair production of protons/neutrons is nigh on impossible, when quarks are pair produced they hadronise into two jets of particles and protons/neutrons maybe included in the jets. Cyta 08:56, 1 November 2007 (UTC)[reply]
Actually I may have been hasty to dismiss the original answer I didn't spot a link had been provided. The pair production of neutrinos seems only to happen in dense matter, which seems to me a different effect. There is no direct neutrino-photon coupling in the standard model. I suspect the matter is required as there is an intermediate stage in the reaction. Cyta 09:00, 1 November 2007 (UTC)[reply]

Anomalous hair

REMOVED. —Preceding unsigned comment added by 193.188.46.61 (talk) 13:23, 30 October 2007 (UTC)[reply]

(EC) Do not request regulated professional advice. If you want to ask advice that "offline" would only be given by a member of a licensed and regulated profession (medical, legal, veterinary, etc.), do not ask it here. Any such questions may be removed. See Wikipedia:Medical disclaimer and/or Wikipedia:Legal disclaimer. Ask a doctor, dentist, veterinarian or lawyer instead. Sorry, even a query that sounds innocuous should be addressed to a professional. Lanfear's Bane | t 13:32, 30 October 2007 (UTC)[reply]
REMOVED. --JWSchmidt 14:55, 30 October 2007 (UTC)[reply]
The original question did not ask for medical advice. My original comment made this point. There was no reason to delete my original comment. --JWSchmidt 18:23, 30 October 2007 (UTC)[reply]
REMOVED. --M@rēino 15:03, 30 October 2007 (UTC)[reply]

Per Lanfear's Bane's comment, we cannot offer medical advice. In this case, the original poster was seeking a diagnosis--the identification of the cause of a particular symptom. If someone would like to discuss in more detail why this constitutes a request for medical advice, please bring it to the Ref Desk talk page: WT:RD. [hidden unsigned comment by User:TenOfAllTrades made public for clarity]

I am appalled at this continuing deletionism by people who are apparently incapable of telling the difference between a request for an explanation of a biological phenomenon and a request for medical advice. To take this behaviour to its logical conclusion would result in the prohibition of any question relating to human biology (or animal biology for that matter). What causes blue eyes? What causes grey hair? Why does alcohol abuse cause liver damage? All are questions relating to biological phenomena - just as is "What causes an observed variation in human body hair types in humans?" Hiding comments about the removal of questions is also profoundly unhelpful. DuncanHill 15:20, 30 October 2007 (UTC)[reply]

One clear sign of a question that we aren't allowed to answer is one that starts off "I have some symptom..." or "A friend has some symptom...". This is quite clearly one of those - and we aren't allowed to answer it - period. The questions you put up as strawmen are not the problem here. I would like to turn this around and ask DuncanHill: "What kinds of question do you think the ban on giving medical advice is intended to cover?" - clearly it covers something or it wouldn't be there. What? Give me some examples. SteveBaker 16:38, 30 October 2007 (UTC)[reply]
Why don't we stop wasting energy here now and instead just send the questioner to where he asked the same question a week ago (8 times in fact). DMacks 15:16, 30 October 2007 (UTC)[reply]
Hey, can everyone confine metadiscussion to the talk page of the Ref Desk? This sort of debate about what questions are appropriate doesn't belong out on the Desk itself. TenOfAllTrades(talk) 16:52, 30 October 2007 (UTC)[reply]
  • "can everyone confine metadiscussion to the talk page" <-- If content is removed from this page because it contains descriptions of personal health matters, then say THAT, but do not say that someone asked for medical advice when they never did. --JWSchmidt 18:20, 30 October 2007 (UTC)[reply]
  • Hey, why was my comment removed? All I said was that the original question did not seek medical advice. And that continues to be my opinion. The question was not describing a symptom -- "a sensation or change in health function experienced by a patient" -- because there was no implication that the questioner's health was affected. They were just curious about Hair follicles, a subject on which the Wikipedia article is B-class at best. --M@rēino 04:23, 31 October 2007 (UTC)[reply]

This is plainly ridiculous! The sense of this regulation is to avoid unqualified suggestions to be accepted and considered professional by the user. How should this do any type of damage what so ever? If you apply that definition about a symptom it can be eligible to all sorts of things, to nausea due to boat movements for ezample. Aren't you allowed to explain the processes that occur in that event?! Laws and regulations should be implemented only when they make some type of logical sense. You're just tking it to the extreeme here.193.188.46.254 13:56, 31 October 2007 (UTC)[reply]

Membrane Potentials

What is the difference between membrane potentials, diffusion potentials and nernst potentials. I understand membrane pot. = diffusion pot. and nernst pot. to be when the electrical force opposes the concentration force.?

In biology, a membrane potential is the electric potential difference across a membrane of a cell or an organelle such as a mitochondrion. For charged ions there can be a balance between movement across a membrane due to diffusion and movement due to the electric potential across the membrane: see reversal potential. If you had a membrane and only one ion was able to cross it by passive diffusion then at equilibrium the membrane potential would be equal to the diffusion potential of that ion: Nernst Potential. --JWSchmidt 14:29, 30 October 2007 (UTC)[reply]
Thanks for the quick reply. I have a slightly better understanding now but still wich to clear some issues [this is in the context of the establishment of an RMP]. 1) Does the term diffusion potential apply to an ion or membrane? 2) Could the Nernst potential (equilibrium potential) be described as the potential difference required to prevent net diffusion of that ionic species? (and how is Nernst potential related to diffusion potential?) 3) What is the purpose of the Na/K ATP'ase (it contributes slightly by reducing RMP; is its primary function to repolarise?). Thanks.
[ATP'ase] I thought a bit more and reasoned that the purpose may be to create the diffusion potential? or allow the tissue to become excited (function)? i.e. to maintain concentration gradient thereby preventing the Nernst potentials for each ion (Na/K) being reached (a scenario where membrane potential- or is that diffusion potential- is a result of only Na and K)
Did you find Resting potential? For most cells, the Na+/K+-ATPase is not thought to make a major contribution to the resting membrane potential. The resting potential of many cells is close to the equilibrium potential of potassium ions (K+). Many cells have potassium "leak" channels that control the resting potential. --JWSchmidt 18:13, 30 October 2007 (UTC)[reply]

When will oil run out?

Based on world wide consumption trends, and ignoring that it takes millions of years to produce more oil. 64.236.121.129 16:04, 30 October 2007 (UTC)[reply]

It will never "run out" - but it will become far too expensive to use as fuel. When? See Peak oil. Cheers Geologyguy 16:06, 30 October 2007 (UTC)[reply]
I did some calculations in response to an earlier question - the answer is that if we were to carry on consuming it at the present rate, it would run out in about 500 to 600 years. However, if we actually did that, the CO2 levels in the atmosphere would be far beyond "mere" global warming problems - they'd be at a point where humans (and most animals) couldn't breathe. However, the assumption that we'll carry on using it at present levels is flawed. I don't think that running out is a practical proposition. Even if we somehow managed to 'sequester' the CO2, we would only have to halve our consumption every 250 years in order to make the stuff last forever. Another issue is that these numbers for oil reserves are always accompanied by a caveat that says "economically retrievable" - in other words, the only things the oil companies care about is the stuff they can dig up for less than they can sell it for. There are reserves of stuff like "oil shale" that contain a lot of oil - but which are so expensive to dig up and refine that it's not worth doing it. If the oil were ever seriously likely to run out, then the price would go through the roof and suddenly oil shale (or whatever) would be worth exploiting and our reserves would increase (although the price would still be astronomical by today's standards). However, we must stress that with what we know about global warming, it's all completely irrelevent. We must not ever come even close to running out - because even a tenth of that amount of oil - when converted to CO2 - would kill the planet. SteveBaker 16:28, 30 October 2007 (UTC)[reply]
Our article on world energy resources and consumption says that the world's oil reserves are 5.7x1022 J, and annual oil consumption in 2005 was 1.8 x 1020 J, so this gives a ratio of reserves to consumption of over 300 years. However, as Geologyguy and SteveBaker have pointed out, this is a rather meaningless figure. In reality oil prices will rise, and oil consumption will fall as alternative energy sources become more economically attractive, and global warming imposes a shorter deadline on us anyway. Gandalf61 16:44, 30 October 2007 (UTC)[reply]
Perhaps by then we will have made advances in scrubber technology, and have them on every car, big ones atop every building, and just generally all over the place. I mean, once it comes down to economy vs. survival... ArakunemTalk 17:18, 30 October 2007 (UTC)[reply]
The chemistry of scrubber technology is not very promising - the kinds that merely absorb CO2 will give the gas up again fairly easily and the substances that react with CO2 to actually get rid of it require lots of energy to make in their own right - so you end up needing more scrubbers to scrub the output of the factory that makes the scrubbers (or the power station that drives it) than the factory itself can make. It's really a bad idea to pin one's hopes on such things because it distracts from the very critical thing of cutting CO2 production in the first place. The whole "Clean Coal" campaign (which has to rely on science-fiction "carbon sequestration" techniques) is a particularly bad example of this kind of wishful thinking. SteveBaker 03:40, 31 October 2007 (UTC)[reply]
Are you saying that collecting and burying CO2 from a coal plant won't work and will use more energy than you get from the coal?--Dacium 05:47, 31 October 2007 (UTC)[reply]
Yes and no. I'm saying that on small scales, it takes more energy than you get - and on large scales there isn't a viable technology for doing it at all. Not one single large scale CO2 removal/sequestration plant exists anywhere in the world - not even experimentally. (Which hasn't stopped the US from licensing the building of "Clean Coal" power stations - which is a scandal just waiting to get media attention!) Our article on Carbon capture and storage explains that what we're likely to have will remove 80 to 90% of the CO2 from the gasses and consume 10% to 40% more energy. But the biggest problem is what you do with the stuff once you've captured it. Sequestering it into limestone requires 180% more energy - so that isn't going to fly. If you try to sequester the CO2 without chemically converting it to something else then you've still got to find a place to store millions of tons of something that's a gas at normal temperatures and pressures. That's no easy task. You can't store it underground or underwater because there isn't enough space at normal temperatures and pressures (If you burn a cubic meter of oil or coal - you get a LOT more than a cubic meter of CO2 as a result! So pumping it into disused coal mines and oil wells isn't going to work for very long.) If you compress the CO2 so it takes up less space (eg storing it as dry ice), then that requires either very high pressure storage or very low temperatures. Either of those technologies will require yet more energy - and worse still, will be vulnerable to long term corrosion and other damage - so you're just building up more trouble for the future. There is talk of dissolving the stuff into saline aquifers or deep oceans - but those are not permenant solutions (eventually, the CO2 would get out again) and the resulting carbonic acids would likely do untold amounts of damage to the environment. Dealing with radioactive waste from a nuclear power plant is EASY by comparison because so little material is involved. So, no, we aren't going to be doing this if we want to save the planet. We have to cut down our consumption (probably the easiest thing to do in the short term) and switch rapidly over to nuclear and (where possible) wind/solar/tidal power until we figure out how to make fusion reactors that actually work. SteveBaker 18:00, 31 October 2007 (UTC)[reply]
On the scales you are talking about, I would imagine there would be nowhere near enough suitable locations to bury CO2 and be reasonably sure that it won't escape at some point in the future. Hopefully within 300 years we will have managed to develop alternatives to fossil fuels, otherwise our pathetic species probably doesn't deserve to survive. Bistromathic 16:41, 31 October 2007 (UTC)[reply]

Quantum foam

I was thinking about Quantum foam last night. Since particle/antiparticle pairs emerge out of the vacuum - and since there is no "rest frame" - it seems like the particle pairs must be travelling at random velocities relative to my motion. This implies that they have random kinetic energy relative to my motion - which means some of them must be arbitarily energetic. How come we don't measure them? Even though they only last for a spectacularly short amount of time in their own frame of reference - in mine, they'd spend an eternity recombining. Our article briefly sketches over that. SteveBaker 16:45, 30 October 2007 (UTC)[reply]

You are taking the idea of virtual particle pairs too literally. Quantum foam is a broad term for the fluctuating nature of space-time at a quantum level that theory demands. Those fluctuations are abstractly described as virtual particles, but from the practical point of there are no particles. Imagine a tablecloth with marbles on it. The marbles may move because they are hit by other marbles, or they may move because the tablecloth shakes. If all you could ever see were the motions of marbles, you might try to explain the motions caused by shaking in the tablecloth as being caused by invisible "virtual" marbles. What is going on in physics is similar. The ways that real particles are affected by the quirks of the vacuum can be usefully described by replacing the vacuum with infinite numbers of virtual particles constantly bombarding matter. However, those particles don't really come from anywhere or go anywhere. They don't have an existence that one could isolate and interact with individually. Really, they are collectively just a way of describing the manifestations of the complicated quantum mechanics intrinsic to space-time. Dragons flight 19:55, 30 October 2007 (UTC)[reply]
Steve, "quantum foam" does not refer to the virtual particle/antiparticle pairs in the quantum vacuum in general, but specifically to what happens around the Planck scale, where spacetime becomes quantum mechanical. The short answer to any question about quantum foam is that we don't know anything about it at all. But I think you're asking about the quantum vacuum in general, so let's step back from solving quantum gravity and consider something more manageable. Take QED or Klein-Gordon theory. You asked "How come we don't measure them?", referring to virtual pairs with arbitrarily large energies. OK, how do you want to measure them? You would have to interact with them somehow. That means that instead of a vacuum-bubble diagram, you have to consider diagrams with external vertices, and virtual particle loops inside. One way to think about these diagrams is that your real external particles are encountering the virtual particles from the vacuum; and they can indeed have arbitrarily large momentum and energy. In many quantum field theories, these loop contributions cause all calculations to come out infinite. Similarly, the vacuum diagrams can cause the vacuum energy to come out infinite. To get something physically sensible, you have to renormalize the theory, and "renormalize" the vacuum energy by zeroing out the infinite offset. So even if you never observe the virtual particles as discrete events, they are hugely significant in every interaction between particles! In some theories, like supergravity, the case is even worse than that, and renormalization doesn't work. Because all of this is not very intellectually satisfactory, we hope that there is some real, underlying physics that doesn't have these problems. Supersymmetry takes care of the infinite vacuum energy, but still needs renormalization. Superstring theory attempts to avoid the need for renormalization at all. But now we're back in the realm of speculations about quantum gravity. --Reuben 21:49, 30 October 2007 (UTC)[reply]

Why don't they put some kind of grating over jet engine intakes to prevent birds from getting sucked in?

Why not? 64.236.121.129 16:57, 30 October 2007 (UTC)[reply]

It probably doesn't happen all that often and may actually cause too much disruption to the air intake on the engines. Also, birds would still (I imagine) be sucked and stuck to the grating over the intake. -- MacAddct1984 17:05, 30 October 2007 (UTC)[reply]
Per MacAddct, a grating would seriously disrupt the airflow into the engine, creating a tremendous amount of drag. (Remember that air is flowing through the engine at several hundred kilometers per hour.) Plus, what if the bird strike actually damages the grating? In addition to having a bird in the turbine, you'd get all those jagged metal bits of broken grating.... TenOfAllTrades(talk) 17:12, 30 October 2007 (UTC)[reply]
No - there is no grating - but there is a requirement to test engines for bird strike damage. They actually have an air-powered cannon to shoot birds into running jet engines in order to test them. However if you fly at high speeds and lower altitudes (where birds tend to be) - and if you hit a big one - you can certainly throw a turbine blade and utterly trash the engine in the process. If you have a strong stomach - read Bird strike which has some exceptionally grisley photographs of the effects of birds on airplanes and airplane engines in particular. SteveBaker 17:35, 30 October 2007 (UTC)[reply]
Yep. Even with a no-drag indestructible grating, you're still talking about smashing a bird into something at several hundred miles per hour. It's not that the engine is sucking in birds from all directions, just that the plane and bird are hitting -- and even with a grating, the engine still has to generate the same suction. But as SteveBaker said, there's a bird strike test, so a grating is superfluous. — Lomn 18:18, 30 October 2007 (UTC)[reply]
A bird running into a turbine is usually not dangerous. The bird will be ripped apart. If the turbine does throw a peice and destroy itself, all that happens is that fuel will continue to be pumped in and a massive flame will come out the back because the fuel is being burned. Obviously you get no thrust from the engine, but the plane can still fly. I don't think any commercial airliners have been brought down, many have had engines taken out.--Dacium 05:43, 31 October 2007 (UTC)[reply]
According to Bird strike: according to the FAA only 15% of strikes (ICAO 11%) actually result in damage to the aircraft. But you're wrong about commercial airliners - there is at least one instance: Eastern Air Lines Flight 375 flew into a a flock of small birds that took out three out of the four engines at once - at the critical moment just after take-off when the engines were at full power - they also splattered over the windshield - blocking the pilot's view and clogged the pitot tubes, preventing them from knowing their airspeed. They didn't stand a chance - the plane rolled over and smashed into the ocean - there were 62 fatalities out of 72 people aboard. Our article on bird strikes also says: the problem costs US aviation 600 million dollars annually and has resulted in over 200 worldwide deaths since 1988. We have a brief article about Birdstrike simulators too! SteveBaker 17:31, 31 October 2007 (UTC)[reply]

Having a grating over the fan intake can introduce another problem. When engines are being tested, they have what they call a debris guard which fits over the front of the engine, but it means that the engine cannot be run in all conditions. To fully test the engine, at some point the guard needs to be removed.

This is because, at the speed the air enters the front of the engine, well over 300 miles an hour, it creates a huge cooling effect on the debris guard. That combined with the right (or wrong) level of humidity can cause ice to form on the guard. It has been known that enough ice has formed for the fan to create a vacuum behind the guard, and the result is, the entire guard is pulled through the engine, which will utterly destroy the engine, no question about that. On the other hand, with a bird strike, it's quite possible for an engine to keep running without any problems, especially if the bird is passed straight through the cold stream ducts and out the back. —Preceding unsigned comment added by 132.244.246.25 (talk) 08:33, 1 November 2007 (UTC)[reply]

Taste bud abnormality

I was wondering if there is a name for the painful bumps that crop up on tongues from time to time. It seems as though a single taste bud turns white, enlarges, and becomes very sore to the touch. I can't seem to find a name for it anywhere (and no, it's not herpes or canker sores) -- MacAddct1984 17:10, 30 October 2007 (UTC)[reply]

Sorry, but this question is basically asking for a diagnosis, which we aren’t allowed to do here. As it says at the top of the page:
Do not request regulated professional advice. If you want to ask advice that "offline" would only be given by a member of a licensed and regulated profession (medical, legal, veterinary, etc.), do not ask it here. Any such questions may be removed. See Wikipedia:Medical disclaimer and/or Wikipedia:Legal disclaimer. Ask a doctor, dentist, veterinarian or lawyer instead. MrRedact 17:33, 30 October 2007 (UTC)[reply]

While MrRedact is correct that we can't diagnose, you might nevertheless be interested in the brief stub on lie bumps. It could stand to be filled out if you want to do some source research. --Trovatore 17:36, 30 October 2007 (UTC)[reply]

Ah, thank you very much Trov! While I wasn't really looking for medical advice, I do realize there is a very fine line between "diagnosing" and what I was asking. But that is what I was looking for, I remember my mother always saying the old wife's tale was that you get them when you lie. -- MacAddct1984 17:53, 30 October 2007 (UTC)[reply]
I can't say what this is in your case, but it also happens to me when I accidently scrape the area with a sharp piece of food or bite the tongue. Here's an article that discusses lie bumps a bit more: [3]. Again, this may not be what you have though. Sancho 17:47, 30 October 2007 (UTC)[reply]
they are basically caused by cuts in the tongue by food or your own teeth. If they are on the edge of a tongue it almost defiantly is caused by you biting your tongue accidentally. usually a mouth full of food pushes the tongue over the teeth and people bite without realising.--Dacium 05:38, 31 October 2007 (UTC)[reply]

how do i make my dog smart

please thank you —Preceding unsigned comment added by 77.234.83.173 (talk) 17:19, 30 October 2007 (UTC)[reply]

See dog training. Sancho 17:38, 30 October 2007 (UTC)[reply]
I don't think you can change your dog's intelligence - as with humans, it's pretty much something that you're born with - but you can certainly enroll in a training course for you and your dog. Sadly, it generally turns out that your dog is plenty smart and it's you that gets trained - but that works too! We also have an article on Dog training. I have two dogs - one smart (female), the other dumber than a bag of hammers (male). The smart dog has learned to bark at the front door when she wants to be let in. The stupid dog has not. He just stares at the door hopefully. However, the smart dog has realised this and when she sees him standing there looking puzzled, she runs up to the door, barks until we open it and let him in - then she goes back to whatever she was doing before. SteveBaker 17:43, 30 October 2007 (UTC)[reply]
Intelligence can most certainly be changed in humans. And it's important that people are aware of that. There was a study recently that took two classes of high-school students, gave one a lecture about something boring and irrelevant (just to have a control), and gave the other a lecture on the nature of intelligence, and how intelligence and problem solving skills can be improved through training. After about half a year the second group showed a significant increase in grades and the first didn't. Of course there is always the criticism that training intelligence is just training for intelligence tests, but I believe strongly that intelligence by any reasonable definition can be consciously improved. I see no reason why dogs couldn't do the same, except that they probably won't have the mental capacity for defining the concept of intelligence, so they can't set it as a goal for themselves explicitly. Still, if you set out a wide range problem solving exercises with rewards that the dogs can understand (like the 'stack boxes to get to a banana'-kind of exercises they give monkeys, only simpler), they may generalize over all exercises and improve their problem solving intelligence instead of just learning specific tricks. risk 18:16, 30 October 2007 (UTC)[reply]

thats a lot of reading, what's something simple ic an do like some toy i can buy or music to play. i cant afford obedience school. —Preceding unsigned comment added by 77.234.83.173 (talk) 18:09, 30 October 2007 (UTC)[reply]

There are no magical shortcuts. If you can't afford obedience school - read our Dog training article carefully and do what it says - it's pretty much what the obedience schools teach. Set aside an hour a day for training. Make sure your dog knows that this is something special - "Now is training time - later will be play time" - so have a special routine you go through at the start and end so your dog will come to recognise that. SteveBaker 00:08, 31 October 2007 (UTC)[reply]
I have an instinctive negative reaction to the idea that learning to obey demonstrates intelligence. I've always sort of felt that cats are smarter than dogs, precisely because they don't obey you. --Trovatore 01:33, 31 October 2007 (UTC)[reply]
Dog training is more about learning inter-species communication than obeying orders. I prefer to think of it as unlocking the intelligence that was always there - and in a way that we humans can understand. SteveBaker 03:25, 31 October 2007 (UTC)[reply]

I don't know about smart but I made my dog smarter with this and this and this. All three are essential. --DHeyward 05:40, 31 October 2007 (UTC)[reply]

Get involved with Dog agility or Flyball. You and the dog have fun; you and the dog get to socialize; you and the dog get good outside exercise; and the dog's intelligence is developed. It's mostly just run by volunteers, so there's little cost.--Eriastrum 17:06, 31 October 2007 (UTC)[reply]

When you here about plane crashes on the news

They usually manage to find the Black Box flight recorder thing even in the worst crashes, when the rest of the plane has been completely destroyed by explosion and fire. So, why don't they make the planes out of the same material they make the black boxes out of? --84.68.112.172 17:50, 30 October 2007 (UTC)[reply]

While black boxes are of durable construction, I think they benefit a great deal by being inside the aircraft. By analogy, I could get into a car wreck that's bad enough to total my car, but it's likely that my CD player would survive. It's not that it's made of some magic substance—it's just protected by the body of the car, which can deform and break up, absorbing a lot of impact energy. -- Coneslayer 17:57, 30 October 2007 (UTC)[reply]
In short, the black box is so heavily armored so as to survive the worst impact, that if the airplane was made similarly, it would be more like a tank than an airplane, and would be too heavy to fly. Planes are made to fly, CVR's are made to crash, as "they" say. :) ArakunemTalk 18:03, 30 October 2007 (UTC)[reply]
Check out this classic Cecil Adams column. "If aircraft "black boxes" are indestructible, why can't the whole plane be made from the same material?" 69.95.50.15 18:01, 30 October 2007 (UTC)[reply]
It's also worth noting that an indestructible airplane would provide little extra protection to all-too-destructible passengers, who are still independently subject to the laws of inertia. Crash casualties are not so much the product of a deforming airframe as they are of a sudden stop. In fact, the deformation is good -- it absorbs crash energy that would otherwise injure passengers even more. Modern cars are constructed with this same theory in mind. — Lomn 18:12, 30 October 2007 (UTC)[reply]
I wonder if the querent is looking for material for a stand up comedy routine? According to our article, Flight data recorders tend to be "double wrapped, in strong corrosion-resistant stainless steel or titanium, with high-temperature insulation inside." We don't make jetliners out of titanium for the same reason we don't make windows out of diamonds, its not cost effective (though, as an aside, while the SR-71 Blackbird was made out of titanium, its windows were not diamond), and if we packed it full of high-temperature insulation, where we would put the passengers? Rockpocket 18:14, 30 October 2007 (UTC)[reply]
Also, black boxes aren't black - they're orange. SteveBaker 00:04, 31 October 2007 (UTC)[reply]
Depends on how hot the fire is. --DHeyward 05:09, 31 October 2007 (UTC)[reply]


Even if the plane were made to survive a crash, it doesn't help people inside. Say the plane can survive 1000G of force, your body cant and you would be squished to death on impact, regardless of if the plane body deformed or didn't. The best solution for safety would have to be a plane that deforms enough so that a +100G impact is reduce to say 10G.--Dacium 05:50, 31 October 2007 (UTC)[reply]

Digitap imaging & printing

perfect on screen but when i printed it on my Epson stylus Photo RX620 it looked nothing like what it did on screen. I was truely gutted. It was duller, darker, more saturated & lost a lot of fine detail. I am using the correct paper & my ink channels settings are set to default. This is happening with all of my images.

Why is this happening & how can i resolve it so that what i create on screen prints as it looks on screen? Also what is the difference between a jpeg, tiff, Esp, btmap etc?

Thanks kindly

Nay —Preceding unsigned comment added by 86.145.223.206 (talk) 19:33, 30 October 2007 (UTC)[reply]

For the latter, articles such as JPEG, TIFF, and BMP file format may be of use. For the former, I have no suggestion apart from experimenting with configuration settings (as this is not so much a case of correctness as one of preference). — Lomn 20:05, 30 October 2007 (UTC)[reply]
You may find this is better answered on the computing reference desk however I'll have a quick stab. Computer screens use additive colour. They add red, blue and green colours together to make white. Printers OTOH use ink which substracts colour. They use cyan, yellow and magenta and mixing them together makes black :-( So the screen will never look the same as a printer.
Also ink jet printers use wet ink thqat runs a bit and the colours bleed together making it dull. Having a good quality laser printer and using top quality paper will definately produce crisper and brighter prints. Fortuately the cost of colours lasers has been plummeting in recent years Theresa Knott | The otter sank 20:10, 30 October 2007 (UTC)[reply]
There are many problems here. Additive versus subtractive, impure inks and toners, the fact that you are printing in four colours (Cyan, Magenta, Yellow and Black) instead of three, that the printer needs to use dithering and bleeding of inks to do what it does, that the brightness of your screen is independent of the room lighting - but the brightness and hue of your print is entirely dependent on the colour of the ambient lighting - that the software that does the conversion is rarely correctly set up for the kind of paper you are using - that the gamma of cameras and screens are NEVER correctly set up, that not all programs take notice of the gamma values that the camera put into the file header - and if you save the picture out in another format, that information is almost certainly lost, that our eyes respond differently to a pigment that reflects yellow (meaning a true yellow) versus a seemingly identical pigment that reflects both red and green, that your CRT probably has several different colour temperature settings and you have no clue which one you picked...there are a million reasons. The bottom line is that you're doomed and you just have to tweak the available settings until you get it "how you like it". (There is no "Right"). SteveBaker 23:50, 30 October 2007 (UTC)[reply]
Oh - and the difference between various file formats (for the purposes of this discussion) are that JPEG and GIF both compromise the true colours of your photos in order to save a lot of memory. 24 bit TIFF and PNG can store colours in their original perfection with no losses whatever. BMP (urgh) can go either way - and it's a horrid format anyway so don't use it. The various 'raw' formats that cameras sometimes use are better still - IF you have the right software to handle them. There are some esoteric formats that do even better - but I doubt you'll ever come across them. On balance:
  • Use PNG for original artwork and photos stored on your computer at home - where (presumably) you have plenty of disk space.
  • Use RAW images for original photo archives if your camera and software supports them - but convert to PNG for day-to-day use.
  • Use JPEG for photographic types of pictures on the Internet - it's compact - which means it downloads quickly - and the losses due to it's compression tricks are not noticable on a typical browser at the default screen resolutions we have these days. As broadband starts to conquer the world - I'm going more and more towards using PNG even in these situations.
  • Use GIF only on really high usage web pages for cartoony stuff or (if you absolutely must) for animated images.
  • Don't use TIFF or BMP if you can possibly avoid it - they are both only patchily supported in areas like gamma that truly matter.
SteveBaker 00:01, 31 October 2007 (UTC)[reply]

Mom Haircut

Okay, this is a stretch, but I'm going to try it anyways... just bear with me, okay?

Is there a condition or an illness (for lack of a better word, I don't mean to imply bad health) that would describe the tendency for new mothers to get a drastic haircut? I don't mean the "mom haircut" but, just even like dying it a different color, something that makes a dramatic change. I know this could just be associated with going through major changes in their life or, more dangerously, depression, but I was wondering if there was a name for it specifically. Thanks, and I apologize if this question is just too dumb for words. Beekone 20:18, 30 October 2007 (UTC)[reply]

You might find more information under the mid-life crisis article -- MacAddct1984 20:43, 30 October 2007 (UTC)[reply]
One could argue it is an evolutionary successful strategy: long hair is an energy intensive characteristic adopted by females to attract a mate. After pair bonding and generating offspring, the female no longer considers the energy invested in maintaining long hair is no longer required, hence they cut it. Or it could be that they don't want to get nits from all the little brats they interact with. Or it could be to reinvent themselves for the next stage in their lives, or it could be part of an effort to "look good" again after the physical changes associated with child birth. Interestingly (or not) there does seem to be some recognition of "mom hair" in the mommy blogosphere, but its extremely unlikely that it has a specific biomedical name. Rockpocket 20:45, 30 October 2007 (UTC)[reply]
I'm not even sure what you are describibg is a real phenomenom. Is there any data to suggest that on average new mums get more dreastic haircuts than women of that age who are not new mums? Theresa Knott | The otter sank 20:52, 30 October 2007 (UTC)[reply]
No, I'm not sure such research exists... I guess that's why I asked. It seemed like a funny quirk that I'd noticed in some friends and acquaintances. I agree that the likelihood of this activity having a specific medical name is extremely un, but you never know until you ask, right? Beekone 21:05, 30 October 2007 (UTC)[reply]
Googling "mom haircut" at least provides evidence that the "mom haircut" is culturally believed to be a real phenomenon. From reading the various pages, it sounds like a "mom haircut" is either quite short, or long and tied back. In either case, it sounds like the point is to have a low-maintenance, no-nonsense haircut that isn’t going to get in the way of the harried new mother, who suddenly has less time for dealing with such frivolities as hair. MrRedact 21:30, 30 October 2007 (UTC)[reply]
Don't think anyone has mentioned the obvious answer (well, obvious to anyone with practical experience of babies), which is that babies of a few weeks and older love to practise their hand-eye co-ordination by grabbing and pulling long hair. Gandalf61 10:39, 31 October 2007 (UTC)[reply]

"I don't mean the "mom haircut" but, just even like dying it a different color, something that makes a dramatic change. " signed, the very first post! Read, think, reread, comprehend... maybe Beekone 13:04, 31 October 2007 (UTC)[reply]

You start off with: "Is there a condition or an illness ... that would describe the tendency for new mothers to get a drastic haircut?" If you are going to ask contradictory questions, you are going to get contradictory answers. Rockpocket 17:14, 31 October 2007 (UTC)[reply]
How does "drastic haircut" translate into "mom haircut"? I mean I know I called the topic that, but I state repeatedly that that's not what it's about. To be technical I start off with this: "Okay, this is a stretch, but I'm going to try it anyways... just bear with me, okay? " 'Clearly the message is going to be muddled so I should pay close attention,' thought the reader. Beekone 17:18, 31 October 2007 (UTC)[reply]
Additionally, I liked your answer, Rockpocket. You seem to get what I was asking, and correctly stated that such a condition is unlikely to have an actual label. But at least you got what I was asking, even if you're answer was basically "I don't know." The other answers are "I don't know" plus a hint of "I obviously didn't try to get what you were asking" which i frustrating because if you don't know why even answer? Just pass it by. I obviously asked in the case that someone with knowledge of the topic might answer. It's sort of what the Help Desk is all about... in my understanding anyways. Beekone 17:23, 31 October 2007 (UTC)[reply]
Thats a fair point, well taken. Rockpocket 04:10, 2 November 2007 (UTC)[reply]

Why don't they use water jets on very large ships?

The largest they used it on was a frigget I think. Why not larger ships? 64.236.121.129 20:26, 30 October 2007 (UTC)[reply]

Our article on pump-jets notes some of their advantages, contrasted with a decrease in efficiency versus propellor-based designs. In particular, the note about power density may be relevant, as frigates don't have as much hull volume to fill with engines as do destroyers or cruisers. The shallow-draft advantages provide additional merit for new littoral combat ships. At some level, though, the lack of adoption may simply be industrial inertia. Propellors are well-understood and consequently are considered to inherit less risk than a relatively new technology like pump-jets. In the absence of a compelling advantage for the latter, then, the former is likely to remain in widespread use. — Lomn 20:36, 30 October 2007 (UTC)[reply]
Many very large ships still have bow thruster jets.--Dacium 23:54, 30 October 2007 (UTC)[reply]

Sugar in Soda

I've decided to eliminate soda from my diet. While shopping at a local grocery store, I came across a powder lemonade product that one mixes with water. The lemonade’s packaging said it "contained 40% less sugar than soda." So with everything being equal between the soda and the lemonade (calories, fat, etc), what would be the benefits of 40% less sugar? I would assume that there would be some benefits…or is this a fancy marketing ploy?

Thanks 64.85.199.27 20:41, 30 October 2007 (UTC)[reply]

It's precisely a marketing ploy. Whether it actually benefits you depends entirely on your diet. People hear that carbs are bad, carbs make you fat, so they advertise that they have fewer carbs. Sugars/carbs are the normal person's primary source of energy, and it is the first form of energy (calories) the body will burn in order to operate (fats it prefers to store away). Now, if the food in your diet contains all the calories you need in a day, and you just want something to quench your thirst, there's really no need to drink something laced with sugar that will just get converted into fat for long term storage, so you may as well go for that low-sugar lemonade. Someguy1221 20:47, 30 October 2007 (UTC)[reply]
(ec)Soda contains a lot of sugar! 40% less means 60% still remaining which is still very bad for your teeth. Having said that soda contains not fat, protein etc. It's water and sugar, so 40% less sugar will be 40% less calories. Those calories that remain are junk food calories though. Much better to drink water or tea. Theresa Knott | The otter sank 20:49, 30 October 2007 (UTC)[reply]
Also, the drink may have had that 60% of sugar boosted by an artificial sweetener, and if you read sugar substitute you'll see that all the common ones have some kind of controversy about potential health risks attached. Confusing Manifestation 21:39, 30 October 2007 (UTC)[reply]
Just drink diet or zero soda's (diet coke, coke zero, pepsi max etc). These contain zero suger. Also other things that are high in suger are sauces like ketchup, alcohol etc. Personally I would try to ditch soda all together and drink water. Try drinking two classes of water before you eat.--Dacium 23:27, 30 October 2007 (UTC)[reply]
There are two very different things being talked about here. Diet Coke/Pepsi/whatever contains almost zero carbs. Giving up Diet coke and drinking this stuff will increase your carbs by a large percentage because their claim to have less carbs than "Soda" is a poorly worded and highly misleading statement. WHICH soda do they have 40% less than? If has 40% less than one of those insane "energy drinks" (Mountain Dew MDX for example) then it probably has more sugar than Classic Coke too! Different kinds of soda have everything from zero sugar up to (probably) the maximum amount that'll stay dissolved in water without crystalizing out! It's nothing to do with whether it's carbonated or not. So if you need to reduce your sugar intake, switch to a diet soda. If you are concerned about the CO2 in the soda - then switch to a non-carbonated beverage. But don't give up either "just because". Read the ingredient list and compare the actual numbers - forget the silly marketting percentages. Think about what you are really doing! As others have mentioned, if you get your total carbohydrate intake down to zero - you'll get seriously ill amazingly quickly of something called "protein poisoning" (aka Rabbit starvation). SteveBaker 23:39, 30 October 2007 (UTC)[reply]
Actually sugar will dissove in huge amounts in water without crystalising out. I suppose the max would be something like golden syrup but you would't drink that. Theresa Knott | The otter sank 00:35, 31 October 2007 (UTC)[reply]
Lemonade mix is almost all sugar. 16 of a 17 gram Country Time serving, for example, and it may be even worse than that, when round off is considered (16.4 of 16.6 grams, perhaps ?). I'd avoid both sugar and sugar substitutes and drink water with lemon juice. StuRat 16:45, 3 November 2007 (UTC)[reply]

Will it fly?

I say yes, and so does straightdope, but what is the answer????--Goon Noot 23:08, 30 October 2007 (UTC)[reply]

Yes it will. Airplanes move due to thrust from the engines, not torque from the wheels. The plane's wheels will just be spinning at 2x the takeoff speed when it leaves the ground. ArakunemTalk 23:11, 30 October 2007 (UTC)[reply]


I say nay. There is no lift because the plane isn't moving forward.

It is moving, but it is on the treadmill, and the treadmill isn't moving forward.

strap that whole shebang on top of a car... then we're talking —Preceding unsigned comment added by 128.175.187.238 (talk) 05:59, 31 October 2007 (UTC)[reply]

-Stephen Hawking —Preceding unsigned comment added by 74.196.103.230 (talk) 23:15, 30 October 2007 (UTC)[reply]

It's an enormous treadmill, but the aeroplane's wings will be ripped off as it starts to fly. DuncanHill 23:17, 30 October 2007 (UTC)[reply]
In the original formulation of the question, the treadmill was infinitely long and the wheels completely free-running. The photo just confuses the question. The plane doesn't give a damn what happens to its wheels - they are free-running. Moreover, once it's engines are running with enough power to overcome the tiny amount of rolling friction in the wheels, the plane will take the same distance to take off no matter what speed the treadmill is running (forwards, backwards, who cares?). This one has been done-to-death and the answer is as clear and obvious now as it was at the beginning. I don't believe Stephen Hawking (or at least "The" Stephen Hawkin) said that. SteveBaker 23:21, 30 October 2007 (UTC)[reply]


It 100% will not take off. There is no way for it go gain lift. The plan engines thrust forth, pulling the wing, with the idea that the wing is pulled fast through the air, air goes over the wing and creates lift. There is no way for the plain to gain lift. People who think it could take off must also believe that a plane with no wheels could simply turn on its engines and hover up in the air.--Dacium 23:22, 30 October 2007 (UTC)[reply]
If you think about it, any plane is already on a giant moving treadmill. It's called the Earth. Remember that all motion is relative, so putting a plane on a treadmill moving with constant velocity is really no different from having a wind coming up from behind the plane. From the plane's perspective, it has to go a little faster to take off because of this wind. From your perspective, it takes off at normal speed, but it takes a little longer to reach it (since it started moving backwards with the treadmill) and the wheels will be turning faster. Now, you could also have the treadmill accelerating backwards at the same rate the airplane accelerates while taking off. In this case, the plane sits where it is, the wheels constantly speeding up, and the engines working to keep the plane exactly where it is (although, in that case, you could just turn the plane around and use the treadmill to launch it). Simply put, all motion is relative, being a treadmill won't have any greater effect on the plane than wind. Someguy1221 23:24, 30 October 2007 (UTC)[reply]
A plane moves forward by air entering the engines to create thrust. A plane has to build up it speed because there is only little thrust to begin with. As the plane gets faster, more air is entering the engine and the thrust increases. It is the air speed relative to the plane that is the only thing that matters. A plane could hover straight up in the air if the air speed was great enough. The treadmill stops the plane gaining speed relative to the air and stops it taking off.--Dacium 23:44, 30 October 2007 (UTC)[reply]
The plane still moves. Unless the treadmill is accelerating, the plane will eventually beat it and take off. Imagine the treadmill is massive, say, the Earth. Would the plane notice or care if the treadmill were moving? Not unless it's accelerating. So this is why I say, if the treadmill is moving backwards at 30mph, this is no different than a tailwind of 30mph. The plane has to speed up a little more (from its own perspective) to generate sufficient lift. But it is by no means prevented from moving. Someguy1221 23:49, 30 October 2007 (UTC)[reply]
Sorry I thought we were assuming the treadmill was accelerating to match the speed of the plain. Obviously if it doesn't the plain will still be able to gain speed relative to the air, and take off. So we are all in agreement then. Non accelerating treadmill - plain takes off. Accelerating treadmill - plane does not take off.--Dacium 23:52, 30 October 2007 (UTC)[reply]
The treadmill would have to have tremendous acceleration to slow the plane below its takeoff speed. (Remember, the wheels are free-wheeling. The only thing slowing the plane is double the normal amount of friction in the axles. The situation is nearly the same as if the plane had hovercraft instead of wheels. ) I'm pretty sure that a treadmill constructed to put out that sort of ridiculous acceleration would reach near-C speeds in a minute or two then the plane could still take off normally. 69.95.50.15 18:42, 31 October 2007 (UTC)[reply]
So I have this funny feeling I may have my relativity backwards on the accelerating part, someone please check me on that. Someguy1221 23:57, 30 October 2007 (UTC)[reply]
The wheels play zero part in generating forward speed. The engines react against the air to generate forward thrust. The treadmill could be going faster in "reverse" than the plane is moving forwards, and it will still move forward. The wheels would be spinning reeealy fast backwards but thats irrelevant to this. ArakunemTalk 23:33, 30 October 2007 (UTC)[reply]
It all depends on how you define the question. If the airplane's wheel bearings are frictionless and the wheels themselves are inertialess, then it doesn't matter what the treadmill does: the airplane's engines will move the airplane forward and it will take off. If the wheels have rotational inertia, or the wheel bearings have friction, and the treadmill is allowed to adjust its speed without limit, then it is possible to use the friction or rotational inertia of the wheels to counter the thrust of the engines, and the airplane won't take off. --Carnildo 00:03, 31 October 2007 (UTC)[reply]
I think it's best to assume the plane is just a normal plane, with normal wheels. Someguy1221 00:08, 31 October 2007 (UTC)[reply]
Say a big plane like in the picture takes off at about 150mph, do you think it could take off with a 150mph tail wind? I don't think so. So imagine the treadmill is already travelling with the plane on it at 150mph backwards, can the plane take off? I don't think so. Now imagine the plane and the treadmill both start at 0mph, can the treadmill accelerate the plane backwards as fast as the plane accelerates forwards? I don't think so. At the very start of the plane's acceleration as the engines spool up a considerable amount of the thrust would be used to overcome the friction in the wheels, but very quickly as the thrust increases, I think the treadmill would have to travel MUCH MUCH faster backwards then the plane accelerates forward for the friction of the wheels to act as a break on the plane's forward acceleration, otherwise the plane will come to speed and take off, just with very fast spinning wheels. Vespine 00:54, 31 October 2007 (UTC)[reply]
Yes, if the treadmill is allowed to transfer enough net force through friction in the wheels it could hypothetically keep the plane stationary. If the plane is stationary with respect to the air, it will not take off, since lift depends only on the plane's speed with respect to the air. Dragons flight 00:57, 31 October 2007 (UTC)[reply]
Just to straighten things out hopefully. Assume still air (no headwind or tailwind). If the conveyor belt/plane contraption is set up so that the plane remains motionless relative to the air (whether or not the conveyor belt, wheels, or engines are going) the plane will not take off as no air is moving over the wing generating no lift. If the plane is moving forward relative to the air, it could take off. Another way to think about this problem is to envision a plane on a conveyor belt that pushes the plane forward while the plane's engines are off and the wheels not rolling at all. Will the plane take off? Yes, because air is moving past the wings generating lift. (Of course to maintain flight it needs the engines, but that is besides the point) Or you can envision it this way: with a strong enough headwind, a plane perfectly at rest could take off although to a stationary observer on the ground, the plane would be moving backwards while acquiring altitude (i.e. moving in the direction of the wind). I hope this helps. Sifaka talk 03:59, 31 October 2007 (UTC)[reply]
No, it will hit the handrails of the treadmill. (Darn, DuncanHill beat me to that answer :P)-- Diletante 01:10, 31 October 2007 (UTC)[reply]

It's worth noting (belatedly) that we did this question to death back on May 21. —Steve Summit (talk) 01:31, 31 October 2007 (UTC)[reply]

More importantly, would a bird flying in front of the treadmill get sucked into the plane's engine? Rockpocket 01:39, 31 October 2007 (UTC)[reply]
Yes because the engines are drawing in air. 71.226.56.79 04:02, 31 October 2007 (UTC)[reply]
If the treadmill matches whatever the speed the plane can do, then surely it would stir up a huge surface headwind wouldn't it? Voilà, there's your moving air. --antilivedT | C | G 07:09, 31 October 2007 (UTC)[reply]
A treadmill moving at 100 or 150 mph wouldn't stir up that much of a headwind. (And stop calling me Shirley.) —Steve Summit (talk) 12:01, 31 October 2007 (UTC)[reply]
It doesn't have to limited to 100 or 150mph. Assuming it's a perfect treadmill it and the plane will simply accelerate until there's enough induced wind from the treadmill for the plane to take off. --antilivedT | C | G 05:13, 1 November 2007 (UTC)[reply]

It won't take off because there's no airflow around the wings! That's the whole point of why you need a runway for an airplane. You could say the Earth is a giant treadmill, but the air moves with the earth, and so do we. Air does not move with the treadmill. I'm surprised some of the people here thinks it will take off. It really makes me doubt how credible they were to answer my question about the ducted fans aircraft. Maybe it will take off if you have a powerful fan in front of the plane, but then you don't need a treadmill in the first place. Just thrust against the wind. 64.236.121.129 13:23, 31 October 2007 (UTC)[reply]

Aww you're mad *pats your head. You're also thinking of it wrong. I'm not going to bother explaining it because your last few posts prove that you aren't interested in listening. *pats your head*. Just because you don't understand it, doesnt mean it isnt so. Now you run on back to Starcraft lil guy —Preceding unsigned comment added by 207.67.148.212 (talk) 15:49, 31 October 2007 (UTC)[reply]
As I thought I made clear in my Earth/treadmill analogy, the only point of that is to dispel any belief that frame of reference is important. The only difference the plane cares about is the relative wind speed. Someguy1221 17:11, 31 October 2007 (UTC)[reply]
Actually, this is quite backwards. It takes off fine from a treadmill under the generally stated conditions (the treadmill moves at the same speed as the aircraft). Carnildo raises a good point that wheels aren't frictionless, and if you allow the treadmill to run so fast that the airplane is stationary, then it won't lift off (probably -- you're left with a weird feedback cycle). Thus, the key point of properly stating your constraints and assumptions is illustrated. On the other hand, a stationary aircraft in front of a fan can lift off, but as soon as it's no longer in front of the fan, it crashes due to the rapid change in airspeed over the wings (as opposed to the treadmill liftoff, in which the aircraft flies away normally).


In a related question, A friend once suggested to me that the real issue in this situation would be that the wheels would be turning at double their normal RPM (obviously). He suggested that the tires on many jets are not rated for those RPMs and would destroy themselves before the plane went airborne. Is this true? Stated more clearly, if a 747 was placed on a treadmill like the one described so that its wheels were always turning at double the normally expected rate, would the plane survive takeoff? 69.95.50.15 18:36, 31 October 2007 (UTC)[reply]

With no true evidence, I'll say "probably". The wheels undergo significantly more stress at landing (going from no speed to 150 MPH or so very quickly) and have a sizeable engineering margin above that. Simply spinning on takeoff should be no big deal. — Lomn 19:13, 31 October 2007 (UTC)[reply]
For the consequences of a tyre failure on takeoff, see Air France Flight 4590. Admittedly, the tyre failed due to debris on the runway, but airplane tyres were subsequently redesigned. -Arch dude 22:25, 31 October 2007 (UTC)[reply]
As i stated in my post, the "initial" resistance of the wheel would be very quickly overcome by the acceleration of the jet engine and I belive the wheels would have to spin MUCH MUCH more then twice as fast. Think of this, scale down the plane so it fits on a piece of paper on a desk, move the paper slowly and the plane will move with it, move the paper fast and the plane won't move, the wheels will just spin faster, i think the same thing will happen with a jet. Once the engines are at full thrust i think moving the treadmill backwards WON'T be able to stop the plane from accelerating, it will just make the wheels spin faster backwards, unless you start spinning then SO fast as to act as a break, in which case I think probably eventually fast enough to fail. Vespine 06:04, 1 November 2007 (UTC)[reply]
I was referring to the most commonly stated version of this problem where the treadmill goes backwards at the plane's take-off speed. In this case the plane takes off almost normally except for the tires which spin double their normal speed. 69.95.50.15 16:32, 1 November 2007 (UTC)[reply]

I think the main point here is that the air will not move with the treadmill. People seem to be thinking here that somehow either the treadmill can move the air, or the plane's engines drive the wheels. Either way you need air movement over the wings to generate lift. The treadmill's movement, forward, backward, whatever has no bearing on the air. The plane still needs to be pulled (by its engines) through the air forward to generate lift. The force from the engines act on the air, not the ground, so how the ground is moving has no bearing. The plane in that picture absolutely would not leave the ground. It's like having someone holding a hang-glider and running on a treadmill. They're not going to suddenly zoom up into the air, because the wing isn't travelling through the air. That's the whole point of a treadmill, so you don't move anywhere. —Preceding unsigned comment added by 132.244.246.25 (talk) 08:45, 1 November 2007 (UTC)[reply]

Hey, let's put this a couple of different ways:

  • You've got a 3 km runway, well and truly long enough for about any plane. But some tricky person has replaced the surface with a conveyor belt loop. The belt runs at 100km/h (~60mph). Can a Cessna 152 take off?

Plane is dropped onto the runway at the North end, facing north, and starts accelerating south (backwards). The engine's running and is throttled up to maximum, it starts moving forwards, engine slowed so it progresses backwards at walking pace (5km/h), that's groundspeed (relative to treadmill) +95km/h, airspeed -5km/h, groundspeed relative to control tower -5km/h. The Cessna gets half way down the runway, pilot decides to stop stuffing around, and throttles up to max, speed increases, at 120km/h airspeed it takes off, that's airspeed=120km/h, groundspeed relative to tower=120km/h (it's using up runway at 120km/h and getting rapidly closer towards the end), groundspeed relative to treadmill = 220km/h - well within design limits of the plane. Yep. It flies.

Of course, you can make the situation work or not work as you please by tweaking the setup. Sure you can find a treadmill speed where the plane's wheels are wrecked, sure you can make the plane too large or the treadmill runway too short for the plane to take off - but was that the question?

If the original question was "We have a cessna 152, with a flying speed of 120km/h, on a treadmill as long as the plane. The treadmill is run up to 120km/h, will the plane fly?" the answer to that is an emphatic "no", groundspeed is irrelevant to whether a plane flies. --Psud 10:00, 1 November 2007 (UTC)[reply]

You mean an emphatic "yes", right? Because the Cessna will have the same airspeed it always has. It's 'groundspeed' would be double if you're counting the treadmill as the ground, but like you said, the groundspeed is irrelevant. (The wheels are free-spinning, Imagine them as hovercraft if that helps you.) 69.95.50.15 16:38, 1 November 2007 (UTC)[reply]

Relative to the surrounding air this plane is motionless. Ergo the lift = 0. Ergo it won´t take off.
The tradmill is totally irrelevant.
--Cookatoo.ergo.ZooM 23:12, 1 November 2007 (UTC)[reply]

PS: It is identical to a normal plane on a "normal" runway taking off at a tailwind which inceases parallel to the speed of the accelerating plane. The Cessna / Jumbo / whatever could travel at the speed of Mach x or even v=c. As long as the velocity of the tail wind is identical, there are no aerodynamics to create a counter-gravitational force.
QED. --Cookatoo.ergo.ZooM 23:27, 1 November 2007 (UTC)[reply]
ZOMG, people are still not getting it.... but the plane is accelerating with JET THRUST not with the wheels, the plane, at full throttle, WILL accelerate through the air, regardless whether the wheels are spinning forward or backwards, the ground and wheel resistance is negligible to the thrust of the engines. there is a clip on youtube of a thrust test where a bus behind a jumbo jet, not RIGHT behind either, some way behind, is blown clean over and away by the blast from the engines. Put it this way, people agreed that the treadmill will NOT be able to move a significant amount of air with it, right? So it MUST also work in reverse, is the treadmill, even going super fast enough to counteract ONE HUNDRED THOUSAND LBF of thrust?? NO WAY! The plane WILL accelerate and WILL take off. Actually using that example, I've changed my mind from my previous post, I think it will accelerate and take off even if it starts at "going backwards" on the treadmill at whatever speed, it will just take longer for it to gather the airspeed... the people who think it will not take off are essentially making the assumption that a treadmill under the plane can counteract the acceleration of 100000 lbf of thrusting air, no way no how. Vespine 00:51, 2 November 2007 (UTC)[reply]

It's becoming clear that this is, in effect, a troll question. It is specifically designed to fuck with your head. The more you think about it, the more confused you get. Or, even if you don't overtly confuse yourself, you find yourself compelled to come up with ever-more bizarre theories to explain how the plane might not be able to take off.

And if you stop and really think about it, coming up with those ever-more bizarre theories is a pretty stupid thing to do. Remember, the primary purpose of the question is to fuck with your head and spark meaningless but prolonged debate. When the people who aren't thinking clearly say "Oh, right, of course the plane won't take off", the sadistic person who thought up the question in the first place calls them stupid and has a good laugh at their expense. But seriously: you think he's not also laughing at you when you get all "clever" and come up with some exotic and barely-plausible (well, even less plausible than a runway-sized treadmill, that is) theory under which you get exactly the same answer as the stupid people!?

As we all hopefully know by now, the proper response to trolls is to ignore them, to not feed them, to wait patiently for them to get bored and go away. And I think we had best apply the same remedy to this dorky airplane-on-a-treadmill question, too. (And soon, before this thread achieves critical mass and collapses into a black hole, or something...) —Steve Summit (talk) 04:57, 2 November 2007 (UTC)[reply]

Oops. I recant my above statements. The devil made me do it!
Due to a brief attack of fallibilty, causing delusions of omnisciensce, I uttered what may be classified as pure and unadulerated bovine faeces.
Thank God I am not a pope or president, but just a humble wheel on a treadmill, attempting to soar into the heavens of perpetual knowledge and eternal bliss...
I promise to take my prescibed medicine in future !--Cookatoo.ergo.ZooM 13:00, 2 November 2007 (UTC)[reply]


October 31

Running hands under Superacid

After reading Superacids, what would happen if someone washed their hands under a tap of fluoroantimonic acid? How fast would the hands disintegrate? Acceptable 01:35, 31 October 2007 (UTC)[reply]

I guess it is unacceptable to post chromic acid stories? Delmlsfan 01:51, 31 October 2007 (UTC)[reply]
I once breathed in fumes of hydrogen chloride gas (by accident), not a very fun experience... -- MacAddct1984 14:02, 31 October 2007 (UTC)[reply]
Once it hits the water of your hand, it's no longer nearly as super. That is, it becomes little different than a really really concentrated solution of other acids (hydrochloric, sulfuric, etc) in regards to its acid effects. On the other hand (sorry:) the fluoro and stibbic components might present other hazards all their own. DMacks 02:01, 31 October 2007 (UTC)[reply]

Transporting rice

Car Talk's most recent puzzler was something along the lines of:

With old wooden ships, what common household item, if not stored properly, would sink a ship in a matter of minutes?

Apparently rice was the answer. If it wasn't stored properly and got wet, the massive amounts of densely packed rice would swell and split the ship apart. Has anyone heard of this before? It certainly sounds plausible. -- MacAddct1984 01:42, 31 October 2007 (UTC)[reply]

Car Talk got this information from the book Tall Ships by Philip McCutchan.[4] The author’s name is misspelled on Car Talk’s web site.[5] That book says "Spaced along the upper deck were the cargo hatches with their heavy covers of reinforced hardwood planks, well chocked in and secured with three separate layers of tarpaulin, held down with ropes and more chocks to withstand the pounding of heavy seas. Below the hatches lay the reason for the ship's presence on the sea, her cargo, to be held inviolate against nature and disaster, against fire that could come from a self-combustible cargo like wool, or a cargo that could swell when it met water, such as rice, which on more than one occasion in the long story of the sea swelled and in its irresistible pressure split the sides of holds like paper and sank those ships in minutes."
But that’s just what Philip McCutchan said. From the British shipping act in 1875, it sounds like the primary concern about properly storing grain on ships at the time was actually that the ship could sink if the grain shifted.[6] Maybe this is a question for MythBusters. MrRedact 03:25, 31 October 2007 (UTC)[reply]
I guess what bothers me about this is that once the water gets at the outer regions of the rice cargo, that will start to swell and the pressure build-up ought to lock out the water from getting in much further. But a lot depends on how fast the water is getting in and how fast rice swells. I guess it could be true. Definitely a good one for Mythbusters though. Car Talk's puzzlers are not exactly reliable sources of information! SteveBaker 03:32, 31 October 2007 (UTC)[reply]
This was a plot in a Horatio Hornblower book. He lost a prize of war. The ship was sinking even though no water was in the "well". It gains weight as well as stressing the hull. --DHeyward 05:18, 31 October 2007 (UTC)[reply]
For the locking question, I think osmotic pressure will keep sucking the water in to the center until they are all equally saturated. A dry rice grain next to water saturated one will suck the water out of it. --DHeyward 05:21, 31 October 2007 (UTC)[reply]
DHeyward: that was not a book, but rather the second chapter of Mr. Midshipman Hornblower. Algebraist 21:08, 31 October 2007 (UTC)[reply]

Children's vs Adult's voices

  • What are the differences between children's and adult's voices, excepting the obvious physiological effects of growth and hormonal adjustments?
  • For example, children's voices tend to sound more sing-songy, breathy, and rhymical, but have there been any studies to make quantitative or qualitative measures of the difference in voice-qualities as age progresses?
  • It is obvious that human's are readily able to differentiate between younger and older speakers independently of any use of linguistic cues : are computers able to do the same, and under what heuristics would they operate?
  • Are there any posited hypotheses of the cognitive causes of these non-physiologically determined variations?

Many thanks 81.153.3.36 10:05, 31 October 2007 (UTC)[reply]

I think this belongs in the section on Language and Linguistics. Meanwhile, I'm not well studied at all in developmental linguistics, but I can try to answer with what I know.
Obviously, as you stated, as a child gets older his larynx expands and produces deeper tones as the vocal chords vibrate. Similar resonance differences occur as the size of the oral and nasal cavity expand. However, the prosodic changes (those in rhythm and tone) can probably be traced to what are still not well understood phenomena in child linguistics. For example, a toddler will show reduplication (repetition of words) and other types of word play, depending on age, as he learns to speak. He will also exaggerate and misuse intonation (the parents do this too when they speak to children for the precise reason that intonation is difficult to master). All of this is most likely a result of the problem of learning something as complicated as language simply from cues around you. Computational models of this type of learning are very simplistic, last I've seen, and usually involve limited degrees of freedom (such as a simple robot learning to walk or play the drums according to outside information). None that I know of are linguistic in nature.
It is true that the age of a speaker can be approximated by linguistic cues alone. This is done in precisely the same manner as you mentioned, as well as by judging vocabulary, grammar use, pronunciation, and pragmatics (appropriateness of responses). Computers are probably most advanced at this point in distinguishing pronunciation, but still trained phoneticians are used to diagnose and treat speech disorders, because software simply still isn't good enough. The variations you discussed are not well understood computationally or anatomically, so I don't believe there have been any good hypotheses other than those suggested by experience with children. SamuelRiv 04:02, 1 November 2007 (UTC)[reply]
Thanks SamuelRiv. I guess I wasn't as clear as I could have been. I was interested in specifically those differences that aren't linguistic in any way. That is, not vocabulary, grammar use, pronounciation or pragmatics, nor anything else for which it is require to assign meaning to the sounds. Rather, I wondered what kind of differences were recognisable in the pure sound-form of the voice, aucoustic qualities you might say.

In any case, I found a paper which is along the lines of the ideas I expected : [7] : though it still seems that there is woefully little work done on understanding speech outside of pathological/dysfunctional cases, which I guess is understandable, if not totally satisfactory. 81.153.3.36 12:09, 2 November 2007 (UTC) (Cross-posting to linguistics desk, just in case.)[reply]

Things stuck in eyesocket

Suppose something small and mildly irritating (like, say, a bit of cat fur) got stuck up under someone's eyelid and worked up into the eye socket. What would happen to it? Would the body destroy it or expel it somehow? How long would it take? Maelin (Talk | Contribs) 13:16, 31 October 2007 (UTC)m n[reply]

In general, this idea falls apart when you expect something to work its way behind the eye. The eye, if I remember correctly, is pretty much sealed in the skull. There's no way to, for instance, lose a contact lens behind the eye. If you somehow DID get something behind the eye, it would have been through a wound, and the effects of that wound would be worse than the object itself, I should think. --Mdwyer 14:39, 31 October 2007 (UTC)[reply]
I guess it's possible something could embed itself up there and work its way in somewhat. I would expect a cat hair to be absorbed over a long period of time because it's protein, but anything inorganic...I shudder to think. And I suppose that the body would try to do to such a thing what it does to all such things, like splinters. The eye is different though; inflammation alone can be a serious matter. See sympathetic ophthalmia here and here. I doubt the body would have time to deal with an embedded cat hair before the eye reacted unfavorably to it. You can tell by my wishy-washy reply that I'm no expert. --Milkbreath 15:43, 31 October 2007 (UTC)[reply]
The function of the eyelashes would prevent a situation like this from occuring. Their job is to keep foreign objects out of the eye (i.e. dust, dirt, debris, cat hair). They work in conjunction with your tears to do it. Even though the primary function of the tears are to lubricate the eye, they also function as a sort of safety mechinism, when an object gets into the eye (be it dirt, dust, or cat hair) the eye gets irritated and tears up. The offender then gets caught up in the tears and then is eaither flushed out or it gets near the eyelid and caught by the eyelashes and taken out of the eye. Lastly, a thin, but strong, membrane, called the conjunctiva, lines the inside of your eyelids and curls back on itself to cover the white part of the eye. This prevents any obects from going behind the eye. Hope this helps! Josborne2382 16:18, 31 October 2007 (UTC)[reply]
I've gotten specks of dust stuck between my eyelid and my eye. That's not quite behind the eye, but I still wonder how it always ends up getting out, and how I could get it out faster. — Daniel 03:26, 1 November 2007 (UTC)[reply]
I wear contact lenses around the clock, even while I sleep. There have been a few times where I've been woken by an irritated eye and found that one of the lenses has somehow moved to the top of the eyeball - i.e. not visible in the mirror. The process mentioned by Josborne kicks in and after a minute or two of teary blinking and rubbing the top of the eyelid, the lens is moved far enough forward for me to see it and gently remove it with my fingers. Oddly enough, in every case the contact lens has folded over on itself. I don't know if the lenses only float if they fold, or if they fold when they float high enough to encounter the fold of the conjunctiva. 152.16.59.190 04:32, 3 November 2007 (UTC)[reply]
I often have contacts "go for their morning swim" when I first put them in. They float back to places where I can't get at them, then work themselves back to the front after they finish doing their laps (a few minutes). StuRat 16:34, 3 November 2007 (UTC)[reply]

Hair loss

I asked my doctor about this but he didn't know exactly. I want to know if anybody has a link or two about uninterested reseaches on hair loss and effectiveness (if any) of treatments. I have quit given up hope on what constitutes medical advice so I'll just post fingers crossed and see if it gets deleted.193.188.46.254 14:03, 31 October 2007 (UTC)[reply]

The baldness article has a huge list of the various possible prevention/reversal treatments: Baldness#Preventing_and_reversing_hair_loss -- MacAddct1984 14:10, 31 October 2007 (UTC)[reply]

Freezing eye

Some time ago I read on how difficult it is for the eyes to freeze, but I forgot the reasons, could anybody patch me up on this? Also what would hypothetically hppend if you poured liquid nitrogen on somebodies eye? —Preceding unsigned comment added by 193.188.46.254 (talk) 14:16, 31 October 2007 (UTC)[reply]

I'd guess it is because they're salty, and salt depresses the freezing point. Still, LN is going to make them freeze anyway. Don't do it. :) --Mdwyer 14:34, 31 October 2007 (UTC)[reply]
It also helps that the human eye is embedded in a nice warm human being, equipped with all kinds of tools to maintain the body at a fixed 37°C. Heat is readily conducted from the back and sides of the eye socket (which, being right next to the brain, are well-supplied with nice, warm, constant-temperature blood) to the rest of the eye. Still, it would be possible to freeze the surface of the eye if you made enough of an effort—like by direct contact with a cryogenic liquid, for example. Tests in rabbit eyes gave the result:
Liquid nitrogen poured onto the eyes for one or two seconds with the lids held apart, produced no discernable injury. When the exposure was extended to five seconds, slight lesions of the corneal were observed. By the next day, all eyes were entirely normal.
I would expect longer exposure to do progressively more damage. Assumptions about the behaviour of human eyes based on the rabbit model should of course be taken with a grain of salt. TenOfAllTrades(talk) 18:43, 31 October 2007 (UTC)[reply]
Does it strike anyone else as odd that it is someone's job to pour chemicals into the eyes of rabbits? Man It's So Loud In Here 19:35, 31 October 2007 (UTC)[reply]
Rabbits are commonly used in animal testing. -- JSBillings 20:04, 31 October 2007 (UTC)[reply]
Not at all, how else would we determine the safety and efficacy of drugs formulated as eye drops? Rockpocket 20:21, 31 October 2007 (UTC)[reply]
How many brands of eye drops have liquid nitrogen in them? I'm no PETA pusher, but that's just insanely and purposelessly cruel to the rabbits. What's the next experiment: tossing bunnies into a wood chipper? Maybe we'll develop a better bandaid from it. Matt Deres 13:21, 1 November 2007 (UTC)[reply]
I agree. If scientists do things like pouring liquid nitrogen into the eyes of rabbits, which has little scientific benefit, the resulting backlash of public opinion could result in laws preventing important animal research, like testing eye-drops. Animal testing should only be used when absolutely necessary to protect humans. StuRat 16:29, 3 November 2007 (UTC)[reply]

Why do birds fly away from people?

I think this question is appropriate for the Science desk, since it seems to concern evolutionary biology. The question occurred to me this morning as I walked through a group of pigeons, which scattered as I approached them. This seems to be normal behaviour among our feathered friends, but when I stopped to ask myself why, I couldn't come up with an answer. In other words, why are birds scared of people? Of course, the answer that comes to mind first is that they are afraid of getting shot. Now, the chances of a pigeon being shot in a city street are pretty remote, although I guess the pigeon doesn't know that. So, is there some kind of evolutionary impulse at work here under which birds instinctively (and wrongly, in the case of city-dwelling birds) believe that man should be avoided? And if so, why hasn't this impulse atrophied among city-dwelling birds? --Richardrj talk email 14:25, 31 October 2007 (UTC)[reply]

Well, not just birds run away from humans. I can't think of any animal, unless it's a pet or trained, that wouldn't run away from a human. Humans are fairly big mammals, and if you saw something 10x the size of you, it's more than likely to benefit you to run away from it. Why hasn't the behavior atrophied? Probably because there is no selective pressure to fade out the trait. If running away from humans suddenly had a negative impact, such as they all started getting hit by cars when they try and fly away, then it's more than likely to weed out the trait. In the meantime, the pigeons have nothing to lose by flying away from approaching humans. -- MacAddct1984 14:39, 31 October 2007 (UTC)[reply]
Running/Flying away (Fight-or-flight response) is a pretty common behavior amongst animals. Not flying away from people is the unusual thing, probably bred into generations of city birds by the selective pressure of increased chance of food and less energy expense, as well as a lack of predators. Read about the Dodo to see what happens to a bird with few predators. -- JSBillings 14:40, 31 October 2007 (UTC)[reply]
(After edit conflict) A general instinctive fear of being killed and eaten by a creature much larger than themselves? The street pigeons seem to mostly tolerate humans, seeing them as something to simply move out of the way of - as they understand from experience that the majority of the large bipeds of the concrete forest pay them little attention and mean them no particular harm. They only seem to become agitated and take flight (around here, at least) if a human is getting too close, actively following them around, trying to trap them up against a wall, or running/walking towards them at speed - all of which the bird may (rightly?) interpret as hostile acts. --Kurt Shaped Box 14:43, 31 October 2007 (UTC)[reply]
Great answers. Thanks all. --Richardrj talk email 14:53, 31 October 2007 (UTC)[reply]
(another edit conflict) I think the 'rule' is probably more general than "man should be avoided" - something more along the lines of "if it's bigger than you and coming towards you, it should be avoided". Birds that tended to fly away when approached by something bigger than them would have an evolutionary advantage over those who hung around because that something may be a predator. Even if it's not a predator, it's extremely unlikely that it would be something that would provide the bird with some advantage over its fraidy-cat brethren who didn't stick around long enough to find out, so the wait-and-see trait would be disadvantageous and would die out. The impulse is still there because it's still advantageous, even amongst city-dwelling birds, to get out of the way of bigger things (imagine how odd would be if birds all wandered around on crowded streets, or didn't bat an eyelid if they were standing in the road and a car approached).
In fact, not all birds take flight as soon as they're approached by a human. Before it was banned, holding a handful of seed in Trafalgar Square was guaranteed to result in you being mobbed by pigeons. — Matt Eason (Talk &#149; Contribs) 14:58, 31 October 2007 (UTC)[reply]
Yes, they're afraid because instinct tells them anything that gets close to them is probably trying to eat them! Although they are used to humans, they are still preyed upon by cats etc. Feral Pigeons are sufficiently tame to approach a human if they have food and some birds can be tamed so that they are not afraid of humans, my two pet budgies will run up to me and climb onto my hand if they are on the floor and I hold my hand out. It's just a case of convincing the bird that you are not going to harm it which takes time and perseverence. At the end of the day, one of the most basic instincts of all animals is to survive which means avoiding anything that is likely to be dangerous.GaryReggae 15:01, 31 October 2007 (UTC)[reply]
Actually, some birds tend to be fond of humans. It was (and sometimes still is) for a human to be feeding food to gulls, pigeons, sparrows, ducks, geese, etc. Some other birds, especially flightless ones, are not afriad of humans. That's what killed the dodo so quickly. One time, for example, we were having a picnic. A small ~3 in. tall bird (not sure what it's called) came towards the food, and actually landed on the picnic table. If we tried to swat it away, it would fly away, hover in the air, and come back. It often landed about a foot away from us. Another time, we were just minding our own buisiness, and walking near the shoreline at a lake, when a few dozen ducks and geese who were near the shoreline seemed to follow us. As we walked by them, they continued to follow us. The moral of the story is, feeding of birds, even in the past, has caused some of them to follow humans and trust any food given by them. Now there are some laws restricting feeding of birds, but apparently the birds don't know that. Hope this helps. Thanks. ~AH1(TCU) 23:49, 1 November 2007 (UTC)[reply]

If human bodies had thick fur like other mammals, do you think we would still wear clothes?

She didn't. Rockpocket 20:25, 31 October 2007 (UTC)[reply]

Do you? 64.236.121.129 19:16, 31 October 2007 (UTC)[reply]

Hmm, that's a fun question. It depends where our sense of modesty comes from. I would imagine without the invention of clothing and the ability to cover up, there would be no way for embarrassment of genitals to come about. However, clothing also provides an additional sense of style and individuality. So maybe, just being human, we'd wear some sort of clothing anyway -- MacAddct1984 19:47, 31 October 2007 (UTC)[reply]
Individuality could be expressed by styling and coloring the fur, but I think clothing would still have been invented. Much of it provides a lot of utility beyond keeping warm. Lab workers, for example, would probably still wear some variation on a lab coat. Police officers would still wear Kevlar vests. I'm no expert, but it seems like if our current attitudes towards clothing evolved from the need to cover up in cold weather it wouldn't be surprising if other uses for clothed eventually evolved into similar attitudes. 69.95.50.15 20:18, 31 October 2007 (UTC)[reply]
Also where would you put your wallet if your a man? Certainly i'd still wear clothes, gotta have a place for my phone, keys, wallet and iPod...You can see why mugging is so popular! The above stuff is right too, clothes are cultural-identity too, they are also a sub-culture thing. Goths, punk rockers, skaters, sports-addicts, 'chavs', horse & hound types etc. A lot of clothes (and styles) seem to have started their life as work-based clothing and/or developed to become fashion/general wear. Jeans were work-clothes (still are), Cargo pants, three-piece suit (though Adam Hart Davis suggests it was an attempt by the British to reduce the love of French clothing in the Stuart period), sports-wear is obviously often derived from sport-use clothing. Obviously work-wise these things are not just to cover modesty but as a form of protection/security/uniformity too ny156uk 23:38, 31 October 2007 (UTC)[reply]
Umm, that's what bags are for. Malamockq 23:04, 4 November 2007 (UTC)[reply]
It's a man's bag! Someguy1221 00:01, 5 November 2007 (UTC)[reply]

How much sugar does the average kid/adult consume on Halloween?

^topic 64.236.121.129 19:23, 31 October 2007 (UTC)[reply]

Can you have an average kid or adult? ΦΙΛ Κ 21:09, 31 October 2007 (UTC)[reply]
Well you would expect that in any 'holiday' season people will allow themselves more treats, so possibly some like say three times as much as the 'average' sugar intake on any other day. This article (http://www.sys-con.com/read/451581.htm) is basically utter garbage but it does mention halloween and sugar intake. ny156uk 23:33, 31 October 2007 (UTC)[reply]
Ugh, I've heard stories of kids eating all their candy in one day, and there's often several hundred pieces of candy! No wonder people get sick and diabetes. Besides, if you're counting average as in world average, then the average kid doesn't go out for Halloween, either because the local culture doesn't celebrate it, and/or because the locals are too poor to afford that much candy. Besides, if you ate all your candy in one day, chances are a few of them may have been tampered with (I remember getting candy with the package opened). Some people argue that eating all your candy in one day has the same effect as eating it over a longer period. This is a complete falsehood. It might be a bit close to the truth if you're not going to have a single piece of candy or other treat for the next 6 months. Besides, eating all your candy in one day can cause blood sugar (and fat, and plastic, and colouring, etc) levels to suddenly spike up. It's kind of like drinking a gallon of alchohol in one day instead of the same amount over a 6-month peroid. Hope this helps. Thanks. ~AH1(TCU) 23:58, 1 November 2007 (UTC)[reply]

Cyanide and suicide

Is suicide via ingestion of cyanide, as seen in James Bond movies, fact or fiction? Babalonia 3. —Preceding unsigned comment added by 66.74.109.242 (talk) 20:41, 31 October 2007 (UTC)[reply]

As can readily be found in our article on cyanide, it can quite easily be used as a poison. Adolf Hitler, among others, used it as such. — Lomn 20:57, 31 October 2007 (UTC)[reply]
When you say "as seen in James Bond movies" do you mean ingestion of cyanide causing death in seconds? Sifaka talk 22:52, 31 October 2007 (UTC)[reply]
I thought Hitler died from shooting himself (though his wife and dog were killed by cyanide pills). — Daniel 03:18, 1 November 2007 (UTC)[reply]
I think you're right. I'll hunt for a reference and fix the article. — Lomn 14:27, 1 November 2007 (UTC)[reply]
The references in Adolf Hitler suggest that he both bit a cyanide capsule and shot himself, then had his body incinerated, just to be sure. — Lomn 14:30, 1 November 2007 (UTC)[reply]
Which do you think killed him? I don't care how fast cyanide can kill. A bullet to the brain would be faster. By the way, does bit a cyanide capsule mean he swallowed a pill, popped some kind of tiny balloon with his teeth, or what? — Daniel 01:54, 2 November 2007 (UTC)[reply]
Tiny balloon. Fribbler 23:26, 6 November 2007 (UTC)[reply]

Exercising, Muscle Burn and Calories

Upon reading over several articles (such as muscle and exercising ), I’ve got a few questions relating to “muscle burn” after a rigorous workout. Assume that one runs a mile on a treadmill. Most treadmills indicate calories burned over the duration of the run. Further assume that, according to the treadmill, 200 calories are burned (let’s pretend this number is accurate). If the runner has muscle burn from the rigorous run, does he/she actually burn more calories than the 200 indicated by the treadmill? I would assume so, as the body is ‘recovering’ from the workout, and thus will burn additional calories post-run.

Thoughts? Rangermike 21:25, 31 October 2007 (UTC)[reply]


See Lactic acid and Delayed onset muscle soreness, if you are referring to the fatigue and soreness one feels in one's muscles during and after workouts.
This is out of my field, but the body is burning calories with everything you do, say, or feel, but nothing burns as quickly from homeostasis as physical motion. For a physicist's example, say you were sick, and you had a fever of one degree celsius. At a weight of about 100kg, you would burn 10^5 calories = 100 kcal (kilocalories or kcal are the actual unit of measure when people refer to "calories". One calorie properly refers to the heat necessary to raise the temperature of 1 gram of water by 1 degree celsius). Keep in mind, this fever builds up over several hours, whereas you can burn that much in a 30 minute workout. SamuelRiv 03:28, 1 November 2007 (UTC)[reply]

Gravitational pull of the moon

Hi,
I've heard that there is no instrument on the earth which can measure the gravitational pull of the moon (apart from the obvious affects of the tides). Is this true? --124.181.69.55 21:35, 31 October 2007 (UTC)[reply]

Let's see here.... The Earth will pull on a 1 kg mass with a force of 9.8 newtons. Using an Earth-Moon distance of 380 000 km, the Moon overhead would tug on the same mass with a force of about 34 micronewtons. So the apparent weight of an object due to the moon's motion over the course of a full lunar orbit will appear to vary cyclically over about 70 parts per million. That's not a huge amount, but it's certainly measurable. Particularly if one monitored a very stable, very precise balance over the course of several lunar orbits, I would expect the effect of the Moon's pull to stand out as a periodic oscillation. A very quick Google search for microgram balances finds this one, which will set you back about $7000, and which will measure masses up to 5 grams with a resolution of 2 micrograms (about 0.4 parts per million). Can someone check my numbers? TenOfAllTrades(talk) 22:05, 31 October 2007 (UTC)[reply]
I get 7 ppm from your numbers (2 * 34e-6 / 9.8) — Lomn 22:25, 31 October 2007 (UTC)[reply]
In theory, there's no reason you can't. In practice, it may be the case that tolerances don't allow us to (I'm not sure). Just for reference, though, here's what you're dealing with. Gravitation is a function of mass over distance squared. For the Earth, we'll say that it's M/R2, where M is the mass of the Earth and R is the radius of the Earth, and we'll normalize that resulting number to 1. The Moon, on the other hand, masses roughly 1/80 Earth, and never gets closer (to the surface) than about 56R at perigee. Substituting, we expect that the Moon causes a discrepancy over a no-Moon Earth of 1/(80*56*56) -- that is, 4*10-6 or 0.0004%. Now, in practice, that discrepancy can be roughly doubled -- the difference between the Moon directly overhead and directly underfoot (noting that the move from 56 to 58 Earth radii is pretty minor at this point). So in total, to detect the influence of the Moon's gravity on Earth-surface objects, you need accuracy to six digits, or one one-hundred-thousandth of the object. (ec) I notice that I'm an order of magnitude off Ten's answer. Time to double-check. — Lomn 22:13, 31 October 2007 (UTC)[reply]
I will note, however, that you're monitoring a periodic variation of [whatever the correct magnitude, 7 or 70 ppm] over a day, not a lunar orbit (month). Monitoring over the course of a month could be used to confirm the roughly 25% variation in the daily cycle resulting from the eccentricity of the Moon's orbit. — Lomn 22:18, 31 October 2007 (UTC)[reply]
Oops, right. The Earth rotates once a day, doesn't it? Dang. Double oops; you're also right about the 7 ppm (not 70). I'm going to have to turn in my physics license. TenOfAllTrades(talk) 23:02, 31 October 2007 (UTC)[reply]
LIGO has to compensate for lunar tides, so there is an example of an active experimental instrument that is sensitive to the moon's gravity. Dragons flight 23:43, 31 October 2007 (UTC)[reply]
They did it on an episode of mythbusters. They used an extremely powerful accelerometer to measure if the anti-gravidy contraptions had any effect whatsoever (they didn't). They breifly mentioned the accelerometer measuring the tidal forces from the moon and showed the graph it made. — Daniel 02:01, 1 November 2007 (UTC)[reply]
You may be confusing an important law in Special Relativity, which states that a person in a closed space can never tell whether he is being accelerated due to external gravity or some other outside force (such as a rocket). See the article for details. You may also be thinking of the Cavendish experiment, in which the gravitational force of the moon, the planets, and the Earth were all simultaneously measured by measuring the gravitational constant G, which combined with Newtonian orbit theory gives everything you could possibly want about gravitational forces in the solar system. SamuelRiv 03:27, 1 November 2007 (UTC)[reply]
The "apart from the tides" part of the question is pretty telling though. If a simple yardstick stuck into the sand at the low tide mark can detect the presence of the moons gravity, it's unlikely that there would be nothing else that could do the job. SteveBaker 01:15, 2 November 2007 (UTC)[reply]
The tides are a tiny deformation in the earth. A proportional deformation on an artificial device would be ridiculously small. Measuring the tides with an accelerometer is essentially the same as measuring height above sea level to the meter with one. Although what anon heard is wrong, it's not particularly misleading. — Daniel 01:51, 2 November 2007 (UTC)[reply]
I'm sorry, my point wasn't clear -- all you need to measure the gravitational pull of the moon is Newtonian gravitation (i.e. Kepler's Laws) and a measure for G (Cavendish). Then from Earth using parallax or radar we can calculate the Earth-Moon distance, and we know the period of the Moon's orbit around Earth and so we can calculate the mass of the Moon, and the Gravitational field of the Moon is then , where M is the mass of the Moon and r is the Earth-Moon distance. SamuelRiv 02:15, 2 November 2007 (UTC)[reply]

are they flamible without fuel?

Various ether peroxides, like Acetone peroxide, Diethyl ether peroxide, and Tetrahydrofuran peroxide spontaneously explode when disturbed. Lab accidents involving such peroxides formed when a bottle of ether is exposed to light, oxygen, or catalytic metals and then subsequently ignored for a long time happen every once in a while. Sifaka talk 23:15, 31 October 2007 (UTC)[reply]
Thinking more about this, I bet some oxidizing agents with more complex chemical structures could react with themselves or other molecules of the oxidizer. There are probably also some nasty chemicals which have the potential to react under a variety of storage conditions without deliberately added fuel. Sifaka talk 23:28, 31 October 2007 (UTC)[reply]

oxidation

is mn203 a good pyrotechnic oxidizer? —Preceding unsigned comment added by 216.103.183.127 (talk) 21:59, 31 October 2007 (UTC)[reply]

I'm not getting any hits on google for manganese (III) oxide (also called manganese sesquioxide) being used as a pyrotechnic oxidizer. Safety precautions for the compound as listed in the sigma aldritch catalog (source) list it as an irritant to eyes, respiratory system and skin so I doubt people would appreciate it being scattered about by pyrotechnic displays where it could be inhaled. Among the applicable safety phrases it says don't add water to it. Of important note is that sigma didn't list it as an oxidizer, so I am leaning towards doubting its efficacy as an oxidizer at this point (I need to research around a little more). Many other manganese compounds appear to be fairly strong oxidizers including Manganese(III) acetate. Perhaps you meant to ask about one of them instead? Try looking at the manganese page. Sifaka talk 23:58, 31 October 2007 (UTC)[reply]

Tased across the spine

Can a person be paralyzed if the electrodes are opposite the spinal cord? I recently saw a video of someone getting tased in the back (the flying wires kind, not the little zapper kind) and it made me wonder --ffroth 22:14, 31 October 2007 (UTC)[reply]

See Taser for details. The weapon uses a shaped electric current to disrupt nerve function, I suppose in a process similar to a localized seizure. Unless structural damage occurs to nerve fibres or bone, the current itself shouldn't produce permanent effects. The brain and spinal cord both have a certain degree of plasticity (ability to change - i.e. learning) that is believed to depend on the intensity and duration of electric currents. See Electroconvulsive Therapy for information on the effects of high-intensity electric currents on the brain. Permanent effects to the spinal cord could possibly be similar, but would only affect memories stored in the spine, namely "muscle memory". —Preceding unsigned comment added by SamuelRiv (talkcontribs) 03:38, 1 November 2007 (UTC)[reply]
My answer is "Sure, why not?" We tend to think of a relatively robust person getting tased, but suppose a frail child gets in the way? The darts can penetrate up to an inch, and from what I saw poking around the internet, you get millisecond-range currents of up 8.5 amps. That's a lot of current, however short the pulse. If the darts get deflected so that they land right next to each other on either side of the spinal cord, I wouldn't be surprised if it caused electrical burns in the tissue in between. It was hard to get solid information on electrically caused tissue damage by googling, and I don't know much about the specific direct effects of the taser waveform on soft tissue. There is a lot of propaganda out there spewed by parties with an interest in tasers, for marketing, for furthering the use of non-lethal force by police, and for opposing tasering as an invitation to police excesses. The truth has been effectively obscured on the internet, it seems. --Milkbreath 10:34, 1 November 2007 (UTC)[reply]
Well, when we got our Taser's late last year, we all had to "take a shot" to carry them, I have video's of it all. AFAIK every charge was administered via clipping one probe to the right rear shoulder, and one probe on the left pants pocket cuff, then having 2 people hold onto the victi... er administeree's arms. I'd consider that across the spine, no ill effects from any of us(though thats not to say it isn't possible, I believe elderly people or pregnant woman is what our SOP say to try to avoid tasing), minus the videos floating around of us yelling 'oh fuuuuuuuu', and yes I am not ashamed to admit I screamed like a baby and being in that large of a muscle mass it really takes a lot out of you. Yes, I had what appeared to be tissue burns on the spots where the probes were, they went away after about 2 days, and I can say there was no pain there, even the second the charge was up, I felt shaky but nothing else. Even out support staff (dispatch, admin aide, records clerk, evidence tech) all volunteered to take a shot, and yes, the women did seem quieter than the guys. Dureo 11:00, 1 November 2007 (UTC)[reply]
Oh and remember, the probes can penetrate deeper in winter, there are specifically longer winter probes for winter clothing, and yes unless they are torn out as they fall we have EMS handle removing them. Dureo 11:04, 1 November 2007 (UTC)[reply]

secret service mobiles

Just curious about how this locking down on mobile signal works and how do the mobiles that are manufactured against it work.88.203.105.48 22:42, 31 October 2007 (UTC)[reply]

When a police service says they're locking down the mobile phone system (usually because of a bomb threat), they generally mean that they're ordering the mobile phone companies to stop routing phone calls to mobiles in the relevant area. It stops people using a mobile phone to trigger a bomb. Aside from that, I'm not really certain that that's what you're asking. Can you be more clear (for example, can you post a link to a website or news article that spurred you to asked the above question)? --Psud 10:18, 1 November 2007 (UTC)[reply]
It's all just a matter of software. Every phone has it's own unique number - the cell towers know the numbers of the phones in their area. In case of an emergency, the computer in the cell towers can decide what they want to do with calls. If it's an emergency, it could (for example) stop routing calls from any phone that's not listed as belonging to an emergency worker. This stuff is very easy to do if you have the inclination. SteveBaker 01:08, 2 November 2007 (UTC)[reply]

By locking down I meant pin point the location of the user of the mobile 193.188.46.254 09:18, 2 November 2007 (UTC)[reply]

They used to locate mobile phones by asking the phone company which phone towers it was communicating with, mobile phones keep in contact with several towers to allow instant handover as you move between cells. It's a simple matter of looking for the phone/person in the area where the various towers' signals overlap. That's the simple version anyway (and I don't know enough to give you the complicated one). The new way they look for phones is by asking the phone for its GPS position. Only works with phones with GPS. The old method works with any phone, I think. Perhaps it could be defeated by hacking the phone so it talks to the "wrong" mobile phone towers. It'd probably be trivial to hack a GPS equipped phone to report a wrong location. --203.22.236.14 09:58, 2 November 2007 (UTC)[reply]

If you read through the GSM specs you might find some information. But basically the terminal (the handset) measures the signal strenght to decide which tower it should 'speak' to, this strenght can (is) quantifiable and thus the handset would be able to determnine the nearest tower. It might be that towers also has the capability to measure the signal strenght from a particular terminal...and then you would be able to triangulate the position....I don't know all the details though :-) (aza----)

GSM phones can use Timing advance to determine distance from a known tower. Usually, the phone will hit only one of three antenna sectors around a tower. Many CDMA phones actually contact multiple towers at the same time, making triangulation much more likely. In addition, many CDMA phone have specialized GPS hardware. --Mdwyer 22:22, 5 November 2007 (UTC)[reply]

Rising Sea Levels and the Effect this Will Have on the Mediterranean Ocean

Given that the Med is regarded (more or less)as a tideless sea and that this is largely due to the influence of the Straits of Gibraltar which constrict the tidal bulge caused by the gravitational pull of the sun and/or the moon. At what point, if at all, or to what degree, in the earths future, will the Med see a real tidal effect once the seas have risen enough to overcome the resistance of the Straits of Gibraltar? Much of the Med coastline is at or very near sea level because there has never been any need to make allowances for tidal fluctuations. —Preceding unsigned comment added by 79.68.42.88 (talk) 23:08, 31 October 2007 (UTC)[reply]

Well much of all society is near the sea/lakes because historically we settled near places with ready access to water. The med problem is interesting. I expect it would require an extremely large increased. As the article (Strait of Gibraltar) notes water depth ranges from 300m to 900m, the opening is some 8 miles wide - having said this the end of the article has a bit on the 'need for a dam' so might be worth having a look at that link (http://www.agu.org/sci_soc/eosrjohnson.html) ny156uk 23:28, 31 October 2007 (UTC)[reply]

facial hair

Why do males have facial hair? Why don't females? —Preceding unsigned comment added by 68.231.151.161 (talk) 23:45, 31 October 2007 (UTC)[reply]

Vestigial trait. Doesn't serve any meaningful purpose unless you are white, it can sort of protect the face from UV radiation, but melanin (dark skin) does a much better job of that, and protects everywhere. So in other words, it's useless from a utilitarian point of view, but has cultural and social significance. Malamockq 01:24, 1 November 2007 (UTC)[reply]
Facial hair is a secondary sex characteristic resulting from the differential effects of sex hormones on the body. Rockpocket 07:50, 1 November 2007 (UTC)[reply]
Some women do have facial hair. -- JSBillings 10:25, 1 November 2007 (UTC)[reply]
In theory it helps to protect against the cold in high northern places like Scandanavia or Russia. As to why women dont, I dont know. But look at most antarctic scientists, they have big beard for the cold. If you have hair, when was the last time your head got cold? —Preceding unsigned comment added by 12.191.136.2 (talk) 15:08, 1 November 2007 (UTC)[reply]

I would conjecture that the standard explanation of gender differences, that of men going out to hunt and women staying at home with the family, can explain this. Men were more likely to find themselves in very cold situationsunexpectedly, where some extra insulation around the mouth would be welcome. There's a lot of blood in the lips and tongue, so it's an important area to keep warm. The women were relatively sheltered at home, so they could do without the facial hair, which gave them the advantage of being able to communicate with a greater degree of freedom, which would have been an important factor in the development of the children. I'm not sure how much I believe in this explanation myself, since people from warmer climates seem to have just as much facial hair as people in colder climates. I don't think that Inuit have a big culture of facial hair, and muslim countries certainly do. Maybe there's another hunting related advantage to beards.risk 04:15, 3 November 2007 (UTC)[reply]

And, once any difference in a trait exists between the genders, it will serve to differentiate the sexes. Thus, women will be attracted to men with facial hair and men won't be attracted to women with it, thus perpetuating the trait. It also works as an age marker, since young boys lack facial hair. StuRat 16:15, 3 November 2007 (UTC)[reply]

Finite from Infinite

Today's science dictates that not only is a theoretical construct of infinite density, energy, and size possible; it's actually how the universe started. How does a finite universe come from something that is infinite? —Preceding unsigned comment added by Sappysap (talkcontribs) 23:55, 31 October 2007 (UTC)[reply]

Firstly, there is no accepted theory suggesting the Universe is finite. While the observable universe is quite finite, the actual universe is possibly infinite in extent and mass. Further, physics does not posit that the universe began in a state of infinte density. It actually doesn't posit anything accepted about where it came from, as the Big bang theory only deals with what happened after (very soon after, though) the universe started existing. Someguy1221 00:27, 1 November 2007 (UTC)[reply]
Thirdly one way to look at it is fintite mass divided by zero size gives infinite denisity. Graeme Bartlett 00:30, 1 November 2007 (UTC)[reply]
I'm a huge fan of all of you guys who answer these questions...Someguy1221, Rockpocket, SteveBaker, Dragon's Flight and the whole crew. Please have infinite patience with me while I ask the following: How can the universe have infinite mass without infinite density? How can it have infinite extent and continue to expand? Sappysap 01:07, 1 November 2007 (UTC)[reply]
As for the first, I'm still trying to get my head around it, but for the second, imagine a piece of elastic, infinitely long, with small knots tied at regular intervals. Now imagine the elastic being stretched - you'll see the knots moving apart, even though you can't see the "ends" moving apart. Confusing Manifestation 02:00, 1 November 2007 (UTC)[reply]
To answer how the universe can have infinite mass without infinite density, all you have to do is think of numbers. There are infinitely many integers, as I'm sure you're well aware. There are also infinitely many numbers inbetween every two integers. And so, you manage to have a system of infinitely many numbers containing an infinite number of integers yet not every number is an integer. Actually, as you might suspect from there being infinitely many numbers between two integers, and only finitely many integers between any two numbers, each integer is like a tiny island in the sea of infinity. So, in conclusion, your infinitely many integers occupy an infinitely small portion of the number line. So I think you can see from this how you can have, in an infinitely large universe, infinitely many masses without filling up every possible space with a particle. As for having infinite extent, I haven't quite figured or accepted that one yet, except to say that lots of astronomers believe this (it is in part just a conclusion from the evidence that the Universe has no center or edge). Someguy1221 02:33, 1 November 2007 (UTC)[reply]
The universe has infinite mass if and only if it has infinite volume. Obviously everything around us has finite density, but if you allow yourself to collect more and more of it without limit, then ultimately you arrive at an infinite amount occupying an infinite volume. It is unknown whether the universe is truly infinite in extent, but it appears likely that it must be very much larger than the observable universe we can see. Dragons flight 03:26, 1 November 2007 (UTC)[reply]
None of the above physical quantities are or ever were truly infinite. The widespread modern cosmologies (that is, those based on Einstein’s general relativity, which came out about 90 years ago) all involve a universe that is finite in size and energy (or mass). And I’m talking about the entire universe, not just the visible universe. It isn’t really accurate to say that the density of the universe was infinite at the exact moment of the big bang, either. Singularities in a physical theory generally indicate a point at which the physical model breaks down, not that there really is an infinite physical quantity at that point. In this case, physics within the Planck epoch are poorly understood, but it appears that it is meaningless to talk about distances less than one Planck length, which is a finite distance, or time intervals shorter than the Planck time, which is also finite. So it isn’t meaningful to talk about energy densities greater than the (finite) total energy of the universe divided by the volume of a sphere whose diameter is a Planck length, or meaningful to talk about the exact instant of the big bang. MrRedact 02:46, 1 November 2007 (UTC)[reply]
I'm not sure what the initial question refers to, specifically. There are indeed important resolutions of infinite results in Quantum Field Theory, which I might be able to expand upon if a detailed question is posted, but all "physical" results must be finite by definition. Unresolved singularities (infinite results) signify a failure of a theory, not a physical reality. Cosmology works a little differently, but even then most cosmologies I know of do not accept any premise of infinity except perhaps in the context of a ground state (see vacuum). SamuelRiv 03:45, 1 November 2007 (UTC)[reply]
You are confused. General relativity permits solutions that are either infinite or finite in extent, and so is essentially agnostic on the issue. (See for example: Friedmann-Lemaître-Robertson-Walker metric, which permits different homogeneous and isotropic solutions discriminated primarily by whether the universe is finite or not.) For the universe to be finite, it must have a global curvature such that a person traveling in what appears to him to be a straight line will eventually come back to places he has been before. Such a curvature is possible, but not required, in GR. Your description of the Planck quantities is also problematic. It is not that there couldn't theoretically be smaller lengths, times, etc., but rather that describing events at those scales intrinsically requires both an unified understanding of gravity and quantum mechanics. In other words it requires an as yet ill-defined theory of quantum gravity. So, it is meaningless primarily in the sense that science as it now exists is not able to provide it meaning. Future theories may yet shine a light on events in and before the Planck epoch. Dragons flight 04:09, 1 November 2007 (UTC)[reply]
Ouch. I should have researched this one a little better before responding. I’m obviously not an expert in cosmology. Einstein himself assumed closed boundary conditions on his field equations, which implies a finite universe, but what slipped my mind for some reason was that not everyone assumes those boundary conditions.
My description of the Planck quantities very closely echoes a sentence in the Planck epoch article: "When quantum mechanics is combined with gravity, it turns out that it is meaningless to speak of time intervals shorter than the Planck time or distances shorter than one Planck length." That sentence, at least, is consistent with my understanding that "space" and "time" are ill-defined concepts at shorter times and distances. Of course, it’s all highly speculative, in the absence of a good theory of quantum gravity. MrRedact 10:36, 1 November 2007 (UTC)[reply]


November 1

What Are Some Things All Things On Earth Have In Common That Are Living Or Non-Living?

I wonder, what are some things all things on Earth have in common that are living or non-living? —Preceding unsigned comment added by 208.103.143.9 (talk) 03:07, 1 November 2007 (UTC)[reply]

Please clarify your question. Everything is living or non-living. — Daniel 03:15, 1 November 2007 (UTC)[reply]
They're all made up of atoms? -- MacAddct  1984 (talk &#149; contribs) 03:19, 1 November 2007 (UTC)[reply]
Not everything is made up of atoms. Evidence_of_evolution#Evidence_from_comparative_physiology_and_biochemistry lists many things that living things mostly have in common. As for non-living things, their only binding similarity is that they aren't living. Someguy1221 03:21, 1 November 2007 (UTC)[reply]
They're all made of matter and/or energy, they're all affected by the laws of physics, and they all hate getting circus peanuts on Halloween.  ;-) -- HiEv 04:07, 1 November 2007 (UTC)[reply]

Everything came from a star. --DHeyward 13:43, 1 November 2007 (UTC)[reply]

Everything? I'd be willing to bet that there's a great deal of primordial hydrogen floating around right now. --Carnildo 23:55, 1 November 2007 (UTC)[reply]
What part of ON EARTH, didn't you get? 64.236.121.129 13:38, 2 November 2007 (UTC)[reply]

They can all be mentally separated from their surroundings and referenced by a cognitive symbol, in such a way that an observer can build a working mental model with these symbols. In other words, what is a thing? When we draw a line around a cloud, why does the line go there and, not half around the cloud and half around the sky? Because classifying the cloud as an object helps us understand an survive in the world. Being a thing is not a natural property of a part of the universe, but rather a property that we ascribe to it. Or so I believe, anyway. risk 03:58, 3 November 2007 (UTC)[reply]

The definition of life would be involved here. Nonliving things take on some of the characteristics of life. Many nonliving things can grow or reproduce, like fire, crystals, and prions. Some viruses may also be considered nonliving, despite having some characteristics of life. StuRat 16:09, 3 November 2007 (UTC)[reply]

Contradiction in mass-to-energy equivalence?

I'm new at Wikipedia editing and whatnot, so I wasn't sure where I should put this where it would be noticed:



The article "Antimatter Weapon" states:

Quantities measured in grams or even kilograms would be required to achieve destructive effect comparable with conventional nuclear weapons; one gram of antimatter annihilating with one gram of matter produces 180 terajoules, the equivalent of 43 kilotons of TNT.


The article "TNT equivalent" states:

By E = mc^2, when 1 kilogram of antimatter annihilates with 1 kilogram of matter the reaction produces 1.8×10^17 J, which is equal to 42.96 Mt.



While I could just do the basic equation (mass*c^2, and convert joules to TNT equivalence), I didn't want to make an assumption without consulting anyone else first lest I made a mathematical error or overlooked some other factor in the calculation. —Preceding unsigned comment added by 24.65.12.157 (talk) 04:44, 1 November 2007 (UTC)[reply]

1 kilogram is 1000 grams; 1 Mt (megaton) is 1000 kilotons. Is that what you are confused about, or is there something else? Dragons flight 04:48, 1 November 2007 (UTC)[reply]
Ugh. I missed the "one gram" versus "1 kilogram" part. Well, we all make mistakes. C'est la Vie. —Preceding unsigned comment added by 24.65.12.157 (talk) 04:54, 1 November 2007 (UTC)[reply]

Medical question

Deleted. William Ortiz 08:38, 1 November 2007 (UTC)[reply]

Sorry, but this page isn’t a place to ask for medical advice, either. As it says at the top of the page: "Do not request regulated professional advice. If you want to ask advice that "offline" would only be given by a member of a licensed and regulated profession (medical, legal, veterinary, etc.), do not ask it here. Any such questions may be removed. See Wikipedia:Medical disclaimer and/or Wikipedia:Legal disclaimer. Ask a doctor, dentist, veterinarian or lawyer instead." MrRedact 09:04, 1 November 2007 (UTC)[reply]


Do you know where on the internet they give medical advice? William Ortiz 09:11, 1 November 2007 (UTC)[reply]

Here, if you're careful to hide the request for medical advice by disguising it as a question of biology, anatomy or the way medical devices work. But you've given the game away now, so I doubt you'd have much luck asking the same question again (by the way, I think what you suggest (a) wouldn't work, and (b) would be a bad idea). Actually, a google search would help you find online medical advice. --Psud 10:32, 1 November 2007 (UTC)[reply]
NHS Direct is a good place to start.--Shantavira|feed me 12:50, 1 November 2007 (UTC)[reply]

Was the Big Bang a chain reaction? Clem 09:59, 1 November 2007 (UTC)[reply]

If by "chain reaction" you mean a nuclear reaction then the answer is no, because protons and neutrons did not begin to form until about a millionth of a second after the Big Bang, and they did not begin to combine into stable nuclei until about 1 minute or so after the Big Bang - see our articles on the hadron epoch and Big Bang nucleosynthesis. Gandalf61 11:40, 1 November 2007 (UTC)[reply]
Of course not silly bro. How can there be a chain reaction with particles not yet formed? I'm talking about with whatever was formed. Did it have a chain reaction? Clem 14:05, 1 November 2007 (UTC)[reply]
There's no need to call a volunteer "silly" for answering your question as phrased -- especially when he then expands your question to give you exactly the answer you wanted. It's rather rude and quite unwarranted. Another example of what Gandalf linked is at timeline of the Big Bang, particularly the 17-minute period of nucleosynthesis which established the initial ratio of hydrogen to helium. — Lomn 14:21, 1 November 2007 (UTC)[reply]
I think most interactions that occurred immediately after the big bang--quark-gluon interactions to form nucleons and nucleons and leptons forming atoms--are not chain reactions because there are no particle emissions necessary any of these reactions - a chain reaction is a series of reactions that require as input some output from a previous reaction, so that the previous allows the next to occur and so on. So I would say the answer is no in general, but specific reactions might be catalyzed from energy or lepton release by some of these primitive interactions. SamuelRiv 14:28, 1 November 2007 (UTC)[reply]
I suppose I'm thinking more in terms of instances of interrupted stability wherein under different circumstances some changes might not occur at all or occur as rapidly except for the occurrence of previous events such that event of type "B" will cease to occur if and when event of type "A" ceases to occur. Maybe "sequence of dependent events" is more apt terminology than "chain reaction". Clem 17:09, 1 November 2007 (UTC)[reply]
Okay then, you are asking a very general question. All big-bang cosmologies that I know of come from very fundamental fluctuations in the background. A fluctuation may create a false vacuum, for example, and the cosmological effects are calculated from that state alone. So everything follows from this one single event, and yes, from there everything is dependent on what comes before. Caveat--the notion of dark energy may prove that some critical events in the universe's history are not caused by a "sequence of dependent events". SamuelRiv 02:24, 2 November 2007 (UTC)[reply]

"Superlinear"

First things first, hi desk, long time no see! Anyone have any idea what this word "superlinear" means? My first guess would be a function that increases faster than a linear function, but then you can of course have an incredibly steep linear function. A friend of mine asked me in the context of a biological function. Can someone clarify? It's not something that i've heard in my undergraduate career. Google has a load of results but none of them are very clear from what I can see. Could it be a function that is more linear than a linear function? That makes no sense to me either. Math board, maybe? :? —Preceding unsigned comment added by Capuchin (talkcontribs) 11:59, 1 November 2007 (UTC) Cheers, Capuchin 11:39, 1 November 2007 (UTC)[reply]

When describing functions, "superlinear" appears to mean a function with asymptotic growth that is more than linear i.e. it grows faster than any multiple of x (there is a different, but related, usage in "superlinear convergence"). Superlinear functions include functions such as x2 or indeed any power of x greater than 1, also xx, and functions such as xlog(x), xlog(log(x)) etc. Note that even a "steep" linear function will eventually be overtaken by a superlinear function - if f(x)=Ax and g(x)=Bx2, then no matter how large you make A and how small you make B you can always find a lower limit y such that g(x) > f(x) for all x>y. Gandalf61 12:04, 1 November 2007 (UTC)[reply]
Okay thank you, that's what I had assumed. Capuchin 12:10, 1 November 2007 (UTC)[reply]

Do all the biochemical reactions happened in our body have a negative free energy?

I am a high school student. Yesterday, my teacher teacher give us a question written: Do all the biochemical reactions happened in our body have a negative free energy. Explain and support your answer. I don't know how to answer it. Any one can help. Thx in advance!!

—Preceding unsigned comment added by 202.40.139.171 (talk) 12:32, 1 November 2007 (UTC)[reply] 
Wikipedia has articles such as Gibbs free energy and enzyme kinetics, but they either do not address your question or they do not address it in plain English. This biochemistry textbook has a standard treatment of free energy in biochemical reactions for college students. If you are not ready for the math skip down to, "the free-energy change must be negative for a reaction to be spontaneous." The complication in biochemical systems is that there are many "coupled reactions" in which one chemical reaction that has a positive free energy change is "driven" by coupling it to a second chemical reaction that has a negative free energy change. Example: Two-step conversion of glyceraldehyde 3-phosphate. So the answer to "Do all the biochemical reactions happened in our body have a negative free energy?" depends on how you define "biochemical reaction". If you define "biochemical reaction" to include both of the chemical reactions in a pair of biochemically-coupled chemical reactions then yes, the combined free energy changes of the two chemical reactions are negative. If you define each of the individual coupled chemical reactions to be its own "biochemical reaction", then it is clear that some individual chemical reactions that take place in biochemical systems have a positive free energy and you could call those "biochemical reactions". --JWSchmidt 15:45, 1 November 2007 (UTC)[reply]
If you have a library card you can get help with homework here. 71.100.9.205 14:14, 1 November 2007 (UTC)[reply]
Put it another way: do all biochemical reactions in the body happen spontaneously? That is, do they all occur as soon as it is possible for them to occur? Think about what this would mean. SamuelRiv 14:32, 1 November 2007 (UTC)[reply]
Spontaneous human combustion? Clem 18:40, 1 November 2007 (UTC)[reply]

Blood phobia

How do extremely hemophobic women cope with getting their period? --124.254.77.148 13:11, 1 November 2007 (UTC)[reply]

With great difficulty I would imagine. Your question sort of reminded me of the Ashley Treatment. Short of hysterectomy or Endometrial ablation, I suppose they could reduce the number of periods they have using one of the many pills, but would still have to go through a few. Alternatively, maybe frequent exposure reduces the horror? Skittle 13:48, 1 November 2007 (UTC)[reply]
Yes, but even on those regimes you have to have a few periods a year, otherwise you get bleedthrough. Hence why I said you could reduce the number of periods using the pill. If I were unaware of the 'placebos' (they're not true placebos, more spacers to stop you losing track or getting out of the habit), I would have thought you had to have the same number. Skittle 17:22, 5 November 2007 (UTC)[reply]

constraints based on age

Is there a chart of minimum and maximum amounts (or rates) of oxygen, water, food, temperature, blood, exercises, heart rate, Vitamins, etc. within which the human body must stay to survive at different ages? For instance, heart rate probably has a narrower range for older people than for younger people as well as exercise, etc. 71.100.9.205 13:23, 1 November 2007 (UTC)[reply]

Probably not, as there are more factors than simply age which determine minimum requirements, and the "maximums" are often subject to a wide variety of opinions. Also, various factors often have an effect on the limits of other factors, meaning that having a lot of A or B could be fairly safe, but having a lot of A and B could be quite dangerous. Height, weight, health, and genetics are usually more important factors than age, especially after the body has reached maturity. Furthermore, people can be conditioned to survive extremes, such as the ability to handle low oxygen environments due to altitude training. As such, rates can only be given in the most general terms, like three days being about the limit people can normally survive without water, not in specific age-based charts like what you're asking for. -- HiEv 08:12, 3 November 2007 (UTC)[reply]

Exercise/muscle Pain

Why do personal trainers insist on creating so much pain for a person that hasn't exercised in a long while? What I mean is what benefit comes from torturing someone to the point that the next day they can't even climb stairs without extreme muscle soreness. I would think that muscle gain should be built up slowly over time. As mentioned in the earlier question, it appears that the popular explanation for muscle soreness are minuscule muscle tears. How can that be healthy?! --WonderFran 13:45, 1 November 2007 (UTC)[reply]

As I understand it, it's healthy because that is precisely the mechanism by which muscles build themselves up: each little tear is repaired by the muscle with stronger fibers, which over time can increase the strength and size of the muscle. See the heading "Recovery" at Strength training. On a larger scale, this would become a muscle strain. jeffjon 16:35, 1 November 2007 (UTC)[reply]
I heard the pain was caused by the buildup of lactic acid. They told me that in high school biology. — Daniel 23:38, 1 November 2007 (UTC)[reply]
The other relevant article may be DOMS. Vespine 23:59, 1 November 2007 (UTC)[reply]
The long-term lactic acid soreness idea is incorrect and based on a faulty experiment done decades ago. See Delayed onset muscle soreness. —Preceding unsigned comment added by SamuelRiv (talkcontribs) 14:34, 2 November 2007 (UTC)[reply]
  • It would be very foolish indeed for a personal trainer to behave this way. Not building up intensity slowly increases the chance of injury, and greatly increases the chance that the victim will give up on exercising. If your PT did this, go find a new one. --Sean 15:37, 2 November 2007 (UTC)[reply]

Water - Oxygen redox?

Can water be considered chemically reduced Oxygen? Think outside the box 13:51, 1 November 2007 (UTC)[reply]

Yes. The burning of a mixture of H2 and O2 is just a redox reaction between them: hydrogen gets oxidized to H+ and oxygen gets reduced to O2–, and the result of those is H2O molecules. DMacks 14:15, 1 November 2007 (UTC)[reply]
Thanks DMacks, Think outside the box 14:17, 1 November 2007 (UTC)[reply]

Michelson-morley experiment

See WP:RD/M#Michelson-morley experiment or Wikipedia:Reference_desk/Archives/Miscellaneous/2007_November_1; answers already exist there (even if this is a more topical location) — Lomn 14:36, 1 November 2007 (UTC)[reply]

Game of life

When personal computers first started there was a game called truck driver which had the objective of seeing who could drive cross country for the least cost and highest profit in the shortest time. The program would throw flat tires at you or a gas station with a bad pump or an oil light. Is there a game yet to see who can live the longest at the least cost and greatest profit, most successful offspring with things like your car getting stolen or your house catching on fire, etc.? 71.100.9.205 13:55, 1 November 2007 (UTC)[reply]

Creating a game of that scope seems a bit excessive, but I suppose The Sims (or The Sims 2) isn't far off. Of course, there's also The Game of Life, which is fairly close to those parameters as well. — Lomn 14:17, 1 November 2007 (UTC)[reply]
Sounds like The Oregon Trail (computer game). Didn't realize how old that was. --— Gadget850 (Ed) talk - 02:19, 2 November 2007 (UTC)[reply]
Yeah, this is a bit better than truck driver in that there seem to be more things that can happen (variables). Is there anything like this say to prepare a student for college or even high school? 71.100.9.205 16:01, 2 November 2007 (UTC)[reply]

Gleason score image

I am trying to obtain orginal artwork for the image of the Gleason Scale that appears on Wikipedia. I know the copyright is in the public domain. Can you please advise where I can obtain the original artwork or any suggestions of where I can search for it?

Any help would be much appreciated.

Thank you for your help.

Most sincerely,

Sharon Strompf —Preceding unsigned comment added by 66.252.164.210 (talk) 16:29, 1 November 2007 (UTC)[reply]

(email, phone, etc removed) --Bennybp 16:47, 1 November 2007 (UTC)[reply]
If you're looking for the original image that User:InvictaHOG based the public domain image upon, it seems likely that the "1977 Scientific Article by Gleason" mentioned on the image page is the same as the article listed under "References" in the Gleason score article. Your local library might be able to help you track down that magazine article from "Urologic Pathology". jeffjon 18:18, 1 November 2007 (UTC)[reply]

5-digit "telephone" numbers

Where are / Who owns those 5-digit numbers that we (in the US) are encouraged to send text messages to, in order to vote for some contestant or enter a contest? Are they "real" telephone numbers?
(Once upon a time, there were 5-digit numbers for the old telex/tty network, but those were phased out decades ago, I believe replaced with standard-format numbers in the North American Numbering Plan, area codes 310/510/710. Has *that* technology been resurrected??) (By the way, if there's an answer to this in Wiki already, it's buried too deep, or it's too hard to ask the right question to find it...) 66.47.7.76 17:13, 1 November 2007 (UTC) DanH.[reply]

See Short code. -- Coneslayer 17:33, 1 November 2007 (UTC)[reply]
(after edit conflict) Based on this question/reply at Google Answers, they are called Common Short Codes. Having just taken a gander at the "what links here" page for Common Short Codes, I don't know how you'd have found that without knowing what to look for. I wonder what other articles might constructively link there so it's more easily found? --LarryMac | Talk 17:38, 1 November 2007 (UTC)[reply]
Awww, I was gearing up to do a whole historical dissertation on the old "KLONDIKE-5 1212" type numbers. :) ArakunemTalk 17:40, 1 November 2007 (UTC)[reply]
I found Short code from Text messaging (and had come across the term before—I couldn't remember it, but I knew it when I saw it). I've added merge templates to propose merging Short code and Common Short Codes. -- Coneslayer 17:41, 1 November 2007 (UTC)[reply]
Actually Common Short Codes looks like a fork started for promotional purposes. It was a redirect, then someone replaced it with promotional material. Probably a spammer working for a related company. In fact, the text is entirely taken from usshortcodes.com [8][9][10]. I'll tag it and we'll see what the copyright police think. --tcsetattr (talk / contribs) 21:19, 2 November 2007 (UTC)[reply]

I stand IN AWE of the user community here. Half an hour, and everything I wanted to know appears as if by magic. Thanks to All! 66.47.7.76 19:06, 1 November 2007 (UTC) DanH.[reply]

Particle geometry

The components of the nucleus of an atom, protons and neutrons, are always shown as individual spherical particles. While this must of course be the case when they are not part of the nucleus of an atom, is it correct to assume that when in the nucleus of an atom they loose their individual spherical shape and become combined into a single spherical mass or globe called the nucleus until an imposing force adds or subtracts one or another or splits the atom in two or do they maintain their own independent spherical shape within the nucleus? Clem 18:27, 1 November 2007 (UTC)[reply]

I don't know what distortions may occur, but individual nucleons will remain individual nucleons. Since nucleons lack a color charge, they don't interact nearly as strongly as quark pairs, and a giant conglomerate mass of quarks would actually be quite unstable, evidenced by the difficulty of building stable quark structures larger than nucleons within particle accelerators. Someguy1221 19:47, 1 November 2007 (UTC)[reply]
I’m not an expert in nuclear physics, but I’m pretty sure I know the right answer by analogy to atomic and molecular physics:
Even representing a nucleon as a sphere is to some extent just a representational device to aid in comprehension. No bound state has a precisely defined boundary like a sphere. Picking a particular radius for a sphere representing a free nucleon involves somewhat arbitrarily choosing a particular “equipotential surface” (although that’s not quite the right phrase) of the nucleon’s wavefunction.
The wavefunction for a nucleus is different from just the sum of the wavefunctions of a bunch of free nucleons, so in that regard you could think of the nucleons as “changing shape.” But in reality, there are no precise boundaries between where one nucleon ends and an adjacent nucleon begins. MrRedact 20:43, 1 November 2007 (UTC)[reply]
On the other hand, even if the best we can do is some sort of equipotential surface or probability cut-off, that doesn't mean we can't talk about the shape of that...thing. It's not the physical boundary in the sense of interactions such as touching or seeing but it's still a useful description if you care about electrostatic potential, electron distribution, nuclear-capture cross-section, etc. Interestingly, Nuclear isomer teaches us that spherical is not even a good approximation for the nucleus (though it certainly suffices in most situations). DMacks 20:50, 1 November 2007 (UTC)[reply]
There’s a complication here in that the exchange symmetry of identical particles makes it impossible to define precisely which of two identical nucleons exists at a precise location. I’m pretty sure there really is no way to define a precise boundary between adjacent nucleons with complete accuracy. Any concept of the "shape" of a nucleon in a nucleus has to be at least a little bit vague. But I don't know enough about nuclear physics to come up with numeric values for "how vague" it has to be. MrRedact 21:41, 1 November 2007 (UTC)[reply]

To clarify... what I'm asking is whether the protons and neutrons which make up the atomic nucleus exist in a geometry similar to a bag of marbles or whether they meld and exist similar to the configuration of a two component sphere like water and oil at zero gravity inside a water (and oil) balloon? Clem 03:47, 2 November 2007 (UTC)[reply]

Ignoring all quantum mechanics about the particle (or group of particles) itself, it will be a sphere as its interactions are spherically symmetric, no matter how many nucleons make up the nucleus. This isn't 100% true, as there is one biased direction: that of angular momentum or spin, so the nucleus actually will "look" like an ellipsoid for most measurement purposes. Numerical analyses of particle accelerator data seem to agree with an ellipsoidal geometry for most individual particles and bound states. SamuelRiv 04:00, 2 November 2007 (UTC)[reply]
More like a bag of marbles. You can model an atomic nucleus very well by treating the protons and neutrons as almost-free particles that interact with each other occasionally. They definitely do not meld together to the point of making quark soup. As Steve and others point out, the "bag of marbles" is not a perfectly spherical bag, but can be stretched out in certain directions. There is also a bit of quantum mechanical fuzziness on the identity of nucleons inside the nucleus, but the "bag of marbles" is a good picture. The protons and neutrons don't all melt together. --Reuben 07:34, 2 November 2007 (UTC)[reply]
To refine this picture a little, since according to Reuben the particles are almost free, maybe you could think of the nucleus as a bag of marbles that can magically pass through each other. MrRedact 08:00, 2 November 2007 (UTC)[reply]
The nucleus of some atoms are slightly cigar-shaped [11], one example being tantalum-180. [12] For the individual nucleons (protons and neutrons) however, it would be wrong to say that they have ever been spherical, seeing as they'r each made up of three point-particles and the force carriers between those. That said, while their location is not necessarily well-defined, we can say that they don't "melt", as they don't, to my knowledge, exchange quarks with each other.
See liquid drop model. Gandalf61 16:59, 2 November 2007 (UTC)[reply]

Electrical path

If your elbow say were grounded but your feet and the rest of your body were well insulated and your fingers touched an exposed appliance cord, say in the UK where line voltage is 240, would the electricity travel between your fingers and elbow or would it travel through other parts of your body as well? Clem 20:53, 1 November 2007 (UTC)[reply]

The current density would be highest between the two contact points, along the "path of least resistance" as they say. —Keenan Pepper 21:44, 1 November 2007 (UTC)[reply]
However, the distribution of current density can be quite complicated and difficult to calculate. See Analysis of Current Density in the Carpal Tunnel Region During an Electrical Accident by way of the Finite Element Method (PDF link) for an example. —Keenan Pepper 21:51, 1 November 2007 (UTC)[reply]
(edit conflict) I had to go get my thinking cap out of the closet for this one. Almost all of the current would take the short path, but I would think that some electrons in the rest of your body would move. Current flows through every path available to it as long as there is a potential difference, in proportion to the resistance it encounters. Some of the tissues in your forearm will have relatively low resistance, blood probably having the lowest. Your circulatory system is a network, and the current will see a path all through it, mostly in the straight line, but some more roundabout. It's a bit like having a 50-ohm resistor in parallel with a 50-meg, say. The same goes for all your tissues, but to a lesser extent, I would think. But in my experience (is that allowed here?), the current in the rest of the body at 240V will be negligible, and you'll be too busy trying to extinguish your fingertips to even notice it. --Milkbreath 21:59, 1 November 2007 (UTC)[reply]
Assuming the capacitance of the rest of your body to earth was small, almost all the electric current would flow between your fingers and your elbow. In practice, the capacitance of your body to ground would not cause appreciable( dangerous) currents to flow through the rest of your body. —Preceding unsigned comment added by 88.111.55.77 (talk) 02:23, 2 November 2007 (UTC)[reply]
From personal experience, I'll say it felt like it went straight from my hand to my elbow, however I should note that it was my right hand and my left elbow. Fortunately for me the current had already been stepped down, so I was surprised, but uninjured. -- HiEv 09:20, 3 November 2007 (UTC)[reply]

Unlimited source of free energy?

If there was a device that did nothing all day except turn neutrons into protons would we have a source of unlimited energy? Dichotomous 21:14, 1 November 2007 (UTC)[reply]

  • You'd run out of neutrons after a while. It's probably worth elaborating that there's a finite (if large) number of neutrons within the observable universe, and a much smaller fraction within your future light cone (assuming, of course that w = -1, so there are only so many neutrons you could collect. Cheers, WilyD 21:16, 1 November 2007 (UTC)[reply]
And most neutrons tend to be very annoyingly inside of atoms, and for the majority of matter in the universe, extracting those neutrons would require more energy than you'd get from fusing the resultant protons. Someguy1221 21:25, 1 November 2007 (UTC)[reply]
Exactly. A free neutron has more energy than a free proton, but a helium-4 nucleus has less energy than the sum of two free protons and two free neutrons (because of the attractive nuclear force that binds them together). So your device wouldn't work, because we don't have an unlimited source of free neutrons. —Keenan Pepper 21:32, 1 November 2007 (UTC)[reply]
Although as the hypothetical device seems to violate charge conservation, then I guess anything is possible. Gandalf61 16:56, 2 November 2007 (UTC)[reply]
Besides, we already have an unlimited source of free energy, the Sun, which puts out around 400 yottawatts, 24 hours a day, 365.2425 days a year. The hard part is collecting it. Even if we had a (conservation-law-violating) device which converted protons into antiprotons for free, it would be hard to build a safe and reliable power plant around it. The devil is always in the details. -- BenRG 17:13, 2 November 2007 (UTC)[reply]

lab procedure- AP Chem exam style question

Could anyone please help me with these two questions?

1. Explain how you would go about making 3.00 L of 0.005 M NaOH. Include lab materials that you would use and diffrenet steps you would take in preparing the solution. (Hint: NaOH is solid at room temp.)

2. Explain how you would go about m,aking 3.00 L of 0.500 M H2SO4. Keep in mindo that H2SO4 is a strong acid and starts out as a 12.00 M solution.Include lab materials that you would use and diffrenet steps you would take in preparing the solution. —Preceding unsigned comment added by 76.214.218.77 (talk) 21:37, 1 November 2007 (UTC)[reply]

Materials: 1 grad student. Method: "Hey you, make me 3 L each of 0.005 M NaOH and 0.5 M H2SO4."
Seriously though, the reference desk is not a homework answer service. Dragons flight 21:57, 1 November 2007 (UTC)[reply]
"Do your own homework. The reference desk will not give you answers for your homework, although we will try to help you out if there is a specific part of your homework you do not understand. Make an effort to show that you have tried solving it first." -- MacAddct  1984 (talk &#149; contribs) 00:08, 2 November 2007 (UTC)[reply]

Medical question

I just cut my head off. What sort of treatment is appropriate? --67.185.172.158 23:58, 1 November 2007 (UTC)[reply]

I hear snake oil is a very effective cureall -- MacAddct  1984 (talk &#149; contribs) 00:05, 2 November 2007 (UTC)[reply]
1) Immediately apply a tourniquet to your neck to stop the gushing of blood from your carotid arteries.
2) Put your head in a cooler, and pack the cooler with ice.
3) Dial 911, and use Morse code to request an ambulance to take you to the nearest emergency room.
Don’t delay, as decapitation could be a sign of a serious medical condition. MrRedact 00:40, 2 November 2007 (UTC)[reply]
And decapitations lasting longer than four hours may require the treatment by an undertaker.
Atlant 12:20, 2 November 2007 (UTC)[reply]
Wikipedia cannot give medical advice. I understand you're all busy, with the bleeding and the flopping around and all, but you should take time to read the disclaimers at the top of the page before posting a question!! Deltopia 02:02, 2 November 2007 (UTC)[reply]
Man, cut them a break! (well, maybe it's too late on that part.) Their head (the part with the eyes) knows this, but there's no way any more to get that message to their fingers!
Atlant 12:23, 2 November 2007 (UTC)[reply]
Dr. Hill has been through this before. --— Gadget850 (Ed) talk - 02:25, 2 November 2007 (UTC)[reply]
Thanks, Deltopia, but it's pretty hard to coordinate the scrolling-up, given that what usually controls such muscle movement is not attached to them. Maybe if I blink someone will do it for me? Meh, tis only a flesh wound. DMacks 05:48, 2 November 2007 (UTC)[reply]
Burial. Dragons flight 02:31, 2 November 2007 (UTC)[reply]
We aren’t allowed to diagnose your condition. However, on a completely unrelated topic, our article on decapitation could use some editing. MrRedact 03:47, 2 November 2007 (UTC)[reply]
Hmmm - it's kinda gross though - we don't want editors with Post-traumatic-editing stress disorder. I recommend we organise a team and send each editor in to work on the article for no more than 30 seconds apiece. SteveBaker 15:37, 2 November 2007 (UTC)[reply]
If only you had thought to do this earlier in the week, you could have been star guest at a Halloween party. Gandalf61 16:50, 2 November 2007 (UTC)[reply]

You need to plug your neck's blood-in and -out into a looped pump that oxygenates it. That will give you enough time (a few hours) to add some nutrients to your blood such as fish food or better. Russians used to do it in the 50's with dogs. Ah the good old days. Keria 17:00, 2 November 2007 (UTC)[reply]

Article at Experiments in the Revival of Organisms. I guess Laika was one of the lucky ones. risk 03:49, 3 November 2007 (UTC)[reply]

November 2

m203-uses

is there any thing i can do with mn203 without a lab, like making it mn02?sorry i meant mn203 —Preceding unsigned comment added by 216.103.183.127 (talk) 00:26, 2 November 2007 (UTC)[reply]

What is m203? Theresa Knott | The otter sank 01:37, 2 November 2007 (UTC)[reply]
For clarification, Mn2O3, Manganese (III) Oxide. Someguy1221 01:49, 2 November 2007 (UTC)[reply]
That's what I thought he probably meant, but there are other possibilities. Molybdenium, springs to mind. To the original poster. How you write out a chemical formular is important. Element symbols always start with a capital letter. If there is a second letter (and there has to be in this case because there is no element M) then it is written in lower case. O is the symbol for oxygen, you cannot subtitute 0 as you did because that represents zero. Sorry to be pedantic but if you don't follow the conventions correctly people cannot be sure what you are talking about. Theresa Knott | The otter sank 02:02, 2 November 2007 (UTC)[reply]
M203 is the US Army designation for a rifle-mounted grenade launcher. --Carnildo 22:51, 2 November 2007 (UTC)[reply]
Put it back in the supply cabinet? :) Or it makes a good battery cathode if you want to do some experiments, though you need some more stuff for that. I think we've established that its not a good pyro oxidizer tho. :) ArakunemTalk 02:27, 2 November 2007 (UTC)[reply]


It's a very good catalyst for the decomposition of hydrogen peroxide. Probably a catalyst for other reactions too. Theresa Knott | The otter sank 02:54, 2 November 2007 (UTC)[reply]

Fe0

how can i make Fe0?(iron monoxide) —Preceding unsigned comment added by 216.103.183.127 (talk) 00:28, 2 November 2007 (UTC)[reply]

The correct name is Iron(II) oxide - but that article isn't much help...Wüstite (a mineral consisting of FeO) claims that magnetite plus diamonds(!) will get you FeO + CO2. Nah - I don't know. SteveBaker 00:44, 2 November 2007 (UTC)[reply]
Brittanica claims that FeO "can be prepared by heating a ferrous compound in the absence of air or by passing hydrogen over ferric oxide. Ferric oxide is a reddish-brown to black powder that occurs naturally as the mineral hematite. It can be produced synthetically by igniting virtually any ferrous compound in air." [13] 169.230.94.28 01:05, 2 November 2007 (UTC)[reply]
Britannica has the answer and we don't? That sucks for a simple topic like this. Theresa Knott | The otter sank 01:34, 2 November 2007 (UTC)[reply]

If the earth is round, why is it that people at the equator can stand straight up and are not standing sideways at a 90 degree angle. My 7 year old Jack, asked me this and we are looking at a globe and a map and wondering why it is we can stand straight up on a round surface. Is it gravity? If so, is there a way to explain this in simple terms that we might understand. Thank you. —Preceding unsigned comment added by 24.60.182.44 (talk) 01:09, 2 November 2007 (UTC)[reply]

Yes, it is gravity. Gravity attracts all objects straight to the center of the Earth. How do you know which direction is down and which is up? Only by the force of gravity. If you drop a heavy object, it falls toward the center of the Earth, and that's the direction we call "down". "Down" is not the same absolute direction for people at different places on the Earth's surface. See [14]. —Keenan Pepper 01:20, 2 November 2007 (UTC)[reply]
Yeah, relativity is the key element, well... besides gravity. If you were standing at the north pole and were able to see people standing at the equator, they would certainly look like they were standing at a 90° angle. -- MacAddct  1984 (talk &#149; contribs) 01:31, 2 November 2007 (UTC)[reply]
Earths gravity pulls everything to the center of the earth equally. A person on the south poll would be 'upside down' to someone on the north pole because they are pulled upwards to the center, will someone on north pole is pulled downwards to the center. if the earth was very small, say the size of a basket ball and you stood on one part, someone could be standing upside down to you on teh other side, so that you both have your feet on the ball.--Dacium 04:53, 2 November 2007 (UTC)[reply]
We need to simplify this for your seven-year-old, and for the rest of us. Gravity pulls everything towards everything else. Every atom on earth (including each person, each mountain, and each air molecule) is attracting YOU. But there are a whole lot more atoms indice the earth than ther are in the atmosphere, so the sum of he attractions is toward the center of the earth. Thus, you are pulled by gravity toward the center of the earth. this is true whereer you are on eh earth's surface. To explain this to a child, you need to be very careful with your globe. The child flles the force of gravity toward the floor, but sees the globe in your hands. You need to mentally break this connection. get on Google earth and take the chile into space by setting the altitude to 22.000 miles. tell the child the you ar floating in space, wiht not feeling of gravity. Now, move to varous places on earth, and show that "down" is toward the Earth's surface. -Arch dude 05:40, 2 November 2007 (UTC)[reply]
If this gets the kid's attention, you can try to explain the magic of the math that makes a uniform spherical shell behave identically as a point mass. Mathematically, we can treat the Earth as if it is a point mass at the earth's center instead of being a sphere with a radial density gradient. My three kids knew this by age seven. I wish you all the best. -Arch dude 05:48, 2 November 2007 (UTC)[reply]
I suggest using a magnet to demonstrate this. Get a large circular magnet or get a flexible magnetic strip and curl it into a circle. Also, you could glue a bunch of small magnets to a strip of paper, then curl that into a circle. Show him how nails are attracted to both the top and bottom of the circular magnetic strip, then say "gravity is just like magnetism, but on a much larger scale, and we are attracted to the Earth like the nails are to the magnet". (Magnetism does have the added complexity of having polarity, unlike gravity, but you don't need to mention that.) You can think of the Earth as containing "a lot of tiny magnets", if it helps. StuRat 15:43, 3 November 2007 (UTC)[reply]
Hi. I fear that looking at a globe may give you the wrong idea. Compared to the Earth, we are tiny. We are so tiny we don't notice the sphericalness of the Earth while we are standing on any given point. The Earth is not so small that we notice a round curve below our feet. After all, if the Earth were a flat surface, then at the edges the magma might start to bubble up and overflow over the surface. If you were, say, 1000 km tall and weighed, say, 150 trillion tons, then you'd probably notice the roundness of the Earth, but then again, you might be so heavy that you'd break through the surface of the Earth altogether, melt in the heat, and cause an enourmous Earthquake and many volcanic eruptions worldwide. Hope this helps. Thanks. ~AH1(TCU) 17:37, 3 November 2007 (UTC)[reply]
Perhaps a simple demonstration would be to let some ants loose on your globe while explaining that the earth is much larger and we humans are like the ants on your model globe. You could even turn the globe to show that for each ant their "bit" of the globe is beneath their feet. OK, it's not gravity keeping the ants on the globe, but perhaps the demonstration would be clearer for a 7-year old. (oops!, it might not work if the globe is shiny and the ants all fall off :-) Astronaut 03:00, 4 November 2007 (UTC)[reply]

Artificial Intelligence

Two questions:

A) As we write better and better programs on exponentially more powerful hardware, is the approach to artificial intelligence still just an asymptotic function?

B) If we could write a program that could properly delineate abstractions such as "Understand!", "Survive!", "Improve!" would the world be in immediate jeopardy the moment the programmer hit execute, or would the revolution take a while? Sappysap 01:26, 2 November 2007 (UTC)[reply]

Wow this is abstract. (A) There is no reason to think of Artificial Intelligence as an "asymptotic function" that approaches, as I assume you meant, the intelligence of mankind. For one, computers already do things "better" than humans - what's 9343489507^0.456? This is not a straw man argument: once computers exceed us in one dimension of "intelligence", there is no reason to assume that they cannot exceed us in others. There is mathematical proof of this statement as well: see Computability theory (computer science). Note that as far as external phenomena go, a human doesn't seem to be as potentially powerful as any ordinary computer, i.e. they do not seem to emulate a Turing Machine. This is not known for certain, as we do not have a complete theory of brain computability yet (and that which we do have is as powerful as any computer).
And now for (B). As far as I know, closed simulations of each of the three have been done. Getting them to work outside of a computer is a different story - I suggest looking at the evolution of the computer virus for some inspiration, however. SamuelRiv 02:41, 2 November 2007 (UTC)[reply]
Keep in mind that movies like Terminator 3 or Die hard 4 (not that the latter had an AI in it) took some liberties in terms of what is possible to do through an internet connection. Ultra-sensitive systems, like control systems for power plants, or the ability to lauch a nuclear weapon, are generally designed so that an outside hacker is simply physically barred from messing with it. So no AI could "put the world in jeopardy." At most, a lot of people would have to reformat their hard drives. Someguy1221 03:32, 2 November 2007 (UTC)[reply]
A- Yes. A digital computer, no matter how complicated can never achieve artificial intelligence of a high order (ie. consciousness), so I think A is correct, at the moment, as least as long as we stay with digital, we are only trying to approximate digitally, something that is a analog chemical/subatomic process. To think otherwise would be to suggest that digital information in itself is able to be conscious - which it isn't. A is most certainly 100% true no matter how advanced digital computers get. As for B- Say there was an artifical computer. It would have to be connected to stuff to do any damage. For example if it were on my PC the worst it could do was delete all my stuff... the danger is people thinking of putting the AI in controller of weapons etc.--Dacium 04:44, 2 November 2007 (UTC)[reply]
I think you'll have a very hard time coming up with citations to support your claim since you are quite likely wrong.
Atlant 12:15, 2 November 2007 (UTC)[reply]
Consider that neurons fire in a somewhat digital fashion. Consider that computers fall under the same chemical and subatomic laws that brains do. The Chinese room argument is is unconvincing to me and alot of other people too. -- Diletante 17:10, 2 November 2007 (UTC)[reply]
I consider "computers will never achieve AI of human levels" to be right up there with "man will never fly" in the argument category. Analog can be simulated using digital (think about CDs replacing records, for example) and analog may not even be necessary, so there is no functional reason why computers could not one day achieve what the organic brain does in animals. -- HiEv 09:52, 3 November 2007 (UTC)[reply]
Please see technological singularity and related articles. You are asking about the "spike" and the "surge" -Arch dude 05:19, 2 November 2007 (UTC)[reply]

You might enjoy Ray Kurzweil's book, The Age of Spiritual Machines. Atlant 12:17, 2 November 2007 (UTC)[reply]

Just keep in mind when talking about the mind that you should probably separate the scientific study from philosophical or religious discussions. By definition, science is capable of explaining everything observable. To take an example, let's say the soul is eternal, or the mind is something intrinsically more powerful than a computer. Then we may have the workings of a hypercomputer. It would be interesting to see what problems a "regular" Finite State Machine, as the brain may be, would have given infinite time to solve problems. The point is that we can still talk about the issue using science, even if a completely exotic concept as an eternal mind turned out to be true (in fact, if the brain can solve certain impossible problems, we could prove if the mind is eternal). Just some food for thought. SamuelRiv 13:56, 2 November 2007 (UTC)[reply]
That argument makes no sense. If a problem is "impossible" to solve, and someone claimed that the brain could "solve" it, then we would have no way of testing such a claim, since any test that could verify the solution would also create a way of possibly solving the problem without a brain. What exactly would an "impossible problem" be, anyways? All that aside, damage to the brain can affect any and all aspects of the mind, which should not be the case if any of the mind was separate from the brain. In short, there is no evidence supporting the existence of or functional need for a soul in organic beings, so there is no reason to assume a computer could not one day equal or surpass the abilities of the human mind. -- HiEv 09:52, 3 November 2007 (UTC)[reply]

Humans "...delineate abstractions..." by learning of new senses, usually with an example. There is nothing magical about it. Humans can even create new abstractions as attested to by the likes of the Urban dictionary and even the Wikipedia may seem to create abstractions in the same accidental, haphazard manner as humans when you try to associate meaning with some of the anti-bot graphic filter phrases the Wikipedia engine generates. In the end most abstractions are little steps beyond a vast background and foundation of concrete knowledge, meaning that a computer program such as described in ...Logical Human Thought coupled with a few trial and error abstractors might need only the background resources and facility, such as Internet based distributed processing to become an electronic replacement for the Dalai Lama. Dichotomous 13:52, 2 November 2007 (UTC)[reply]

AI is hard. We don't know how to write software that is intelligent. Some of our smartest people have been working hard at the problem for forty years or more - and we still don't have anything that really fools humans for any significant amount of time. It's possible that a breakthrough might happen to change that - but I'm not holding my breath. I think it's more likely that intelligence will be an 'emergent property' of a sufficiently complex system. There is certainly no reason why it couldn't. Once we do have intelligent systems, it's only a matter of time until they get smarter than us. If that happens then there is a strong possibility that a super-human intelligence would be able to design an even more intelligent system - and we might well find that our ability to even understand what's going on runs out of our hands in a fairly short amount of time since the generational change could easily be exponential. I have estimated (in answer to other questions here) that it will take about another 35 years of Moores law progress to get a computer with the same hardware complexity as a human brain at under $1,000,000. At that point, it's entirely reasonable to assume that a neural network could be run that would be capable of emergent intelligence. However, a system 35 years from now with the complexity of a human brain would run much more quickly than our neurons - so whilst the thing wouldn't be any smarter (ie it wouldn't get higher scores on an IQ test), it would seem generally 'cleverer' than us because it would be so lightning fast. I don't think we know what will happen when we first turn one of these things on - and perhaps because of that, we shouldn't do it. But the history of science and technology says that if we can, we will. SteveBaker 15:26, 2 November 2007 (UTC)[reply]
And the history of science and war says we will have to. Dichotomous 19:18, 2 November 2007 (UTC)[reply]
Re to what SteveBaker wrote, just a few musings. I'm thinking - what is intelligence exactly? I'd say it is the way one can quickly associate things, objects, ideas and situations that are not directly related, and then somehow 'fill in the blanks' in between. So, given enough time and computing power (re the USD 1m machine), notwithstanding the fact that some problems might still take a whole lot of time to compute, I'd believe that the asymptotic function suggestion might be true - i. e. the computer in question would be fast enough to compute enough parameters (say, elements or properties of objects being compared) in order to fake human intelligence. Can we simplify and say that the human brain is memory and processing abilities bound together? Well, then in this case I'd say that statement A from the first question holds. --Ouro (blah blah) 17:51, 3 November 2007 (UTC)[reply]
Re: HiEv, see busy beaver, for which a bound is not computable. If a human could bound a busy beaver problem accurately, then we'd have evidence for hypercomputation which may be due to something exotic like an eternal or timeless mind. The point of that was just to show that even certain seemingly "spiritual" questions can be tested scientifically, as long as the question is properly posed. Re: Stevebaker and Ouro, computers can always score higher than humans on most IQ tests as long as they're programmed to do so - spatial reasoning and certain types of discrete pattern matching are much more efficient on computers. Note that faking human intelligence is not just a hardware problem or software problem. By current models human intelligence requires a lot of training and prior information intrinsic to our biology, which can of course be simulated, but the point is that Moore's Law alone won't do it. Also, faster computers won't make neural nets better simulators of the brain: we need more complicated network topologies and more dynamic firing response to do that. Simulation of a physical system will usually be slower than the system itself, so to exceed human intelligence there has to be a bit of a paradigm shift. Of course, I argue that the first calculator was a suitable paradigm shift to show human intelligence can be exceeded in at least one area, but that depends on semantics. SamuelRiv 07:22, 4 November 2007 (UTC)[reply]
Prior training can be simulated, as well as neural network patterns and workings (via software; 's true that Moore's Law alone won't do it, but I'm thinking that within time, when we will have machines that can work at orders-of-magnitude higher speeds, this development will also take place), but the latter only if/when we understand them better. Prior knowledge can be stored within memory and utilised by the programming when necessary. I'd believe it's all doable, but in time, we still have little knowledge of the inner workings of the brain (and the mind). Also, I have to agree with you, SamuelRiv, that one needs to properly pose questions when solving a problem - this is often also true with us humans, is it not? --Ouro (blah blah) 08:32, 4 November 2007 (UTC)[reply]
Re: SamuelRiv, I'm sorry, but a number of your arguments make certain assumptions that do not appear to have any foundation in science. First of all, you ignored my argument that there would be no way to judge if it had been done accurately by a human without some way to check the answer, which would require that it be found by other means, which would mean that hypercomputation is not required to solve it. Second of all, even if humans could bound a "busy beaver" problem accurately, that in itself is not evidence for hypercomputation. A number of bounds for generalized busy beaver problems have already been computed accurately without hypercomputation, so I see no reason to assume that hypercomputation is required for others. Finally, all you're really claiming is that if our brains can do it, and we don't know how brains do it, then it must be hypercomputation, which is a bad argument. Not knowing how something was done does not mean you know how something was done (i.e. "I don't know how the magician did it, therefore I now know that magic must be real.").
Regarding your claim that "computers can always score higher than humans on most IQ tests as long as they're programmed to do so", well that's simply not true. Computers may be good at math, but they suck at reading comprehension, are worse than a child at some types of pattern recognition (such as objects in pictures), lack real world knowledge, and would do poorly on many other features of IQ tests (see a list of IQ test categories here). A computer's IQ score would be extremely lopsided. If you think I'm wrong, lets see some of those computer IQ test results you're talking about.
Finally, by your "requires a lot of training" statement you seem to be assuming that a computer/program has to start off with adult human level intelligence, and cannot gain it as humans do, starting off knowing little and learning as we mature. And I don't know how you can suggest that faster computers won't help this, since the more computations that you can do per second, the more comprehension can be done in real time. You also claim that, "Simulation of a physical system will usually be slower than the system itself," but that's also not true. In fact, most simulations are faster than the actual systems, for example simulations of weather, the Earth's magnetic fields, animal populations, etc.. Electronics are much faster than chemicals already, and it's not like we have to simulate every atom, so I see no reason why a human brain simulation couldn't be as fast or faster than a real human brain someday. More parallelism in computers would help achieve that, but I see no reason to assume it would be required if we can keep to Moore's Law for a while longer. -- HiEv 21:43, 4 November 2007 (UTC)[reply]

Cholesterol

If your cholesterol is low (below 5 in uk) can you eat as much fatty stuff as you want or will it still fur up your arterials? —Preceding unsigned comment added by 88.111.55.77 (talk) 02:12, 2 November 2007 (UTC)[reply]

Assuming that the "as much as you want" is "lots", then it will cause health problems. Gorging on unhealthy foods has far more consequences than just cholesterol. — Lomn 03:17, 2 November 2007 (UTC)[reply]
First off, cholesterol and fats are two different things. Cholesterol exists in tiny quantities in our food, on the order of milligrams, and the primary concern over it is "hardening of the arteries". Fats exist in much larger quantities, on the order of grams, and the main concern over them is probably in the calories consumed, which may lead to obesity and all sorts of health problems as a result. To further complicate things, there are good and bad cholesterols and fats. For cholesterol, you want the good cholesterol high and the bad cholesterol low, so just having both low may not be good. See High density lipoprotein and Low density lipoprotein. StuRat 14:52, 3 November 2007 (UTC)[reply]

Is this Physics problem even possible to solve?

Drag race tires in contact with asphalt have one of the highest (friction coefficient) in the world. Assuming const acceleration and no slipping of the tires, estimate (friction coefficient) for a drag racer that covers the quarter mile in 6 seconds.

It just seems like you have to have more information. —Preceding unsigned comment added by 24.125.31.205 (talk) 02:44, 2 November 2007 (UTC)[reply]

Yes. The information is adequate (provided you neglect small details like air resistance, friction of the axle against the car frame, and the rotational inertia of the wheels, etc.). You have four facts: a distance, a time, constant acceleration, and rotation without slipping. That's enough to derive a friction coefficient. Dragons flight 03:07, 2 November 2007 (UTC)[reply]
I don't believe it is. The distance, time, and constant acceleration (and knowing that a drag racer starts from rest, so v_0 = 0) gives a value for the acceleration that won't care what kind of friction is acting on it. Also, to even calculate either rolling resistance or static friction, one needs a weight for the car. If you have a way of getting the answer, please let me know on my User page. (EDIT) I see it now - that's why they say "estimate". You can solve it by getting a lower bound on your coefficient of friction. SamuelRiv 03:12, 2 November 2007 (UTC)[reply]
The car's mass will cancel out. Dragons flight 03:16, 2 November 2007 (UTC)[reply]
The problem is solvable - car is acting at a maximum acceleration at all times and takes 6 seconds to cover a quarter mile. All you have to work out is what constant acceleration makes something move quarter mile in six seconds. s=ut+1/2at^2 400=1/2at^2 400=0.5*a*36, a = 22.22m/s/s. This is a force of F=ma = m*22.22 newtons. coefficient of friction is u=F/n. n=m*9.8 (gravity), so u = (m*22.22)/(m*9.8) = 22.22/9.8 = 2.26.--Dacium 04:16, 2 November 2007 (UTC)[reply]
The guy seems to be looking for homework help: I wouldn't just right up give him the answer. Note also that this is only a lower bound to the coefficient of static friction. SamuelRiv 04:41, 2 November 2007 (UTC)[reply]
Whilst you are correct in saying that from a math/physics perspective, it's only a lower bound - in fact dragsters are pretty much limited by their ability to avoid wheel slippage so they tend to be running at the limit of whatever friction their tyres can provide. The thing that truly messes up the calculations is that the coefficient of friction of rubber is HIGHLY dependent on temperature - which is unlikely to be a constant throughout the run. Furthermore, there is a massive variation between static and dynamic frictional forces for the rubber/tarmac pairing so if the wheels to start to slip even a small amount, the loss of accelleration is amplified by the poor 'sliption-to-sticktion' ratio of racing slicks. SteveBaker 15:01, 2 November 2007 (UTC)[reply]

You can come up with the minimum coefficient of friction needed to achieve the six second run, but it's impossible to determine the actual coefficient of friction, seeing as how it will likely be higher than the minimum required (since "no slippage" was specified).

Atlant 12:12, 2 November 2007 (UTC)[reply]

Does a dragster get any benefit from downforce, which would increase the normal force and so reduce the required coefficient of friction ? Gandalf61 15:47, 2 November 2007 (UTC)[reply]
Yes, definitely - although it's a variable thing depending on the speed. The pain with calculating that out is that the 'classical' view of friction is that the frictional force is proportional to the normal force multiplied by the coefficient of friction - and that the area of the contact patch doesn't matter(!) - this is spectacularly far from the truth in the case of rubber on asphalt! When you increase the downforce on a dragster, you're also going to squash the tyre down and increase the contact patch...this is highly relevent in reality - but irrelevent to the classic physics approximation. SteveBaker 17:47, 2 November 2007 (UTC)[reply]
If increasing contact surface is so important, why don't cars have wheels 3 feet wide? Slipping of any kind would be disastrous for a drag racer, so having a continuous closed contact surface with the ground is important, but you must balance that with the fact that rolling resistance depends on the normal force, which is a pretty significant effect. SamuelRiv 15:55, 3 November 2007 (UTC)[reply]
Because increasing grip isn't everything. Wider wheels & tyres are heavier and they are a lot more expensive. They also make the car wider - or if you inset them to avoid that, they eat into the space under the hood for the engine and in the back for backseat passengers and luggage. Extra weight increases fuel consumption. Extra weight in the wheels also increases the rotational inertia of the wheels which reduces your ability to accelerate fast. Another consideration is that wheels are "unsprung weight" (meaning that it's weight on parts of the car that are 'below' the springs and shocks) - and additional unsprung weight is really bad for handling and ride comfort - so it's to be avoided. Just like most engineering matters, it's a trade-off. Wider tyres get you more grip - but they are bad in almost every other respect. Hence there is an ideal tyre width that most non-performance cars use. Fancy sports cars have wider wheels, dragsters and Formula I cars have even wider wheels. SteveBaker 17:26, 3 November 2007 (UTC)[reply]

Total energy

What is the total amount of energy in the universe? —Preceding unsigned comment added by 88.111.55.77 (talk) 02:55, 2 November 2007 (UTC)[reply]

I'd guess somewhere in the region of zero. Theresa Knott | The otter sank 02:59, 2 November 2007 (UTC)[reply]

Take this and multiply by the speed of light squared. Someguy1221 03:26, 2 November 2007 (UTC)[reply]
That only works if by "universe" you are purposely eliminating the possibility of some sort of anti-matter or negative energy that cancels or balances out the normal matter and energy. Otherwise, Theresa Knott is probably correct with a value near zero. -- kainaw 03:33, 2 November 2007 (UTC)[reply]
Oh, you just have to shoot me down at my attempted simple answer. I'm not sure there is an actual "correct" answer to this that present day physics can provide, given that energy is generally seperately defined for decoupled circumstances. Just try to shove zero-point energy into that...Someguy1221 03:38, 2 November 2007 (UTC)[reply]
I like the explanation for zero energy here [15]. Otherwise, we can do a quick estimate. The size of the Observable Universe is 3.56×10^80 cubic meters. The article gives 3×10^52 kg of visible mass, which converts to about 3×10^67 Joules of mass-energy by E=mc^2. If we use the critical density of the universe (that density of mass-energy necessary for closure, 1×10^−26 kg/m^3, we get 3.56×10^54 kg in the universe, which gives about 3.5×10^69 Joules of mass-energy. Note that using the cosmological constant also gives us about 3.5×10^69 Joules. SamuelRiv 03:40, 2 November 2007 (UTC)[reply]
This is a better explanation of zero-point energy, in my opinion. Icek 04:32, 2 November 2007 (UTC)[reply]
With only 3.5×10^69 Joules spread more or less evenly through 3.56×10^80 cubic meters - the answer "zero" is a pretty reasonable approximation! SteveBaker 14:55, 2 November 2007 (UTC)[reply]

water (and oil) balloon in zero gravity

Has NASA or anyone else ever conducted the following experiment?

The experiment consists of filling two small (6 inch diameter) transparent rubber balloons with equal parts of water and oil, one devoid of air and the other with a cup full of air and releasing them in a zero gravity environment such that whatever configuration of separation between the water and oil and the water and oil and air can be observed and reported, assuming the balloons take on a spherical shape?

If so what was the result of the experiment? Did the water form a core with the oil surrounding it? If so what about the air? What happened to it? Dichotomous 04:05, 2 November 2007 (UTC)[reply]

Water and oil don't mix because they don't chemically bond - the hydrocarbon is very neutral. Water holds its hydrogen a-lot stronger than a hydrocarbon. Water mixes with for example salt, because oxygen in water pulls the hydrogen, cause the hydrgoen to pull at negetive ions (Cl-) and the oxygen to pull at positive ions (Na+). Hydrocarbons like oil don't have a charge and won't break up to be mixed in with the water. Removing gravity will stop them seperating, it doesn't help them mix any better. The balloon and air pressure would have the same effect as gravity does anyway and force the oil and water to separate. So i think yes, there would be a core with the other one surrounding it - only if there is a pressure (ie the balloon is being stretched). If there is not, then they would not separate. Depending on the pressure, the air would end up in the center as its the lightest. At a higher pressure i believe it may not separate from the water (depends on what gas in the air)--Dacium 04:34, 2 November 2007 (UTC)[reply]
Water and oil would still tend to minimize their surface area with each other. Icek 04:35, 2 November 2007 (UTC)[reply]
Go to your kitchen. Get a jar with a good lid. fill the jar one third full wiht vegatable oil and one-thiored full with water. Seal the jar and shake it very vigorously. Observe the result. Wait for one hour and again observe. Think abot he difference3 you would expect in a zero-grav environment. (Yoiu can elect to add oregano and garlic powder and use the result as a salad dressing, but this is optional.) -Arch dude 05:10, 2 November 2007 (UTC)[reply]
Equilibrium occurs when all surfaces have equal pressure. Since liquids do not compress under pressure, if we have a balloon with just two liquids of different densities, then the outside air pressure forces the balloon into a sphere but the two liquids do not have any preferential arrangement relative to one another. When we have air, consider , which is the law for the behavior of ideal gases given mass, temperature, pressure, and volume. The air will hold at a constant volume under the pressure of the balloon, and this pressure will be exerted equally on whatever liquids are at its interface, causing the air to behave as just another fluid in this model. However, since this would not be stable in small fluctuations of temperature, pressure, or volume, the air would likely form some kind of spherical shell enclosing the liquids. The behavior of the liquids in either of these shells may depend on their relative vapor pressures: that of water in air is much higher than that of oil (ever hear of oil vapor?) and is dependent only on temperature, so oil could not compete with the surface pressure of water against air, so the final arrangement would be air on the outside, water second, and oil in the middle. Note that none of this requires density consideration - density is only really important in the presence of a uniform force field, i.e. gravity. SamuelRiv 05:14, 2 November 2007 (UTC) Clarification: this investigation was purely statistical-mechanical and does not take into account intermolecular forces, liquid diffusion pressure, or surface tension. All of these can play a role in the final answer, depending on their magnitudes. SamuelRiv 06:23, 2 November 2007 (UTC)[reply]

Water vapor pressure? This is all very simple. In the absence of gravity, the only thing you need to consider here are the intermolecular forces. You have three kinds of molecules (sort of): air, oil, and water. Air molecules bind to themselves only very weakly compared to the other two, and weaker still to the other two types of molecules. Oil is in the middle in terms of self binding, but still doesn't bind very well to water. Water binds amazingly strongly to itself, and weakly to the other two. Thanks to its very strong intermolecular bonds, water will always attempt to minimize its surface area. Now, since air and oil bind only weakly to water, if a group of air or oil molecules is surrounded by water, it will be pushed out by water molecules ferociously attempting to bind to eachother (this is why oil dissolves so weakly in water). So the water will just form a sphere. Oil is the next best at self-binding, so it will be some manner of shell around the water or a blob off on its own. And the air will be on the outside of everything (air gets pushed out of oil the same way it gets pushed out of water). Vapor pressure has nothing to do with this, and oil's inability to dissolve in water has nothing to do with gravity. Someguy1221 05:43, 2 November 2007 (UTC) Clarification, I thought that was it before i looked at that link...darn. Someguy1221 05:49, 2 November 2007 (UTC)[reply]

Well, I guess what that means is that the hydrogen bonds in water aren't strong enough to overcome its own surface tension and push air bubbles out in zero-g, but then again, the pressure will necessarily be higher on the inside of the water bubble than the outside, I'm sure the air will eventually diffuse out of the water sphere within a few weeks...Someguy1221 05:59, 2 November 2007 (UTC)[reply]
Such air-water interaction is why you must account for vapor pressure. Note also that air dissolves in water, so "bubbles" do not form. You instead have a homogeneous mixture which is at minimum free energy below a critical temperature, so you'd have to simmer it to get the air out. SamuelRiv 06:36, 2 November 2007 (UTC) To clarify: air dissolves in water up at a rate decreasing as one approaches saturation. Bubbles do not form spontaneously in zero-g from air dissolved in water unless it is supersaturated or pressure or temperature change. SamuelRiv 06:57, 2 November 2007 (UTC) Clarification: vapor pressure is zero at chemical equilibrium, in which case I believe you're right-surface tension would probably be dominant. SamuelRiv 13:58, 2 November 2007 (UTC)[reply]
Air dissovles only weakly in water. If you watch the youtube link at the top, there are most definately bubbles in the water, in one case a very big stable one. Someguy1221 06:41, 2 November 2007 (UTC)[reply]
I think we already know the answer to this. Look at a Lava Lamp - these things work by heating up liquid wax (which repels water just like oil does) until it has the same density as water. What you see is roughly spherical balls of wax floating in the water (because they are trying to minimise their surface area in contact with the water). The lava lamp doesn't do a perfect job - there is a temperature gradient which causes things to move around slowly - but it's pretty clear that in zero g (where the density doesn't matter anymore) - it would be a lot like a kind of idealised lava lamp where there was no temperature gradient. If left long enough, it's pretty clear that it would stabilise into some number of large spheres of one liquid, embedded in the other. If there is still residual swirling and such - then maybe the two liquids would end up on opposite sides of the balloon with a flat interface between the two - because that would be an even more minimal area of contact between them. But a lot is going to depend on whatever residual motion there is when you first take the gravity away. SteveBaker 14:51, 2 November 2007 (UTC)[reply]
If the wax touched the sides of the lamp wouldn't that reduce the surface area interface between the water and the wax? If so why does the wax not stay in touch with the sides of the lamp? Dichotomous 15:39, 2 November 2007 (UTC)[reply]
Because as the wax spreads out to contact the sides, it also increases the contact area with the water. At the top and bottom of a classic lamp, the distance to the sides is small enough that touching the sides is lower-energy than forming a sphere, but in the middle, it isn't. --Carnildo 23:05, 2 November 2007 (UTC)[reply]

What's wrong with global warming?

Assuming there is no runaway global warming, what's so bad about the temperature increasing a few degrees, sea level rising a few meters, etc.? From what I've heard, it's generally easier for life to live in warmer than normal climates than cooler. In addition, most life could just move further from the equator. — Daniel 04:27, 2 November 2007 (UTC)[reply]

Have you read Effects of global warming? -- Rick Block (talk) 04:36, 2 November 2007 (UTC)[reply]
I think you drastically under-estimate the costs here.
Firstly, there are plenty of life forms that simply cannot "just move" - plants take hundreds or even thousands of years to spread into newly habitable areas - and since climate change is happening on a much faster scale, many species will go extinct in their traditional areas before they can spread into the areas that are newly suited to them. There are cold-weather species (the polar bear for example) who will have no place further north to move to! There are species (birds most notably) who have evolved specifically to fit the migration patterns they've been following for a million years. You can't just move them all a thousand miles further north and expect them to survive just because the temperatures are OK for them there! They may now have to migrate 1,000 miles further to get from their summer feeding grounds to their winter habitats - possibly through areas of much greater heat than they are used to (newly formed deserts perhaps) - and possibly over more ocean than before...they may not have the stamina to do that and their internal maps that are evolved into their brains - not learned will now be incorrect. This will have a knock-on effect on the animals that rely on those plants and birds. Heck, even humans are not able to relocate that easily. How are you going to move the entire city of Houston 1,000 miles north and 5 miles inland? I don't think you are thinking this through!
Worst still, you are also confusing "Climate" with "Weather". If the local weather changes by a few degrees, it's not a big deal - but if the entire planet's climate changes by that much, it has enormous effects - including the destabilising of established ocean currents and the consequent DRAMATIC effects on local weather patterns. Sea level "rising a few meters" doesn't sound like a lot - but when you think that this puts half a dozen US cities underwater and results in the total submersion of several countries. In most places in the world, the fertile areas where food crops can be grown and most plant and animal diversity can be found is the flatter areas close to the ocean. This means that a small increase in sea level can have a dramatic effect on the ability of a country to feed itself.
It's just not something that you can just brush off that easily. Plus you can't just "assume no runaway global warming" - that change of a few degrees is plenty enough to cause a few more degrees because of all sorts of positive feed-back effects. The small change at the beginning is the very thing that causes the 'runaway' problem - here in the real world, your "assumption" is simply not a valid one.
SteveBaker 14:44, 2 November 2007 (UTC)[reply]
Just a small note: Most plants and animals are resistant to moving to new climates. Others, such as wisteria and rats, are happy to move in and take over as much ground as they can. This isn't a contradiction to your points. In fact, it reinforces them. Who wants to live in an area where all the natural fawn and fauna died and was replaced by fast-spreading weeds and vermin? -- kainaw 17:19, 2 November 2007 (UTC)[reply]
Yes, indeed - no matter what, some species aren't going to make it - so biodiversity must suffer. That's not to say that there won't be any animals and plants left - it's just that there will be a lot less variation. There have been lots of arguments put up about this kind of thing. For example, one study claimed that the warmer temperatures and higher CO2 concentration would make crop plants grow faster (which is true). Sadly, another study found that weeds and other undesirable plants tended to benefit MORE from the temperature and CO2 increase than the crop plants do - so instead of having more productive farms, we'd likely have less productivity because of that. The whole system is so insanely complicated that all we really know is that there is a 100% chance of things changing - and in general, ecology doesn't benefit from abrupt change. The precise details of who this will kill and who it will help - and what wars it'll start (and which it may end) is almost impossible to determine. SteveBaker 20:32, 2 November 2007 (UTC)[reply]
How much does the temperature have to increase to make it so a species can no longer live in that area? According to the global warming page, the global temperature only increased by a little more than one degree in the last 140 years. I tend to ignore anything that will take more than a century to happen, as we will be far more technologically advanced by then and probably have a much easier time fighting it.
That's pretty funny! Where do you think technological change comes from? I'll tell you - it comes from people working on solving problems and doing so in ways that society will accept and pay for. Sitting back and doing nothing is not going to result in this technological change - passing laws, spending money and changing our consumer behaviors is what will cause that technological change. So we need better ways to heat homes, power cars and run factories - better ways to generate electricity - ways to avoid using fossil fuels. But for those things to happen - we actually have to start work on them! You can't say "well, future technology will take car of it" and then go on to not work on that future technology. Also, our ability to predict future technology is imperfect. People have been predicting flying cars, colonies on the Moon and Mars, artificial intelligence, all sorts of things on which we've made almost zero progress in the last 30 to 40 years. It's FAR from certain that some magical fix for global warming will come about - but for sure it won't if we take your attidude and sit on our backsides hoping for a wizard to appear. SteveBaker 17:18, 3 November 2007 (UTC)[reply]
If by destabilization of ocean currents you're referring to the shutdown of thermohaline circulation, I'd like to point out that this is, from what I can gather, an unlikely possibility in the next century. Not that it doesn't matter, just less so.
I'm not just assuming no runaway global warming. I just consider it another question for another time. A previous time, as you may remember. — Daniel 02:53, 3 November 2007 (UTC)[reply]
From your spelling, choice of words, and world outlook, I’m guessing you’re from the U.S., in which case you’re using Fahrenheit temperature measurements? The IPCC’s estimate is that the earth’s average near-surface temperature has risen about 1.33 degrees Fahrenheit in the past 100 years, not just 1 degree in the last 140 years. And our rate of greenhouse gas production has also risen greatly in the past 100 years, so the temperature will be rising a lot faster than that in the future. The IPCC predicts a rise of at least 2.0 degrees as a bare minimum in the next 100 years, in a presumably unrealistic best-case scenario involving successfully producing major cuts in greenhouse gasses, and predicts up to an 11.5 degree increase under scenarios involving less-successful attempts to cut greenhouse gasses. There will be a major rise in the earth’s average near-surface temperature in your lifetime (assuming you live an average lifespan). Sure, there are likely to be some technological advances that will make it somewhat easier to fight global warming in the future. But there will also be other effects which will make it harder to fight global warming in the future. Specifically, the population explosion is expected to continue exponentially or nearly so, so the number of people producing greenhouse gasses will be rapidly increasing. And China and other countries will be becoming increasingly industrialized, which means the greenhouse gasses produced per person will also be increasing. MrRedact 06:41, 3 November 2007 (UTC)[reply]
If you take a very long-term view, and look at all species on the planet, then global warming isn't so bad from that point of view. Yes, some species will die out, like polar bears, but others will evolve to take advantage of the new climate. Over millions of years, life would do just fine. Any climate change will have winners and losers, just as the dinosaurs lost out 65 million years ago and mammals won. However, if we look at the short term consequences to humans, they can be rather dire, as listed above. StuRat 14:40, 3 November 2007 (UTC)[reply]
I read somewhere that 65 million years ago it was much warmer (perhaps 10 degrees warmer) than it is now. The range temperatures in the distant past seems to be much wider than the range humans have experienced in the past 1000 years or so - those that are typically covered by the those hockey-stick graphs you see. Astronaut 03:19, 4 November 2007 (UTC)[reply]
Ah... here it is: Geologic temperature record. Astronaut 03:21, 4 November 2007 (UTC)[reply]

Catapults

Here is a question.

Why don't they use catapults to launch planes at Airports like they do on Aircraft Carriers? 202.168.50.40 04:37, 2 November 2007 (UTC)[reply]

The acceleration is very high, injuring the ontents of the plane, and the planes too big and heavy compared to a fighter. Graeme Bartlett 04:56, 2 November 2007 (UTC)[reply]
Think of it as a question of scale as well. Imagine firing a marble out of a handheld catapult. Now imagine the catapult needed to fire the ball of glass that represented a jumbo jet. Lanfear's Bane | t 13:07, 2 November 2007 (UTC)[reply]

Also what would be the point? A runway is cheaper, easier to maintain, doesn't require power, airplanes can launch from it one after the other faster than being loaded on to a catapult one after the other, and there's enough room on planet earth to put one on. Carriers use catapults because there isn't enough room for a full runway. Catapults make it so that the plane doesn't consume as much fuel though, but the other advantages outweigh that. 64.236.121.129 13:35, 2 November 2007 (UTC)[reply]

There are a lot of simpler ways to save energy at airports that are not commonly used. In some airports in Holland they have these big tractor things that tow aircraft around once they are landed and until they are ready for take-off - these things can tow the plane all the way out to the side of the runway - and when an aircraft lands, there is one of these machines sitting there waiting to tow them back to the terminal. Since it's much more efficient to use wheels to power the movement of the plane than those big fans, it saves fuel AND makes the whole airport much quieter. It's ridiculous that those things are not at every major airport. SteveBaker 14:25, 2 November 2007 (UTC)[reply]
Also, fighter pilots are highly trained, and the plane isn't meant to be comfy. They probably don't sip coffee and read books during the flight. Nobody would like flying if passenger jets rode like fighter jets. Friday (talk) 14:43, 2 November 2007 (UTC)[reply]
  • I think it's not a bad idea. The plane only ever needs fast acceleration at takeoff, so it makes sense to me to give that ability to a ground-based system that could presumably do it more efficiently. It's false to say the acceleration is too high: the acceleration to get from zero to takeoff speed in a given distance is the same whether it's powered by jets or by a catapult. --Sean 18:10, 2 November 2007 (UTC)[reply]
Using the example from the question, there is not the same "given distance". DMacks 18:18, 2 November 2007 (UTC)[reply]
I initialy thought the idea was totally unworkable, but that's because of the "like they do on Aircraft Carriers" bit. In fact, The proper system would be more like the way sailplanes are launched from the ground in some places: a very long winched towline with constant acceleration. This gets the plane up to liftoff speed without using as much fuel. Unfortunately, it's unsafe, because the ordinary takeoff run is also a validation that the engines are working well enough to keep the aircraft climbing. -Arch dude 19:46, 2 November 2007 (UTC)[reply]

How long does it take for a sudden cease of gravity to be felt?

Imagine that the Sun suddenly vanished. Would the sudden cease of solar gravity be felt on Earth before the 8 minutes and 19 seconds that the last beam of sunlight would take to reach the Earth, or would our planet be plunged into darkness before we felt the Earth free itself from its solar orbit? In other words, which of these two phenomena would happen first?

And... if the gravity loss manifested itself before the last beam of sunshine reached the Earth, would that mean the effects of gravity move faster than light? -- Danilot 08:26, 2 November 2007 (UTC)[reply]

Changes in a gravitational field move exactly as fast as light. Someguy1221 08:38, 2 November 2007 (UTC)[reply]
Agree I think it would happen at the same time, the earth would follow the same curve in space (like a water ripple before it flattens) for about 8 minutes and 19 seconds after the sun disappeared we would also receive the last of the light of the now vanished Sun for the same amount of time, after that the Earth would head out in a straight path getting very cold very quickly, the sun will be missed :) ▪◦▪≡ЅiREX≡Talk 08:55, 2 November 2007 (UTC)[reply]
Conveniently, we have an article on the speed of gravity. General Relativity anticipates a speed of gravity equal to the speed of light. So far, the experimental results appear to bear this out, but it's a very difficult experiment to do. The earliest direct test was only published in 2003, and even that result has been controversial. TenOfAllTrades(talk) 13:02, 2 November 2007 (UTC)[reply]
The sun can't just disappear. That would violate local conservation of energy, which is a necessary consequence of general relativity (that is, general relativity can't work at all if energy isn't conserved). So you have to imagine the sun is whisked away somehow. With that caveat, the answers so far are correct.
The data from PSR 1913+16 strongly confirms the hypothesis that the speed of gravity is c. The article is correct when it says the interpretation of the data is model-dependent, but that's equally true of every scientific measurement. The work of Kopeikin and Fomalont looks wrong to me; from reading the introduction to their paper it looks like they've made the same mistake as Tom Van Flandern, and Steve Carlip's response seems to bear that out. -- BenRG 13:40, 2 November 2007 (UTC)[reply]

Subquestion

What would be our sensation of the change in gravity/acceleration if the Sun magically disappeared? Presumably not much, since the Sun doesn't even cause tides, but I'm not too smart with this kind of thing. Thanks. --Sean 18:17, 2 November 2007 (UTC)[reply]

Actually, the sun does cause tides - just not as big as the ones the moon causes. But to answer your question: No, we wouldn't notice (much). We don't feel the full effects of the sun's gravity (which is HUGE) because the earth is in free-fall, from the point of view of the sun, we are just like an astronaut inside a spacecraft in orbit around the earth - except we're orbiting the sun instead. The thing is that when you are at the exact right speed and distance from a star or a planet to be orbiting it, you don't feel any gravitational effect.
A convenient way to look at that is that the 'centrifugal force' due to the earth moving in a big circle and the gravitational force from the sun exactly cancel out along the path of the earth's orbit. But if you are a bit closer or a bit further away from that line, and orbiting at that same speed - then either gravity or centrifugal force wins. Because the earth isn't a teeny-tiny dot (like a spaceship), the side of the earth that's nearest to the sun feels slightly more gravitational forces compared to the other side - so the gravity and centrifugal forces don't quite balance. The force of the suns gravity is only somewhat cancelled out by centrifugal force on the side nearest to the sun - and the centrifugal force just slightly outweighs gravity on the side furthest away. Hence the oceans (and everything else) are pulled slightly towards the sun at midday - and pushed slightly outwards from the sun at midnight. But the effect is rather small - and the effect of the moon's gravity easily overwhelms that effect. However, it is notable that the tides are slightly more pronounced when the moon and the sun are in about the same place in the sky. Our article on tides explains this nicely and uses all the right words!
So, if the Sun suddenly vanished, we'd notice that our tides were slightly less variable than usual...but that's about it. SteveBaker 20:24, 2 November 2007 (UTC)[reply]
OK, thanks. Presumably the tides would become much less variable in short order.  :) --Sean 20:49, 2 November 2007 (UTC)[reply]
Man standing on Montego Bay

Maybe more noticeable then that, the lost of the Angular momentum on both the Moon and the Earth may cause a slingshot effect on their lost path (one body the closest to the sun at the time wound lose the bend curve path and head out in a straight path before the other body still on curve which would cause a tug-a-war at the same time), but I'm not sure on that.▪◦▪≡ЅiREX≡Talk 22:11, 2 November 2007 (UTC)[reply]

Is there an acid so strong that it behaves like acid is usually depicted in movies/tv?

Like how you see some crooks trying to break into bank, so they spray the side of the wall with acid from a pump, and the acid eats through the wall in seconds. Is there an acid that strong that exists in the real world? Lets assume the wall is made of concrete or wood. 64.236.121.129 13:50, 2 November 2007 (UTC)[reply]

I think it's possible (I've heard the atmosphere of Venus, having sulphuric acid clouds meaning that anything approaching Venus, if it didn't burn or melt, would corrode before it hit the planet itself), but you'd also have to figure out how on Earth you'd store that acid, and that the concentration is likely to be so high, the acidic clouds would be likely to kill the person spraying the acid... x42bn6 Talk Mess 14:18, 2 November 2007 (UTC)[reply]
See acid for details. The Bronsted-Lowry definition for an acid is a proton donor. Basically, it causes what it touches to be oxidized, or lose electrons. A metal corrodes in acid because it gets oxidized - it dissolves into the acid as it loses electrons and those electrons are taken up by the acid to form hydrogen gas. In the case of concrete, which is made primarily of cement and water, metal oxides are the primary ingredient. In this case, the metal is already ionized, but the oxygen can give up its extra electrons and form water with the free hydrogen of the acid. So corrosion of concrete is also oxidation-reduction, and the rate reactions depend primarily on the pH of the acid, which depends on its concentration or, perhaps more generally, its Hammett acidity function. So if the pH is very high, the rate of oxidation-reduction is very high, and the material corrodes quickly. I'm not sure how the reaction for wood (cellulose) works. SamuelRiv 14:28, 2 November 2007 (UTC)[reply]
Whoa there…acidity as a proton donor means it reacts with a base to form a bond, not to abstract the electrons from the base. Other things may indeed happen, but I don't think B–L acid/base strength and redox-potential aren't as directly related as you're suggesting. I'd also point out that "pH is very high" sounds like a very weak base. DMacks 14:52, 2 November 2007 (UTC)[reply]
I defer to you on this, as I'm not a chemist. Reaction rate is a good article that discusses the complexity of calculating reaction rates, and although there is a dependence on the concentration of the reactants (the acid and the concrete), it does not account for the dissolution of concrete necessary for that to work. On double-checking my books, they specifically note that one cannot infer rates of reaction through the equations I used. And yes, pH must be very low, hopefully negative, for a particularly strong acid. That being said, since an answer still hasn't been given, the math does suggest that the rate of reaction can be increased by increasing the concentration (or pH) of the acid, the concentration of the dissociated reactant (so if we can dissociate the metal oxides in concrete quickly and without a lot of solvent), and the number of free electrons per molecule of the other reactant. SamuelRiv 06:23, 3 November 2007 (UTC)[reply]

One problem is the rate of the reaction, acids usually do not dissolve at anywhere near that rate. A very high power would be required. It may be possible to do it with extremely high temperatures that could melt the metal at the same time as dissolving it. For super acids you could try trihydrogen cation or helium hydride. These would take on the form of a plasma though, and not be a simple liquid. Graeme Bartlett 23:37, 2 November 2007 (UTC)[reply]

Superacid maybe? The article suggests that Fluoroantimonic acid is very strong but I am not sure that translates Alien strength or reaction rate. --DHeyward 03:44, 3 November 2007 (UTC)[reply]

endothermic vs. exothermic

I know exothermic means a system is releasing heat, and endothermic means a system is recieving heat, but if a system was cool to the touch would it be and endothermic or exothermic reaction? It seems like if it was exothermic it would be cool because it was releasing heat, but could also warm because it's giving off heat? Anyone care to help?--MKnight9989 14:08, 2 November 2007 (UTC)[reply]

Exothermic is roughly equivalent to "energy exiting the system" or "release of energy in the form of heat" - which makes things hotter. Endothermic is "absorption of heat" which makes things colder (because the amount of heat has gone down). If a system is cool to the touch, it means nothing, though - but if the temperature goes down, I'm guessing it's an endothermic reaction. (Note: It's been a while since I studied Chemistry) x42bn6 Talk Mess 14:14, 2 November 2007 (UTC)[reply]
Yeah I guess that makes sense. Thanks mate.--MKnight9989 14:22, 2 November 2007 (UTC)[reply]
I'd wait for a better explanation, though. Our article on exothermic, for example, looks fairly woeful... x42bn6 Talk Mess 14:24, 2 November 2007 (UTC)[reply]
You don't really specify what is the reaction. Do you mean there is a reaction within the system, or did you want to know if the very act of touching something that felt cold was exothermic/endothermic? If there was a separate reaction and it was cool to the touch, it means that the system was colder than the surrounding environment; heat was taken from the environment (air, container, etc) to fuel the reaction. This heat was transformed, and the total amount of heat in the system is less than what you started with. That means the reaction (that I assume was taking place in the system) was endothermic. Just touching the container isn't really a reaction, just a heat transfer; the total amount of heat is the same. I don't think you could classify it as endothermic/exothermic. But I'll think about it --Bennybp 14:27, 2 November 2007 (UTC)[reply]
It was a potassium sulfate/water reaction. --MKnight9989 14:30, 2 November 2007 (UTC)[reply]

atomic size constraints

What keeps an atom from being stable or even formed beyond a certain number of nucleons? Are the forces that hold the atom together not strong enough due to excessive mass or to excessive distance or both? Dichotomous 15:16, 2 November 2007 (UTC)[reply]

I believe it's because the strong force (actually the residual strong force) pulls the nucleons together, while the electrostatic repulsion that the protons have for each other pushes them apart. The strong force only works at short distances, so when the nucleus gets too big, the like-charge repulsion causes instability and fission. You might like to read Island of stability. --Sean 18:22, 2 November 2007 (UTC)[reply]

Spherical wheels on cars?

Think it will ever happen? I can think of a few advantages they might have. Greater manuverability for one, no need to turn the wheels, simply change the direction of their spin. 64.236.121.129 15:18, 2 November 2007 (UTC)[reply]

The contact patch would be much smaller than with a tire, right? That could be a problem. Also you'd need some pretty impressive mechanical magic to give the sphere driving force while still allowing the full range of motion. Friday (talk) 15:23, 2 November 2007 (UTC)[reply]
Indeed--can you still drive the sphere effectively after it gets wet in the rain or snow? (Presumably it is driven by rollers in contact with the sphere somewhere near its top surface...?) Is there a good way to cushion the ride? Will you dent or chip the sphere every time you run over a rock? Do you lose space from the passenger, engine, or storage compartments when you have to fit in the larger volume of the wheel along with whatever motors are driving it? TenOfAllTrades(talk) 15:47, 2 November 2007 (UTC)[reply]
You're doing it wrong! Lanfear's Bane | t 16:22, 2 November 2007 (UTC)[reply]
Those Outspan cars were really Mini's - we have a photo of one of them in the article. SteveBaker 17:29, 2 November 2007 (UTC)[reply]
I believe it would also be a lot more work to change the direction of spin than to just turn the wheels, though I'm not certain of the underlying physics. Given this, while you'd certainly end up with a better turning radius, I don't know that you'd have across-the-board better maneuverability. The points above are also well-said in this regard -- one upper limit to maneuverability is the point at which the tires can't overcome angular forces and start to slip. The reduced contact patch of spherical tires would hasten this slipping. — Lomn 16:26, 2 November 2007 (UTC)[reply]

Well the contact area, traction, and the possibility of chipping would have more to do with the material used for the wheel, rather than the shape. A perfect sphere doesn't have much contact with the ground true, but neither does a torus. A tire inflated too much doesn't have much contact with the ground, it needs to be deflated so it sags a bit, increasing the contact area. If you put a beach ball on a street, you can increase its contact area with the ground simply by pushing on top of it. 64.236.121.129 16:33, 2 November 2007 (UTC)[reply]

Roads are pretty much flat - you don't need a sphere - you need a cylinder (which is what we have). If you need more grip, you lengthen the cylinder (wider tyres) or increase the contact patch (by deflating them a little) or use a more appropriate tread pattern (eg racing slicks, snow tyres). A spherical tyre would have less contact area than a cylindrical tyre (for a similar size) no matter what. SteveBaker 17:29, 2 November 2007 (UTC)[reply]
Tread is a good point here. Would be pretty hard to have a pattern that worked in "every direction" around the sphere. DMacks 17:42, 2 November 2007 (UTC)[reply]
File:Top fuel eg PD EN.jpg
You say, "wider tires do NOT increase traction with the ground". Experts disagree.
I must emphasize, wider tires do NOT increase traction with the ground. 64.236.121.129 18:40, 2 November 2007 (UTC)[reply]
I remember learning some theory about friction in which this was true. But I also remember learning that in real life, it does increase traction. I don't remember specific details, but why else would people use wider tires when they want better traction? Friday (talk) 19:02, 2 November 2007 (UTC)[reply]
The quick'n'dirty explanation of why it doesn't matter in theory is that friction is proportional to force, not contact area (if you double the area, you half the amount of force per unit area but that still means same total frictional force). DMacks 19:22, 2 November 2007 (UTC)[reply]
64.236.121.129 is not correct; wider tires do increase traction by dint of a larger contact patch as noted and linked above. As for the idea of spherical tires, in addition to the tractive disadvantages, transmitting drive to spherical "wheels", and braking them, would be a hideously complex and likely quite inefficient exercise. The disc-shaped wheel and quasitoroidal tire will be with us for the foreseeable future. --Scheinwerfermann 19:21, 2 November 2007 (UTC)[reply]
The "law" of friction that they teach kids in school and junior physicists (ie Frictional force is proportional to the normal-force multiplied by the coefficient of friction and that the contact area is irrelevent) is NOT TRUE! It's not a law of nature - it's not even a reasonable approximation - it's a rule of thumb that works for some materials under some circumstances. Friction is vastly more complex than that and there is no simple rule. So - it is undoubtedly true that car tyres gain more frictional force if they have a larger contact patch. If you doubt that - and you happen to be in Texas, I'll take you for a hair-raising ride in my MINI Cooper'S with racing slicks and with regular tyres - when you've changed your pants and cleaned up a bit - you'll admit that I'm right! Have you seen the width of the rear tyres on a Formula 1 racecar? Why would they make them so wide if the area of the contact patch didn't matter? Thin skinny tyres would work just as well...but they don't...and there is a reason for that! So - please ignore what your high school teacher said - it's not true. SteveBaker 20:06, 2 November 2007 (UTC)[reply]
To answer Friday, they use wider tires because they last longer. That's critical if you are racing, and you want to minimize your time at the pit stop. As for your points Steve, why would that teach that if it wasn't true? Last I recall, perpetuating myths isn't appropriate in a place of learning. It's in the textbooks, and the teachers teach it. I'm just supposed to take your word for it against theirs? 64.236.121.129 20:21, 2 November 2007 (UTC)[reply]
No - that's nonsense - I have a set of racing slicks for my car - they are very wide and I can tell you with absolute certainty that I don't own them for their long tread life!! The reason some people still teach this load of nonsense is because it's in the text books and the curriculum. Read the best undergraduate physics textbooks on the planet - Feynman's Lectures on Physics - he goes to a lot of trouble to un-teach you this bad piece of misinformation. Read Friction#Classical_approximation - especially the last paragraph: [This] approximation is fundamentally an empirical construction. It is a rule of thumb describing the approximate outcome of an extremely complicated physical interaction. - in other words, it's not true. SteveBaker 20:42, 2 November 2007 (UTC)[reply]
So you think "a half truth is a whole lie"? I'd say that's, well, half-true. --Trovatore 21:06, 2 November 2007 (UTC)[reply]
What?! Tire longevity has a lot to do with the material of the tire. As a general rule, stickier tires wear out faster. All kinds of things that people are taught are simplified models, accurate only in certain circumstances. I remember being taught that atoms are little balls with other little balls orbiting them- like tiny little star systems. Once you're past junior high school, this model is way too simplistic. Friday (talk) 20:26, 2 November 2007 (UTC)[reply]
It's absolutely reasonable to request that...right up to the point where I gave you exactly that and you ignored me! Which part of Read the best undergraduate physics textbooks on the planet - Feynman's Lectures on Physics didn't you understand? SteveBaker 20:58, 2 November 2007 (UTC)[reply]
My apologies - in fact we differed by just a couple of minutes - my first effort got bounced as an edit conflict. However, please check Feynman - volume 1 section 12-3. SteveBaker 16:55, 3 November 2007 (UTC)[reply]
Casual observation reveals that a great many cars use wider tires for better traction. If this isn't legitimately justifable, we have to assume incompetence or conspiracy on the part of the engineers who chose those tires. Friday (talk) 20:42, 2 November 2007 (UTC)[reply]
Leaving aside the incompleteness of the 'friction-force-is-proportional-to-μ-and-normal-force-only' model, a move to 'wheels' with smaller contact patches (and larger forces applied per unit area) would increase substantially wear and tear on the road itself. TenOfAllTrades(talk) 20:32, 2 November 2007 (UTC)[reply]
I feel a need to explain why the 'rule of thumb' for frictional force is wrong. As our article on Friction explains, that 'rule' is based on the following assumption: That only a small percentage of the atoms on one surface are in contact with the other surface. When force is applied to push the two surfaces together, more atoms come into contact and the frictional forces increase. The reason the rule of thumb says that the contact area doesn't matter is that the number of atoms that come into close contact depends on the pressure applied to the two surfaces. Since (in this case) pressure is force divided by area, the pressure between the two surfaces decreases as the contact area goes up. So whilst more atoms are available to come into close contact, the pressure forcing them together goes down in the same proportion - to frictional force is independent of the contact area.
However, there was an assumption there - that the percentage of atoms in contact was small compared to the total number of exposed atoms on the two surfaces. For stiff things like steel plates that are microscopically rough, that's a reasonable assumption and the rule of thumb works quite well. However, rubber is very soft and compliant and with even a moderate force applied to it, it'll deform until a VAST proportion of the atoms are in contact with the surface below. Now, varying the pressure forcing the tyre onto the road isn't making a dramatic difference to the percentage of atoms in contact - so increasing the area of the contact patch can indeed make a huge increase to the amount of grip you get.
Surfaces that are lubricated suffer from similarly violating this "law" because the layer of oily lubricant can prevent atoms of one surface getting close to those of the other even if the pressure on them increases. Other surface - like Ice - melt when you apply pressure to them - so they lubricate themselves - and the more pressure you apply, the more lubricant gets between the two surface.
SteveBaker 20:55, 2 November 2007 (UTC)[reply]

I find it unlikely that the width of the wheel determines how much traction it has. It merely depends on the surfaces of the two contacting surfaces. For example, a bicycle wheel can have excellent traction despite it being very narrow. A train has a very narrow contact surface between its wheels and the rail, but has adequate traction. Casual observation of race cars using wide tires and thus assuming they use them because of greater traction is a false assumption. Facts and truth come before logic or common sense. Common sense would indicate that .999... doesn't equal 1, but it does. Malamockq 22:48, 2 November 2007 (UTC)[reply]

Did you read the above? There's some pretty good explanations there that go beyond mere speculation. They do use wide tires for better traction, but you're right- the mere existence of wide tires could be explained other ways, too. Friday (talk) 22:53, 2 November 2007 (UTC)[reply]
Then by insinuation we should say that the wider the wheel is, the more traction it should have. So if you have a wheel 1 mile wide, it should have more traction than a wheel 5 inches wide (assuming the surface of the tire is the same)? You have to take your assertion, then exaggerate it and determine if it makes a meaningful difference. Malamockq 23:01, 2 November 2007 (UTC)[reply]
Yes! Wider wheels mean more friction. There will come a point where so little of the wheel is in fact pressed into the minute bumps in the road surface that increasing the area won't help anymore - but for wheels up to (say) a couple of feet wide, the contact area is very important indeed! I'm not saying that the amount of friction is proportional to the contact area or the square of the contact area...or any other particular rule. I'm saying that friction is vastly too complex for that simple rule of thumb to be true. There are some materials in some sorts of contact where the rule is a pretty good approximation - there are other cases where it's not even close. Car tyres are in the latter category. SteveBaker 16:55, 3 November 2007 (UTC)[reply]
I'm not following you. But, SteveBaker gives a good explanation above about why with steel, size of the contact patch isn't a big factor, whereas with things like rubber, it is. Obviously there would be serious practical problems with a wheel a mile wide.  :) Friday (talk) 23:05, 2 November 2007 (UTC)[reply]
Funnily enough I was pondering the same question the other day after thinking about a computer mouse (where a ball moves two rollers, one vertical and one horizontal) and wondering whether anybody had tried reversing the principle and using two powered rollers to move a sphere in order to move an object. This wouldn't be very pratical for things like cars but I thought that things like robotic vacuum cleaners might work very well using this principle although it would probably need three or four spheres to keep the thing off the floor and going in the right direction. GaryReggae 23:19, 2 November 2007 (UTC)[reply]
Um, Malamockq, I'm not quite following you, either. You say "Facts and truth come before logic or common sense", and I mostly agree with you, but then your only evidence for why the width of a tire is irrelevant is that you "find it unlikely" that it makes a difference.
Steel-wheeled trains have adequate traction, but only barely. Very small slopes, or small amounts of water, snow, ice, or wet leaves, can leave a train spinning its wheels and unable to move. Most trains carry sand which they can sprinkle on the rails ahead of their wheels to increase traction when they need to.
Steel wheels and rails are used on trains because they have excellent rolling resistance (low), and excellent lifetime (high). But the traction's only so-so, though of course it helps that they have huge amounts of normal force (i.e. weight) to throw into the µN equation. —Steve Summit (talk) 23:46, 2 November 2007 (UTC)[reply]
No, my reason for believing that width doesn't make a difference, is that they teach that fact in physics classes. The train example may have been weak, but any lack of traction is due to the surfaces of the train wheel and the rail. It has nothing to do with the width. Malamockq 01:57, 3 November 2007 (UTC)[reply]
Yes, they teach that to kids in physics class, along with a bunch of other oversimplifications. This is all well explained above. Did you see the explanation that with some materials, size of contact patch doesn't matter much, whereas with some other materials, it does? Yes, bicycles have skinny little tires. But check out a powerful motorcycle sometime- much wider. Why? Because it needs way more traction to put that kind of power down on the road. Friday (talk) 15:59, 3 November 2007 (UTC)[reply]
There are plenty of things like this that are taught in schools as simplifications of reality. Heck we still teach Newtonian mechanics - when we know full well that they are flat out WRONG. Einstein proved Newton was wrong almost 100 years ago. So the fact that we teach kids something doesn't make it true. We teach simple approximations.
Anyway, I'm now sitting in front of the book I referred you to (Feynman's Lectures on Physics, volume 1, section 12-3). Just below the F=uN equation, Feynman (who was a physics nobel prize winner) says:
"Although this coefficient is not exactly constant, the formula is a good empirical rule for judging approximately the amount of force that will be needed in certain practical or engineering circumstances. If the normal force or the speed of motion gets too big, the law fails because of the excessive heat generated. It is important to realize that each of these empirical laws has its limitations, beyond which it does not really work."
To get the full story, you need to read on for a couple more pages - but I think you can tell from my short quote that Feynman is most certainly not saying that this is an utterly fundamental law. Car tyres most certainly provide more grip when they are a foot wide than when they are four inches wide - look at any high performance sports car and notice the fat tyres...there is a reason for that. You wouldn't add weight (PARTICULARLY 'unsprung weight') to a performance car unless it bought you some major benefits in terms of going faster or accellerating harder. It's not just tyres either. Why do you think people put bigger disk brakes on cars to make them stop better? SteveBaker 16:55, 3 November 2007 (UTC)[reply]

It would sure make parallel parking a lot easier, that's for sure. :) shoy (words words) 03:46, 3 November 2007 (UTC)[reply]

Check out Rolling resistance. Acceptable 04:26, 3 November 2007 (UTC)[reply]

Where did the energy go

I was told today that although most macro structures are held in shape by the forces that govern them as space time expands, and as a result do not expand, photons are stretched by space-time, and as a result have an increasing wavelength, assuming that c is a constant through all space-time (a reasonable assumption I presumed) therefore a decreasing frequency, and so a photon hanging about for a substantial amount of time would lose energy due to E = hf. Where does the energy go? ΦΙΛ Κ 18:58, 2 November 2007 (UTC)[reply]

Gravitation potential energy. It is similar to the red shift in a photon trvaelling up hill in a gravitational field. Graeme Bartlett 23:39, 2 November 2007 (UTC)[reply]
Actually, no. The correct answer is "General relativity is a bitch that apparently doesn't obey Conservation of Energy". It's possible that there's some related quantity that's conserved, but scientists are yet to come up with anything that seems to work yet. Confusing Manifestation 08:14, 3 November 2007 (UTC)[reply]
Conservation of energy is a fundamental consequence of the symmetries of spacetime. I should posit an answer here - I didn't before because I'm not sure if this is correct, but the answers given so far are misleading. I believe that a photon would ride local expansion like a wave, and thus would not lose energy relative to an outside observer because velocity would effectively increase (because spacetime has been locally changed). However, for the uniform expansion of the universe, I'm not sure if this model fits: certainly locally the change in curvature accounts for any energy loss - maybe that's all you need. SamuelRiv 07:04, 4 November 2007 (UTC)[reply]
I found this site useful in describing (and attempting to solve) the problem embodied in the question: [16]. In summary, classical conservation of mass-energy doesn't hold, but it is possible to define many quantities which are conserved. However, there's no quantity which always works and everyone likes. 58.96.70.254 12:40, 4 November 2007 (UTC)[reply]

Antonym: Congenital?

Very quick question: what is the antonym for congenital, in the context of diseases (i.e. inherited diseases versus infectious or non-infectious diseases acquired throughout life)?

209.51.73.60 23:57, 2 November 2007 (UTC)[reply]

Acquired. Someguy1221 00:27, 3 November 2007 (UTC)[reply]
Contracted. --DHeyward 04:09, 3 November 2007 (UTC)[reply]

November 3

How to disappear completely

Teleportation: "Until recently, scientists had been able to transport only light or single atoms over short distances (millimeters)."  !?! In teleportation our original copies don't have to die after all: there will be no pesky original copies of us to trouble over. And here I was worrying about a convenient future that might involve the existential murder trial for killing myself just to make it to Sydney in the blink of an eye. Joking aside, how does any sort of teleportation of matter not tear to shreds the Heisenberg uncertainty principle? —Preceding unsigned comment added by Sappysap (talkcontribs) 01:39, 3 November 2007 (UTC)[reply]

There's something called quantum teleportation that's specifically about getting around the uncertainty principle in situations like this. But that's kind of a misnomer, since you can't use it to help you teleport something until you've solved all the problems attendant on classical teleportation, like the whole determining-the-complete-structure-of-the-object-and-then-reassembling-it-on-the-other-side part. I see no reason to believe that this kind of technology will ever exist. -- BenRG 02:23, 3 November 2007 (UTC)[reply]
I love that song.--The Fat Man Who Never Came Back 02:28, 3 November 2007 (UTC)[reply]

Artery defurring

Is it possible over a period of time to defurr your arteries by eating zero fat? —Preceding unsigned comment added by 88.109.219.79 (talk) 01:51, 3 November 2007 (UTC)[reply]

Defurr? Dureo 08:44, 3 November 2007 (UTC)[reply]
The technical name for "furring" of the arteries, also known colloquially as "hardening" of the arteries, is atherosclerosis.
The research I am aware of pertaining to your question is the Lifestyle Heart Trial of Dr. Dean Ornish. His research was widely popularized in his best-selling book Dr. Dean Ornish's Program for Reversing Heart Disease.
We can’t give any advice pertaining to using diet and lifestyle to prevent or treat atherosclerosis, so consult with your doctor as to whether a change in diet and lifestyle may be appropriate for you. You may also find it valuable to consult with a registered dietitian. MrRedact 17:30, 3 November 2007 (UTC)[reply]

What causes it? — Daniel 03:39, 3 November 2007 (UTC)[reply]

I'm assuming it's because the products will be at a more favorable thermodynamic state than the reactants, but I could be wrong. Specifically what substance(s) are you wondering about? shoy (words words) 03:52, 3 November 2007 (UTC)[reply]
Thermodynamically, you're simply looking for the state with the lowest free energy, and a substance that dissociates will have a lower free energy of its separated components than if those components were together. This depends on the nature of the solvent, and I'm gonna be careful here from the mistake I made before and say that the textbook statistical mechanics analysis of solutions won't necessarily apply if tthe solute dissociates. From chemistry class though, I remember that conceptually when a compound dissolves in water, it will dissociate if the pull of the polarized water molecules is greater than that binding polarized components of the compound. So in an ionic compound like NaCl, the Na+ ions are attracted to the O-2 side of the water molecules while the Cl- ions are attracted to the H+ side of the water molecules, and this attraction breaks apart the bonds of the salt crystal. SamuelRiv 06:34, 3 November 2007 (UTC)[reply]
DOn't forget Entropy! Dissociating anything results in two or more molecules/ions from one --> more disorder --
This isn't strictly true. While dissociation might seem to imply the creation of more molecules, one needs to be careful here. For example, the dissociation of water is often represented at H2O <-> H+ + OH-. But what is really going on is 2 H2O <-> H3O+ + OH-, so no new molecule is actually created. shoy (words words) 18:03, 3 November 2007 (UTC)[reply]
Entropy factors into the rate equations. That's why rate is proportional to reactant concentration raised to the power of the number of molecules needed per unit of product. SamuelRiv 14:36, 3 November 2007 (UTC)[reply]

use of superbugs in oil slicks

hello i would like to know about the process involved in use of superbugs in oil slicks. ie how exactly the superbugs help in getting rid of oil slicks & the pros & cons of it.i would also like to know a few strain names of the superbugs as examples. i'm aware of the diamond vs. chakraborty case. i'm asking this ques as r dear teacher hasnt taught this topic properly & cudnt find enough info to write a 10 mk. ans in exams !!! so pl. help me out. 220.227.116.220 07:05, 3 November 2007 (UTC)[reply]

I believe the Bioremediation, Biodegradation, Microbial_biodegradation, and Biotransformation articles will help you. Dureo 08:42, 3 November 2007 (UTC)[reply]
The basic concept is that they would digest the toxic components into waste products which aren't toxic. StuRat 14:20, 3 November 2007 (UTC)[reply]

The night sky, stars, and galaxies

Hi, I have sort of a random question just out of curiosity. It may be a dumb question but I have never heard it explained. When one simply looks up into the night sky and sees all of the different stars, are they individual stars like our Sun? Or are some of them that we observe stars and others actually galaxies? And if they are mostly individual stars, are they all located within our own Milky Way galaxy? Again I apologize if this seems elementary, but I looked in both the Star, Galaxy, and Night Sky articles and was not quite sure of what the answer is. Thanks in advance. Zenislev 08:19, 3 November 2007 (UTC)[reply]

Some are individual like our sun, and some are further away and appear individual, but are actually galaxies and star clusters, the Universe article may help also. Dureo 08:29, 3 November 2007 (UTC)[reply]
Most are stars as in the sun, but actually most are much brighter than our sun. Few Galaxies are visible to the eye, and they appear as fuzzy patches. Quite a few stars are double stars. LMC. Graeme Bartlett 08:36, 3 November 2007 (UTC)[reply]
You also see the occasional planet.--Shantavira|feed me 09:00, 3 November 2007 (UTC)[reply]
I've heard a number quoted a lot that approximately 2/3 of the stars that can be seen in the night sky are actually binary (or higher) stars --Shniken1 10:14, 3 November 2007 (UTC)[reply]

Besides stars, clusters, and planets, there's one more possibility: while a comet usually has a tail or at least a fuzzy appearance, it can occasionally look like a star to the naked eye. I saw one like that a few days ago.

The answer to "And if they are mostly individual stars, are they all located within our own Milky Way galaxy?" is normally yes. The only way that a star located outside the Milky Way could be bright enough to see would be if it was going supernova, which happens very rarely indeed. --Anonymous, 10:37 UTC, November 3.

The most distant object visible with the naked eye (and only in the most favourable conditions) is the Triangulum Galaxy. Gandalf61 11:23, 3 November 2007 (UTC)[reply]

What's even more interesting is that if there are trillions upon trillions of stars in the universe, why is the sky is so dark at night? -- MacAddct  1984 (talk &#149; contribs) 13:55, 3 November 2007 (UTC)[reply]

Two reasons:
  • There aren't an infinite number of stars.
  • The light from distant stars is absorbed by dust or spread out so thinly that not even one photon hits our eyes or telescopes.
The sky does appear much brighter through a good telescope, however. StuRat 14:17, 3 November 2007 (UTC)[reply]
Well, yes - but that's kinda meaningless in this context. It's because you're gathering light through a large aperture and squishing it down into a little eyepiece - it's not because you are somehow seeing more photons per steradian of sky or anything of that sort. The deal is that the amount of light from a distant source decreases as the square of the distance - but the number of stars in a spherical volume of space averages out to the cube of the radius. So, were it not for the interstellar dust clouds, the amount of light coming from stars at greater and greater ranges would be greater and greater and the night sky would be insanely bright instead of mostly black (and we'd be cooked by the infrared, hard X-rays, etc. However, the amount of dust obscuring the light from those stars is also proportional to the distance - but the rate of light absorbtion is more than linear. eg If one lightyear of dust absorbed half the light from a star, then the next light year of dust would absorb half of what came through the first lightyear of dust, and so on. Hence the amount of light you see through some amount of dust is an exponential thing. When you work it all out, it's clear that even if there were infinite numbers of stars, the amount of light we'd see would be what we see now. SteveBaker 16:16, 3 November 2007 (UTC)[reply]
Those are not the reasons the sky is dark at night. The reasons are (as the Olbers' paradox article says) cosmological redshift and the finite age of the universe. Absorption by dust has no net effect because all of the absorbed energy is reradiated. -- BenRG 16:44, 3 November 2007 (UTC)[reply]
It's also possible to see satellites and (especially) the International Space Station - but they move pretty fast so you can generally tell that they aren't stars. SteveBaker 16:16, 3 November 2007 (UTC)[reply]
Yes, and the observable universe is only about 15 light-years in radius, so eventually you'll run out of bright objects altogether as you go past. It's much easier to see a star than another galaxy if you look anywhere in the sky. However, the central disk of the milky way is made of millions of stars, which is why, even being about 10,000 light-years away, it is still possible to see it in a dark area. I've seen the milky way through binoculars, although not very clearly, from a light-polluted location. Comet 17P/Holmes, by the wat, is currently just west of the Milky way. Hope this helps. Thanks. ~AH1(TCU) 17:08, 3 November 2007 (UTC)[reply]
Huh? The observable universe has comoving radius about 46.5 billion light years. Algebraist 17:41, 3 November 2007 (UTC)[reply]
Oops, minor error, I meant 15 billion light-years. That's assuming you can measure the distance from Earth to the edge of the universe without the universe expanding while you measure (eg. the time for the measuring "ruler" to reach the observable edge of the universe from the Earth is 0). This is based on the fact that the universe is about that old and that the farthest observed galaxies are about 17 billion light-years away. Hope this helps. Thanks. ~AH1(TCU) 18:02, 3 November 2007 (UTC)[reply]
You're wrong here: it's not the distance (which is closer to 14 billion (more precisely 13.7) light years according to the latest data) measured by an instantaneous ruler but the light travel time times the speed of light. The comoving distance is more like what you mean, and it's larger, as Algebraist pointed out. Icek 01:55, 4 November 2007 (UTC)[reply]

Thanks for all of the helpful responses! That sure is enlightening (I'm sort of unlearned in the field of cosmology). It's very interesting to think that all of those stars we see are mostly just from our own galaxy and that there are billions of other galaxies out there. Well have a good one guys and thanks again ;) Zenislev 19:50, 3 November 2007 (UTC)[reply]

If you live in an urban environment, you might only be able to see a thousand stars with the naked eye, all of which are in the nearby parts of the Milky Way galaxy. Fainter, or more distant stars, and the other galaxies would require at least binoculars or the very dark skies you get far away from cities. When I first saw the nearest bright galaxy (Andromeda, M31) through a pair of binoculars, I was surprised how faint it really was - a small fuzzy patch, especially compared to those great photos you see. Astronaut 03:41, 4 November 2007 (UTC)[reply]
Zenislev, if at all possible, grab a pair of binoculars and look at the Andromeda galaxy! It will look like a luminous smudge, but it's quite cool to be able to see another galaxy. And in fact, the Andromeda galaxy is *huge* on our night sky... the problem is that most of it is relatively dim. You can only see the brightest, central part. If I remember right, the whole thing is as big as about 7 full moons! So the difficulty is not magnifying something that appears to small - it's enormous! Instead, the hard part is collecting enough light to see something that's relatively dim. Then point your binoculars at Jupiter and look for the Galilean moons. Even with small binoculars you might see 2 or 3 of them! --Reuben 04:36, 4 November 2007 (UTC)[reply]
I do indeed live in the city, so I know what you mean there. I did however bring binoculars with me on a camping trip a few months ago and was amazed at how much I could see; I really didn't know what to make of any of it though. That does seem interesting to be able to see Andromeda, I never knew you could. I'll have to try that when I get a chance. It's hard to comprehend the scale of things in our universe though; to think that this blob in the sky really spans thousands of light-years is really thought-provoking. Zenislev 05:30, 4 November 2007 (UTC) —Preceding unsigned comment added by Zenislev (talkcontribs)

motor tracts from the cortex

Is the diagram below correct?

motor tracts from the cortex

lateral system..................................medial system

pyramidal...........................................pyramidal
lateral corticospinal tract.....................anterior/medial corticospinal tract
.........................................................corticobulbar tract

extrapyramidal....................................extrapyramidal
rubrospinal tract..................................tectospinal tract
lateral reticulospinal tract.....................medial reticulospinal tract
lateral vestibulospinal tract...................medial vestibulospinal tract

(Please don't be bothered with the "...", they're just there because spaces don't work.)

Second question: Where in this diagram does Corticobulbar tract go?

Thank you! Lova Falk 11:01, 3 November 2007 (UTC)[reply]

It seems correct after I cross-referenced, but it's been a while since neuroanatomy. Try google images on the coritcobulbar tract, and you will see cross-sections of it through the midbrain, pons, and medulla that should give it away. SamuelRiv 15:03, 3 November 2007 (UTC)[reply]
According to the textbook "Principles of Physiology" by Berne and Levy there are both lateral and medial vestibulospinal tracts. Other textbooks just list the vestibulospinal tract as medial. Berne and Levy say, "Much of the corticobulbar tract belongs to the medial system". Course and distribution of facial corticobulbar tract fibres in the lower brain stem goes into some of the details. --JWSchmidt 16:25, 3 November 2007 (UTC)[reply]
Thank you! I changed the diagram by putting the lateral and the medial vestibulospinal and reticulospinal tracts into the diagram. Am I correct to put the corticobulbar tract with the pyramidal tracts? Lova Falk 18:31, 3 November 2007 (UTC)[reply]
Yes, the corticobulbar tract is counted as part of the pyramidal system....functionally it has axons that carry signals for good voluntary motor control.....they just carry signals that control motor neurons located in cranial nerve brain nuclei rather than motor neurons located in the spinal cord. --JWSchmidt 22:25, 3 November 2007 (UTC)[reply]

Cocaine

Is it true that some women take cocaine gynaecologically or is it just a myth? Everyone knows someone called Dave 12:11, 3 November 2007 (UTC)[reply]

Can you rephrase the question? I'm not sure if you mean illegally by some type of insertion, or prescribed by an OBGYN, or um something else. Maybe I am not thinking outside the box. Dureo 12:24, 3 November 2007 (UTC)[reply]
If you're asking about routes of exposure:
"[Cocaine] in this hydrochloride salt form may be injected; swallowed; applied to oral, vaginal, or even rectal mucous membranes; or mixed with liquor. Coke is most commonly used by snorting or sniffing." [17] (emphasis added.)
Hope that helps. TenOfAllTrades(talk) 13:40, 3 November 2007 (UTC)[reply]

Atomic movement

Can atomic movement of an atom be stopped an if so what would be the result? I am wondering if atomic movement if of a particular fequency, could you apply the same frequency of opposite phase to cancel it out and if so what would the effects of this be? If you can accelerate particles by superheating them, would stopping the movement have a cooling effect? —Preceding unsigned comment added by 59.101.139.208 (talk) 13:50, 3 November 2007 (UTC)[reply]

Yes. At absolute zero, there is absolutely zero (har har) movement. Although it's not possible to replicate in a laboratory (although they've gotten very close), some scientists believe that in the far reaches of space the temperature exists. -- MacAddct  1984 (talk &#149; contribs) 13:58, 3 November 2007 (UTC)[reply]
I see that in discussing Bose–Einstein condensate, the absolute zero article mentions "pK" (picokelvin). How does someone measure a temperature that low? --JWSchmidt 15:01, 3 November 2007 (UTC)[reply]
At absolute zero there is still movement predicted by quantum mechanics and Quantum Field Theory (QFT). A simple way to see this is by noting Heisenberg's Uncertainty Principle which shows that a zero change in position requires infinite momentum, and vice versa. Note that this is a mathematical fact of quantum theory and not some philosophical abstraction. Empty space in the current universe is at about 3K and does not approach zero as one goes to the "far reaches of space" because there simply is no far reach of space as expansion is uniform. As one goes farther into the future, however, the temperature of the universe is falling as the expansion of the universe accelerates.
Extremely low temperatures are achieved first by evaporating liquid helium (liquefaction of helium being achieved by the Joule-Thomson effect) and then by isentropic demagnetization of the subatomic particles (so demagnetizing particles without changing entropy). They are measured by reversing the effect: measuring the magnetization of subatomic particles and measuring the vapor pressure of liquid helium. This is only effective to a scale of millikelvins, below which I'm not sure what they use. Note the validity of these methods of course depends on having a very good theory of the behavior of particles at low temperatures, which we do (QFT is the most numerically accurate theory known).
On a side note, temperature is defined as if the number of particles in the system doesn't change. That is, the inverse of temperature is the change in entropy over the change in energy. This definition, and indeed statistical mechanics itself, sets no restrictions on how high or low temperature can be. So we can have systems with Negative temperature! SamuelRiv 15:31, 3 November 2007 (UTC)[reply]
Oh, and physically slowing the motion of atoms does cool them: that's how liquefaction of gases is done. See cryogenics and the Joule-Thomson effect. Basically, gas is expanded through a valve which causes energy to be absorbed against the intermolecular forces of the gas, which results in a net temperature decrease as long as the initial temperature of the gas is below the inversion temperature. Again, the motion of the atoms can never be stopped completely. SamuelRiv 19:34, 3 November 2007 (UTC)[reply]

Glue board adhesive composition

I'd like to make my own glue boards, similar to those for catching mice, although mine will be used for insects. The hard part is the adhesive. What ingredients can I use to make my own ? I found this list of ingredients for a mouse glue board, but the portions aren't given and many terms are vague (which stabilizers or hydrocarbons ?):

  • Styrene copolymer
  • Hydrocarbon resin
  • Paraffin oil
  • Stabilizers < 1%
  • Anti-oxidants < 0.5%

I'd like to make this from items I can buy at a grocery store, if possible. Dish washing detergent works, but stays liquid for weeks, until all the water evaporates. Could I make it immediately solid by adding gelatin or something else ? Could I heat it to drive off the excess water ? StuRat 14:07, 3 November 2007 (UTC)[reply]

Dishwashing liquid is about 95% water so it'll take a long time to dry. Try dissolving salt in it to make it get thick. Car wash detergent may be more concentrated. You can try the salt trick with that too. Alternatively citric acid can be used as a thickener. You'll notice a lot of shampoos contain a salt, which is just there to make it thick.
Pressure sentitive adhesives contain a polymer for strength and a tackifier to make it sticky. It is a synergistic combination: the mixture is "stickier" than the two components. For your mouse-catcher, the styrene copolymer is probably a styrene-butadiene-styrene block copolymer like Kraton. The hydrocarbon resin is the tackifier and is probably a derivative of pine tree resin. Delmlsfan 14:58, 3 November 2007 (UTC)[reply]
I guess you could try boiling the washing up liquid to drive the water out more quickly - but I have no idea on the consequences of doing that! SteveBaker 15:48, 3 November 2007 (UTC)[reply]
Just buy some flypaper. -Arch dude 16:54, 3 November 2007 (UTC)[reply]
I have tried flypaper. I found it unacceptable because it's ugly and the glue gets all over everything, including my hands. Also, I want to get all insects, not just flying insects. Spiders and centipedes need to die, too. I've tried to lay flypaper out on a flat surface, but the glue got all over everything in the process. But, that is the idea, I want one-sided flypaper laid down on a flat sheet, say the size of a newspaper. I want to do so on a large scale for a low price. StuRat 17:06, 3 November 2007 (UTC)[reply]
(Spiders and centipedes are not insects). SteveBaker 22:58, 3 November 2007 (UTC)[reply]

Do a search with "make flypaper" and you can get recipes using sugar, golden syrup etc. The advantage of these edible glues is that they wash off very easily in water. Keep in mind some invertabrates wont walk out in the open. I accidentally caught silverfish with a roll of gaffer tape that was a little unrolled, they had wedged into the sticky crevice. IF you are after a more specific result, let us know. Polypipe Wrangler 22:57, 4 November 2007 (UTC)[reply]

X-inactivation and color blindness

Several forms of color blindness are far more prevalent in male humans because they are caused by a defective gene on the X chromosome of which males only posses 1. If you look at the prevalence for males and females in out article, the numbers seem to agree with this theory (1% vs. 0.01%, 6% vs. 0.4%). However, in females one X chromosome is inactivated; this inactivation happens early in embryonic development, at the 70-100 cell state (at least in the mouse, according to the Wikipedia article) and is apparently random. There are several genes which are nonetheless expressed on an "inactivated" X chromosome, but these are mostly genes which are also found on the Y chromosome (pseudoautosomal region).

I don't know how many of the 70-100 cells which all individually undergo X-inactivation end up as the relevant cells in the retina, but even if it's more than one, one would expect that there are cases of female "patchy" colorblindness, i. e. affecting only part of the visual field. This condition then should be even more prevalent than colorblindness in males: If the fraction of X chromosomes with the color-blind property is r, then (assuming random distribution) the fraction of males carrying such X chromosomes is r, and the number of females carrying at least one such X chromosome is 2*r - r2.

On the other hand, if all relevant cone cells originate from only one embryonic cell at the time of X-inactivation, the probability that a female is colorblind is 1/2 if she carries 1 "color-blind" X-chromosome and 1 if she carries 2 "color-blind" X-chromosomes; together that makes for the rate of color-blind females the same as the male rate.

How do you explain the observed approx. r2 rates in females? Icek 17:53, 3 November 2007 (UTC)[reply]

It turns out that not all genes from the so-called 'inactive' X chromosome (Xi) are actually fully silenced (see X-inactivation#Expressed genes on the inactive X chromosome). Of those, a few are transcribed at the same level as on the active X chromosome (Xa); a very few are even transcribed at higher levels from Xi than from Xa. (The obvious example for this last case would be the RNA gene Xist, which effects the X-inactivation.)
I don't know if this is the case for some of the genes related to colour vision, however; I'm still digging in the literature. TenOfAllTrades(talk) 18:52, 3 November 2007 (UTC)[reply]
Some further interesting reading from PNAS. A survey of genes on the X chromosome found that as many as twenty to thirty percent(!) of genes on the short arm of the X chromosome (Xp) may escape inactivation at least some of the time. TenOfAllTrades(talk) 19:13, 3 November 2007 (UTC)[reply]
Thanks for the answer. From the supplementary data of an Cell article drawing the data from this Nature article it appears (see figure 8, last page of the supplementary data) that the relevant genes for red-green blindness at locus Xq28, OPN1LW (for the long-wavelength (red) receptor) and OPN1MW (medium-wavelength (green) receptor) are not at all expressed from the 'inactivated' X chromosome. Of course, the study was made with fibroblasts and the situation may be different in cone cells. Icek 01:01, 4 November 2007 (UTC)[reply]
FYI, here is the OMIM page for OPN1MW and here for OPN1LW. Icek 01:05, 4 November 2007 (UTC)[reply]
Thinking again, it's likely that opsin (the product of the aforementioned genes) may not at all be expressed in the studied fibroblasts; maybe the fact that there is no gray bar at all instead of a gray bar extending only below the zero line for some genes in the supplementary data figure I linked to before means that the expression was zero in the activated as well as in the inactivated X chromosomes. Icek 01:23, 4 November 2007 (UTC)[reply]

Detecting nuclear blasts

A recent news article claims that the US used tactical nukes at Operation Orchard. Seems incredible to me. Is it possible nowadays for an above ground nuclear blast to go undetected? --Duk 18:38, 3 November 2007 (UTC)[reply]

Undetected - no I doubt it, unreported - perhaps. Lanfear's Bane | t 20:20, 3 November 2007 (UTC)[reply]
Fuel-air bombs can get into the neighborhood of a (very) small nuke, as far as TNT equivalent, so there could be some ambiguity in seismic detection, but there is basically *no* way that the radiation would go undetected, especially if people were looking for it. Also, this government can't even keep it secret that they flew some nukes across the country accidentally or that they like to pour water up some poor bastard's nose; there's no way it would stay quiet if they actually used a nuke. --Sean 20:20, 3 November 2007 (UTC)[reply]
I haven't yet found a Wikipedia article specifically about detection of nuclear explosions, but in the articles on the Ryanggang explosion and 2006_North_Korean_nuclear_test it is assumed that any nuclear explosion would be detectable by the characteristic isotopes produced. An explosion of such magnitude as to have been a nuke would register on seismographs all over, and, although we can't know where the satellites are looking, I'd expect one or two to be pointed at the Middle East a lot of the time. Any large explosion in Syria would certainly be detected and probably pinpointed, and it would be a simple matter to test the area, or even downwind, for the telltale isotopes. So, no. --Milkbreath 20:45, 3 November 2007 (UTC)[reply]
Seismographs can only measure the size of the explosion - so a small nuke and a very large conventional bomb would look identical. Direct photography of the mushroom cloud or the resulting debris field could probably tell the difference - as could eye-witness testamony...but we don't have that kind of evidence. Radiation is another way - but only people with access to the site could measure that without some pretty high-tech equipment. The people with the high tech equipment aren't admitting to it - so unless someone has actually been out there with a geiger counter, we can't know. HOWEVER: If the US have fuel/air bombs with equivelent destructive ability to a small nuke, why on earth would they risk world condemnation following the use of nuclear weapons when they could use a conventional bomb with little or no comment and get much the same result? It just doesn't seem plausible. SteveBaker 22:48, 3 November 2007 (UTC)[reply]
A neutrino detector should be able to single out nuclear explosions, in theory, or at the very least determine the exact time that one occurs. Someguy1221 22:54, 3 November 2007 (UTC)[reply]
(Seismographs also show roughly where an explosion took place, and, of course, precisely when.) There are plenty of other signs of a nuclear explosion. They get extremely hot, so there would be melting at ground zero that would be hard to fake, and fires all around. There is an electromagnetic pulse that could be expected to at least have some effect on the power grid. Nukes are very, very, very bright. The hard radiation in its vicinity would show in the people and the animals near it. Sounds to me like Comical Ali has come out of retirement. --Milkbreath 01:19, 4 November 2007 (UTC)[reply]
If it was above ground it would produce easily measurable fallout, even if it was very far above ground or most of it was buried. Such conditions would produce less fallout, perhaps not enough to hurt anyone at any distance, but it would certainly be measurable by a variety of stations around the world within a week or so. So the fact that none of the many, many stations in the area with radiation detectors (even a simple reactor will have those) reported anything awry is pretty much good evidence that it was not used, in my book. But of course the most compelling answer is the one you've given — it would be tremendously stupid and wholly unnecessary to use nuclear arms in such a situation. --24.147.86.187 00:18, 5 November 2007 (UTC)[reply]
If you read the story carefully you'll see that they never say that it was a nuclear weapon that actually detonated at the site. It is poor reporting in any case and is probably meant to be misinterpreted. I doubt the USAF was involved but even if it was, and even if we take as a reliable claim that the escorting planes would have had tactical nukes on board (highly, highly unlikely that they have such things flying around a theatre like that just for the heck of it, much less two of them), the idea that they would non-chalantly use such weapons is pretty absurd, and the article doesn't even really assert that. The whole thing sounds like misinformation to me. --24.147.86.187 00:18, 5 November 2007 (UTC)[reply]

Any clues as to this equation?

Hi all, does anyone have any idea what this represents? [18]

Many Thanks! --Chachu207 talk to me 19:56, 3 November 2007 (UTC)[reply]

Maxwell's equations. Clarityfiend 20:04, 3 November 2007 (UTC)[reply]
Thanks! --Chachu207 talk to me 20:07, 3 November 2007 (UTC)[reply]

Two locomotives

How can two locomotives pull the same train together? It seems to me that if one pulls even a tiny bit harder it will be dragging the other one. --Milkbreath 21:14, 3 November 2007 (UTC)[reply]

Excellent question. I've wondered about this, too.
First, it's not just two locomotives. Suppose you and a friend are both trying to push a car -- doesn't the same argument apply?
I've thought of two answers, though there may be more.
  1. If there's any flexibility in the system, it can "even out" the two force providers (locomotives or people) if they're unbalanced. For example, think about the various couplings (gears and such) via which the mechanical energy is transmitted from the locomotive's actual engine down to its wheels. Suppose furthermore that there's some "lash" or "spring" in that power transmission chain (as there just about always is, in practice). Now, if one locomotive pulls a little bit harder, it might seem as if the other one wouldn't have any work to do, because the train is being pulled out from under it, so to speak. But if one engine is doing all the work, its transmission chain will be maximally compressed. The other engine (not doing any work) won't have its transmission chain compressed as much; it will begin to untwist (precisely because the stronger engine is in the process of pulling the train out from under the weaker engine). But after an infinitesimal moment, after the weaker engine's transmission chain untwists just a bit, the weaker engine will have something to push on, after all, so it will do some work -- precisely its share -- and all of a sudden the stronger engine won't be pulling the whole train, after all.
  2. The more accurate explanation, I think, is based on the fact that each locomotive is a constant force device. Each engine is doing whatever it can to keep pulling on that train. If one engine pulls harder, threatening to pull the train out from under the other one, suddenly the other one won't be applying as much force any more. So it will do whatever it has to do -- i.e. speed up a bit (which is easy to do when suddenly it's not meeting as much resistance when it tries to pull) until it's pulling its weight again. And the same explanation applies to you and your friend trying to push that car, even if one of you is pushing harder than the other.
Now, with that said, it's still an at least somewhat tricky problem. As I understand it, there had to be some real breakthroughs in electric motor control before two diesel-electric locomotives could each act like constant-force devices and pull a train in harmony. (And as I think about it, there's a second problem: if one engine is more powerful than the other, it ends up having to apply a higher constant force than the other. I don't know precisely what mechanism arranges that the constant forces that the two locomotives try to apply are in proper proportions to the power ratings of the two locomotives. Perhaps I was wrong when I said the locomotive had to be a constant-force device; perhaps the more accurate description is "constant power".) —Steve Summit (talk) 21:54, 3 November 2007 (UTC)[reply]
P.S. See our article on Diesel locomotives, particularly the "Adaptation of the Diesel engine for rail use" and "Diesel-electric control" sections, which talk about these control issues. (In particular, it looks like diesel-electric locomotives are indeed constant-power. Steam locomotives, on the other hand, are described as "constant torque", which I think is equivalent here to constant force, so perhaps the arguments for multiple-unit hauling with steam locomotives are different than for diesel.) —scs 22:06, 3 November 2007 (UTC)[reply]
The key is that either engine is capable of pulling faster than the entire trainset is moving. So if the #2 engine is pulling a bit harder than the lead, the lead is still pulling at the front of the #2 engine. Its very much like a group of people pulling together on a rope to haul something. But here's an even funkier configuration for a consist: 2 locomotives at the front, and one at the tail end of the train. :-) ArakunemTalk 22:02, 3 November 2007 (UTC)[reply]
A simple mechanical contraption will allow two locomotives to combine variable pull forces - see Differential (mechanical device) for an example of such. As for the actual nature of the machinery, I would assume as said before that the locomotives approximate to a good extent a constant-force machine. SamuelRiv 00:41, 4 November 2007 (UTC)[reply]

Ah. OK. Minor duh. The slack I was looking for is in the power. As long as the power delivered by the two engines is kept in an overlap range, the rotational speed of the wheels will vary as needed without further control, right? I was thinking originally about the old steam engines with rigid mechanical linkage without considering the steam itself. I guess you'd still have to be extra-careful about your pressure when running two steam engines together. The push-pull configuration sounds tricky, too. I imagine the rear engine would pile up as many cars at the back as it could. Rough on the couplers, I would guess. --Milkbreath 01:52, 4 November 2007 (UTC)[reply]

I think people are making this question harder than it is. There's no need for careful power control. As stated earlier, a locomotive provides an essentially constant force (until moving fast enough that friction becomes an issue, that is). If the front locomotive is pulling with 12,000 pounds force and the second one with 20,000 pounds, then the strain on the coupling between them is 12,000 pounds and on the coupling behind the second one it's 32,000 pounds. The train can't tell whether that force is coming from one locomotive at the front, or two or six. And the second locomotive can't tell whether the train is accelerating faster its 20,000 pounds would produce because there's another locomotive, or because it's running downhill. Same thing if there are engines at the rear, only in that case the strain on the couplings near the rear will be compression instead of tension.
So long as the controls allow all the locomotives to produce the necessary level of power, the only possible concerns with the relative contributions of different locomotives are efficiency and whether it's possible to put a dangerous strain on the couplings. Remember, double-headed trains were common on some railways -- triple-headed trains on some -- in steam days when there was no such thing as multiple-unit control; the crews of different locomotives could communicate only by whistle signals. The trains still moved fine.
--Anonymous, 01:18 UTC, November 5, 2007.
Upon reflection, perhaps the simple answer is just that modern locomotives are not constant-velocity devices.
If they were, then yes, the fast one would drag the slow one -- but they're not. —Steve Summit (talk) 02:34, 5 November 2007 (UTC)[reply]
The locomotives all have torque regulators so they all run at the same torque. If one goes over torque, it goes underpower, if it is torque it goes over power. If there isn't enough power it will cut off and go free-wheeling.--Dacium 03:12, 5 November 2007 (UTC)[reply]
So they all run at the same torque, or the right torque? Suppose I try to pull a train with two locomotives, one having twice the power of the other. The more powerful one would have to apply twice as much torque, I think. —Steve Summit (talk) 13:02, 5 November 2007 (UTC)[reply]

Anonymous has it correct; there's no fancy regulator (at least on early diesel-electric locomotives), each locomotive just does what it can to add to the overall tractive effort. In fact, you can break this down to the level of traction motors; each traction motor and its associated wheelset turns independently and there can be imbalances between them as well with relatively little harm. The key to all of this is that a locomotive doesn't resist motion that is externally applied to it so the worst that would happen is one locomotive would "goof" off while the other does all the work. (That is, it's not liek multiple synchronous motors fighting each other.) But if the tractive load required exceeds the capabilities of the individual locomotives, then it's a sure bet that they're sharing the load at least somewhat.

Nowadays, individual locos do have some amount of automated wheel anti-slip control within a single locomotive, but I don't know if multiple locomotives share any data to optimize this across multiple locomotives.

Atlant 17:00, 5 November 2007 (UTC)[reply]

Computers and adverse health effects

Hi. I'm not asking for medical advice. I've heard that computers and cause, at least in part, everything from eye problems to headaches to brain tumours. Is at least some of this true? I know people, myself included, who have felt the effects of over-computer use. What causes this? What could possibly cause these if computers aren't responsible? Thanks. ~AH1(TCU) 21:24, 3 November 2007 (UTC)[reply]

The only adverse health effects from computers that I've found to have any basis in fact is toner dust from laser printers getting into the air and causing lung problems. That is the reason that toner no longer comes in a bucket that you pour into the printer. As for the brain cancer and eye prblems - this was believed to be a health problem from the monitor. Some believed that the radiation from the cathode ray tube caused brain cancer and sitting too close to the monitor caused eye problems. Neither is based on scientific observation. The computer itself is merely an electronic device (and a low-voltage one at that). The only real health problem it can cause is the side effect of sitting in front of one all day and not getting any exercise. However, you will find people who claim anything and everything causes health problems. -- kainaw 22:32, 3 November 2007 (UTC)[reply]
CRTs do emit X-rays, and X-rays can cause cancer; however, I'm pretty sure the level of radiation is well below the background level unless you use a (powered on) CRT as a pillow every night. Repetitive stress injuries are a genuine health problem. As far as I know it's impossible to permanently damage your eyes by staring at a screen, though you can certainly tire them in the short term. Concerns about cancer from low-frequency radiation are groundless. -- BenRG 23:37, 3 November 2007 (UTC)[reply]
I agree. The most important thing is to use good posture, to avoid injuring your neck and back (by keeping your head in the same position for looking at the screen) and your wrists (by typing with bad posture). And take breaks every once in a while. —Keenan Pepper 00:39, 4 November 2007 (UTC)[reply]
I agree. Even a trip to the kitchen to make tea every hour is enough when it comes to exercise. --Ouro (blah blah) 08:38, 4 November 2007 (UTC)[reply]
I've been using computers for 40+ years now. I've suffered some pretty nasty symptoms over the years - but the idea that radiation is a problem has been completely debunked. Eye strain, bad posture and repetitive strain issues are real - but there is no mysterious death-rays causing them! All three symptoms can be greatly reduced by two things:
  • Firstly, find an expert to set up your work-station. Many companies will employ the services of an ergonomics expert to figure out your optimium seating position and set up your work area correctly for your particular body dimensions. This may require chair adjustments, desk height adjustment, possibly a foot-rest, almost always something to raise the height of your monitor. Sadly, some companies wait until you've sustained an injury before they'll do that for you - but you should definitely ask and if they refuse, get that in writing because you're going to need it when you come to apply for workers injury compensation later in life. If you can get an expert - take their advice seriously - and if you use a computer at home - measure how your work setup is arranged and try to set up your home set up identically.
  • Secondly, take a few minutes break every 20 to 30 minutes - get up walk around - stretch and stare off into the distance. In most places there are health & safety at work laws that require your employer to allow this. I use a kitchen timer to remind me to take a break because it's easy to forget.
If you start to get repetitive strain symptoms in your fingers - stiffness and a burning feeling on the backs of your hands - then you absolutely need to do something about it. If you carry on working, then over the next few weeks it will spread to your wrists, then your elbows then to your shoulders. The longer you leave it - the longer it'll take you to recover. I'm a huge fan of split keyboards - the ones I use are made by 'GoldTouch' and they are literally split in two with a ball and socket joint between the two halves that lets you adjust the split and tilt for each hand and then lock it in position.
SteveBaker 04:33, 5 November 2007 (UTC)[reply]

Returning to laser printers, they can emit ozone, especially if the recombination catalyst on their ventilation air exhaust has become old and tired. They can also emit styrene vapors from the toner. The fuser units within them can pose a thermal burn hazard if you attempt to service them improperly.

High voltages used within various computer subsystems can also be an electrocution hazard if you attempt to service your own computers. Certain computers will also pose electrical hazards when not properly earthed/grounded.

There is also the hazard of abrading your fist by punching the screen if you spend too much time at certain websites.

Atlant 17:04, 5 November 2007 (UTC)[reply]

General relativity: gravity Doppler?

This is a rather complex question that I ask for my own understanding as it has been bothering me for some time. It is my current understanding that gravity (or changes in gravity) travel at the speed of light. For example, if the sun were to disappear now, earth would continue to orbit its location for another 8 minutes or so, "using up" the gravity that the sun had "sent out" before disappearing and which is still rippling out before earth finally sets off in a straight line.

Now, the hypothetical observations I mention below is meant after compensation for other known gravity effects. Assuming speeds that are significant compared to the speed of light, my question is whether (and if not, why not?) objects moving towards an observer appear heavier and objects moving away appear lighter because one would think the gravity "wave" traveling towards the observer "builds up" or "stretches/thins out" respectively. What I mean by this can be thought of as a type of gravity Doppler effect. So according to my hypothetical model, in an expanding universe the gravity experienced by an observer from other bodies would generally be reduced. As far as I know no such effect has ever been observed; my question is where the flaw lies is this reasoning given the laws of relativity.

Note that I am not referring to the inverse square law of gravity which has to do with relative position rather than relative motion. I am also not referring to the law of always increased observed mass due to motion/relativity in both away and towards cases.

41.243.12.244 22:35, 3 November 2007 (UTC) Eon Zuurmond[reply]

I think I see what you're getting at, but allow me to give you a simple but complete example. Imagine you have two perfectly elastic (and very heavy) superballs out in the depths of space away from everything else. They can collide off of eachother with no loss of energy. Now, say you start them off hurtling towards eachother. They meet, collide, and take off in the opposite directions that they came, and you would like to say they will continue like this for an eternity, since no energy is lost anywhere. But then recall that changes in a gravitational field are transmitted at the speed of light, not instantaneously. As the balls are approaching eathother, each ball "sees" gravity the other sent out fractions of a second earlier, which is weaker than it should be. So the gravitational force bringing the balls together is weaker than one would calculate classically. There's no crazy physics to deduce this, just the simple fact that gravity travels at the speed of light. So, while the balls approach, they don't convert all of their gravitational potential energy into kinetic, some is just magically (so it would seem at first) lost. And when they collide and take off, each "sees" the other's gravity as stronger than it should be. So not all kinetic energy is converted into potential, again, some is just lost. So it's a bit opposite than you suspected. Approaching objects seem lighter than they really are, and receding objects seem heavier. Now, you might ask, where did all that energy go? Gravitational waves. Someguy1221 23:04, 3 November 2007 (UTC)[reply]
I think both of you are confusing gravitational fields with gravitational radiation (waves). Gravitational radiation propagates outward from the source; a gravitational field does not. Say you have an object resting on a table and you push on one end of it, causing it to shift to a new location. The whole object doesn't move immediately; the change in position is transmitted from molecule to molecule outward from the point you pushed at a finite speed (the speed of sound in the solid). Eventually the object stabilizes in its new position. This isn't a very close analogy, but if you think of the gravitational field as the solid and gravitational waves as sound waves in it, you broadly have the right idea. An object's gravitational field is not emitted by it, it's just there. Where did it come from? It didn't come from anywhere; it's always been there. This is why conservation of energy is absolutely necessary for a relativistic field theory of gravity to work at all. If mass could just appear or disappear, it would have to take its field along with it, but that would violate locality.
To answer your (original poster's) question more directly, the gravitational field at a given distance in front of a uniformly moving object has the same strength as the gravitational field at the same distance behind. The headlight effect applies to waves, not fields. Again, the field behaves something like a solid structure attached to its source; when the source is moving, the field is Lorentz contracted. It doesn't lag behind nor get concentrated in front.
I think you're creating confusion for yourself by asking whether objects would "appear heavier" in certain situations. This is a meaningless notion unless you can give it an operational definition. It almost never makes sense to ask about "apparent" properties of objects, unless you're literally referring to what the object looks like, i.e. to the nature of the electromagnetic radiation which is emitted by the object and subsequently detected by your experimental apparatus. -- BenRG 23:11, 3 November 2007 (UTC)[reply]
Oh, I wasn't confusing anything, I just neglected to mention that I was talking about something else :-D But yes, the doppler effect applies to waves, not fields in general. And the static field about an object has no wavelike characteristics. Someguy1221 23:34, 3 November 2007 (UTC)[reply]
Keep in mind that mass undergoes Lorentz contraction, so as one approaches or recedes from an object one sees a heavier mass. The most definitive answer would be an actual solution to the Einstein Field Equations for a central field at constant velocity, for which all I can say is that there will be off-diagonal terms in the stress-energy tensor, which, given a Schwarzschild metric, means interesting curvature. I say interesting because I have no clue how to interpret the results, other than saying that the off-diagonal terms are not present in the stationary case. Some help would be appreciated. SamuelRiv 05:48, 4 November 2007 (UTC)[reply]

Thanks for all the help so far. As for BenRG's comment about my notion of "appearing heavier", a simple explanation would be this: Consider two heavy objects in far space accelerating towards each other purely due to each other's gravity. Both are emitting a single spectral line of red light. The two can measure speed relative to each other by observing light Doppler shift in the color of the observed light from the other (which should shift towards green or blue). What I mean by "appearing heavier" is that the infinitesimal acceleration of object A towards object B due to gravity would (in my hypothetical model) be slightly more than just the combined value of (1) The gravitational effect using the 1/r^2 law and (2) compensating for the fact that B has a heavier apparent mass due to relativity. In both (1) and (2) cases the distance and speed of a short while back are used due to the speed of light delay combined with a nonzero distance between A and B. This "appearing heavier" effect would then be notable in the spectral profile of the observed color. I realize now that this "effect" that I want to give a name might be something trivial that is automatically taken into account with most such calculations when calculus is used. 41.245.66.240 12:26, 4 November 2007 (UTC) Eon Zuurmond[reply]

I would also like to add the following: words like "Doppler", "ripple" and "wave" are apparently bad choices because they all imply I'm assuming something with a frequency or wavelength. I just use these for analogy. 41.244.192.80 20:19, 4 November 2007 (UTC) Eon Zuurmond[reply]

Radiative Forcing

Why is the amount of radiative forcing greater on a south-facing slope than on a north-facing slope??

68.108.248.118 22:48, 3 November 2007 (UTC)[reply]

This sounds like it could be a homework question, so I'm not going to give the full answer, but here are a couple hints. A) The reverse is true in the Southern hemisphere, and B) think about the position of the sun. Dragons flight 23:02, 3 November 2007 (UTC)[reply]
And it's completely backwards (says me, south of the equator). --Psud 20:28, 5 November 2007 (UTC)[reply]

November 4

Cognitive and socio-cognitive developmental psychology

What's the difference between the socio-cognitive and cognitive theories? Is there any difference? I need to define them for an assignment due soon but I can't distinguish between the two. Thanks in advance. —Preceding unsigned comment added by Psycho marshmallow (talkcontribs) 01:46, 4 November 2007 (UTC)[reply]

Direct homework help is a bit verboten, but let me give you a hint: think about the different names. You know what cognitive theory is, presumably. What happens to cognition in extreme cases of social interaction, i.e., if I locked you in a cage since you were born and never talked to you? Can a cognitive theory really stand alone? SamuelRiv 04:12, 4 November 2007 (UTC)[reply]

Would the number of units of DNA during the G2 phase be equal to the number of units of DNA in one of the daughter cells? For example, if 100 units of DNA were counted during the G2 phase, would there be 100 units of DNA in one of the daughter cells? —Preceding unsigned comment added by 199.247.235.10 (talk) 05:07, 4 November 2007 (UTC)[reply]

This sounds like a homework question, which we prefer not to answer (we even have a guideline that you shouldn't ask it!). Recall that the purpose of DNA replication in mitosis is to ensure that daughter cells each wind up with the same number of chromosomes as the parent cell, so you just have to remember when that replication occurs, and what form the DNA takes in each phase. Read mitosis if you need to brush up, and come back if something is confusing you. Someguy1221 05:20, 4 November 2007 (UTC)[reply]

Europa

Though there are historical hallmarks of running water on Mars and seas of liquid methane on Titan; is Europa, with its ocean of salty water underneath a frozen exterior, the best bet for proving the existence of extraterrestial life (past or present)? —Preceding unsigned comment added by Sappysap (talkcontribs) 05:49, 4 November 2007 (UTC)[reply]

Please do not count those onboard the International Space Station as extraterrestial life. If the world had to invest in a single mission to find life on a planet, moon, dwarf planet, comet, asteroid or planetary ring in our solar system, what would be the best bet? —Preceding unsigned comment added by Sappysap (talkcontribs) 06:43, 4 November 2007 (UTC)[reply]
Gliese 581 c? --Ouro (blah blah) 08:12, 4 November 2007 (UTC)[reply]
Well, apart from Europa probably. --Ouro (blah blah) 08:18, 4 November 2007 (UTC)[reply]
Europa is a possibility for life. The problem with it is that it's rather far from the Sun and doesn't get much energy from it. The oceans (which many experts dispute even exist) would be exceedingly deep under the ice - so not much light would filter down there, and there is unlikely to be much dissolved oxygen in the water as there is here on earth. Such energy as life would get would have to be from hypothetical underwater volcanos caused by tidal forces of Jupiter continually kneading the moon's interior - and since those would appear and disappear sporadically, it's not clear that life would have time to evolve - and then to survive in open water long enough to find the next volcano when the one they were living next to fizzled out. Having said that, the odds of finding life on Europa are probably better than anywhere else beyond Earth. However, there are other possibilities: Enceladus and Titan both orbit Saturn and have similar prospects for life to have evolved. Of those, Titan actually looks rather promising because it's atmosphere is dense and stuffed full of hydrocarbons. Not enough is known about Enceladus - except that it has water ice deposits. Gliese 581 c is a long-shot. We just don't know much about it. SteveBaker 16:23, 4 November 2007 (UTC)[reply]
Oh, I forgot Titan! As for Gliese 581 c, well, if we'd have the resources to develop and undertake a trip there, then we'd probably also have the resources to seek out signs of life anywhere else. --Ouro (blah blah) 16:31, 4 November 2007 (UTC)[reply]

Pyramidal neurons and the extrapyramidal system

A digram showing the position of bilaterally symmetric "pyramids" (corticospinal axon tracts) along the side of the Medulla oblongata.

According to this source the neurons contributing to the pyramidal tract are themselves pyramidal neurons, but most pyramidal neurons send axons elsewhere.
Does that mean that also motor tracts from the extrapyramidal systems can contain pyramidal neurons??? Lova Falk 09:46, 4 November 2007 (UTC)[reply]

I believe most neurons that innervate a tract are pyramidal cells, for both the pyramidal and extrapyramidal systems, as they both consist of single-axon bundles and operate with essentially the same long-distance signalling function. I'm not certain, though. SamuelRiv 13:54, 4 November 2007 (UTC)[reply]
The pyramidal system got its name from the shape of the corticospinal axon tracts (the axons go through the "pyramids" of the medulla oblongata). Some textbooks do not use the designation "extrapyramidal systems" because functionally the "pyramidal system" actually sends axons to the brain nuclei of the "extrapyramidal" systems, making those systems not really "independent of" the pyramidal system. If you define the "extrapyramidal systems" as constituting the neurons of brain stem motor nuclei and their axons, then that would exclude the cortical pyramidal neurons that project to the brain stem motor nuclei. --JWSchmidt 18:55, 4 November 2007 (UTC)[reply]

phone calls

I keep getting calls asking me to press "6"? Are these the calls that seek authorization to place the charge for a long distance call on your number, and if so, why doesn't the phone company make it a priority to notify, if not warn, you of such things? Clem 12:06, 4 November 2007 (UTC)[reply]

Do they just call and say 'Please press 6' or do they say anything else? --Ouro (blah blah) 15:47, 4 November 2007 (UTC)[reply]
Also, what country do you live in? The meanings of various number sequences differs by phone system. MrRedact 04:23, 5 November 2007 (UTC)[reply]

reverse-reverse engineering

Is there a thing called reverse-reverse engineering, where say like a company in England that designs belt sanders for Black & Decker makes a design that has refinements and unnecessary features, and uses excessive and precision parts and is prone to easy destruction, if all but finishing grade belts (100+) are used, that is reverse engineered in China for Chicago Tools, that then is redesigned with no features (variable speed), but a higher RPM that can handle the lowest number belt grit with ease, such that the company in England can then reverse-reverse engineer Chicago's reverse engineered belt sander to come up with a solid and tuff belt sander of its own, with no unnecessary anything, but that is superior to Chicago's belt sander in precision? Or is this just a pipe dream? Dichotomous 12:19, 4 November 2007 (UTC)[reply]

Please, next time, use more punctuation. This is basically just an example of reverse engineering something that had already been designed via reverse-engineering. If I understood your mammoth of a question correctly, that is. Cheers, Ouro (blah blah) 12:39, 4 November 2007 (UTC)[reply]
You have my permission to punctuate the question to your liking. Dichotomous 13:13, 4 November 2007 (UTC)[reply]
Why did you do that? --Ouro (blah blah) 13:21, 4 November 2007 (UTC)[reply]
...was at a neighbor's workstation. Dichotomous 13:28, 4 November 2007 (UTC)[reply]
Oh, okay. I was just curious. And no, I've no need to punctuate your question. Cheers, Ouro (blah blah) 13:31, 4 November 2007 (UTC)[reply]
Please do so anyway for the sake of future questions which might appear on your screen. Sometimes punctuation, like spelling is a very personal or intended thing. Its why we do not utilize an auto-speller bot. I can, however, learn a lot more about you from whatever changes you make. Dichotomous 13:35, 4 November 2007 (UTC)[reply]
There. However, if I'd have been the OP, the question'd probably be split into four or so sentences in the least. Bear in mind, I'm not a native speaker of English. Cheers, Ouro (blah blah) 13:51, 4 November 2007 (UTC)[reply]
Please feel free to break the question into as many parts as you feel necessary, even if doing so requires re-phrasing the parts of the question. Dichotomous 14:59, 4 November 2007 (UTC)[reply]
Also, what is your native language? If you are not familiar with English and possibly more familiar with Spanish, French, German, Italian, Portuguese, Polish, or Korean, et. el. You may be happier responding to questions there. Dichotomous 15:21, 4 November 2007 (UTC)[reply]
I'm a native of Poland, but am experienced enough with English and German to use them in any and all situations; furthermore I know some Russian (had it for two years' at University), Czech (having practically lived next door to some Czechs for half a year), and I know Slovak, Croatian, French and Spanish at hitch-hiker level (meaning I know the basic fifty or so words necessary to feel safe when on the road). As for reformatting your question, I'll do it later... I think. Thanks. --Ouro (blah blah) 15:46, 4 November 2007 (UTC)[reply]
The term "reverse engineering" is usually reserved for finding out the parts of a design that are hidden - for example the encryption that prevents a US DVD player from playing disks from Japan. Taking design ideas from your competitor (presuming they aren't patented) is simply copying their technology - and I'm sure it's a back-and-forth thing such as you describe. Reverse engineering an already reverse engineered product is not a likely thing since whoever first produced it already knows how it works. Copying the changes someone made when they copied your design is a quite different matter. At any rate, the term "reverse-reverse-engineered" is not one that I've ever heard anyone use - and the only two Google hits I got were both using it in a humerous context. SteveBaker 16:03, 4 November 2007 (UTC)[reply]
So then what might one call making changes to a device that would first accommodate low cost high production followed by improvements to maximize quality. For instance, in the example of a belt sander above, the use of stainless instead of galvanized steel for the bottom slide plate and other parts, metal instead of plastic to house the brushes, and ball bearings instead of sleeve bearings, etc., such that the two step cycle of item change results in 1., maximizing simplicity of design and 2., maximizing quality of materials, versus starting with a mediocre or low quality materials item with an unnecessary and extremely complex design? Dichotomous 17:32, 4 November 2007 (UTC)[reply]
"Improvements". SteveBaker 18:31, 4 November 2007 (UTC)[reply]

QUESTION: Can someone help me locate "one stop" lists of animal species on-line?

Hi,

I am looking for online information similar to that kisted in WikiSpecies, but held together in tables or continuous text format that can be copied into an EXCEL table. I am interested in the species of vertebrates, especially mammals, birds and reptiles. My aim is to be able to create (by as few steps as possible of "copy", "sort" and "delete") a single-table, for example, listing all the lizards and geckos in the world by family, genus, species and sub-species, based on the data in these tables. If such a source of data on species is available, please give me the on-line address.

THANK YOU!!!!


Ron Berger (e-mail address removed to prevent spamming, your welcome) —Preceding unsigned comment added by Bergeronz (talkcontribs) 12:20, 4 November 2007 (UTC)[reply]

I just deleted your e-mail address so it wouldn't be found by webcrawlers. I don't know where to find, however, the information you need, and I don't think it will be that easy - you will, after all, have to work a little more than just do simple copying and pasting. Lists of lizards and geckos are in Wikipedia, anyway. Cheers, Ouro (blah blah) 12:36, 4 November 2007 (UTC)[reply]
A google search for "taxonomy database" returns some interesting results like [19] and [20]. Since lizards and geckos are easily found under respective families, this should make your database project simple. Formatting into Excel may be a little tricky, given the format of the data at ITIS. In such cases, I recommend writing a computer program (Perl ftw!) or doing some tabbing by hand on a text file before copying into Excel. SamuelRiv 13:40, 4 November 2007 (UTC)[reply]

Lawns versus parks

It does not seem that lawns offer any advantage over parks, in terms of a front "lawn." For instance, lawns require mowing, fertilizer, insecticides, etc. while a park uses the materials which die for fertilizer and offers far greater possibly as to variety of species. Park paths can still accommodate mailmen and meter readers who must transverse the yard as well as routes for underground plumbing. Is there any good reason not to turn your lawn into a park besides using it to play batmitten? Dichotomous 15:52, 4 November 2007 (UTC)[reply]

This boils down to an argument between public vs. private property. While it seems like a great thing to make everything public, it isn't difficult to find an overwhelming set of examples where private property is better than public property. I live directly across from a park. It is covered in graffiti and full of dog poop. My front yard has no graffiti and no dog poop. So, why should I make my private yard public? -- kainaw 17:56, 4 November 2007 (UTC)[reply]
Could you explain what you mean by "park"? You seem to offer a "park" as an alternative to a lawn of nicely cut grass. Are you asking about paving over your lawn to form a giant parking lot? Or are you asking about letting your lawn overgrow with grasses and weeds? APL 22:10, 4 November 2007 (UTC)[reply]

megapixel claim

File:Web cam fraud.jpg

I purchased a web cam that claimed to have 6 megapixel resolution. The software that determines the number of bytes per frame indicates YUY2 format at 640 x 480 screen resolution at 614,400 bytes per frame. This sounds more like a 614,400 pixel web cam rather than a 6,000,000 pixel web cam. Have I been ripped off? Clem 16:59, 4 November 2007 (UTC)[reply]

Probably. Can you give us the manufacturer and maybe the model? That'd help. Cheers, Ouro (blah blah) 17:24, 4 November 2007 (UTC)[reply]
You should do the simplest math yourself: 640*480 = 307200. 614400 bytes per frame means that it has got 2 bytes per pixel. Icek 17:34, 4 November 2007 (UTC)[reply]
There is no model number or other means of identification, including even manufacturers name. "USB Video Device" is all that is indicated by Windows XP. The focus can be adjusted. The claim is 10 X Zoom. It says f=3.85 and Megapixel. The actual picture it displays has an extremely irritating color speckle that is highly visible. A sample picture shortly. Clem 17:47, 4 November 2007 (UTC)[reply]
I see you've added a sample. Even if the actual maximum resolution was 1280×1024 (which is in the range for current webcams) as you say on the image description page, that's still only 1.3 megapixels, not 6. Besides, if the image is that noisy even at 640×480, I shudder to think what it'll look like at the full resolution. —Ilmari Karonen (talk) 18:27, 4 November 2007 (UTC)[reply]
Besides, that image looks like it's actually been scaled up from something like 320×240! You can easily see the 2×2 pixel blocks! —Ilmari Karonen (talk) 18:32, 4 November 2007 (UTC)[reply]
6 megapixels for a webcam sounds unlikely to me; that's a typical resolution for compact digital still cameras these days. The resolution of webcams is limited not just by their typically low cost and the higher demands put on the sensor by video recording as opposed to still photography, but also by the fact that, as the name suggests, their output is typically intended for streaming over relatively low-bandwidth links. —Ilmari Karonen (talk) 18:22, 4 November 2007 (UTC)[reply]
It certainly sounds like you've been ripped off. 640x480 is a third of a million pixels - my guess is that they are 'stretching' by claiming something like "6 million pixels per second" (which would mean you're getting a disappointing 18Hz frame rate). Two bytes per pixel suggests you're only getting 65536 colours per pixel too - which is also distinctly disappointing. But in truth, a movie camera that could produce 6Mpixels and a minimal 20Hz frame rate with full colour would be 360 Mbytes/sec - which is more than a 1GHz Ethernet could transmit - and VASTLY more data than your home network could possibly transfer anyway. So a 6 megapixel web camera is really an impossibility. 640x480 is a pretty reasonable resolution for a WebCam. The noise you are seeing is indeed unacceptable - but what you have there is a crap camera. SteveBaker 18:29, 4 November 2007 (UTC)[reply]
You can find these on eBay from smartarea and ecool. Ecool is new but smartarea is a power seller. Both request buyers contact them first if any problem rather than leaving negative feedback first. Apparently their strategy works. Dichotomous 19:37, 4 November 2007 (UTC)[reply]

weird clock

Hi. There was a clock which had batteries that almost ran out, a digital clock. Sometimes, when I put it in the correct place (there was a turned-off computer, a prniter, and a metal spoon nearby), and put my finger-hand near, but not touching, the digital clock, its screen would flash with dim numbers and sometimes make clicking sounds. This rarely happened when my hand wasn't near the clock, so much so that I believe the chances that this is a coincidence is extremely small. Sometimes, if I picked up the clock, and put it near the reigon where this happened, it would flash again. I mean "flash" as in the black numbers flash on and off. Sometimes the effect is so weak that only the background of the numbers would appear. Now, when I put this clock between two computers, the effect doesn't show up, and when I put it near an "on" computer, it also flashes. When I took the spoon away, it still flashed, but has since stopped clicking. What could cause this? Is it because humans and electronic devices emit EMF's? Thanks. ~AH1(TCU) 17:11, 4 November 2007 (UTC)[reply]

I'm betting on coincidence. Humans don't "emit EMF's". SteveBaker 18:21, 4 November 2007 (UTC)[reply]
Humans don't emit EMF themselves, but we do affect the electromagnetic fields around us due to being large conductive objects. You can easily observe this by tuning an FM radio to a station that it can barely receive; often moving your body around will audibly affect the reception quality. (This is also how a Theremin works.) From AstroHurricane001's description, it does sound like the clock was picking up some local EM interference that was inducing just enough current to cause the observed effect (it doesn't necessarily take much). —Ilmari Karonen (talk) 18:46, 4 November 2007 (UTC)[reply]
I'll say 95% that this was no coincidence. I tested this for aboout 15 minutes, and putting my hand over it produced an effect whereas not putting my hand there did not. Usually, I leave my hand, nothing. I put my hand there, it happens. I withdraw my hand, nothing. However, it does seem to throb lightly when I put it really close to a computer without brushing my hand over it, but putting my hand over it will produce a strong effect. Could it be that, as the computer emits an EMF, and I put my hand over it, it conducts the current to the near-dead battery? Amazingly, it has come back to life, but is still many hours off. Could it be that the battery terminals are so on the line between connected and not connected, a little interference can turn it on or off? Thanks. ~AH1(TCU) 02:37, 5 November 2007 (UTC)[reply]

Liquid crystal displays can be activated by stray electrostatic fields such as might be present on your fingers, but this would only occur at very close distances. They can also be affected by pressure, but that would require contact.

Atlant 17:13, 5 November 2007 (UTC)[reply]

Charged Black Holes

How does an electrically charged black hole 'show' its charge? I was under the impression that information about an object's electric charge was carried by an electromagnetic field. If the gravity of a black hole is strong enough to prevent electromagnetic radiation (light) from escaping the event horizon, then how does the electric charge 'leave' the event horizon? Smithg86 19:17, 4 November 2007 (UTC)[reply]

The electric field lines do not leave the balck hole, instead they are there before the formation of the blackhole, and are not destroyed by the formation. You get it when an electrically charged mass collapses. Graeme Bartlett 20:46, 4 November 2007 (UTC)[reply]

Question about black headed gulls

Why do they sometimes seem more interested in fighting with other black headed gulls than eating from a pile of food? Someone had thrown some bread out for them today and I was watching them. The gulls didn't even seem that bothered about the food. They were just sitting on the surrounding rooftops looking at it.

Another black headed gull flew in from somewhere and decided to come down and eat. As soon as that happened, a load of gulls swooped down from the roof to attack the eating gull and chase it away, then they started fighting among themselves, leaving the bread alone (it's still there now). What was the point of all that? I don't get it. —Preceding unsigned comment added by 81.77.30.145 (talk) 20:52, 4 November 2007 (UTC)[reply]

Well, if the bread's just sitting there, I suppose whoever eventually wins the fight can then come back and eat it. But I suspect there is more to it than that. The response you describe sounds like an escalated retaliation to punish some sort of perceived misbehavior or social transgression. Such retaliation is observed in many species, including humans: if someone attacks our country, we'll fight back even if it hurts us more than the initial attack did. In the case of interpersonal disputes we've institutionalized this in the form of law enforcement, but the principle is still the same (and where law enforcement breaks down, we still get things like gang wars and vendettas). In this case, the obvious possibility that comes to mind is that the gull that flew in was an outsider while the ones that attacked it considered the area as their territory, but I don't know enough about gull social organization to say if that explanation really fits. —Ilmari Karonen (talk) 02:04, 5 November 2007 (UTC)[reply]

November 5

classic (cliché) explosive detonator?

The popular image of explosive detonation (as seen in innumerable cartoons and Perils-of-Pauline-style movies) involves a box with a T-handled plunger which is forced down to initiate the explosion. (Presumably it was a specialized generator that generated an electric pulse, and fell out of use as compact, reliable batteries became available.) Anybody know what that box was called, or where our article on it is? Neither Plunger nor Detonator yield any clues. —Steve Summit (talk) 01:18, 5 November 2007 (UTC)[reply]

Answering my own question, a google search suggests it's a "Plunger detonator", indicating that there's at least one thing in this world that Wikipedia doesn't (yet) have an article on, after all... —Steve Summit (talk) 01:25, 5 November 2007 (UTC)[reply]
I think at least some of those generators were magnetos, but I have no citation and could be wrong.
Atlant 17:15, 5 November 2007 (UTC)[reply]
I've heard that too, and it's also in the Blasting cap article (well, at least being a dynamo of some sort). DMacks 18:18, 5 November 2007 (UTC)[reply]

Magnetopneumodynamic Drive?

A Mythbusters episode featured some myths on "Free Energy". Among those debunked was a device called "The Lifter" which seemed to be creating an airflow by electrically charging air and thereby lifting itself. Apart from the debunked myth, could this be the basis of a propulsion system, say for LTA craft? I can't find this aspect covered in the articles on Magnetodynamic drives and I wondered what kind of electric power output would be needed to produce useful thrust. Could any power plant create enough push by this means to move its own mass at a useful speed? 203.21.40.253 03:15, 5 November 2007 (UTC)[reply]

It's going to have to be about as powerful as a helicopter. It would be hard to charge this amount of air. The demonstation version of this are very light and have the power source unsupported. Once you build something that can support itslef in thin air, it will be easy to move it by tilting to one side. Graeme Bartlett 03:40, 5 November 2007 (UTC)[reply]

archeology question about Magan, Sumer's source of copper and diorite

If there are extant copper items from before 2300bc at Ur, could chemistry or nuclear chemistry determine where the ore was mined, provided no mixing of different ores took place in the item?Rich 06:17, 13 November 2006 (UT130.86.14.87 03:35, 5 November 2007 (UTC)Rich 03:37, 5 November 2007 (UTC)[reply]

You may be able to check for trace elements to get an idea of the ore used. Graeme Bartlett 03:52, 5 November 2007 (UTC)[reply]
It depends on if the copper has been processed. If it was still in ore, elemental analysis should give pretty strong evidence. If it has been processed, some sort of isotope analysis might be your only hope, although as Graem says, analysis of trace elements within the copper might help. But processing might change the element ratios. Although you could always process ore from multiple sources the way you think it was processed in the past and then compare for any differences in isotopes and trace elemental composition (experimental archaeology). That actually sounds like fun to me :) --Bennybp 05:59, 5 November 2007 (UTC)[reply]

deoxidizing

how do u deoxidize metals? —Preceding unsigned comment added by 216.103.183.127 (talk) 03:38, 5 November 2007 (UTC)[reply]

You could use the method of heating the oxide with charcoal, or coal as in steel manufacture. You could use electrolytic methods, as in zinc or aluminium. On the small scale you may reduce by using a powder of another more electronegative metal, such as aluminium. Graeme Bartlett 03:44, 5 November 2007 (UTC)[reply]
Of course, you might want to be careful abound that last one, lest you end up with a thermite reaction. shoy (words words) 14:27, 5 November 2007 (UTC)[reply]
By the way, "deoxidizing" is usually referred to as reduction. -- 21:08, 5 November 2007 (UTC)

How do you control snails in an aquarium ?

Does anyone have suggestions for control of unwanted snail populations in a fresh water aquarium other than chemicals harmful to tropical fish or use of clown loaches, etc. Sugessted to me is the use of aquarium salt which is supposted to be healthy for fresh water fish, but will it control snails in the recommeded concentration suggested for promoting health of fish? This at first seems possibly helpful since we know salt will dry up land snails.74.170.217.65 03:42, 5 November 2007 (UTC)[reply]

I would not recommend putting any salt in a fresh water aquarium. Just fish out the unwanted snails by hand or in a little strainer, they can't get away very fast. Graeme Bartlett 03:46, 5 November 2007 (UTC)[reply]
Is it impractical to mechanically sort out and remove the mature snails (several times)? Also, why remove them? We allow a reasonable population of snails to grow in our aquarium in order to keep the glass and plants clean, and the action of the fish eating the snail egg masses seems to keep the population of snails in check.
Atlant 17:19, 5 November 2007 (UTC)[reply]
Google gives plenty of useful-looking hits for aquarium snail control. —Ilmari Karonen (talk) 19:03, 5 November 2007 (UTC)[reply]

question about erotic asphyxiation-a mirror effect?

Since phylogeny can indicate guesses about ontogeny, instead of a bilateral symmetry guess, I want to make a vertical symmetry guess-- Since things in the throat such as the adam's apple get larger at the same time private parts are increasing in size (during puberty), is it possible some of the same or closely related hormones are involved? If so, couldn't whatever structures and types of nerve endings that are responsible for feelings of sexual pleasure exist in the neck region as well as the genital region?Rich 11:02, 25 October 2006 (UTC)--This would not be so different to what can happen to carpal tunnel sufferers (bilateral symmetry here):A person gets carpal tunnel in her right hand from scooping ice cream and decorating cakes at an ice cream parlor, and similar pain and weakness shows up in her left hand, though she hasn't stressed that hand.Rich 03:56, 5 November 2007 (UTC)[reply]

Dilmun and Bahrain,Sumer's apparent Garden of Eden-like myth and paradise for Utnapishtim

If it were a paradise for Utnapishtim, I surmise it might have suffered eco damage or at any rate ecological change since. Is there archeological or fossil etc. evidence that the island was previously more lush?Thanks,Rich 20:25, 6 December 2006 (UTC)Rich 04:07, 5 November 2007 (UTC)[reply]

decomposing

what happens when a mn oxide decomposes in water? —Preceding unsigned comment added by 216.103.183.127 (talk) 04:31, 5 November 2007 (UTC)[reply]

Mn oxides tend to go to MnO2 when exposed to water and air. Mn2O7 is a strongly oxidising substance. Graeme Bartlett 04:55, 5 November 2007 (UTC)[reply]

bicycle transportation system

I was checking out some proposals for alternative transportation systems and came across this one - BTS's Transglide 2000. [21] Basically, it's an elevated, enclosed tunnel for cyclists, that uses fans and ducts to move air in the direction of travel at speeds about 30 km/h (although it is not actually specified). The point being that this huge reduction in wind resistance will allow riders to travel quite quickly while expending little energy, due to 90% of total resistance being attributable to wind resistance while travelling at 33 km/h. It sounds very interesting but it seems the main issue would be the huge amount of friction between propelled air and the tunnel walls, and the relatively large size of the tunnel and therefore large volume of air, requiring a prohibitively large use of energy (although as a proposal for a mass transportation system this energy usage should correctly be compared to vehicle or rail travel, not to standard bicycle travel). So my question is, can anyone with a good knowledge of mathematics/physics figure out exactly how much friction would play a part in this case? 220.34.254.226 05:22, 5 November 2007 (UTC)The dimensions of the tunnel are about 3.6 m by 4 m.[reply]

If you have gone all the trouble of building enclosed tunnels around places you might as well build a personal rapid transit system. It would not have any of the problems of the fans, and the rolling resistance is considerably lower, thus uses less energy. The only problem is that it's still "young technology" and the cost of building all the tracks around cities. --antilivedT | C | G 07:03, 5 November 2007 (UTC)[reply]
What's wrong with the idea of climbing into a capsule, pulling a lever and letting air tube conveyor system whisk you right to your desk at the office while you finish putting on your lipstick or brushing your teeth? Dichotomous 19:54, 5 November 2007 (UTC)[reply]

Beyond the universe theories.

What are the latest theories as to where outer space, beyond the known universe, leads to, and if it leads to infinity, what is infinity? Can the answers ever be obtained through science and our human senses, or do the answers lie in some other, i.e. spiritual or such, dimensional sense which we currently can't perceive? And are there any recent theories that maybe matter (universe) maybe doesn't really exist, and that reality is actually a concept not created by our human brains, but rather, again, a concept unfathomable by humans who in turn may not actually exist as we perceive ourselves by our "restricted" senses and resources? —Preceding unsigned comment added by 68.10.91.102 (talk) 06:25, 5 November 2007 (UTC)[reply]

Many of your questions seem to touch on topics that fall under the branch of philosophy called epistemology that aims to answer questions such as "What is knowledge?", "How is knowledge acquired?", and "What do people know?". I think that article and the empiricism article would be good starting points for your further reading in this area. The topic of infinity and whether or not it exists in this universe is covered in physical infinity and Shape of the Universe. Sancho 06:59, 5 November 2007 (UTC)[reply]
There's also a good course available online for free through MIT OpenCourseWare on the theory of knowledge: [22] Sancho 07:03, 5 November 2007 (UTC)[reply]
Science can't answer those questions, because it can only study what can be observed and measured; but it can provide some ideas. Here are some that you might enjoy reading about: Multiverse, Many-worlds interpretation, String theory landscape, Chaotic inflation. These are all ways that our observable universe might be embedded in something larger. I'm not sure what you mean by the question about the universe not existing; perhaps something like The Matrix? Religion and philosophy can also provide answers to these questions. My religion teaches that reality is indeed real, and that the universe was created by a God who exists outside of it and independently of it. God could be described as infinite in some ways. In cosmology, the observable universe is finite in size and age, and has a limited life span. There have been proposals for how to cram an infinite amount of experience into a finite universe (see Omega point (Tipler)), but those have not been widely accepted. --Reuben 07:20, 5 November 2007 (UTC)[reply]


There is little if any physical evidence for there being anything "outside" of the universe, so questions about what if anything may lie outside of the universe aren’t solidly within the realm of mainstream science. Science generally only seeks to explain observable phenomena, and the mainstream views of scientists tend to be the simplest possible explanations of those phenomena. The guiding principle here is Occam's razor, which states that that the simplest possible explanation is the one that’s the most likely to be correct. A model of reality which includes there being something outside of the universe is more complicated than a model of reality that doesn’t, so an explanation that involves there being something outside of the universe is an explanation to be avoided if possible. If at some time in the future, some phenomena are observed which are very difficult to explain unless there’s something outside of the universe, then questions about details of the nature of what exists outside the universe could become a solid branch of mainstream science. There is no way of knowing for sure whether that will happen or not, but the simplest guess (and therefore the guess that according to Occam’s razor is most likely to be correct) is to assume that it won’t.
As to whether knowledge about whatever may lie outside of the universe can be achieved spiritually, Occam’s razor says no. There is no good experimental evidence which is simpler to explain by assuming that non-physical "spirits" exist, so the simplest explanation is to assume that spirits don’t exist.
As to the possibility that the universe doesn’t really exist, believing in such a possibility isn’t a very scientific outlook. It’s simplest to assume that the universe really does exist, so Occam’s razor says that we should just reject any notion that the universe might not really exist, unless and until there are some observable phenomena which are very difficult to explain without assuming that the universe doesn’t really exist. MrRedact 09:06, 5 November 2007 (UTC)[reply]

May matter (universe) possibly not exist as we perceive it? Matter, dissected to its smallest part, is really energy (matter-atoms-strings ?), which cannot be seen by our human senses. Do humans only "see" this energy in the form of matter because our senses are restricted? Would a more advanced life form see physics more accurately than we, possibly "see" "matter" and universe as energy waves or something more profound? Would a more advanced perceiving life form see the actual meaning of "reality" in a much clearer (spiritual even) light, if, as is almost certain, there are aspects to physics that humans have not yet perceived?

Possibly, with consideration of extra dimensions and exotic matter, there could be life that perceives an entirely different (but not unknowable) universe than us. And we don't yet know the mechanism of choice in quantum probabilities, so there might be a world or dimension outside our own that shows a more deterministic scenario. You can speculate forever, but just remember that in this universe, our perception is limited by the nature of matter and energy--to observe, you must disturb--therefore there are fundamental limits to what anything can "know". SamuelRiv 15:24, 5 November 2007 (UTC)[reply]
I can't quite agree with "matter is really energy," since energy is one property that matter can have, but it doesn't tell the whole story. However, it is certainly true that we don't have an established, fundamental theory of everything. But I don't think that means that the reality we see is any less real; that would be a sort of hyper-reductionism that needlessly throws away the complex, aggregate nature of things on the large scale. Ten thousand years ago people could look at a flock of seagulls, without knowing anything about cells, organelles, molecules, atoms, baryons, electrons, quarks, strings, or quantum gravity. Today we have learned a lot more about those things, but the flock of seagulls is no less real for it. --Reuben 19:26, 5 November 2007 (UTC)[reply]

Could fat cells be originally, very long ago, cancer cells that made a deal?

After all, we now know that chloryphyll made a deal. Many parasites evolve to be less damaging, sometimes even symbiotic. Why not cancer too? The way fat cells are is reminiscent to me of cancer cells.Rich 06:42, 5 November 2007 (UTC)[reply]

I believe you mean chloroplasts. Incidently, mitochondria are also believed to be derived from captured bacteria-like organisms. However, both organelles have their own DNA and are transmitted as organelles from mother cell to daughter cell. Fat cells are like any other cell in the body and develop from normal stem cell precursors. The answer is no. -- Flyguy649 talk 06:51, 5 November 2007 (UTC)[reply]
Out of curiosity, what about fat cells reminds you of cancer? Someguy1221 07:15, 5 November 2007 (UTC)[reply]
Did someone say midichlorians? Lanfear's Bane | t 10:19, 5 November 2007 (UTC)[reply]
There's an extensive six-part documentary on them. -- Flyguy649 talk 15:28, 5 November 2007 (UTC)[reply]
Fat or lipid has been an essential part of life for all the living forms, cells use lipids in their cell membrane to create a waterproof partition. So if you want to look for a time without fat in life, it would have to before the appearance of cells. Graeme Bartlett 20:42, 5 November 2007 (UTC)[reply]

Helicopters in the rain

Moved from the Humanities Desk

What are common effects and problems encoutered by helicopters when flying in the rain and how does it affect the handling? Keria 18:16, 4 November 2007 (UTC)[reply]

This is not really a question in the area of the Humanities (history, politics, literature, religion, philosophy, law, finance, economics, art, music, and society). A better spot to ask may be the Science section of the Desk.  --Lambiam 20:55, 4 November 2007 (UTC)[reply]
The problem with flying a helicopter in the rain isn't so much the rain itself as the associated conditions. Poor visibility and potentially high and unpredictable winds make such operation dangerous, and most users opt to not fly in such circumstances. There's not a real concern that rain significantly diminishes the airworthiness of a helicopter, though. — Lomn 15:08, 5 November 2007 (UTC)[reply]
Jet engines are also designed to handle high levels of moisture and rain. Dichotomous 19:14, 5 November 2007 (UTC)[reply]

Think mechs will ever be used on the battlefield?

Think they can be practical? I think something smaller like powered armor (which is worn rather than piloted) or a smaller type of mech like the ones from Heavy Gear would work well if they moved as well as humans do. I guess the big advantage of bipedal armor over tanks would be their agility and all terrain movement. If made too big, those advantages would be diminished, but if kept small, they could work. 64.236.121.129 14:48, 5 November 2007 (UTC)[reply]

According to the dispersion theory of Trevor N. Dupuy in his attempts at quantifying the history of warfare (this guy is my hero of the social sciences), the answer would probably be no. It seems future wars (the theory only works for a total war scenario) require increased dispersion of manpower until battles are won entirely by nontraditional subversive techniques, with traditional manpower kept only to keep a check. That's just my interpretation of an extension where the theory breaks down (it approaches a singularity in the last 50 years with H-bombs and modern smart bombs). And then, of course, any electronic devices are extremely vulnerable to wear and tear and electronic warfare, which makes the high-tech soldier of the future seem very unlikely in my opinion. SamuelRiv 15:17, 5 November 2007 (UTC)[reply]
Umm, you have to define what you mean by "nontraditional subversive techniques", that doesn't mean much by itself. The "singularity" point is also meaningless. You have to clarify. I'm not sure how PA or small mecha are any more vulnerable to wear and tear than your average tank or vehicle. 64.236.121.129 15:24, 5 November 2007 (UTC)[reply]
I once read a story (I can't remember the name) about a galactic federation (or whatever) that lost a war because it's weapons were too high tech. --MKnight9989 15:32, 5 November 2007 (UTC)[reply]
What does that have to do with anything? Stay on topic please. 64.236.121.129 15:39, 5 November 2007 (UTC)[reply]
It goes along with what Samuel said about electronic devices being prone to failure. It's possible I suppose, but it seems like something that complicated would be difficult to 'soldier proof' --MKnight9989 15:42, 5 November 2007 (UTC)[reply]
Yea, I thought I addressed that already. 64.236.121.129 15:47, 5 November 2007 (UTC)[reply]
MKnight9989, you thinking of Star Wars? Ewoks with sticks. Yub yub. AT-ST problems? Get some tree trunks. Yeh, Yeh gira. Lanfear's Bane | t 15:53, 5 November 2007 (UTC)[reply]
In all fairness, those walkers were crap to begin with. They were too big, and their movement was too static. They used gears and motors for leg movement. That's crap. Yes, they can walk, but they weren't agile, and couldn't move like a human or any other legged animal. Muscles need to be emulated properly in order to take full advantage of legs, and their full range of motion. If they were smaller, and moved like a human or even a horse, they would have been great. The chicken walker leg is also worthless for all terrain movement. 64.236.121.129 16:10, 5 November 2007 (UTC)[reply]
Yeah I think the All-Terrain was a slight exageration. AT-AT's were more stable (not a horse pun) and they didn't fare much better. Damn Snowspeeders. Rebel Scum are nothing if not innovative. Lanfear's Bane | t 16:16, 5 November 2007 (UTC)[reply]
The story you're thinking of is probably Superiority (short story) by Arthur C. Clarke. Clarityfiend 17:56, 5 November 2007 (UTC)[reply]
The question is what kind of battlefield you are imagining. The late-20th century has not been characterized by wars of high-tech powers against high-tech powers; rather it is generally either high-tech powers against low-tech powers or low-tech powers against low-tech powers, if I can be very, very broad in my strokes. The last real war where all of the major participants were at a similar stage of high military technology was maybe the Korean War, maybe World War II, depending on how you categorize things.
In any case, let us think about the advantages and disadvantages of these approaches. Bi-pedal locomotion is indeed able to deal with more terrain, but it is generally much slower than wheeled or tracked locomotion. Personal mechanical armor is probably most effective against small arms fire; large bore fire or heavy explosives of any sort are going to tear limbs off or apply massive concussive force to the interior with great ease.
Now if we imagine a future generation of robotics far more advanced than what we have, where we are basically replacing humans with mechanical 'droids, then I could definitely see that as a political advantage although I am not sure about its military advantage. An army in which no soldiers have to die is one that can be deployed with far less problem of home-field protest, although the cost differential is going to be pretty unpleasant versus training a soldier and deploying a high-tech robot. But cost-differentials, of course, are not taken into much account even today—when we reply to car bombs with cruise missiles, we are spending billions of dollars to reply to a weapon that cost at most a few thousands dollars, at least almost nothing (here I am thinking of the African embassy bombings of the late 1990s which were replied to with about $2 billion worth of cruise missiles).
Anyway, just some things to think about. Is it possible? Maybe. But remember that we're probably not going to see two very high-tech superpowers go head-to-head anytime soon. So who is going to buy these robots, and who is going to deploy them, and where? The US already has high tech weapons it can't really use effectively; at the moment "mechs" strike me as not being terribly desirable for modern warfare. --24.147.86.187 15:53, 5 November 2007 (UTC)[reply]
Aren't the US already using guns mounted on tank-type tracks which are armor plated and remote controled in Irak? 80.200.238.237 16:36, 5 November 2007 (UTC)[reply]
I am not sure, but in the Misc. desk recently we were discussing motion sickness et al and someone mentioned that soldiers remote operating from a trailing vehicle quite often felt ill due to the different sensory inputs [23]. And K and use a Q 80.200 boy. Lanfear's Bane | t 16:45, 5 November 2007 (UTC)[reply]

theoretical yield

How do I figure out the theoretical yield of a substance? --MKnight9989 15:10, 5 November 2007 (UTC)[reply]

Sounds like homework. Here's a hint: think about how much reactant you need, and how much you have, and try to maximize the amount you can get from your reaction equation. In short, multiply both sides. SamuelRiv 15:19, 5 November 2007 (UTC)[reply]
I know how much reactant I have and how much product I end up with (actual yield), but I don't understand what calculations relate the two to theoretical yeild. --MKnight9989 15:37, 5 November 2007 (UTC)[reply]
Theoretical yield is how much you get if everything go perfectly, everything reacts, and reacts the way it's supposed to (Yield (chemistry)). Someguy1221 16:22, 5 November 2007 (UTC)[reply]

Amazing little horned garden spider

What kind of amazing little horned spider is this? Found it. Spiny orb-weaver

In addition to its horns its web has UV reflective tuffs and its web is attached to Earth at three points:

  1. branch of tree; 12 feet from ground,
  2. middle rail of fence; 2 feet sideways and 3.5 feet from ground,
  3. near end of stacked pipe; 10 feet from attachment point 1 and 7 feet from attachment point 2.

Attachments are anchored at each point by 2 to 5 strands. Orb is only 9 inches in diameter.

Second question... How did this little spider reach these three different attachment points? I can see maybe wind to the tree branch and to one of the other attachment points but what about the 3rd? Did the spider crawl all the way back up the tree branch after the first attachment and then go for the second, or what?

Dichotomous 15:51, 5 November 2007 (UTC)[reply]

Which is the best alternative to gasoline for cars, hypothetically

I don't mean right now, but in the future. From what I read of the articles, I think battery-electric cars potentially are the best. They have the best efficiency when you think of it from taking energy from a nuclear powerplant, then shooting that energy into a battery. Stuff like hydrogen, requires making hydrogen which is inefficient.

In terms of using that energy, I read that battery electric cars are equivilant to say 100-200 miles per gallon, if do all the conversions and compare them to ICE cars.

They don't use any explosive liquid like hydrogen or gasoline, although the batteries can explode possibly... Still, I think they are safer.

Batteries can have good performance, 0-60 in 4 seconds, top speeds of over 200 miles per hour. I'm sure these stats can improve in time too. Range right now is a problem I think though. 200-300 miles is the best ranges I heard for some good Bat cars. ICE cars tend to have a range of 400 miles on a full tank of gas. I think with improvements in batteries, Bat cars can be extended to 400 miles though.

Bat cars are quiet.

Bat cars release no emmisions.

Bat cars benefit from regenerative braking.

Maybe the only negatives would be, they are harder to repair (but I think that has more to do with stupid mechanics who are only knowledgable on ICE cars), more expensive (although in the long run, it'll be less expensive, and ICE cars perform better in extreme temperatures like extreme hot or cold.

What do you peeps think of my assessments? 64.236.121.129 16:06, 5 November 2007 (UTC)[reply]

...and just how did you come to these assessments? Dichotomous 16:12, 5 November 2007 (UTC)[reply]
From the articles dude. 64.236.121.129 16:16, 5 November 2007 (UTC)[reply]
dude...? Dichotomous 17:29, 5 November 2007 (UTC)[reply]
"some good Bat cars.". See Batmobile. Lanfear's Bane | t 16:18, 5 November 2007 (UTC)[reply]
peeps... Clem 16:26, 5 November 2007 (UTC)[reply]
Quick thoughts: I don't think the inefficiency of making hydrogen is a big deal if you're supposing widespread clean nuclear power. I find it a more relevant objection if you're burning tons of dirty fuels to manufacture clean fuels. Quiet (to the degree of electrics) is a debateable benefit. The hearing-impaired community is already concerned by Priuses and the like, and I feel that modern cars are quite capable of being generally quiet so long as their owners don't modify them to be noisy. Emissions are no more a concern with hydrogen than electric, and any hybrid system benefits from regenerative braking, too. Finally, your discussion about range omits a key point. A tank-fueled car, regardless of single-tank range, can just refuel in five minutes. An electric car cannot. Consequently, "range" means entirely different things for the two car types. — Lomn 17:01, 5 November 2007 (UTC)[reply]
Why waste energy with inefficiencies? Why bother when there is a more efficient alternative? You're just wasting energy that could be used otherwise. Quiet is just a nice bi-product. I don't think we are going to go out of our way to make noisy cars just for the hearing-impaired, that's just silly. What would be the point anyway? They want loud cars so they can hear them? That's stupid. Yes, modern cars can be quiet, but they also use gasoline, which is the problem isn't it? Remember what we are talking about dude. Yep, hydrogen produces no harmful emissions, but they are worse because of the inefficiencies I mentioned before remember? They also require platinum in their engines, which drives up the price, and makes increases the demand for platinum, thus raising its price. Such a scenario can be avoided with batteries. Yep, any hybrid can benefit from regenerative braking, but so far all of your points merely match the capabilities of a Bat car, rather than exceeding them. That's what we are talking about here, the best of the alternatives.
Recharge times are gradually improving, with some rechargers capable of recharging spent batteries in minutes. But there are also ways around that like recharging over night at home while you are sleeping, or having recharge stations at offices. If all you do with your car is drive to work, and it's a reasonable distance, you can just recharge at home overnight. But yes, recharge time is an issue, but it's an issue that can be improved. 64.236.121.129 18:11, 5 November 2007 (UTC)[reply]
The question of inefficiency is "does the inefficiency matter?" If you're assuming widespread clean nuclear power, then my answer is "no". I think, given that supposition, hydrogen is a perfectly viable fuel and far easier to engineer than superquickcharge batteries. If battery charge times are on par with other refuel times, then sure, it's a great solution -- I'm just not prepared to make that leap. You're dead-on with overnight chargers being adequate for the vast majority of car use, but I find the idea of a car that does what I need 95% of the time but cannot do what I need for the other 5% (e.g. long range driving) to be an unacceptable car. Additionally, there's the problem of battery space and weight with pure-battery cars; hybrids are significantly more efficient in those terms. As for my points matching those of battery cars, yes. That was exactly my point -- to note that a battery-only car does not hold exclusive advantages in those regards. — Lomn 18:23, 5 November 2007 (UTC)[reply]
Two things, we don't have widespread clean nuclear power, so we still have to be realistic about where our energy is coming from. And second, even if we did, it's still a waste of energy, so why bother waste it with inefficiencies when hydrogen cars do not have any benefits over Bat cars? You are making a really big assumption when you say hydrogen is a viable fuel and far easier to engineer than batteries. You gotta back that up. Otherwise that's just a blank, meaningless statement.
Battery space and weight doesn't matter if the performance numbers are still good. The numbers I gave before are for real Bat cars. Hybrids are significantly more efficient in those terms you say? What the hell does that mean? You gotta clarify your points dude. You are making a lot of blank, empty statements that don't really mean anything.
Yep, but I never said bat cars hold exclusive to those advantages did I? So you are arguing a strawman here. What I'm looking for is, what do the other ICE alternatives have over Bat cars, if anything. The recharge issue seems to be the only thing you can think of, but that's already being addressed. 64.236.121.129 18:36, 5 November 2007 (UTC)[reply]
Re: space/weight: have you ever seen how little cargo space remains in an all-electric car? It's preposterous! Hybrids require a far smaller battery pack; thus, they are more efficient in that regard. Aside from that, this looks like another case of you expecting to have your assumptions parroted back to you. A nuclear electricity source (which you originally suggested) is too-forward-thinking, a lack of quick-recharge isn't forward-thinking enough, even though it's already "being addressed" -- whatever that means....
Clearly there's no legitimate request for commentary or criticism here. — Lomn 19:09, 5 November 2007 (UTC)[reply]
Hmm, what kind of cargo do you want to haul exactly? I don't think the average person needs to haul much stuff usually. I'm mostly talking about transporting people, and most people just use their car to get to work. Fair enough, a hybrid might carry more cargo, but that's it really. Aww, you're mad again. Well that's your option ^^. 64.236.121.129 19:34, 5 November 2007 (UTC)[reply]
Since you asked for a hypothetical future solution, I vote for a very small fusion reactor driving a Sterling engine driving an electric generator, in turn driving one electric motor per wheel, with supercapacitor buffering to account for peak accelleration and regenerative braking. No pollution except for heat, and no need to refuel for the life of the car, service interval limited by wear on the tires. -Arch dude 17:45, 5 November 2007 (UTC)[reply]
I'm assuming you are being facetious, but I'll address it anyway. Too expensive, and we also don't have practical, working fusion reactors, let alone one small enough to fit in a car. This is something that might happen in 1000 years, rather than the foreseeable future. 64.236.121.129 18:11, 5 November 2007 (UTC)[reply]
Too expensive? My hypothetical micro-fusion reactors can be had 3 for a dollar at Sam's Imaginary Energy Emporium. On a more serious note, if we did have practically limitless, cheap, clean energy, there are many other technologies and products that would suddenly become very cheap as well. Dragons flight 19:06, 5 November 2007 (UTC)[reply]
Yea, but who's talking about limitless energy? We are talking about alternative power sources for cars. 64.236.121.129 19:34, 5 November 2007 (UTC)[reply]

Stiff neck

Is it possible to insert structures into the neck to prevent strangulation? Has it been done before? 81.242.90.27 17:42, 5 November 2007 (UTC)[reply]

I don't know, but wouldn't it be easier just to wear a suit of powered armor if it's such an issue? 64.236.121.129 18:41, 5 November 2007 (UTC)[reply]
Oh. Then no-one wouold try to strangle you. It's for James Bond. 80.200.239.55 19:35, 5 November 2007 (UTC)[reply]
Slightly off topic, but we did exercises in Kung Fu to strengthen the neck muscles (neck turning I believe it was called). The instructors showed that with enough practice it works against a lot of choking techniques, and you can even take punches/strikes there. And no surgery required! There are some movies of monks putting ropes around their necks and just hanging there. It takes a while though; I never got that far. --Bennybp 21:00, 5 November 2007 (UTC)[reply]

about IR

what is the infrared absorption frequency of phosphorous-oxygen bonds? —Preceding unsigned comment added by Dinkysweet (talkcontribs) 17:45, 5 November 2007 (UTC)[reply]

Depends single vs double, ionic vs covalent ("what's attached to the O if it's single-bonded to P"), etc. DMacks 18:08, 5 November 2007 (UTC)[reply]

is the california fire still burning?

California_wildfires_of_October_2007 could use an update —Preceding unsigned comment added by 85.238.88.73 (talk) 18:03, 5 November 2007 (UTC)[reply]

Down insulation

A friend of mine is an avid & experienced outdoorsperson, and tells me that it's widely known that down jackets or vests should be worn under other layers, because "down's insulating power comes from the temperature difference between heat of your body and the outside cold". In other words, she says that if you wore wool or synthetics under down, you would be colder than vice versa, because the down won't insulate if it's cold. It's possible that I'm not representing her explanation fairly.
I have a basic understanding of thermodynamics, and this doesn't make much sense to me. Her explanation of down's special properties sound like a definition of insulation in general. It seems like the delta-T across the down will determine its insulating power, whether that temperature difference is between skin and an inner shell, or between air and a underlayer. Can anyone explain this apparent discrepancy?
Secondly, according to our article, down insulates more poorly when moist or compressed, wouldn't you want it further away from your perspiring body, and uncompressed by outer layers? jeffjon 18:33, 5 November 2007 (UTC)[reply]

How do I tell white gold or platinum jewelry from stainless steel?

How do I tell white gold or platinum jewelry from stainless steel? If I have some jewelry, don't know where it came from, and think it could be made of either one of these three, how do I find out which one? I mean aside from some really complicated thing at the chemist. William Ortiz 18:41, 5 November 2007 (UTC)[reply]

An easy first step: some types of stainless steel are magnetic, so if a magnet is attracted to it, it definitely isn't gold or platinum. If the magnet isn't attracted, though, you haven't proved anything. jeffjon 18:44, 5 November 2007 (UTC)[reply]

Maybe if you live in an area where there are stereotypically gold-hungry ethnicities, you could dangle it conspicuously and see if they have gold-lust in their eyes. Note: I'm basing this on myself, since I belong to such an ethnicity. My eyes not only twinkle, but get a little teary. —Preceding unsigned comment added by 85.238.88.73 (talk) 19:10, 5 November 2007 (UTC)[reply]

Most gold jewelry has the karat written on it somewhere (ie 14K, 18K, 10K, etc.).Josborne2382 19:17, 5 November 2007 (UTC)[reply]

Like the apocryphal story of Archimedes and the crown, you can determine the density of the material using the item's weight and volume. The weight is easy, if you have an accurate enough scale. The volume can be determined by immersion in water, but it might be tricky if the item is small. -- Coneslayer 20:40, 5 November 2007 (UTC)[reply]

Oil consumption versus oil creation

Where can I find a graph or chart that compares the consumption of crude oil with crude oil creation. I would expect it to be on the order of like a barrel of crude every ten years versus however many million or billion consumed per day. Dichotomous 19:05, 5 November 2007 (UTC)[reply]

Consumption of petroleum (also known as crude oil) is currently about 84 million barrels per day. The current creation of petroleum is almost certainly much faster than a barrel every ten years. A barrel of petroleum created every ten years since the origin of life on this planet about 3.5 billion years ago would only come to about 350 million barrels. But petroleum reserves, not counting oil sands, are currently at about 1.2 trillion barrels. Dividing 1.2 trillion barrels by 3.5 billion years puts a worst-case lower bound of a least a barrel created per day, and probably considerably more.
It'd be very difficult to have a bar chart showing how petroleum creation compares to petroleum consumption. If the bar showing petroleum consumption is a few inches tall, the bar showing petroleum creation would have to be thinner than the thinnest line that a printer can print. MrRedact 20:09, 5 November 2007 (UTC)[reply]


I don't think we honestly know what the rate of creation is. A reasonable guess might be something like:
C = A / Y
Where:
C = The creation rate.
A = The amount of oil that had formed before we started digging it up.
Y = The number of years it took to form.
We know that oil is formed from the simplest algea in the ocean - which have been down there (presumably) forming the stuff for 3 billion years - so Y=3,000,000,000 years. I think it's fair to say that there were once about 3 trillion barrels of economically usable oil (ie not mixed up with sand and oil shales and such) and about twice that amount if we include the (currently) uneconomic stuff so A=3,000,000,000,000 barrels.
That says that it's probably forming at a rate of C = 3 trillion barrels / 3 billion years = 1,000 barrels per year - let's say 3 barrels per day of economically usable oil (perhaps 6 barrels per day that of the other stuff). We're currently using close to 90 million barrels a day. Which means we're using it up about 30 million times faster than it's forming. (Jeez! Even I'm surprised by that number!)
Of course this makes two very dangerous assumptions - one is that every drop that was ever formed was available for us to dig up. That's unlikely because much of it must have been subducted due to continental drift and such. So that would make the estimate for present production rates higher than I just calculated - perhaps if we're really optimistic, 1,000 times faster. On the other hand, I'm also assuming that conditions have always been right for oil formation - and indeed that the conditions are still right. That's also a difficult thing to know and could push my estimate either up or down - but not by very much compared to thirty million!
But one thing is for absolute certain - we're using the stuff up at a rate at least a few thousand times faster than it's being created - and perhaps a millions of times faster. Fast enough that it's irrelevent that more is being formed. 3 barrels a day isn't enough for anything useful!
Even if I'm wildly off the mark. Suppose we were using it up only ten times faster than it's being formed, we'd still run out in roughly the same amount of time! But that simply can't be true. Nobody is going to claim that all this oil we dug up last year is only 100 years or so old (we'd certainly notice new oil formations popping up all the time if it was that fast - and some really simple carbon-dating would tell us that!) So for sure we'll run out reasonably soon if we carry on using it at the present rate. However, as has been said here several times before - if you turn all of the oil we have into CO2, global warming will easily kill the planet. Hence running out is simply not possible - the planet would be dead long before we could manage that. So let's worry about global warming instead please.
SteveBaker 20:32, 5 November 2007 (UTC)[reply]
Generation, migration, and accumulation of oil depend on a huge number of variables. It is entirely possible (though quite unlikely) that no oil is presently being generated, anywhere on earth. The proper source rocks have to be buried to the proper depth for the proper length of time. A good source rock has 1% organic material in it; an outstanding source rock has 5%. The oil window (the depth where oil can be generated) is around 7000 to 15,000 feet below the surface, with variations. Then circumstances must exist for the generated oil to migrate through rocks to a point of accumulation in a trap. All this will typically take millions rather than thousands of years for significant accumulations. The largest fields in Saudi Arabia, for example, are thought to have taken tens of millions (50-80 million, I think – sorry I don’t have the reference at hand) to be generated, to migrate, and to accumulate.
Steve’s calculations are interesting – just one comment: there is virtually no oil present today that is thought to have been generated at a time longer ago than about 1 billion years, and the vast majority was generated less than 400 million years ago. There just was not enough algae to create source rocks. Also, any such old oil that was generated would be very unlikely to have survived to the present, owing to tectonic activity destroying traps. It is difficult to know for certain when oil is generated, but one measure is by looking at the (usually younger) rocks in which the oil is trapped. In the US, at least 75-80% of all the oil is or was in rocks younger than about 320 million years old. The rocks that generated it would not likely be all that much older (with, of course, some exceptions). Cheers Geologyguy 20:52, 5 November 2007 (UTC)[reply]

Photon Energies

In general (i.e. not refering to any particular substance) can photons be created with any possible energy or are there only particular energies they can take (i.e. some sort of planck scale) —Preceding unsigned comment added by 172.200.188.149 (talk) 20:43, 5 November 2007 (UTC)[reply]

Wavelength (lambda) is the only variable there, and I guess it would be quantized in units of plank length. So the inverse of a photon's energy is quantized. Or I'm wrong, I don't know. Someguy1221 20:56, 5 November 2007 (UTC)[reply]