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November 26

Fire Accelerants

Are there any fire accelerants that aren't immediately lethal when consumed in considerable quantities? I know drinking a cup of gasoline will be unpleasant, but is there something that doesn't kill you, unless you then swallow a match or something... 74.230.231.13 (talk) 00:04, 26 November 2007 (UTC)

It depends on your definition of fire accelerant. From the article, "an accelerant is any substance or mixture that "accelerates" the development of fire", I think bottles of pure oxygen will accelerate a fire very quickly, but won't immediately kill you. --antilived^{T | C | G} 02:12, 26 November 2007 (UTC)

Define "considerable" - almost anything will kill you if consumed in high enough quantities. Many spirits are flammable, as demonstrated by party tricks such as flaming sambuca; and I expect something like an overproof rum would serve as an accelerant - according to our article, a mix of water & ethanol with over about 50% ethanol is flammable, so a spirit which is in the 60-70% ABV range should go up easily enough (although it's drinkability is another matter...) -- AJR | Talk 02:14, 26 November 2007 (UTC)

Some substance such as vegetable oil, ghee, or glycerol are actually food items and will also accelerate a flame. Others such as wax may not be food, but are fairly harmless to eat. Graeme Bartlett (talk) 07:56, 26 November 2007 (UTC)

Murder

In response to the answers above, my friend says, "OK, so we are going to get someone to OD on it and then shoot them with a flaming arrow." But I don't think that would work. Would the [alcohol, glycerol, ghee, oil] remain flammable after being introduced into the stomach? HYENASTE 23:19, 27 November 2007 (UTC)

In sufficient quantities to perpetrate arson? I doubt that, but also doubt that the substance would catch on fire, would there be enough oxygen present to allow it? Also, the substance in question, when being reacted on by the stomach acid may not be flammable either. 68.39.174.238 (talk) 03:45, 28 November 2007 (UTC)

Balancing a chemical equation with silver nitrate copper and water

Alright I have this problem. I disolved .0185 moles of silver nitrate into a unkown amount of water to form a solution. I then dissolved 2.81 grams of copper oixide water into the solution until there was 1.25 grams left. I need to write the chemical equation for it if the copper is suppose to be +2 and +1 in charge. Does anyone know how to do that? —Preceding unsigned comment added by 70.249.230.252 (talk) 00:26, 26 November 2007 (UTC)

It sounds like homework or schoolwork help. The question also seems to have a typo ("2.81 grams ... until there was 1.25 grams left"). Just try writing out your reaction equation: the question is asking to use two forms: cupric and cuprous oxide, which are CuO and Cu₂O, respectively. SamuelRiv (talk) 02:03, 26 November 2007 (UTC)

Kinetic energy released by expansion of space

If space is expanding everywhere then the space in which all solid bodies exist must be expanding. As it expands the bodies themselves do not because the atoms that make up the body are attracted to each other and the nuclear forces keep the particles at a constant distance.

But this must mean that the atoms, if effect, move downhill a to keep that constant distance and, as such, there is a conversion of potential energy to kinetic energy.

Has anyone calculated the rate of kinetic energy being imparted to the earth as a result of the expansion of space?

Also, where does this energy come from? Or is this question completely off base?

Doug Moffat

209.5.192.16 (talk) 00:34, 26 November 2007 (UTC)

Energy doesn't come from anything. It cannot be created or destroyed but can only change energy types of change into matter. I'm not sure what you mean when you say there is a conversion of potential energy to kinetic due to space expanding. I'm not sure if that actually occurs, because space expanding has little effect on the bodies occupying that space, it just causes the bodies to move away from each other, yet still retain there current position in space. Space (which is not matter) is expanding and doesn't really effect the matter. —Preceding unsigned comment added by 70.249.230.252 (talk) 00:55, 26 November 2007 (UTC)

No, the point is a fair one. Anon is basically saying that if we take two massive bodies stuck to each other by gravity, inflation will try to pull them apart. If it succeeds, then you suddenly have gravitational potential energy that can be exploited if you stopped inflation for a second. Unfortunately, stopping inflation is the only scenario in which that energy can be exploited, so the end effect is just an effectively lower force between objects, if I'm reading this right. That is part of the reason we need dark matter: we find that certain clusters are not being pulled apart like they should be, so obviously the gravity in the cluster is higher than that due to visible mass. SamuelRiv (talk) 02:37, 26 November 2007 (UTC)

Space is only expanding between galaxies that are far apart and have very little gravitational influence on each other. Where matter exists, space does not expand (I think). Although your point is still valid (I think) because there would still be some gravitation force between them however small. Does the energy come from vacuum energy? Shniken1 (talk) 06:07, 26 November 2007 (UTC)

I don't believe that's correct. From what I understand, space is expanding evenly throughout the universe. In fact, if portions did not expand evenly, then that would either warp space or require that other sections expand faster to make up for the non-expanding portions. Expansion is slow, and gravitation can usually hold objects together despite space expanding. Of course with gravity, the closer together the objects are, the stronger it is. So, far apart objects, like galaxies, are more affected by expansion than the objects in a solar system would be. Essentially, gravity helps prevent the objects from expanding, but not space from expanding. While I don't see that explicitly stated there, you might try looking through metric expansion of space for more information on this topic. -- HiEv 14:39, 26 November 2007 (UTC)

Absent a cosmological constant, the expansion is nothing more or less than objects moving away from each other. There's no outward pull on anything; it's just inertia. Space is only "expanding" in areas where things are still moving apart (i.e. far from galactic superclusters). I suppose you can think of the cosmological constant as adding a ubiquitous outward pull, but all this does, like any other (sufficiently small) source of tension, is perturb the object into a different equilibrium state. So maybe the ground-state energy of hydrogen is slightly (undetectably) different than it would be without a cosmological constant, but that can't be used as a source of energy because there's no lower energy state to push it into. -- BenRG (talk) 15:42, 26 November 2007 (UTC)

I think you're confusing the expansion of the universe with inflation. The universe is still expanding, but inflation, if it happened at all, ended 13.7 billion years ago. -- BenRG (talk) 15:42, 26 November 2007 (UTC)

The kinematics of dark energy can be reasonably well approximated by adding an extra force to the universe such that every object experiences an apparent F_dark = D*M*x, where D is a small constant, M is it's mass, and x is its distance from the observer. In other words, the apparent force is trivial at short range and large at great distances. It also follows that adding a small constant force, doesn't generate additional energy for an object like the Earth which is held together by much larger forces. Dragons flight (talk) 11:30, 26 November 2007 (UTC)

Two Point Function

I was reading this book awhile ago and it talked about something called the two point function. It was caused by two flucuations in a vacumm in space diverging until they became so close that their energy density matrices became infinite. Thus causing for the equation having to be renormalized, and this somehow caused an expansion in space-time. Does anyone know what I'm talking about or does this sound like nonsense? —Preceding unsigned comment added by 70.249.230.252 (talk) 01:01, 26 November 2007 (UTC)

Could it be Zero-point energy, and the related cosmological constant problem it apparently poses? -- Finlay McWalter | Talk 02:16, 26 November 2007 (UTC)

My bet is that it's the vacuum fluctuations of Edward Tyron that describes how the universe may have been created out of nothing. There is an excellent nontechnical overview of this here [1]. It could also be bubble nucleation of a false vacuum, which is another common pre-inflationary scenario. SamuelRiv (talk) 02:26, 26 November 2007 (UTC)

Reading the link you provided SamuelRiv Inflation for beginners paragraph 4 reads:

If the Universe starts out with the parameter less than one, O gets smaller as the Universe ages, while if it starts out bigger than one O gets bigger as the Universe ages. The fact that O is between 0.1 and 1 today means that in the first second of the Big Bang it was precisely 1 to within 1 part in 10⁶⁰). This makes the value of the density parameter in the beginning one of the most precisely determined numbers in all of science, and the natural inference is that the value is, and always has been, exactly 1. One important implication of this is that there must be a large amount of dark matter in the Universe. Another is that the Universe was made flat by inflation.

Now 1^st sentence make sense. Then: how do we observe it to be smaller than 1? If it is anything from 0.1 to 1 (does it mean it's not precisely determined or that it varies localy?) today how do we calculate it would have been precisely close to 1 in the first second of the Big Bang? "the value is, and always has been, exactly 1", hang on didn't they just say it would be anything between 0 and 1? I'll carry on reading. Keria (talk) 10:34, 26 November 2007 (UTC)

Qualitatively, it is like O(t2) is approximately O(t1)^(s(t2)/s(t1)) where O(t) denotes O at time t, and s(t) is the size of the universe at time t. If O is approximately 0.5 now, then when the universe was 1/10 this size, O would have needed to be 0.5^(1/10) = 0.93. To allow for an O roughly near 1 today, it implies that O was very, very near 1 in the distant past when the universe was very small. An appealing solution is to posit that O is exactly 1 at all times. Incidentally, if O is much different than 1, it would imply that the ultimate fate of the universe would already have been realized (either through collapse or run away expansion). Hence O approximately 1 can also be looked at through the anthropic principle since we could not exist in a universe that was otherwise. Dragons flight (talk) 11:06, 26 November 2007 (UTC)

Query related to Urine

Is there any harm (or what are the results), if a girl drinks a man urine mistakenly / willingly. —Preceding unsigned comment added by Ashish.k.garg (talk • contribs) 06:57, 26 November 2007 (UTC)

Amazingly, we have an article on this. Urophagia. Someguy1221 (talk) 07:05, 26 November 2007 (UTC)

If the person in question doesn't have any diseases and is healthy, it should be ok. Urine straight out of the body is sterile. But it can be contaminated, and that's what causes that urine smell. 64.236.121.129 (talk) 15:53, 26 November 2007 (UTC)

I think the urine smell is due to the ammonia in the urine. 128.163.170.161 (talk) 17:39, 26 November 2007 (UTC)

There is no ammonia in urine. If there was, it would be unsafe to drink, which it isn't. Ammonia is converted into urea before it is excreted. 64.236.121.129 (talk) 14:44, 27 November 2007 (UTC)

The whole point of urination is to remove toxic substances from the blood e.g. excess salt, urea, uric acid, creatinine etc., so drinking it cannot be healthy in any large quantity (although it is sterile). And I once heard that cat's urine has ammonia Tomi P (talk) 22:51, 27 November 2007 (UTC)

Other than possibly consuming too much salt, there is nothing dangerous about urine. See Urophagia. Btw, we are talking about human urine, not cat urine. 64.236.121.129 (talk) 16:42, 28 November 2007 (UTC)

Actually as the article mentions (admitedly with a citation needed tag), you probably should take care if the person is taking medications Nil Einne 08:32, 1 December 2007 (UTC)

gide us

Dear sir,

We want to make one controlling project for university and we need some information and help for designig a pc bord or programing a one plc with this specification:

• voltsge source:12 V
• it sould be have 40-60 Switchs
• and this equipmet sould be control with progaram and it's capacity is 2Km.
• please send us yor guids and name of some company that can help us. —Preceding unsigned comment added by 91.184.66.107 (talk) 07:53, 26 November 2007 (UTC)

So to attempt to clarify, you want to remotely control something 2 kilometers away, by operating 40 to 60 switches at the remote position. What do you want to switch at the remote location - do you want relays? Are you willing to run a copper pair or optic fibre between your controlling point and the remote device, or do you need a wireless system? Graeme Bartlett (talk) 08:01, 26 November 2007 (UTC)

GREATMIND Q ABT. DOG

can I detect pregnancy in dogs by a Hcg hormone pregnancy tester used in human females? —Preceding unsigned comment added by 59.95.178.103 (talk) 08:41, 26 November 2007 (UTC)

No, human pregnancy tests are useless in dogs. Dogs are an estrous species, rather than a menstrual one: dogs undergo the same hormonal changes whether pregnant or not. Dog pregnancies are traditionally "diagnosed" by ultrasound or palpation.... there is a blood hormone that is elevated in pregnant dogs, called relaxin, and a blood test is available for this, but it's useful only later in pregnancy than the human tests we're used to are. - Nunh-huh 08:56, 26 November 2007 (UTC)

Does being cold and wet, increase your chances of getting a virus like a cold or flu?

Why or why not? 64.236.121.129 (talk) 15:51, 26 November 2007 (UTC)

There have been studies that have found a correlation between being cold (or cold and wet) and catching a cold. I haven't seen such studies on the flu. They are usually dismissed due to poor management of the control and test groups (or a complete lack of a control group). In the U.S. NIH book, being cold or wet is not listed as a cause for the cold or the flu. However, both are listed as "seasonal" - meaning that they occur predominantly during a certain time of the year. Anyone who has kids knows that a lot of things pass from children to parents. In the winter, we send kids to school where they share all kinds of things and then bring them home. So, it is pretty much a no-brainer as to why there are more cold/flu issues during the children's school-year. -- kainaw™ 16:03, 26 November 2007 (UTC)

Why do these viruses exist during one time of the year, but not the others? 64.236.121.129 (talk) 16:07, 26 November 2007 (UTC)

Keeping everyone in close quarters (in the winter) makes it easier to spread viruses around. Plus, the lower relative humidity probably also makes it easier to become infected from a given number of virions.

Atlant (talk) 16:39, 26 November 2007 (UTC)

I don't know about that. I don't see complete strangers huddling around just because it's colder outside. How does lower humidity make it easier to become infected? 64.236.121.129 (talk) 17:03, 26 November 2007 (UTC)

You don't need to huddle, you just need to spend more time breathing in recirculated air. Haven't you ever seen waydowntown??

Atlant (talk) 17:32, 26 November 2007 (UTC)

No. No I have not. And people spend time in the same building through other seasons too. Whether you go to school or work, you are still spending time in a building with other people, through all the seasons. 64.236.121.129 (talk) 18:50, 26 November 2007 (UTC)

Did you read the first reply above? It is colder in the winter. The school year tends to be in the winter. Children spend more time around each other during the school year. So... children are closer to more children when it is colder outside. It is all about proximity. The viruses don't gather super-virus strength from the cold. They still need people to be close to one another to travel from host to host. -- kainaw™ 17:43, 26 November 2007 (UTC)

The school year is during part of the summer, fall, winter, and spring. It's not mostly in the winter. The only time school is out is during the summer, but the school year does extend to parts of the summer. Also, I question whether your assumption is correct. You are assuming children are the source. You are also assuming children are closer together during the winter. That's just speculation. Also how does one get the virus in the first place? In order to catch it from someone else, someone initially had to catch it. 64.236.121.129 (talk) 18:47, 26 November 2007 (UTC)

I know that the Wikipedia rules insist that we assume good faith, but I'm starting to feel like you've simply come here for an argument.

Atlant (talk) 19:01, 26 November 2007 (UTC)

Nope. Assume good faith. I'm just asking questions. 64.236.121.129 (talk) 19:31, 26 November 2007 (UTC)

So if people catch colds more during the Winter because they spend more time indoors with others, does that mean that in places such as Phoenix, Arizona where people spend more time indoors in the Summer, people get more colds in the Summer? Deli nk (talk) 18:56, 26 November 2007 (UTC)

You have a bit of a point there. Contrary to the OPs assumption that I'm just speculating, it is my job to manage health data for millions of patients. There are exceptions, but the rule is that cold/flu cases spike in September. That is when children go back to school. They slowly go down until January when kids come back from the winter break (smaller spike than sept). Then, they keep dropping and dropping until there are no significant number of cases by summer. However, there are many cases of summer colds and flus (just not enough to be significant). Comparing desert regions to non-desert regions, the percentage of people with summer cold/flu cases is higher in the desert regions. While I don't have enough Phoenix patients in my database to draw a real conclusion for that particular city, I do have over three million patients Arizona and New Mexico - which is to my knowledge mostly desert. So, you have some data to back up your claim that desert-dwelling people have higher rates of summer colds and flus.

To the OP... you appear to believe that cold/flu viruses go away and then come back. They don't go away. In any large population, there is always someone with a cold/flu virus. Most often, it is the children (again, I can look at the data and see that the younger the person the higher rate of having cold/flu diagnoses - so this is not just speculation). To catch the cold/flu, you must be around someone who has it and have the virus physically travel from the other person to you and then successfully make it past your body's defenses and start multiplying. At that point, you will risk infecting everyone around you. The more people you have around you, the higher chance you have of infecting someone else. That is why having people near each other is the key to spreading the cold/flu virus. -- kainaw™ 19:14, 26 November 2007 (UTC)

What about anecdotes like that one President of the United States who gave a really long inaugural address out in the cold and then died a few weeks later? Also, wasn't there an American football coach who was doused by the customary, celebratory cooler of Gatorade only to get sick and die afterward? Besides frostbite and hypothermia, can exposure to the cold give you other problems?--The Fat Man Who Never Came Back (talk) 18:52, 26 November 2007 (UTC)

For the U.S. President, see William Henry Harrison.

Atlant (talk) 19:01, 26 November 2007 (UTC)

This is an interesting discussion (though I'm sure it's been discussed since time immemorial). Can being cold and wet give you, say, pneumonia or other conditions?--The Fat Man Who Never Came Back (talk) 19:38, 26 November 2007 (UTC)

Weakening your immune system when you have a cold/flu can lead to further complications - such as pneumonia. So, the question is, "Does being cold and wet weaken your immune system?" I did a quick AMA search and found nothing on that topic. I'm sure you can find many hits on Google - but not necessarily proper medical studies. -- kainaw™ 19:56, 26 November 2007 (UTC)

A lot of it is to do with heating systems, viruses like warmth as much as humans and any systems that recirculate air (such as that found in large buildings) is going to ensure the microbes get maximum circulation. The more people that get these bugs, the more carriers there are to ensure they keep spreading. The start of the heating season always brings the bugs out. GaryReggae (talk) 21:43, 26 November 2007 (UTC)

You might want to try reading The Straight Dope article "Why is winter the season for colds, flu, etc.?" It explains that being cold or wet does not increase your chances of getting sick, and also notes that some "seasonal" illnesses are actually encountered at various times throughout the year, and that some factors such as cold stress, which can cause cold/flu-like symptoms, and seasonal psychological stress, which can weaken the immune system, may be adding to the winter cold/flu stats. Hope that helps! -- HiEv 22:50, 26 November 2007 (UTC)

Psychological stress can weaken the immune system? Has this been proven? My other question is, if it is true that one usually catches it from another person, how did patient 0 catch the cold in the first place? Just incompetance by touching dirty objects, then putting their hands in their mouth/nose? 64.236.121.129 (talk) 16:46, 27 November 2007 (UTC)

Yup, stress hormones from chronic stress can suppress the immune system. See Ask A Scientist - Stress and immune system. As for your "patient 0", that person might have existed centuries ago, and the virus just keeps circulating through the population. Or, as is the case with new flu viruses, existing viruses co-mingle and/or mutate in a host (pig, chicken, duck, human, etc.) and produce a new virus. There are lots of ways viruses are created and spread, so there isn't just one answer to that question. -- HiEv 19:47, 27 November 2007 (UTC)

Well, the being cold - catching cold connection may have some fact behind it. The thing is, when you're cold, your body tries to conserve heat, and keep the heat in, leaving less heat at the extremities. This can include your head, if it is not well covered. If you are wet, there is a good probability your hair will be too. When your head becomes cold, it starts to restrict blood flow, but of course not hinder it altogether. When this happens, your immune system in the cold area can decline, since the blood carries white blood cells. The area, in this case your head, becomes cold, but still warm enough for the viruses to develop. There are viruses all around us, but usually they are kept at bay by your white blood cells. The viruses and bacteria can now reproduce, because there aren't as many white blood cells to attack them. When you come back inside, the viruses are still there, but now more heat is induced, so the body no longer has to conserve, and the viruses multiply even faster, giving it time to spread, but now since the blood flow is increased, the white blood cells come back to attack the viruses. Now, your immune system will take care of the rest, so you usually don't get an automatic cold that way, but you may see the early symtoms of one: high head temperature, when the blood flow increases to get rid of the germs; runny nose, when the mucus attempts to wash away the viruses in your nose; sneezing, to get rid of viruses in your nasal system; coughing, to get rid of viruses in the throat; stuffy nose, caused by excess mucous and nasal activity, etc. So, the correlation may be there, but usually isn't as direct as you may think. Now, if you wear too much clothing outside on a cold day, it will increase your body heat substantially, but since your head is usually exposed, the temperature at the surface of your face may go down. This prevents large amounts of white blood cells from entering the area, but produces enough heat so that the viruses can multiply profusely. Plus, the excess clothing causes you to sweat, and wind can allow the cold to enter your body, potentially degrading your immune system. The numbing cold near the surface of your face, supplied with feeling by your warm interior, can cause pain in your face and sinuses, and possibly cause a headache. A risk of not wearing enough clothing in the cold is more of hypothermia than getting a cold. Plus, if you're not wearing excess clothing because you're going to be excersizing strenously, it could cause further complications. Excersize causes sweat to be produced, and the sweat, along with the wind and cold, can allow coldness to enter your body. I don't think this has been proven, but not tucking your shirt in can expose your stomach areas to the cold, and supplied with enough body heat, can allow invasive viruses to develop in the stomach and intestines, potentially causing diarrhea. Also, the sudden warming of extremities after coming inside might explain why my hands, for example, often feel warm instead of uncomfortably cold to the rest of my body. Remember that this is mostly theoretical, and science seems to want to reject any "myths" based on experience rather than scientific fact. Hope this helps. Thanks. ~AH1^(TCU) 21:59, 27 November 2007 (UTC)

Ok so now some statements are contradicting each other. Astro said viruses are around us all the time, so that means we can get infected even without being close to someone who is already infected. Which means the so called, patient 0 can be anyone who happens to catch a cold from the enviroment. But HiEv said patient 0 existed since the cold virus was first contracted, and has simply spread from person to person since then. This implies that one does not catch the cold from the enviroment, but rather from people. Who is correct? 64.236.121.129 (talk) 16:40, 28 November 2007 (UTC)

Well, the thing is, if someone sneezed in a building, the viruses do not go away immediately, correct? So, if someone walked through the sneezed area, there will be a higher concentration of viruses in that particular area, which means there can be viruses in the air, especially in an enclosed public building. Also, viruses can live on uncleaned surfaces for days, so if you touched the surface and rubbed your nose, some viruses might enter your nose. What I also meant was, there can be viruses inside you, although they are usually controlled so you don't get sick. Sorry if you misunderstood, and I'm not really an expert, so make sure you understand that when you're deciding the general answer to your question. Hope this helps. Thanks. ~AH1^(TCU) 18:45, 28 November 2007 (UTC)

Let's clear up a few things. Some viruses, like HIV need bodily fluids to survive and are destroyed on contact with air, while other viruses can live briefly outside of the body, like cold and flu viruses, some of which can survive up to about 48 hours depending on conditions.[2] Also, "patient zero" generally refers to the first/main person who caused the disease to spread within a population, which is not necessarily the person who spread it to you. Finally, viruses cannot reproduce without a host. What this means is that any viruses in the environment come from another host, though that host may not be a human. This is not a contradiction, it's just neglecting to mention that the "viruses around us" come from other living things. -- HiEv 03:39, 2 December 2007 (UTC)

I didn't read the whole thread but a few points to mention. It has been suggested although as far as I'm aware is completely unproven that people may have slightly weaker immune systems in cold weather and also that viruses may last longer in the environment which are some of the reasons why cold and flus are more common in winter. In any case, it's generally accepted that the closer contact between people etc in the winter season are the primary reasons. Bear in mind it AFAIK isn't just winter, cold and flus tend to be more common in the rainy season in tropical countries when it's colder (although to a far less degree) and people also spend more time in close quarters. Nil Einne 08:40, 1 December 2007 (UTC)

Reflection of voltage/current ???

Hi, I was reading web(including wikipedia) articles on what the characteristic impedance aka 75 ohms mean, and some related stuff about transmission. I've seen and how generic waves reflect a part of their energy (at the interface) when they enter a medium of different 'elasticity' than in which they were traveling.
But I don't understand how electricity(or any wave for that matter) can reflect off an interface of different impedance. Can some one direct me to a wiki/web page which deals with this sort of reflection of current/voltage as i don't know how any technical terminology to search with. I haven't so far dealt with electric fields inside conductors and their role in conduction and I find it confusing to think about infinitely long conductors and effect of transmission being non-instantaneous. 59.93.3.188 (talk) —Preceding comment was added at 15:55, 26 November 2007 (UTC)

See impedance matching, standing wave, and standing wave ratio. These aren't much help, admittedly. Remember that it's the impedance of the load we care about, and that depends on frequency. If it matches, all the energy will be taken up. You can see the load sort of like the physical equivalent spring-and-mass system. If the load fails to use up the power, it has to go somewhere, so it reflects back and forms standing waves in the transmission line. A wave in a jumprope is an honest-to-goodness wave just like an electromagnetic one, at least as far as power transmission goes. This is a dumbed-down version, not because I think you're dumb, but because I have forgotten most of it. --Milkbreath (talk) 16:21, 26 November 2007 (UTC)

Howdy mean, Milky, they aren't much help? I find them quite enlightening. Do you have any suggestions on how they might be improved? —Preceding unsigned comment added by TreeSmiler (talk • contribs) 01:42, 28 November 2007 (UTC)

I was hoping for a down-and-dirty walkthrough of power reflection, which was the original question. The impedance matching article blows right by that as if it's self-evident. Power is reflected?!!! What the hell does that mean, we ask ourselves, and we're left to imagine it. I used to be a First Phone, but even back when I had a clue, transmission lines and antennas seemed like magic to me. "Thou shalt build the antenna 2.7 cubits by 3.31 cubits by the square root of five cubits...." It's pure math, pure applied theory. So I can tell when I'm not being told something, but that's about it. --Milkbreath (talk) 03:00, 28 November 2007 (UTC)

Seen reflection coefficient yet? —Preceding unsigned comment added by TreeSmiler (talk • contribs) 03:03, 28 November 2007 (UTC)

Rather than thinking about electrical signals, you might want to think about electromagnetic energy of a shorter wavelength: light. When it is travelling in a material of one impedance (which optics folks tend to call refractive index) and it meets material of a different refractive index, some of the light is transmitted into the new material but the rest of the light (that wasn't transmitted) is reflected back. Electrical impedance works the same way; when the impedance changes, a reflection occurs. For radical changes of impedance (to an open circuit or a short circuit), the entire electrical wave is reflected back to the source. You might also enjoy our articles about time-domain reflectometry and the time-domain reflectometer.

Atlant (talk) 16:33, 26 November 2007 (UTC)

Possible to destroy a virus by smashing it?

If you knew a virus was on a stone, and you took a sledgehammer and you smashed the area it is in, is it possible to destroy it? 64.236.121.129 (talk) 15:58, 26 November 2007 (UTC)

Localized heating of the impact area might denature your virus, destroying it. But I think it would be a very chancy thing, with a good chance of aerosolizing virions as well, so if you thing there is, say, some Captain Trips on the rock, why not just walk away?

Atlant (talk) 16:28, 26 November 2007 (UTC)

Of course you should just walk away. But that's not what my question is. 64.236.121.129 (talk) 17:00, 26 November 2007 (UTC)

Viruses are very, very small. So small that when you are talking about mechanically smashing them, you have to think about how tight a seal it is going to be. Your hammer, no matter how smooth it might seem, has lots and lots of imperfections and the odds are that you're not going to smash it. Even very small bugs (e.g. fleas) are incredibly hard to smash for this reason (along with the fact that they have slippery shells that are meant to make it hard to smash them). Something as small as a virus, I would say that your odds of actually making contact with it are very minimal. --24.147.86.187 (talk) 16:57, 26 November 2007 (UTC)

I agree. Hmm. But if you had, say a nanomachine on the scale of a virus, and it went up to the virus and ripped it up, it would die then right? 64.236.121.129 (talk) 17:00, 26 November 2007 (UTC)

It's arguable whether the virus was "alive" before you smashed it. But yes, if it can move xenon atoms around to spell IBM[3][4], then something like an atomic force microscope or scanning tunneling microscope could probably "dismantle" a virion.

Atlant (talk) 17:25, 26 November 2007 (UTC)

Your problem will not just be one virus, but there could easily be 1000000 viruses on your stone. Even if you destroy 90% you still have 100000 infectious particles. Graeme Bartlett (talk) 20:09, 26 November 2007 (UTC)

The goal would be making a manmade virus that is attracted to a certain virus and then restructures that virus to clone the manmade anti-virus virus. So, every time the manmade virus meets a virus it is supposed to kill, it actually makes another one of itself to fight off the entire virus population. I wonder if there'll be enough D&D fans on the team that invents this to give a name that references the charm spell used to make enemy monsters fight for you. -- kainaw™ 20:23, 26 November 2007 (UTC)

Sounds like a really big prion. (sorry, not a D&D fan.) Someguy1221 (talk) 20:50, 26 November 2007 (UTC)

Well, viruses can't reproduce by themselves - they need the facilities of a host cell in order to reproduce (think like the aliens in the movie "Alien"!) - that's why they are arguably not 'alive' at all. So if your manmade "virus" is truly a virus and not some other form of science-fiction nanotechnological assembler - then it can't really make copies of itself at will like that. But even a nanotech assembler would need energy and raw materials in order to make a copy of itself. That would likely be a time-consuming thing compared to the time a virus needs to reproduce (if it has a host). So whilst it may one day be possible to build tiny robots that can shred viruses mechanically, I doubt they'd do it by duplicating themselves and then committing suicide in order to take down their opponent. However, since we have no way to build such machines - nor any real certainty that they'll actually be possible at all - it's tough to speculate. SteveBaker (talk) 21:05, 26 November 2007 (UTC)

Smashing? Well, no, but then smashing isn't really good at destroying things anyway. Smashing a piece of wood and you get lots of little pieces of wood. Smash a rock and you get smaller rocks, etc. But smashing does little to change the basic nature of the substance. However if we are going to try sterilize with machine shop tools, I bet an arc welder would be pretty effective. More ambigously, I wonder how well an angle grinder would do at killing virii? If you grind down a surface, I'd lay good odds that most of the virii that were removed (at the least). Dragons flight (talk) 21:25, 26 November 2007 (UTC)

Why do I get cold after I eat?

Sometimes it's said that people get drowsy after lunch (or any big meal), because 'the blood goes to the stomach to aid digestion'. I don't know if that makes sense or not. I do know that I seem to get cold after I eat, which corresponds with my recollection that it always seems colder outside when one goes back on the ski slope after lunch. So, the question is, does it make sense, physiologically, that the body sends extra blood to the stomach to aid digestion, and this takes blood away from the task of keeping me warm? Are there other physiological functions that would be impacted in the same way, after eating? I know of someone who claims that its harder to keep an erection after eating a big meal...would it be the same story? —Preceding unsigned comment added by 213.84.41.211 (talk) 16:30, 26 November 2007 (UTC)

The sensation of warmth and cold is most directly related to how cold your skin is. So if the body does indeed direct an overly large amount of blood to your digestive system to the detriment of blood flow to your skin, then your skin will certainly feel colder, though this says nothing about your internal body temperature. Someguy1221 (talk) 16:41, 26 November 2007 (UTC)

One factor in getting sleepy when you eat is serotonin, which comes from 5hydroxytryptophan which comes from tryptophan which as is widely known comes from food. But, everybody gets it wrong; tryptophan is an amino acid and comes from protein, but it's relatively rare with respect to the other amino acids so that the presence of a lot of protein to be digested causes a lot of competition and saturates the transport sites and it actually gets taken up less. In fact, more carbohydrates in the meal causes serotonin to go up, and drowsiness. So it's not the turkey on thanksgiving. Judith Wurtman at MIT did a bunch of work on this for the defense department, who were interested in questions like how to make sure the folks with their fingers on the triggers in the missile silos wouldn't fall asleep. she found some folks who were so sensitive they couldn't stay awake after the highcarb lunch, period. other folks had a self-medicating thing, i.e. got really antsy midafternoon if they couldn't get a carb break. here's a couple of general public type refs: http://www.fastcompany.com/magazine/06/diet.html http://www.healthsystem.virginia.edu/uvahealth/news_mindbody/0610mb.cfm Gzuckier (talk) 21:50, 26 November 2007 (UTC)

Metal case enclosure heat transfer coefficient

What would be a good way to calculate the capacity of a metal case to dissipate energy by natural convection.

I'm thinking of building a case 20 x 15 x 10 cm in aluminium and would like to know how much heat can be produced inside without overheating. The internal temperature would probably be 40-45 degrees and the maximum external temperature 35 degrees.

Most of the websites I've seen don't offer a simple way to calculate the convective heat transfer coefficient of the walls.

I'm guessing 50W might be the limit at 35 degrees outside and maybe 80W at 20 degrees. --Jcmaco (talk) 16:40, 26 November 2007 (UTC)

It would depend a lot on whether the internal and or external air is stirred by a fan or left to natural convection. For the convective situations, it would also depend on the orientation of the enclosure. Normally, to dissipate 50W, you'd want some method to directly couple the heat source to the aluminum enclosure. You can experiment with all of this rather easily by using resistors as the heat source and your favorite thermometer (thermocouples, thermistors, infrared thermometer, whatever) as the measuring device(s).

Completely off-the-cuff, managing to dissipate 50W and achieving a 10C temperature differential using only natural convection sounds optimistic to me; I'll bet you'll at least need cooling fins on the outside of the case.

Atlant (talk) 16:45, 26 November 2007 (UTC)

The heat transfer coefficient is not easily calculable from what you have, so you may have to run an experiment first using the relevant equations from that article. We can do an estimate, but to start we need two things: Newton's Law of Cooling $T=T_{e}+(T_{0}-T_{e})e^{-kt}$ with k predetermined for aluminum in air (try google or experiment, or just set = 0.5), and the Heat law $Q=\sum mc\Delta T$, with the specific heat c being widely available for both aluminum and air (both are about 1 J/g/K). Now we take the time derivative and get $dQ/dt=P=\sum mc(-k(T_{0}-T_{e})e^{-kt})$ for our power dissipation, then set that equal to $dQ/dt=hA(T_{0}-T_{e})$ and we get our heat transfer coefficient $h=-(k/A)e^{-kt}\sum mc$.

Now let's say we're adding power P0 to the system, from whatever's heating your case. Equilibrium then occurs where $dT/dt=0$, or $P0=dQ/dt=hA(T_{0}-T_{e})$, which you can plug in values for at t=0. SamuelRiv (talk) 21:45, 26 November 2007 (UTC)

What metal is the worst conductor of heat?

^Topic 64.236.121.129 (talk) 17:04, 26 November 2007 (UTC)

There's a chart here: http://www.engineersedge.com/properties_of_metals.htm --JDitto (talk) 19:10, 26 November 2007 (UTC)

According to that chart, it's stainless steel - which suprises me a lot - but in any case, that's kinda cheating because there is a ton of (non-metallic) carbon in steel so it's really not a pure metal. In terms of pure metals, Lead wins the prize - which is about what I'd expect - lead is nowhere near as "cold to the touch" as most other metals (which is a good quick way to guess what the thermal conductivity of a material is). That chart seems to cover only the more common engineering metals - whether you'd find that something weird like metallic liquid hydrogen or one of the transuranics had a lower coefficient is hard to guess. SteveBaker (talk) 20:54, 26 November 2007 (UTC)

Mercury has a lower thermal conductivity, though I don't know whether that would hold true when it was solid. In either state, I'd guess it's not practical for any application you'd be considering. jeffjon (talk) 21:41, 26 November 2007 (UTC)

According to this list of the thermal conductivities of all elements, neptunium is the worst conductor of heat amongst the metal elements, followed by plutonium and manganese. --Bowlhover (talk) 00:27, 27 November 2007 (UTC)

How does stainless steel compare to say... Concrete or stone? 64.236.121.129 (talk) 14:41, 27 November 2007 (UTC)

Concrete conducts heat MUCH less well than stainless steel. The coefficient of conductivity for steel is around 14 to 16, for concrete it's 0.8 to 1.3. For comparison, metals like copper and silver that conduct heat very well have coefficients up around 400. But check our article List of thermal conductivities - it has an extensive list. SteveBaker (talk) 17:01, 27 November 2007 (UTC)

Quartz

Hello

My queastion is as follows. I found a geod near the town I live in, upon opening it the cyrstal inside was a light, medium brown. I looked through the entire area of this site on quartz cyrstal and did not see the same crystal. So i would like to know if there is a possability that someone could tell me what variety it is. Thank you. --63.245.189.4 (talk) 17:18, 26 November 2007 (UTC)

Your talking about a geode right? This site, has several brown crystal geodes, each one described as calcite crystals. Check out the 2nd picture in the calcite article to see if it's similar to yours. --JDitto (talk) 19:08, 26 November 2007 (UTC)

Quartz in geodes is often coated by iron oxide, making it look brown. Cheers Geologyguy (talk) 22:04, 26 November 2007 (UTC)

Smoky quartz is brown. DuncanHill (talk) 09:04, 27 November 2007 (UTC)

Spacetime Dipping

According to general relativity all masses make a dip in the sort of sheet of space time, making a potential well, and thats how masses have gravity. But if two massive particles, say a low energy electron and positron annihilate, what happens to the 'dips' in space time that they both have, seeing as the energy is carried away by two photons which do not have mass anywhere equivalent to the electron positron pair. ΦΙΛ Κ 20:23, 26 November 2007 (UTC)

I'd suspect the mass would be converted into energy. No mass, no gravity well. --Kjoonlee 20:41, 26 November 2007 (UTC)

Actually, the photons will have mass exactly equivalent to the electron/positron pair (if we're talking about relativistic mass, that is). Someguy1221 (talk) 20:48, 26 November 2007 (UTC)

Ok, almost exactly. Maybe some gets radiated as EM waves, maybe a tiny bit as gravity waves. Someguy1221 (talk) 20:49, 26 November 2007 (UTC)

But if there is some loss, does this require some instantaneous movement of space time to account for this, to just sort of instantly pop into a different shape to coincide with the annihilation ΦΙΛ Κ 21:13, 26 November 2007 (UTC)

Not at all. At the the "moment" right after annihilation has two photons with the same mass and approximate position as the two particles that preceded them. As far as spacetime curvature is concerned, nothing has really changed. When I referred to loss, these would be gradual losses as the two particles approach, not at the moment they annihilate. Someguy1221 (talk) 21:18, 26 November 2007 (UTC)

Mass-energy is conserved, and mass-energy is the source of spacetime curvature in GR, so the shape of spacetime immediately before the interaction is the same as its shape immediately after. —Keenan Pepper 21:46, 26 November 2007 (UTC)

Heh, I should have known better. :) --Kjoonlee 22:47, 26 November 2007 (UTC)

Calcium carbonate from breathing into a solution

I recently performed a scientific experiment. We were, in class, given a filtered solution of what I believe to be some kind of "chalk water". This was put in a glass, and once one breathed into it through a straw, calcium carbonate would form and settle at the bottom (with time). My question is then, what was the solution I added CO2 to? And I assume that, in order to balance the equation, H2O must be added (so it is CO2 + ? = CaHO3 + H20)?

This is not a homework question, I've simply forgotten what the equation went like =) Also, my teacher balanced it the wrong way, saying in fact that calcium carbonate is CaHO2. Thank you for your help! 81.93.102.185 (talk) 22:30, 26 November 2007 (UTC)

Your equation can't be right - you have a carbon atom on the left side - but no carbon on the right. Oh - I see. You said that Calcium Carbonate was formed...that's CaCO3. Anyway, the answer is in our Calcium Carbonate article Ca(OH)2 + CO2 → CaCO3 + H2O - so the missing ingredient is Calcium hydroxide. Woohoo! Steve gets to answer a chemistry question! With a more than 1% chance of being correct! SteveBaker (talk) 22:54, 26 November 2007 (UTC)

(after edit conflict) The unbalanced equation would then be CO₂+Ca(OH)₂

--> CaCO₃ (note that calcium carbonate is CaCO₃, not CaHO₂). Since water is neither created nor used up during the reaction, I don't think it is not part of the equation; it simply helps the reaction. --Bowlhover (talk) 23:03, 26 November 2007 (UTC)

Sorry, water is created during the chemical reaction. The balanced equation would then be Ca(OH)₂ + CO₂ --> CaCO₃ + H₂O. --Bowlhover (talk) 23:17, 27 November 2007 (UTC)

Otherwise it would act similar to spectator ions. 81.158.37.78 (talk) 23:14, 29 November 2007 (UTC)

What is the biological basis for appreciation of music?

That is, what is it in our genes that makes us react to a pattern of arbitrary sounds? —Preceding unsigned comment added by Xhin (talk • contribs) 22:50, 26 November 2007 (UTC)

Honestly, I'd be surprised if anyone had anything other than speculation and a few unconnected tidbits of information on this topic at this point in time. The brain and genetic expression are probably the two biggest mysteries left in explaining how organisms work. Toss in the rather subjective appreciation of auditory beauty and you're asking for something that is at the far boundaries of human knowledge. Good luck on your quest for an answer. -- HiEv 23:08, 26 November 2007 (UTC)

I agree that we don't know for sure - but one thing that's interesting is the mathematical basis of many musical systems. The fact that we like music where there are simple mathematical relationships between the frequencies and durations of the notes cannot be a mere coincidence. I suspect this has something to do with it - but precisely what is uncertain. SteveBaker (talk) 23:16, 26 November 2007 (UTC)

Unless the mathematical aspect of music is simply Emergent. -=- Xhin -=- (talk) 23:27, 26 November 2007 (UTC)

Try music theory. The mathematics of musical perception are a subject of continuing research, but it is fairly well understood how thirds, fifths, octaves and inversions make up the geometry of music. Physically, our cochlea does something similar to taking a Fourier transform of the sound waveform, giving us a range of frequencies that harmonize according to mathematical rules. Thus chords sound like the fundamental, etc. Chord progressions are not well understood, but there is some very promising research by Dmitri Tymoczko on their geometry based on 2, 3, and 4 dimensional topological mappings of the chord symmetries. See [5]. SamuelRiv (talk) 23:52, 26 November 2007 (UTC)

I worked on a research project a few years ago: "Can a computer tell the difference between pleasing and non-pleasing music?" Yes - it can. This was based on the realization that many "zipfian" distributions exist in "pleasing" music. Since the same balances exist in pleasing poetry, paintings, and throughout nature, it is possible that our brains pick up the natural balance and "appreciates" it. If you are interested in the research, it is here. -- kainaw™ 23:54, 26 November 2007 (UTC)

The only problem with that is the subjectivity of music appreciation -- for example, I've heard songs which somehow got record labels, but sound like cacophany to my ears. Anyway, you guys are helping -- slowly. Keep it up! -=- Xhin -=- (talk) 00:16, 27 November 2007 (UTC)

It probably poses more questions than it answers, but Oliver Sacks' book, Musicophilia: Tales of Music and the Brain offers a fascinating insight into the neural coding for music and its appreciation. Here is a podcast of Sacks talking about it (he addresses the question of subjectivity too). Rockpocket 00:22, 27 November 2007 (UTC)

I've read reviews of that book, and I plan to be delightfully surprised to find it in my Christmas stocking. Another very good one is Music, the Brain and Ecstasy by Robert Jourdain. Re cacophony, I was going to make the point that it's fascinating how a computer can distinguish between pleasing and non-pleasing music but many humans seem incapable of so doing (and here I'm thinking of things from post-Schoenbergian squarks to heavy metal) - but then, it's all subjective, and what I enjoy would be rubbish to someone else, I guess. -- JackofOz (talk) 00:33, 27 November 2007 (UTC)

Appreciation of a particular sequence of pitches? Doubtful. But given that an appreciation of music something found in countless cultures throughout time and location (and I do believe Brown included music in his list of Human Universals). It seems likely that there is a biological basis for this universal appreciation.--droptone (talk) 02:21, 27 November 2007 (UTC)

How do you derive the properties of chemistry (ie, flowing liquids) from physics?

Unloaded question, but much help appreciated ! -=- Xhin -=- (talk) 22:51, 26 November 2007 (UTC)

You solve the quantum mechanical equations for material...Computational Chemistry may help.Shniken1 (talk) 23:24, 26 November 2007 (UTC)

By far the greatest success in understanding physical chemistry has been statistical mechanics for molecular structure combined with quantum mechanics for atomic structure and electrodynamics for interaction. Flowing liquids, for example, are described quite well by Bernoulli's principle, which can be derived statistically. The ideal gas law is very simply derived from first principles in stat mech as well. For a good, thorough book on the subject, see Thermal Physics by Kittel and Kroemer. SamuelRiv (talk) 23:56, 26 November 2007 (UTC)

Failing audio cables

I have a lot of trouble with audio cables 'breaking'. I use various types of these a lot for connecting musical instruments, amps, speakers, mics, computers etc such as mic leads, jack leads, RCA leads anbd various combinations. For example the 3.5mm stereo jack to double mono RCA lead I use to connect my laptop to my main PC's speakers (as the laptop's speakers are vey poor) has just started playing up, I only so much have to breathe on it and one side of the stereo signal cuts out, when I wiggle it near the jack plug, the dodgy side cuts in and out although the other side is OK (I have been testing it by shifting the balance to the affected channel only).

I know the easy answer is simply to buy a new cable which in this case is fairly cheap but when £10 mic leads fail it is no joke and it seems a waste to keep buying new leads. I just wondering what actually causes these leads to fail? I presume they are made of copper wire which somehow breaks but why is it such a problem with audio/video leads? I have never had this problem with mains leads or any other type of lead, ie USB, parallel, firewire etc and as the latter types are digital rather than analogue, I would have thought broken wires would cause more problems than with analogue leads. I don't exactly pull on the leads, I coil them up when not in use but not tightly so I can't think that I am doing anything beyond what they are designed for.

Secondly, is it possible to repair them? Obviously, this depends on what the actual problem is! GaryReggae (talk) 22:53, 26 November 2007 (UTC)

Mine always fail from other bending so I suspect it is something to do with this. Perhaps a little more slack would reduce wear, or perhaps higher-quality cable would be more sturdily built. I was told to loop them together up/down against each other to stop wear. When it's occurred to me it has always been near the headphone socket/headphone so I guess it must be strain/stress and bending that is causing it. Not sure if it is repairable, I would expect for the cost it wouldn't be worthwhile. ny156uk (talk) 23:09, 26 November 2007 (UTC)

Generally, it's just because the wire is bending a lot. If you take a piece of wire like a paperclip and bend it back and forth over and over, eventually, it'll break. Copper is pretty flexible - but eventually, it goes. Probably the wire at your laptop end broke because it's being moved more often (or plugged and unplugged more often) than the ends at the speakers or whatever. You can mend those wires reasonably easily - there are two approaches. Firstly, if you can tell where it's broken (usually within an inch or two of the connector) then cut the wire an inch or so beyond the break - get a wire stripper (or a pair of scissors if desperate) and now you need to re-attach the connector. There are three approaches:

1. Cut the wire on the other side of the break, strip those wires - then twist the ends together and wrap them up with a few inches of electrical tape so they don't short out. This isn't 100% the best thing - and if the break is too close to the connector, you can't do it this way - but it's very easy and gets you going with no tools more sophisticated than scissors and electrical tape!
2. Carefully remove the plastic shroud on the connector and thread it onto the cut/stripped side of the wire. De-solder the short bits of wire from the connector, solder on the new ends and replace the shroud. Of course this assumes you have a soldering iron (and possibly a desoldering gun) along with the necessary skills to do it. I suspect you don't or you wouldn't be asking...but hey - we all had to learn sometime!
3. Go to your local Radio Shack (or whatever national equivelent you have) - locate the rows of little, beautifully labelled drawers - find a connector that looks like the one you cut off BUT WHICH HAS SCREW CONNECTIONS instead of solder joints. Now you can wrap the stripped ends of wire around the screws and tighten them up. This is easy and has one HUGE advantage over the other two ways. When the wire breaks AGAIN (as I'm sure it will), you can just trim off a bit more wire and fix it again with no more tools than a screwdriver and a wire stripper (or scissors if you are careful!).

Good luck! SteveBaker (talk) 23:13, 26 November 2007 (UTC)

Thanks Steve, I will try fixing it later, the break seems to be right by the jack connector and it's a moulded plastic one so I think I'll just go to Maplin's and buy a new plug with screw fittings if they have them as I'm not very good at soldering. As you say, the next time it breaks, probably in the same place near the connector, it will be easy to fix! GaryReggae (talk) 09:07, 27 November 2007 (UTC)

Actually, if it does break often, some kinds of connector come with strain relief widgets. These take the form of a long plastic or rubber sleeve that covers the wire out to a distance of a couple of inches out from the connector. Others are like a stiff spring that you thread over the wire+connector. They take some of the load off the wire and prevent it from bending unnecessarily. If you are buying a new connector - you might look to see if any of them have superior strain relief. SteveBaker (talk) 16:51, 27 November 2007 (UTC)

November 27

Lighting yourself?

Continuing the thread about fire accelerants above, would it actually be possible to light yourself on fire by drinking gasoline and then swallowing a match or something? I am certainly not contemplating doing this, but I'm just wondering. bibliomaniac15 00:34, 27 November 2007 (UTC)

No, there is no oxygen (or at least not enough) —Preceding unsigned comment added by Shniken1 (talk • contribs) 01:08, 27 November 2007 (UTC)

If it was possible, then either they would've either done it on Jackass (or similar) by now, or someone would've done it whilst trying to get on Jackass (or similar) and we'd have read about it in the papers... :) --Kurt Shaped Box (talk) 01:16, 27 November 2007 (UTC)

Or Darwin Awards. --antilived^{T | C | G} 01:28, 27 November 2007 (UTC)

Since they can make fuel from pigs and chickens I guess they could make it out of us too. Its the third way between burial and classic cremation: get cremated in an internal combustion engine. "Help your children and become a galon of petrol". Keria (talk) 11:55, 27 November 2007 (UTC)

blah blah blah

OMG i can change the internet!

Rechargeable batteries

Why do rechargeable batteries have only 1.2 volts while other batteries have 1.5V? I'm referring to the common types: AA and AAA.

And I've heard that overcharging a rechargeable battery reduces its life. Is this actually true? Why is the life shortened? --Yanwen (talk) 03:21, 27 November 2007 (UTC)

The chemicals which make up the electrodes of a battery determine the open circuit voltage it produces. Alkaline batteries or Carboin Zinc have a higher voltage in each cell than rechargeable Nicad batteries because of the chemicals used for electrodes. Lead-acid batteries are rechargeable but have a higher voltage per cell than alkaline. Overcharging a battery is a bad idea because it can cause it to heat up and can cause the electrolyte to evaporate. Edison (talk) 03:37, 27 November 2007 (UTC)

You may also be interested in our article about nickel-cadmium batteries (i.e. rechargeable batteries). ›mysid (☎∆) 05:50, 27 November 2007 (UTC)

Overcharging can reduce the life of a rechargeable battery by driving water out of the cell, either directly as water vapor or as hydrogen and oxygen gases resulting from the electrolysis of the water within the cell. Most batteries contain a "recombination catalyst" that will burn small amounts of evolved hydrogen and oxygen back into water but serious overcharging will overcome the ability of the catalyst to cope with evolving gases.

Atlant (talk) 13:14, 27 November 2007 (UTC)

Actually, it depends on what kind of battery chemistry is being used - some of the rechargables have 1.3v, 1.4v or 1.5v. I used to mess around in the Lego robotics community - and the Lego computer runs off of 6 AA's to get 9v total - if you used NiCd rechargeables, you only got 7.2v and the computer couldn't run on so little. Since these things chewed through a set of disposable batteries in a couple of hours it was an expensive hobby. However, rechargeable Lithium batteries produced something like 1.4v and that was enough to run the computer. The subtleties of recharging them also depends on the chemistry of the battery. Some can be overcharged and need special rechargers that detect the fact that the battery is full and turn off - some cannot. Some batteries NEED to be fully discharged before you recharge them (NiCd's, certainly), others need to be kept fully charged as much as possible (Lead-Acid batteries like in your car), others don't care. With the wild profusion of battery types out there, you need to be careful that the 'rules' you are following are the right ones! SteveBaker (talk) 16:45, 27 November 2007 (UTC)

Endosteum: soft tissue?

Would the endostuem be considered, technically speaking, a soft tissue? Thanks. 75.42.209.73 (talk) 03:22, 27 November 2007 (UTC)

Yes. Its bascically mesenchymal cells in a collagen matrix that lines the cortical bone at the corticomedullary junction in long bones. Its function is to conduct nutrient blood vessels to the medullary surface of the cortical bone, and also contains undifferentiated osteoprogenitor cells that are recruited in Bone_healing. The article on endosteum is a bit light on detail, but you may find it helpful. Mattopaedia (talk) 13:05, 27 November 2007 (UTC)

Okay thanks that's what I thought! 75.42.209.73 (talk) 17:50, 27 November 2007 (UTC)

Cats, the hunting instinct, and feline psychology

My domestic cat brought a number of live mice into my home and I started putting the ones I rescued into a habitat so the cat would have mice around without letting them loose in the house. Now she spends a lot of time watching the mice, actually strongly resembling a human sitting in front of a television. (Feline reality TV! She's even gained weight with her new couch-potato lifestyle!) My question: Is the presence of mice giving the cat hours of lively entertainment, or is it horribly cruel to expose her to mice that she will never be able to catch? Thanks for any insight. Peter Grey (talk) 08:30, 27 November 2007 (UTC)

Kudos to you for not knocking the mice on the head or necking them when the cat brought them in - but is it really fair on/healthy for them in terms of stress to be placed on constant display in front of a fearsome (to them) predator many times their own size? --Kurt Shaped Box (talk) 08:42, 27 November 2007 (UTC)

I have heard (paraphrasing somewhat) that, while for a human being stalked by a carnivorous predator a hundred times their size would probably lead to post-traumatic stress disorder or worse, for animals like mice it would just be an average day. Plus I'm guessing a habitat with unlimited food, even with a cat nearby, is still preferable to being mauled to death. Peter Grey (talk) 12:17, 27 November 2007 (UTC)

If your cat was bringing in live mice for you and has stopped, she might be waiting for you to eat them or give one to her. If she's not desperately trying to open or get into the habitat, then she's probably not stressed about it. If she's stressed, she'll let you know (with piercing whines, for one). The mice may very well be under stress, but likely then they wouldn't be eating or drinking (I had a cat who had to live with a big dog for a couple months. She lost a lot of weight because she would only eat when the dog wasn't around, i.e. taken on walks). SamuelRiv (talk) 12:52, 27 November 2007 (UTC)

Still, it couldn't hurt to stick the mouse cage up on a shelf, table or something. You don't want a mess on the floor/massacre when the cat finally gets hungry, do you? --Kurt Shaped Box (talk) 17:25, 27 November 2007 (UTC)

Mice are hard-wired to feel fear when they smell cats, so it's possible that the mice in the habitat are suffereing from constant fear, unless they happen to be the genetically altered mice with no fear of cats. -- JSBillings 13:32, 27 November 2007 (UTC)

Don't worry about the cat. Cats are evil. They were created by Satan in mockery of the blessed Dog, who can be trained not to defecate in your house. Proofs of the cat's malignancy abound: cat scratch fever, fur balls, spraying, the infernal yowling when they copulate, that murderer's blank stare. They are often found in the company of witches, where their true nature as a mere receptacle for an otherwise incorporeal demon is revealed. Your "cat" will be content to gloat over your captive rodents and revel in their delicious terror at his presence. --Milkbreath (talk) 14:00, 27 November 2007 (UTC)

And a bunch of Animal lovers is hampering the efforts of Italians to rid the world of evil. Keria (talk) 16:20, 27 November 2007 (UTC)

Hence I don't want to inadvertently annoy my evil housemate. Peter Grey (talk) 17:12, 27 November 2007 (UTC)

That's why I prefer gulls to cats - gulls have honesty. Compare your average Herring Gull to your average HouseCat. The gull is every bit as loud, aggressive, raucous and cold-hearted as the cat (and - to bring up an old running joke, could probably equal it in a vs. battle) - but it never pretends to be anything different. --Kurt Shaped Box (talk) 17:54, 27 November 2007 (UTC)

You've never fed a gull? They can be very good at begging and acting nice if they learn it gets them fed. I remember one year my mother leaving out fish (leftovers from our cats — how's that for a tie-in?) for a young herring gull that had fallen out of its nest. For the rest of the year, until the weather got cold and the gulls flew south, we basically had a semi-tame pet gull. He (or she) learned to recognize my mom by sight, and would wait on the rooftops among the other gulls for her to come out. Then, if no other people or gulls were in sight, he'd glide down and land some distance away, all very quiet, and then walk up to my mom and stand there looking at her and giving her the "puppy eyes". Probably would've let us pet him if we'd wanted to. He may not have been physically as cute and cuddly as a cat, but he was most definitely on his best behavior towards my mother, even despite our efforts to convince him that there were other, more appropriate food sources for a gull. —Ilmari Karonen (talk) 19:13, 27 November 2007 (UTC)

I'd heard that cats watch you and decide that somehow you never seem to catch any mice. This elicits a behavior similar to how they train their kittens to hunt. They catch a mouse - deliberately not killing it and bring it to you so you can practice on it. If you don't succeed with live mice, they'll bring you ones they've wounded to slow them down so you'll have a chance to get some practice in. If all else fails, it's on to dead mice. If you pick up the dead mouse and chuck it out somewhere, the cat figures you've finally gotten the idea - so it's back to live and wounded ones. Yeah - dogs rule. They may be stupid - but at least they know it. They don't pretend to be smart like cats do. SteveBaker (talk) 18:46, 27 November 2007 (UTC)

So what does it say when my neighbor's cat decided to show up at my doorstep with half a mouse? (It was the back half, for what its worth.) Dragons flight (talk) 18:56, 27 November 2007 (UTC)

I'd guess it's either "here, I've saved the best part for you" or "here, I ate the best part, you can have the rest if you like", but, without firsthand knowledge of that particular cat's idiosyncratic dietary preferences, I can't really tell which. Or perhaps she was just halfway through her meal and taking a short break when you interrupted her. —Ilmari Karonen (talk) 19:19, 27 November 2007 (UTC)

I don't think it was accidental. A) My doorstep is not really a normal place for her to visit and B) She seemed very enthusiastic about delivering the half a mouse to me. Dragons flight (talk) 19:23, 27 November 2007 (UTC)

I think this cat has gotten to the point where it suspects you are an especially slow learner and you aren't really ready for an ENTIRE dead mouse - so perhaps you'd better start with a half of one and work your way up. SteveBaker (talk) 22:11, 27 November 2007 (UTC)

There seem to be different schools of thought on this behaviour, essentially coming down to a) the mouse is for the human and b) the mouse is for the cat, she just wants to get away from other cats while she toys with it. In this case, she isn't bringing me the mice - I've had to fight her for them. Peter Grey (talk) 07:40, 28 November 2007 (UTC)

There probably isn't a lot of research in this area. I'd guess that, as long as your cat is well fed, having a habitat for mice that the cat can watch is probably no more cruel than giving the cat toys that it can't eat. In other words, it's probably not a problem. I often see people trying to anthropomorphize animals by attributing human-like thinking to them, but they don't try to see if their hypothesis is correct or think about simpler explanations that would produce the same behavior. Don't overthink things. If your cat looks entertained, then it's probably entertained. If it looks stressed, then it's probably stressed. However, as long as your cat isn't starving, threatened, sick, injured, lacking sleep, or too hot/cold/wet/dirty, then it probably isn't under much stress. -- HiEv 20:33, 27 November 2007 (UTC)

The mice, on the other hand...may be nervous. SteveBaker (talk) 22:11, 27 November 2007 (UTC)

Who can say? Mice always look a little twitchy to me. And, for the dog lovers/cat haters who seem to have used this question as a dumping ground for their vile hatred towards nature's most perfect animal, I can only say this: Yes, a cat might shit inside your house, but even it, in contrast to a dog, is smart not to EAT cat shit. For all the tricks they can learn, even the smartest dog has struck me as being no smarter than a bug. Matt Deres (talk) 02:34, 28 November 2007 (UTC)

Besides that, my cat, indeed all cats I've ever owned have never shit inside my house at least not for long instead they do it outside and don't leave it all over the place for you to step on like dogs do... Nil Einne 12:00, 3 December 2007 (UTC)

This seems to be the right answer - I get to ask a vet and it's roughly the response I got. Peter Grey 23:48, 30 November 2007 (UTC)

acting out of character

Does anyone know in how many cases where someone has started acting selfish, self-centered and strange have refused to cooperate even with the police that something has gone wring with them physically internally. I am not refereeing to the brain purse but rather to things like some form of body cancer that has finally reached the stage that it is noticeably to the individual interfering with body functions or weakening the individual to the point of representing a clear threat of death where the individual does not know the cause but only that their life is threatened and at risk unless they do something even to the point of causing a public disturbance while refusing to cooperate in any way? The second part of my question is if the police were aware of this possibility - a feeling on the part of individual of a life threatening situation, the cause to them unknown, would the police act less violently toward the individual, especially to the extent of causing the individuals death? 71.100.0.58 (talk) 08:37, 27 November 2007 (UTC)

Hypochondria? Munchausen syndrome? ›mysid (☎∆) 12:26, 27 November 2007 (UTC)

We can't give professional advice. Seek medical help. --Sean 14:59, 27 November 2007 (UTC)

You are on an, "Every question someone asks that a doctor could answer is a request for medical advice." kick. If you are going to work the science desk I suggest you get off this kick. 71.100.0.196 (talk) 23:43, 27 November 2007 (UTC)

This topic was in the Canadian news yesterday [6], although specific to mental disorders. Conditions such as hypoglycemia and electrolyte imbalances (which themselves could be as a result of pancreatic beta-cell cancer or any variation of renal diseases, respectively) could cause changes in mental status. (EhJJ) 16:10, 27 November 2007 (UTC)

The second part of this question assumes that police normally act in a violent deadly manner because it asks if the normal violent deadly action would be changed. What is the basis for assuming that police normally act in a violent deadly manner? -- kainaw™ 16:13, 27 November 2007 (UTC)

Normally the police try to match, if not exceed, any potential level of violence with which they feel they may be faced. In the US it is common for the police to aim their weapons for their own protection on mere suspicion or "just in case" their subject has a gun.

If the police tailor their actions to match the situation exactly and the situation is as described above then the police will escalate their reaction to be greater than the subject can overcome - including choke holds, placing too much weight on the subject's chest, tazerring the subject for too long or at too high a current setting, or not calling for medical help if the subject stops breathing or begins turning purple - all on their own assessment of being necessary to do their job and to protect their own lives while the subject may have been resisting at a level he felt necessary due to his own internal conditions to protect his life.

You Tube has a video example of this happening to an individual in an airport resulting in the individuals death by tazer. 71.100.0.196 (talk) 23:43, 27 November 2007 (UTC)

The most common cause for people to act out of character is mental illness, particularly bipolar affective disorder and other schizoaffective disorders. There are other causes as well, such as electrolyte disturbances, some cancers (this is rare though), and drugs. The other thing that often happens in these circumstances is the person loses insight, and so may not necessarily be aware of a need t o act in a certain way, or feel there is anything wrong with their behaviour. They may believe there is some threat to their life, but this may not be factual. As for the police, although IANAcop, I would expect them to only use reasonable force to preserve their own safety and the safety of any bystanders. It is a sad thing, but some people who are mentally ill do end up severely injured or dead as a result of an altercation during an acute phase of their illness, but suicide in the mentally ill is certainly a far bigger problem. It must be devastating for police officers to find they'd injured of killed a person whose actions were driven by mental illness. But that risk ultimately is a part of their job, which is a good reason to treat them with the utmost of respect, rather than criticising them for doing their job. Mattopaedia (talk) 05:38, 28 November 2007 (UTC)

I'm suggesting more in terms of life threatening illness generated paranoia (only it's not paranoia, but real) to the extent of causing total lack of cooperation and in fact opposition to anyone or anything versus the suicide-by-police due to mental illness syndrome. 71.100.0.196 (talk) 11:40, 28 November 2007 (UTC)

Building services

what are the environmental impact of the use and abuse of building services? —Preceding unsigned comment added by 81.199.59.84 (talk) 15:10, 27 November 2007 (UTC)

We don't normally do homework. Is this a homework question?

Atlant (talk) 16:44, 27 November 2007 (UTC)

This is a quite wide-ranging question that does sound like a homework essay so I am not going to give you anything more than a few bullet points to get you started:

Energy use (therefore carbon emissions) is the primary environmental cause for concern from building services

Heating, ventilation and air conditioning (HVAC) services are the primary use of energy in buildings. Natural ventilation is far better from an environmental perspective than mechanical ventilation. The fuel used for heating is also a consideration, generally coal is worse than oil which is worse than gas. Consider forms of renewable energy such as using photovoltaic panels to heat water, wind turbines or geothermal ground pumps. Combined Heat & Power also helps as it is more efficient than using fuel solely to produce heat.

Lighting is also a big user of energy. Natural lighting should be maximised in buildings where possible and energy efficient light fittings are preferred.

Refrigeration systems, for example cooling in air conditioning often contain gases that can contribute significantly to climate change. While CFCs (a major contributor to the depletion of the [[[ozone layer]]) have been all but phased out, their replacements are not perfect. Refrigerant gases should be treated with care and not allowed to escape to the air. Care must be taken when disposing of redundant refrigeration equipment, particularly older stuff.

Old electrical transformers contain Poly Chlorinated Biphenyls which are a major carcinogen.

Many older buildings (generally prior to the 1980s) may contain asbestos, particularly in heating plant and insulation to pipes.

Older heating systems often utilised underground oil storage tanks and these have a significant chance of leaking into the ground. Newer storage tanks for oil must be bunded, that is located in an enclosure that can trap the contents if the tank leaks.

Water usage is also key. Some cooling systems such as cooling towers can use a large amount of water.

Also consider wastewater disposal. It is better to have separate drains for foul and surface water as surface water can generally be returnerd straight to rivers while foul water has to be trated, which used energy. Where surface water is drained from car parks or roads, oil interceptors should be used to ensure water is not contaminated.

This is only scratching the surface but hopefully it will inspire you to come up with some more ideas. Let me know if you have any specific questions. GaryReggae (talk) 22:53, 27 November 2007 (UTC)

Gadget Show phone comparison

Last night's Gadget Show featured a comparison of three multimedia phones. To compare the music capabilities of the devices, identical MP3 files were loaded onto the phones, then each phone was connected in turn to a sound desk in a music studio and the track was played through the studio's speakers. Since the link between the phones and the sound desk was presumably digital, I expected there to be no difference at all in the quality of the playback - it seemed it should be just like plugging three different makes of USB flash drive into your laptop. Indeed, I failed to see what the point of the test was. Yet the testers claimed to notice differences in the quality of the sound depending on which phone the track was "played" on. What am I missing here ? Gandalf61 (talk) —Preceding comment was added at 15:12, 27 November 2007 (UTC)

Why do you presume that the connection from the phone to the sound deck was digital? My guess would be that it was analog, but I did not see the show. -- Coneslayer (talk) 15:27, 27 November 2007 (UTC)

Good point. Because they were using a fancy sound desk, I assumed that the connection would be something more sophisticated than a 3.5mm jack plug - but if was just an analogue link, the whole test makes a lot more sense. Gandalf61 (talk) 13:38, 28 November 2007 (UTC)

The available output of the phone would be the limiting factor; they could have had the super-whiz-bangiest desk in the universe, but I'd be surprised to find a mobile phone with a digital output. --LarryMac | Talk 20:12, 28 November 2007 (UTC)

Why do amps kill you, but volts don't?

Like when being hit with an electroshock weapon. Why is it that a large number of amps can kill you, but a large amount of volts do not? 64.236.121.129 (talk) 15:56, 27 November 2007 (UTC)

I usually simplify electrical stuff to water and plumbing. I feel this helps people understand it - even though it is not a perfect analogy. Voltage is roughly equivalent to water pressure while amperage is roughly equivalent to the amount of water that is flowing. I can take a squirt gun and hit you with high pressure water, but not hurt you at all because there is very little water flowing. Alternately, I can hit you with 500 gallons of water at very low pressure and you'd definitely feel it. So, you can see that water pressure can be felt, but it is the quantity of water that is required to cause damage. Similarly, high voltage can be felt, but is high current that kills. -- kainaw™ 16:08, 27 November 2007 (UTC)

Regarding the analogy you and others have used: I know you admitted the analogy is not perfect, but I should point out that small streams of very high pressure water can cut through titanium. See the water jet cutter article. Can very high voltage with low amperage do damage in the same way? -- HiEv 20:56, 27 November 2007 (UTC)

That's more or less the kind of explanation I was going to attempt. All I can think of to add is that we have articles on electric current and voltage which also explain the difference, altho not necessarily specifically in the context of injuries they cause. Oh, it looks like Electric_shock is the article most relevant. Friday (talk) 16:11, 27 November 2007 (UTC)

Yes - the electric shock article also covers what kind of physical damage high electrical current can cause - which I didn't mention in any way. -- kainaw™ 16:15, 27 November 2007 (UTC)

(Darn - beaten to it by an edit conflict!)

The Hydraulic analogy often helps here: Voltage is like the pressure of the water, Amps (current) is like the amount of water flowing. Ohms (resistance) has to do with the diameter of the hole or pipe through which the water is flowing (small pipe - lots of resistance, big pipe - less resistance). Ohms law says V=IxR (V=Voltage volts, I=Current in amps, R=Resistance in ohms). In water-analogy terms, more pressure (voltage) comes about when there is a lot of resistance to the flow or when a lot of water is flowing.

When a pipe bursts, it's the volume of water that causes the damage to your basement (When you touch the bare wire it's the Amps that kill you). But even if the pressure in the pipe is huge - if it's squirting out through a tiny pinhole - it doesn't bother you much because not much water comes out. (If the voltage is high, then so long as the resistance is high, you don't get many amps). But if the pressure in the pipe is high and there is a HUGE hole (so not much resistance), then lots of water is going to flow and you're in trouble. (If the voltage is high and the resistance is low then the current is high and you're in trouble). So voltage does matter. The other part of the analogy is that if whatever is pumping the water has a limited capacity - so even if there is a big hole in your water pipe, if there isn't much pressure then not much water ends up in your basement (So if the power supply is a little 1.5v AAA battery, then even if the resistance is low, not much current will flow and you'll be OK).

The pressure of water can be high (like inside a coke can that you just shook up), and the resistance can be low (like you suddenly pulled the tab on the can) but because there is only 8 ounces of liquid in there - the resulting high current won't flow for very long. This is like one of those static electricity demonstrations where there is a million volts (lots of pressure!) built up in a nice shiney dome - and as you touch it, a spark ionises the air (making a low resistance path) and you get a very short, sharp 'zap' of current - then it's all discharged (the coke can is now empty).

SteveBaker (talk) 16:19, 27 November 2007 (UTC)

I see, thank you. How do Watts fit in? Is there a water analogy for it too? 64.236.121.129 (talk) 16:38, 27 November 2007 (UTC)

Watts represent power, so there are still water analogies. A huge river (high current) can be flowing by you at quite a low pressure (voltage) yet still represent a large amount of mechanical power. A fire nozzle or water cannon, spraying a relatively small volume of water (low current) but at very high pressure (high voltage) can also represent a large amount of mechanical power. But a very large volume of water that's just sitting there with no pressure (zero volts) isn't doing any work (right now) nor is a tank of water at very high pressure with no outlet (zero current).

Atlant (talk) 16:48, 27 November 2007 (UTC)

(edit conflict) "Voltage" is nothing. It is "potential difference", that's all. Nothing happens. Some electrons over there want to get over here, and how bad they want to is what we call voltage. It's when the electrons actually move that the party begins. That's current. Current generates heat and disrupts the body's electical stuff, the heart's most importantly because that can kill you quick.

It does seem counterintuitive that 5,000 volts can leave you unharmed and 300 volts can kill you, but the truth is that the harmless kind of 5,000 volts is not really 5,000 volts at all. A power supply, be it a circuit or transformer or battery, can only provide so much current all at once, not infinite current. As soon as you exceed its capacity, its voltage goes down. A supply rated at 5,000 volts at one milliamp will only stay at 5,000 volts if you draw one milliamp or less, and will fall in voltage as necessary to maintain that maximum level of current if you try to draw more. Or smoke, pop, and melt. Or blow the fuse. On the other hand, if you get across 5,000 volts from one of those power company transformers that looks like a refrigerator, we're talking closed casket.

Another thing to bear in mind is that the human body has a fixed resistance, and that will dictate the maximum current that can be drawn at a given voltage. This means that no matter how many 12-volt batteries you strap together in parallel, they can't hurt you if you get across them with your hands, even if their current capacity is a zillion amps. At the body's 50,000 ohms, you get 240 microamps, period. With dry hands, one hand on each pole, it takes about 300 volts at 100+ milliamps to have a good chance of stopping your clock. Wet hands or something like taser darts bring the lethal voltage way down, because the resistance is less and the lethal current of about 100 milliamps can be more easily produced. So in this respect you can say that it is the voltage that kills you.

When I said that "nothing happens" with voltage, that was for voltages we're likely to encounter in our everyday lives. Something like lightning is a bit different. Electrical current creates a field in its vicinity (you can feel the effects of a field by holding a hairy arm near the front of a TV screen). If the field is strong enough and sudden enough it will make the electrons in your body move violently all by itself. People get knocked out or even killed by near misses of lightning all the time. --Milkbreath (talk) 17:00, 27 November 2007 (UTC)

So if you dump water or salt water on someone, it will decrease his electrical resistance, and thus increase the amount of amps he recieves? Even if he is shot by the same exact electroshock weapon? 64.236.121.129 (talk) 17:37, 27 November 2007 (UTC)

Yes. Of course, if it's the kind that sticks barbs into your flesh, it won't matter how wet or dry you are. But the cattle-prod kind will be more effective on a salt-water soaked suspect because that will negate the insulating effect of any clothing in the way. Bear in mind that these weapons are pulsed, that is, they provide current in many very, very quick bursts, which is a different ball game from a constant current. We can take a lot more current in little doses. Also, the current mostly runs between the contacts, and so doesn't get to the heart or brain so much, meaning that such weapons can run a higher voltage than a person could take right across the chest. --Milkbreath (talk) 17:56, 27 November 2007 (UTC)

What if it's an electrolaser? 64.236.121.129 (talk) 18:58, 27 November 2007 (UTC)

Then it's insult to injury. Once the holes are burned clean through you, you'll welcome the anaesthetizing effects of the electricity. --Milkbreath (talk) 19:29, 27 November 2007 (UTC)

Big Idea of Particles, Mass Conservation

Hi im afraid that i have three q.that i would like to ask. I would firstly like to know, what is the big idea of particles? I h ave searched in books and on the internet and i have so far found nothing that can help me. I would also like to know as to how mass is conserved in a reaction between an antacid tablet and the hydrochloric acid in ones' stomach. I would finally also like to know about a particle diagram.

Thank you for your help in advance.

P.S. Please try to explain this in simple terms.

I believe mass is conserved as long as it is converted into another form. Either another form of matter, or into energy. 64.236.121.129 (talk) 16:42, 27 November 2007 (UTC)

That's a pretty general question. What do you mean by particle--are you asking about the physics concept of subatomic particles, or about something else? Also, mass is not conserved--conservation of mass is a historical theory which, like Newton's physics, is close enough in most situations. In the same way that Newtonian physics start to fail obviously at high velocities and tiny scales, mass becomes very obviously not conserved in nuclear reactions. What is conserved is energy; a bit of mass is converted into energy, released as heat and light. What you're talking about seems to be stoichiometry--does that article help at all? And can you provide an example of what you mean by a "particle diagram"? grendel|khan 17:18, 27 November 2007 (UTC)

Let's be careful not to confuse the OP with teeny-tiny details here. Mass IS conserved to a truly spectacular degree of precision through almost all 'everyday' events. Utterly, vanishingly tiny amounts of mass are interchanged with energy in rather obscure ways relating to relativity and other stuff - but if we are talking about normal, mundane, day-to-day stuff like antacid pills in your stomach - then it's accurate to say that mass is conserved. This is a valuable principle that one should not toss out in ones zeal to be modern and utterly correct. In a chemical reaction - whatever atoms you started with are the atoms you have left at the end. None appear from nowhere or vanish abruptly...unless you are looking inside a particle accellerator or a nuclear weapon or something much more bizarre. As far as 'classical' physics and chemistry is concerned, mass is conserved. SteveBaker (talk) 18:26, 27 November 2007 (UTC)

When antacid is mixed with acid, you get a chemical reaction that produces some salt, some water and maybe some carbon dioxide or something - but no mass is lost. The mass of the antacid plus the mass of the acid is PRECISELY equal to the total mass of the byproducts you are left with at the end. The atoms that were present in the antacid and the acid merely rearranged themselves. Sodium bicarbonate (a common antacid) is NaHCO3 which means that each molecule contains one sodium atom (Na), one hydrogen (H), one carbon (C) and three oxygen (O). The acid in your stomach is hydrochloric acid (HCl) with one hydrogen (H) and one chlorine (Cl) atom. When an HCl molecule meets an NaHCO3, you get CO2 + H2O + NaCl - which is Carbon dioxide, water and common table salt. No more acid, no more antacid - just salty water and a small (but hopefully polite) <burp>! But at the end, you still have one sodium, two hydrogens, one carbon, three oxygen and one chlorine atom...they just rearranged themselves. Nothing was lost or gained in the process. SteveBaker (talk) 18:17, 27 November 2007 (UTC)

If you take a sealed, air tight, container and put acid and antacid (a base) in it but not mixed, and weigh it, then mix them and allow them to react and then weight them again, the weights will be the same. That is concervation of mass. It's all still there just rearranged. (Note: this assumes the containter doesn't explode or something because of gas pressure, that could be dangerous). RJFJR (talk) 19:08, 27 November 2007 (UTC)

Inability to Scream

Is there a medical term for someone who has the inability to scream? I am not referring to any dreams nor any type of sleep or awakenings. --WonderFran (talk) 17:42, 27 November 2007 (UTC)

That seems like a rather unlikely condition - can this person talk? Talk loudly? Shout? Shout loudly and incoherently? Shout "Aaaaaaarrrrggghhhhhhh" with steadily increasing pitch? 'Cos if so, they are screaming. It's hard to imagine any kind of condition that would strike selectively at one part of that process. SteveBaker (talk) 18:08, 27 November 2007 (UTC)

Spasmodic dysphonia can inhibit certain vocalizations while not affecting others. As I recall, Scott Adams was able to give prepared speeches, yet unable to speak conversationally. I don't know about an inability to scream, specifically. -- Coneslayer (talk) 18:20, 27 November 2007 (UTC)

Actually, it is me. I have no problem speaking in normal tones. I am in a girl in my late 20's but I have an unusually high pitch, soft voice. My mother had the same thing, too. The problem is that no matter what I try to do, I cannot sound my age. When I speak, people assume that I am much younger or dim witted and it's hard for me to get respect. My dentist had mentioned that the roof of my mouth was unusual but I don't know if that affects it. All my life, I cannot scream at all. It's wierd. I remember my teachers in elementary school trying to get me to scream for a school play and I couldn't. I tried to scream lot's of times, even tried to scream into a pillow and I cannot scream at all. It almost sounds like I am exhaling really loud. --WonderFran (talk) 18:24, 27 November 2007 (UTC)

This seems as though it would be general dysphonia, not spasmodic. There is no ICD9 or ICD10 code for an inability to scream. It is classified as dysphonia-unspecified. As a rule, we do not diagnose your specific problem or offer treatment. Dysphonia is merely the medical term for a voice disorder - not a diagnoses of the reason for the disorder. If you are concerned about this, seek advice from a medical professional. -- kainaw™ 18:32, 27 November 2007 (UTC)

Correction - I have just been told by an ENT surgeon that it is adductor spasmodic dysphonia and that it is not very rare. -- kainaw™ 18:36, 27 November 2007 (UTC)

Argh! You blew it. Now we know it's you - this is a medical diagnosis question and we aren't allowed to answer it. If it bothers you - see a doctor. But - did you ever wonder: perhaps your 'exhaling really loud' is sound so high in pitch that you can't hear it? SteveBaker (talk) 18:37, 27 November 2007 (UTC)

How long do you think it will take for someone to delete this whole thread even though no diagnoses were given? -- kainaw™ 18:42, 27 November 2007 (UTC)

You guys are funny! I was looking for a medical term, not a medical help. The only reason why I gave more info was because I wanted to give Steve Baker a better unerstanding of what I was referring to. --WonderFran (talk) 18:49, 27 November 2007 (UTC)

Well, I'm still curious. So indulge me - let me ask some questions - without diagnosing anything. You can shout - right? Can you shout the word "Aaaaaaarrrrrggggggggghhhhhhh!"? If not, what words can you shout? But if so, can you just gradually increase the pitch and volume of your voice? If so, then at this point, I think we have what I would call a scream - but if you can't do that then at some curious point in this process, there was something you were unable to do - some kind of threshold. I can't get my head around where that point is exactly. SteveBaker (talk) 22:03, 27 November 2007 (UTC)

This is a touchy subject because some editors are far too sensitive to the whole "medical advice" issue and some are happy to hand out medical advice without any expertise in the area at all. Everyone else is stuck in the middle between some editors blanking threads and others diagnosing and prescribing treatment. -- kainawBPG7WY 18:52, 27 November 2007 (UTC)

Seriously, she's just asking for a term. Don't be a rules nazi. 64.236.121.129 (talk) 21:17, 27 November 2007 (UTC)

Stop gratuitously insulting people (or is "Nazi" a term of endearment to you?). Kainaw was speculating that other people would think she had violated the rules, and would therefore delete this thread. Seeing as how Kainaw provided a very useful response, it is clear that he or she does not personally believe that a rules violation occurred. -- Coneslayer (talk) 21:57, 27 November 2007 (UTC)

Voice Immodulation Syndrome? :) shoy (^words _words) 23:41, 27 November 2007 (UTC)

Make sure you dont have The Tingler inside you! [7] ARRRRGGHHH! —Preceding unsigned comment added by TreeSmiler (talk • contribs) 02:48, 28 November 2007 (UTC)

By asking for a diagnosis, how can this question be anything but asking for medical advice? 199.76.152.229 (talk) 04:41, 28 November 2007 (UTC)

It is far too common for editors here to have a complete misunderstanding of what a "diagnosis" is. This person asked for the medical term for a physical property. This is no different than asking for the medical term for an ear ache or the medical term for a broken toe. It is not a diagnosis. If, however, she said she couldn't scream and asked what was causing it, it would be a request for a diagnosis. Of course, a person can present a diagnosis without asking for one. She could have said that she was in a car accident and asked if that could lead to an inability to scream. Again, it is not a request for a diagnosis, it is a question about the possibility of one event causing another - similar to asking if I can get liver damage from drinking too much alcohol. I hope that it will be possible for editors to eventually understand what is and what is not a diagnosis. -- kainaw™ 13:08, 28 November 2007 (UTC)

This is just to answer Steve Baker's question. I can shout but not that loud and sometimes it feels very straining to do so. Butwhat came to mind that men in general do not have the ability to do a high pitch scream like some women can, especially in horror films. I don't know if this helps, but I also have a horrible singing voice. Its sounds like cats drowning or mating or something.... :) --WonderFran (talk) 17:45, 28 November 2007 (UTC)

That didn't stop Yoko Ono from singing (sorry - couldn't help it). If you believe that this is purely physical and you have no health problems, you could work with a voice coach. They specifically handle this sort of problem. -- kainaw™ 18:56, 28 November 2007 (UTC)

Can both of you have your discussion about the Desk's policies on the talk page, and leave the Desk for answering questions? TenOfAllTrades(talk) 19:45, 28 November 2007 (UTC)

I wouldn't be sure that most woman can scream in a high pitch just because they seem to do in horror movies. It might be just a pre-recorded effect by some other actrees, like the Wilhelm scream. – b_jonas 16:23, 29 November 2007 (UTC)

What happened to the Michael Pursinger page?

Awhile ago Wikipedia had an excellent page on Dr. Michael Pursinger, a scientist who researches how electromagnetic fields affect the human brain. I'd like to know why this page was taken off Wikipedia; Dr. Pursinger's work is at the cutting edge of psychological research and so it definitely merits to be put back online. ````M.P.

Michael Pursinger does not seem to have been deleted. Is it spelled correctly? RJFJR (talk) 19:01, 27 November 2007 (UTC)

Michael Persinger? Dragons flight (talk) 19:03, 27 November 2007 (UTC)

Thanks a million! I've seen his name spelled differently on other websites (namely the CBC's) that's what was throwing me off... best 216.123.137.85 (talk) 20:00, 27 November 2007 (UTC)M.P.

What was the other spelling? Perhaps we need a redirect. Graeme Bartlett (talk) 02:38, 28 November 2007 (UTC)

We did need one, and we got one, which is why the two links are now both blue. -- 02:41, 28 November 2007 (UTC)

Accessory cell (immunology)

What is an accessory cell (in immunology)? --Seans Potato Business 19:18, 27 November 2007 (UTC)

My medical dictionary says: SYN antigen-presenting cell. (EhJJ) 20:16, 27 November 2007 (UTC)

That's what I had gleaned from other sources too. If anyone knows otherwise, they should go to the antigen presenting cell page and remove the addition I made (and also delete or rewrite the redirect). --Seans Potato Business 20:54, 27 November 2007 (UTC)

Liquidfy Diamonds? Solidify oxygen?

Is it ever possible, in our universe somehwere, to liquidy diamonds? Or to solidfy oxygen? --WonderFran (talk) 21:05, 27 November 2007 (UTC)

Diamond is an allotrope of carbon. Presumably, carbon can be liquid, but if it's liquid, it's not diamond anymore. Oxygen presumably has a freezing point, but I don't see it given in the article. - whoops, it's there, under "melting point". Friday (talk) 21:16, 27 November 2007 (UTC)

To rephrase on the point above (in simpler terminology), the name Diamond specifies a how the carbons are bonded together. So by definition, diamond must be solid. Someguy1221 (talk) 02:59, 28 November 2007 (UTC)

According to our article, carbon has a melting point that's higher than it's boiling point - so the only way to get it to liquify would be under huge pressures (it says 10 megapascals in the infobox). Normally, it'll presumably sublimate - which is to say go straight from a solid to a gas (imagine how 'dry ice' (solid CO2) turns into CO2 gas without ever becoming a liquid...that kind of thing). I don't think it matters whether it's diamond, graphite, amorphous or buckyball carbon. "Somewhere in the universe" could mean deep in the heart of a planet - with ridiculous temperatures and pressures - so probably there is liquid carbon somewhere. SteveBaker (talk) 21:50, 27 November 2007 (UTC)

Certainly going from the sublime to the ridiculous there Steve! —Preceding unsigned comment added by TreeSmiler (talk • contribs) 02:40, 28 November 2007 (UTC)

I remember answering a somewhat similar question on Usenet once... ah, here it is: "Boiling diamonds?" (The first link in the message is broken; the article it's supposed to go to, IIRC, is this one, but unfortunately the full text there is subscription-only. You could just Google for "phase diagram of carbon" instead.) —Ilmari Karonen (talk) 18:45, 28 November 2007 (UTC)

Or just see this article (Ghiringhelli et al., "Modeling the Phase Diagram of Carbon", Physical Review Letters 94, 2005). —Ilmari Karonen (talk) 18:52, 28 November 2007 (UTC)

Oyxgen evidentally has quite a number of different solid forms. See [8] and [9] Nil Einne 11:43, 3 December 2007 (UTC)

Why doesn't lightning damage antennas when it hits them?

When lightning hits a tree, it can make it explode sometimes. But when it hits an antenna at the top of a sky scraper, no damage occurs. Why is this? 64.236.121.129 (talk) 21:33, 27 November 2007 (UTC)

Lightning is just electricity. When electricity flows through metals, it passes quite easily due to their low resistance. When electricity flows through a tree - the resistance is considerable - so a lot of electrical energy is lost - and converted to heat. Hence the tree gets very hot. If the tree has lots of sap, it may even boil - and when that happens, a lot of steam has to go somewhere in a hurry and KABLAMMO! (That's a scientific term :-) SteveBaker (talk) 21:43, 27 November 2007 (UTC)

So if lightning hits a person covered in salt water, it'll prolly kill him, but won't make him explode. But if he's wearing some thick coat, he might explode? 64.236.121.129 (talk) 21:46, 27 November 2007 (UTC)

Probably not. Humans are rather good conductors (mostly salt water all through), and they also do not resist expansion well enough for a good explosion anyways. Also, as far as I know (but I may be wrong), much of the current will travel along ionized air on the outside of a human. --Stephan Schulz (talk) 22:04, 27 November 2007 (UTC)

( edit conflict) Lightning is unpredictable. Sometimes it mainly flows over the surface, sometimes it does you like a Presto Hotdogger. Antennas often have a spark gap at the base of the mast consisting of two points (> <), one connected right to the transmission line, and the other to ground. The points are just far enough apart not to arc when transmitting, but close enough to arc when lightning strikes. See lightning, lightning rod, spark gap, and lightning safety.--Milkbreath (talk) 22:12, 27 November 2007 (UTC)

Since Lead is a poor conductor of electricity, what would happen if lightning hit a chunk of lead? 64.236.121.129 (talk) 16:47, 28 November 2007 (UTC)

All metals are "good" conductors. Silver, copper, gold, and aluminum are great conductors. Lead has no problem conducting in your car battery. Too, there's lightning, and then there's lightning. It can range in intensity from a piddling couple of million volts up to a billion volts or more. The snappiest strikes can melt a copper antenna no problem, but they are rare, and they tend to hit the mast instead of the antenna, anyhow, because the mast is grounded. I would expect a piece of lead of reasonable size to disappear in a puff of incandescent lead vapor with any decent strike. Lightning is scary stuff. --Milkbreath (talk) 17:05, 28 November 2007 (UTC)

The article on lead said that lead is a poor conductor of electricity. Did it just mean, compared to other metals? Also since it has low resistance, shouldn't it be unharmed by the lightning? I thought lightning only damages objects that have high resistance. 64.236.121.129 (talk) 17:37, 28 November 2007 (UTC)

Poor compared to other metals. Lead has an electrical resistance of 208 nano-ohms per meter. Iron is at 96 and Silver is 16. So, yeah - lead is a poor conductor compared to iron or silver. But take a typical non-metal like (say) Sulphur and it has a resistance of 2x10¹⁵ ohms per meter (that's 10²⁶ times worse than lead!!) Compared to sulphur, lead is a spectacularly good conductor! SteveBaker (talk) 20:23, 28 November 2007 (UTC)

I just did some math, and if a lightning bolt carrying 300,000 amps of current passes through a lead stick one centimeter in cross-sectional area and one meter long, you get 189 million watts dissipated in it. That should be enough to turn it into plasma, don't you think? --Milkbreath (talk) 20:44, 28 November 2007 (UTC)

Don't you also have to factor in voltage? 64.236.121.129 (talk) 21:08, 28 November 2007 (UTC)

The voltage across the lead stick is determined by the current from the bolt; the current from the bolt is determined by the total voltage of the bolt (which doesn't depend at all on what you fire it through) and the total resistance in the bolt's path; the resistance in the bolt's path is almost entirely within the air, not the stick, so the stick has a somewhat negligable effect on the total current of the bolt. It might affect how much of that current goes in, across, or around the stick, but it won't really have an affect on the current (and thus, on the voltage across the stick). Someguy1221 (talk) 21:19, 28 November 2007 (UTC)

Right. All the current would probably go through the stick because its resistance is very much less than that of the air around it. As for the "voltage", power is calculated by P=IE (current times voltage). Algebra and Ohm's law let us do away with voltage if we know the current and the resistance, giving P=I²R. --Milkbreath (talk) 21:29, 28 November 2007 (UTC)

If lightning hit glass, it should shatter it right? But would it penetrate it, and hit whatever was behind it too, or would the glass stop the lightning from advancing further? 64.236.121.129 (talk) 17:43, 28 November 2007 (UTC)

The lighting will keep going until the current reaches the ground. As far as the lighting is concerned, the glass is just another, bothersomely-high-resistance medium to get through on its way to the ground, just like air and trees. --Chris, 20:48, 28 November 2007 (UTC) —Preceding unsigned comment added by 63.138.152.238 (talk)

Lightning is unpredictable. If it wants to go through glass, it will, blowing it to smithereens in the process, I suppose. Thing is, lightning follows a leader, and the leader wouldn't form through glass in the first place. --Milkbreath (talk) 20:57, 28 November 2007 (UTC)

I think the lightning strikes before the leader makes it all the way to the ground, and just arcs through the air. In any case, the lightning isn't going to just stop and have the ions sit next to the glass. — Daniel 02:23, 29 November 2007 (UTC)

I wouldn't say that lightning is unpredictable - it's HARD to predict - but it's not some magical thing, the laws of physics that it follows are well understood. I imagine that the kind of thing that'll happen with a window in (say) an aluminium frame - is that the lightning will be trying to find the lowest resistance path to the ground. That's not the glass - it's the frame - so the frame gets hit - a bazillion amps of current flows - and the frame duly melts, then boils in a very little amount of time. This heat will shatter the glass because it can't expand evenly enough - just as if you pour boiling water into a cold jamjar. As soon as the glass opens up a small crack - the lightning can head through (presuming there is a lower resistance route to ground inside than outside). Your bathroom window is probably next to some nice copper pipes that lead in exactly the right direction - so if your house got struck by lightning, this is a very possible outcome. SteveBaker (talk) 03:05, 29 November 2007 (UTC)

Lightning hasn't read those physics books you put so much faith in. It can hit a chicken standing on a rubber mat two feet away from a 300-foot-high grounded tower if it wants to. The underlying physics may be known, but real-world chaos makes the event unpredictable. --Milkbreath (talk) 03:38, 29 November 2007 (UTC)

That's nonsense - and you know it. "if it wants to"?!? It doesn't "want" anything - it's an application of field theory - and it most certainly follows the same laws of physics as the rest of the universe. If you set up an FEM model of the tower, the air and the chicken with the appropriate laws of physics built into your mesh and cycled it at a small enough time step, you'd be able to predict pretty accurately where any given lightning bolt would disperse to. It's not a chaotic system at all. Not all systems that are complicated are also chaotic (and not all chaotic systems are complicated) - that's a horrible misconception. People can and do model the behavior of lightning. My first job out of college was doing research into telephony systems - and there were people where I worked who calculated how a lightning strike on the telephone exchange would propagate out into the phone lines - and vice-versa. SteveBaker (talk) 04:21, 29 November 2007 (UTC)

Lightning is unpredictable not because it has volition. I'm not a voodoo witch doctor, in case you thought that. Not chaotic? Are you saying that there is a model for the progress of a leader through the atmosphere that can predict its path? That is absurd, and you know it. "According to my calculations, it will strike right over there", he declared confidently, only a moment before a rogue bolt leapt from a thunderhead 10 miles away and vaporized his pointing finger. --Milkbreath (talk) 11:51, 29 November 2007 (UTC)

I didn't say it would be easy to predict - you'd need some insane amount of data about the air on the way down - but it's not SENSITIVELY DEPENDENT on that data in a way that would make it chaotic. Maybe you need to read some more about chaos theory. Not all complicated, hard-to-predict things are chaotic. SteveBaker (talk) 14:10, 29 November 2007 (UTC)

Who said no damage would occur to an antenna from a direct strike? Induced currents can however be mitigated by connecting the mast to earth. I beleive this is normal practice in the professional and ham radio world. —Preceding unsigned comment added by TreeSmiler (talk • contribs) 03:46, 29 November 2007 (UTC)

Heres a link [10]--TreeSmiler (talk) 04:03, 29 November 2007 (UTC)

Theropod dinosaurs and birds

Only Theropod dinosaurs are considered to have evolved into birds right? Other dinosaurs like Sauropods and Cerapods aren't. Correct? Hmm, if this is true, can we consider Sauropods and Cerapods to be 100% reptiles, while Theropods to be part reptile, part bird?64.236.121.129 (talk) 21:37, 27 November 2007 (UTC)

We have an article on bird evolution about this. (And yeah it looks like theropods are considered their ancestors.) I don't know that the second half of the question is very meaningful- we lump things together as reptiles or birds because it's a useful classification. These classifications may be fuzzy around the edges. Friday (talk) 21:47, 27 November 2007 (UTC)

Of course it's meaningful, otherwise we wouldn't classify animals at all. 64.236.121.129 (talk) 21:54, 27 November 2007 (UTC)

We classify, and often reclassify as our understanding improves. Theropod classification is described in the article. They are considered dinosaurs, which are considered reptiles. Friday (talk) 22:03, 27 November 2007 (UTC)

Perhaps what Friday meant by "meaningful" was monophyletic, while your suggestion of some species being "part reptile, part bird" would be paraphyletic along the lines of saying something is "part fruit, part apple". --Sean 00:15, 28 November 2007 (UTC)

The sauropods almost completely died out before the Cretaceous, and the final species died out in the Cretaceous–Tertiary extinction event. Other interesting suborders are the ornithopods (bird-feet, with bird hips and duck bills) which also died out, and don't seem to have connection to modern birds besides analogous evolution. SamuelRiv (talk) 23:17, 27 November 2007 (UTC)

Sauropods didn't die out before the Cretaceous. Argentinasaurus lived in the middle of the Cretaceous for example. 64.236.121.129 (talk) 16:32, 28 November 2007 (UTC)

Why does Carbon nanotube have stronger properties than buckyballs?

They are made out of the same matter right? So why the differences? Would a buckyball have better compression strength due to its shape? 64.236.121.129 (talk) 22:00, 27 November 2007 (UTC)

Buckeyballs contain both hexagonal rings and pentagonal rings, whereas carbon nanotubes ideally contain only hexagonal rings. Hexagonal rings are significantly more stable, and thus require greater strain to break. Indeed, pentagonal rings of carbon are naturally under stress (see ring strain) already due to their non-ideal bonding configuration. Adding further to that, forcing carbon rings to form spheres or cylinders instead of sheets also strains the bonds, and I suspect this to be greater in a buckeyball than a nanotube. And finally, buckyballs aren't resonance stabilized, whereas nanotubes are. Someguy1221 (talk) 10:51, 28 November 2007 (UTC)

Is there any shape that is potentially stronger than a hollow cylinder, since a hollow cylinder is weak in compressive forces? 64.236.121.129 (talk) 17:39, 28 November 2007 (UTC)

I don't know much about that specific question, but it pays to be cautious when considering questions like this. It sounds like you're assuming that what's true on a large scale would also be true on a molecular scale, which is always a dangerous assumption. Friday (talk) 21:24, 28 November 2007 (UTC)

I'm well aware of that. 64.236.121.129 (talk) 20:38, 29 November 2007 (UTC)

When people talk about carbon nanotubes being strong, they're referring to tensile strength, not compressive strength. — Daniel 02:15, 29 November 2007 (UTC)

Yea, I know. They have terrible compressive strength because they are hollow cyclinders. It's naturally bad at compression. That's why I'm asking if there's a better shape to handle that kind of stress, while maintaining CNT's other strong properties. 64.236.121.129 (talk) 20:38, 29 November 2007 (UTC)

How come the buckyballs aren't resonance stabilized? Don't they also have delocalized electrons like the tube or graphite? – b_jonas 16:03, 29 November 2007 (UTC)

How much would someone weigh on mount everest

I weigh 130 pounds and for science class i need to know the formula to calculate how much i would weigh on the summit of mount everest? Any Help???? —Preceding unsigned comment added by 71.219.144.66 (talk) 22:23, 27 November 2007 (UTC)

See Newton's law of universal gravitation RJFJR (talk) 22:28, 27 November 2007 (UTC)

On Mount Everest, your weight will not change much. You will weigh the same as much you weigh here. Though, if you want to be more precise, the value of 'g' will slighly change, because distance between you and the centre of earth has increased, but this will be very insignifacnt because radius of earth is much more than the additional 8 KM that you would add if you reach Mount Everest, so there is not much difference in your weight here and on mount everest. However, your weight will drastically change, if you reach moon. The reason is because of the difference in radius and mass of the moon, that account for your weight on its gravitational field. For further clarification, read article posted by RJFJR and other related articles. DSachan (talk) 22:37, 27 November 2007 (UTC)

If you really want to be pedantic, you can try adding another quite small term for the gravity caused by Everest itself. Historically, trying to directly measure the gravity of mountains was one of the methods used when attempting to quantify Newton's constant, G. Dragons flight (talk) 23:02, 27 November 2007 (UTC)

A gravity map of earth's surface I found here, suggests that the large mass of earth beneath you largely outweighs the decrease in gravity caused by the altitude gain. So you would weigh more on everest then somewhere else. 71.214.181.11 (talk) 03:52, 28 November 2007 (UTC)

Actually, you are misinterpreting a poorly explained graphic. What that shows, I'm virtually certain, is the gravity anomaly from "normal" gravity that general models would predict for each particular latitude and elevation. Not variations is the acceleration of gravity in an absolute sense; not variations from some constant value. These anomalies are much smaller in magnitude than the variations with latitude and elevation. Gene Nygaard (talk) 15:49, 29 November 2007 (UTC)

See Earth's gravity#Altitude (Why doesn't this link work right?). --Milkbreath (talk) 22:43, 27 November 2007 (UTC)

Corrected your link for you. (EhJJ) 22:56, 27 November 2007 (UTC)

For the purposes of your science class: the fact you need to know is that gravity decreases in strength in proportion to the square of the distance you are from the center of gravity of the source. So - your weight at sea level is proportional to one over the radius of the earth squared. Your weight at some altitude is proportional to one over the radius of the earth at sea level PLUS your altitude - all squared. We can lose the annoying constants of proportionality by dividing one equation by the other. Hence your weight at altitude A as a percentage of your sea level weight is: 100 x R²/(R+A)² - so go to Earth and look up the radius of the earth, go to Everest and figure out it's height - get the percentage increase and add that to your present weight. Problem solved.

BUT: We have to whine and complain a bit - because this is the reference desk and we all want to show how complicated things really are! So:

• The earth isn't spherical. It's an oblate sphereoid. You might think this is negligable - but the difference between the polar radius and the equatorial radius is about 20km. Everest is less than 9km tall. So the weird shape of the planet has more to do with where you are in latitude than where you are in altitude!
• The mass of mount everest itself is not negligable. Normally one may assume that the center of gravity of a spherical body is at it's center. But not so when you are standing really close to a GIGANTIC chunk of rock. So while your altitude will reduce your weight a bit...the mass of Everest itself will increase it some. Is this negligable? I don't know and I can't be bothered to work it out.

SteveBaker (talk) 01:04, 28 November 2007 (UTC)

"Increase", I like that.

A really complete, nitpicky solution would also allow for the variation in one's weight with latitude, due to the Earth's rotation.

On the other hand, a counter-nitpick is that if the initial weight is given as "130 pounds", it's a number with only two significant digits and therefore the answer should be rounded o the same precision. In which case it's going to be 130 pounds.

--Anon, 02:54 UTC, November 28, 2007.

I think that the amount of mass underneath a mountain should be about the same as the amount of mass under non-mountain areas, because the crust floats on top of the mantle, so as with all things that float, the crust must displace its equal weight in the underlying mantle. So if the mountain were not there, it would be filled with the same weight of mantle. --Spoon! (talk) 11:59, 28 November 2007 (UTC)

This is true only on the large scale (e.g. ~100 km or so), on the moderate to small scale, the crust is rigid enough to support local anomalies (i.e. mountains) that deviate from bouyancy. That is what the gravity map that an anon linked above is getting at. Dragons flight (talk) 17:50, 28 November 2007 (UTC)

The gravity map was made by a pair of satellites. It doesn't depict the gravity at the elevation of the ground - it measures it at the altitude of the satellite. So it's not taking account of the altitude of the mountains at all. Hence we don't know (without knowing what the colours on that map mean numerically - and doing some more math) whether the effects of the mass of mountains outweighs (hehehe!) the effects of their altitude. SteveBaker (talk) 20:16, 28 November 2007 (UTC)

An even bigger but than Steve Baker's above: But for real-world purposes such as the medical sciences and sports, in contrast to his "for purposes of your science class", your weight on Mount Everest is the same as it is anywhere else: 130 lb, or in the units used for this weight throughout the world including many hospitals in the United States, 59 kilograms. Gene Nygaard (talk) 14:06, 29 November 2007 (UTC)

I also find it rather strange that so many teachers would choose to try to make a point about the strength of the gravitational field, involving that jargon meaning of the word "weight". It is Mount Chimborazo which is the highest mountain on Earth in both ways relevant to this question, something like 2,000 meters farther from the center of mass of the Earth, and hundreds of kilometers farther from the axis of rotation.

Reminds me of an example Aviation Week used to have on their aviationnow.com website (apparently lost in reorganization), concerning the "weight" in the physics jargon sense (even though they used kilograms for the units of this weight, and the kilogram-force is no longer an acceptable unit) of an elephant at the Denver Zoo and at the San Diego Zoo. Problem is, they got it all wrong. An elephant of the same mass will exert more force due to gravity at the Denver Zoo than at the San Diego Zoo, not less. The author of that piece, like the teachers who use Mount Everest as an example, failed to understand the simple fact that latitude is a more inportant factor in this regard than elevation above sea level is. Gene Nygaard (talk) 14:21, 29 November 2007 (UTC)

It should be noted that "pounds" can refer to "pound mass", which is the same regardless of gravity, or to "pound force", which is dependant on gravity. From the original question I think we can assume that by "pounds" they meant pound force, not pound mass. -- HiEv 04:06, 2 December 2007 (UTC)

November 28

Nasal cauterization

When I was quite young I underwent a few treatments of nasal cauterization. I don't remember much about them except that they turned my nose a deep shade of purple that needed to be washed off later (?). What method of cauterization was this? Would it have hurt? (I don't remember—I was quite young, just the purple nose remains in my memory and the notion that it was cauterization of some sort.) --24.147.86.187 (talk) 04:52, 28 November 2007 (UTC)

Its all a bit vague, but here's some possibilities:

1. The purple stuff was an antiseptic, not a cautery agent
2. Whatever the purple stuff was for, the only thing I can think of that might fit the description is potassium permangenate (KMnO₄). Once upon a time it was used as an antiseptic agent, and as I recall it also has some caustic properties, and so may have been used as a chamical cautery agent.
3. Chemical cautery is most often performed with silver nitrate (AgNO₃). Phenol is another popular agent. Not usually very painful.
4. Electrocautery is painful.
5. Cryocautery is only mildly painful
6. If the method they were using was considered painful, its more than likely they used some local anaesthetic prior to the procedure.

Mattopaedia (talk) 05:59, 28 November 2007 (UTC)

So mine is electrocautery eh? Unanaesthetised it felt someone's trying to poke a hole through the middle of my nose. --antilived^{T | C | G} 10:57, 28 November 2007 (UTC)

Also, silver nitrate is a lovely shade of grey, but might conceivably look purplish depending on your skin colour and your memory. Check out this picture; I could see someone calling that "purple". Matt Deres (talk) 11:53, 28 November 2007 (UTC)

I am not a doctor, but I've been used/abused/experimented on by many. Many invasive procedure (meaning that you've been opened some place you shouldn't be open, or that they are going to open you, etc) start with "Scrub area thoroughly with Iodine solution". Betadine is a common brand for such. The solution is a reddish-brown, and under some light, it might look purple. -SandyJax (talk) 15:52, 29 November 2007 (UTC)

Electrochemistry

what is the difference between Formal and standard reduction potential? —Preceding unsigned comment added by 59.163.146.11 (talk) 11:22, 28 November 2007 (UTC)

Standard reduction potential is the reduction potential at 25^oC, 1M concentration of both anion and cation. Formal reduction potential is the actual reduction potential at whatever specific conditions you're concerned with. Someguy1221 (talk) 15:23, 28 November 2007 (UTC)

Brain activity: reading vs TV

What are the differences in active areas of the brain or general brain activity between watching TV and reading a book/article? Have McLuhan's hot and cold medium any basis in brain activity as I think he claims? What are the resulting differences between watching TV or reading for 4 hours a day? Keria (talk) 12:29, 28 November 2007 (UTC)

• The brain has different centers for processing images and text. Each activity thus stimulates a different part of the brain, though invariable there will be some overlap. - 131.211.161.119 (talk) 14:47, 28 November 2007 (UTC)

whey protein

I wasn't sure if this should go in science or misc., but here goes: How many servings of whey protein should I have a day? When should I take it (morning and night, before working out, after, etc.). Should I drink it every day, or just the days I work out? All help would be greatly appreciated. --MKnight9989 13:51, 28 November 2007 (UTC)

The answers to your questions could depend on a number of factors, such as your age, gender, current diet, exercise regime, training goals, medical conditions, etc. Your best bet is likely to speak with a physician, dietician or athletic trainer. Our article on whey protein also has several external links which may be able to provide some insight. (EhJJ) 14:36, 28 November 2007 (UTC)

I'll probably ask my doc, but I'm 17 and weigh approx. 140lb. I want to be reasonably strong (I'm enlisting in the USMC after highschool, so I want to be able to survive boot camp), but by no means do I need to have arms a foot in diameter. I usually take in 2000-3000 Calories a day. --MKnight9989 14:45, 28 November 2007 (UTC)

I'm sorry if I gave you the impression that we would be able to answer your question if you gave us more information. Rather, at the Wikipedia Reference Desk, we can not give you medical (including dietary) advice. We are usually more than happy to aid your understanding of a concept or interpret information that you do not understand (in fact, one of the main reasons this reference desk exists is to find which Wikipedia articles need improvement), but we can not give you regulated advice. Essentially, think of us as volunteers at your local library. I would never even think of asking my librarian for medical, legal or (perhaps especially) body-building advice (I've never met a "built" librarian, but they could be out there). Some of us may be doctors or lawyers (I, for example, am a medical student), but law (and Wikipedia policy) prohibits us from giving you medical or legal advice online. Good luck! (EhJJ) 20:17, 28 November 2007 (UTC)

We may decide not to answer the question, but whether we do or not, it won't be because it's medical advice, since it involves neither diagnosing nor treating a medical condition. Perhaps we should simply change our prohibition to "we don't give advice". - Nunh-huh 20:20, 28 November 2007 (UTC)

As far as I'm concerned, you or anyone else is free to tell him whatever you'd like, but I wouldn't listen to it relating to my personal health. Do you know how tall he is? What if I told you he's 4'10" and 140lbs. Is your advice different from 6'2" and 140lbs? What if he has phenylketouria or any number of other health conditions? Now, I have no problem providing information but I don't feel comfortable providing specific instructions to him. That said, I did point him to our whey protein article and it's external links plus said that he's probably best of talking with a trainer. Was there something wrong with that? (EhJJ) 21:31, 28 November 2007 (UTC)

No, the only thing you've done wrong is suggest that advice on nutritional supplements is medical advice. If it were, thousands of health food store clerks would be in shackles. Not all advice with medical implications is "medical advice". - Nunh-huh 22:37, 28 November 2007 (UTC)

Well, I hope it's OK to direct you to the National Institutes of Health's recommendation, contained somewhere in here. I don't think it specifically mentions whey protein, but it's the amino acids that are important. Someguy1221 (talk) 20:51, 28 November 2007 (UTC)

Chemicals Released while burning...

Hi,

I have read through your articles about Chrome and Nickel Plating, they helped answer my question on how plating is done and the chemicals involved therein but the one thing I have not yet found out is if there are any chemicals (especially piosonous ones) released when an item plated in either Chrome or Nickel is burnt in an open flame. For instance a grill on a fireplace/Barbecue. The companies that I have spoken to that do the plating say that although they have never done any tests themselves they do plate for a certain brand of Barbecue and have never heard of any issues.

Your assistance is much appreciated, Tyron —Preceding unsigned comment added by 41.243.189.182 (talk) 14:01, 28 November 2007 (UTC)

Chrome or nickel plating is safe enough for cooking on even in an oven or barbeque. The chemicals used in plating are toxic, but of your new product is well made and washed it should be safe for cooking. There would be limits to a safe temperature to heat it to, and once the grill is left outside and it rusts, the plating could come off in sharp fragments that will be nasty to eat. Graeme Bartlett (talk) 20:10, 28 November 2007 (UTC)

In terms of toxicity, I would care far more about soot from the fire. Icek (talk) 20:46, 28 November 2007 (UTC)

Do you know what the temperature limit that it could be burn to would be? As an open flame can reach at least 400°C. Jet (talk) —Preceding unsigned comment added by 41.243.189.182 (talk) 05:59, 29 November 2007 (UTC)

The articles Chrome plating and Electroless nickel plating might be of help. shoy (^words _words) 17:13, 29 November 2007 (UTC)

Nickel has the sentence "Nickel is reacted with carbon monoxide at around 50 degrees Celsius to form volatile nickel carbonyl". So if CO comes into contact with hot nickel a poisonous gas is formed.Polypipe Wrangler 08:23, 1 December 2007 (UTC)

N-glycosylation

Can someone point me to one or more methods I can use to identify the sugar residues on the N-glycosylation sites of a viral protein? I considered using specific enzymes (endo-proteases?) to cut the outer residues in order to find out the attached sugars, but that would be quite a laborious process. Has anyone got a better idea? - 131.211.161.119 (talk) 14:45, 28 November 2007 (UTC)

Matrix-assisted laser desorption/ionization mass spectrometry (MALDI-TOF) is one such technique. 142.20.217.152 (talk) 22:01, 28 November 2007 (UTC)

How would I know which masses are from the protein and which ones come from the glycans? Would this mean a lot of preparation to obtain a suitable sample to put in the mass spec? - Mgm|^(talk) 23:11, 28 November 2007 (UTC)

There are many published articles that describe methods for analysis of oligosaccharides such as this one. One approach is to use a glycosidase to cut off the oligosaccharides and then use some form of chromatography to isolate the oligosaccharides prior to mass spectrometry. --JWSchmidt (talk) 00:37, 29 November 2007 (UTC)

• I'm not sure why this article didn't show up on my pubmed search, but at first glance it appears to be just the sort of thing I'm looking for... Thanks to both of you for your input. I'm still open to more ideas, though. —Preceding unsigned comment added by 131.211.161.119 (talk) 08:30, 29 November 2007 (UTC)

Thermodynamics of heating your house

What's more energy efficient?

• turning the heat down to 58ºF every evening and turning it back up to 68ºF in the morning
• not turning it down too much so that it doesn't have to re-heat the whole house from scratch every morning

This is a practical question of mine. Basically we have the heating set to go down to 58ºF every evening starting at 10PM, and then at 6AM it is set to rise back up to 68ºF. Is this an optimal way to do things (assuming that we want it at 68ºF during the day and not freezing at night)? It seems like it has to do a lot of work in the morning to raise the temperature again—is it better to keep the temperature a bit higher so that it doesn't have as much work to do? Is there a more optimal heating solution? Surely one of you science geeks will have a good answer for this. ;-) And though I know it sounds like a textbook homework question, it isn't—I don't do homework anymore, thank goodness. --24.147.86.187 (talk) 15:27, 28 November 2007 (UTC)

If you have a very simple furnace, it will necessarily require more energy to keep the house warm at night than to cool it up in the morning. Over the course of the night, the house will release, say, X units of heat into the environment when the furnace is off. So the furnace has to generate X units of heat during in the morning. If you leave it on at night, the house will initially be losing energy at the same rate. But hot things lose heat energy faster than cold things (my simplest way of putting Newton's law of cooling), so the house will lose more than X units of heat over the course of the night, so the furnace will have to generate more energy overall to keep the house warm. This could be complicated by a furance that can run at multiple "speeds" as with many new gas furnaces. Such a furnace would have a different energy efficiency for the two situations (ie, it would waste a different percentage of energy heating a warm house or a cold house). Someguy1221 (talk) 15:39, 28 November 2007 (UTC)

Interesting. Well, it's not a new furnace at all (it is quite an old boiler) so it is probably the case that turning it down and up again in the morning is the better approach. Sigh. --24.147.86.187 (talk) 03:33, 30 November 2007 (UTC)

Trust the mass on a bottle label?

I often purchase small amounts of relatively pure chemicals, usually 50mg or so. I need to make solutions of these chemicals at a known concentration. The thing I have never been sure about is whether chemical supply companies take care to put as close as possible to 50mg (or whatever is on the label) into the bottle. You see, life is easier if I can assume that the right amount is in there, and I can add water directly to the bottle to disolve it and make my solution. If I have to transfer it, inevitably I won't be able to recover some from the bottle or the weigh paper, and additionally my balance isn't the best. What do you think, is the stated mass very close to the actual mass in the bottle? ike9898 (talk) 15:31, 28 November 2007 (UTC)

It's going to depend, a great deal, on the particular product and the particular supplier. Usually you won't be getting shortchanged, but the margin over and above the label amount can vary. How precisely do you need to know the quantity and concentration? Is there some sort of assay you could perform (spectrophotometric, etc.) on the stock solution to determine its concentration to an acceptable precision? TenOfAllTrades(talk) 17:12, 28 November 2007 (UTC)

Concurred with TenOfAllTrades: customers would make a hell of a noise if they often found themselves paying for something that wasn't all there. When you say you "need to make solutions of these chemicals at a known concentration", does that mean you have to hit a pre-set (specific) concentration, or you need to be near some value and also know exactly what it is? The latter is much easier, since you can make and then assay. Or else you could weigh the full bottle, then dissolve in the bottle (so you don't lose material) and transfer into another container. Wash the bottle, reweigh it, and now you know the tare and hence the exact (to your instrumental ability) amount of stuff in the solution (you lose a diluted drop or two rather than a few grains of pure material). DMacks (talk) 17:32, 28 November 2007 (UTC)

I wouldn't be able to assay afterwards because this solution is supposed to be the analytical standard. I like the idea of weighing the bottle before and after. ike9898 (talk) 18:53, 28 November 2007 (UTC)

I suppose this goes without saying, but if you're getting your chemicals from a single supplier, you probably only have to weigh one bottle once and then use that as your tare in the future. Obviously, that would depend on how accurate you need everything to be, but if it was essential, I'd presume you'd have a more precise scale for weight and not need to ask! Matt Deres (talk) 02:40, 29 November 2007 (UTC)

The mass of small reagent vials from the same lot usually vary on the order of a few milligrams unless they are specifically matched. ike9898 (talk) 15:04, 29 November 2007 (UTC)

Power lines, and electric shock

If you put one hand on a power line, you won't get shocked right? If you put two hands on the power line you still won't get shocked right? But if you put your two feet on the power line, then put your hands on another power line parallel to it, you will get shocked then right? 64.236.121.129 (talk) 16:37, 28 November 2007 (UTC)

You can get a shock from touching one power line, if you're in good electrical contact with the earth ("grounded"). -- Coneslayer (talk) 16:44, 28 November 2007 (UTC)

What if you are standing on a brick? 64.236.121.129 (talk) 16:49, 28 November 2007 (UTC)

It depends on the electrical conductivity of the brick. I would guess that that might depend on whether the brick is damp. -- Coneslayer (talk)

Don't most shoes have rubber soles? Wouldn't that prevent grounding? Also if you are hanging from the powerline what would happen if you were hanging from one hand, and two hands. 64.236.121.129 (talk) 17:01, 28 November 2007 (UTC)

Also aren't powerlines covered in some kind of rubber usually? Wouldn't this prevent shock? 64.236.121.129 (talk) 16:50, 28 November 2007 (UTC)

According to Overhead powerline#Conductors, most overhead power lines are uninsulated. -- Coneslayer (talk) 16:56, 28 November 2007 (UTC)

You should certainly consider them to be uninsulated, even if they might be insulated. The power company guys certainly think this way: the only time they think it's safe to touch a power line is after they've clamped big grounding cables onto it and proven that the line is then grounded. Before that, they strictly do all their work with their hot stick.

Atlant (talk) 17:56, 28 November 2007 (UTC)

Yes although there is a fairly recent development for LV OH cables called semi-insulated cable. This means they dont trip the breaker when they momentarily touch together. But I wouldnt advise any one touches them. In fact the answer on higher voltage cables is that it is a waste of time insulating them as youre going to be dead before you touch them, and most of the electric field will still be from the outside of any 'insulation' to earth (because of the capacitive divider effect).--TreeSmiler 01:54, 1 December 2007 (UTC)

Oh, look: we didn't have an article on hot stick yet. I've stubbed one out; help me fill it in! (And somebody tell Rockpocket.) —Steve Summit (talk) 01:33, 29 November 2007 (UTC)

You ONLY Get shocked if you bridge the gap between two large differences in voltage. One hands or two hands or one hand and one foot on the same power line will have NO result, as long as no other part of your body is touching something else with different voltage potential like the Earth, another power line, etc. Shoe rubber is a fairly unreliable insulator, unless you are wearing something specifically designed to protect you from shock. Disclaimer: DO NOT TRY THIS AT HOME TO PROVE ME WRONG --Jmeden2000 (talk) 17:55, 28 November 2007 (UTC)

This is where I get confused though. The current is still passing through you if you are touching the power line with limbs touching the same power line right? Or... maybe it isn't. So how come a bird doesn't get shocked by this, but a bird can get shocked by a lightning bolt hitting it, when it is flying, and isn't grounded. 64.236.121.129 (talk) 21:16, 28 November 2007 (UTC)

If you were made of aluminum, you'd have a problem. As it is, though, your resistance hand-to-hand is about 50,000 ohms. So, worst case, if you stretch out your arms and grab the line, your 50,000 ohms will be in parallel with about a meter of aluminum as thick as a pool noodle at roughly zero ohms. I wouldn't advise doing that, though, because I'm not sure what role skin effect plays at very high voltages in a case like this. I'd keep my hands close together. I think you'd feel the induced current from the 60-cycle (or 50) field as an unpleasant whole-body sensation if the voltage was high enough.

When a bird is hit by lightning, it's because it was unlucky enough to have been in the path of the leader when it flashed over. The bird becomes part of the conductor, which is a line of ionized air between the cloud and the ground or between two clouds. --Milkbreath (talk) 00:53, 29 November 2007 (UTC)

Power lines (the high voltage ones at least) do not have electrical insulation because electrical insulators also tend to be good heat insulators and it's essential that these lines remain cool. They also have to remain flexible and there are a bunch of other constraints on them that make insulation impractical. SteveBaker (talk) 19:53, 28 November 2007 (UTC)

(edit conflict) Read Ohm's law and alternating current. They are pretty straightforward, but alien to everyday affairs. Once you get them under your belt, you can answer these questions yourself.

Whether you get shocked or not touching a bare conductor will depend on many things, principally the voltage. High-tension lines can have more than million volts on them. The rule of thumb for arc distance in dry air is 10,000 volts per inch, so a million volts will jump 100 inches, or about eight feet. If something more conductive than air enters that zone between a high-tension line and ground, like YOU, for instance, the arc will see you as a shortcut to ground and use you as part of the path. Think how cool you'll look with a blue arc as big around as your arm entering the top of your head and incinerating your Air Jordans on the way out. Your wafer-thin shoes would present no obstacle to that kind of power. Even the best insulator has a breakdown point, and the current would be able to pass through the air around the soles, too.

Another thing is that even if you could approach a million-volt line without being near ground, like in a balloon, the line is going to want very much for you to be at the same potential as it is, and it won't wait for you to touch it to accomplish that. The power line repair guys use great big gloves and a long wand to draw the arc that causes, mostly because they want to live through the experience.

A lot of things are dangerous, but you can get away with messing with them sometimes. Bears, for instance, or ex-girlfriends, or cliffs. Electricity has its own agenda and its own rules, and you get one chance not to screw up. --Milkbreath (talk) 20:02, 28 November 2007 (UTC)

A bird can perch on an uninsulated electric wire and not get shocked, because no current flows through it, because there is no place for the current to flow to.

An ungrounded human can do the same, for the same reason.

There are several ways a human can be ungrounded:

• being in a balloon or helicopter
• jumping in midair
• wearing rubber-soled shoes (though as noted above, their insulating properties vary)
• standing on something insulated, like a wooden bench (though again, the insulating properties vary)
• standing in the basket of an insulated-boom cherry picker

And, there are also several ways for a human to be grounded, and these are why electric shocks are possible (and rather common, among the careless):

• being barefoot, especially while standing on concrete or in water (or on a brick)
• accidentally touching something else grounded, like a water pipe
• being in a bathtub
• accidentally touching someone else who is grounded

And, not only do the insulating properties of various materials differ, but the amount of insulation you need obviously depends on the voltage you're worried about. For "ordinary" household voltages, you can engage in risky behavior and not get electrocuted too much of the time, because (for example) a good pair of tennis shoes will usually withstand 120V. (Although I'm not sure about the 220V that's common in Europe. Can you tell there's some OR here?) But when you get up into the high hundreds of volts or more, the rules are rather different, and unless you know exactly what you're doing and are very careful and use all the right equipment, you can easily get not only a lethal shock, but also electrocute innocent bystanders who try to rescue you by grabbing your energized carcass, or incinerate them in the fire your charred carcass starts.

Finally, there's the matter of inrush current. Even if you're perfectly ungrounded, when you touch an energized wire, there's a relatively small, instantaneous flow of current which brings you up to the potential of the wire. The bigger you are, and the higher the voltage is, the larger this current is. So I think there's a voltage at which the birds are fine, but a human would get a painful (if not fatal) shock from the inrush current. —Steve Summit (talk) 00:13, 29 November 2007 (UTC)

There is also the field problem. Even at low frequencies, polarized molecules like water can be moved by electric fields. They move more at resonant frequencies like in the microwave oven (think what happens to food in a perfectly insulated plastic box. Put a metal fork in that plastic insulated box and you'll really see what electric fields can do, but I digress.) IF the voltage is high enough and the person close enough, this movement can damage internal structures of living organisms even at low frequencies and without even touching the wires. Birds actually die before landing on these wires. I have only read about the bird deaths occuring in South American transmission lines but even in North America, it appears birds find the very high tension lines to be an unpleasant experience and seem to stay off them. A cool video is to watch how they repair remote high tension lines by helicopter. They first electrically attach the helicopter to the wire and there is usually a three foot arc when the pole gets close. --DHeyward 07:09, 4 December 2007 (UTC)

Cool Video of guy touching high tension wires after being delivered by helicopter --DHeyward 07:18, 4 December 2007 (UTC)

Future of Everyday Travel

Is there a reasonable possiblity that there will be Sky Highways in next 30 years? Meaning flying cars or hover crafts. Is it possible to use solely solar power to fuel these crafts or that would be impossible? --WonderFran (talk) 16:47, 28 November 2007 (UTC)

It's another one of those things where when engineers say "It'll be here in 5 years", it'll probably be here in less. When they say "It'll be here in 20 years" they mean "I have no idea at all when it'll be here - if at all". So, yes, we'll have sky highways within 20 years!

But seriously - the fuel requirements for a flying vehicle are unlikely to ever be as good as they are for a ground vehicle - because you have to generate lift as well as overcome friction and drag. With the pressure to make more fuel-efficient transport, using flying vehicles seems wasteful. Having said that, the ability to travel in a straight line to your destination, the fact that there is more space up there (so no traffic jams) could override that - and I suppose we could get our lift from balloons instead of using motors.

There is also the safety issue. If you have a fender-bender - or any kind of an accident at 30mph on less - or if your car craps out on you on the freeway - you aren't going to die. You'll probably walk away without a scratch. At 10,000 feet, any problem at all will probably be fatal. The people who are building these things say that they'll have multiple redundant engines and that kind of thing - but I've seen cars on the road with TWO of those 'skinny spares' on them driving at 70mph! I've met people who actually wore out a skinny spare bald because they drove on it for 10,000 miles! What happens when your flying car is that poorly maintained?

As for solar power - there simply isn't enough area on the roof/hood/trunk of a car to capture enough energy to propel it through the air at the kinds of speeds that we are used to. The best solar cars go slowly and have interiors that are stripped down to basically being bicycles with supports for the solar panels. No amount of clever engineering can cover for the fact that there simply isn't enough area. Then what about driving in overcast conditions - or at night? At best, you might have a 'Solar Hybrid' that runs on batteries that can be recharged from the sun with a gasoline or hydrogen engine as a backup.

But I doubt we'll see flying cars in our lifetimes. SteveBaker (talk) 19:50, 28 November 2007 (UTC)

The NASA Pathfinder could fly on solar power, and the Zephyr could even store enough energy to fly through the night (but they had to strip out the pilots to make them light enough, though). Maybe Solar Impulse will change that, but I'm not sure if this will ever be, well, safe (or economical, or not a pain in the ass). And in addition to all of the speculations on what fantastical yet-to-be-invented-inventions we'll all be using in the future, remember that it was once speculated that there'd be an autogyro in every American garage by now. Someguy1221 (talk) 20:31, 28 November 2007 (UTC)

Well, yes - but you wouldn't describe an amazingly flimsy plane with a wingspan bigger than a Boeing 747 and a top speed similar to a bicycle that can only fly in calm air with no rain as "practical"! There is a really fundamental problem here - in order to have enough solar cells, you need HUGE wings. In order to get off the ground with so little power, the plane has to be super-lightweight. Super-lightweight with HUGE wings means two things: Obviously, it's going to be very, very fragile because you can't afford the weight to make it strong. Secondly, it's going to have a very low "wing loading" and planes with low wing loading are extremely vulnerable to turbulance. These are not characteristics you want for a mass-produced civilian plane - but what's the alternative? Even 100% efficient solar cells wouldn't gather enough power to do this properly - so there is a really fundamental limit. Increasing the area increases the weight - but increasing the weight increases the power requirement which forces you to increase the area. There is no way out of that! SteveBaker (talk) 21:43, 28 November 2007 (UTC)

I'll just maintain that "pain in the ass" = um, all of what you just said ;-) Someguy1221 (talk) 21:56, 28 November 2007 (UTC)

I agree - but some things are just not meant to be. There is this blind faith that science can always improve things and make magic happen - but sometimes there just isn't. We MIGHT someday make solar cells that are 99% efficient and we MIGHT make stronger, lightweight materials - but the wing-loading issue is still there as is the fact that it's never going to fit into an urban environment with 200' wings. These are all pretty fundamental problems that put practical solar powered cars and aircraft permanently out of reach. Sadly, these incredibly flimsy machines give the public an altogether different idea - which is a shame. 100% computer controlled (but gas-guzzling) helicopters are probably the closest we'll get to flying cars - and those are going to ALWAYS be a lot more expensive than a mere ground-car - and given that we need to be driving fuel consumption DOWN, not UP, that's not a smart way to go. Perhaps hydrogen-filled, hydrogen-fuelled, mini-zepplins might work. SteveBaker (talk) 22:16, 28 November 2007 (UTC)

A minor nitpick: According to the second law of thermodynamics, solar cells are not going to be more efficient than 96% (the temperature of the solar radiation is about 5780 K, the ambient temperature for an aircraft at least about 220 K). Icek (talk) 22:38, 28 November 2007 (UTC)

Solar cells are not heat engines. The Carnot efficiency limit does not apply. There are technical and physical challenges to achieving high efficieny, but that is not one of them. Dragons flight (talk) 01:36, 29 November 2007 (UTC)

Eh - I think you are both wrong. Or maybe you are both right...I'm not sure which! The second law doesn't only talk about heat - it actually says that you can't convert 100% of heat into work - which includes electricity - so the second law definitely says you can't be 100% efficient...or at least it would apply if solar cells converted heat into electricity - but they don't. What solar cells are doing isn't converting HEAT into electricity - they are converting LIGHT (electromagnetic waves) into electricity. Photons knock electrons out of the silicon. However, they certainly can't be 100% efficient. Some photons pass right through the silicon, some reflect off the surface, others don't have enough energy to produce an electron so they turn into heat. SteveBaker (talk) 04:04, 29 November 2007 (UTC)

The Carnot efficiency limit does apply. Else I could simply surround the heat source usually used in my Carnot engine with a vacuum and solar cells. Light does have a temperature. If it's monochromatic, the temperature is zero, being analogous to mechanical energy. If it has a Planckian spectrum, it has the according temperature. Icek (talk) 11:45, 29 November 2007 (UTC)

Temperature (yes, sort of), but it's not a closed system. The Carnot limit is a statement about entropy conservation in closed systems. Not to mention that at no point is the solar cell in thermal equilibrium with sunlight anyway (compare to the cycle of a heat engine where the operating fluid itself varies in temperature from the high to low temperature). Yes, there are physical limits to the efficiency of solar cells. Some even relate to temperature (e.g. the spontaneous recombination of electron-hole pairs), but a solar cell is simply not a heat engine. Dragons flight (talk) 16:37, 29 November 2007 (UTC)

The Carnot limit only needs a warm reservoir - the incident radiation (or the surface of the sun if you like to view it like that) - and a cold reservoir - the surrounding air or whatever; no closed system is needed. It's a very general statement about entropy, and does not depend on the details of the energy conversion, as far as we know. A relevant Wikipedia article: Exergy. Icek (talk) 17:56, 29 November 2007 (UTC)

I think the Carnot efficiency applies, but not quite in the way you suggest. Solar radiation at the Earth's surface is actually more like 100°C: that is, an object at that temperature cannot be heated by sunlight because it emits as much radiation as it could absorb. However, radiation that is streaming in a particular direction (rather than suffusing an enclosed space) cannot be said to be in thermal equilibrium with itself (because it does not have appropriately random velocities), so it does not have a true thermodynamic temperature. This is good: it means that the Carnot restriction only operates on the source temperature, and we get your 96% number instead of something pathetic like 17%. But I'm not entirely sure that Carnot applies at all; a heat source in a vacuum is also not in thermal equilibrium (because it has not equilibrated with the vacuum), so it may have more free energy than the corresponding equilibrium situation with the same "temperature" (of the object). --Tardis (talk) 18:26, 29 November 2007 (UTC)

How did you come up with 100 °C? The greenhouse effect can occur in macroscopic solid objects as well, and I doubt that you couldn't get more than 100 °C. Anyways, that's beside the point, the solar radiation does in no sense have such a low temperature. I don't quite get what you are trying to say: Even if a solar cell was enclosed within a sphere or bubble of gas of a temperature of 5780 K, the solar cell would work as long as its temperature is kept low enough. Icek (talk) 22:24, 29 November 2007 (UTC)

100°C comes from the solar constant (I used 1kW/m² for sea level) and the Stefan-Boltzmann law. The greenhouse effect can only happen if the (say) glass is not at the same temperature as the contents, which again means that you do not have a single object in thermal equilibrium with itself, so its temperature is not well-defined. My point was that the solar radiation (here, away from the sun) does not have a temperature at all, for similar reasons; I'm not sure what your statement about a solar cell in an oven is supposed to establish. --Tardis 17:36, 30 November 2007 (UTC)

My statement about the solar cell surrounded by a thermal radiator was a response to your statement that the radiation does not have appropriately random velocities/momentums ...

You are defining temperature by the power of the radiation/equilibrium temperature of an object, I am defining it by the spectrum of the radiation. But however you put it, the Carnot limit with the spectrum-defined temperature does certainly apply. Icek 20:52, 30 November 2007 (UTC)

SteveBaker made some excellent points about efficiency, but there's more than just that. Flying cars exist; and if you paid me enough I would design a custom one for you. But most flying cars still need to take off from airports. How are you going to get to the airport? For a flying car to be practical it would have to be VTOL, that is, take off vertically from the top of your house. This adds considerably to the complexity of the aircraft, and therefore its costs. It also takes a lot of energy to take off vertically. And the safety concerns addressed above also are important: if a car's electrical system totally craps out, meaning the spark plugs aren't firing, the driver can pull to the side of the road. General aviation aircraft with internal combustion engines have totally redundant systems: dual magnetos with dual spark plugs in each cylinder, completely independent of one another. This is true of every system in an aircraft; safety through redundancy. It makes them very expensive. You could buy a flying car, and afford to operate it, if you were filthy rich, but in order to have an air highway system lots of people would have to be able to afford it. moink (talk) 02:50, 29 November 2007 (UTC)

Ok it wouldn't be as Zipadidooda but mini Zeppelins are relatively safe and fuel efficient, no zipping around though. Keria (talk) 09:20, 29 November 2007 (UTC) Maybe a blimp is a more appropriate term. Keria (talk) 15:32, 29 November 2007 (UTC)

The difference between a blimp and a zepplin is that the zepplin has a rigid structure where a blimp is held in shape purely by the pressure of the gas. The original name "blimp" comes from a British term for a particular kind of barrage balloon (designed to prevent enemy planes from flying low over cities) called a "B-type Limp Balloon" (limp as opposed to rigid). This got shortened to "B-limp" and the rest is history. I wonder what became of the Alimp?

Anyway - the cool thing about hydrogen as a lifting gas is that it's cheap, renewable and provides more lift per cubic meter than helium. You can also use it to fuel your motors - so you have this ENORMOUS gas tank and as you consume it, you are forced to gradually lose altitude such that you are always on the ground LONG before you run out of fuel...a handy built-in safety thing since a Blimp without engine power can be exceedingly dangerous! But more interestingly, the fact that you'd continually be using and replenishing your lift gas would go a long way to avoiding the key problem with hydrogen balloons - which is that oxygen can slowly diffuse into your tanks and if too much accumulates: KABOOM! With care, that could be completely avoided here since you can consume hydrogen AND the dissolved oxygen in your engines and always refill with pure hydrogen - so the oxygen levels can never build up to a dangerous degree so long as the vehicle is used reasonably regularly. A practical machine would use hydrogen fuel cells to make electricity and use electric motors to drive the thing around. When you "park" it at home, sensors could monitor the oxygen levels and when they approach the danger level, start the hydrogen fuel cells to consume it safely - putting the 'spare' electricity into your home or back into the electrical grid.

The huge problem is the size of the things and the top speed. Unless we're going to have HUGE parking lots - we'd need ways to tether and stack them. Also, 'runaway' blimps would be a serious matter in strong winds. SteveBaker (talk) 20:44, 29 November 2007 (UTC)

What happens during the cooking process (and what are those black bits)

So when you slap a piece of meat on the grill, it usually comes off the grill in some shade of brown, with black lines wherever it touched the grill. My general understanding of what happens when stuff is heated up is as follows:

1)various condensation reactions occur, releasing water 2)H2 gas comes off and burns, leaving unsaturated carbons behind. (this is why aromatic/carbon-rich compounds give a sooty flame - there isn't enough H2 given off to facilitate the burning of the carbon itself)

My theory is that those black strips are essentially carbon/aromatic compounds that form from the protein on the surface of the meat. If so, wouldn't those black bits be carcinogenic? Maybe it's not the consumption of red meat itself that causes cancer, but the consumption of those black bits. If so, would cooking on a surface that doesn't leave black bits be healthier/no cancer? (I also notice a shitload of powdery, black residue on the bottom of oven-cooked pizzas)

thoughts? 18.60.12.185 (talk) 20:14, 28 November 2007 (UTC)

Start with Browning (chemical process) --Mdwyer (talk) 20:20, 28 November 2007 (UTC)

And then hit The Straight Dope. -- Coneslayer (talk) 20:28, 28 November 2007 (UTC)

For some of the black products, see polycyclic aromatic hydrocarbon. Icek (talk) 20:54, 28 November 2007 (UTC)

Eating just before bedtime

Many people believe that eating before bedtime would increase weight gain. My thoughts are that (calories in) = (calories out)and that what matters is how many calories you consume in a day, not when you ingest them. Is there a physiological reason why eating at bedtime would increase fat/weight gain? Are there references available to explain this?

David Winkelaar —Preceding unsigned comment added by Davidwinkelaar (talk • contribs) 22:24, 28 November 2007 (UTC)

As far as I know, the thinking behind not eating big meals before bedtime is that the digestive system slows down significantly while we are asleep..however, it is often easier to sleep after having a big meal as the body diverts energy to digesting the meal (until you actually fall asleep!) GaryReggae (talk) 22:34, 28 November 2007 (UTC)

The process is (assuming you're not on the Atkins diet): Sugar enters your body. Sugar that your body needs is put into the blood and taken up by cells. Sugar your body doesn't need is turned into glycogen. When your sugar starts running low, your body converts that glycogen back into sugar (glucose, specifically) so your cells can use it. Glycogen is simultaneously being converted into fat. When you start running low on both sugar and glycogen, your body could just start burning that fat, but instead it decides to be a whiner and make you hungry so it'll get more sugar. The idea is that by eating at night, you increase the proportion of sugar intake that gets converted into fat, calories that your body would rather leave be and make you eat more than actually eliminate. If you actually make sure to balance your calorie intake with exercise, this won't be a problem, as your body will have no choice but to consume the fat once the glycogen runs down. Someguy1221 (talk) 22:59, 28 November 2007 (UTC)

Machine lubrication oil contaminates kitchen...

Yesterday, my special needs brother cut open a full can of general purpose lubrication oil (for machines and weaponry) in the kitchen, causing it to spray all over the kitchen counters and on a lot of our dishes and cookware. The label on the can says that it's toxic and should not be used on or come in contact with any surface used to prepare food.

My mom seems to think that a simple washing of everything will correct the problem, but I fear otherwise. My question is, what materials must be thrown away and which ones can be cleaned? The oil contaminated things made of glass, metal, wood, plastic, and rubber, as well as metal cooking implements with wood or plastic handles and non-stick pans. It also got on sponges and dish rags.

If any of these things CAN be cleaned, should I use any methods beyond dish soap and water?

Thank you. --69.207.99.230 (talk) 22:41, 28 November 2007 (UTC)

At the risk of offering something that may sound like medical advice, I agree with your mom. Soap and water will clean the stuff off of hard surfaces very well. Depending on the fabric, a good detergent should also be able to get it out of many cloth items. While motor oil must (in today's climate) be labeled as "toxic", it's not like Arsenic or Plutonium or anything.

For peace of mind, though, you'll want to discard any contaminated food. And if the oil soaked into anything porous (such as wooden cooking implements), you may want to discard them as well, because you probably won't be able to get all the oil out, and the odor and discoloration will be unpleasant even if the contamination doesn't kill you. —Steve Summit (talk) 23:28, 28 November 2007 (UTC)

Thanks. My boyfriend told me that the oil would be absorbed by and thus ruin all things that are plastic, but I don't know if he's being paranoid or not (one of the plastic things covered in oil belongs to him). He might just be looking for an excuse for my mom to replace it, though. Unfortunately for me, now I need to do the dishes... —Preceding unsigned comment added by 69.207.99.230 (talk) 23:38, 28 November 2007 (UTC)

Plastic tends to be nonpolar, just like oil, and there are some polar substances which can more-or-less permanently attach themselves to and/or intermingle with plastic molecules. You can notice this with some plastic cookware, especially those ubiquitous semitransparent food containers -- they eventually get discolored by some of the foods they come into contact with. (Oily foods reheated in the microwave are definitely the worst. I've seen this happen both with spaghetti sauces containing tomato products and beef or sausage oil, and Oriental foods containing chili oils.) But while this could be a problem in your situation, I doubt it would happen noticeably unless there were also heat involved. If after a good detergent wash, the plastic thing in question is still slippery or smells like oil, then it may be indelibly contaminated, but if not, it's probably fine. (But far be it from me to intervene in this heaven-sent opportunity for the bf to guilt-trip you, if that's how the stars have aligned here...) —Steve Summit (talk) 00:33, 29 November 2007 (UTC)

If you are looking for something stronger than dish soap, then I would recommend Simple Green. It is essentially an industrial strength degreaser, but also non-toxic, so unlike most comparable products, it would be safe to use on food surfaces. We use it in a variety of applications around our lab. It is sold in some grocery stores and most hardware stores. Dragons flight (talk) 01:09, 29 November 2007 (UTC)

Clearly the stuff can't be that toxic. People handle things lubricated with that oil all the time - it gets onto their hands and - yes - people do eat without washing their hands and they don't get sick as a result. I'd toss out things made of wood or cloth just because you'll never get them clean. Anything that's made of plastic or metal - run through the dishwasher on it's hottest/longest cycle. Anything that's metal or plastic that won't fit into the dishwasher, wipe off the excess oil then scrub down with some detergent - washing up liquid will be fine. SteveBaker (talk) 01:25, 29 November 2007 (UTC)

Just to reiterate a basic point... go ahead and clean the stuff you can (wipe off, then wash, etc.) and then use your brain. Look at the stuff that got splattered and ask yourself whether it looks, smells, and feels like it did before or whether it is different. If it's different: toss it. Blowing twenty bucks on some replacement Tupperware is a pretty small price to pay to avoid even a potential problem. In other words, why risk it? Matt Deres (talk) 02:51, 29 November 2007 (UTC)

November 29

How many SigFigs?

When performing the following simple addition:

    -260
   + 273.15

How many sigfigs should the answer, 13.15, carry? Intuitively, it should be 13. However, when this is inputted into the Science Tools Significant Figures calculator on a TI-84+, the answer says 10 (1 sigfig)

Which is correct? Thanks. Acceptable (talk) 00:09, 29 November 2007 (UTC)

Your calculator. The 260 has the tens place as its lowest significant figure, so the final answer should only be specified to the tens place. Someguy1221 (talk) 00:28, 29 November 2007 (UTC)

That's BOGUS. Sure, if the 'true' value is 261 and it was written as 260 then you have 2 digits of precision - but you don't know that. Perhaps the 'true' value was really 260.00000000 - it's written as 260 which is THREE significant digits. The fact is that with normal arithmetic symbolism, you simply can't tell whether it's 2 digits or 3. So you must calculate out to three digits in order to be sure you aren't losing information. It's not a "real" problem though - in any real situation, you know where the data came from and therefore how precise it is - and you know how much precision you need in the result. SteveBaker (talk) 01:16, 29 November 2007 (UTC)

Um, steve, that's what the period is for, dude. I'll agree with you though that this doesn't generally apply to "real" problems. Someguy1221 (talk) 01:22, 29 November 2007 (UTC)

OK - so you weaseled out of that one (damn!) - but I don't give up debunking bogosity that easily! How about this sum?

 -2600
+ 1234.45

Does 2600 have 2 or 3 significant digits? What about 26,000,000 ? You have no way to know - you are left guessing. If I tell you that the population of France is 26,000,000 people - you might guess that I gave you only two significant digits - but you don't know that there are really 26,012,345 people living there and I gave you 3 sig digits. In truth, if you aren't specifically told - you don't know. Teaching people that there are rules that tell you this stuff is BOGUS. SteveBaker (talk) 03:19, 29 November 2007 (UTC)

If it's possible for it to be ambiguous, you use scientific notation. :-p neener neener neeeeenerrrrrr. Anyway, I'm happy to say that I use plain old means and standard deviations when I do analytical experiments :-) Someguy1221 (talk) 03:23, 29 November 2007 (UTC)

Excuse me!?! I've never seen the population of France represented in scientific notation - NEVER! That's an utterly useless rule. SteveBaker (talk) 03:35, 29 November 2007 (UTC)

Population of France: 2.60×10⁷. Now you have! Dragons flight (talk) 04:51, 29 November 2007 (UTC)

Oh no - now you've gone and done it. Here, let me fix that number for you: 02.60x10.00^07.00 ...there, much better! SteveBaker (talk) 06:22, 29 November 2007 (UTC)

In the scientific-notation version, the "10" and "7" are considered exact numbers, giving them an infinite number of significant digits. --Carnildo (talk) 01:16, 30 November 2007 (UTC)

So the '260' in the example up top had infinite precision according to you? You'll probably say "No" because you happen to know that 10 and 7 were integers and you happen to believe that the 260 is not. My point being that you have to somehow know. If you aren't told then you're guessing. Not good. SteveBaker 23:54, 1 December 2007 (UTC)

By the way...how many significant digits does '2' have? You're going to say "just one". So if I want to add 2 to 101. - the rule is that I have to do it to one significant digit so the answer is only 100 - unless I write 2.00 + 101. then I get 103. ?? That's something else I've NEVER seen anyone do? What's relevent in addition is the number of digits before/after the decimal point - not the number of significant digits. This supposed formalism is utterly useless in practice - and it's definitely going to cause horrible mistakes. Why are we even bothering to teach it to people? SteveBaker (talk) 03:44, 29 November 2007 (UTC)

For an amusing real-world conundrum involving trailing-zeros ambiguity, see the "Measurement" section of our article on Mt. Everest. —Steve Summit (talk) 03:53, 29 November 2007 (UTC)

Oh sorry, that'd my mistake, I guess I forgot to mention that. (-260) has 2 sigfigs, while (273.15) has 5. Acceptable (talk) 01:19, 29 November 2007 (UTC)

New question: how do I inform the calculator that this particuloar 260 accurate to three significat figures, if "260" is two significatn figures, and "260.0" is four significant figures? -Arch dude (talk) 01:05, 29 November 2007 (UTC)

And 260. has three sig figs ;-) Someguy1221 (talk) 01:06, 29 November 2007 (UTC)

AFTER EC: Oh, I thought when adding/subtracting sigfigs, the answer should have the decimal places of the value with the least decimal places. Acceptable (talk) 01:07, 29 November 2007 (UTC)

No it doesn't. It's one less. If you are really adding 100.6 to 100.6 but we do the sum accurate to three significant digits, you get 101. + 101. = 202. - but the real answer is 201.2 - so your answer isn't accurate to 3 significant digits anymore - it's now only accurate to 2. When you add two inaccurate things - the result is less accurate than the two numbers you added. Real math isn't done with 'significant digits' - it's done with error bars. So the right way to do this is to say 100.6 can be represented as 101 plus or minus 0.4, so when you add 101 (+/-0.4) to 101 (+/-0.4) you get 202 (+/-0.8) - which means that you know the true answer is between 202.8 and 201.2...which is good. This whole business of "significant digits" is just a VERY rough way to represent error bars and to make sure people don't over-specify the precision of the result. But to pretend that this is some kind of mathematical formalism is nonsense (but totally typical of the nonsense they fill school-kid's heads with when they could be teaching them REAL math). It's main purpose in practical math, science and engineering is to allow you to do 'back of envelope' calculations and not be ridiculously over-precise in your results. When it matters, you use proper error bars with all the formalism that entails. SteveBaker (talk) 03:33, 29 November 2007 (UTC)

If you do as scientists often do and treat your errors are normally distributed, then the standard error propagation tells you that 101 +/- 0.4 + 101 +/- 0.4 = 202 +/- (0.4*sqrt(2)) = 202 +/- 0.6. Though, for that matter, I can't imagine anyone looking at 100.6 and rushing to 101 +/- 0.4. The default position would be to ask the uncertainty, but if that is not available the customary assumption (at least back in my lab classes) is to assume the last digit is meaningful but on the threshold of not being so and go with 100.6 +/- 0.5, which would give the sum as 201.2 +/- 0.7. Dragons flight (talk) 04:48, 29 November 2007 (UTC)

Indeed, for many sorts of error, a normal distribution is an appropriate error model - and I agree that if you use that then no answer is "wrong" because it could simply be off in the long tail of the distribution. Hence if you calculate the answer to my little sum to be 202 +/- 0.6, you aren't saying that the true answer must lie between 201.4 and 202.6 (as indeed, it does not) - you are merely saying that the odds are much better that the true answer lies within that range than outside of it. However, it's not always the case that a normal distribution is appropriate. For example if you are using a computer to crunch your numbers, the error due to finite precision machine arithmetic is sharply delimited and equiprobable - it's certainly not gaussian.

But whatever - the point is that if you actually care about the precision of your results, you do something about it. The whole "significant digits" thing is vague, poorly determined and patchily implemented. It's simply a rule of thumb to prevent gross over-specification of precision. Once people start getting dogmatic about how it's treated, the system simply falls apart because of those weaknesses.

The computer geek in me is particularly upset because the rate at which precision is preserved, amplified or casually discarded by this so-called system depends entirely on what base of arithmetic you use! If you do your math in binary - then knowing how many significant binary-digits the number has preserves precision better than knowing how many significant decimal digits it has - and MUCH better than doing it in hexadecimal or radix 50 or counting the number of bytes (radix 256). Since the choice of a base-10 number scheme is ENTIRELY arbitrary, anything which bandies about "significant digits" is doomed to failure in the face of proper numerical analysis. SteveBaker (talk) 06:40, 29 November 2007 (UTC)

Significance arithmetic summarises quite a lot of what has been said here. 130.88.79.39 (talk) 14:30, 29 November 2007 (UTC)

One of the references in that article [11] says: The whole notion of significant digits is heavily flawed; see section 9 for more on this. Anything that can be done by means of significant digits can be done much better and more easily by other means. People who care about their data don’t use significant digits. There are plenty of important cases where following the usual "significant figures" rules would introduce large errors into the calculations. ...which is precisely my point (although I'll admit to VERY informally using the significant figures "rule" when I'm doing strictly back-of-envelope math. SteveBaker (talk) 20:31, 29 November 2007 (UTC)

This addition problem doesn't really involve the number of significant digits anyway. That's a multiplication/division factor. What is inolved here is that we don't know that the digits of one of the numbers is significant beyond the tens place (it might be significant in the units place, but not beyond there or it should have included significant digits after the decimal point. Gene Nygaard (talk) 22:02, 29 November 2007 (UTC)

Particle Accelerators use electric fields to accelerate and magnetic fields to bend particles?

Hi, in reading about the Large_Hadron_Collider and the article on Magnetism, I'm wondering some simple questions, like... 1) If a Proton at rest sees an electric field, the proton accelerates towards the negative voltage. As the proton attains the speed of Light (c), the electric field starts looking like a magnetic field, right? 2) The Large_Hadron_Collider has a round track 26.6 km long that bends the moving protons using magnetic fields. As the proton attains the speed of light (c), doesn't the proton see that magnetic field more and more as an electric field?

Just curious. Thanks! --InverseSubstance (talk) 02:43, 29 November 2007 (UTC)

(1) not quite, and (2) no. To be more specific, it isn't that an electric field "looks like" a magnetic field - put simply, when you have some charged particles and you are in a frame of reference where they are at rest, they have an electric field and no magnetic field, that will exert a certain force on another charged particle. When you are in a different (intertial) frame of reference, those particles will be moving, but by the rules of special relativity the force they exert on that other particle should be the same, even though length dilation means that the separation between them appears to have changed, and thus their electric field is different. The magnetic field is essentially the difference between the two possible electric fields. So yes, if you start with a "pure electric field", then as you accelerate you'll see some of the electric field "become" a magnetic field. However, if you start with a magnetic field, then as you accelerate the magnetic field will change, as will the associated electric field, but the one won't completely turn into the other. Confusing Manifestation(Say hi!) 02:53, 29 November 2007 (UTC)

One thing - the proton doesn't attain the speed of light - it can get close but it can never quite get there. SteveBaker (talk) 03:09, 29 November 2007 (UTC)

Extended/Extra Optical Zoom?

I bought Panasonic Lumix DMC-FZ18 recently, I have heard that its got "Extended Optical Zoom" but it works with lower resolution output only. http://panasonic.co.jp/pavc/global/lumix/fz18/18zoom.html . Does anyone know whats the difference between Extended/Extra Optical Zoom and Digital Zoom?

Thanks --Spundun (talk) 03:59, 29 November 2007 (UTC)

It's just precropped digital zoom. Uh, so basically, it's digital zoom :-) Someguy1221 (talk) 04:09, 29 November 2007 (UTC)

EOZ isn't truly digital zoom - it doesn't "stretch" pixels like digital zoom does. It is only used when you have chosen to take smaller pictures than your camera can take - for example, to save space on your memory card. Normally in this case the camera "squashes" pixels. However, if you wish to zoom in, the camera will use only pixels from the middle of your sensor - no "stretching" required. 138.38.151.57 (talk) 12:32, 29 November 2007 (UTC)

So in other words, it takes a full picture and crops it - that's "electronic zoom" - calling it "optical" is a nasty deception! Electronic zoom is by far inferior to optical zoom - but it's cheaper. If this were an optical zoom, the optics would have reconfigured to cover the entire sensor - then if a lower resolution were demanded by the user, "squashing" (as you put it) could be performed. However, by merging together a larger number of sensor pixels to get one stored pixel, you improve the quality of the image. SteveBaker (talk) 13:53, 29 November 2007 (UTC)

While I agree with you that "digital zoom" is vastly inferior (personally, in fact, I'd go farther and say that it's no zoom at all), based on the evidence at hand, I don't think we can quite prove malfeasance in this case.

Consider my camera (a 3.2 megapixel Canon PowerShot A510). I can shoot in four different resolutions:

L	2048x1536	(3.14MP)
M1	1600x1200	(1.92MP)
M2	1024x768	(0.78MP)
S	640x480	(0.30MP)

Normally I shoot in M1 mode, because my pictures aren't that great and mostly I use them on-line where anything bigger is just overkill. (And it doesn't even bother me that I've effectively castrated my already-puny 3.2 MP camera down to 1.9, because it's no contest for me.)

Obviously, the camera's internal CCD is 2048x1536, and when you shoot in one of the lower-resolution modes, the camera downsamples by a factor of 1.28, 2, or 3.2. Thus, the M1 images (for example) require 60% of the storage of an L image.

Now, suppose there was a way to divide the raw CCD image up into thirds, both horizontally and vertically. Suppose the camera downsampled the outer 8 of the resulting 9 subimages as usual, but left the center one at full CCD resolution, and stored this composite somehow. For M1, this would require 8/9 x 1.92MP + 1/9 x 3.14MP = 2.06 MP, or 65% of full-resolution storage. So for a 5% increase in storage (above normal M1, that is), you'd retain the ability to zoom back in to full CCD resolution, as long as you limited yourself to the image's fovea, so to speak.

I doubt you could store this "composite" image using a standard image file format, although I suppose you could arrange to use a higher JPEG quality factor on the center of the image than the periphery, and gain something. (Or perhaps you could store a 2048x1536 image as a 2048x1536 JPEG, with normal JPEG quality in the center and a much lower JPEG quality at the periphery, and then make a note to always display the image shrunk by 78%, so that the coarseness around the edges wouldn't show, but so that -- again -- you could digitally zoom in on the center without loss of sharpness.)

I have no idea if this is actually what the DMC-FZ18 is doing, but it's not completely outside the bounds of possibility. —Steve Summit (talk) 01:30, 30 November 2007 (UTC)

Yes, EOZ still isn't as good a thing as true optical zoom. It's still much better than standard digital zoom - it lets you zoom in without upsampling. 138.38.149.208 (talk) 08:56, 30 November 2007 (UTC)

Time to fall

If a point mass of mass m is released at some distance d above the surface of a perfect sphere of uniform density, with mass M and radius r, then, using Newtonian universal gravitation and assuming there are no other forces acting within the system and ignoring the effect of the point mass' gravity on the sphere, how long does it take the point mass to fall to the surface of the sphere? This question occurred to me a while back and I have been trying to figure it out, but I have no idea how it would be calculated, since the distance travelled at any time depends on the velocity/acceleration, but the acceleration is constantly changing and depends on the distance from the centre of the sphere. --superioridad ^(discusión) 04:57, 29 November 2007 (UTC)

The answer is calculus (mostly integral) - consider acceleration, velocity and radius/position all as functions of time, use calculus to find an expression for the position, and solve to find the two time points. As for an actual answer ... give me a second (or perhaps someone else will take care of it). Confusing Manifestation(Say hi!) 05:55, 29 November 2007 (UTC)

We do have an article on this. It's not the most apparent method of approaching this, but it starts with more general conditions. Someguy1221 (talk) 06:07, 29 November 2007 (UTC)

And I suppose you could use that solution, then transfer to a co-ordinate system where the second mass is stationary, if you wanted the "Earth/bigger object doesn't move" solution. Confusing Manifestation(Say hi!) 06:12, 29 November 2007 (UTC)

(e/c)... OK, time for some TeX. Acceleration is given by Newton as

$a=-{\frac {GM}{r^{2}}}$

But $a={\frac {dv}{dt}}$, and using a little fiddling we get

$-{\frac {GM}{r^{2}}}={\frac {dv}{dt}}={\frac {dv}{dr}}{\frac {dr}{dt}}=v{\frac {dv}{dr}}$

And if we integrate with respect to r, we have

${\begin{aligned}\int \limits _{r_{1}}^{r_{2}}-{\frac {GM}{r^{2}}}dr&=\int \limits _{r_{1}}^{r_{2}}v{\frac {dv}{dr}}dr\\\left[{\frac {GM}{r}}\right]_{r_{1}}^{r_{2}}&=\int \limits _{v_{1}}^{v_{2}}v\ dv\\{\frac {GM}{r_{1}}}-{\frac {GM}{r_{2}}}&=\left[{\frac {1}{2}}v^{2}\right]_{v_{1}}^{v_{2}}\\&={\frac {1}{2}}{v_{1}}^{2}-{\frac {1}{2}}{v_{2}}^{2}\end{aligned}}$

... and then you use the fact that $v={\frac {dr}{dt}}$ and do some more calculus, and plug in your known values, and get some kind of answer. I have to go now, but I'm sure you or someone else can happily work away at this step. Confusing Manifestation(Say hi!) 06:12, 29 November 2007 (UTC)

${\sqrt {d^{3} \over {2GM}}}$ - Dragons flight (talk) 06:28, 29 November 2007 (UTC)

Drunken vampire bats

If a vampire bat sucks the blood of a drunk person, does the bat become drunk too? --Candy-Panda (talk) 09:55, 29 November 2007 (UTC)

I don't believe it would because of the way it can quickly excrete it.--58.111.143.164 (talk) 12:50, 29 November 2007 (UTC)

Also consider how the body dilutes alcohol that's ingested. You can drink 95% pure alcohol, which your body will dilute down -- half a percent of alcohol in your blood has a 50% chance of killing you. So if the bat is drinking blood, chances are good that it's drinking at most a 0.5% alcoholic drink, which in the U.S. is the upper limit to be considered a non-alcoholic drink. So, given a lot of other unfounded assumptions about a bat's metabolism and response to alcohol, the answer is: the bat becomes about as drunk as you could get on O'Douls. jeffjon (talk) 14:34, 29 November 2007 (UTC)

Using carbon nanotubes to transport electrons?

I read on the article on CNT's that certain CNT structures are excellent conductors of electricity, while others are semi conductors. Could CNT's be used instead of metal wires to transport electricity? Are there any advantages to this over what we currently use? 64.236.121.129 (talk) 14:43, 29 November 2007 (UTC)

You might enjoy our article about field emission displays for one such use of carbon nanotubes. I think a lot of the possible applications haven't yet been explored because of the relative shortage of long nanotubes.

Atlant (talk) 16:42, 29 November 2007 (UTC)

CNT's could certainly do that (and they are also very strong - which would be handy). However, the longest ones we can make are a few millimeters long - and the longest ones we can make in industrial quantities are utterly microscopic in length. So the technology just isn't there yet. However, there is huge interest in making long buckytubes - they have HUGE implications for materials science and a lot of very smart people are trying very hard to crack the problem of making them. I think our kids are going to have a lot of fun with these things. SteveBaker (talk) 20:16, 29 November 2007 (UTC)

If you fill an airtight car with helium, will it become lighter?

^Topic 64.236.121.129 (talk) 14:52, 29 November 2007 (UTC)

• No, you can only pump Helium in if the air can be displaced. With the car airtight, there is no place for the air to go, so no more room inside for the helium. - 131.211.161.119 (talk) 14:54, 29 November 2007 (UTC)

That's not what I meant. I meant the car is already filled with helium, and is sealed airtight to prevent the gas from escaping. 64.236.121.129 (talk) 15:05, 29 November 2007 (UTC)

• A car filled with helium would weigh less than a car filled with air, all else being equal. From a practical point of view, helium is a very small molecule and can often escape from containers that are "air-tight" (for example, certain types of balloons). ike9898 (talk) 15:00, 29 November 2007 (UTC)

I see. If this is true, why not fill compartments on airplanes (seperate from the cabin which contains people) with helium to make the plane lighter. 64.236.121.129 (talk) 15:05, 29 November 2007 (UTC)

As stated - the helium will escape. Also, to get any substantial effect, you would need to create a helium cavity the size of a zeppelin - which would greatly inhibit the aircraft's ability to fly. All in all - the benefit is not worth the effort. You'd get a much better benefit by banning all luggage from aircraft. -- kainaw™ 15:08, 29 November 2007 (UTC)

Ok lets say the airplane is a military fighter. We have a couple of lightweight tanks filled with helium. Once the helium escapes, it can drop the tanks. Wouldn't this help? 64.236.121.129 (talk) 15:11, 29 November 2007 (UTC)

• To get any significant benefit you would very large helium tanks with very lightweight walls. Maybe some sort of fabric or plastic walls. Then you want to make sure your centre of lift is above your centre of gravity, for stability - so put your crew cabin underneath your big helium tanks. No need for a fuselage or those heavy wings - just stick a rudder and elevators onto your big helium tanks. What you now have is called a dirigible. Gandalf61 (talk) 15:25, 29 November 2007 (UTC)
• No. It's very unlikely that you can build tanks that can withstand military-style maneuvers and still have positive buoyancy even if you assume they contain a vacuum. Also, you would still increase mass and, presumably, air resistance. The lift you can get from the surrounding air is fairly low - about 1.2 kg/m³ at sea level. --Stephan Schulz (talk) 15:34, 29 November 2007 (UTC)

Ahh! Is it time for our next round regarding "vacuum balloons"? If we went with the version inflated by photons, perhaps they could perform double-duty by both providing buoyancy and, later, allowing them to be dropped on the enemy as a sort of photon torpedo? This would be especially effective if they were inflated with some nasty photons such as gamma rays.

Atlant (talk) 16:45, 29 November 2007 (UTC)

Hmmm, here is an idea: A balloon made from a very thin film with a very high static charge. If you can charge it enough, it should keep the balloon expanded even against air pressure. Of course it needs a bit of material science handwaving ;-). --Stephan Schulz (talk) 12:46, 30 November 2007 (UTC)

On the topic of vacuum balloons, I saw one when I was watching an episode of Pushing Daisies last night on television. The protagonist had a flashback to a childhood boarding school's science lab, and he recalled an incident with a friend who had been an expert at constructing paper airplanes. One of the characters filled a balloon from a benchtop gas hose barb; once inflated and tied off, the balloon floated away. (Appropriately enough for this thread, I believe the balloon was actually used to provide lift to a paper airplane.) There was a very quick cut that showed the gas valve as it was being turned on, it bore the legend VAC. I thought that it was a very clever touch. TenOfAllTrades(talk) 17:47, 29 November 2007 (UTC)

Imagine a helium balloon. Think of how little a helium balloon can lift. Those little plastic tags they put on the end of the string are usually sufficient to keep the balloon from flying away. Even if your helium 'tanks' are made of the same lightweight material as the balloon you got at the fair the 'tanks' would only make the plane lighter by the weight of that little plastic tag. Not very useful. The problem is that air is already so incredibly light that replacing it with something even lighter does not achieve much. That is why even small, four passenger blimps have gas-bags the size of an office building. 72.10.110.107 (talk) 15:50, 29 November 2007 (UTC)

Also - aircraft flies at an altitude where there is almost no buoyancy offered by helium. -- kainaw™ 16:07, 29 November 2007 (UTC)

Not to mention that all your passengers would suffocate. shoy (^words _words) 17:24, 29 November 2007 (UTC)

No they wouldn't. In my first question, I didn't specify people were in the car, and in my second question, the helium would be in either compartments seperate from the cabin, or in tanks. But I agree with the other users that they prolly wouldn't contribute much lift to the airplane. 64.236.121.129 (talk) 17:42, 29 November 2007 (UTC)

This is definitely a non-starter. The trouble with the aircraft idea is that very little volume can be enclosed within the plane and the additional weight required to get it airtight (seals and whatnot) would easily outweigh the benefits. One liter of helium can only lift just over a gram. Take an F16 fighter - it's 14m long and the fuselage is about 3x3 m - the wings are 27 sq.m in area and about 20cm thick. So we have a total volume of about 130 cubic meters - which is 130,000 liters. If the ENTIRE plane was filled with nothing but helium (ie we take out the engine, fuel tanks, instruments, landing gear, internal bracing...the lot), it would be lighter by 130,000 grams - 130kg. Fully fuelled, the plane weighs 12,000kg. So at VERY best, you'd get about a 1% weight saving. But in truth there is very little free space inside an F16 - perhaps a hundedth of it might be utilised giving us an 0.01% weight reduction overall. The hassle involved is simply not worth it. Worse still - if that gas is pressurised to sea level then as you take off and the air gets less and less dense, the helium produces less and less lift. In a zepplin or a blimp, the lift bags expand to take account of that - but you can't do that in these confined little spaces. So in the end, this is certainly a lot more trouble than it's worth - and in all likelyhood would actually make the plane heavier - not lighter. A similar argument applies to a car. My MINI Cooper'S has 1.47 cubic meters of interior volume, that's 1470 liters - so 1.47kg of lift from filling it with helium. Sadly, it weighs 1200kg - so now we are looking at 0.1% reduction in weight. But its hard to imagine you could seal up all of the little holes and escape routes with less than 1.47kg of material - so in practice, you wouldn't save anything. SteveBaker (talk) 20:07, 29 November 2007 (UTC)

Indeed hmm. Is it possible to just shove more gas in there by force, and increase the density of helium in the tanks. Just hypothetically, assume the tanks are invincible, and won't explode. 64.236.121.129 (talk) 20:17, 29 November 2007 (UTC)

Helium atoms still have positive mass (just lighter than the molecules that make up air). If you shove more helium into the tank, increasing its density, you make it heavier, not lighter. -- Coneslayer (talk) 20:21, 29 November 2007 (UTC)

Going by what Alant said, if you had tanks that were filled with nothing, there's a vacuum in them, they would provide more lift than helium? 64.236.121.129 (talk) 20:27, 29 November 2007 (UTC)

Yes, vacuum (nothing) is lighter than helium (something). But it requires a much stronger, and hence more massive, container. -- Coneslayer (talk) 20:36, 29 November 2007 (UTC)

Yep - vacuum balloons sound like a great idea - but air pressure is 15psi - that's fifteen pounds pushing in on every square inch of the surface. It takes a pretty heavy construction to resist that much force - and that's going to be heavier than a fairly small amount of helium. SteveBaker (talk) 00:25, 30 November 2007 (UTC)

That's why we have to inflate them with an enormous flux of massless photons! (In case anyone's wondering, yes, I'm joking, at least until perfectly-reflective unobtainium becomes available.)

Atlant (talk) 12:53, 30 November 2007 (UTC)

Steve said 1 liter of helium can lift 1 gram. How much can 1 liter of vacuum lift? 64.236.121.129 13:56, 30 November 2007 (UTC)

Based on the densities listed in Lighter than air, one liter of helium has a buoyancy of 1.11 grams, minus the weight of its container, if immersed in sea-level air. A liter of vacuum has a buoyancy of 1.29 grams, minus the weight of its (by necessity much heavier) container. The difference is very small, since helium is already very much lighter than air. --mglg_(talk) 17:22, 30 November 2007 (UTC)

I see. But the buoyancy will increase if the air density is higher right? So if we were on Venus, it would produce more lift? 64.236.121.129 18:55, 30 November 2007 (UTC)

Yes. Closer to home than Venus, see Density altitude.

Atlant 22:27, 30 November 2007 (UTC)

ionic,covalent,coordinate bonds

Which one of the following compounds contains all the three ionic, covalent and coordinate compounds?? A-hydrogen cyanide B-ammonium nitrate C-potassium permanganate D-sulphuric acid —Preceding unsigned comment added by Knowiteverything (talk • contribs) 15:54, 29 November 2007 (UTC)

This a really easy homework question!--Stone (talk) 16:53, 29 November 2007 (UTC)

One of three relevant wiki articles gives the right answer in a nice picture!!!!--Stone (talk) 16:57, 29 November 2007 (UTC)

ornamental millet

Yesterday I saw a plant at a Nursery named as an ornamental millet. It was a feathery gras like plant with a deep puple head. The staff was unable to give me any information on it and I can't find any reference to ornamental millets in Western Garden Book or on line. Can you help? —Preceding unsigned comment added by 69.3.233.29 (talk) 17:54, 29 November 2007 (UTC)

Tall, dark and handsome describes this purple-leaved millet. Young plants are green-leaved, direct sunlight induces the purple leaf color. Capable of growing 3 to 5 feet tall, the plants are embellished with 8 to 12 inch flower spikes. The immature spikes can be cut and used dramatically in floral arrangements. Left on the plant, the millet seed spike attracts birds that snack on seed. 'Purple Majesty' is very easy to grow and is very tolerant of heat and low moisture. The purple leaf blades and spike are distinctly different from all other ornamentals.

Purple Majesty is best started outdoors when temperatures are consistently above 60 degrees or indoors in warmed seed trays. Chilly weather will stop plant growth or weaken its stems, so don't start your seeds too far in advance. Germination is very quick -- just 3 days! -- and plant growth is rapid and vigorous under good conditions. If you begin seeds indoors, the plants will remain green until set outside, then burnish a lovely violet within several days! You might like to buy some from here, [12]Richard Avery (talk) 18:15, 29 November 2007 (UTC)

Largest marine animal in captivity

Quick question: what is the largest marine animal ever kept in a zoo or aquarium? 68.23.172.215 (talk) 18:36, 29 November 2007 (UTC)

I think Japan has an aquarium with a whale shark in it. Whale sharks are larger than orcas which should come in 2nd. 64.236.121.129 (talk) 19:45, 29 November 2007 (UTC)

The Georgia Aquarium also has whale sharks. --Sean 23:45, 29 November 2007 (UTC)

How big would your arms have to be in order to fly/glide?

Lets assume you weigh 140 pounds, and you want to glide by stretching out your arms like a plane and running forward off of a cliff. If your arm width is normal, how long would they have to be in order for the air moving around it to provide enough lift to glide? 64.236.121.129 (talk) 20:24, 29 November 2007 (UTC)

I doubt you'd ever glide, no matter how long they are. Ignore your body for a moment, and just consider the arms. If you make your arm a foot longer, then that extra foot of arm has to produce at least enough lift to support itself, or you're no better off. But an arm, being roughly cylindrical and fairly dense, is not a very good airfoil. If you think about objects similar to a piece of arm, like a baseball bat or slender log, they don't "glide" if you throw them. That is, they don't produce much lift compared to their own weight. Thus, I don't think that making your arms longer is going to produce enough lift to support your arms, let alone your body. -- Coneslayer (talk) 20:34, 29 November 2007 (UTC)

I guess, we will have to change the shape of the arms also in addition to changing the size only. Airplave wings have very specific shapes, which provide the airplane enough lift to fly. Once the shape based on small prototypes has been determined, we will accordingly fix the size depending upon our weight. But our lower body also has to be in accordance with the aerodynamic structure. So, I guess if you have distended belly for example, you will have more problem than those having 6 pack :). DSachan (talk) 20:37, 29 November 2007 (UTC)

Shape helps a lot, but I think air moving over any object should eventually provide lift, depending on the speed of the airflow, and the weight of the object. Hmm, that's why cars have rear spoilers. When they go fast, the air moving around it can cause lift, which makes the tires lose traction with the ground. The spoiler keeps the car in contact with the ground. 64.236.121.129 (talk) 20:43, 29 November 2007 (UTC)

I guess, this is a wrong statement to make that air moving over any object should eventually provide lift, depending on the speed of the airflow, and the weight of the object. It will be possible only when the body itself is such that it is able to create the pressure difference on its two sides. This will eventually depend on the shape, and texture of the body. Of course, speed of the airflow plays a role in it, but only a part of the whole. DSachan (talk) 20:51, 29 November 2007 (UTC)

I read that the pressure difference thing is a fallacy when it comes to wings and lift. The articles here say that something as simple as a flat board, angled up will cause lift. 64.236.121.129 (talk) 20:56, 29 November 2007 (UTC)

A flat board can certainly generate lift. I've seen a couple of radio controlled planes that flew reasonably well with flat 'planks' of foam polystyrene for wings. The pressure difference thing isn't exactly a fallacy - it's just a very minor part of what's going on. Many aerobatic aircraft have symmetrical cross-sections so they'll fly just as well inverted as the right way up. They get their lift because of the angle of attack of the wing to the air flow. SteveBaker (talk) 04:39, 30 November 2007 (UTC)

(edit conflict) Why do you say "any object"? You're contradicting yourself. If air moving over "any object" caused lift, then the air moving over the spoiler (or, say, a Formula 1 wing) would cause lift, rather than downforce. If you have a perfect cylinder in air (as a model for your arm), why should airflow cause lift? You could flip the whole scene upside down, and it would look the same. -- Coneslayer (talk) 20:52, 29 November 2007 (UTC)

Right, I should say any object that isn't shaped at an angle so that the airflow pushes the object down. 64.236.121.129 (talk) 20:56, 29 November 2007 (UTC)

To elaborate, another example on wiki also says that if you stick your arm out of a moving car and angle it up, the air flow will push your arm back and up, causing lift. Factors that affect lift are airflow, weight of the wing, shape of the wing, and wing area. 64.236.121.129 (talk) 20:59, 29 November 2007 (UTC)

'Glide' can have many meanings—how much lift will you settle for? Take the Space Shuttle orbiter—it glides in to a landing, but it lands very steeply and very quickly compared to a commercial airliner, owing to its great weight and short, stubby wings. (It's been said that the Shuttle 'glides like a toolbox'.) See also our article on lifting body. TenOfAllTrades(talk) 21:48, 29 November 2007 (UTC)

Take a look at a hang glider - now you have your answer. SteveBaker (talk) 00:04, 30 November 2007 (UTC)

Actually, Wingsuit flying is just about...kinda/sorta gliding. It's all a matter of what glide ratio you anticipate getting! SteveBaker (talk) 01:55, 30 November 2007 (UTC)

Why doesn't boyancy work in solids?

It works in liquids, and gas... Why not solids? 64.236.121.129 (talk) 20:40, 29 November 2007 (UTC)

It does. Dragons flight (talk) 20:43, 29 November 2007 (UTC)

Ok lets say, we cover a person in lead, then let it solidify. Since he weighs less than the lead, his corpse should come to the top? 64.236.121.129 (talk) 20:47, 29 November 2007 (UTC)

Most solids have enough internal rigidity to resist the bouyant force, but on very large scales or long timescales, the same principle applies. For example, salt diapirs. Dragons flight (talk) 20:54, 29 November 2007 (UTC)

Or the ballast in the basement of the Kansai Airport to prevent it from floating, perhaps? Or am I remembering it wrong... --Mdwyer (talk) 22:27, 29 November 2007 (UTC)

• If you vibrate a container full of balls made of wood and others made of lead, all the same size, presumably the wooden ones will float to the top due to buoyancy. --Sean 23:51, 29 November 2007 (UTC)

Natural and Phenomena and Natural Resources

People from Europe came to the New World, noe the United States, for the natural resources.What kinds of resources or products did they ship back to Europe? —Preceding unsigned comment added by Felicia25 (talk • contribs) 22:44, 29 November 2007 (UTC)

Colonization_of_America#Economic_immigrants. Someguy1221 (talk) 23:35, 29 November 2007 (UTC)

November 30

MO theory

Hey, I'm in Chem II and we are doing MO theory. We did easy ones in class like O₂ and N₂ and the n=2 elements. But I have no clue how to do like Al₂³⁺! Does the d orbital even have anything to do with MO theory? Thank you! —Preceding unsigned comment added by HALPpls (talk • contribs) 02:04, 30 November 2007 (UTC)

I assume you're referring to molecular orbitals. However, I still have no idea what exactly you're trying to do. Someone versed in the field might understand your question as asked, but it might be a good idea to explain your goals more precisely. For example, are you looking for a qualitative statement or a quantitative calculation? Algebraist —Preceding comment was added at 03:11, 30 November 2007 (UTC)

d atomic orbitals are quite important for the molecular orbitals for some bonds and molecules. It would be hard to get 10 (!) bonds from the central iron atom in ferrocene with just s and p. At the level I think you're working, I think you do Al₂³⁺ the same way you have done other examples (that's how things are usually taught, no?): list the atomic orbitals of each atom, hybridize and mix'n'match them by symmetry. Perhaps you could tell us how far you've gotten on your own and what specific problem you're having? DMacks (talk) 03:19, 30 November 2007 (UTC)

It's a way of telling what electron configuration of a molecule is most likely, and how stable it is, if it exists at all. It uses quantum mechanics. —Preceding unsigned comment added by HALPpls (talk • contribs) 03:31, 30 November 2007 (UTC)

Aluminium doesn't have d electrons!!! Valance electrons are 3s2, 3p1. So... You will have a filled bonding sigma orbital and a fill antibonding sigma orbital. Plus one filled pi orbital. But that is for Al₂.

Don't know what you mean by Al³⁺ Is it a bond between Al and Al³⁺? —Preceding unsigned comment added by Shniken1 (talk • contribs) 04:00, 30 November 2007 (UTC)

Al₂³⁺ doesn't say anything about where the charge is…that's one of the things that MO theory can help determine. You just line up the molecular orbitals in order of energy, then start adding as many electrons as you have in total. Each Al would contribute 3 valence electrons if this thing were neutral; it's a net +3 charge, so it has overall 3 less electrons than that. DMacks (talk) 04:13, 30 November 2007 (UTC)

Even though Al doesn't have any d electrons, it can indeed use d orbitals to make bonds. Anyway, however I have only known aluminum to do this when forming complex ions. Just treat like you'd treat anything else, but with three electrons. Someguy1221 (talk) 08:04, 30 November 2007 (UTC)

An example of an answer:

O₂ Bonding electrons: 8 Anti-bonding electrons: 2

Likelyhood: 6 - very likely

O₂^1- Bonding electrons: 8 Anti-bonding electrons: 3

Likelyhood: 5 - likely

O₂⁴⁺ Bonding electrons: 4 Anti-bonding electrons: 2

Likelyhood: 2 - semi-likely

I don't know how to show my work on here, because it involves pictures...--HALPpls 22:29, 30 November 2007 (UTC)

You're problem is using the electrons on only one atom to make the bonds. You must use all valence electrons on both atoms. And remember that a diatomic molecule can make as many molecular orbitals as it can atomic orbitals, so this also means dioxygen has twice as many available orbitals as a lone oxygen atom. Think about this, and maybe you'll see where you're going wrong. Someguy1221 22:37, 30 November 2007 (UTC)

I take that partially back, I'm not sure where you got your total number of electrons from. Just sum up all the valence ones...Someguy1221 22:45, 30 November 2007 (UTC)

I'm looking for the likelihood of binary molecules that have a d orbital.--HALPpls 12:02, 1 December 2007 (UTC)

Light green slime

What metal would react with human sweat/salt to give a light green colored slime. i just scraped some off my glasses frame wher they contact the side of my head.--TreeSmiler (talk) 03:34, 30 November 2007 (UTC)

Anything with copper in it will produce a green "slime" when it is in contact with sweat. You mention glasses though. It is common for glasses to go day after day in contact with the skin (dirt and sweat) without being cleaned. So, just about anything can grow on them. I found that boiling my glasses once a week helped a great deal - more often during football season. -- kainaw™ 03:39, 30 November 2007 (UTC)

Hmm dont seem to look like copper, looks more like nickel. Could be some form of cupro-nickel I suppose--TreeSmiler (talk) 05:06, 30 November 2007 (UTC)

Planet Venus

If the poisonous gases and voilent weather did not exist on Venus, what would happen to you if you landed on Venus?--WonderFran (talk) 03:40, 30 November 2007 (UTC)

You would be crushed by the atmosphere that is 300 times as dense as the Earth's. Oh and it is a bit warm there so pack your shortsShniken1 (talk) 03:43, 30 November 2007 (UTC)

What would the pressure be like compared to the deepest part of the ocean? --WonderFran (talk) 03:45, 30 November 2007 (UTC)

Read the article on Venus. This is a direct quote from it: "...a pressure equivalent to that at a depth of nearly 1 kilometer under Earth's oceans." -- kainaw™ 03:46, 30 November 2007 (UTC)

The New York Times today ran a science article [13] saying that the latest results from observations by the Venus Express probe imply that it had origins very similar to Earth, and that it once could have had oceans. The two planets were described as "twins separated at birth." Venus is about the same size and mass as Earth. Its present inhospitable atmosphere took eons to develop, as the water evaporated and the hydrogen dissipated into space, leaving the oxygen combined into carbon dioxide. So what might Venus have been like on its most Earthlike day? Could it have been like early 20th century sci-fi writers and scientists imagined it, with tropical jungles? Edison (talk) 04:28, 30 November 2007 (UTC)

Still, we are doing our best to make the twin planets identical again.  :-( SteveBaker (talk) 04:33, 30 November 2007 (UTC)

Venus does not have (and there is no sign that it once had) a magnetic field strong enough to block radiation. It would be difficult for life to survive while constantly bombarded by high amounts of radiation from space. Perhaps it could develop in the deep water (as it did on Earth), but not on the surface. -- kainaw™ 04:35, 30 November 2007 (UTC)

It's basically a myth that the Earth magnetic field is directly critical to life. The Earth's field collapses to near zero every once in a while, and there has never been an unambiguous impact on the fossil record. The atmosphere itself is far more critical to blocking radiation on our planet. Dragons flight (talk) 08:15, 30 November 2007 (UTC)

While Venus's lack of a strong magnetic field certainly contributed to its current situation, getting twice as much sunlight just might have played a big role. Someguy1221 (talk) 08:18, 30 November 2007 (UTC)

It's not how much sunlight you GET that matters - it's how much you KEEP. With no green plants turning its CO2 into Oxygen, CO2 buildup from things like volcanoes is no small matter. With the ultimate in global warming and absolutely nothing there to reverse it - this is what you get. As the temperature rises, the oceans evaporate - water vapor is even more potent as a greenhouse gas than CO2 - so once things get just so hot, you get to a point of no return. At high enough temperatures, the water dissociates into hydrogen and oxygen - the hydrogen drifts away and the remaining oxygen gets turned into CO2 too. CO2 is a lot heavier than air - and pressures also increase with temperature - so you get an amazingly hot, dry planet with off-the-chart atmospheric pressure. If green plants had appeared on Venus, the extra sunlight would have been tolerable. After all, there is a much more than 2:1 variation in amount of sunlight we get at the equator than at the higher latitudes. SteveBaker 18:18, 30 November 2007 (UTC)

How much sunlight you get does matter when you consider the effects it has on the atomospheric conditions of a planet with large water oceans :-p Don't take my word for it, read the nytimes article linked to above. Someguy1221 19:35, 30 November 2007 (UTC)

Can dolphins and whales REALLY talk in some way that humans do?

I am a talking dolphin / whale sceptic, big time. I don’t believe that whale songs that repeat long and complex sequences over long periods of time constitute “talk” in any way that we humans understand it. Human language involves transmission of complex and finely nuanced information. To do this, we have labels (words) for just everything we have encountered. With this, we can convey very detailed information using those words in a structured way. The other day, I saw a Chinese newsreader on ethnic TV, and she was going full blast. I don’t have a word of Chinese, but the sound of it was like no animal noise. Common sense would tell you that the more you repeat the same phrase, the less information you are giving out, not vice versa. If whales could really communicate, they would have organised a boat tipping sortie in the old days and capsized the rowboats with their harpooners. I honestly cannot see how, historically, whale reaction to whale hunts has been any different to the reaction of any game animals that have been hunted by humans.

And I am puzzled as to why data on this question seems to be so murky. There are any number of good experiments I can envisage which could go a long way to answering the question, but no one else seems to be bothered. For example, you could have a pair of dolphins in one section of a pool which has another adjoining section. One dolphin has to swim to the second section where he can see three apertures in a wall, all designed to look different, e.g. one looks like a mound of seaweed, the second a rock, and the third a ball. A light appears above the aperture which holds the food treat. The only way the dolphin can get the treat is to swim back to the other dolphin (who cannot see into the second pool section) and “tell” her which aperture holds the food. Then the second dolphin is allowed to swim into this section and touch the food aperture, releasing the food. But she only has a certain amount of time to do this, and if she nudges the WRONG aperture, the treat is forfeit. Can the dolphins communicate to the extent of “select the seaweed thingy”, or not? Because if they can’t, then I would be loathe to say they can communicate in any way analogous to humans, who have, in every culture and for at least 60 thousand years been able to do things like that from eqarly childhood.

There are many such possible experiments. Is there any consistent research been done on such topics? I am sick and tired of hearing New Age types waxing fulsome on cetaceans being another intelligent species, with very little evidence to show for it. Myles325a (talk) 05:13, 30 November 2007 (UTC)

I don't think that many argue that cetaceans have anywhere near as complex a language than humans do. I don't wish to imply that intelligence is unconnected with language, but language is not the end-all, be-all of intelligence. These matters and others are discussed in our article on cetacean intelligence.--Fuhghettaboutit (talk) 05:34, 30 November 2007 (UTC)

First of all, there are a number of ways in which your experiment could fail, not least of which would be the utter confusion of being placed in an unfamiliar experiment setting. Specific instructions can also be difficult linguistically, as if the situation is less common then the potential for miscommunication or misunderstanding becomes very high. I was very briefly involved in research on dolphin communication, and everything was done with data from wild dolphins who were observed while communicating. In the first stage of analysis, the data was going to be analyzed in a blind fashion to see how different sounds corresponded to different behaviors. There would then be a breakdown of signals into principal features which would then be analyzed to see if there is a component structure to the communication. The last thing to analyze would be the arbitrariness of the components to determine if the language was in fact symbolic. So far, I think it's still at the first stage. SamuelRiv 15:12, 30 November 2007 (UTC)

Your post about language not being everything reminded me of this bit from a Douglas Adams book:

Man had always assumed that he was more intelligent than dolphins because he had achieved so much — the wheel, New York, wars, and so on — whilst all the dolphins had ever done was muck about in the water having a good time. But conversely, the dolphins had always believed that they were far more intelligent than man — for precisely the same reasons.

 :) --Sean 16:18, 30 November 2007 (UTC)

I don't know about dolphins, but my understanding of whale "communication" is that it has some features of what we would call language but missing some other parts that are vital (and I think the experiment laid out would fail to show any sort of communication). Personally, I would call what they have "proto-language" or "pre-language" since it is more than the biologically ingrained noises (such as that of dogs or cows) but does not have the complex (and arguably ingrained) features of human language. If I recall correctly, whale researchers have found that different pods will have different "song dialects" and that closely related pods will have similar dialects (see historical linguistics for the human language analogue).

The main problem with our understanding of whale songs is that if there's any meaning behind them we really don't know what they're ever saying. Ƶ§œš¹ _{[aɪm ˈfɻɛ̃ⁿdˡi]} 01:40, 1 December 2007 (UTC)

I doubt any other animal REALLY talks in the same way that humans do, to use the OP's words. But whether other animals "talk" depends on what one means by talk, and language. Dolphins and whales certainly seem to be contenders. The Orca article (a dolphin-related species) has a few teasers on this. Many bird species also seem contenders, depending on how you define "language". See Crow#Behavior for example. And crows are neophytes compares to some birds. In my personal experience, crows seem to communicate some fairly complex ideas between one another. Of course none of these examples are talking like humans REALLY talk. But they clearly, to my mind, indicate intelligence and rather sophisticated communication skills. Pfly 03:47, 1 December 2007 (UTC)

It could be that humans have frontal lobes, and no animals do. The frontal lobe is what helps you use logic. They probably communicate, but it's probably more like voice tone communication. RJRocket53 02:59, 2 December 2007 (UTC)

Long acting injectable drugs

I had a couple questions about long acting injectable drugs. Lets say you have testosterone enanthate or cypionate.

1. Why do they have to suspended in oils? 2. What happens if its not in an oil? Like in a liquid like bacteriostatic water 3. What happens if its not put into a muscle and just under the skin? —Preceding unsigned comment added by 76.167.132.90 (talk) 05:58, 30 November 2007 (UTC)

1. The oil is the means by which the long-acting effect is achieved. 2. It would last a considerably shorter time. 3. Again, it would last a considerably shorter time, and would also hurt like hell. - Nunh-huh 13:09, 30 November 2007 (UTC)

1 & 2: Considering testosterone, I guess it's not much better soluble in water than cholesterol (0.095 mg/l), and esterification with relatively hydrophobic acids won't change much. So if you have got a solution of water and testosterone it has to be very dilute or the testosterone will separate (and it's solid at room temperature). Icek 13:16, 30 November 2007 (UTC)

I just discovered that testosterone can practically be applied as a suspension in water[14], so forget my previous comments. Icek 18:31, 1 December 2007 (UTC)

Vandalism?

Is there really a medical problem called "Seradocious Lupius canaralium", or is this (as I suspect) vandalism of James Francis Smith? Clarityfiend (talk) 08:00, 30 November 2007 (UTC)

Since google shows nothing medical related for even one of those words, I'm going to guess vandalism, and about the oldest lingering vandalism I have ever seen. Someguy1221 (talk) 08:11, 30 November 2007 (UTC)

Unless Leonhard Euler is in fact known for "loving toast real good", the vandalism wouldn't be out of character for that IP. jeffjon 13:50, 30 November 2007 (UTC)

Either way - if there isn't a good reference for it, you are justified in removing it on suspicion alone. SteveBaker 17:59, 30 November 2007 (UTC)

Why does poliovirus preferentially infect motor neurones

The article on Polio states that motor neurones are preferentially infected - why is this? —Preceding unsigned comment added by 81.179.82.224 (talk) 12:02, 30 November 2007 (UTC)

This is an area that is incompletely understood, but poliovirus receptor and host factors, as well as type I interferon response, are believed to play roles in the neurotropism of virulent strains of poliovirus. [15] - Nunh-huh 13:07, 30 November 2007 (UTC)

Interesting. Thank you for your answer!

splitting atoms

How do you go about splitting an atom and what are the results of doing so? (aside from the obvious of coarse) —Preceding unsigned comment added by 59.101.238.31 (talk) 12:31, 30 November 2007 (UTC)

Have you read our article about Nuclear fission?

Atlant (talk) 12:57, 30 November 2007 (UTC)

The easiest way to split an atom is separating one or more electrons from it, creating an ion. But you probably mean splitting an atomic nucleus; Atlant already linked to the relevant article. Read it and come back if you have further questions. Icek (talk) 13:02, 30 November 2007 (UTC)

It's pretty much what you'd expect. You whack the atom with something really hard and fast (like a high speed charged particle in a particle accellerator) - it breaks - and you end up with one or more smaller atoms and some debris. That debris is in the form of left-over neutrons and such. If you hit a big, fat atom then the resulting matter will weigh less than the original atom and because of e=mc² you'll get some energy out too. (You could also put this another way and say that some of the debris you got out was in the form of photons...light, radio waves, etc). When you hit a Plutonium atom with a fast-moving neutron, you get energy AND some more fast moving neutrons. If there is enough plutonium around, those neutrons will probably hit more plutonium atoms and make yet more energy and yet more free neutrons - and before you know it...KERPOW! SteveBaker 17:57, 30 November 2007 (UTC)

No No No. A hammer and chisel will do.

Why does lightning shock you even if you are ungrounded?

Still a little confused on this, so if someone can explain in layman's terms, that would be great. You can touch a power line, and not get shocked if you are ungrounded. But if you are hit by lightning you will be shocked whether you are grounded or not. Why is this? 64.236.121.129 13:58, 30 November 2007 (UTC)

"Grounded" isn't an all-or-nothing concept; whatever insulator is keeping you ungrounded can break down if the voltage is high enough. For example, the rubber soles of your sneakers might (might!) keep you ungrounded against your 110 V home outlet, but could break down and provide a path to ground if you touched a high-voltage transmission line. Similarly, if you were hovering 10 feet above the ground, that 10 feet of air might keep you ungrounded if you touched a transmission line, but a lightning bolt is capable of ionizing and traveling through that air. (After all, it can travel through air the whole way from the cloud to the ground.) -- Coneslayer 14:14, 30 November 2007 (UTC)

What if you shoot a taser or electrolaser at a bird or someone who happens to be floating in the air (for whatever reason). 64.236.121.129 14:52, 30 November 2007 (UTC)

The relevant potential difference in a taser is between the two terminals of the taser. It does not require an earth ground to operate. (Same idea as a battery, for example. If you short the terminals of a 9 V battery with a paperclip, you'll see a spark. No connection to earth ground is required.) -- Coneslayer 14:58, 30 November 2007 (UTC)

I see. What about an electrolaser? 64.236.121.129 15:12, 30 November 2007 (UTC)

I don't know anything about them, but I think the article answers your question directly:

To complete the electric circuit, there must be either a second laser beam, or a ground return, from the target to the last transformer in the step-up series.

-- Coneslayer 15:19, 30 November 2007 (UTC)

No, there doesn't. It depends on the frequency of the source and the impedance match of the target. Energy will be directed and reflected back. The plasma channel supplies both the forward (S12) and return (S21) path. --DHeyward 15:44, 30 November 2007 (UTC)

It's a matter of completing the circuit from source to ground. Let's imagine three cases:

1. You are wearing a jet-pack and you fly up to a 1000 volt overhead power line and grab hold with one hand. To be clear, this is not one of those insanely high voltage wires...but it's enough to kill you if you mess up.
2. You do the same thing but grab hold with both hands.
3. You are flying along in your jet-pack and you get struck by lightning.

In the first case, the wire doesn't have enough voltage to arc down though the air to hit the ground - we know that for sure.

When you touch the wire, you are at 0 volts, the wire is at 1000 volts and very briefly, a small (hopefully) current flows to bring you up to 1000 volts - but it can't go anywhere because it's still not strong enough to break through the air to the ground - so no massive amounts of current flow through your body and whilst you might feel it - it won't kill you because this is like static electricity. If the voltage on the wire is really high though, that initial transfer to get you up to the voltage on the wire could still kill you. When you touch the wire with both hands, electricity could theoretically flow up one arm, across your chest and back down the other - but it's not going to do that unless your body has a higher electrical conductivity than the wire itself - which it doesn't. The current prefers to take the easy route...so not much voltage passes through you and you're OK still.

When you get struck by lightning, the critical differences are that the voltage is more than enough to allow the lightning to break down the air and reach the ground - so there is no problem with the lightning hitting your head, passing through your body and exciting at your feet - then carrying on to complete the circuit through the air to the ground. The other thing is that your body (being mostly made of salty water) conducts electricity better than air - so unlike case (2), the electricity prefers to flow through your body than through the air - so you are definitely dead. SteveBaker 17:36, 30 November 2007 (UTC)

Interesting, thanks steve. Few questions. You said that the wire is 1000 volts. If you touch it with one hand, it will bring you up to 1000 volts. But how many amps is the wire, and how many amps are flowing through you if you are ungrounded and touch with one hand? How many volts and amps would flow through you if you touch the same wire while being grounded? If your body for some reason had higher electrical conductivity than the wire, and you touched with two hands, how many volts and amps would pass through you then? 64.236.121.129 18:50, 30 November 2007 (UTC)

(I just picked 1000 volts - real power cables have all sorts of different voltages, many are MUCH higher than that!)...I don't know how many amps the wire could deliver - but will be a lot more than it'd take to kill. If you are holding onto the wire and are not grounded then ZERO current is flowing through you. If you are grounded then we can use Ohms law (V=IR) to calculate the current (I, in amps) from the Voltage (V in volts) and the resistance of your body (R in ohms). Most people's bodies have a resistance of about a half to two MegaOhms - but most of that resistance is due to your skin - beneath the skin you are made of salty water - which conducts electricity extremely well. Since even a brief jolt of 10milliAmps to serious damage to you - and only 100 milliAmps will kill - you can tell that a voltage of only 500,000 x 0.01 = so 5,000 volts is a problem. However, that assumes your skin is intact. If your skin is broken, then your body resistance is very low indeed - 100 ohms perhaps. Then 100 x 0.01 ...ONE VOLT is enough to do it! If this seems unlikely to you, check this out: Courtesy of the Darwin Awards [man electrocuted by 9v battery]. This is why just 110 volts from a US wall outlet is enough to kill you if you take it for long enough to damage your skin. SteveBaker 22:00, 30 November 2007 (UTC)

That's interesting how you can have voltage, but not current (amps) running through you. If you were to use the water analogy, how would that work? It would be like there's a mysterious pressure running through the ground, but no water... Malamockq 00:28, 1 December 2007 (UTC)

Take a stream -- that's your wire. Dig a small hole beside it -- that's you. Dig a shallow trench connecting the two -- that's your hand. The brief inrush of water is the "ow shit" feeling you get as you grab the wire, but once the hole fills with water, nothing more flows -- you're now at the voltage potential of the wire. --Carnildo 01:25, 1 December 2007 (UTC)

You know, now that I look back on this question, http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Science#Power_lines.2C_and_electric_shock Steve Summit said that there will be a small influx of current when you touch a powerline while ungrounded. So maybe the current isn't quite zero? Malamockq 00:34, 1 December 2007 (UTC)

It isn't quite zero. There's a small transient current, corresponding to the self-capacitance of the human body. But that capacitance is extremely small, so the transient current approaches zero very quickly. --Trovatore 00:40, 1 December 2007 (UTC)

(ec) Current only flows when there is a voltage (potential) difference across you. Current flows when you're grounded because the ground and the wire are at a different potential, and it decides to use your body to make the trip. When you touch a wire, like Steve says, your body is at a different potential, so current will flow. But it only takes a teensy tiny amount of charge to bring your body to an equal potential. You get electrocuted when that tiny amount of charge can then leave your body, and more comes in behind it (quickly). Someguy1221 00:44, 1 December 2007 (UTC)

Oh, I hadn't noticed that the original question specified a 1000-volt overhead power line. That's alternating current, not direct. So I was wrong -- the current is not actually a transient that approaches zero; there's a tiny current that will flow in and out of your body for as long as you hold on to the wire. Presumably the impedance article will tell you how to calculate it, once you find a numerical value for the self-capacitance of the human body. --Trovatore 00:49, 1 December 2007 (UTC)

I would like to point out a detail that I'm surprised this wasn't mentioned in the earlier thread. Yes, if you touch a live object, and you're not grounded, there will be a brief influx current as your body matches the voltage. But if that object is live with AC, like a typical power line, its voltage is constantly changing, and your body has to keep matching it. So the influx current continues flowing back and forth as long as you're touching it. As previously noted, we are talking about a very small current, nothing like what you'd experince if you were grounded, but I don't know how small. --Anonymous, 00:54 UTC, December 1, 2007.

I would beg to differ with the original questioner in that it is, in my opinion, quite possible to get shocked/killed etc whilst 'ungrounded' but swinging from (or being in close proximity to) a high voltage overhead power line. This is because the electric field strength (esp on 132 kV and above) can be great enough to cause lethal currents through the body. This is also why linesmen working on live overhead cables wear conductive equipotential suits to ensure that no dangerous current flows within their bodies.--TreeSmiler 01:37, 1 December 2007 (UTC)

That's true - and we discussed that in detail in an earlier question. I deliberately chose to use a (possibly mythical) 1000v powerline in order to avoid that complication. The OP is already pretty confused and we need to break the explanation down into easily digested chunks. So what you say is true - but not for my deliberately simplified example. SteveBaker 05:52, 1 December 2007 (UTC)

how tall is the talest dwarf

Wikipedia article dwarfism says: "The Little People of America (LPA) defines dwarfism as a medical or genetic condition that usually results in an adult height of 4'10" (147 cm) or shorter." So the answer to your question is 147 cm - that is, if you mean living people and not a fantasy creation. Lova Falk 15:19, 30 November 2007 (UTC)

...such as Carrot Ironfoundersson, the 2m tall dwarf (adopted, fictional). Gandalf61 15:35, 30 November 2007 (UTC)

Or the seven-foot dwarfs of the Kingdom of Loathing. Algebraist 04:56, 1 December 2007 (UTC)

Bone marrow transplant and DNA

Yesterday we watched a whodunit on tv in which the rapist/killer had got a bone marrow transplantation as a child, and because of this, the dna of his sperm was different to the dna of his blood. The sperm had his own dna, but the blood had the donor's dna. Is this really possible?? Lova Falk 15:32, 30 November 2007 (UTC)

I don't know about the transplant case, but it can happen due to having an undeveloped twin: see Lydia Fairchild, chimera, teratoma. It seems plausible. --Sean 16:28, 30 November 2007 (UTC)

This is absolutely true. The circulating cells in the blood all come from the stem cells in the bone marrow; see haematopoiesis for details of this process. If a patient's bone marrow cells are wiped out using chemicals and/or radiation (for example, to treat leukemia: a cancer of the blood) and replaced through a transplant, the patients blood will contain almost exclusively the donor's DNA. TenOfAllTrades(talk) 16:50, 30 November 2007 (UTC)

I thought red blood cells don't have DNA in them, so this is a non-issue. – b_jonas 09:08, 2 December 2007 (UTC)

What's the largest galaxy that we know of

^Topic 64.236.121.129 18:43, 30 November 2007 (UTC)

Galaxy IC 1101. Someguy1221 19:07, 30 November 2007 (UTC)

Zero-point energy and manipulating objects

I read the article on 0 point energy, and I don't understand most of it. In the game Half-Life 2, a zero-point energy device can manipulate, levitate, and throw objects without touching them. Is this a real potential ability zero-point energy can do? How can it do this? 64.236.121.129 19:00, 30 November 2007 (UTC)

In a word, "no". For more words, see Zero-point energy.

Atlant 19:01, 30 November 2007 (UTC)

Haha, being terse I see. I don't really understand the article, I already said that. 64.236.121.129 19:06, 30 November 2007 (UTC)

Zero-point energy is the lowest amount of energy a quantum mechanical system can contain. Since it's the lowest amount possible, you can't remove it from the system, and thus you can't ever do anything with it (well, you can do some fun things with the Casimir effect, maybe). As for the zero-point energy device, it was in severe violation of the law of conservation of momentum. It has no basis in reality, some guy (but not this guy) just made it up. Someguy1221 19:28, 30 November 2007 (UTC)

Some people think that it may be possible to use the Casimir effect to extract actual useful energy from absolutely nothing. Serious scientists get palpitations at the mere thought of it. Hence science fiction writers like to use it as a plot device to allow spaceships to zip around the place without needing tediously difficult things like fuel. As irresponsible as this is of sci-fi writers, video game writers are even less likely to be bothered with reality. So if you want to describe how some handy in-game gadget works, feel free to grab any set of random science-words and string them together. I want one of those nanotech quantum gravity tachyon crossbows - it fires this pinkish-green ball lightning stuff and turns anything it hits into blue Play-Doh! (See how easy that is? I can (and indeed, do) do this kind of thing all day!) SteveBaker 20:51, 30 November 2007 (UTC)

Along the lines of sci-fi and video-game dialog and "explanations", folks may want to see our Treknobabble article.

Atlant 22:36, 30 November 2007 (UTC)

Is there some connection to The Incredibles here? Pfly 04:23, 1 December 2007 (UTC)

Sure. The movie, like many other sci-fi things, uses the term "zero point energy" to mean "scientific magic". The whole point is that real-life zero-point energy has little to nothing to do with how the term is used in science fiction. It's the modern version of the term "atomic energy" which was used to basically explain away any magical technology in works of the 1940s, 50s, and 60s before it became more mundane and less mysterious. Obviously there was such a thing as "atomic energy" but it had nothing to do with the ways the term was often used in fiction. --24.147.86.187 19:45, 1 December 2007 (UTC)

White holes attract matter?

The article on white holes says both black and white holes attract matter. But white holes spit out matter. I don't really understand the statement in the article. How can it attract, and spit out matter? Does it just mean, it attracts matter because it has gravity, but it also spits out matter from its event horizon? So does that mean, the matter it spits out, it also attracts with its own gravity? 64.236.121.129 19:05, 30 November 2007 (UTC)

The article makes it clear that matter is attracted to the white hole but never crosses the event horizon. -- kainaw™ 19:20, 30 November 2007 (UTC)

Yea, but it doesn't make it clear as to how this works. Does the white hole look like a donut then? Does it spit out matter similar to a polar jet? Since it spits out matter, and attracts matter, does it suck it's own matter back into it? What would it look like? 64.236.121.129 19:24, 30 November 2007 (UTC)

A lot of physicists say it doesn't look anything different from a black hole. Maybe if you hopped in you'd find out, but I wouldn't recommend it. Someguy1221 19:29, 30 November 2007 (UTC)

As exciting as this topic may be - it is vitally important to point out that we've never found any hint that white-holes actually exist. The only reason we even bother to think about them at all is that they show up in some of the relativity equations. I think most cosmologists would say that they do not - nor cannot - actually exist. SteveBaker 20:42, 30 November 2007 (UTC)

Yes, the matter that the white hole spits out is attracted to the hole. So is Hawking radiation, but it escapes to infinity anyway. To understand where the distinction between black and white holes comes from, consider the graph of $r(t)={\sqrt {k+(ct)^{2}}}$ for different values of $k$. ($c$ is the speed of light and $r$ is a kind of radius.) For $k>0$ this is a nice smooth curve (half of a hyperbola). This is the surface of an ordinary gravitating spherical object in Kruskal-like coordinates. For $k=0$ it reduces to $r=|ct|$, which has a sharp 90-degree turnaround in the middle where $r(t)$ goes from "inward at the speed of light" to "outward at the speed of light." The inward half is the white hole horizon and the outward half is the black hole horizon. All of the usual "black hole" solutions to general relativity (like Schwarzschild and Kerr) are really gray holes with both black and white hole horizons in them. More realistic classical models have only the black hole horizon. But the whole business seems rather artificial -- maybe the correct description is more like the $k>0$ case, where the white-black distinction is inherently absent. Hawking radiation makes this more plausible -- maybe it is the time reversal of absorption in quantum gravity. -- BenRG 23:55, 1 December 2007 (UTC)

Black holes attract matter, which passes through a wormhole and exits via a white hole. However, Einstein's gravitational field equations only predict this if the mass of the black hole is 0, which is obviously impossible since black holes form from star collapse. Also, if anything with any mass enters a massless black hole, its associated wormhole and white hole will immediately cease to exist. In other words, white holes don't exist. --Bowlhover 16:12, 2 December 2007 (UTC)

The most destructive thing in the universe?

Are supermassive black holes the most destructive thing in the universe? 64.236.121.129 19:13, 30 November 2007 (UTC)

You must first define "destructive". Do you mean the ability to rip a spaceship apart? Do you mean the ability to destroy planets? Do you mean the ability to tear atoms apart or smash them together? -- kainaw™ 19:16, 30 November 2007 (UTC)

Uhh I'm not sure. I guess the ability to destroy planets, stars, and if applicable, galaxies. 64.236.121.129 19:18, 30 November 2007 (UTC)

Well, there's not much else that can consume a galaxy. A hypernova, or a very powerful gamma ray burst could do a nice job wiping out all life in a galaxy. And if you believe in the big rip, then dark energy is the most destructive force in the galaxy. Alternatively, you could say the hypothetical big crunch, caused by gravity, is just as destructive, although this might recycle everything into a new universe. Someguy1221 19:22, 30 November 2007 (UTC)

How long does a hypernova last? Like from the beginning to the end, when the hypernova is considered "over". 64.236.121.129 19:27, 30 November 2007 (UTC)

Well, we obviously don't have data yet, but this paper [16] shows a light curve for a very massive blue supergiant supernova, which shows about a 100 day peak from rise to fall and another 300 day-long tail "plateau" of intense emissions, and after that the explosion itself continues indefinitely with a long tail of emission from the ejected matter. The actual detonation, though, lasts only a few seconds, depending on how you define the beginning and the end (a hypernova is believed to emit a "long" >2sec gamma ray burst on detonation). SamuelRiv 20:23, 30 November 2007 (UTC)

Is it possible for a Terrestrial planet to be as large as a gas giant?

All the terrestrial planets that we know of are quite small compared to gas giants. Is it possible for a terrestrial planet to be as large as a gas giant? Like say, Jupiter? Is there anything preventing this from happening? 64.236.121.129 19:17, 30 November 2007 (UTC)

This article [17] is about a newly discovered planet: "With a mass of only 14 times the mass of the Earth, the new planet lies at the threshold of the largest possible rocky planets", which suggests (but doesn't explain) that there is a limitation for the size of a terrestrial or rocky planet. jeffjon 20:02, 30 November 2007 (UTC)

Well, as far as I can tell, nothing catastrophic would occur. Gravity would be much stronger (about 3-5g) on a giant terrestrial planet of the same size as Jupiter because Jupiter is only slightly denser than water, whereas Earth is composed mostly of heavy elements (excluding the atmosphere). Another comparison: Jupiter is 1000 times larger by volume than Earth but only 300 times more massive. Then there's radiation and tides: the amount of pressure in the center of such a planet would give you a large, dense, solid core that may have significant effects on the magnetic field. There would be a lot more heat radiated from the planet (Jupiter radiates more heat than it receives from the sun) as residues of its formation, and any object nearby (such as another planet or moon) would feel significant tidal effects (something that causes friction against their rotation and revolution - so something like the Earth's moon would end up always orbiting above the same spot very quickly). Surface pressure would be significant due to a large, thick atmosphere and stronger gravitational pull, so I would estimate it at around 10atm. Finally, the rotation of such a planet would probably be very slow, because, all else being equal, angular momentum must be conserved so the angular velocity must fall with the square of the radius. This makes avoiding getting burned or frozen much more of a hassle.

As far as formation goes, there you would have problems in the currently accepted Nebular hypothesis of planetary formation. Basically, terrestrial planets only reach a certain size in the primordial accretion disk due to some basic nucleation properties of the swirling matter and the chaotic gravitational interaction of nearby protostars and Jovian planets. They end up clearing all matter in their orbit eventually, but there's only so much dense matter than can be taken up and the sphere of influence of a planet is only so large, as it drops with the square of distance from the center of the planet. Finally, if such a planet existed, its density would be such that we would probably have seen it through current extrasolar planet surveys. SamuelRiv 20:00, 30 November 2007 (UTC)

Just a small note, Earth's atmosphere is composed of heavier elements than Jupiters. Ours is O2 and N2, Jupiter's is H2/He. Shniken1 13:06, 1 December 2007 (UTC)

Manipulating objects without touching them

Lets say you have a plastic ball on a table. You want to pick it up, move it around, spin it around, bounce it on the ground, and put it back on the table without ever physically touching it. Is there any force in science that can possibly do this? 64.236.121.129 19:35, 30 November 2007 (UTC)

If you're looking for the force, no. Objects can only be realistically manipulated from a distance using electromagnetic force. So actually, if you allow the ball to actually be built into a small radiocontrolled helicopter, then there's a definite yes, but I'm going to assume that's not what you were looking for ;-) You can use static electricity to push or attract an object from a relatively short distance, and the same with magnets (these both act through specific aspects of the electromagnetic force). So I can concoct a system of magnets within and outside a plastic ball to make it hop up, spin around, and fall back down, but nothing that would work in general on any given object. Someguy1221 19:52, 30 November 2007 (UTC)

Isn't plastic diamagnetic? 64.236.121.129 19:56, 30 November 2007 (UTC)

Everything possesses diamagnetic character, but that doesn't help you unless all you want to do is push it. You can't get more complex motions out of it, as the sole reaction of a diamagnetic object to a magnetic field will be to feel a force in whatever direction the field weakens most quickly (down the gradient of the field (you have to ignore the directional component of a magnetic field to define a gradient, but that's ok (nested perentheses, wow))). Someguy1221 20:01, 30 November 2007 (UTC)

(EC) It should be diamagnetic (but not everything is diamagnetic - they can also be ferromagnetic, ferrimagnetic, or paramagnetic). That means with a strong enough magnetic field, you can repel any diamagnetic object by diamagnetism or attract any paramagnetic object by paramagnetism, in spite of what you might think intuitively. See Magnetic levitation. SamuelRiv 20:06, 30 November 2007 (UTC)

Use gravity! Take a black hole, pour electrons into it until it builds up a nice large negative charge. Now you can move it around using some handy positive charges you might have in your garage. Now, suspend it above the ball. Gravity will pull the earth, the table and the ball towards the black hole because of its gravitation - but tidal effects will make the ball fall faster towards the hole than the other things (because it's closer) - so gradually, the ball will lift off the table - as it does so, move your star-stuff further away and you've managed to grab the ball without touching it. ('Caution: Wikipedia would like to point out that this is a THOUGHT experiment and any actual attempt to perform this trick could be considered harmful.)

Personally, I find connecting the hose of your shop-vac vacuum cleaner to the output instead of the inlet hole is a great way to play with a ping-pong ball without touching it - and it's a lot easier than the first way I suggested! It's amazing that the ball stays held in the air stream rather than being blown off to one side.

20:29, 30 November 2007 (UTC)

There is a very new and exciting technology that can accomplish exactly this. Friday (talk) 20:46, 30 November 2007 (UTC)

Moving air could probably assist here, especially when combined with the Bernoulli effect used for attractive purposes. Depending on your definition of "not touching", a pool cue may also be useful, especially when combined with chewing gum. Failing all that, work on developing telekinesis.

Atlant 22:45, 30 November 2007 (UTC)

You can do what magicians and movie people do: put a little piece of very fine thread on it, the attach said thread to your glove or something. Poof, magic, just like in the movies! --24.147.86.187 03:29, 2 December 2007 (UTC)

Hire a janitor, and ask her to put the plastic ball back after cleaning the table. Spinning and bouncing around is left as an exercise to the reader. – b_jonas —Preceding comment was added at 08:51, 2 December 2007 (UTC)

Andromeda-Milky Way collision

Hi. I have a few questions about this theory. First of all, it says that the chances of some stars colliding are remote. However, any lifeforms within the two galaxies exsisting at the time will likely be more prone to extinction than before the collision. The thing is, the new galaxy will likely have about 600 billion stars within the main galaxy, surrounding clusters, and orbiting satellite galaxies. Surely at least a few thousand, if not a few million will collide? The merging of the galaxies will probably cause countless new gravitational changes that were not present previously, correct?

Now, if some stars are flung into space, it will likely slow down before it leaves the local group altogether, as there is an enormous envelope of dark matter surrounding both galaxies at the moment. Now, when the galaxies collide, the dark matter will probably be attracted toward the new galaxy, correct? So, if it is attracted toward the new galaxy, it will probably "push" the parts of the galaxies towards each other, and cause a farther merging. If the galaxies are to merge, then the cores will be affected by each other. This should bend each core's path around each other, potentially locing each other's core in an orbit around each other held together by the surrounding envelope of matter.

If a large galaxy usually swallows up an approaching smaller galaxy, why is it that when two large galaxies of about the same mass are expected to collide, they expect the galaxies to miss each other altogether? I know that in three-dimentional space, two objects approaching each other, in this case the galactic cores, rarely, if ever, hit each other head-on. So, since the stars near the centre of the galaxies are more tightly closer together, and the ones near the edges are farther spread apart but still orbit as quickly, is it accurate to say that the stars near the centre of the galaxies have a better chance of collision?

Now, if there are lifeforms living within the galaxies at the time, a higher chance of supernovae nearby and dense intergalactic clouds should cause a higher extinction rate, correct? Now, the Triangulum galaxy is not far in the sky from the Andromeda. Also, it is closer to us than the Andromeda. This places it roughly between the two galaxies. So, when the Milky Way and Andromeda collide, is there a good chance the Triangulum could join in as well? The Milky Way, Andromeda, and Triangulum galaxies are the three largest in the entire local group! In addition, the three combined probably have a few dozen orbiting companion galaxies. So, if the three galaxies merge before they have a chance to disperse, might their companion galaxies, as well as some of the dark matter, start moving toward, rather than away, from the new galaxy, due to a relatively sudden increase in gravitational attraction?

If they eventually do start to "fall" in towards the trio, they will probably come close to the centre. Could this, along with the surrounding dark matter, which is about 10 times more massive than the entire cluster of merging galaxies, trap the group of galaxies inside each other? Now, would some of the heaviest stars in the galaxies now be at the centre, since they would be more attracted by the dense centre that has now formed? So, we can now assume that the densest material will now be at the centre of the new large galaxy. So, the cores will likely still be trying to escape each other through centrifigual force, giving time for more nearby surrounding galaxies to start to be attracted by this increase in gravity, merging into the gravity feild of the group.

However, as gravity only travels at the speed of light, and decreases by the square of distance, only the nearest galaxies will likely become attracted into the group. Eventually, might we have perhaps a dozen galaxies, big and small, within the new galaxy, which will likely now be elliptical, as the new forces will likely have destroyed its former arms. So, as the galaxy is still being held tight, the galactic cores will likely rotate around each other, spinning more and more rapidly, but probably increasingly being attracted toward each other with the new material caving in. By now, I think tens of millions of less affected stars will likely have left the galaxy, perhaps being absorbed by surrounding galaxies, affected and unnafected, while proabably tens of thousands more will already have either collided with other stars, or destroyed by the immense gravitational forces and nearby eruptions.

Now, I imagine that several hundred black holes, both supermassive and mini-sized, will have been caught by the increasing gravitation at the centre. As more massive stars will likely be at the centre, and since areas close, but not too close, to large black holes are good formation centres for massive stars, due to the more concentrated gravitational fields, the leftovers from massive stars that have decayed over this time, some of which are bllack holes, will likely keep joining the ever-increasing gravitational centre, bringing yet more stars and very close companion galaxies and their dense cores toward the centre.

By this point, will the gravity have increased so much that even centrifugal force can no longer stop it? If so, the now-very prominent core, illuminated by occasional explosions, will likely start collapsing in on itself, much like the way that the cores of massive stars and very close binary stars start allowing gravity to beat centrifugal force, tearing matter toward each other rather than away. Now, eventually the supermasive black holes will likely be the closest to each other from the reasons that they have more gravity on each otehr and are the first to merge into this group in the first place. What will happen then? Will the cores of the most supermassive black holes start moving closer and closer together, reaching their orgasm climax of gravity, and eventually colliding so that they explode in an enormous explosion, wiping out the stars and black holes closest to the centre?

Will the explosion have enough energy to travel at nearly the speed of light, pushing intergalactic material, such as nearby dwarf galaxies, stars, and dust clouds, superheating them and adding them to the shell of explosion? How bright would this be when it reaches a planet, say, 1 billion light years away? Would it be visible to the equvalent of the naked (human) eye? Would the now diluted gas and dust and radioactive matter cloud brighten the sky at this site, or perhaps interfere with lifeforms, or would it just be too dilute at this distance for any of this?

So, what do you think, is some of this plausible? Why has the scientific community not considered all these potential factors? Do I just have an overactive imagination? Are there too many "what-if?'s" in this scenario to be plausible? Would Hubble shift prevent some of this from occuring? Or would hubble shift not have increased that much within this timeframe? Are the mentioned bodies (objects) too far away from each other to have much gravitational effect on each other before hubble shift combines with centrifugal force to tear the objects away from each other? However, if this is the case, why did they consider the collision of the two galaxies in the first place; wouldn't the hubble shift along with dark energy dominate over the non-increasing gravitation between the two galaxies and tear them apart before they start moving towards each other?

If even the meeting of the two galaxies in the first place is plausible over such a great distance in non-increasing non-densifying gravity, shouldn't the events mentioned above be plausible under smaller distances and continously increasing and densifying gravitation? Or has too much speculation been put into this senario? Am I asking too many questions (the way I usually do)? Is the theory of the collision and related matters too early in its infancy to approve of all these potential future fators? Are there just too many possible factors that anything could happen? Thanks. ~AH1^(TCU) 22:35, 30 November 2007 (UTC)

Okay, the overarching question has to do with supermassive black hole collisions, which could in principle happen, with enough counteracting matter to slow them down, but this process would take a very long time (given the momentum of the black hole). You seem to be ignoring the fact that when one object is moving and gets attracted to another object by gravity, it goes into orbit. Unless there is friction, it will not collide (unless its orbit intersects the object's radius). Black hole coalescence on smaller scales may be the cause of gamma ray bursts, but we don't know.

Now let's take this from the beginning. When galaxies collide, and they have many, many times in our universe, we have never, not once, seen stars collide. The stars are much more disperse than galaxies, even in the densest regions (you are looking at over 10^6 stellar radii cubed of empty space around each star, versus 10^1 galactic radii of empty space around each galaxy in a cluster). If they did collide, believe me, we would know it.

Current simulations of galactic collisions with dark matter show basically the same result as without dark matter. Just remember that if dark matter works how we think it might, it will not "push" other objects together. It will, by definition, not interact significantly in any way other than gravitationally. Stars do get flung out into space as large, irregular groups after galactic collision: these are called irregular galaxys.

What I'm leading up to here is that it is very likely that Local Group galaxies collided or came close to colliding at one point in time, and will likely collide again in the future (by the way, the Andromeda Galaxy is closer than the Triangulum Galaxy). This probably resulted in the Magellanic Clouds and a newly-discovered third irregular galaxy around us, all of which orbit harmlessly around the galaxy as it is now. Again, let me stress that gravity causes moving objects to orbit, not collide.

Life on Earth is about 3.5 billion years old. If a galactic collision occurred since then, then it didn't do much to life. It would likely result in an increase in large star formation and thus supernova activity, which may have caused mass extinctions in the past, but nothing of the scale that you're talking about.

At this point, pretty much everything afterwards is invalidated. Hypothesizing is good, but just remember: always check your assumptions. In this case, your assumptions about gravity's attractive nature were the fatal blow to this line of thought. SamuelRiv 23:31, 30 November 2007 (UTC)

A back-of-the-envelope calculation shows why galaxies can collide, but stars hardly ever do. Consider a sun-like star falling through the core of the Milky Way, which is about 3,000 light years across with an average density of about 2 stars per cubic light year. The star is 1e6 km across, which is about 1e−7 light years. In falling through the core, it can collide with stars whose centers lie in a cylinder of volume about 1e−10 cubic light years, so it expects to hit on average about 2e−10 stars on its journey through the core. Its odds of hitting another star are billions to one against. (And passing through the core is very much the worst case: the spiral arms are several order of magnitude less dense.) Gdr 12:31, 1 December 2007 (UTC)

Hi. Thanks for your helpful answers, but there are still a few things. I know that gravity causes not collision, but orbitation. In fact, I know that gravity usually cuases objects to depart rather than attract each other via centrifugal force. However, the stars near the core are more dense, so they should have some attraction between the cores, possibly stabilising the orbit, and possibly attracting the cores gradualy and very slowly toward each other, instead of away from.

Also, could the Triangulum galaxy eventually start slowly orbiting the Andromeda, and eventually start orbiting the newly collided galaxy, and perhaps eventually join? The Milky way is already expected to swallow the Magellanic Clouds within 10 billion years if it is not disturbed, indicating that the gravity of the Milky way alone is enough to allow closeby galaxies to move toward, rather than away from, the Milky way. So, if the two galaxies collide, the total gravity should be stronger, and there will likely be more companion dwarf galaxies in total, so eventually a few galaxies should fall into the new galaxy. Well, if the two galaxies collide, the resulting new galaxy should probably be elliptical because it will attract its former arms more inward. So, being elliptical, it will likely be denser.

Also, your calculation of billions to one against, still do not mean that no stars at all will collide. In fact, if the orbits of the two cores hold each other and stabilise enough, some stars might expect to pass not one, but perhaps dozens of times through the centres. So, let's say that there are about 500 billion stars within the elliptical galaxy alone, and since their orbits will likely be distrubed by the new gravitation in opposing forces, and because the stars might pass through the galaxy's centre many times, and because the central reigon will likely be denser than it is today and there is more than one dense area, let's say that about 50 billion stars will find itself passing through the core. Of course if one star collides, the other star it's collided with has also collided. So, assuming that the centre is quite dense, and a star might pass through the core more than once, due to the gravity of the centre which will likely keep in an orbit (albeit a very haphazard one), the chances of a random one of these stars colliding is a billion to one. So, that means about 50 stars will have collided.

Also, by now the triangulum galaxy will likely be caught up within the gravitation of the new epliptical, and so will some smaller dwarf galaxies, that some stars might end up passing through dense reigons while outside the centre. However, probably more stars will have been thrown away than collided. Since there might be many centres, and this tug-of-war might last billions of years if the galaxies catch each other, and because whatever star a star collides with is also collided, let's raise the total number of collided stars to 300, before either the galaxies escape each other or their cores get caught up and collide. So, assuming that there will be about 100 000 stars within 500 light-years of the collided stars, which if those stars harbour solar systems which harbour life, would cause extinctions. So, assuming that there are about 2000 out of those 100 000 stars that have lifeforms, and that there are an average of 10 000 lifeforms per star, and that roughly an average of 1/10 of the lifeforms will go extinct within this reigon, we could be looking at roughly 2 million extinctions with the collision of just one star.

Now, multiply that up, and we could have 200 million extinctions caused by collisions alone within the two galaxies. Now, factor in all the other more common causes, such as supernova explosions, dust clouds, new asteroids, etc, and we could have about 10 billion extinctions related to the collion of the two galaxies. So, maybe it isn't so safe after all, astronomy within 3-5 billion years? So, if gravity causes the galaxies to orbit, will the orbit eventually become larger and larger before escaping each other's grasp, or will they tighten before eventually a collision occurs?

I now understand this isn't very likely, but if the galaxies were to collide, and their centres were to collide as well, about how bright would it be if it were viewed from 1 billion light-years away, factoring all the triggered collisions, stars destroyed by the burst of explosion, clusters of stars and whole galaxies pushed by the shell of gas? If this were to happen, would the now dispersed shell of gas brighten the sky at this distance? Keep in mind that a marshmellow dropped on a neutron star will release about as much energy as an atom bomb on Earth. Collide two neutron stars, you get even more energy. Collide two black holes, yet more energy. Collide two or more supermassive black holes, and let the explosion push and swallow stars, yet more energy.

Would the radiation at one-billion ligh-years, in the unlikely scenario that this were to happen of course, cause extinctions, or would the radiation be too dispersed, decayed, and old, to cause any damage? Could the heavier particles collide with the observer's planet's atmosphere and cause a radioactive meteor shower? Keep in mind that the most common type of uranium, a common radioactive element, takes 4.5 billion years to halflifeanate, about the same time it would take for this hypopthetical shell of radiation and matter and stargas to reach the observer's home planet.

Do any epliptical galaxies right now look like they might have been caused by the fusing of two or more spiral galaxies? Again, would hubble shift quickly tear the orbits apart and prevent this from happening? As a side note, after half an hour of searching using a detailed star map book, I finally found the Andromeda galaxy through binoculars. It was so cold and windy that, although I was wearing gloves - two pairs of them - my fingers got so cold they hurt down to the bone. Well, I don't think I'm going to be able to do that tonight: nearly a foot of snow may be expected by the end of the storm coming tonight! I found the Andromeda by rotating my head to Cassiopia looked like an M, then followed it "up" so that there are three stars in a row "above" the M, with the one farthest "left" "higher" than the one farthest "right"; I looked "down" from the middle star, leaning my FOV toward the "right" to another star, continuing "down", this time leaning "left" before I saw a fuzzy patch. Well, it looks like I always need to keep changing my countless hypotheses to suit new information. So, can someone answer my few remaining questions? Thanks. ~AH1^(TCU) 16:40, 1 December 2007 (UTC)

Please read my previous answer again. I need to make this abundantly clear. No star collision has ever been observed in galactic collisions (which are quite common), and it would be very obvious if we had observed it. Orbits are stable. They do not fly apart or come closer together. They stay in the exact same position forever. It's the same reason why planets do not collide with the sun. It is centripetal force, not centrifugal force, that counteracts the attractive force of gravity - just note that the two are different manifestations of the exact same force, so they counteract exactly in a complete orbit. The centers of galaxies are not very dense. Again, 1 star has about 10^6 radii cubed of clearance, even in the center, so the chances of collision are about 1/10^18, or one in a billion times a billion. Galaxies are generally moving fast enough relative to one another and have enough gravitational influence from other nearby galaxies that they do not form stable rotations. (Andromeda and Triangulum are moving on the order of 100km/s toward us after correcting for the rotation of our galaxy (with some additional unknown tangential component, so they are probably not on a direct collision course)). SamuelRiv 18:26, 1 December 2007 (UTC)

I don't disagree with your point that no collisions have ever been observed, but I don't think your calculation of the probability of collision is correct. It's the size of the swept path that needs to be considered. So you can't cube the clearance, you can only square it. Gdr 18:37, 1 December 2007 (UTC)

You're right, my bad. I was considering it from a stochastic motion perspective, not a linear motion, which works for collisions in condensed matter but not in astrophysics. SamuelRiv 20:58, 1 December 2007 (UTC)

Supposing that vampires existed and had the same basic daily calorific needs as regular humans...

How much human blood would a vampire need to consume per day in order to remain healthy? --Kurt Shaped Box 22:50, 30 November 2007 (UTC)

According to [18] and [19], it takes ~600 Calories to regenerate the constituent parts of a pint of blood (a remarkably large number). Assuming your vampires could extract energy from it with perfect efficiency, they would only need 3-4 pints per day. I will note though that most vampires of legend also ate traditional food in addition to drinking blood, so it would appear that blood was not envisioned as a primary food source.

So, Kurt, how many gulls would a vampire abstaining from humans need to eat? Dragons flight 23:03, 30 November 2007 (UTC)

Dunno. Has anyone here ever read the nutritional information on those little bottles of supermarket gull blood? ;) So, an exclusively blood-drinking vampire would need to kill a *lot* of people, as I thought (one every two days or so?). Not exactly going to remain inconspicuous for long, is he? Even moreso if he has to hunt for virgin blood (I know that some vampires are very specific about this). --Kurt Shaped Box 23:39, 30 November 2007 (UTC)

He could just hunt children in that case. He could also store unconsumed blood in a cooler. Yet further, knowing that Vampires are cold-blooded and spend much of the day motionless in a coffin, they may have considerably lower daily caloric requirements than a live human. Someguy1221 00:46, 1 December 2007 (UTC)

Well, wiki tells us that the birth rate is ~20 per 1000 people per year. And the death rate is about 9 per 1000. So that gives an excess of 11 snackable humans per 1000 per year to maintain a fixed population size. So, in order to eat one every two days, each vampire would need a feeding ground of at least 16,500 people. If your diet calls for only nubile virgin girls, then perhaps at least 50,000 people per vampire. So a major metropolitan area could easily support twenty or more vampires. However, they would probably need to work out a way to dispose of their bloodless victims. (Lots of neck wounds might look suspicious.) At that end, going into the mortuary business might be a good idea. What's one more corpse at a funeral home? Not to mention the convenient supply of beds. Dragons flight 01:19, 1 December 2007 (UTC)

Well, in Asian supermarkets they do sell cow blood and stuff. If a vampire was running short on virgins, they could always disguise themselves and buy themselves a pint or two. bibliomaniac15 01:23, 1 December 2007 (UTC)

...but can't vampires just turn themselves into vampire bats and find their own cows rather than having to rob the dead or actually run a funeral home or do whatever else to pay with cash? 71.100.1.143 07:27, 1 December 2007 (UTC)

December 1

TV set circuit mystery

What circuit in a TV set goes at (approx) line frequency (and creates bursts of hf radiated noise) but is not locked to it?--TreeSmiler 02:19, 1 December 2007 (UTC)

If you run from a VCR or DVD player the frequency is not locked in precisely and can drift around iwth local oscilators or mechanical speed in the play back. Graeme Bartlett 02:54, 1 December 2007 (UTC)

(edit conflict) Vertical deflection. It goes at exactly line frequency, as I recall, but it syncs on the signal, not the line. I wasn't able to find a Wikilink for it. There doesn't seem to be an article on the innards of a television. --Milkbreath 03:11, 1 December 2007 (UTC)

Ah!. But isnt the vertical sync signal synced to the line freq at the transmitter? Otherwise you would have some bars rolling up or down the screen would you not?--TreeSmiler 03:21, 1 December 2007 (UTC)

I don't think so. You'll see clocks that sync off the line to keep time. The power company tries their best to keep the frequency steady, and they do a darned good job of it, but not perfect. Why would the TV station rely on them when it's a simple matter to generate your own, rock-steady 60 cycles? The vertical oscillator in the set is synchronized by the sync pulses in the signal. Hum bars are a symptom of trouble in the set, like a dry filter cap in the power supply. --Milkbreath 03:55, 1 December 2007 (UTC)

Hmmm! (sorry) That would explain it -- as this article I just found does. Thanks. [20] .Do we have an article on Hum bar Apparently not.--TreeSmiler 04:26, 1 December 2007 (UTC)

I think such a system couldn't have worked in western Europe, because they were using the PAL system which goes 60 Hz, while the power supply goes on 50 Hz. – b_jonas 08:44, 2 December 2007 (UTC)

Also the picture is self is strongly modulated by the line frequency as it tends to be similar from line scan to line scan. Then as the whole picture switches scene the noise suddenly stops (or starts). Graeme Bartlett 07:07, 1 December 2007 (UTC)

That seems to be similar what Im picking up on my scope. ie the amplitude of the bursts seems to vary with time (presumably when the scene changes- altho the tv is in another romm so i cant be sure). Do you have any further info on that phenomenon?--TreeSmiler 14:06, 1 December 2007 (UTC)

I expect that "modern" (1941 and after) TV transmitters use a crystal controlled oscillator rather than the utility frequency for the deflection. Early TV systems, such as that of E.F. W. Alexanderson in the 1920s, used mechanical scanning and had a motor synchronized to the power system. The receiver was likely powered by a line locked in synchrony with the power line at the transmitter, so there would be only a slight phase difference between them which was easily adjusted and fairly stable, to prevent the picture rolling. Edison 03:00, 2 December 2007 (UTC)

CD4+CD25bright

Is CD4+CD25bright another way of saying CD4+CD25high? --Seans Potato Business 03:08, 1 December 2007 (UTC)

CD25 bright is CD25 high, and CD4+ is just CD4 positive, in case that wasn't clear also. Someguy1221 00:15, 2 December 2007 (UTC)

Strong Interaction Exploitation for Macro-Engineering Hobby Project

Having nothing else scheduled for the day (the day in question being sometime late in the 100th century AD), I'd like to send a von Neumann machine off into space to have a family and then spend an epoch or two transforming some uninhabited planets into a hybrid Dyson Sphere/Ringworld (which I imagine as an oblong terraced sphere rotating on its long axis to generate artificial gravity on the interior) and am deeply concerned that scrith may not be available on eBay at that time, so I may be at a loss for a material strong enough to make it big enough to also be interesting. I'd like to carry as much of the load as possible under tension, so I'm imagining an innumerable number (which is to say that I have decided not to count them) of spokes that cross the span from one side to another. I need strong wires- much stronger than carbon nanotubes. I guess I have a little while to work out some of the details but here's the best I've come up with. (Here, have a complimentary grain of salt before continuing). By manipulating infinitesimal black holes in ways yet unspecified, confine a stream of protons (or, as long as we're playing around with confined spcaes, super-protons made from decay-proof top and bottom quarks ;-) into the extremely confined space-time geometry of the tiny black hole's interior, and then, taking advantage of very clever timing, trajectories, and an Acme(r) Maximum Luck Field Generator, the black hole's own tendency to evaporate due to Hawking Radiation as a way of opening up an end in another place (white hole? Aaaargh), hope (against hope) to hold the geometry open with the string of protons held inside. Then, inspire the string of protons to behave as a single macro nucleus- only with all its strength focused on a single one-dimensional axis- by providing an appropriately charged negative charge at the ends (simulating the presence of electrons), which also, by the by, provides the means for attaching the ends of the "wire" to the things it needs to hold together. Being particularly naive from this particular technological vantage point, I'm hoping to cheat on the distances required- make the strings short, but connect points that are very distant. Essentially, I think I'm imagining a narrow-throated Einstein-Rosen Bridge held open, not by some exotic anti-energy, but by the presence of the protons themselves. At the end of the day (when it is appropriately dark), my question is singular. How strong could a "confined geometry strong force tension member" be? I'd be happy with anyone's best guess, considering. DeepSkyFrontier 05:51, 1 December 2007 (UTC)

Ringworlds and Dyson spheres have one exceedingly difficult problem. They are unstable. If one part of a ringworld gets a little bit closer to the star than it should be - and (consequently) the part on the opposite side is a bit too far away then because gravity gets stronger as you get closer to the star, the side closest to the star is attracted more strongly than the part on the far side - which serves to accentuate the problem. Hence you have a system that just won't stay put. SteveBaker 06:29, 1 December 2007 (UTC)

I'm sure you could manage some kind of stabilizing propulsion with all that energy you're collecting from the Dyson Sphere ;-) Someguy1221 11:14, 1 December 2007 (UTC)

What are you using for reaction mass? Remember, your sphere weighs at least as much as all of the planets, moons, asteroids and other debris in your solar system. SteveBaker 23:38, 1 December 2007 (UTC)

The step-terrace scheme will fail, I think. How would you prevent each latitude ring from falling toward the equatorial plane? For your strong-force cables, I think we have a little problem with electrical charge. For your proposed schedule, forget it. The universe will have converted itself to pure computronium long before that time. -Arch dude 13:36, 1 December 2007 (UTC)

To actually answer the question, I think you're confusing the nature of these geometric (actually topological) structures, as they do not extend in any spatial dimension. They are, from our perspective, 0-dimensional, and have another dimension in hyperspace. What this means is that you can't use it as a tension or compression member because it doesn't extend out in any real direction - it is just a hole is space. This is if I understand your question correctly.

Furthermore, these holes cannot be stabilized with photons. I'm afraid you need exotic matter - no photon is able to penetrate the hole before it collapses.

And to answer Steve, lots of things are unstable, like Maglevs, LaGrange points, and modern fighter jets. That's why we have corrective feeback loops. SamuelRiv 14:07, 1 December 2007 (UTC)

Yes - but it takes energy (which I'll grant you have plenty of) and reaction mass. As you throw off mass, you have no way to recover it - so gradually, your system 'evaporates'. The mass of a Dyson sphere is immense - it's estimated that you'd need all of the mass of our solar system to form even a skinny ringworld. You aren't talking about pushing a spaceship around - you're talking about pushing the combined masses of Jupiter and Saturn around. For a full 1 AU sphere - thousands of them! A 1 AU sphere would rapidly cook it's inhabitants without some kind of unobtainium heat radiators...so it's gonna be a lot bigger than 1 AU. The power involved in moving this thing is decidedly non-trivial! It's pretty well established that these things are impossible for a bunch of practical reasons...but I find instability to be the biggest. SteveBaker 23:38, 1 December 2007 (UTC)

I answered the question, but asked for clarification. You cannot use such bridges as tension or compression members - the new space that you create will just stay the same as you move your holes closer or farther apart. There is a possibility that you'll lose effective mass due to gravity waves if they are accelerating relative to each other, but I don't see a way for them to act as a tension member in the way you're describing. SamuelRiv 16:29, 2 December 2007 (UTC)

Protons, for one thing. Yes, thank you. You essentially answered the question by saying, "You can't ask this question this way." What I'm asking has, at its root, nothing to do with black hole geometry. I'm asking about how strong the nuclear strong force is. If it could be exploited- by whatever means- what would be the result. Thank you for addressing the question I asked. Your edit interfered with my own restatement of my question that replaced the comment that you responded to. DeepSkyFrontier 16:45, 2 December 2007 (UTC)

Most desirable place to live

Aside from the fact the individuals may have different preferences, where is the most desirable place to live, strictly as an environment and in total if all Friends, possessions, etc. could be relocated their, or is this more a matter of personal history such as where you were born, where you grew up, where you had the most rewarding experiences? Second part of the question is where can I find an online questionnaire that will ask me different questions like "What kind of climate do you prefer: warm, mild, cool, varied...?" and then narrow down the places that meet the criteria I entered? 71.100.1.143 07:02, 1 December 2007 (UTC)

It depends entirely on the criteria you choose to rate. There are plenty of attempts to rank the best places to live, both globally and in specific countries. Here are links to some of them: [21] [22] [23] [24] [25] As for an online questionnaire, try this if you are interested in the USA. Rockpocket 08:11, 1 December 2007 (UTC)

Perhaps it would help to work backward somewhat and decide where the least desirable places to live are and why. For example beyond mere climate considerations, living near an active volcano might not be a good idea. (Also consider places that experience lots of earthquakes, hurricanes, floods, etc.) Besides natural issues there are also man made hazards to consider like areas of high crime, or other armed activity. Along man made lines, also important factors would be local health care and whether or not one could drink the local water without getting sick. Anynobody 08:52, 2 December 2007 (UTC)

Clanks

I just read the Girl Genius series, and there's something I can't help wondering. Just how far can clockwork robotics be pushed? I know some remarkable things were invented in the old days using nothing more than gears and pulleys, up to and including small programmable robots, but just how much is possible? Black Carrot 11:17, 1 December 2007 (UTC)

The non-computer parts of a robot are mostly "clockwork", so you are actually asking about the computer parts. It is theoretically possible to perform any computing task using a mechanical computer, because you can build a mechanical Universal Turing Machine. In practice and using mechanical compones built with today's technologies, an elaborate purely mechanical robot is not feasible. A "mechanical" computer becomes possible with nanotechnology. At that size scale, a mechanical computer may be smaller than an equivalent electronic computer because electrons are hard to confine to small volumes. -Arch dude 13:20, 1 December 2007 (UTC)

If you want to be amazed, look up examples of medieval and Renaissance automata. It's really quite amazing what you can do with the technology; it really just comes down to how many gears you can fit into a given space and how much precision you can have with the gears. I've seen some amazing videos of automatons that could write, play complicated instruments, etc. Obviously they aren't intelligent in any sense but they can fake some pretty fine motor skills. --24.147.86.187 18:08, 1 December 2007 (UTC)

I was too hasty earlier. Sorry. There is a class of non-digital control mechanisms that can be used to create analog robots. You can get really complex behavior from relatively simple analog electronics. A lot of very interesting devices of this type were developed prior to the development of the first digital computers. A famous example was the Nordon bombsight. The mechanical equivalent of an analog circuit is perfectly feasible and easy to implement. By contrast to digital, the analog approach is much more compatible with a Victorian/steampunk worldview. -Arch dude 22:22, 1 December 2007 (UTC)

There is a mathematical theorem called "The Church–Turing thesis". It says that in principle, any minimally capable computer can do anything that any other computer can do - providing it has enough time and memory (and of course the right peripherals - printers, robot motors, whatever). This minimal computer is called a Turing machine. So if you could build a clockwork Turing machine - then it could do anything the PC in front of you right now could do - if there were enough memory and time. Looking at the architecture of the Turing machine, it's quite plausible that someone could make a clockwork machine like that. Here, for example is a "Turing machine" made from toy trains. Well, those could be clockwork trains...so there you are.

OK - so that's the theory...but in practice? Well, no. Turing machines are S-L-O-W - and a clockwork machine would be HIDEOUSLY slow. It would need winding up A LOT! If you look at something like Babbages' Analytical engine, it's full of gears and such - and it's equivelent to a Turing machine - so the Church-Turing thesis says it's a proper computer. It's still pretty slow - and I think a literal clockwork motor might be a little impractical as a power source - but a good sized steam engine would work. It's huge - and still pretty slow. If it had ever been completed (it never was) it would have been able to store 50,000 decimal digits - which is about 200 kbits of memory. That's more than the first computer I used in my first job (an Intel 8008 with 4kbytes of memory). It would have been able to multiply to twenty digit numbers, in three minutes. That would take about a billionth of a second on a modern PC - so this machine would take thousands of years to something as (seemingly) simple as displaying the page you are reading right now.

So - yes, in theory, mechanical computers could work - but they absolutely could not be practically useful.

There is a caveat to that - which is that if nanotechnology lives up to some of it's more extreme promises, we'll be able to make very small mechanical systems that would actually be faster than modern electronics. If we get to that point, then mechanical ("clockwork") computers will be back in vogue.

SteveBaker 22:47, 1 December 2007 (UTC)

(What was I thinking?! No discussion of mechanical computers could be considered complete without mention of the one I happen to own!) Mechanical computers do actually exist - you can buy one called the Digi-Comp I (I own one - it's the one in the photo in that article). This thing is made of plastic (if you have an original one) or cardboard (if you have the modern version). You put it together yourself in about an evening. It's pretty minimal - it has a three bit memory and it's power source is you wiggling a lever back and fourth. It does one 3-bit calculation each time you move the lever across and back - maybe one per second (if you go too fast it jams up)...but it can play the game 'Nim', count and add a 1 bit number to a 2 bit number to produce a 2 or 3 bit result! Yes, it's horribly limited and pretty complicated for a cardboard computer. You program it by moving little bits of drinking straw around - you input data by directly switching it's 3 bit memory - and the output "display" is three little windows that show '1' or '0' depending on ths state of the memory. There was once a Digi-Comp II which used marbles rolling down a pinball-like machine with little plastic levers. It had a dozen or so bits of memory and could do multiplication (eventually) - there was a little electric motor driving the mechanism that recycled the balls to the top of the track. SteveBaker 23:11, 1 December 2007 (UTC)

Ah, yes, the Digi-Comps. The Digi-Comp II my uncle gave me for Christmas when I was 10 years old or so was certainly a formative experience. And I do still have it, somewhere; I'll have to haul it out and take a picture of it for our article.

It's quite an amazing machine. It's got a 7-bit accumulator, a 4-bit addend register, and a 3-bit multiplier register. There was even a way to rig it up to do long division, but it required about a hundred marbles to filter through, all without error, and it was virtually impossible, but fun to try. (And mine had this great bug where a marble would occasionally bounce out of the track it was supposed to be in, roll all the way down to the bottom of the board, bounce back up, fall into an unrelated hole, and start a calculation over again that was supposed to be terminating. Loads of laughs.)

I never heard of an electric motor, though; that may have been some sort of field upgrade. Mine, you had to recycle marbles back up to the top yourself if a long multiplication or division required more marbles than you had. —Steve Summit (talk) 02:01, 2 December 2007 (UTC)

Are you interested in selling it? I've always wanted one and I'm no too fussed about the condition it might be in. I have frequently considered building one out of Lego - but without complete details, it's hard to figure out exactly how it works. Contact me at my talk page if you are interested. You're obviously right about the lack of a motor - I have never actually seen one of these gizmo's actually working. I assumed there must be a motor or something because the description of them always talks about an 'Automatic' mode of operation versus 'Manual' - which I took to mean that there must be some kind of automatic ball feeder...but there was certainly only one version of the Digi-Comp II so if your's has no motor then that's that.

The Digi-Comp I (which I have) was made in polystyrene a year or two before the II came out (around 1963 I think). However, they were flimsy and broke easily - and hence they are much rarer than the Digi-Comp II (in working condition at least). A few years ago an enthusiast made a replica of the Digi-Comp I out of heavy cardboard and is now selling these replicas in kit form. However, so far, nobody has ever built a replica of the Digi-Comp II - and whilst it's mechanically easier to understand than the Digi-Comp I, it's actually the more computationally powerful of the two machines. Of course, neither of them are 'Turing-complete' - so in a sense, they are not true computers. However, they are both programmable and can perform arithmetic, so in another sense, they are computers.

For the technically minded, the Digi-Comp I is essentially just three flip-flops which are clocked synchronously. The "wiring" between them is handled by 'programming' the machine by putting short sections of drinking straws onto the front of the machine where they can block parts of the machine from moving. You can also insert mechanical analogs of "AND" and "OR" gates in front of the inputs to the flip-flops (with certain constraints). Because there is only a three bit memory, if you want a program to actually step through multiple instructions, you have to use one or two of the bits as your "program counter" - which leaves you with very little left for doing actual work! But you can set it up to do things like count from zero to seven (which is an impressive sight actually!) - and it can play a minimal version of the game of Nim using seven objects in one heap...which is actually the only program for it that's actually any use for anything! Programming the thing to do ANYTHING is absolutely mind-bending - you need a very good knowledge of what flip-flops can do. The idea that it might ever have been a kids toy is ridiculous!

SteveBaker 05:52, 2 December 2007 (UTC)

Hypothalamus and mammillary body

Reading Hypothalamus#Nuclei gives me the impression that the mammillary nuclei (and thus the mammillary body) are a part of the hypothalamus. Is this true? Lova Falk 14:15, 1 December 2007 (UTC)

Neuroanatomists have often categorized the mammillary bodies as part of the hypothalamus. One anatomy text that does so says, "synthetic descriptions may suggest that all hypothalamic nuclei are well delimited structures" and then it describes how hypothalamic nuclei transition gradually into surrounding brain regions. --JWSchmidt 15:14, 1 December 2007 (UTC)

Thank you! As I usual do with your explanations, I quoted you in mammillary bodies. Could you change the reference to a proper one? Lova Falk 16:34, 1 December 2007 (UTC)

You want me to admit how old my neuroanatomy textbook is? Ow. The quote I gave above is from Human Neuroanatomy (8th edition) by Malcolm B. Carpenter and Jerome Sutin (1983) ISBN 0-683-01461-7.--JWSchmidt 19:50, 1 December 2007 (UTC)

That's almost prehistoric. But it's fine for an uncontroversial question like this one. Maybe I should never have thrown my 1977 book called Psychophysiology... —Preceding unsigned comment added by Lova Falk (talk • contribs) 08:40, 2 December 2007 (UTC)

McDonalds Fries

"because there's too much moisture in the potatoes for them to crisp during baking, I needed to release some of the water first. I ran home at record speed (for me), went directly to the kitchen, and tried parboiling the fries before baking them" - am I not correct in thinking that boiling (par or otherwise) will increase the amount of water in the potato?--Seans Potato Business 19:30, 1 December 2007 (UTC)

What this refers to is commonly called blanching - and is required to make crispy fries. It is the reason that most home-cooked fries aren't as good as fast-food fries. Boiling them makes the water in them expand. They are pulled out and quickly cooled to stop the cooking process. Since the outer layer is mostly cooked - it is no longer very permeable and the cold water has great difficulty soaking back into the potato. The end result is a reduction in water inside the potato and a pre-crisped outer layer. This is highly effective in most "wet" foods that you wish to deep fry. The trick is timing the boiling process so that the cold water won't soak back in. -- kainaw™ 21:14, 1 December 2007 (UTC)

Of course the classical way to make fries is to fry them twice - once to a light brown, then take them out of the oil, reheat it, and do the second stage frying at a somewhat higher temperature until they are golden brown. This should have a similar effect. The result also depends on the kind and age of potato, of course. Very fresh ones are not as good for frying as potatoes that have been stored for a while (in a proper cool, dry environment). --Stephan Schulz 22:35, 1 December 2007 (UTC)

• For a wonderful survey of these sorts of nitty gritty food details, check out On Food and Cooking, by Harold McGee. --Sean 00:03, 2 December 2007 (UTC)

I read a recipe decades ago, by Julia Child, in which she said to cook the fries at the high temperature first rather than second. A home deep fryer usually does this rather automatically, because the initial fat temperature drops shortly after the fries are added, probably because the moisture flashing to steam takes heat from the oil. A restuarant fryer has so much heat capacity in the larger amount of hot oil, as well as a more powerful heating element that the temp stays more constant as things are lowered into the oil. Edison 02:55, 2 December 2007 (UTC)

Some people salt the chopped potato a while before cooking and then dry them with a paper towel just before tossing them into the oil. I presume that the purpose of this is to draw water from the fries via osmosis. SteveBaker 14:49, 2 December 2007 (UTC)

Jesus illusion

What's the general name for optical illusions of the kind of the famous "Jesus illusion"? Icek 20:19, 1 December 2007 (UTC)

I'm still looking, but if you like that, you'll love the lilac chaser. --Milkbreath 20:44, 1 December 2007 (UTC)

It's called an afterimage, explained by Ewald Hering. Sadly, it is not called the Hering illusion, though. This site calls it a "stare negative". --Milkbreath 21:08, 1 December 2007 (UTC)

With a dose of pareidolia thrown in for fun. Matt Deres 02:07, 2 December 2007 (UTC)

snowing

hello im shima ghadiri im from iran and its my first time that i meet your site and now im a new member of your site and i have really strong confidence to be areally active member im 15 years old and i study math in grade two in high school i have alot of question in my mind and i thouth this is a best way to find some one in your sitemaby some abroad iranian at the end my english is not raelly good but i can express my idea thank u alot. my first question is:why when is snowing we dont have any i dont know exact word for example light in sky but when rainig we have alot of light in sky that in persion we say radvabargh?

Do you mean a rainbow? --Reuben 20:54, 1 December 2007 (UTC)

For language questions, try our Wikipedia:Reference desk/Language. Of course we do have a Wikipedia that's written and edited entirely in Persian/Farsi here - it has over 28,000 articles (including one about rainbows [26] !) - but that's only a small number compared to the number we have in English. Perhaps a good way to practice your English would be to translate some articles from the English Wikipedia into Farsi - there are plenty to choose from! SteveBaker 22:12, 1 December 2007 (UTC)

Well, if the question is "why do we see rainbows when it rains, but not when it snows?" that's certainly a science question. The answer is that raindrops are close to spherical in shape while snowflakes are not. This matters because a sphere is very symmetrical and therefore when light from the Sun hits a raindrop at a certain angle it will always emerge at a particular angle (for the same color of light) -- the same angle from one raindrop to the next. What makes a rainbow is the rays of light emerging from different raindrops in the same direction. Snowflakes have complicated shapes and the flakes in a mass of snow are all facing different ways. This means that the light from different snowflakes does not point in the same direction, so it does not form a rainbow.

Another possible form of water in the sky is ice crystals. These have shapes more complicated than raindrops but simpler than snowflakes. In some conditions they become aligned (facing the same way) and then you can get an effect somewhat like a rainbow. There are different types; two of them are called sun dogs and halos.

--Anonymous, edited 22:29 UTC, December 1, 2007.

I just wanted to say that that is a delightful and very informative answer... - Nunh-huh 22:51, 1 December 2007 (UTC)

December 2

High temp superglue

Is there a version of super glue that will withstand soldering temperatures (say 250 C) for a short time without failing?--TreeSmiler 01:30, 2 December 2007 (UTC)

I would try some kind of epoxy instead. —Preceding unsigned comment added by Shniken1 (talk • contribs) 02:09, 2 December 2007 (UTC)

Many electronic components (transistors, capacitors, etc) are sealed in plastic or epoxy of some type and clearly are able to withstand soldering temperatures for a short time. RTV sealants (room temperature vulcanizing, I think, is what the RTV stands for) can withstand high temperatures (used to attach space shuttle tiles, as I recall) and comes out of a tube in liquid form. Edison 02:50, 2 December 2007 (UTC)

Modern printed circuit boards are assembled with components on both sides and use "interesting" mass soldering techniques. In some processes, the components are glued to the board before the actual soldering step. -Arch dude 11:18, 2 December 2007 (UTC)

ant trails

Do ants leave trails the same way cats mark boundaries by spraying at certain spots and if so what determines the spot or decision criteria at which to spray? Also is there a die that will reveal the trail chemical and can ant be genetically programmed to release a visible spray? 71.100.1.143 01:39, 2 December 2007 (UTC)

See ant. It clearly states that ants leave a pheromone trail for other ants to follow. They make a trail - not a random spot here and there. As for genetically altering ants to make their pheromones visible, it is possible but not probable. As for being able to see pheromones - I know of nothing in current science that makes pheromones visible (except for Hollywood CGI magic, of course). -- kainaw™ 02:21, 2 December 2007 (UTC)

Ever hear of bread crum trails? Be it bread crumbs or pheromones they are only a mark and a rather scarce mark at that which requires they not be laid end to end but at worst only at decisive intervals. 71.100.1.143 05:07, 2 December 2007 (UTC)

(ec) I think the trail ants leave is formic acid (certainly there's a well-established association between that acid and those insects), but our Ant article just says "pheromones". —Steve Summit (talk) 02:23, 2 December 2007 (UTC)

• I vaguely remember that in one of Richard Feynman's memoirs he describes somehow smearing ant trail stuff in patterns of his choosing, which caused the ants to walk in those squares and circles and things. :) --Sean 03:43, 2 December 2007 (UTC)

Yep - I was going to mention the exact same thing. He found that ants were getting into his larder and rather than killing them or spraying with nasty chemicals, he simply devised a way to have their scent trail lead away from the food and back outside. It's described beautifully in several of his biographies. It's a wonderful example of how you can be an amateur scientist, discover interesting things by yourself and apply them to your own benefit. (Feynman is my personal hero BTW). SteveBaker 05:24, 2 December 2007 (UTC)

QUESTION ABUOT TEH TERMINATOR LOLZ (seriously!)

"I HAVE BEEN SENT HERE TO DESTROY-- I MEAN, CONFUSE HUMANITY."

Hi wikipeoples! What are the minimum and maximum ground speeds of THE TERMINATOR of Earth??? Kreachure 02:33, 2 December 2007 (UTC)

The max speed would be at the equator - regardless of axial tilt. That is the point with the greatest circumference since the Earth bulges out in the middle. The speed of rotation at the equator is 465.11m/s. The minimum would be when the tip of the terminator hits one of the poles. It stops traveling and becomes 0m/s. -- kainaw™ 02:45, 2 December 2007 (UTC)

...OP says, "of" Earth, not "on" Earth. 71.100.1.143 05:10, 2 December 2007 (UTC)

Yes - but complaining about small grammatical slipups on the part of our questioners is not considered cool. We know what was intended. SteveBaker 05:19, 2 December 2007 (UTC)

Wait, is "of Earth" really ungrammatical? Maximum ground speed of terminator of Earth = maximum speed of Earth's terninator, while maximum ground speed of terminator on Earth = max speed of the terminator that's on Earth or max speed of terminator while on Earth. Both are appropriate, I think. --Bowlhover 16:07, 2 December 2007 (UTC)

Thanks Kainaw. But geez, you guys always find a way to make a questioner feel, um, inadequate about his/her questions, huh? The definition of terminator in the article is, "the line between the illuminated, day side and dark, night side of a planetary body." So, no mistake as far as I'm concerned! Kreachure 16:36, 2 December 2007 (UTC)

HIV (AIDS)

I want to know if a person (girl/boy) willingly or unwillingly drink a person urine, is there any chance of getting HIV infection from the same. —Preceding unsigned comment added by Ashish.k.garg (talk • contribs) 03:35, 2 December 2007 (UTC)

• Didn't you just ask this last week? May I suggest that you stop drinking people's urine until you've discussed this practice with your doctor? --Sean 03:48, 2 December 2007 (UTC)

As for the question of any chance. Blood-to-blood transmission is known to be a method of HIV infection. If there is blood in the urine and open wounds in the mouth, then there is a chance. Of course, you should see a doctor if you are concerned about HIV. The test is simple and rather accurate. -- kainaw™ 03:59, 2 December 2007 (UTC)

• How dare you attempt to trample on this user's inalienable right to engage in Urophagia;-)--Fuhghettaboutit 05:23, 2 December 2007 (UTC)

• S/he can engage in coprophagia for all I care, as long as s/he asks a doctor about it rather than the Wikinerd Collective. --Sean 16:52, 2 December 2007 (UTC)

Chicxulub crater and Plate tectonics

It occurred to me that the theory of an impact event at Chicxulub 65 million years ago should be deformed by continental drift. I was reading up on Pangaea and linked to one of its references, a USGS site. Which discusses its break-up and the formation of the geographic Earth as we know it today. If this image is accurate then there are two plates pushing north, one to the ne and the other nw, south of Chicxulub. Surely I must've missed something, as I'm not a geologist, but I'm just wondering what that is.

This also prompts the question of our ability to accuratly document IEs even earlier. Since Pangaea wasn't the first super continent, just the most recent it seems like craters, mountains, volcanoes, etc. could have been Etch-a-sketched away by continental drift/collision the further in the past they occur. Anynobody 08:31, 2 December 2007 (UTC)

65 million years is not a long time span in the context of plate tectonics. Pangaea was already pretty much split up by the late Cretaceous, and since then its constituent plates have just moved further apart. And remember that the interiors of the plates do not deform - spreading, subduction and mountain building happen at the boundaries between plates. The main reason that it is very difficult to clearly identify craters from older impact events is the removal of surface features by erosion, which happens over shorter timescales - the Grand Canyon, for example, is less than 6 million years old. Gandalf61 11:07, 2 December 2007 (UTC)

This and that book approach

In university, I'm faced with books with titles such as

• Computer Architecture: A Quantitative Approach
• Software Measurement and Estimation: A Practical Approach
• Electromagnetic Field Theory: A Problem-Solving Approach.

I find these approach titles so abstract and, reading in these books, I cannot find any difference in the style of writing among them. Can someone try to come up with definitions or rule of thumbs about what an author wants to tell the readership when he decides on an XYZ Approach title? Thank you, --Abdull 10:10, 2 December 2007 (UTC)

Without actually answering the question, I'll point out that it's the publisher that chooses the title of a book (although normally in consultation with the author). Which means they take marketing into account. --Anon, 10:41 UTC, Dec. 2, 2007.

Gah, edit conflict with Anon, but I'm inclined to agree - here's what I was in the process of posting...

I've always assumed those kinds of books mean the author thinks s/he has a new, useful, or particularly interesting method of approaching the subject. In the examples you gave above, the author of the software measurement book obviously believed that the approach taken was a particularly practical one (therefore implying, I suppose, that other books take a less practical approach). I really think it's mostly down to the author's point of view, and maybe just a little bit of marketing. After all, which would be first to catch your eye in the bookstore - 'Software Measurement and Estimation: A Guide' or 'Software Measurement and Estimation: A Jellybeans-and-donuts approach'? ;) --Monorail Cat 10:48, 2 December 2007 (UTC)

"Computer Architecture: a quantitative approach" was the first mainstream book on the topic that used real measured data (mostly from the VAX) to explain the differences and trade-offs (things like "backward branches are taken 9X% of the time, forward branches 50%", "90% of of the time is spend in 10% of the code", locality of reference). It still is an excellent book by some of the early RISC pioneers. --Stephan Schulz 11:14, 2 December 2007 (UTC)

fibre transfer forensics

82.7.252.205 11:36, 2 December 2007 (UTC)Can anyone tell me when the use of fibre transfer forensics was first used and who first used it ?

• Fibres are generally considered trace evidence. That term might give you better search results. Trace evidence is based on Locard's exchange principle, but I wouldn't be surprised if someone used it before he (Edmond Locard) made his postulation. It would be entirely possible for such evidence to have been used a few times in ancient Rome, Greece or Egypt. However, processing of most trace evidence relies on microscopy, so if you go that route, you wouldn't need to go any further back than the 1600s when Antonie van Leeuwenhoek made the first useful microscope. - Mgm|^(talk) 13:57, 2 December 2007 (UTC)

According to Lens, the oldest lens is dated to c.640 BC - so whilst microscopes certainly weren't around in any numbers until the late 1600's - simple magnifying glasses would have been around much earlier. These wouldn't allow you to identify fibres with the exactness demanded by modern judicial standards - but they'd certainly have been enough to give a lynch mob the information they needed to go on if any of them took the time to check. So I don't think this is a matter of when the technology became available so much as when the methodical collection of evidence and control of the crime scene became commonplace. SteveBaker 14:41, 2 December 2007 (UTC)

Forensic science#History of forensic science has some examples of the early use of trace evidence in criminal proceedings. Gdr 15:00, 2 December 2007 (UTC)

Flushing toilet

Hi. I've heard that flushing the toilet can cause fecal coliform bacteria to land on your toothbrush, and even enclosed toothbrushes get exposed to it (MythBusters). However, I know a few people who close the toilet lid when they flush. Does this help keep the toothbrushes free of f. coli bacteria? Or, will peeing and pooing in the toilet cause water torise up, carring fecal germs around the room? To approximately what efficiency (percent) will closing the toilet lid have with preventing contamination of toothbrushes? Thanks. ~AH1^(TCU) 16:23, 2 December 2007 (UTC)

I know I'm supposed to Assume Good Faith, but before I answer, do you mind if I ask: how serious are you in asking this question? Are you truly concerned, or do you just enjoy talking about peeing and pooing, and egging on other people to do the same? (Me, I've never heard of any concerns about flush toilets and toothbrushes, although not being an OCD sufferer, perhaps I've been sheltered.) —Steve Summit (talk) 16:39, 2 December 2007 (UTC)

• If you google "Template:Websearch", you'll see it's not a novel concern. --Sean 16:56, 2 December 2007 (UTC)

Hi. This is intended to be a serious question. I want to know if closing the lid will prevent toothbrush contamination. If yo watched that episode of MythBusters or read the book you will see that they confirmed the presence of f. coli bacterial after a few days of bathroom use. Thanks. ~AH1^(TCU) 17:07, 2 December 2007 (UTC)

Yeah - but they also proved the stuff is EVERYWHERE. it's just a natural part of the habitat in which humans live. The Mythbusters finding was absolutely not that toothbrushes closer to the toilet had a bigger does - it was that it didn't matter where you put the toothbrushes (including in the kitchen, 50 feet from the bathroom) - the results were substantially the same. However, we humans have perfectly good defenses against these things. You are doing yourself far more damage from stressing out over it. SteveBaker 17:35, 2 December 2007 (UTC)

Who will die earlier?

Two equally healthy people are put on the test. One is without water and the other is without food. Who will die earlier and why? Also, How long is going to be the difference between their deaths, approximately? DSachan 16:47, 2 December 2007 (UTC)

If the food contains enough fluids, the one without water might die of old age. Lova Falk 17:30, 2 December 2007 (UTC)

You can't survive without water more than a couple of days - you can go without food for weeks. SteveBaker 17:30, 2 December 2007 (UTC)

Camera settings for astrophotography

Hi. Say I wanted to photograph Comet 17/P Holmes sometime in January using a digital camera, a 4 - 6 inch Equatorial Reflector in light-polluted skies (although there are no lights very close to the location of the comet), and Registax. Now, I want to ask approximately what camera settings I should use. I'm thinking of photographing it while it is near either M34 or Algol, under approximately 50x. Which is a better choice for a better image? The camera is capable of up to 10-, 13-, and 15- second exposures. How many images in total should I take, before stacking them? For the aperture, should I use a focal length of 2.8, 3.2, 3.5, 4.0, 4.5, 5.0, 5.6, 6.3, 7.1, 8.0, or should I let the camera set the aperture automaticly? Should I hold the camera using my hands and telephoto, or should I use a special camera adapter to hold the camera steady? Should I use the best resolution and compression avalible? Should I store the images as JPEG or RAW, and what's the difference? Should I leave the focus frame in the centre, or adjust it according to where the comet's nucleus is? Will I be able to visually view the comet's coma in January, using this type of telescope, under light polluted skies, prior to taking the images? Should I zoom in so the FOV of the eyepiece completely fills in the FOV of the camera? What light metering system should I use: evaluative metering, center-weighted averaging, or spot metering? Should I use strong contrast, sharpness, and saturation? What ISO speed should I use: 50, 100, 200, or 400? Thanks. ~AH1^(TCU) 17:02, 2 December 2007 (UTC)