Wikipedia:Reference desk/Archives/Science/2008 July 10

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July 10

PMS citation needed

Dear Mr. editor,

I assume only male editors of a certain sort (who don't know women well, or aren't even interested) would feel that, under the description of PMS symptoms, "increased arousal and sex drive" needs citation as it might be "challenged"--by other males, no doubt. There is such a strong desire in the patriarchy to see generally normal conditions women go through (conditions that are counterproductive and not useful to Wall Street, or the male and extremely virile Spirit of Business) as 'diseases' in need of cures, that it has blinded many young men from discovering women for themselves. It makes many of us women who are self-aware and highly educated feel uneasy.

As a 27 year old woman who "suffers" from increased arousal and sexual desire before the onset of the period, and who knows many others who do, I find the feeling of the editor offensive.

pissed off in Brooklyn! —Preceding unsigned comment added by 69.112.205.26 (talk) 00:40, 10 July 2008 (UTC)

...did you have a question somewhere in there?

Oh, never mind, I'll answer one I am going to pretend you asked: just about every piece of information in Wikipedia should be cited in one way or another. (A lot of the time this doesn't happen, but that's still the ideal we strive for.) Why? Because it means that the information we put in here is verifiable, which means that if you read it, you can go back and see that it's not just something somebody on the internet made up.

Let me put it this way: a male editor of a certain sort, who doesn't know women well or isn't even interested, and is definitely kind of a dim bulb -- you may have met people like this -- may believe any kind of insane and stupid crap about women he happens to read on Wikipedia, even if it happens to be written by some malicious and misogynistic jerk. He has no way of knowing whether that information is any more real than the information about increased arousal and sexual desire before the period, and if there are no references cited, he has no easy way of checking whether this it's true. Whose purposes does that serve? (You may argue that people aren't that stupid, and I can only point out that a lot of people are, in fact, exactly that stupid.) If the facts we put in are verifiable, that means he can check the source for the information and see whether it's true. -- Captain Disdain (talk) 01:04, 10 July 2008 (UTC)

_______________________________ Thanks, Captain, for replying.

The question was implicit, which you responded: why the need for citation. While I completely agree with you, what ticked me off was that that was the ONLY 'symptom' the editor FELT needed citation, which I didn't emphasize above. So why not all of them?

For any of us who have even cursory knowledge of how women's sexuality and completeness as human beings has been denied throughout the ages ('penis-envy,' 'frigidity,' dispute of the existence of so-called 'G-spot'--basically calling into question anything that suggests women's affirmation of their 'virility'), a need for citation on particularly THAT point can appear as a symptom of something in need of overcoming. And I don't mean 'overcoming' as in practicing political correctness and affirmative action, that doesn't work. What works, and what every intelligent man knows, is paying attention to women's experience of their own bodies, beginning with one's girlfriend.

It's acceptable to tag any uncited statement with "citation needed" based on only a feeling. The policy on Wikipedia is to assume good faith; if an editor's actions can reasonably be explained as an attempt to improve the encyclopedia, we should treat it as such. A quick glance at the external links in premenstrual syndrome revealed that none reported increased sexual desire as a symptom. In fact, http://familydoctor.org/online/famdocen/home/women/reproductive/menstrual/141.html even listed "not feeling as interested in sex" as a characteristic of PMS. [1] and [2] list only changes in libido, implying libido sometimes increases and sometimes decreases. It's very probable that the editor who inserted the "citation needed" tag actually did believe that the statement regarding sexual drive is false, for there are plenty of reasons for that belief.

69.112.205.26, please sign your posts using ~~~~, like so: --Bowlhover (talk) 04:24, 10 July 2008 (UTC)

Another reason it may have been tagged is because the symptoms section is globally cited with 3 sources. I quickly browsed those sources and the 'symptoms' you mentioned aren't on any of them (I didn't check the others individually, just your fun ones). Maybe the people who originally posted those symptoms recognized a new one that wasn't from their citations and wanted it cited separately. Also, as a guy I'll admit reading that part of the article had a "really?" effect on me. Browsing around I saw various sources comment that there are more than 150 listed symptoms for PMS, so it must be listed in the literature somewhere if you want to go to the trouble of improving the article. --Shaggorama (talk) 06:04, 10 July 2008 (UTC)

The {{fact}} tag appears to have been added by a female editor.[3] Assumptions are always dangerous, in real life and here on Wikipedia. Franamax (talk) 08:32, 10 July 2008 (UTC)

The user is asking an excellent question. There is a good deal of information about the subject, though it may not be easily obtainable without a trip to a good academic (university) library. Here are two scholarly references that summarize a great deal of prior information.

Hedricks, Cynthia A. 1994 "Sexual behavior across the menstrual cycle: A biopsychosocial approach." Annual Review of Sex Research, 5: 122-172.

Wood, James W. 1994 Dynamics of Human Reproduction: Biology, Biometry, Demographics. NY: Aldine de Gruyter. Chapter 7: "Fecundability and Coital Frequency". See especially pp. 305-314, "Empirical patterns of intercourse."

Summarizing, the pattern of arousal and intercourse across the menstrual cycle is not simple. One pattern is a sharp decrease in intercourse during menses, followed by a sharp increase, a decrease, another increase around mid-cycle, another decrease, and another increase toward the end of the canonical 28 day cycle. Other women enact a different pattern, with a large mid-cycle increase, and still others have an increase toward the end of the 28 day period, as the questioner points out. All of these patterns are normal, meaning that they occur in women in good physical health and are not the result of physical or mental illness. Note that no single pattern characterizes all or most women.

Both Hedricks and Wood go into a great deal more detail, and give figures and citations to this body of work. The questioner might find the Hedricks and Woods' references very interesting.

Concerning the issue of verifiability and anger about what the questioner took to be knee-jerk masculine disdain and ignorance, I would like to apologize to her on behalf of Wikipedia. I also feel that Captain Disdain owes the questioner an apology for an uncivil, indeed, impolite answer.

Once again, it is an excellent question with very complex answers. I hope that I've at least supplied a starting point for more investigation and reading.

Timothy Perper (talk) 16:31, 10 July 2008 (UTC)

I would have to disagree with your apology suggestion. I don't find Captain Disdain's reply at all rude. In fact, the original questioner was rather rude and appears to be going on a long off topic rant that has nothing to do with the reference desk and didn't even ask a clear cut question in the first instance. Not only were the somewhat offensive assumptions of the questioners pointless, they have also been shown to be likely incorrect which just goes to show why you should never make pointless assumptions (the questioners could have simply 'asked' why it was the only sentence fact tagged without the long rant.) Captain Disdain did an admirable job under the circumstances. If the question asker had bother to do a bit of reading on policy and a bit of research, she could have improved the article herself and maybe even helped "blinded young men" and "self-aware and highly educated (women)" by producing a better quality and properly referenced article. Nil Einne (talk) 18:01, 10 July 2008 (UTC)

I quite concur. It's not the easiest thing in the world to be simultaneously angry, misandrist, and homophobic, but our original questioner manages it as though it were second nature to her. Perhaps because it is. - Outerlimits (talk) 18:06, 10 July 2008 (UTC)

Dang, I just lost out on being first to say that Captain Disdain has nothing to apologize for, they did their best to answer a (non-)question neutrally, the user went on to thank them for their answer. And the question was about referencing, not asking for the references themselves. Nil Einne, I'm not too happy though with you discussing a "pointless off-topic rant", the OP did have a point on a particular topic, they just didn't make it very clear. Franamax (talk) 18:05, 10 July 2008 (UTC)

Well, thanks. I wasn't going to anyway, though; I don't think I was being uncivil. A little blunt, sure, but certainly no more so than the original post. (Which, for the record, I didn't find to be especially uncivil, either. Angry and a kinda knee-jerky, sure, but I understand why someone would wonder about that.) -- Captain Disdain (talk) 09:13, 11 July 2008 (UTC)

Nonetheless, this is a reference desk, not a ranting desk. Or maybe that's just me lording my patriarchal power over Wikipedia. --LarryMac | Talk 18:10, 10 July 2008 (UTC)

Wow, are all the 14-year-old boys logging on today? If you can't help answer the question, don't post to the thread. This is a reference desk, not a place to retaliate in kind to people asking questions. If the question doesn't make sense, ask for a better question. We're supposed to welcome people here. Franamax (talk) 18:55, 10 July 2008 (UTC)

Yes, by calling them immature boys, for example. It's quite appropriate to point out that misandry and homophobia are not welcome here. This is a public forum and here one should conduct oneself decently and without bigotry. - Outerlimits (talk) 03:49, 11 July 2008 (UTC)

Not so much fun when you think the comments are aimed at you, is it? Outerlimits, you made what looks to me like a direct personal attack on a particular person, please review WP:NPA and don't repeat such actions. I, on the other hand, while being somewhat POINTy, did not name anyone, and in fact I was referring to the immaturity of some of the responses. There is a difference between commenting on the contribution, and on the contributor. Franamax (talk) 07:15, 11 July 2008 (UTC)

It's not a matter of fun. Denigrating all 14-year-old boys is certainly not better than actually confronting a specific behavior of a specific person. That your comments were nebulous doesn't improve them. That they assault a group rather than the person that irks you is a mark of the immaturity you claim to eschew. And directing people to policy pages as though they weren't familiar with them is not much better. I contend that WP:NPA supports informing people that racial, sexual, homophobic, ageist, religious, political, ethnic, or other bigotry is never acceptable. As for the original questioner: when someone spews their homophobia in public spaces, I will continue to note that it is inappropriate. Sorry you object. - Outerlimits (talk) 00:39, 12 July 2008 (UTC)

This is a place to answer questions with courtesy and the assumption of good faith. I won't respond to the other comments. They have nothing to do with the question asked. We are here on Wikipedia as a public service, and not to insult people. If the references I supplied helped, well, good, that's what an encyclopedia does. I am not here to call people names, like angry, misandrist, homophobic, pointless, and so on. Leave that home when you sign on to Wikipedia. Timothy Perper (talk) 19:39, 10 July 2008 (UTC)

Good faith can't be assumed when bad faith is already in evidence. It's quite appropriate to point out that bigotry is not to be tolerated here. We must maintain an editing environment that doesn't cater to popular prejudices. - Outerlimits (talk) 03:49, 11 July 2008 (UTC)

Trying to provide a good-faith answer: PMS/PMDD has a very loose, variable set of symptoms, some of which are contradictory from woman to woman(or even within the same woman's experience). For instance, one might lose her appetite and even become nauseous, while another might have cravings for food she normally dislikes. So if you're dealing with a list of symptoms that anyone-- including any woman-- can edit, there will be "O RLY" reactions from some woman as well. (I'm a woman, by the way, and I was a little surprised at your entry "increased need for emotional closeness", because in my experience I usually want to be left alone at that time.)207.233.86.98 (talk) 20:49, 10 July 2008 (UTC)

Personally, I found the OP's comments dubious. I am sexist, yet I've never before heard of men denying that women experience sexual arousal. I understand that she found it unusual for one symptom in a list, instead of the list itself,to be tagged with "citation needed", but her attribution of this to misogyny was unresearched speculation.

It's appropriate to inform the OP that her behaviour violates Wikipedia's policies, but since she was a new editor, it could have been done less harshly. --Bowlhover (talk) 19:45, 11 July 2008 (UTC)

High Frequency Electric Signal Over a Buffered Solution

When measuring the capacitance of electrodes in a buffered solution, why is it advisable to operate the circuit at a high frequency (above 1 MHz)? Is this to avoid electrolysis? —Preceding unsigned comment added by 140.247.153.158 (talk) 02:12, 10 July 2008 (UTC)

Do trees ever grow up ?

Say one of those lovely Druids nailed somebody to a tree so they would wiggle less while being disemboweled, would that nail still be the same height from the ground (assuming the soil level hasn't changed) ? I'm guessing that for a normal tree the height wouldn't change, but for bamboo and other similar fast-growing trees that it would be higher. StuRat (talk) 03:31, 10 July 2008 (UTC)

For any tree that is still increasing in height, you would expect that nearly all portions of its trunk would continue to get higher over time. Upward growth in trees includes not just growth at the tips, but pushing and stretching along nearly the entire length of the tree. Dragons flight (talk) 04:59, 10 July 2008 (UTC)

Hmm. I just erased my post saying that trees definitely do grow upward over time, based on my experience seeing fences lifted way above where they were originally strung. I know for sure that young trees grow upwards and I recall seeing fence-lifting on more mature (8-10" dia.) trees too. I'm stuck at this question though: most of a mature tree trunk consists of dead cells, other than the heartwood and cambium if I'm not mistaken (and I probably am) - so how would those dead cells grow upward with the tree? Do they separate and become less dense? Franamax (talk) 08:14, 10 July 2008 (UTC)

Mmmmh - 'fresh' cells can 'grow' by stretching ie elongation. At 8-10" maybe the cells still have some stretch left. By the time the tree is say 1foot diameter or more I'd imagine that the only way it can grow is out ie around.87.102.86.73 (talk) 10:50, 10 July 2008 (UTC)

Apologies if this would be a second question, but wouldn't they also grow out, stretching the remains over time? I guess it depends on the type of tree, I'm thinking of giant oaks here...209.244.30.221 (talk) 14:14, 10 July 2008 (UTC)

Bamboo is a grass, not a tree. — DanielLC 16:34, 10 July 2008 (UTC)

Bamboo is definitely a grass, but I think it meets our definition of tree as well. "Tree" is a growth form, not a clade. But then our definition of tree, requiring apical dominance, seems odd to me. A typical open-grown oak (surely a tree) doesn't have apical dominance, does it? (And I hope somebody knows the answer to Franamax's question.) --Allen (talk) 17:49, 10 July 2008 (UTC)

A tree doesn't "grow" in the way that you or I grew (where our head slowly moves higher and higher from the ground). Any given point within the envelope of a tree will remain in exactly that same place (+/- some bending and bugs) forever more. What happens is that every year, an entirely new layer of tree is deposited on the entire outside of the old tree.

The upshot is that branch you walked into mowing the lawn ten years ago is probably still there in exactly the same place to whack you on the head this year, but it's now a thicker, stronger branch. Well, unless it got shaded by things above it in which case the tree, forever optimizing its exposure to sunlight, may have discarded the branch entirely.

So if you're nailed to a tree and the nails were inserted dead level (horizontally), you'll be located at about the same height up the tree for as long as it takes you to fall to pieces.

Atlant (talk) 20:15, 10 July 2008 (UTC)

You may be right about trees after a certain stage of development, but your statement certainly doesn't apply to seedlings, so maybe a better way to ask the question is, "When, and how quickly and completely, do trees lose the capacity for stem elongation?" --Allen (talk) 21:45, 10 July 2008 (UTC)

I would imagine the transition point is that point when the tree contains dead wood (heartwood) within it. Those dead wood cells certainly aren't going to change form very much.

Atlant (talk) (yesterday relative to 12 July 2008)

Exactly. It's really hard to envision people stringing a fence, nailing it to a tree, and while doing that stretching it up a foot above the ground, so tight it's pulling up the adjacent posts. It's a little easier to think that the tree grew upwards after they nailed the fence onto it. Now we have two open questions, Allen's and mine. Plus maybe StuRat's :) Franamax (talk) 06:58, 11 July 2008 (UTC)

It is possible the ground has subsided in such a case, but then I'd expect to see exposed roots. StuRat (talk) 20:35, 11 July 2008 (UTC)

The reason I listed a possible exception for bamboo, whether it's a tree or not, is that it seems to me they must grow at all the joints along the trunk, not just at the tips, to achieve their rapid growth upward. Does anyone know if this is indeed the case ? StuRat (talk) 20:35, 11 July 2008 (UTC)

Ammonium methanide

Is ammonium methanide an existing chemical compound? - Xxxx00 (talk) 07:28, 10 July 2008 (UTC)

By methanide do you mean CH₃^- or maybe carbide - either way the compound doesn't exist because both anions are too basic and will deprotonate the ammonium giving ammonia and a hydrocarbon.87.102.86.73 (talk) 09:47, 10 July 2008 (UTC)

There's also 'methanide' which is a generic tradename for a compound see http://www.chemindustry.com/chemicals/573762.html and http://www.flexyx.com/M/Methanide.html

This compound is already a ammonium type compound see quaternary amine87.102.86.73 (talk) 10:30, 10 July 2008 (UTC)

Thanks. --Xxxx00 (talk) 16:44, 10 July 2008 (UTC)

Decibel question

I apologise in advance if this is a stupid question: is it possible to convert between two different types of decibels? Specifically, from dBu to dBm. Thanks! Xenon54 11:03, 10 July 2008 (UTC)

Not without further information, since the dBm is a unit of power, while the dBu measures voltage. Algebraist 11:24, 10 July 2008 (UTC)

What do you mean by further information? If it helps any, I need to convert 56 dBu to dBm. Xenon54 12:07, 10 July 2008 (UTC)

You need to know the impedance. Is it 50 ohms, 75 ohms, 300 ohms or 600 ohms? —Preceding unsigned comment added by 79.76.204.247 (talk) 14:14, 10 July 2008 (UTC)

Seventy-five ohms. Xenon54 15:11, 10 July 2008 (UTC)

+56 dBV is one hell of a voltage especially in 75 ohms. Dont you mean -56 dBV? —Preceding unsigned comment added by 79.76.204.247 (talk) 15:36, 10 July 2008 (UTC)

I meant dBμV (aka dBμ and dBu). Sorry I didn't make that clear. Disregard this question, I found the answer anyway. Xenon54 15:45, 10 July 2008 (UTC)

And electric power is voltage times current/ --Bowlhover (talk) 11:32, 10 July 2008 (UTC)

GAS TO ELECTRIC POWER

QUESTIONS: 1. PLEASE,HOW IS GASES BEEN CONVERTED TO GENERATE ELECTRIC POWER?

2. WHAT IS THE BASIC PRINCIPLES BEEN APPLIED? —Preceding unsigned comment added by 41.205.163.220 (talk) 12:07, 10 July 2008 (UTC)

This really sounds like a homework question, so I doubt anyone's going to just give you the answer. You may want to check out our articles on natural gas and gas turbine. Also, I know that caps lock is cruise control for cool, but it'd still be nice if you didn't SHOUT. -- Captain Disdain (talk) 12:35, 10 July 2008 (UTC)

You should check out this website: http://www.howstuffworks.com/search.php?terms=gas+electric+car. Maybe this will help you. Sonic99 (talk) 01:53, 11 July 2008 (UTC)

pharmacy education in europe

i was studying b.pharmacy 3year i listen in europe some countries offer free or lessfee education where i can get total information —Preceding unsigned comment added by Anil kakkerla (talk • contribs) 12:29, 10 July 2008 (UTC)

Some European countries offer free or subsidised tertiary education to their citizens and (often) long term residents; students from abroad generally pay hefty fees (indeed many British universities are heavily dependent on the large fees they extract from foreign students). -- Finlay McWalter | Talk 12:39, 10 July 2008 (UTC)

Several Nordic countries, such as Sweden and Finland, offer free tertiary education for everyone; there was a fairly long thread about that on the Miscellaneous refdesk some weeks ago. However, these countries also tend to have relatively high costs of living, so what you gain in free tuition you may lose in expensive food and housing etc. —Ilmari Karonen (talk) 19:30, 13 July 2008 (UTC)

Exercise

With which of the following exercises would I use more energy?

• A: Cycling for 30 minutes at a heart rate of 130 bpm
• B: Increasing the resistance/speed so that my heart rate reaches about 150 bpm but I'll only be able to last about 13 minutes.

I cycle at the gym so air resistance isn't a factor at higher speeds. Thanks, Zain Ebrahim (talk) 13:03, 10 July 2008 (UTC)

According to the manual for the bikes in our PT room, the formula for calories burned (energy used) is: calories/hr = (VWR÷100+.0083V³)×7.2. V is the average velocity of the bike (if it wasn't stationary). W is the weight of the rider. R is the resistance setting. I assume that resistance settings change from bike to bike, but you can see that heart rate is not part of this equation. -- kainaw™ 13:17, 10 July 2008 (UTC)

I think that's based on some kind of average of human muscular energy expenditure. If you go with increased resistance you'll probably build a bit more muscle and stronger muscles use more energy (So, I'd personally go with B) -LambaJan (talk) 13:48, 10 July 2008 (UTC)

I don't know the exact relationship, but I do know anecdotally that heartrate is typically considered very closely tied into the level of exertion and the number of calories burned in an exercise. Can anyone confirm/deny? The article on aerobic exercise is mute on the issue. EagleFalconn (talk) 13:51, 10 July 2008 (UTC)

An increase on 20 bpm isn't very much at all. You aren't burning many more calories per minute at 150 bpm than at 130 bpm. If burning calories is your goal you are better off working out on setting A for 30 minutes than at setting B for less than half that time. Plasticup ^T/C 16:28, 10 July 2008 (UTC)

I think the pertinence of heart rate is that you want to work hard, but aerobically. If your heart rate during exercise is outside of a particular range, your muscles don't receive enough oxygen and becomes more reliant on an anaerobic metabolism resulting in acidosis. In other words, watching your heart rate allows you to do more work longer. --Shaggorama (talk) 18:56, 10 July 2008 (UTC)

African Americans and Sickle Cell Anemia

What will happen if a African American from USA who has Sickle Cell Anemia contracts a bout of malaria? --Anthonygiroux (talk) 13:04, 10 July 2008 (UTC)

Those with sickle-cell anemia should have double sickle-cell genes (one from each parent), which would make them quite resistant to malaria. So, if they did manage to contract it at all, the symptoms should be quite mild. StuRat (talk) 13:44, 10 July 2008 (UTC)

Non-steam coal engine technology?

Is there a type of technology that would use plenty of coal to burn, but does not generate steam? I am thinking of a hypothetical renaissance or industrial society that has access to mining and lots of coal (or a similar fossil fuel??) but very little water. (I'm assuming that steam engines require a quantity of water to boil.) Could the smoke alone from the coal or other primitive fossil fuel be used instead of steam to power heat engines, gears and primitive machines, possibly with the help of bellows, etc.? --Sonjaaa (talk) 13:51, 10 July 2008 (UTC)

A majority of energy and electricity generating technologies work on precisely the principle you describe. By burning the coal, water (or in the case of solar thermal energy a high boiling liquid) is boiled and the steam used to power a turbine. I guess you could use alternatives such as some other liquid that boils lower, such as ethyl acetate or acetonitrile, I personally don't know if there are any particular aspects of water that make it more desirable than other liquids besides its ubiquity on Earth and non-toxicity (which are pretty good reasons). FYI, even nuclear reactors operate on this principle...one of the major challenges and one of the really cool things happening in alternative energy is moving away from the turbine model because it is somewhat wasteful (heat transfer to the turbine, moment of inertia requirements for rotation, losses due to friction). Among the technologies that do this are solar panel, though even wind power is just a turbine turned by wind instead of boiled water. EagleFalconn (talk) 14:16, 10 July 2008 (UTC)

If you were to use the gases resulting from the burning of the coal alone, I guess in theory, you could generate power that way, but it'd be so inefficient as to be completely pointless. Heating water gives you a lot of steam to work with, and high-pressure steam is pretty powerful. (Steam engine explosions can be pretty devastating.) The reason for this is that when a liquid turns into a gas, it expands a great deal and thus generates pressure. You can harness that power.

Just burning coal, on the other hand, generates heat and smoke, but not all that much pressure; you get some gas out of it, sure, but nowhere near enough to generate a meaningful amount of pressure -- it's essentially just carbon and whatnot carried on the hot air that rises up from the fire. Using bellows wouldn't really make the smoke pressure any higher, it'd just help the fire get hotter by delivering more oxygen into it. (I guess you would get some additional pressure from the act of pumping in the air, but then you could just pump air in the first place and skip the part with the coal altogether.) Of course, an additional problem would be that in order for the coal to burn in the first place, it would need two airways -- one to remove the carbon that would choke the fire (a chimney, in other words) and another to let in fresh oxygen. This means that there wouldn't be a great deal of pressure building up, since there would be nothing to contain it. I guess you could rig up some sort of a wind mill type of thing operated by the hot gases surging up the chimney, but it wouldn't be efficient. The pressure would just be too weak. -- Captain Disdain (talk) 14:25, 10 July 2008 (UTC)

Two paths - one for oxygen, one for exhaust, need to increase pressure. I think we already have this solved - internal combustion engine. Now only need a way to get the coal into the chamber - coal gas is one way and was available in the time of steam engines. Rmhermen (talk) 15:24, 10 July 2008 (UTC)

That's an interesting idea. I'm not an engineer, so I have no idea how viable that would be, or how well it would fit in with the fairly primitive technology parameters given by the original poster (it strikes me as more advanced than I inferred the question to allow, but maybe I'm wrong)... but I dig it. -- Captain Disdain (talk) 15:57, 10 July 2008 (UTC)

A Stirling engine may be what you're looking for. It's a type of engine that works off of two gas chambers one heated and the other cooled. Except for leakage, no fluid is lost, and the fluid involved is typically just air anyway. APL (talk) 14:31, 10 July 2008 (UTC)

I've heard of steam engine trains having some device to condense some of the water and reuse it that was used when the train went through large areas with no water. This probably was't the most efficient way, but they just needed it to be efficient enough, and didn't want to have to build a new engine. — DanielLC 16:29, 10 July 2008 (UTC)

This was a major issue for oceangoing steamships. Seawater was aggressively corrosive to most equipment of the day, and was a nuisance to use in boilers because it would leave solid salt deposits on evaporation. Obtaining frequent freshwater refills is even more difficult at sea than on long stretches of wilderness railroad. Consequently, closed-cycle systems were developed that condensed the steam (after expansion and extraction of most of its energy) back into liquid water.

I will also note that one of the products of combustion of hydrocarbons is water vapour; if you had a cold surface you could fairly readily condense some of that water out for use from the exhaust of your system. (If you look at the tailpipe of a car right after starting the engine on a cold day, you'll often see drops of liquid water.) TenOfAllTrades(talk) 17:51, 10 July 2008 (UTC)

The oldest part of what is now the London Underground system was originally operated using condensing steam locomotives, but it was done to keep steam out of the tunnels, not to economize on water. In fact when the route now called the Circle Line was completed, the condensing became a major issue because the condensing tanks had to be drained and refilled with cold water in order to keep working, and once the route was a continuous circle there was no place where a long enough stop was possible. So after that the drivers sometimes had to release steam into the tunnels after all. (And clearly this is irrelevant to the original question, for which some other sort of engine is wanted.) --Anonymous, 22:52 UTC, July 10, 2008, link dabbed later.

Both World War II Germany and apartheid-sanctions era South Africa made extensive use of coal to produce a direct substitute for gasoline. Franamax (talk) 18:50, 10 July 2008 (UTC)

You can run a diesel engine or a gas turbine engine on powdered coal. There are problems with doing so that have never been fully solved however, specifically the abrasive nature of the ash, although powdered coal was one of the target fuels for the original diesel engine invention. The USSR also had a powdered-coal MHD plant near Moscow but this was reported to not have been a success. --BenBurch (talk) 19:00, 10 July 2008 (UTC)

You may be interested in the Thermoelectric effect (Peltier effect). Series of wires of different metals heated on one side and cooled on the other can cause a current to flow - no fluid needed. The efficiency and scalability is not all that great, but as an added benefit, they can function equally well as Peltier coolers: add current, and get a temperature gradient. -- 128.104.112.147 (talk) 19:33, 10 July 2008 (UTC)

Note: you could still operate a steam engine in the conditions you describe by having a closed system with condensors - so that no steam is lost - you'd need some water - but only at the beginning - such a set up would be better suited to a stationary engine since those condensors would probably take up a bit of room..87.102.86.73 (talk) 19:42, 10 July 2008 (UTC)

Static build up on my cable modem?

I frequently need to power-cycle my cable modem to get it to work -- once every two or three days. The technician says that this is most likely because of static buildup. (1) Is this really the most likely explanation? (2) Could I solve this by grounding the box in some way? Like, say, attaching a wire between the metal end of the coax cable and the ground pin of a wall outlet?

Thanks! — Sam 13:59, 10 July 2008 (UTC)

It would depend on the symptoms you're experiencing. My best guess says probably not, most networking hardware that I'm familiar with is grounded, particularly stuff that your internet service provider is going to want back. The way to know if its already grounded is to see if it has one of those 3 pin plugs (if you're in the United States). From my experience with tech support, power cycling on a cable modem is usually necessary due to voodoo (which is to say, theres no discernable logic as to why you really need to...you just do). If static charge really is the answer, grounding the box should solve the problem. I would recommend against using the ground pin on the wall outlet though, you'd be better off (its easier and safer) to attach the modem via a wire to something else that is already grounded (the power supply on your computer, for example). EagleFalconn (talk) 14:23, 10 July 2008 (UTC)

Oh, and an afterthought. I don't know that unplugging your cable modem and then plugging it back in would actually get rid of a static charge on the thing. If its already grounded, that shouldn't be the problem. If its not, unplugging the thing should result in it maintaining the charge because it'll have no where to go (air is a very good insulator). EagleFalconn (talk) 14:29, 10 July 2008 (UTC)

I've come across a lot of networking stuff (albeit I've never used a cable modem) with simply a power supply brick. Unplugging it in itself may not help with static buildup but if the case is made of metal when you touch it and you're grounded you're likely to discharge the thing. If this is the case then simply going through the same motions as when you unplug it without unplugging it should do the trick though Nil Einne (talk) 17:40, 10 July 2008 (UTC)

inequalities for one-dimensional Schroedinger equation (moved from Math desk)

For concreteness, I assume that I have the one-dimensional, time-independent Schroedinger equation on a finite interval (from -1 to 1, say) with two potentials U and U' which go to plus infinity at both boundaries -1 and +1, so that both Schroedinger equations have a purely discrete spectrum. Further I assume that U' > U in the whole interval -1 < x < 1. What then can be said about the energy eigenvalues? Specifically, is it true that E'_n > E_n for all n? In which book the answer to this question can be found? —Preceding unsigned comment added by 193.144.64.159 (talk) 14:39, 10 July 2008 (UTC)

This isn't very helpful (yet) as I'm still working out the problem. But I can tell you that using proof by contradiction that E_n != E'_n, though I can't tell you whether its greater or less than (yet). I'll post something more complete when I've got a full answer. You may want to study the particle in a box, as its essentially the problem you're proposing here. EagleFalconn (talk) 15:05, 10 July 2008 (UTC)

I've hit something of a roadblock. The problem is that we don't know a specific form for U or U', and so its somewhat challenging to really try to evaluate the energy. I've got a good guess, but I'd seek verification. E'_n must be greater than E_n if U' > U because if the potential were to increase then for the particle's wavefunction to exist in the space of the increased potential its energy must increase. My basis for this is quantum tunneling, but I don't know how accurate that really is. I'm going to do the obvious thing (that I should've done before) and move this question to the Science desk. EagleFalconn (talk) 15:56, 10 July 2008 (UTC)

Well, my intention is to find lower bounds for the energy eigenvalues E'_n of an ugly potential U' by the energy eigenvalues E_n of a nice potential U (concretely the box potential). So the fact that the specific form of U' is not known is unavoidable. I think that it is easy to prove that the inequality holds for the lowest eigenvalue, E'_1 > E_1. The question is what can be said about the higher eigenvalues. —Preceding unsigned comment added by 193.144.64.159 (talk) 16:13, 10 July 2008 (UTC)

almost certainly the same behaviour as for the lowest enegry solution, is the box potential flat? 87.102.86.73 (talk) 17:34, 10 July 2008 (UTC)

The box potential is an arbitrary function U'(x). I feel like the OP is perhaps trying to come up with a method to describe an arbitrary potential in the form of a known potential? OP, is that correct? If so, I feel like you might be better off making it the subject of your PhD thesis...EagleFalconn (talk) 17:46, 10 July 2008 (UTC)

The "box" potential is the infinite square well potential. And yes, I want to use the known energy eigenvalues of the Schroedinger equation of the square well potential to find bounds for the eigenvalues of the ugly potential U', which is known only numerically, but which can be bound by the infinite square well. —Preceding unsigned comment added by 193.144.64.159 (talk) 18:14, 10 July 2008 (UTC)

The potential inside the box. What is that? can it be anything, or a flat finite value? or something else?87.102.86.73 (talk) 18:26, 10 July 2008 (UTC)

note if the potential inside the box is constant at G then the energy eigenvalues have a simple relationship E_n'=E_n-G , if the potential inside the box is not flat then you will almost certainly be looking at a completely different wavefunction solution. Did any of that help?87.102.86.73 (talk) 18:35, 10 July 2008 (UTC) (that was for U'=G, U=0)

My intuition is that what you are trying to prove is actually not guaranteed to be true for any eigenstate other than the first one, however I don't have any rigorous counter examples.

That said, if you know the U' potential numerically, you can apply to eigenstates of the U potential (i.e. sine waves) plus perturbation theory to estimate the energy change. Through second order:

$E_{n}'\approx E_{n}+{1 \over 2}{\int _{-1}^{1}\psi _{n}(U'-U)\psi _{n}^{*}dx}+{1 \over 4}\sum _{k\neq n}{|{\int _{-1}^{1}\psi _{n}(U'-U)\psi _{k}^{*}dx}|^{2} \over {E_{n}-E_{k}}}$

Where E_n' is the energy of the n-th eigenstate of U' and ψ_n and E_n are the n-th eigenstate and eigenenergy of U.

In an expansion like this, one can easily see that the leading order correction is negative (more binding energy) if U' < U everywhere. However the second order terms are only negative everywhere that E_k > E_n, i.e. if k > n for the box potential. It strikes me as possible that one could construct potentials U' where higher order terms gave rise to a net perturbation for the high order eigenstates that was opposite in sign from what you expect even with U' < U everywhere. Regardless, if you know the numerical form of U' then application of perturbation theory is probably a good bet. Dragons flight (talk) 19:01, 10 July 2008 (UTC)

Hi, finally I believe that I solved the problem by myself. Here it goes, tell me if you find an error. We introduce the Hamiltonian

$H(s)=-{d^{2} \over dx^{2}}+U+s(U'-U)$

and study its eigenvalue problem

$H(s)\psi _{n}(s)=E_{n}(s)\psi _{n}(s)$

Acting with

$\int dx\psi _{n}^{*}(s)$

on that expression and performing the derivative w.r.t. the parameter s leads to

${d \over ds}E_{n}(s)=\int dx\psi _{n}^{*}(s)(U'-U)\psi _{n}(s)\geq 0$

from which the statement follows. What do you think? —Preceding unsigned comment added by 193.144.64.159 (talk) 19:20, 10 July 2008 (UTC)

No, that doesn't work.

$\int dx\psi _{n}^{*}(s)$ gives $E_{n}(s)=\int dx\psi _{n}^{*}(s)({-d^{2} \over dx^{2}}+U+s(U'-U))\psi _{n}(s)$

$=-{\int dx\psi _{n}^{*}(s){d^{2} \over dx^{2}}\psi _{n}(s)}+{\int dx\psi _{n}^{*}(s)s(U'-U)\psi _{n}(s)}+constant$

The derivative is:

${d \over ds}E_{n}(s)={-{d \over ds}\int dx\psi _{n}^{*}(s){d^{2} \over dx^{2}}\psi _{n}(s)}+{\int dx\psi _{n}^{*}(s)(U'-U)\psi _{n}(s)}$

However the first term is not guaranteed to be zero, unless ψ commutes with the x derivative, which in general it will not. Dragons flight (talk) 19:41, 10 July 2008 (UTC)

Well, I think it does work. I shall rewrite the proof with more details.

${d \over ds}E_{n}(s)=\int dx({d \over ds}\psi _{n}^{*}(s))H(s)\psi _{n}(s)+\int dx\psi _{n}^{*}(s)({d \over ds}H(s))\psi _{n}(s)+\int dx\psi _{n}^{*}(s)H(s)({d \over ds}\psi _{n}(s))=$ $\int dxE_{n}(s)({d \over ds}\psi _{n}^{*}(s))\psi _{n}(s)+\int dx\psi _{n}^{*}(s)(U'-U)\psi _{n}(s)+\int dxE_{n}(s)\psi _{n}^{*}(s)({d \over ds}\psi _{n}(s))=$ $\int dx\psi _{n}^{*}(s)(U'-U)\psi _{n}(s)\geq 0$

where I used

$\int dx\psi _{n}^{*}(s)({d \over ds}\psi _{n}(s))=0$

which follows from the normalization of the eigenfunctions. —Preceding unsigned comment added by 193.144.64.159 (talk) 09:53, 11 July 2008 (UTC)

$\int dx\psi _{n}^{*}(s)({d \over ds}\psi _{n}(s))=0$ is false in general.

Suppose $\psi (x,s)={(sx^{2}+i) \over {\sqrt {2+{2 \over 5}s^{2}}}}$, where i is the imaginary constant.

You can verify directly that $\int _{-1}^{1}{\psi (x,s)^{*}\psi (x,s)dx}=1$ for all real valued s, hence it is normalized.

However $\int _{-1}^{1}{dx\psi ^{*}(x,s)({d \over ds}\psi (x,s))}={-5i \over 15+3s^{2}}\neq 0$. Dragons flight (talk) 16:42, 11 July 2008 (UTC)

Along with that, $\int dx\psi _{n}^{*}(s)H(s)({d \over ds}\psi _{n}(s))\neq \int dxE_{n}(s)\psi _{n}^{*}(s)({d \over ds}\psi _{n}(s)).$ Dragons flight (talk) 17:23, 11 July 2008 (UTC)

Right, the last part should read instead: "where I used

$\int dx\psi _{n}^{*}(s)({d \over ds}\psi _{n}(s))+\int dx({d \over ds}\psi _{n}^{*}(s))\psi _{n}(s)=0$

which follows from the normalization of the eigenfunctions." But the conclusion remains the same. —Preceding unsigned comment added by 193.144.64.159 (talk) 18:18, 11 July 2008 (UTC)

I disagree with your last comment and insist that

$\int dx\psi _{n}^{*}(s)H(s)({d \over ds}\psi _{n}(s))=\int dxE_{n}(s)\psi _{n}^{*}(s)({d \over ds}\psi _{n}(s)).$

H(s) is a hermitian operator. More explicitly, perform two partial integrations such that H acts on $\psi _{n}^{*}$ instead of $\psi _{n}$

Process Engineering

1. WHAT IS PROCESS ENGINEERING? —Preceding unsigned comment added by 41.204.224.41 (talk) 16:06, 10 July 2008 (UTC)

• See Process engineering. --jpgordon^∇∆∇∆ 16:10, 10 July 2008 (UTC)

some naive hypothesis at odds with basis of physic----- what wrong with them  ?

is true for low and high velocities but is not true for super high velocities when (C^2-V^2)^0.5 is comparable with another kind of velocities :Э=((f*m0)/r0)^0.5 which belong to the object in movement and has to do with gravity. If we ad in denominator this velocity , regardless it may be negligible small,we will have another meaning : the formula became determined and fall the absurdity of infinity. The revised formula will be :S0= C/((C^2-V^2)^0.5 + Э0). When V=C we'll have S0= C/Э0 that is not infinity. Further: the common particle in movement gain mass and shrink in radius via the formula: m=m0*s and r= r0/s all the way that velocity go up. So one component in denominator go dawn the other go up , when they equal we'll have Smax. After this V go 0, Э go C, S go 1 and m go M , r go R that is the common particle became an unique particle. The formula in dynamic supposed to be:

S=C/((C^2-V^2)^0.5+Э0*S0) = 1/((1-β^2)^0.5 +γ0(1/((1-β^2)0.5+ γ0))) here γ0=Э0/C For electron Э0=1.4687204*10^-11 cm/sec, S0=M/m0=2.0411812*10^21. For proton Э0=2.969794925*10^-8 ,S0=1.111662*10^18

III-- The formula for unique particle is M= e/f^0.5 gr. [The formula for common particle is m= f*M^2 / (r*C^2)] Here :"e"is the electric charge = 4.8032041*10^-10 cm^3/2*gr.^1/2*sec^-1

f^o.5 supposed to be gravity charge = 2.5832150*10^-4 cm.^3/2 *gr^-1/2 *sec^-1. let name f^0.5=i we'll have +i,-i

It is known that electric charge are two kind :plus and minus the different sign attract each other the same repel.

The gravity charge too has two sign plus and minus . Here the same sign -- attract the different-- repel.

It came that the unique particle has fore combination of charges : M-I(-e,-i), M-II(+e,+i), M-III(-e,+i),M-IV(+e,-i)

As a charged particle the single unique particle never stand free, it is always in liaison with the same mate.

We'll have ten possible " par combination " depending by each particles charges.Let see them.

The electric force of two unique particles will be:F= (F^o.5)^2= (e/r)(e/r). where "e" can be plus or minus.

the gravity force will be: F= (F^0.5)^2= f*M^2 / r^2 = ((f*M/r)/f^0.5)^2=(Э^2/f^0.5)*(Э^2/f^0.5)= (Э^2/i)(Э^2/i).

Here Э^2=C*Э0. The forces inside the particle will be: e^2/r^2= (CЭ)^2/i^2,or:(e/r)(e/r)=(CЭ/i)(CЭ/i). —Preceding unsigned comment added by 70.133.175.56 (talk) 16:09, 10 July 2008 (UTC)

I'm afraid I'm having a lot of trouble following your work. If you could point me to a previous post or some kind of reference where you start at the beginning, I might be able to help, but as it is a lot of your symbols are for things I don't think you're defining and I feel that perhaps we're stepping in in the middle of your derivation. Is anyone else having more luck following this math? EagleFalconn (talk) 17:54, 10 July 2008 (UTC)

I think it's a continuation of this thread. I can't make head nor tail of it myself. Franamax (talk) 18:46, 10 July 2008 (UTC)

We really need an introduction to what all your symbols mean. So far I only get V and C (velocities?) What sort of particles is this for? are they hypothetical ones? could be be more clear as to what result is at odds with what property of physics.87.102.86.73 (talk) 12:27, 11 July 2008 (UTC)

Clawed Lobster Population

How many clawed lobsters (as opposed to Spiny, Reef, etc) are there in the world.

68.166.120.162 (talk) 19:22, 10 July 2008 (UTC)Andrew

I suspect many lobsters are called Claud. Some have other names :) —Preceding unsigned comment added by 79.76.184.112 (talk) 00:26, 11 July 2008 (UTC)

I did quite a bit of searching with no success. I would say that if you do get a reliable number for the population, subtract two from the total after I have dinner tomorrow night :-) cheers, 10draftsdeep (talk) 16:37, 11 July 2008 (UTC)

quantum mechanics, laplacian

For three dimensions the laplacian operator is used in the schrodinger equation. Why is a linear sum of the partial derivatives used and not the square root of a sum of the squares?87.102.86.73 (talk) 19:29, 10 July 2008 (UTC)

See Laplacian. Essentially, its that the Laplacian is the del operator dot the del operator. EagleFalconn (talk) 19:54, 10 July 2008 (UTC)

If I take it to be related to 'momentum squared' I can see the reason for a linear sum..

But if I take "d²/dx² of wavefunction" to be proportional to E_x then I expect a 'pythagorean sum'.. hence the confusion - one way of looking at it gives different results to the other.?87.102.86.73 (talk) 21:15, 10 July 2008 (UTC)

Energy is a scalar, not a vector. Hence the idea of treating at as the magnitude of an energy vector would make no sense. Dragons flight (talk) 21:19, 10 July 2008 (UTC)

Oil from plant cuttings and wood chips

LS9, Inc. is a company that claims to have microbes capable of eating biomass and producing crude oil. Naturally with the price of crude oil headed toward $$$ per barrel you would expect many such claims. The question is whether such claim are bogus. If not what microbes can do this? -- adaptron (talk) 20:31, 10 July 2008 (UTC)

It is possible that they are a scam, but its not certain. Note that they do not claim to produce crude oil (although they misuse the term petroleum), but rather "a portfolio of drop in compatible hydrocarbon-based fuels and chemicals". Something like this was recently on the Nature Podcast, so it is at least not outright impossible. Whether it is cost- and energy efficient and work on an industrial scale is a different question... --Stephan Schulz (talk) 20:48, 10 July 2008 (UTC)

From a TED talk I remember Craig Venter stating that there are octane-producing bacteria, and saying that it's not yet practical. Icek (talk) 03:53, 11 July 2008 (UTC)

I heard about this a couple months ago on NPR. Apparently it's mostly a problem of scale. The bacteria have only been able to produce a drop or so of diesel. Dismas|^(talk) 03:58, 11 July 2008 (UTC)

There's one reliable source that corroborates LS9's claim: http://www.timesonline.co.uk/tol/news/environment/article4133668.ece by The Times. --Bowlhover (talk) 07:56, 11 July 2008 (UTC)

I don't think it's a scam - but it is a big investment gamble - there will be numerous start ups all claiming to have the 'magic bacteria' - and most will be little more than biochemistry labs with no experience of scale up etc. It's unlikely that you would benefit by investing (rich risk takers only)87.102.86.73 (talk) 12:23, 11 July 2008 (UTC)

Two questions about calorie use

1. According to an online calorie calculator, one can burn about 75 calories by walking at a moderate pace for 30 minutes. About how many more calories are burned if one is carrying a ten-pound load(say, a backpack) for the entire duration of that walk? Is there a formula that determines additional calorie use when carrying burdens of various weights?

2. Does "intellectual activity" use up any significant amount of calories? Say that two people of the same weight, health and metabolism are both lying still and watching TV. However, one is paying close attention to the show, making lots of mental notes on the activity, while the other is just kind of taking it in, without much conscious thought. With that in mind, could one lose weight by doing difficult mental math problems if conventional exercise is not an option? 207.233.86.98 (talk) 20:42, 10 July 2008 (UTC)

from: http://malaysia.answers.yahoo.com/question/index?qid=20070311174445AAWxTft&show=7

"Energy consumption in the brain is related to learning. In other words, once you've learned something (like mastering that chess game), the energy consumption goes down."

"Energy consumption by the brain is 230-247 calories, based on 17 calories/gram and human brain sizes of 1,350-1,450 grams. During periods of peak performance, adults increase that energy consumption by up to 50%, according to psychology lecturer Mark Moss, of the University of Northumbria.

While this may not seem an extraordinary amount of energy, the brain may use 30% of a body's total energy, while being only 2–3% of total body mass." —Preceding unsigned comment added by 68.166.120.162 (talk) 20:50, 10 July 2008 (UTC)

"Energy consumption by the brain is 230-247 calories, based on 17 calories/gram and human brain sizes of 1350-1450 grams."

17 x 1350 ≠ 230 and 17 x 1450 ≠ 247. According to [4], the human brain uses 20 W of power, which is equivalent to 0.00478 calories per second.

As for the first question, this paper reports an increase of 2.55±0.25 W in energy expenditure per kg of load. Combine with [5], which is a table correlating body mass and speed to energy expenditure, and you can easily calculate the power necessary for walking. --Bowlhover (talk) 03:33, 11 July 2008 (UTC)

Closed Timelike Curve question

My question is basically this: Suppose you receive a bottle from the future containing Carbon-14 and Nitrogen-14, and a piece of paper listing their proportions. You then measure the proportion for yourself, and come up with an answer. You write it down and send it and the bottle to the past. However, for this to be the same (as is required to be a stable time loop), the Carbon-14 must not have undergone any radioactive decay in the period between when you received it and when you sent it back. How does this work? --Zemyla^t 23:15, 10 July 2008 (UTC)

I think people who believe that this sort of time travel makes sense would answer that you just can't do this one. Just like you can't have a closed timelike curve that involves you killing your grandfather. --Allen (talk) 23:52, 10 July 2008 (UTC)

Here we go: Novikov self-consistency principle. --Allen (talk) 23:54, 10 July 2008 (UTC)

Since the process starts with you receiving something from the future, an event you have no control over (at least, not yet), you can't force such a cycle to happen since you can't sent the bottle back if you didn't receive it in the first place. Therefore, it's simple enough to just assume no such bottle would ever arrive from the future so the problem will never come up. The alternative is to assume one of the theories of time travel involving alternate realities so the bottle doesn't go back in time to your universe, it goes to an alternate one and the paradox disappears - I've never found those theories convincing, though. --Tango (talk) 23:57, 10 July 2008 (UTC)

Well, radioactive decay is probabilistic, so technically you could be lucky enough to have a bottle containing only atoms that never decay - or there could be random effects that somehow "refresh" the matter in the bottle before you send it back. Confusing Manifestation(Say hi!) 04:47, 11 July 2008 (UTC)