Wikipedia:Reference desk/Archives/Science/2007 December 2

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December 2 edit

High temp superglue edit

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)[reply]

I would try some kind of epoxy instead. —Preceding unsigned comment added by Shniken1 (talkcontribs) 02:09, 2 December 2007 (UTC)[reply]
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)[reply]
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)[reply]
So what kind of glue is used? That would pretty well answer the question. Edison 21:42, 2 December 2007 (UTC)[reply]
Don't know. I was working from a memory of hallway conversations. Look at wave soldering and reflow solder, but they still do not cite the specific glue. -Arch dude 01:01, 3 December 2007 (UTC)[reply]
The plant I got a tour of a while back (A Philips TV factory) used a method where components like connectors that have pins that stick into the board were placed by a 'pick and place' robot first. Then solder was flowed over the entire board (There being a 'solder resist' layer silk-screened onto the board beforehand to prevent the solder sticking everywhere. That process filled the 'via' holes between the layers, soldered in the components with pins (via capilliary action) and left little solder 'bumps' on all of the pads where the 'surface mount' components would make contact (the ones that don't have pins that poke into the board). The board was then kept warm (just below the melting point of the solder) with infrared lamps. Then a second 'pick and place' machine placed the components onto the board by blasting the area around each one with hot air as it placed the component onto the (now wet) solder. This was important because the surface tension of the solder was what performed the final, accurate positioning of the component - keeping it in place as the board moved off to be gently cooled. There was no glue involved - yet with careful process control, I don't see any reason why this shouldn't work for double-sided boards as well. I kinda doubt they ever use glue because I've removed components from mass-produced boards (in order to repair or modify them) and I've never seen any kind of residue left behind. I guess it's possible that some companies use something like a wax that would boil away in the end...but I doubt it. SteveBaker 20:54, 3 December 2007 (UTC)[reply]

ant trails edit

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)[reply]

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)[reply]
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)[reply]
(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)[reply]
  • 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)[reply]
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)[reply]
Different ant species leave trail pheromones in different ways, but most of them touch the tip of their abdomen to the ground where there is a gland called the sternal gland that contains the trail pheromone. Some, like the African weaver ant, have the ability to rotate their terminal abdominal segment, thereby exposing other glands that contain different signals. The sensitivity of ants to trail pheromones are remarkable. For example, Atta texana trail pheromone is known to be methyl-4-methylpyrrole-2-carboxylate. Only 0.08pg/cm of this is deposited to establish a trail (that is 3.5 x 108 molecules/cm) Scientists have shown that with the amount of trail pheromone in a single Atta texana nest, the ants could leave a trail reaching around the world three times. They don't make a decision to "spray" the chemical, its more likely that the pheromone releasing behaviour is an innate response to the detection of food, thereby forming a trail back to the nest and leading others to the food source.
The other thing to consider is that these chemicals are volatile, (which would make marking them visually a bit of a challenge). Some trail pheromones, such as those from fire ants, last only a few minutes before evaporating away, while other species persist for days. The ants don't detected the trail itself, but the diffusing volatile. Thus they travel through a "vapor tunnel". Finally, its interesting to note that there is no evidence that ant trails are directional (there doesn't appear to be any information in a trail that tells ants which direction to go), instead they rely on visual cues for that. This is in contrast to snails which do produced directional trails, though the exact mechanism is unknown. This information is sourced from Wyatt, Tristram D. (2003). Pheromones and Animal Behaviour: Communication by Smell and Taste. Cambridge: Cambridge University Press. ISBN 0-521-48526-6, and the references therein. Rockpocket 21:38, 3 December 2007 (UTC)[reply]
  • Wow, great answer! --Sean 13:35, 5 December 2007 (UTC)[reply]
Thank you, though this is the area of my professional expertise, so I could waffle on for days about it! Rockpocket 08:09, 6 December 2007 (UTC)[reply]

QUESTION ABUOT TEH TERMINATOR LOLZ (seriously!) edit

 
"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)[reply]

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)[reply]
...OP says, "of" Earth, not "on" Earth. 71.100.1.143 05:10, 2 December 2007 (UTC)[reply]
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)[reply]
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)[reply]

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)[reply]

Yep, exactly why I complained to 71.100.1.143. SteveBaker 17:37, 2 December 2007 (UTC)[reply]
Yeah, don't let Mr. 71 get you down. There's nothing wrong about asking about the "terminator of Earth", as opposed to Venus, the Moon, etc., which also have terminators. -- Coneslayer 00:46, 3 December 2007 (UTC)[reply]

HIV (AIDS) edit

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 (talkcontribs) 03:35, 2 December 2007 (UTC)[reply]

  • 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)[reply]
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)[reply]
  • 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)[reply]
This was just covered in the Savage Love Column in The Onion a while back (sad that I remember this!): Link to article. Good luck with that. Saudade7 03:19, 3 December 2007 (UTC)[reply]
He is asking a question within the guidelines of wikipedia's reference desk. Either answer it, or ignore the question. *talking to sean btw* 64.236.121.129 14:32, 3 December 2007 (UTC)[reply]

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)[reply]

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)[reply]
And indeed, the two small plates in the image you refer to, the Cocos and Caribbean, are impinging on the southern edge of the formerly isolated Yucatan microcontinent. Where that impinging occurs, largely in southern Guatemala, we would expect features like an impact crater to (eventually) be significantly modified. The north side of Yucatan, where Chicxulub lies, is part of the Yucatan craton that is not being deformed. That side of Yucatan is a passive margin related to the opening of the Gulf of Mexico; the worst that happens there is simple normal faulting, and as Gandalf61 indicates, Chicxulub is actually on the interior of a continental plate. Cheers Geologyguy 17:54, 2 December 2007 (UTC)[reply]

This and that book approach edit

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)[reply]

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)[reply]
"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)[reply]
I may risk soapboxing, but authors of textbooks on the same subject work very hard to differentiate themselves, because much of the material is common between all of the books, and would be obvious to anyone who has already mastered the material. To disguise this effect, an author or publisher may add gimmicks or "twists" on the subject matter - sometimes with valid reason, and sometimes not. If instead of textbooks, we all used wikibooks, the base material in common could be kept in a single place, and additional insights (such as the "practical" or "quantitative" approaches) could be added in separate sections. However, this would make it difficult for an individual author to take credit for an entire body of work. Nimur 21:46, 2 December 2007 (UTC)[reply]
Continuing from Stephan Schulz, the "…Approach" tag is usually best for comparing/contrasting within a set of works on the topic. If you want to learn about electromagnetic field theory, you could study the theoretical (quantum mechanics, etc) origins, you could study the applications to other realms of science, you could study the practical aspects for engineering, you could just learn how to solve specific problems related to it without delving deeply into the underlying causes. Different texts for different folks. DMacks 07:04, 3 December 2007 (UTC)[reply]

fibre transfer forensics edit

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 ?[reply]

  • 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)[reply]
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)[reply]

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)[reply]

Flushing toilet edit

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)[reply]

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)[reply]
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)[reply]
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 dose - 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)[reply]
(ec)
Well, I didn't see that episode, so I don't know what they found, nor can I comment on their methodology.
My own opinion (but please note that this is not even original research, but rather, armchair speculation) is that everybody poops, and fecal coliform is everywhere. Our immune systems can handle reasonable amounts of fecal coliform quite well -- they'd be grossly inadequate immune systems if they couldn't, given how much of the stuff there is guaranteed to be floating around in the environment! As I understand it, it's not the mere presence of fecal coliform that constitutes a public health threat, but rather, elevated levels of fecal coliform. And in fact, coliform isn't dangerous in itself; the only reason we worry about it so much is because it's an indicator bacterium.
I've never noticed my toilet at home spraying anything. I have noticed some spray from the high-pressure (tankless) toilets used in many commercial buildings. I bet there's a significant difference there.
Certainly, if you're worried about spray, closing the lid when you flush can't hurt. If you're male, I would guess that sitting while urinating would help a lot, too, although I gather that this is found unacceptably effeminate by some. Certainly, regular cleaning of any bathroom is indicated, especially if you're worried about germs. —Steve Summit (talk) 17:42, 2 December 2007 (UTC)[reply]
So based on SteveBaker's gloss of the Mythbusters episode above, I think we can safely say: yes, closing the lid will absolutely prevent untoward toothbrush contamination. However, it will also prevent Kirlian photography, and if you have any perpetual motion machines in your bathroom or nearby, those are likely to stop working, too. You might also want to consider the effects the lid's position might have on your bathroom's Feng shui. —Steve Summit (talk) 18:10, 2 December 2007 (UTC)[reply]
Just about any operation that involves vigorous agitation at a liquid/air interface (as with a stream of urine into the bowl water, or the mixing and agitation that takes place during a flush) will generate aerosols. Aerosols are droplets too small to be seen with the naked eye, that will take seconds, minutes, or hours to settle out of the air under the influence of gravity (larger particles settle faster; droplets are also cleared gradually by air exchange via exhaust fans and such). This is a serious consideration in medical and research settings when individuals work with biohazardous materials—there is often a great deal of careful planning involving in establishing safe protocols for where, when, and how liquids can be handled, mixed, transferred, and dispensed in laboratory environments (Google for 'biohazard' and 'aerosol' for a whole bunch of links). Certainly the violent, vigourous flushing action of tankless institutional toilets is likely to result in greater aerosol generation than a more sedate home toilet, but no toilet will be completely aerosol-free. TenOfAllTrades(talk) 18:20, 2 December 2007 (UTC)[reply]
Do people actually pick up tooth brushes and but them in their mouths without even rinsing them first?--Dacium 03:00, 3 December 2007 (UTC)[reply]
I did not read all of these responses, but just to say that if you are in the bathroom alive and breathing, anything that could possibly get on your toothbrush is already in your mouth. You can keep your toothbrush in one of those sanitizing things that they sell on TV late night in the United States, but you are still going to be sticking the autoclaved toothbrush into your relatively dirty mouth. That's okay! That's why we have immune systems. It's just life on Earth. Don't worry about it. It is actually unhealthy to be too clean! And let's face it, if the few microbes you are worrying about are going to put you off your toothbrush, you'll have to give up on all but virtual forms of sex. I'm reminded of when Neil deGrasse Tyson gives his lecture on why "intelligent design" is a misnomer, saying that putting sex organs next to the (ends of our digestive tracts) is like putting an amusement park in the middle of a sewage plant! Think about that one! Saudade7 03:37, 3 December 2007 (UTC)[reply]
There is certainly evidence that it's bad to be "too clean" - especially when you are very young. We have an article on that (of course!): The Hygiene hypothesis. SteveBaker 20:34, 3 December 2007 (UTC)[reply]


isnt there a gum that says Dirty mouth? Clean it up! YA just buy that!!!

Who will die earlier? edit

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)[reply]

If the food contains enough fluids, the one without water might die of old age. Lova Falk 17:30, 2 December 2007 (UTC)[reply]
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)[reply]
As Lova said, it is possible to get water from food, but it is not possible to get food from water. So, if given moist foods, it is probable that the person without water will be able to get enough water from the food to easily outlive the person who is only taking water without nutrients. -- kainaw 18:12, 2 December 2007 (UTC)[reply]
That's ridiculous - there is no way you can get enough water from food. Even if you ate nothing but watermelon. Some animals can do it - but we're not adapted to doing that. Our article on Water says "Also noted is that normally, about 20 percent of water intake comes from food, while the rest comes from drinking water and beverages"...so you'd have to exist on one fifth of the usual amount of water. It also says that you need to drink "1 milliliter for each calorie of food" - so eating five times more (in order to get more water) doesn't help because you need more water in order to digest it. (Of course for the literalist - according to the OP, the one that gets food "is without water" - so they're getting dehydrated food!) SteveBaker 20:09, 2 December 2007 (UTC)[reply]
How is it ridiculous? I have known many people who never drink "water". They drink fruit juice, coffee, coke, etc - but not water. So, to use your example - the person has plenty of food, including fruit (ie: watermelon). The person can easily get 8-10 cups of water (with various impurities from the food) per day. The food doesn't have to be eaten in whole. If given a watermelon, the person can opt to only drink the juice. Finally, taking the question literally, the person does not have to survive indefinitely. He only has to outlive the person who does not have any food at all. So, even if the person lacks the daily requirements of water intake, by replacing most of the water needed every day, the person can easily outlive the 3 or 4 days quoted as the limit for living without water. -- kainaw 20:24, 2 December 2007 (UTC)[reply]
That's a very misleading answer. We can read the OP's question in several ways:
  1. Person A gets water and no food, Person B gets food - but dehydrated so he gets no water whatever. Person A definitely outlives person B.
  2. Person A is allowed to drink water, Persons B is allowed to eat some kind of a normal diet - but not drink. This is the interesting one. A "normal" food diet doesn't contain enough water. 20% of what you need is what our article says.
  3. Person A can drink whatever he wants, Person B can eat whatever he wants. Both of them survive into old age - the first living on a variety of soups and juices, the second living on watermelon and oranges.
  4. Person A drinks water, Person B is allowed to consume large amounts of oranges and watermelon (at least five times more than a "normal" diet would contain according to our article). In that case, person B will clearly outlive A.
The only way person B wins is to take the last of these four interpretations - but I maintain that this is a very strange reading of the original question. SteveBaker 01:16, 3 December 2007 (UTC)[reply]
I simply took "without water" to mean that the only restriction was no access to plain water and not a restriction to a "normal" diet. I do agree that I don't interpret questions normally. I tend to always look for the loophole. -- kainaw 03:14, 3 December 2007 (UTC)[reply]
The rule of 4s, that I was taught in a Wilderness First Aid class a few years ago, was: 4 minutes without air, 4 days without water, 4 weeks without food. This is probably a very rough estimate, but should give you an idea. Unfortunately, our articles on fasting related topics don't appear to provide any numbers. (EhJJ) 19:10, 2 December 2007 (UTC)[reply]
I actually learned it as the "Rule of Threes" as mentioned in survival skills. Either way, a rough estimate. --Bennybp 19:33, 2 December 2007 (UTC)[reply]
Is there a supertough scout troop somewhere who is trained: 6 weeks without food, 6 days without water, 6 minutes without air? As I heard it from someone who had attended "survival school" the three rule also included three hours without shelter (presumably in freezing weather). Edison 21:40, 2 December 2007 (UTC)[reply]

Ignore any "rules of threes or fours". People are different. A fatass will obviously live longer without food than some emaciated little girl. As for water, about a week tops, although there are exceptions. 64.236.121.129 14:38, 3 December 2007 (UTC)[reply]

Maybe that's why both people stating those rules noted that they are rough estimates. -- Coneslayer 14:47, 3 December 2007 (UTC)[reply]
Also, the original question states that "Two equally healthy people are put on the test." (Granted, "equally healthy" doesn't necessarily mean "equal" or "healthy", but it was clearly implied). (EhJJ) 15:48, 3 December 2007 (UTC)[reply]
They aren't acurate estimates. People are too different to give a general estimate like that. 64.236.121.129 18:35, 3 December 2007 (UTC)[reply]
So why did you give a general estimate of a week? -- Coneslayer 19:06, 3 December 2007 (UTC)[reply]
Because it's a more acurate estimate. Being a fatass means you can last longer without food. The same can't be said for water. 64.236.121.129 20:50, 3 December 2007 (UTC)[reply]
You actually gave an upper bound, not an estimate. And to give a general estimate is to guess where the mean or median is, which is always valid in this sort of problem. As for how variable the situation is, that's just raising the issue of how large the variance is, which is completely unknown. But the variance does not diminish our ability to provide an estimate of the mean/median, only the likelihood of its being correct for a random individual. Someguy1221 21:01, 3 December 2007 (UTC)[reply]
Yes, I gave an upper bound. Coneslayer said I gave an estimate, so blame him for the terminology. An average is fine, unless the variance is too high, and in this case, it is. One guy once went on a hunger strike for over 300 days without food. I don't know if he was a fatass or not, but it gives an idea as to how dispersed these ranges are. Giving an estimate or upper bound for water is a little more consistent from person to person because it doesn't rely on how much blubber you have in your stomach, butt, etc. 64.236.121.129 21:14, 3 December 2007 (UTC)[reply]
My point was that the existence of a high variance doesn't make the average any less true. It just needs to be pointed out. Someguy1221 21:49, 3 December 2007 (UTC)[reply]
Well, here is one excellent source of data in the 1981 prison hunger strike in Northern Ireland. The variance there is that one person died after as little as 46 days - another survived after 70 days, another died at 73 days. According to Hunger strike "The Guinness Book of Records lists the world record in hunger strike (without forced feeding) as 94 days.". But these were people in prison - they didn't have to exert themselves - they weren't suffering temperature extremes - they were getting plenty to drink (even intravenously after they slipped into coma) and they were getting good medical attention. That means that this is probably close to the maximum possible. If you were in a 'survival' situation where you have to find water, make shelter and at least search for food - you're burning through energy more quickly - so you're not going to survive as long. In light of the Guinness book statement - and the evidence that nobody else seems to have survived to even ONE hundred days - the THREE hundred day number that 64.236.121.129 claims is going to need some kind of a reference (that's a polite way to say bullshit!) - the problem here is that there are variations in what is considered to constitute a hunger strike. Can they drink things other than water? Was force-feeding applied? Was water given intravenously after coma set in? SteveBaker 03:26, 4 December 2007 (UTC)[reply]
Continuing: [1] says that one guy survived 123 days on hunger strike - but (a) he occasionally drank coffee and (b) he broke his strike and ate for a few days in the middle - so it wasn't really a continual strike. The trouble is, everything I can find out about him dates from June 2004 and they ALL say "still alive after 123 days so far" - we never seem to find out whether he died or gave it up - and after how many days. But even if he is 64.236.121.129's "300 day" guy - the fact that he "cheated" make him a less than valid scientific test. SteveBaker 03:37, 4 December 2007 (UTC)[reply]
OK - nope, he was force-fed after 130 days[2] - then started eating of his own accord - so no 300 days here either. SteveBaker 03:41, 4 December 2007 (UTC)[reply]
Easy SteveBaker, I said 300 days, but I might be wrong. 130 days sounds about right. 64.236.121.129 15:21, 4 December 2007 (UTC)[reply]
Clearly it depends on things like: the climate, the amount of exertion is required of you, how appropriately clothed and sheltered you are, how much fat deposits you have, how much food/water was in your stomach at the outset, your age, your state of health, what medical facilities there are to help you recover when the ordeal is over...there are bound to be lots of other variables. So any number we tell you here is 100% guaranteed to be wrong for some combination of circumstances. However, "a few days" without water and "a couple of weeks" without food are not bad "rule of thumb" estimates. SteveBaker 20:23, 3 December 2007 (UTC)[reply]
Really, it depends on if you are this guy! (Just kidding, don't try this at home!) Saudade7 13:22, 4 December 2007 (UTC)[reply]

Camera settings for astrophotography edit

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)[reply]

The thing about the night sky is that it is very dark. Even with light pollution. Stars and comets just don't provide much light here on Earth. So, you're going to need a long exposure, which means you definitely can't hold the camera with your hands. The aperture should be wide open, f/2.8 if that's the best you have. There's no reason to stop it down, because everything in the photo is practically at infinite distance, so you don't need any depth of field. The metering systems of everyday digital cameras are not designed for this, so just put the camera in manual mode. If the sensitivity only goes up to ISO 400, then it's probably very noisy at that sensitivity. On the other hand, if you turn the sensitivity down, you need more exposure time, so there's a tradeoff. —Keenan Pepper 19:43, 2 December 2007 (UTC)[reply]
I'm pretty sure the OP noted they were using a telescope, so the bit about holding the camera steady probably doesn't really apply. One thing they don't note is whether they have a tracking mount or not; if not, long exposures with high magnification will suffer from motion blurring just from the Earth's rotation. Since they're going to be stacking the images anyway, I'd concentrate on minimizing the exposure time per frame, setting maximum ISO, maximum aperture and an exposure just long enough for some recognizable detail to emerge from the noise for the alignment algorithm to home in on. Oh, and I'd definitely put the camera in fully manual mode for this; you can use the automatic focus and metering to find suitable values, but set them manually before starting the actual shoot. Can't really say much about the zoom, except obviously not to use "digital zoom". I guess if you can zoom the camera in while retaining focus and a reasonable aperture, why not? I'll just give you that much extra magnification. If you can, shooting in RAW mode is probably best. Also disable automatic dark frame subtraction if your camera lets you, and take your own dark frame(s); I believe registax can handle dark frame subtraction during the stacking. Mind you, I've never really more than dabbled in astrophotography myself, so take my advice with a grain of salt. Oh, and you might be interested in our astrophotography article — or at least the links and references at the end. —Ilmari Karonen (talk) 20:26, 2 December 2007 (UTC)[reply]
A typical rule of thumb, at least for me, is that the combined image will not look more than 70% better than the original images. So if the details in the original photos are barely louder than noise, the noise will be a huge problem when it comes to combining the photos. --Bowlhover 23:01, 2 December 2007 (UTC)[reply]
Oh, I see he did ask about holding the camera in his hands. I suppose, depending on how everything else works out, that it might work if the exposure time per frame is short enough — any shake from his hands will just get added to turbulence and other sources of blurring and hopefully filtered out by registax. But if you have or can reasonably obtain a camera mount, I'd strongly recommend it.
By the way, if your camera has a video mode, you might want to give it a try; that'll give you a large set of low-resolution frames, exactly what programs like registax are written to work best on. —Ilmari Karonen (talk) 20:37, 2 December 2007 (UTC)[reply]
Right, holding the camera simply doesn't work. Neither does touching it during the exposure or less than 3 seconds before the exposure, so I would strongly recommend using the camera's self-timer. I don't think videotaping will work because the minimum exposure time for an adequate-quality photo of the comet, using my Nikon D40x at ISO 1600, was 2 seconds several days ago. Combining the photos is no miracle cure for a low signal-to-noise ratio. --Bowlhover 23:01, 2 December 2007 (UTC)[reply]
Hi. Your responses are very helpful, but I still have a few questions and comments. I did say equatorial reflector, which means I will be using sky tracking. However, I'm quite unfamiliar with this, so I'm not sure if I should use an AC/DC adapter and an outdoor AC line, or some kind of battery. Well, If I can get a camera adapter, how much will it cost? Note that my camera lens isn't that long (only about an inch), and actually moves backward and forward again when I zoom it in. Well, I heard using a larger aperture will fade the background, and the stars pretty much are the background, or are they far enough to prevent this? If I use the movie setting, which is avalible on my camera, I will not be able to set such things as the ISO speed. Will it work if I get the final stacked images for both the long-exposure shots and the movie images, and combine them into one final image? Does registax offer image processing? If I am able to use an accurate equatorial mount and camera adapter, can I use longer exposures (eg. 15 secs)? If I do the long exposure times, and take several of these photos, how many will I need? If I use the movie setting, should I use the higher resolution, and how long should the movie be? Can registax cut out undesirable images of a movie sequence, such as if I didn't have a camera adapter and had to occasionally take the camera off the telescope? Does registax give you image processing tools like those in Photoshop? Would digital zoom interfere with my images? How does RAW mode improve the images? If I am to upload the final result onto Wikipedia and/or Commons, what file endings (eg. bmp, gif, jpg) are suitable? What file ending to registax final images come out as? Will using good settings prevent the photo from oranging? What white balance should I use? Thanks. ~AH1(TCU) 21:20, 2 December 2007 (UTC)[reply]
(1) I recommend using the camera's battery if it can last long enough for your photography session, because the wind will blow your DC adapter line around if you use one, moving the camera and blurring the photo. By all means bring extra batteries because long exposure times are very energy-demanding.
(2) Try doing some research on the Internet on adaptors that have the appropriate output voltage. Adaptors should be very cheap, around 4 USD.
(3) Larger apertures fade the background because it leads to a narrow depth of field. This means a small difference between the distance of an object and the distance of the object your camera focused on will cause a large difference in how well-focused the "unfocused" object is. Since both the comet and the stars can be considered to be at the same distance--that is, infinite distance--the aperture has no effect on how well-focused the stars are. They'll be as focused as everything else in the photo.
(4) Registax offers some image processing like rotating and brightness adjustment. Try it out. You'll definitely need another image processing software to edit your photos.
(5) A 15-second exposure time should be no problem. But since I have no experience using your telescope and your equipment, the best way to find out is to try taking astrophotos with varying exposure times, with varying ISOs and the maximum possible apperture. --Bowlhover 23:01, 2 December 2007 (UTC)[reply]
(after edit conflict & answering your first set of questions) Since you have a tracked equatorial mount, take as many photos as your computer can combine without running out of RAM. The more the better because the final image's random thermal noise decreases with the square root of the number of photos combined.
Brightness of the comet: according to this graph, Holmes will be magnitude 5.2 on January 1. Extrapolating from this angular size chart, it will be 101 arcminutes on the same day. This gives a surface brightness of 23.9 mag/sq. arcsec. By comparison, the darkest skies have a surface brightness of 23 mag/sq! You may be able to see the inner coma, but I highly doubt you'll manage to see, or even photograph, the outer ring.
When photographing, be sure to zoom in as far as you can without cutting off the comet. Zooming in will give you the same number of pixels for a smaller area of sky, thus increasing the amount of detail in your photos.
Finally, make sure to take the photos in RAW format rather than JPEG. The camera records sensor data directly to the file when shooting in RAW but compresses the image before storing it when shooting in JPEG. The compression involves taking pixels of similar brightness and colour and setting them to the same brightness and colour, creating compression artifacts. JPEG files are around half the size of RAW files but with a lower quality. --Bowlhover 22:20, 2 December 2007 (UTC)[reply]
One last thing: The high ISO settings on most cameras are there for marketing purposes only, and the noise you get from them is absolutely unacceptable. Try using the second-highest or third-highest ISO. --Bowlhover 23:01, 2 December 2007 (UTC)[reply]

Racing across the universe while sitting on the couch edit

Hi wikipeoples! I think it's gonna be hard for me to ask this question correctly (even though I saw and successfully passed three "physics for engineers" classes at my university... but that was a long time ago...), and this was probably asked somewhere before, but I'm gonna try anyways, so please don't bite me. In a nutshell, it's this:

What is the absolute velocity of an object resting in place on Earth?

Now that might sound pretty easy or even stupid, but I'm just getting started. Considering a (static) inertial reference frame outside of Earth (I would consider the sun, but maybe I'm complicating myself too much), any object on the surface of the Earth (I like to imagine someone sitting on a couch, for example) would have several velocities, like Earth's rotational speed (so for you picky fellas the guy on the couch would be somewhere on the equator), and the speed given by the orbital rotation of the Earth around the sun. So you got two velocity vectors already; what's the velocity of the object considering these two vectors? (Am I on the right track? Is there something extra I should be considering already?)

So far so obvious, right? Now, I don't know if I should keep going, but, here goes: since according to its article the Sun is moving at 217 km/s around the galactic center, then there's yet another vector to add up. What is, then, the velocity of an object on Earth from the frame of reference of the galactic center of the Milky Way? Is it simply 217 km/s because the other vectors are already negligible? What if I said that this speed isn't considering the bobbing movement of the Sun across the galaxy's disk? What would the speed be with this?

Finally, according to the Milky Way's article, some estimates say that the galaxy is moving at 600km/s from the frame of reference of other galaxies. It also says that the Earth itself would be moving at 51.84 million km/day, but I'm not sure what vectors is it considering. Or, since the magnitude of the galaxy moving would make other vectors negligible, is this 'Earth's speed' all that's needed to consider the velocity of an object on it?

I hope I didn't sound like a moron asking this (though I'm sure my considerations about Earth's speeds around the Sun are wrong somehow). And yes, I'm asking for velocities, not speeds.

So to sum up, I'm asking for three velocities of an object on the surface of the Earth: viewed from the FoR of the Sun, viewed from the FoR of the galactic center (including the bobbing movement if possible), and viewed from outside the galaxy (if the last one isn't already correctly answered). I guess that's it. Thanks in advance! Kreachure 18:13, 2 December 2007 (UTC)[reply]

Some of these things are pretty easy to calculate. For example, the average orbital speed of the Earth around the Sun is just 2 pi AU / year in km/s. Similarly, if the circumference of Earth is about 40,000 km, the rotational speed of a point on the equator is 40,000 km / day in m/s. These two vectors don't have a constant angle between them because they rotate at different rates. Earth is a "prograde" planet, which means it rotates in the same direction as it orbits the Sun, so the vectors add at midnight (by some reckoning of time) and subtract at noon. —Keenan Pepper 19:25, 2 December 2007 (UTC)[reply]


Unfortunately, your first question is self-defeating, as it asks for the absolute velocity with respect to the "fixed reference frame" of empty space outside our galaxy. No frame is fixed! You can pick any "frame" and call it fixed; then, you can calculate the relative velocity of some other frame. Because of special relativity, the velocities must be less than c, so all you can safely say is this: "on your couch, your relative velocity with respect to any other object in the universe has magnitude between 0 and c." If you want velocity with respect to a specific object, (such as your later questions), then you can of course calculate it as Keenan Pepper explained above. Nimur 21:34, 2 December 2007 (UTC)[reply]
Note that you are looking at speeds over a wide range of scales here. Rotational speeds about the centre of the Earth are of the order of 400 m/s, which is very much smaller than the 29.8 km/s orbital speed of the Earth around the Sun. This in turn is an order of magnitude smaller than the 200 km/s orbital speed of the solar system relative to the Milky Way. So if your frame of reference is the Milky Way, you can get a good enough ballpark answer by ignoring the first two components. Gandalf61 10:42, 3 December 2007 (UTC)[reply]
There is no fixed reference frame, as others have pointed out, but about the best you can do is use the CMB's frame; see its footnote 2 for a velocity. --Tardis 16:48, 3 December 2007 (UTC)[reply]

Name of a medieval herbal medicine manuscript? edit

Hey I was wondering if anyone could possibly direct me to the page of a certain medical manuscript whose name I cannot currently recall and after searching both Wikipedia and other resources cannot find. It was an illuminated manuscript which, if I recall correctly, mostly dealt with herbal medicine. It was a European manuscript based off of an Islamic one and I believe it was from the 11th or 13th century. Thanks for any possible help.

Cracksells 20:05, 2 December 2007 (UTC)[reply]

Herbalism mentions some old manuscripts. --JWSchmidt 21:21, 2 December 2007 (UTC)[reply]
Maybe this one? Tacuinum Sanitatis. Saudade7 03:56, 3 December 2007 (UTC)[reply]

Strategies for modern cities to become increasingly energy efficient edit

Can anyone please help me find some strategies for energy efficiency? All I could think of is using alternative sources of energy and using energy efficient products. Please help, I need these for my homework. It would be cool if you provide some links too. Thanks in advance. 202.137.21.176 20:50, 2 December 2007 (UTC)[reply]

Alternative energy sources are not necessarily more efficient. In fact, they are often not (solar cells today have a theoretical maximum of about 33% efficiency). Be more creative - look at how much energy we dissipate as heat or friction into the air when we brake, for example - that's the principle behind hybrid cars. SamuelRiv 21:07, 2 December 2007 (UTC)[reply]
Thanks for answering. So do you have any suggestions? I'm really in need of help! 202.137.21.176 21:21, 2 December 2007 (UTC)[reply]
A modern city with good mass transit would be more energy efficient than one where everyone drives everywhere in a car or SUV. A way for a city to get people to ride buses, subways, trains, or light rail would be to make sure everyone is within walking distance of a transit line, and that the transit runs on a schedule which meets the needs of people needing to get somewherre: the things run frequently and are safe and affordable, and they do not stop running at some inconvenient time like 6 pm. A way to add some 21st century technology would be to use the GPS info the buses and trains often transmit to their control centers to the internet, where a commuter would be able to tell from his home computer or mobile Blackberry just when the bus/train will reach his stop. The same info could allow a display at the train/bus/trolley stop to indicate approximate minutes until the next one arrives, based on GPS info. It is frustrating to reach a bus stop 30 seconds after the bus left with 30 minutes until the next one arrives. Less energy is used per passenger mile on a reasonably well loaded mass transit vehicle than if each person drives. Edison 21:33, 2 December 2007 (UTC)[reply]
Relatedly, Geographic information systems (GIS) can be (and are) used for things like route optimization for buses, garbage trucks, etc. There are a huge number of other ways GIS can be used to increase efficiency, especially in combination with technologies with GPS and wireless networks / mobile computing. Pick anything that involves people or goods moving from one place to another and odds are it could be made more energy efficient (and probably cheaper too). It's hard to pick an example because nearly everything a city government does can be improved in this way. So.. say, I don't know, building inspectors going into the field to inspect buildings -- use the above technologies to optimize schedules and driving times, minimize the need to return to the office, etc. Pfly 22:02, 2 December 2007 (UTC)[reply]

Thanks for all the suggestions guys! I need at least three strategies and so far I've got two on my hand: utilizing energy efficient products on residential and commercial sites, and improvement on mass transit and other transportations. Anyone has one more idea? Or have resourceful links? Thanks in advance! 202.137.21.176 22:16, 2 December 2007 (UTC)[reply]

Also relatedly, mass transit can only work (or can only work efficiently) when people's housing and their workplaces are sufficiently dense and sufficiently close to each other. If everybody wants a house out in the wide-open suburbs, mass transit doesn't work well. If in order to find affordable housing the "wide open suburbs" require a one-hour commute each way, it works even less well. So (just as for any hard problem) some of the solutions for the energy problem will have nothing directly to do with the energy problem. —Steve Summit (talk) 22:19, 2 December 2007 (UTC)[reply]
It's a radical one - but population reduction would work too. If there were half the number of people in the world - we'd use half the amount of energy. The problem is: How to get there from here...but that's a problem with many energy reduction strategies. SteveBaker 00:43, 3 December 2007 (UTC)[reply]
A city is a system. You need to think in term of the system and not of its separate parts. In a city, you can make efficient use of cogeneration. You can cover a city with a dome. mass transit options are much more economical. Economies of scale change the economics of almost all of energy use. -Arch dude 00:51, 3 December 2007 (UTC)[reply]
In a relatively low polution city you could encourage food cultivation inside the city to minimise daily transport from the outside. Keria 10:20, 3 December 2007 (UTC)[reply]
This may or may not help. The trouble with such plans is the cost of localised food production can sometimes be vastly more significant then food production in other areas so even with the added cost of transport and perhaps refridgeration, it can still be cheaper to bring food in. However on a related issue, and somewhat radical as well, greatly reducing meat consumption would help. I've seen some studies which suggest some meat consumption may in fact be more environmentally friendly in some areas anyway because the land is not particularly suitable for most crops but even if this is true and it wouldn't in fact be best to just import food the amount most people eat is definitely way too high. Nil Einne (talk) 13:41, 5 December 2007 (UTC)[reply]
I've heard of a few interesting systems that use the city's sewage as an alternate fuel source. You treat the water so as to conserve it, and you use bacteria for anaerobic digestion on the leftover sludge which produces heat and methane gas, then you use pyrolysis, gasification, or other methods to turn what's left into solid and gaseous fuels, which can then be burned for more energy. This also has the benefit of reducing the amount of waste material that has to be disposed of. Even if you only break even energy wise, the reduction in waste makes it worth it. Of course, these systems are still being developed, so they can be expensive and experimental right now, and there are a number of different methods being tried. See the biomass and related articles for more information, or some examples like this one in Germany, or this one in Philadelphia, PA USA, or even this system using animal waste parts from food production. Hope that helps! -- HiEv 13:47, 3 December 2007 (UTC)[reply]

Not related to cities per se but increasing the energy efficency of homes will obviously help. For example, increasing the usage of energy efficient lighbulbs usually CFL or other flourescent lightbulbs although maybe LED based ones in the future. Increasing thermal insulation of homes and hot water cylinders, particularly in temperate countries. Increasing the usage of solar hot water heaters. Nil Einne (talk) 13:34, 5 December 2007 (UTC)[reply]

Electrical Resonance edit

What would cause the peaks in frequency/current graphs of electrical resonance in an LCR circuit to come out thinner than predicted. with current measured as voltage across the resistor, the peaks have come out much wider and less tight than the peaks predicted by the maths. What could cause this? —Preceding unsigned comment added by 172.203.50.108 (talk) 20:52, 2 December 2007 (UTC)[reply]

I'm afraid I don't understand what you've written; you seem to be saying two opposite things. But have you considered Q? Higher frequencies cause a coil to have lower Q because of skin effect. Certain types of capacitor will exhibit very low Q at high frequencies. --Milkbreath 21:22, 2 December 2007 (UTC)[reply]
As the resistance in a RLC circuit resonant circuit increases, the peak should broaden. See Q factor. The article LC circuit also says that in a series resonant circuit, a high inductance and low capacitance makes for a narrower peak, while the converse is true for a parallel resonant circuit. Edison 21:27, 2 December 2007 (UTC)[reply]
The variations can happen because of stray capacitance, mutual inductance with things that are not inductors, skin effect causing higher resistance than measured at DC. Eddy currents induced due to the magnetic field. Even your measuring equipment could be adding an unexpected load to your circuit. Usually Q gets lower and the peaks broaden. What Frequency are you working with? Graeme Bartlett 01:00, 3 December 2007 (UTC)[reply]
You left out the R, or had a much lower value of R in your experiment than you thought.--TreeSmiler 01:13, 3 December 2007 (UTC)[reply]
On the other hand, if your peaks are too wide, you have more resistance (or less inductance) in the circuit than you thought. Were you inadvertently putting dc through the inductor as well as ac?--TreeSmiler 02:15, 3 December 2007 (UTC)[reply]
I'm not quite as foolish as to have fallen at such basic errors as tree smilers suggestions! But Q factor would seem to account for this, with eddy currents and rogue inductances as such. —Preceding unsigned comment added by Philc 0780 (talkcontribs) 19:49, 3 December 2007 (UTC)[reply]
I didnt intend to insult your intelligence, but you left us wondering about the form of circuit (series or parallel)

, frequency of opeation and how much you actually knew about resonant circuits etc. --TreeSmiler 02:48, 4 December 2007 (UTC)[reply]

Other responders have asked the frequencies tested. I remember from microwave lab in school how a resistor, inductor or capacitor might change in its characteristic as the frequencies used increased from audio to radio frequency to UHF to microwave. Capacitor or resistor leads can become inductive at high freq, and inductors may have appreciable resistance at any frequency. Edison 03:16, 4 December 2007 (UTC)[reply]

electricity consumption of heater edit

I have an electrical heater. It has a label with 1500W on it. Does this mean it will consume 1.5 KW/h for every hour that I use it?217.168.3.246 22:28, 2 December 2007 (UTC)[reply]

Yes, if it is running constantly. -- Coneslayer 22:45, 2 December 2007 (UTC)[reply]
It may have a thermostat - which would make it turn off once the temperature gets high enough - then turn on and off as needed to keep the temperature high enough. In that case, the peak usage would be 1.5kW - but the average may be much less depending on how well insulated the room is. If it doesn't have a thermostat and you leave it turned on at its maximum setting - then it'll use 1.5kW all the time. Without more information, it's hard to know which you have. SteveBaker 00:33, 3 December 2007 (UTC)[reply]
Nitpick: the second unit in your question should be the KWh (pronounced kilowatt-hour) not the KW/h (kilowatt per hour). Algebraist 02:02, 3 December 2007 (UTC)[reply]
Even smaller nitpick: the symbol k for kilo- is not capitalized. You mean kWh. --Anon, 09:35 UTC, Dec. 3.
Thank you for the answers.217.168.3.246 02:54, 3 December 2007 (UTC)[reply]
Further, somewhat important nitpick: your heater should have a voltage rating. It will consume 1500 watts only at the specified voltage. If the voltage is increased, a resistive heater will consume more power (and produce more heat) with thc converse true if the voltage is lower than the rating. For instance: a heater rated at 120 volts, but operated on 110 volts (a very realistic scenario), will only consume 1260 watts (assuming the resistance of the heating element remains about the same, a reasonable assumption). If a 1500 watt heater rated at 110 volts were operated at 120 volts, it would consume 1785 watts. Edison 03:11, 4 December 2007 (UTC)[reply]
The trouble with that is that ratings like 120 volts and wall-socket power at 110 volts are applicable to AC current, not DC - so these numbers are RMS values, and the math is more complicated. Also, a simple application of Ohms law doesn't work because resistance of most heaters varies with the temperature...it gets ugly, fast! SteveBaker 03:52, 4 December 2007 (UTC)[reply]
AC RMS versus DC has no effect on the wattage variation with voltage I stated for a resistive heater. RMS is the heating effect equivalent AC quantity which equals the heating effect of DC. The resistance of the heating element should not vary much over the narrow temperature range involved: not like a light bulb filament going from room temperature to incandescence. As a real-world instance of the variation of heat output with voltage, a kitchen oven was rated for 208 to 240 volt operation, but the designedr told me it was really built for 240. At 208 volts, the element did not get hot enough to broil meat., although eventually it would bake it. Beware of this in any "one size fits all" specification. I have seen electric appliances from India (fans) with a 110 volt nameplate, and wondered the same factory exported heating appliance to the U.S. designed for 110. Edison 05:28, 4 December 2007 (UTC)[reply]