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April 30

Likely that Antarctica has large amounts of oil and gas reserves?

Is it likely or unlikely that antarctica has large amounts of oil and gas reserves? Are most major energy resources already found? Pass a Method talk 00:20, 30 April 2013 (UTC)[reply]

There's definitely coal to be found there, as the land mass once had a much milder climate (before Australia broke away from it). Where there's coal, there's often oil as well. More research needed, though. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:59, 30 April 2013 (UTC)[reply]

Actually, the article you linked describes Antarctica's natural resources. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:04, 30 April 2013 (UTC)[reply]

According to [1] it is likely that it has some reserves, but they would be very difficult, both technically and politically, to exploit. 202.155.85.18 (talk) 06:23, 30 April 2013 (UTC)[reply]

Agreed. The problem with drilling through glaciers is that they move quickly enough to destroy the well. So, you'd need to drill where there are no glaciers, like a rock outcrop, or where the glacier isn't moving, like in a crater. As for the environmental risk, any oil spilled won't be cleaned up by natural forces any time soon, like it would in a more temperate climate where bacteria can get at it. So, you might have an oil slick on the glacier for thousands of years, until it eventually drains to the sea. StuRat (talk) 10:10, 1 May 2013 (UTC)[reply]

I know, let's warm up the atmosphere to melt all those fangled glaciers, by releasing a whole lot of carbon dioxide. That should solve both problems. Plasmic Physics (talk) 08:14, 2 May 2013 (UTC)[reply]

Brilliant, let's suck up the last splosh of hydrocarbons to be really sure that the planet fries and color the ice beaches black while at it ;-) Electron9 (talk) 00:18, 5 May 2013 (UTC)[reply]

Abwehr

In the article "Operation Mincemeat", it says that "The Germans had the means to read the letter without opening the envelope". How did they do it? 24.23.196.85 (talk) 06:09, 30 April 2013 (UTC)[reply]

I wouldn't know, but I can think of a way: I have made etched copper electronic circuit boards from masters printed in electronics magazines. Such masters have been printed in ordinary black ink, with the text on the other side of the page printed in green or blue ink. Making the etch circuit involves coating the copper with a photo-sensitive varnish, placing the magazine page on top, and then exposing it to the sun for 30 minutes or so. The sun's ultraviolet light is blocked to a usefull degree by the black ink but passes through the paper and green ink, and hardens the photo-sensitive varnish. You then wash it with an alkaline solution and only the varnish not exposed to unltraviolet remain. You then dunk it in acid to etch away the copper not protected by varnish. The result is not all that good as the paper causes some blurring, and the contrast is not really sufficient, but it works.

Therefore the Germans could have read letters by exposing photographic film to ultraviolet light (or the suns' rays) passing through the envelope. Making the paper wet would help, and using an ultraviolet filter to block visible light & infrared would have been a good idea.

Ratbone 124.178.140.70 (talk) 06:45, 30 April 2013 (UTC)[reply]

"Darned" if I know for sure, but one method floating around is to slide two knitting needles under the flap, one on each side of the letter and twirl them so that the paper becomes rolled up tightly enough to remove without disturbing the seal too much. I saw this on some TV show or movie. Another claim is that liquid freon can be used to make the envelope transparent for a short while without leaving a trace afterwards. Clarityfiend (talk) 07:01, 30 April 2013 (UTC)[reply]

I've actually used that once. I was down to my last envelope, I needed to post a check, stuck it into the envelope and sealed it...then realised that I'd forgotten to sign it...and it was my last envelope, so I couldn't tear it open. It was really easy to extract the check from the envelope, using a pencil with the sticky strip from the top of a post-it note wrapped around it sticky-side-out. You put the pencil in through the small gap at the top corner of the envelope where there is no adhesive...roll the paper around it into a tight cylinder - then remove the pencil. Putting the the (now signed) check back into the envelope was just as easy. I don't think the "tampering" would be evident...although maybe the check would still have some curvature to it when it arrived at the destination that might alert a suspicious victim...maybe not though. SteveBaker (talk) 20:25, 30 April 2013 (UTC)[reply]

Letter-unsealing without leaving obvious traces was a very well-developed skill by 1940. The British opened and read a large portion of mail going from the US to neutral countries in Europe, and used the info the found to uncover German espionage rings using mail drops in Portugul and elsewhere. (Original research) I tried "steaming" an envelope open using a teakettle, but the paper is quite soggy before the adhesive lets go. A sharp knife, even heated also did not open a letter without tearing the paper. Apparently the spies who received the mail did not detect that it had been opened, so WW2 counterespionage folks must have had really good teakettle skills, or other special chemicals.E-How results on Google just suggest the teakettle routine. It would seem like a good idea to make the letter "tamper evident" by using ink which would smear or blot if the envelope were steamed. Suggestions that the envelope can be opened when it has been in a deepfreeeze run counter to real life experience that letters which have been in a mailbox at -20 F are still securely sealed. Edison (talk) 19:08, 30 April 2013 (UTC)[reply]

It's possible that the material used for envelope adhesive has been improved since the 1940's. I don't see any indication of that in our articles though. SteveBaker (talk) 20:25, 30 April 2013 (UTC)[reply]

The article goes on to describe how the Spanish read the documents - "The Spanish removed the still-damp paper by tightly winding it around a probe into a cylindrical shape, and then pulling it out between the envelope flap, still closed by a wax seal, and the envelope body." and later returned them to the envelopes - "The documents were re-inserted into their original envelopes, reversing the process by which they were removed,". This sounds similar to the knitting needle technique mentioned by Clarityfiend. Note that the British were aware that the documents had been tampered with. Mikenorton (talk) 21:03, 30 April 2013 (UTC)[reply]

What I don't understand is how they could get the paper flattened out afterward so it wasn't apparent it had been rolled up/tampered with. Stick it under some heavy books perhaps? Clarityfiend (talk) 02:03, 1 May 2013 (UTC)[reply]

What I don't understand is how anyone is supposed to get even the faintest hint of what this question is about from the heading "Abwehr". I think I'll run some courses in creating useful and meaningful headings. -- Jack of Oz ^[Talk] 19:22, 1 May 2013 (UTC)[reply]

If only there were some sort of readily-available comprehensive online reference work that might help you to resolve such difficult situations....

Abwehr as a section title makes perfect sense, as the question is about techniques used by the World War II German intelligence service of that name. If there were any confusion about the context or meaning, the OP helpfully linked to the article, Operation Mincemeat, that prompted his question.

Could the OP have used a longer or more detailed section header? Sure—but it still names a relevant subject area, it has the virtue of brevity, and it's a damn sight better than the ever-popular generic "Question". Is knowing the name of a German intelligence service a bit of specialist knowledge? Sure—but it's the sort of semi-specialist knowledge one might expect a reader who could answer the question to have. Someone who doesn't recognize the term Abwehr almost certainly isn't going to be able to write knowledgeably about WWII-era German spycraft. TenOfAllTrades(talk) 20:41, 1 May 2013 (UTC)[reply]

You’re parsing the issue completely backwards. You’re looking at the question and understanding why the OP chose the word Abwehr for the header. We can pretty much always work out how people come up with their strange headers, e.g. why the previous question was headed "Likely"; or why a question that's about the numbers of people killed and injured in a mining disaster in Alabama in 1917 would be headed "America". That's not rocket science, but does that help anyone to know what the question is actually about? Well, not at all. The header is supposed to tell us what the question is about, at least broadly. Searching for this question in 5 years time, who is going to remember that it's headed "Abwehr", given that that word never appears in the question? You might say the header meets the "broadly" criterion. But it doesn't. The question was framed in the context of Operation Mincemeat (a historical event), but it was placed on the Science desk because it’s primarily about the technical aspects of espionage, and only marginally about the history of the Abwehr. Some respondents have shared their own relevant experience of the science involved, and I very much doubt any of them got that experience from the time they spent as members of the Abwehr. -- Jack of Oz ^[Talk] 21:29, 1 May 2013 (UTC)[reply]

(un-indent) Thanks for the input, everyone! So I gather that the knitting-needle trick was the method actually used -- but if I ever use this in my novel writing, I think I'll go with the UV projection technique because it has more of a sci-fi flavor to it. 24.23.196.85 (talk) 00:00, 3 May 2013 (UTC)[reply]

Current = Conductance * Potential difference

V for potential difference, R for resistance, I for current, and G for conductance.
We know, I ∝ V. After removing proportional sign, it is I = G*V. If we write, V ∝ I, then after removing proportional sign, it is V = R*I. Please, check whether my above statement is right or wrong. Scientist456 (talk) 07:55, 30 April 2013 (UTC)[reply]

Yes, you'll see all that at Electrical resistance and conductance. Dmcq (talk) 08:29, 30 April 2013 (UTC)[reply]

Energy in Prokaryotic Cells without mitochondria

We know that a eukaryotic cell gets its energy requirements from a cell organelle called mitochondria. But what's about a Prokaryotic Cell, from where it get its energy requirement?Scientist456 (talk) 09:10, 30 April 2013 (UTC)[reply]

All prokaryotic cells are capable of some form of cellular respiration (many if not most are capable of aerobic respiration, just like eukaryotic cells). They simply don't have specialized organelles in which to carry out the reactions. This is similar to how prokaryotic cells can carry out transcription and DNA replication even though there is no nucleus. If I removed all the interior walls from your house, you would still have a bed to sleep in, even if you could no longer call it a bedroom. Someguy1221 (talk) 09:22, 30 April 2013 (UTC)[reply]

Are you aware of the Endosymbiotic theory? Under that theory mitochondria were originally separate prokaryotic organsisms that formed a symbiosis with another organism. I know this isn't directly relevant to your question, but as you are thinking about related things I thought it might interest you. Equisetum (talk | contributions) 09:44, 30 April 2013 (UTC)[reply]

Binomial plant names contain special character in url (×) and give a 404 if you put a regular 'x' instead

This question has to do with all plants on wikipedia that are hybrids, and therefore can contain a special symbol called a cross (×) as part of their binomial name. While it's correct to have the special character '×' in the binomial name of a hybrid, it's neither consistent across all plants on Wikipedia, nor is it terribly useful for users trying to find a certain plant when Wikipedia gives a 404 if I put an 'x' instead of an '×'.

For example, I can reach the plant Rubus × loganobaccus with this special character (×) (http://en.wikipedia.org/wiki/Rubus_%C3%97_loganobaccus), but not with an actual 'x' (http://en.wikipedia.org/wiki/Rubus_x_loganobaccus).

Could you tell me how I can make links to plants that contain this special character where a person could put a normal 'x' instead?

Thanks Mskogstadstubbs (talk) 12:25, 30 April 2013 (UTC)[reply]

You can do this using a redirect. I've done it for the requested page, to do so on other pages (after reviewing the linked guideline) just create a page with #REDIRECT [[Page name]] Jebus989 ^✰ 12:43, 30 April 2013 (UTC)[reply]

I've taken the liberty of raising this issue at the Village pump (technical) [2]. Wnt (talk) 14:32, 30 April 2013 (UTC)[reply]

...and I've now filed a bug request for someone to update the search. Andrew Gray (talk) 15:09, 30 April 2013 (UTC)[reply]

Unsolved numericals related to electricity

Today, I solved hundreds of numericals related to electricity, but I got stuck at these three -
1. If electrons are caused to fall through a potential difference of 10⁵ volts, determine their final speed if they were initially at rest. (ANS: 23 * 10⁷ m/s)
2. An electric bulb connected to a 220 volt supply line draws a current of 0.05 A. Calculate the amount of coulombs per second flowing through the bulb. (ANS: 0.15 C/s)
3. Three equal resistors connected in series across a source of e.m.f. together dissipate 10 watts of power. What should be the power dissipated if the resistors are connected in parallel across the same e.m.f? (ANS: 90 watts)
This is not my homework. I tried my best to solve these, but I couldn't. Someone give me hints so that I can solve these. Thank you very much! Scientist456 (talk) 16:47, 30 April 2013 (UTC)[reply]

Please show us your work, and how you arrived at a different answer from the stated one, or how far you got before you were stumped, rather than just giving the stated problem and the stated answer. Then folks here will be delighted to help you. We would even point out if a stated answer is incorrect! Edison (talk) 18:47, 30 April 2013 (UTC)[reply]

I'll take number 3. E across each resistor (in series) = E_total/3. Current through each resistor R is the same (in series) so call that I; where (E_total/3)/R = I. so the power for each resistor = (E_total/3)*I And so we have original total power = (power for each resistor)*3 = (E_total/3*I)*3=10, says the problem.

Now, if you connect the resistors in parallel rather than series, each resistor sees E_total, therefore the current through each resistor is I*3 (because the new voltage, E_total, is 3 times the original, E_total/3, therefore the current will be 3 times as much) = ((E_total/3)/R)*3. so the power for each resistor will be E_total*(I*3), and the total for all three will be (E_total*(I*3))*3 which is nine times the original power ((E_total/3*I)*3) which was defined as 10 in the problem, so the new power equals 90.

or you could do it by looking at the total. in the original you have E_total across a total resistance of 3R (where R is the resistance of 1 resistor), giving you a current of I = E_total/3R, and power=E_total*I = E_total*(E_total/3R)=10 watts. then you've switched to E_total with a total resistance of R/3 (because they're in parallel) so you've got a current of I = E_total/(R/3), and power=E_total*I = E_total*(E_total/(R/3))=90 watts, again substituting from the original power.

or, once you get an intuition of where it derives from and how it works, you don't need to break it up into each resistor/current any more, you just use power=(E_total squared)/R_total; E_total remains the same, R_total goes from 3R to R/3, so new R_total = old R_total/9 so the power goes up 9 times. (That's a general rule, by the way; if you have N identical resistors in series and you switch to parallel, the total power goes up by N squared. Or down by N squared, if you switch from parallel to to series; if you understand the above you'll see how that works, the voltage and current on each resistor each go up N times) Gzuckier (talk) 18:49, 30 April 2013 (UTC)[reply]

Falling electrons: volts are joules per coulomb. You know how many volts you have, how many coulombs per a single electron, therefore you can calculate the kinetic energy the electron must obtain. You know its mass so you can solve for velocity. I suspect there's an easier way, but that will do it.

An ampere is a coulomb per second, so the answer you parenthesize there mystifies me entirely.

For resistors in series, the resistance adds up. For resistors in parallel, the conductance adds up. Just think of them intuitively as narrow channels with water flowing through them. Wnt (talk) 19:53, 30 April 2013 (UTC)[reply]

The book answer for 1 has the electron accelerating to 3/4 the speed of light. Is it still ok to ignore relativistic effects at that speed? I agree that the book answer to 2 is too high, and 3 is pretty straightforward given Ohm's law and the formula for power and for series versus parallel resistances. Edison (talk) 22:05, 30 April 2013 (UTC)[reply]
- I think that the answer to 2 may depend on the lamp being AC, but I can't see how to get a value of 0.15 C/s either. The _net_ number of coulombs per second is zero (assuming there's no DC component), ~~the _total_ number of coulombs per second is 0.283 (peak current = rms current * sqr(2), total C/s = peak current * 4)~~. Tevildo (talk) 23:20, 30 April 2013 (UTC)[reply]
  - Sorry, not thinking properly. 0.283 is the number of coulombs per _cycle_, so the number per second will be 14.14 at 50 Hz and 16.97 at 60 Hz. Tevildo (talk) 23:29, 30 April 2013 (UTC)[reply]

Our article on kinetic energy gives both Newtonian and relativistic formulae, so the OP can work them both and compare the results if desired. (because this is a homework question I've been holding off on doing the calculation) But our article on electron says it has a rest mass of 511000 electron volts, and we've just handed it 100000 eV in kinetic energy - if half its mass is kinetic energy it should be pretty relativistic. Wnt (talk) 23:41, 30 April 2013 (UTC)[reply]

I have written here the same questions and answers written in the book. I also think the second Q is wrong as the Wnt pointed. For the first Q, I used the formula: charge (q)* potential difference (V) = Half mass (m)* velocity (v) squared. But after using this, I found that the answer differs a little bit. Then, what should I do? Scientist456 (talk) 00:24, 1 May 2013 (UTC)[reply]

According to [3] (the formula under "thus the work expended" and/or "for higher speeds") 1/sqrt(1-v^2/c^2) = (total mass/rest mass). The rest mass is 511 keV, the kinetic mass is 100 keV. So (1-v^2/c^2) = (511 keV/611 keV)^2 = 0.837; v/c=sqrt(1-0.837) = 0.405 * 3*10^8 m/s = 1.21*10^8 m/s. Hmmm, but they got 2.3*10^8? I'll leave it to you to figure out where I was off - since it's a homework question and all, yeah, that's why. :) Wnt (talk) 01:10, 1 May 2013 (UTC) (some numbers corrected per response below)[reply]

Electron's mass is 511 keV, not 51.1 keV, making that electron safely non-relativistic. Dauto (talk) 14:03, 1 May 2013 (UTC)[reply]

Wait a second, the problem states 100,000 eV, not 50,000 eV which makes the electron borderline relativistic. The classical answer will be off by about 10 to 20%. Dauto (talk) 14:55, 1 May 2013 (UTC)[reply]

Dang it, that should teach me to try to do math in a rush when I'm half-hearted about whether I ought to! Worst part is, I corrected my math above, and now it's way off from the book answer at the beginning. Wnt (talk) 15:58, 1 May 2013 (UTC)[reply]

DNA

In a DNA double helix how do you tell the Template strand from the Information strand? Are they different in some way? — Preceding unsigned comment added by GurkhaGherkin (talk • contribs) 17:02, 30 April 2013 (UTC)[reply]

From our DNA article: A DNA sequence is called "sense" if its sequence is the same as that of a messenger RNA copy that is translated into protein. The sequence on the opposite strand is called the "antisense" sequence. Both sense and antisense sequences can exist on different parts of the same strand of DNA (i.e. both strands contain both sense and antisense sequences). Both strands can contain 'template' sections. AndyTheGrump (talk) 17:31, 30 April 2013 (UTC)[reply]

You might also consider during replication, when you can tell the newly synthesized strand from the old strand by labelling it with 13C/15N (the Meselson-Stahl experiment; the technique is actually still used more than 50 years later [4]) There are also some differences even in naturally grown cells in DNA methylation and histone posttranslational modification when the strand is first synthesized, and occasionally afterward. Wnt (talk) 19:46, 30 April 2013 (UTC)[reply]

The short answer is that the location of the transcription start site tells you which is which. The TSS and associated sequences (like the TATA box) are sequence specific, and not identical to their reverse complement, so on only one strand will it be of the correct sequence to start transcription. As Andy mentions, either strand can (and does) contain TSSs for various genes, but for any particular gene the initiation sequences will be on one strand or the other. -- 205.175.124.30 (talk) 01:07, 1 May 2013 (UTC)[reply]

Thanks guys. I appreciate it.GurkhaGherkin (talk) 14:08, 1 May 2013 (UTC)[reply]

May 1

Can Middle-Easterners be of mixed racial heritage?

Besides Middle-eastern are considered white can some be mixed, like East Indian. If not all Middle-eastern are white, where are middle-eastern most likely to be mixed with? Is most North African people whites? Do people live in Egypt and Morocco include white peoples? Is India/East Indian considered as Asian? I always remove East Indians/India from being considered as Asians because they don't look Oriental. Is Southeast Asia like Malaysia, Singaporean, Thailand ethnic diverse. Like some people living in these countries have Chinese, Vietnamese, Filipino, East Indian. I thought people in Thailand are just Thai, I didn't hear they have Vietnamese, although I hear Thialand people ahve some Chinese in their ethnic groups. Because asiannation.org don't present the interracial group with those categories.--69.233.254.115 (talk) 00:08, 1 May 2013 (UTC)[reply]

I count six questions and three comments there. Could you pick one or two questions from your list? We can't deal with all that in a useful way. Looie496 (talk) 00:18, 1 May 2013 (UTC)[reply]

The easiest way to start might be to abandon the notion that there is such a thing as a pure or standard white person in the way that one can define a pure shade of green on the RGB color map. For example, who are whiter, Irishmen? or Poles? There are racial differences that exist in a multi-dimensional continuum that is much more complex than anything that can be represented on a two-dimensional map. The best one might do is map phenotypes and say that, by average facial appearances, perhaps, Burmese look intermediate to Thais and Bengalis, while Australians and Icelanders do not. Whether Greeks or Finns are white depends on how you want to define your terms. You might enjoy reading Cavalli Sforza and looking at some of his genetic maps.Note both Africa and Europe are green on this one. μηδείς (talk) 01:55, 1 May 2013 (UTC)[reply]

Additionally, if you just look up our articles on the countries you are interested in, such as Thailand, Malaysia and Singapore, they usually have the info you are after right in the lede, such as About 75% of the population is ethnically Thai, 14% is of Chinese origin, and 3% is ethnically Malay;[1] the rest belong to minority groups including Mons, Khmers and various hill tribes. The country's official language is Thai. The primary religion is Buddhism, which is practiced by around 95% of the population. . If you want more info, those articles will either have a section on demographics, or even their own whole page such as the Demographics_of_Thailand. Vespine (talk) 06:21, 1 May 2013 (UTC)[reply]

Racial identity and Passing (racial identity) might be of use here. OsmanRF34 (talk) 15:08, 1 May 2013 (UTC)[reply]

Corynebacterial porin B

Is Corynebacterial porin B an enzyme? Revolution1221 ^{(talk · email · contributions)} 01:01, 1 May 2013 (UTC)[reply]

Believe it or not, Corynebacterial porin B is an article; it sources to [5]. I didn't see anything obvious in NCBI about porin B having known enzymatic activity or an active site for any identified reaction - as a simple channel it doesn't need to catalyze ATP breakdown - but it is not uncommon for a protein to have an as yet unidentified enzymatic activity. Wnt (talk) 01:28, 1 May 2013 (UTC)[reply]

Why does the same temperature feel so much hotter at the beginning of summer than once it's been summer for a while?

I live in Minnesota and within a week recently we went from having snow on the ground to experiencing 73 degree Fahrenheit temperatures. The other night it was only maybe 68F yet it felt absolutely scorching. I know that within a month or two that will feel like a cool temperature. What's going on inside my body that causes me to adjust? Is it blood related? Thank you. NIRVANA2764 (talk) 02:57, 1 May 2013 (UTC)[reply]

You get used to it. μηδείς (talk) 03:19, 1 May 2013 (UTC)[reply]

The concept is called Acclimatization. Wikipedia's article isn't that great, but if you want to do serious research into the concept, you'd use that word to help you along. --Jayron32 05:16, 1 May 2013 (UTC)[reply]

Relative humidity is also likely to be a relevant consideration. High temperatures are uncomfortable, but so too is high relative humidity. In many climatic zones it is likely there will be more water vapor in the air in spring than in summer. If this is true for Minnesota it is likely that 73F in spring is accompanied by higher relative humidity than 73F in summer. A day of 73F with high relative humidity could be very uncomfortable whereas 73F with lower relative humidity would not be so uncomfortable. It is this effect that causes us to say that a night where the temperature is 75F is unbearable whereas a day where the temperature is 75F is comfortable. As temperature falls the relative humidity rises, and vice versa. Dolphin (t) 05:42, 1 May 2013 (UTC)[reply]

Yes, in the NYC and Phila areas, especially more inland, a hot day in muggy June is much worse than an equally hot day in dry August. μηδείς (talk) 16:09, 1 May 2013 (UTC)[reply]

See "Human Adaptations to heat and cold stress", a nice scientific paper on the subject, produced by the US army: [6]. It covers several physiological responses, and discusses what types of temperature exposure are needed to induce reactions. This and other sources basically uphold the notion that your "comfort zone" is based on what you were exposed to in the past few weeks, and that matches your description as well. SemanticMantis (talk) 13:01, 1 May 2013 (UTC)[reply]

Its not just perception either. The same heat wave occurring in May (at the start of the season) tends to be much more fatal than one with equivalent conditions occurring in August (late in the season). The Spatial Synoptic Classification program at Kent State provides heat wave warnings to major cities based estimates of dangerous conditions, and they literally require higher temperatures in order to consider a heat wave dangerous at the end of the season than at the beginning. Part of this is psychological. People perceive the heat and make decisions about managing it differently when it follow months of warm temperatures, than when it comes early in the season. However, there may also be a physiological component involved in adaptation that influences how much mortality can be expected during a severe heat wave. Dragons flight (talk) 13:59, 1 May 2013 (UTC)[reply]

2-Oxabutan-3-one or ethyl formate

Is the activation barrier for the reaction of ethanol with carbon monoxide small enough for the reaction to be spontaneous? Plasmic Physics (talk) 07:30, 1 May 2013 (UTC)[reply]

Not without a catalyst. 24.23.196.85 (talk) 23:49, 1 May 2013 (UTC)[reply]

More specifically, a direct reaction between ethanol and carbon monoxide is a specific case of the Oxo synthesis, which requires a rhodium catalyst and produces (in this case) propionaldehyde:

C₂H₅OH + CO + H₂ → C₃H₆O + H₂O

which can then be reduced with hydrogen to produce n-propyl alcohol. If you want to produce ethyl formate, though, then Fisher esterification (what, no article?) of ethanol with formic acid will work much better for you. 24.23.196.85 (talk) 05:31, 2 May 2013 (UTC)[reply]

You mean Fischer esterification? But I'll add a redirect from an obvious/easy-typo of his name... DMacks (talk) 05:39, 2 May 2013 (UTC)[reply]

Yep, that's what I meant. Thanks! 24.23.196.85 (talk) 23:54, 2 May 2013 (UTC)[reply]

I was trying to logically extend hydroxyaldehyde isomerisation, such as exists for saccharides, to higher bond orders. It's doesn't seem to be extendable. I was inspired by the synthesis of 2-oxabutan-3-ol from ethanol and formaldehyde. Plasmic Physics (talk) 05:41, 2 May 2013 (UTC)[reply]

One important thing to remember about carbon monoxide is that it is technically a carbene, so if you're trying to understand its chemistry it helps to look at it by analogy to other carbenes. That lone pair on the carbon reacts the same way other carbenes do, so you've got all of the same sorts of "singlet" and "triplet" reaction possibilities. I've not sat down to do work out the reaction you bring up here, but I suspect that useful mechanisms would be those that take carbene-like chemistry into consideration. --Jayron32 13:21, 2 May 2013 (UTC)[reply]

I was thinking more about a carbonyl insertion reaction. Plasmic Physics (talk) 05:35, 4 May 2013 (UTC)[reply]

Isotope enrichment

Concerning the mass spectrometric method: why isn't the element converted into a soluble salt instead of ionising it, to save energy costs? Plasmic Physics (talk) 07:45, 1 May 2013 (UTC)[reply]

Are you proposing using some method other than mass spec to separate isotopes? Or...well I'm not quite sure how just making something a soluble salt instead of ionizing it (aren't salts ionic?) would help MS? DMacks (talk) 14:52, 1 May 2013 (UTC)[reply]

It requires more energy to ionise a substance, doesn't it? Salts are ionic, but that is their natural state. Plasmic Physics (talk) 21:21, 1 May 2013 (UTC)[reply]

By definition, it always requires the same amount of energy to ionize a given atom, but the actual "cost" of it could obviously be different. By "ionising it", I assume you mean transiently, using some sort of electrical discharge, laser, etc.? So right there is another difference: if you enrich a salt, you get a salt; if you enrich a neutral metal (or whatever other elemental form), you get a neutral metal. If you want the neutral compound and you convert it into a soluble salt to enrich it, you then have to convert it back to the neutral form. Process design might be more sensitive to the number of steps and amount of processing than solely to the energy cost. But ultimately, you're on the right track, as our Isotopic_enrichment#Electromagnetic article notes, that MS separation is not economically practical and is rarely done unless actual cost is not a concern. If I'm still not understanding your question fully, please give us a specific example of a chemical you have in mind. DMacks (talk) 21:55, 1 May 2013 (UTC)[reply]

Yes, but you only need to convert the isotope to a salt once and back once in a cascade. It should remain in a saline condition between consecutive runs. To be sure that what I'm suggesting could make a significant difference, which part of the MS separation process is the most costly? Is it the transient ionisation?

Here is an example: suppose you want to enrich cobalt, take the sample and convert it to a solution of hexaamminecobalt(3+) chloride in liquid ammonia. Plasmic Physics (talk) 00:23, 2 May 2013 (UTC)[reply]

Yes, I'm aware that the counter-ion will curve in the opposite direction in each run, but certainly there is a way to rectify that at the target? Plasmic Physics (talk) 00:27, 2 May 2013 (UTC)[reply]

I've thought about this before too. Basically, why can't you construct a hydrocyclone where ionic species are deflected with a magnet? Or even a Zippe-type centrifuge that works in aqueous phase. 202.155.85.18 (talk) 03:00, 3 May 2013 (UTC)[reply]

A hydroclone is altogether different. How do you intend ficilitate magnetic fields in a hydroclone? I think that, using a liquid phase in a Zippe-type centrifuge will put too much strain on the system, it will make it too heavy to opperate at those high speeds. Plasmic Physics (talk) 13:20, 3 May 2013 (UTC)[reply]

What if you just had a soluble uranium salt with the natural isotopic ratio in a beaker with a stirrer, and applied a voltage using a cathode in the middle of the vortex and the anode somewhere else? The U-238 should be in lower abundance in the middle of the vortex so with multiple repetitions U-235 should be concentrated on the cathode. Obviously, it doesn't work since every 2 bit dictator isn't doing it, but why not? 202.155.85.18 (talk) 01:21, 6 May 2013 (UTC)[reply]

That is not a hydroclone, hydroclones filter out solid particulates, not solutes. Both isotopes would be concentrated at the cathode, that setup will achieve no-to-very poor enrichment. Plasmic Physics (talk) 02:42, 6 May 2013 (UTC)[reply]

Single iterations of Zippe-type centrifuges achieve poor enrichment too. That's why they're run in series again and again. Also, while what I described above is not a hydrocyclone, what you said about hydrocyclones is total non-sense. They don't filter anything and they are applicable to the separation of liquids of different densities, as noted in the lede of wikipedia's article on the topic. 202.155.85.18 (talk) 06:57, 6 May 2013 (UTC)[reply]

I did say very poor, and how do you suppose collecting the hypothetically enriched isotope. 'Filter' is a loose definition of 'sorting' or 'separation', which is also mentioned in the same lede. Furthermore, the solute in question is not a liquid, so the fact that a hydrocyclone can also sort or separate liquids are of no relevance. Plasmic Physics (talk) 07:40, 6 May 2013 (UTC)[reply]

Time period of a pendulum when placed in water

What is the formula (and its derivation) of time period of a pendulum when it is placed in water? Assume that pendulum is able to oscillate to and fro even its motion is being resisted by water. Concepts of Physics (talk) 08:58, 1 May 2013 (UTC)[reply]

Wouldn't the "period" change as it slows ? And, since the resistance in the water varies with the velocity, we could expect the pendulum to slow more at the bottom of the swing, where the pendulum is going faster, than at the tops of the swing, as well as more when the pendulum is first released. StuRat (talk) 10:00, 1 May 2013 (UTC)[reply]

Perhaps do it for a superfluid like liquid helium? Dmcq (talk) 10:17, 1 May 2013 (UTC)[reply]

I think perhaps this is supposed to be a straightforward question where one just considers the weight as opposed by buoyancy. Unfortunately the effective mass isn't such an easy factor. Dmcq (talk) 10:56, 1 May 2013 (UTC)[reply]

StuRat is back after a short absence (where have you been, Stu?), but he's still posting about things he should check first.

Regardless of whther a pendulum is in a vacuum (no friction), in air (a little bit of viscous friction), or in water (a large amount of viscous friction), the basic period is fundamentally the same - being determined by gravity and the pendulum mass-centre length. However at each swing, the peak deflection reached is a little bit less due to the viscous loss. This means that on the swing out from centre, decelleration is faster than in a vacuum, so it takes less time to reach the peak, and while on the down swing it accelerates a bit less, less energy is lost on the upswing because of the shorter distance. So the overal time for an upswing and a downswing is less than it would be for a vacuum.

We can thus conclude that the greater the viscosity, the shorter the period. It is analogous to an resonating electrical tuned circuit comprising an inductance and capacitance in parallel. Practical inductors and capacitors have electrical resistance, which, by dissipating a small amount of energy each cycle, causes the resonation to be a slightly higher frequency.

It is also analogous to the oscillations of a clock or watch balance wheel. Air viscosity acting on the hairspring and on the wheel cause an energy loss in each swing, as does the bearing oil viscosity. The result is a reduction in amplitude and a shortening of the oscillation period - the clock runs faster, NOT slower.

The physics math in all these cases is the same, only the units change.

Ratbone 124.178.42.184 (talk) 10:50, 1 May 2013 (UTC)[reply]

The Pendulum#Period_of_oscillation is dependant only on the pendulum's length, and the local gravitational acceleration. It is not dependant of the mass of the pendulum. CS Miller (talk) 12:42, 1 May 2013 (UTC)[reply]

Error in my post above corrected. Ratbone 60.228.254.67 (talk) 12:45, 1 May 2013 (UTC)[reply]

You can find an exact treatment in the case of laminar flow in the book on fluid dynamics by Landau & Lifschitz. The exact solution involves an integro-differential equation, the effective friction force at any time t depends on the trajectory x(t') for t'<t of the pendulum. Count Iblis (talk) 12:43, 1 May 2013 (UTC)[reply]

I found this article related to my question, but it explained in a complicated manner. Could someone explain the same in a simple language? Concepts of Physics (talk) 15:15, 1 May 2013 (UTC)[reply]

He considered that the pendulum is acted on by a notional gravity, this being the Earth's gravity reduced by the displacement of water by the pendulum - this causes an upward force on the pendulum. Because the notional gravity is less than standard gravity, the acceleration of the pendulum is lowered, and the time for one complete swing cycle of the pendulum is longer than normal. However, measurement showed that the oscillation time was different to what would be calculated by this alone. He accounted for this difference by taking into account momentum of the water, some of which follows the pendulum in its swing due to viscosity.

It would seem that my post above is wrong in part because I forgot about the momentum of the water dragged about. He essentially assumes the container of water is not full and open at the top, so that the water level rises by its displacement by the pendulum. I assumed that pressure was unchanged, which means the water was not displaced, which is possible in a sealed container but in practice not a likely situation. Serves me right for slinging off at StuRat, though my comment is still valid.

His taking into account the momentum of the water dragged about is reasonable, but I don't accept his calculated result (mass of moving water half the displaced mass - doesn't sound right, and his measured results support that he got it wrong. Consider a pendulum in oil - the displaced mass will be about the same, but the viscosity considerably greater, and more fluid will move. So the effect on the pendulum will be different.) Ratbone 60.228.254.67 (talk) 15:40, 1 May 2013 (UTC)[reply]

In the last part of the article, he gave a formula. Should I use the same formula to calculate the time period of a pendulum placed in water? Concepts of Physics (talk) 16:01, 1 May 2013 (UTC)[reply]

Yes, if you accept his reasoning and results. Note the use of the notation m* and g* - these designate the modifified mass and modified "gravity" as determined by formulae given earlier. You need to do some substitution to actually do the calculation. As stated above, I don't accept his theory. His result does not look right, and he has ignored viscous frictional loss. Do you need further explanation of why? Ratbone 121.215.62.31 (talk) 00:33, 2 May 2013 (UTC)[reply]

I agree with the reasoning there, it looks about right to me, the extra momentum of the water is what I meant by it being difficult to calculate the effective mass, kudos to them figuring it out. The surface of the water doesn't matter. Dmcq (talk) 13:26, 5 May 2013 (UTC)[reply]

What does "[O]" mean?

In a simple chemical reactions, I saw the use of [O] in the equation. Here, oxygen is lying between large brackets. I want to know what does "[O]" mean? What it is denoting? Concepts of Physics (talk) 09:12, 1 May 2013 (UTC)[reply]

I assume they mean a single oxygen atom. The reason for the brackets would be that this is not very stable, and will soon form diatomic oxygen (O₂), ozone (O₃) or some other molecule. StuRat (talk) 09:56, 1 May 2013 (UTC)[reply]

StuRat should stick to topics he understands or has checked before posting. The normal use of square brackets in chemistry is to indicate concentration. For instance, the rate of reaction is often of the form:

R = [X][Y]k

where R is the rate of reaction, [X] is the concentration of a chemical X, [Y] is the concentration of chemical Y, and k is a rate factor that is dependent on temperature, usually assumed to be determined by the modified arrhenious equation. Sometimes the square bracket term is shown in the form [X]² (or some other power) meaning that the reaction rate is determined by concentration squared (or raised to some other power)

Where the chemical is a gas, the square brackets indicate the gas density.

Where you see square brackets used in equations in chemistry, they are not chamical equations such as you probably learnt in junior high school, they are equations of or depending on concentration.

In standard chemistry notation, single atoms (such as atomic oxygen) are always shown without a subscript, as in O, if the diatomic or polyatomic form is meant, you must show the subscript, as in O₂, O₃, etc.

Ratbone 124.178.42.184 (talk) 10:35, 1 May 2013 (UTC)[reply]

If [O] is used above a reaction arrow in a chemical equation, it means oxidation or the addition of an unspecified oxidant. Similarly, [H] refers to reduction or to the addition of an unspecified reducing agent. 148.177.1.210 (talk) 11:45, 1 May 2013 (UTC)[reply]

Electromagnetic absorption by water

Hi, I was reading this article (http://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water) and came across the vibrational spectrum section. I noticed that the first paragraph was cited as coming from the book Tunable Lasers by F.J. Duarte and wondered if you could tell me which page numbers this came from as I cannot find it in the book of which I have a copy.

Any help would be much appreciated.

Thanks Roan — Preceding unsigned comment added by Vazzy044 (talk • contribs) 11:50, 1 May 2013 (UTC)[reply]

The citation is only in respect to defining the atmosphere infrared window. But I could not find it in my 2nd Edition either. Ratbone 60.228.254.67 (talk) 12:57, 1 May 2013 (UTC)[reply]

projectile motion

A bullet fired at an angle 30 degrees with the horizontal hit the ground 3km away.By adjusting its angle of projection can one hope to hit a target 5km away?assume muzzle speed to be fixed and neglect air resistance. — Preceding unsigned comment added by 197.250.192.22 (talk) 14:56, 1 May 2013 (UTC)[reply]

Please do your own homework.

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. DMacks (talk) 15:03, 1 May 2013 (UTC)[reply]

You can use the V_h=V_tsin(a) and V_v=V_tcos(a) to determine the vertical and horizontal components of the trajectory (assuming ballistic, with no air-resistance - in high-school physics these are valid assumptions). s = ut + 0.5 at² gives the time-of-flight. These can then be manipulated to relate i) initial speed, ii) launch angle and iii) distance travelled, for a given height difference. CS Miller (talk) 15:13, 1 May 2013 (UTC)[reply]

Note that in a vacuum fired directly from a flat surface, a 45° angle will go the farthest. However, in the real world, air resistance alters the calculations, as does the height of the gun above the ground and the topography of the ground. StuRat (talk) 18:53, 1 May 2013 (UTC)[reply]

The answer, under the given constraints (which, as some people seem to have missed, explicitly assume negligible air resistance), is obviously "yes" (because you can assume a target that is at a lot lover elevation than the 3km target). --Stephan Schulz (talk) 20:53, 1 May 2013 (UTC)[reply]

For those who are curious, here's a link to lover elevation. StuRat (talk) 03:42, 2 May 2013 (UTC) [reply]

Hey, I was hunting a memory leak all day (and half the night). One little spelling mistake is nothing. Found the sucker, too, at 1:30 am (local time)... --Stephan Schulz (talk) 07:01, 2 May 2013 (UTC)[reply]

Good jab Stu Rat, and good argument by Stephan Schultz, too.

This topic is fight! I'gonna try to put a hint here how to attack this kind of gunnery problem without too much pain,as knowledge is half the battle:

To OP, this is one of the problems which can be solved using an upper bound. If the angle is very low (say, below 10 degrees), the range is roughly proportional to the angle. If it isn't, it can still serve as a crude upper bound. As you know, 45 degrees is the angle for maximum range. Now, do the math, and you'll get a good result.

That's how you can solve that kind of problems (unless the question is more tricky and asks a close enough case to defeat the upper bound).

217.255.147.75 (talk) 05:53, 2 May 2013 (UTC)[reply]

This is straightforward. The range of a projectile fired at elevation θ along level ground (ignoring air resistance, curvature of Earth etc etc) is proportional to sin(2θ) - see Projectile motion#The maximum distance of projectile. You know the range at an elevation of 30 degrees, so you can calculate the range at any elevation - in particular, you can calculate the maximum range, which is achieved at an elevation of 45 degrees. Gandalf61 (talk) 08:23, 2 May 2013 (UTC)[reply]

To explain Gandalf's point further: for an elevation angel (oops, another case of "unclear fusion") of x, the range is

(sin x) (cos x) R,

for some length R. One useful trigonometric identity states that (sin x) (cos x) = sin(2x) / 2, which has a well-known maximum. The upper bound method is not really needed as long as no air resistance is involved. And it's a lower bound. The range is less than 0.1km per degree for angles >30°, so a range of 5km cannot be achieved at angles <=50°. Then again, you'd have a negative derivative at 50°, so you'll know that there is no angle >50° with a range of 5km either. Mathematics For The Win.

I must be getting old. Some IP user beat me to a "pain" topic. - ¡Ouch! (^{hurt me} / _{more pain}) 06:32, 3 May 2013 (UTC)[reply]

Oops sorry. It's an upper bound on the range or a lower bound on the angle needed to achieve the range. It can be seen as both. - ¡Ouch! (^{hurt me} / _{more pain}) 06:39, 3 May 2013 (UTC)[reply]

Are we made to process decimal numbers?

I mean mentally, can be calculate with them faster? OsmanRF34 (talk) 15:03, 1 May 2013 (UTC)[reply]

Faster than what? If you are asking whether 10 is the optimal base, the usual explanation of the reason why we use base 10 is that we have 10 fingers. Looie496 (talk) 15:37, 1 May 2013 (UTC)[reply]

I'am aware that on the physical level having ten fingers makes it appear that's easier to add numbers up to ten using your fingers. But, could something be discovered in our brains? In the same way that our brains are somehow limited to remember smells (dogs have a much bigger area for that). Just imagine we could subitize up to 12, maybe we would use the base 12 then. Maybe the base 10 is not the product of having 10 fingers, but the product of 2x what we can subitize. OsmanRF34 (talk) 16:25, 1 May 2013 (UTC)[reply]

The question is still: faster than what? If you compare base 10 to binary, it's clear that base 10 is easier for us, because it allows us to memorize tables (which our brains are pretty good at) rather than having to maintain a representation of long arrays of 1s and 0s. But if you compare base 10 to, say, base 12 or base 8 or base 16, the answer is far from clear. Looie496 (talk) 16:33, 1 May 2013 (UTC)[reply]

I mean comparing with another plausible base: 12, 16 or~even 60. OsmanRF34 (talk) 18:24, 1 May 2013 (UTC)[reply]

Well, it's certainly easier to add .140625 + .048 than to add 9/64 + 6/125. On the other hand, it's easier to multiply 1/32 * 1/64 than .03125 * .015625. Gzuckier (talk) 16:07, 1 May 2013 (UTC)[reply]

I think the question might be asking about base-10 numers, rather than positional notation. That is, is it easier to mentally calculate something like 145 + 346 or 0x91 + 0x15A. My guess is that it's probably a function of what you've been taught. The Babylonians used base-60, the Aztec and the Mayans used vigesimal (base-20), and I don't think they suffered from it (if you click through the "positional systems by base" list in the info box on those articles, you can see how various cultures used different positional bases). There's also "strange" mixed systems like the yan tan tethera sheep counting systems, although I don't know if any of those were used with arithmetic (e.g. if one group had lezar sheep and the other had borna, would the Teesdale shepard know that there were bumfit sheep in total, or would he have to start counting from yan again). -- 71.35.109.118 (talk) 16:31, 1 May 2013 (UTC)[reply]

The Babylonian system was based on a base 10 then a base 6 to form each 60, and the vigesimal systems were typically based on 5 then 4, you can see that in the larger numbers in the yan tan tethera system. I think the number systems in use are best for remembering everyday quantities, say up to a couple of hundred, rather than for speed, if we had to frequently contend with larger numbers without writing them down I believe we'd spend time on learning to use a larger number base. People doing memory tricks typically use a base of 100 or even a thousand to chunk numbers up, to a person like that a twelve digit number is only four digits long. Dmcq (talk) 17:14, 1 May 2013 (UTC)[reply]

I'm am of an era before pocket calculators were invented. Mental arthritic with fractions was quick and easy before the International System of Units came along, because you had so many common denominators. One could transpose your head. Yet, having everything in base 10 brought is own advantages. With big numbers, you sometimes, only have to mental subtract/add the powers and a little metal arithmetic to get you answer. On balance therefore, think the modern way is better. If you're interested in math, it is worth learning all forms of notation. But I think I know what the OP means... Being able to use factions to come up with the answer before my nephew could switch his calculator on, was not to my mind explaining math but showing off as that as an Uncle I was a smart ass - (where old age and cunning will always triumph over youth and skill..). But fractions were not part of his curriculum – so he would not let me explain it to him. Are are modern schools failing us? We where taught by rote all the common signs and tans (so we could guesstimate without reference to tables), most every day problems. We now have an education system of “one step forwards and two steps back”. No wonder China is now leading the world.Aspro (talk) 18:48, 1 May 2013 (UTC)[reply]

In view of those comments, maybe "mental arthritic" wasn't a spell check error after all. Very Freudian; almost Jungian. -- Jack of Oz ^[Talk] 19:14, 1 May 2013 (UTC) [reply]

Aspro is the typo-spell-champion: "metal arithmetic", "factions", "we where taught". No wonder China is leading the world.OsmanRF34 (talk) 19:29, 1 May 2013 (UTC)[reply]

Quite. We didn't have spell cheque back then to re-right watt we rote: but...

I have a spelling checker

I disk covered four my PC.

It plane lee marks four my revue

Miss steaks aye can knot see.

Eye ran this poem threw it.

Your sure real glad two no.

Its very polished in its weigh,

My checker tolled me sew.

A checker is a blessing.

It freeze yew lodes of thyme.

It helps me right awl stiles two reed,

And aides me when aye rime.

Each frays comes posed up on my screen

Eye trussed too bee a joule.

The checker pours o'er every word

To cheque sum spelling rule.

Bee fore wee rote with checkers

Hour spelling was inn deck line,

Butt now when wee dew have a laps,

Wee are not maid too wine.

And now bee cause my spelling

Is checked with such grate flare,

There are know faults in awl this peace,

Of nun eye am a wear.

To rite with care is quite a feet

Of witch won should be proud,

And wee mussed dew the best wee can,

Sew flaws are knot aloud.

That's why eye brake in two averse

Caws Eye dew want too please.

Sow glad eye yam that aye did bye

This soft wear four pea seas.

What would have helped me would have been touch typing courses so I could look at what I was typing without looking at the keyboard instead. Like what one does when writing with a fountain pen -a communication tool from long before your time. BUT as word possessors hadn’t been invented back then; so I am having to cope the best I can with these modern contrivances -that insist on correcting things (like spelling) for you. You too, will one-day be old, and have the piss taken out of you because you came into the technology late. “Why don't I 'proof read' your own text before hitting enter”, I might next hear you say... My typist did that for me, and she earned about a third or fourth in gross pay to my take home pay after tax. So why, should I get out of this habit when I contribute to Wikipedia Pro bono ? Anyway, I already have you to correct things for me now – funny how the world works, isn't it : ¬ ) Aspro (talk) 21:41, 1 May 2013 (UTC) [reply]

There are reasons to suppose that base 12 and base 60 have advantages. 10 is only evenly divisible by 5 and 2. 12 is divisible by 2,3,4 and 6 - so division is typically easier in base 12. However, there is a downside. In base 10, we belabor our kids to memorize the multiplication tables from 1x1 to 10x10 - which is 55 separate multiplications - of which the 1x and 10x tables are trivial - so really, 36 non-trivial facts have to be rote-memorized at a tender age. If you go to base 12, then you have 54 non-trivial facts to remember. That said, when I was a kid living in the UK, we were forced to memorize multiplications up to 12x12 because with pre-decimalized base-12 currency, you frequently had to use base 12 calculations in order to do arithmetic with money. So clearly it's possible for children to master base 12 multiplication tables...but since the number of things you have to memorize is proportional to the square of the number base - it gets unmanageable fast!

On that basis, I think we could work a little faster in base 12...but it's commonly believed that we do better in base 10 because we have 10 fingers.

But let's push this to the limits:

* Base 60 would be even better from a division perspective (you can evenly divide by 2,3,4,5,6,10,12,15,20 and 40) - but memorizing the 60x60 multiplication table would be horrifyingly bad! But if you *could* memorize them, then the number of digits you'd have to multiply for larger numbers would be tiny. 37x41 is an ikky calculation to do in your head - but in base 60, you have that memorized - so it's no harder than 7x8...so doing the base 60 equivalent of 3741x4137 is no harder than 78x87 in base 10.

Base 2 means that the multiplication tables would only demand that we remember 1x1=1, 10x1=10 and 10x10=100...which is entirely trivial. But multiplying 100010x10101 is *nasty* to do in your head - and almost every division by a non power of two results in a nasty recurring decimal!

So clearly, going to either extreme is bad - a happy medium for mental arithmetic is somewhere between 2 and 60. There is a trade-off between the amount of rote memorization and the reduction in the number of digits and ease of division. Hence, I very much doubt if we'd do better in bases much smaller than maybe base 8 or in bases much higher than maybe 12...but I think 12 would have a clear benefit over 10...if we had two more fingers!

SteveBaker (talk) 20:19, 1 May 2013 (UTC)[reply]

Incidentally - I think a bigger improvement could be had from having the ten symbols we use represent the numbers -5 to +4 rather than 0 to 9. Suppose we use an apostrophy to indicate the negative symbols: So -5 in our system is 5', -4 is 4' and so on. Counting up from zero, you count: 0,1,2,3,4,15',14',13',12',11',10,11,12,13,14,25',24',23'...and so on up to 44,15'5',15'4',15'3'...and so forth. This is clearer if you don't use 5',4',3',2',1' - but some other symbol entirely. Anyway, there are numerous advantages to doing arithmetic like this. Firstly, negative numbers are no longer "special". You add them the same way you add positive numbers. One thing I like is that when you're faced with a long column of numbers, you can go down cancelling out the all of the 4s with the 4's, the 3s with the 3's and so forth - without having to remember how many 10s to carry. It's a little like the roman numerals system where the number 9 is IX (ten minus 1) and 8 is IIX - just as 9 is 11' and 8 is 12' in this system. Of course it's horrible to learn once you know regular arithmetic...but that doesn't make it a bad system! SteveBaker (talk) 20:30, 1 May 2013 (UTC)[reply]

There's more on this in signed-digit representation. TenOfAllTrades(talk) 20:55, 1 May 2013 (UTC)[reply]

Its a bit like the qwerty keyboard in this respect. If we started from scratch, then this notation would be the way to go. How does it work on big numbers? There does not seem to be any problem with getting computers to out put this instead of base 10 as it is just math.Aspro (talk) 22:04, 1 May 2013 (UTC)[reply]

Thanks for that link TenOfAllTrades! I think you provided it the last time we talked about this too! I can never remember the name of the system.

And yes, it has the same problem as switching from QWERTY...or fixing the spelling of English words. We're stuck with a path dependence issue here.

There are no special problems with large numbers - in the signed digit representation some numbers have an extra digit (9 becomes 11') but you don't need a minus sign (-4 is just 4'). Computers would have no problems with using this format to interact with humans. The "two's complement" method for doing arithmetic that all modern computers have built into their hardware is, in effect, a base-256 signed-digit representation - so you could reasonably claim that computers already use this trick precisely because it's so much better than our normal notation. SteveBaker (talk) 14:00, 2 May 2013 (UTC)[reply]

(edit conflict) There's some speculation that early numbering systems were base 12, and that they were derived by counting the phalanges on one hand using the thumb of that hand. Three segments for four fingers is 12. The Wikipedia article on the Duodecimal notes this, indicating that such counting systems still exist today, using the right hand to count to 12, and the left hand to keep track of iterations of 12. That gives one likely source of the Babylonian Sexagesimal system (12x5=60). Now, when written in cuneiform, the Babylonian system was written in base-10 (the 60 numbers were six groups of ten rather than five groups of twelve), but the counting system itself is likely centuries older than the writing system, which could explain the difference. --Jayron32 20:31, 1 May 2013 (UTC)[reply]

The techniques we use to do mental computations are not what the brain uses to process data. In recent years it has been found that the extraordinary skills of savants are actually present and used by the brains of ordinary people, albeit when processing lower level information. Unlike savants, we normally don't have access to this processing capability (we only experience the results). By supressing the temporal lobe, one can turn ordinary people into savants, see here.

When we do mental arithmetic, you have to compare this to using a computer that only runs a word processor. To run that word processor, the computer of course performs many computations, but you can't access those facilities. Then you can still do calculations in an extremely cumbersome way, by e.g. use the word count facility of the word processor. So, while you may be struggeling to multiply two 3 digit mumbers mentally, your brain routinely does much harder computations at lightning speed all the time. Count Iblis (talk) 00:10, 2 May 2013 (UTC)[reply]

I agree with everything Steve Baker said, except this statement "but memorizing the 60x60 multiplication table would be horrifyingly bad!". I don't think it would be that hard. I really wish we still used the sexagesimal system. Dauto (talk) 14:44, 3 May 2013 (UTC)[reply]

You don't even need to memorize it. There are other ways to multiply in sexagesimal with about the same speed as decimal multiplication, e.g. using reciprocal divisors or mixed-radix multiplication. Double sharp (talk) 05:20, 20 November 2013 (UTC)[reply]

To use that example from above: here's how I would calculate 37×41 mentally in sexagesimal, with the convention that 1'00 means sixty. 37×41 = (40×41) − (3×41) = (2/3 × 41'00) − (3×40 + 3) = 27'20 − 2'03 = 25'17. The key point is to note that sixty has the "axes" of divisors 1-60, 2-30, 3-20, 4-15, 5-12, and 6-10: thus multiplying by 15 can be effected simply by multiplying by 60 (trivial in sexagesimal) and then dividing by 4. When faced with coprimes like 37×41, you can break one of the numbers up to a number with a nice relationship to 60 (here I used 40, which is two thirds). Armed with these relationships, all you really need to remember from the multiplication table are the products less than or equal to 1'00. These are: the 2× row to 2×30, the 3× row to 3×20, the 4× row to 4×15, the 5× row to 5×12, the 6× row to 6×10, and the 7× row to 7×8. The main reason why this becomes a practical method for sexagesimal is that almost all these rows involve just factors and are thus easy to remember, and there aren't that many products left (29+18+12+8+5+2=74, not too far off from the decimal or duodecimal tables). If you have a gap between divisors, this becomes a less practical method to use. The other main beneficiary of this method is base 120; the table is as big as vigesimal's, but substantially easier to memorize. Double sharp (talk) 10:35, 8 March 2015 (UTC)[reply]

Gastroesophageal reflux disease Herbal Remedies

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

Disinfectants that don't stink ?

So, I'm happily eating my dinner at a restaurant, when the busboy starts spraying down the adjacent table with an ammonia-based disinfectant. The rest of my dinner now tastes like ammonia. Is there an inexpensive odor-free disinfectant they could use instead ? StuRat (talk) 18:57, 1 May 2013 (UTC)[reply]

That is when I say "excuse me I'm eating here. please spray the rag instead of the table" One time I had a busboy vigorously wiping the adjacent table so hard that he was literally flinging pieces of food from the rag onto my table gross!165.212.189.187 (talk) 19:18, 1 May 2013 (UTC)[reply]

That's why they call them busboys. My brother-in-law actually wrote a mock paper on the "laws of busboys". But that was before the internet. μηδείς (talk) 19:24, 1 May 2013 (UTC)[reply]

They could use something based on a quaternary ammonium compound - very effective and odourless. see: http://www.henryschein.com/us-en/Medical/ResourceCenter/SurfaceDisinfectantsCategories.aspx Richerman (talk) 19:36, 1 May 2013 (UTC)[reply]

Is it inexpensive ? StuRat (talk) 19:45, 1 May 2013 (UTC)[reply]

Not that much more expensive, however it does need a longer contact time that products based on, say, bleach - which are cheap but smelly - but not that bad if diluted properly (think about the stuff they sell for sterilising babies' bottles - it's called Milton in the UK). Actually the correct way to disinfect a surface is to clean it first and then use a disinfectant, as organic matter tends to make disinfectants ineffective. This is less of a problem with quats and they can be made into a product that cleans and disinfects. Richerman (talk) 20:10, 1 May 2013 (UTC)[reply]

I wonder if the management of this restaurant are aware of this practice? In Europe we (and the health authorities) would not stand for such an 'irritant' (which is what ammonia is) being sprayed around. As Richerman above correctly states: organic matter tends to make disinfectants ineffective. Therefore, the busboy should first wipe the crud off the table, then finish off with a 'clean cloth' or wipe. What your restaurant Management use is up to them (and they most probably know the States Hygiene Guidance inside out). In Europe a lot of restauranteurs ( ok – before anybody says - “but my mum and dad never used it” were not talking about your mum and dad) use a dilute solution of distilled vinegar (1/2 to a cup to a bowl (UK units)), (which bulk bought form the wholesaler is dirt cheap and does not leave a residue. The advantage, is that it leave tables sparkling and works to the satisfaction of our health authorities. Relying on other detergents doesn't leave such a finnish. Was this restaurant packed out with people cueing out-side to get in? The management should value your input. If they don't – then there are always other restaurants to pig out in (or should that be dine out in?). Whatever! --Aspro (talk) 23:47, 1 May 2013 (UTC)[reply]

A lot of anti-bacterial sprays for use in commercial kitchens are marked "low taint".[7] Alansplodge (talk) 23:50, 1 May 2013 (UTC)[reply]

So it only smells a bit like a taint then ? StuRat (talk) 03:38, 2 May 2013 (UTC) [reply]

According to its data sheet its just cheap (expensive) isopropyl alcohol – so it wouldn’t leave any residue. BUT LOOK AT THE PRICE. £4.00 per litre! I would not consider a restauranteur to have control of his cost (and thus the quality of the food) if the table attendant started spraying that around. That's what, £40 per litre wholesale!!! That's one of a hell of an very expensive plastic spray dispenser if you ask me.--Aspro (talk) 00:24, 2 May 2013 (UTC)[reply]

Am I missing something? How do you make £4 per litre retail into £40 per litre wholesale? We use very similar stuff at work - a charity - to keep the public health inspector happy. You get an awful lot of squirts out of a litre by the way. Alansplodge (talk) 18:24, 3 May 2013 (UTC)[reply]

With the greatest respect Al, I do think you are missing something. You highlighted a brand above... Quote:[8]. Data sheet say's it contains only between 5 and 10% Isopropyl-Alcohol. Which is £4 English pounds per litre. Look at [9] add water (expensive bottled water if you like) and do the math. I think I have been over generous. I think your charity are buying ruddy expensive plastic spray bottles with a fancy label. What do you think? Please don't think of me as a smartass, I was a commercial buyer at one time - it was my job.Aspro (talk) 22:56, 3 May 2013 (UTC)[reply]

Don't you think now, that these spray-things are just for housewives, who husbands just shell out the money to keep them happy? If you are involved with a charity, every £ counts. Inform the charitably committee that the charity is better off by following the practices of commercial organizations. Volunteer Beatrice et al, might think its the bees knees in their kitchens but are they paying for the over-cost for its use in the charities kitchens? Aspro (talk) 23:26, 3 May 2013 (UTC)[reply]

Speaking of things that stink, your attitude to domestic deities seems somewhat sexist and antiquated. -- Jack of Oz ^[Talk] 00:20, 4 May 2013 (UTC)[reply]

Not at all. Consider this: if your doc told you that your wife, that she was to convalesce in bed for two weeks (– making you the house-husband), meaning that had to do the weekly shop down at the Mall. With all the other things you have to do during this time, you would probably only have time to whizz round and grab bottles off the shelf, without a mind to what they contained or their value for money. So doesn't it follow, if your happy getting in the shop, then you are happy with your wife's efforts to run the home. The fact that you probably (I don't know) leave it to the wife look after the kids ( “ah but I took the kids to the park only last month – Oh was it January -how time flies”), scrub the floors, clean up after you, and do your washing, (not now dear, I'm editing Wikipedia – tell me about it tomorrow!). I think that comment is the pot calling the kettle black when it comes to sexism.Aspro (talk) 21:16, 4 May 2013 (UTC)[reply]

You referred to products that are "just for housewives", as if these are sub-human creatures who are fit only for the worst quality goods imagineable. Welcome to the 21st century. Also, there are actually households containing more than one adult, who sometimes do naughty things with each other when they're not scrubbing the floors or editing Wikipedia, and there's not a wife or even a woman in sight. -- Jack of Oz ^[Talk] 21:56, 4 May 2013 (UTC)[reply]

Would you have been happier if I said house-people? Think you're just trying to make a mountain out of a mole hill, when the average reader would get my gist. This type of nitpicking on WP ref desk reminds my of Troll_(Internet) and Criticism#Psychopathology_of_criticism. Over to you now, as I'm sure you will want to be true true to form and let off some fireworks. 22:41, 4 May 2013 (UTC)

Not at all. I've made my point. You've reacted. It's over. -- Jack of Oz ^[Talk] 22:52, 4 May 2013 (UTC)[reply]

Does anyone know this mushroom?

It's growing in my backyard, but I've never seen it. (But I guess it must be quite common). Txs! Joepnl (talk) 21:04, 1 May 2013 (UTC)[reply]

Holy shit, you've got Morels in your yard! --Digrpat (talk) 21:14, 1 May 2013 (UTC)[reply]

I'm no mycologist, but I've eaten my fair share of Morels, and I concur that these are likely they. I still wouldn't eat these, however, as while Morels are quite tasty, then penalty for being wrong about the kind of mushroom it is could be, um, unpleasant, as Morchella#False morels indicates. --Jayron32 21:43, 1 May 2013 (UTC)[reply]

What or who would be the most reliable source for determining whether a mushroom is toxic or not? ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:12, 2 May 2013 (UTC)[reply]

Well, a mycologist, preferably one with graduate degrees, and with experience foraging and identifying fungi in the local region. You might get some guidance from a university extension, but even the experts are fairly reticent to give formal ID and eating recommendations, especially to people appearing "off the street". Something about ethics and liability. The experts are far more forthcoming if you sign up for a class. Much like our "no homework" policy, they will teach you the mycology you need to make informed decisions, but nobody can do your homework for you. In practice, people also learn from books, clubs, elders, lore masters, etc. The first thing anyone should learn about are the risks involved in eating found mushrooms. SemanticMantis (talk) 04:57, 2 May 2013 (UTC)[reply]

I recall a computer science problem in college - which was to write a piece of software that would take a list of arbitrary characteristics for some objects within a more general class - and to deduce a set of rules to divide them into two groups - and to provide a 'confidence' level for that decision. So, for example, given a description of a tree, find a rule that determines whether it's deciduous or not given things like the shape of the leaves, the height, the branching ratio, the trunk diameter, the smoothness of the bark, the soil pH it prefers and so on. Hand the software a new tree description and it'll tell you whether it's deciduous or not - and the confidence level of that decision - and it'll list the rules it uses (eg, if the length to width ratio of the leaves is greater than 10 then it's not deciduous unless adults of the species are less than 5 feet tall...or something like that).

One of the dozen or so data sets that we were given to test our software with was a long list of the characteristics of several hundred species of mushroom (size, color, surface patterning, shape, smell, taste, location, etc) - with the goal to deduce which were poisonous and which were not. It turns out that there is famously absolutely no rule or even complicated set of rules to allow you to determine this with any confidence level better than chance. The only solution is to come up with one unique rule for each type of mushroom. That fact prevents you from knowing whether a mushroom that you've never seen before is poisonous or not - other than by feeding it to someone and observing whether they drop dead or not! This was an excellent test of our software!

What I learned from that is that the only possible way to be entirely safe is to learn to recognize every species of mushroom that grows in your area - and to do so with perfect accuracy...which backs up what SemanticMantis says.

SteveBaker (talk) 13:37, 2 May 2013 (UTC)[reply]

I think that overstates the problem a bit. People who collect wild mushrooms do learn to recognize a few specific types, and stay away from anything else. The same thing really applies to wild plants of all sorts. It's much easier to learn to recognize things when you have a teacher and a variety of genuine examples in front of you. For example, I can take anybody into our local hill-park and teach them to recognize poison-oak in ten minutes. Learning to recognize it from pictures and descriptions is quite difficult, though. Looie496 (talk) 14:43, 2 May 2013 (UTC)[reply]

And is the method to identify poison oak in 10 minutes to apply it to your skin and see if it starts to itch and burn ? (Personally, I would consider choosing such an identification method to be a rash decision.) StuRat (talk) 06:57, 3 May 2013 (UTC) [reply]

Of course you're joking, but to give a serious answer, one of the big problems with poison oak is that in a person who is not sensitized, it takes several days for the rash to appear (and then you get a month of misery). Even in a person who is sensitized by repeated exposure (like me, unfortunately), it takes a day or so. So learning to recognize it by experience is nearly impossible. Looie496 (talk) 14:34, 3 May 2013 (UTC)[reply]

From Morchella I understand that it's hard to grow Morels. I guess we've managed to apply the right amount of neglect to the yard to fool them. Thanks for the info! I won't be eating them btw :). Joepnl (talk) 17:37, 2 May 2013 (UTC)[reply]

Yes, never eat an unidentified 'shroom...that's the morel of the story. StuRat (talk) 06:57, 3 May 2013 (UTC) [reply]

I am not sure about the US, but in France all pharmacists have a significant amount of training on recognising edible and dangerous mushrooms (they are required to have that by law), so we always check with the local village pharmacist if we are not sure about some of the muchrooms. (I guess you would usually call them drug store owners, not pharmacists). --Lgriot (talk) 08:14, 3 May 2013 (UTC)[reply]

There are old mushroom hunters. There are bold mushroom hunters. However, there are no old bold mushroom hunters.196.214.78.114 (talk) 12:17, 3 May 2013 (UTC)[reply]

May 2

How do scientist/astronomer figure out the star loss mass at giant/supergiant tip

I have heard when higher mass main sequence runs out of hydrogen and fuse helium by the time they get to giant or supergiant tip they lose roughly 7/8 of their mass. I don't know how do scientist figure out the variables, if the variable are guaranteed to be right, or they can as well guess out. I hear when sun gets to giant tip it will lose roughly 1/3 of the original mass, previous calculations shows 1/4. Do scientist even know how much mass sun will have lost when it gets to RGB and AGB tip, or they just group all the stars together and find a pattern. Or the variables they present is at least 70% cavity guess. Is there ways to say how much accuracies and errors academic research documents contains. Or the best answer is some academic paper contains more error, some academic paper may contain more accuracies. Is Academic research paper (=) dumping information they have available (+) fill in the gap (any missing and uncertain cavities they make wild guess). The thing is most people cannot see the guesses research documents have.--69.233.254.115 (talk) 01:30, 2 May 2013 (UTC)[reply]

Oops, everything is found over [10]. I forgot the query type so long ago.--69.233.254.115 (talk) 01:43, 3 May 2013 (UTC)[reply]

OK, glad you found it. I'll mark this Q resolved. StuRat (talk) 07:00, 3 May 2013 (UTC)[reply]

Resolved

Temple of the rats

See Karni_Mata#The_legend. Apparently rats are protected and fed there. So, what controls their population ? StuRat (talk) 03:53, 2 May 2013 (UTC)[reply]

From what I understand, they're only protected as long as they remain in the temple. When they venture outside, they are are at the mercy of their natural predators. Plasmic Physics (talk) 03:56, 2 May 2013 (UTC)[reply]

If I brought a few pet cats in, would that be trouble? ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:05, 2 May 2013 (UTC)[reply]

I imagine 20,000 rats would make short work of "a few pet cats".--Shantavira|^{feed me} 15:26, 2 May 2013 (UTC)[reply]

That might depend on the sizes of the cats and the rats. ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:38, 2 May 2013 (UTC)[reply]

A rat terrier would be a lot more trouble. Looie496 (talk) 16:58, 2 May 2013 (UTC)[reply]

Whatever it takes. ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:13, 2 May 2013 (UTC)[reply]

Hemorrhoid infections

Why don't hemorrhoids instantly become infected or cause serious infection that would surely result if an open wound on some other body part were so directly exposed to feces?68.36.148.100 (talk) 04:09, 2 May 2013 (UTC)[reply]

They are covered with "anoderm" (kinda like skin) - that presumably prevents fecal material from entering and infecting. SteveBaker (talk) 04:13, 2 May 2013 (UTC)[reply]

Butt, some break don't they? The blood in stool article says they are the #1 reason for it.68.36.148.100 (talk) 04:18, 2 May 2013 (UTC)[reply]

Only in New Jersey. Otherwise, see the section about fissures. ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:39, 2 May 2013 (UTC)[reply]

Wow, I wanted to ask this exact same question a few days ago. Yes, it certainly seems that if blood gets into stool then stool might get into the blood. I wonder if there is some special amped up immune response in that area, and, if so, if we could learn to extend that the the rest of the body, when needed. StuRat (talk) 04:49, 2 May 2013 (UTC)[reply]

Best typo ever. Evanh2008 ^{(talk|contribs)} 01:54, 3 May 2013 (UTC)[reply]

Found [11], [12] but neither is quite what I'd call an answer. Wnt (talk) 01:37, 3 May 2013 (UTC)[reply]

Deballocker

Is it true that in World War 2, the Germans used a type of landmine that was specially designed to castrate its victims rather than kill them? 24.23.196.85 (talk) 05:36, 2 May 2013 (UTC)[reply]

Our article Anti-personnel mine explains that such mines are usually intended to injure, not kill, their victims. I guess that the German S-mine could earn a reputation as designed for castration since it was launched about a metre into the air before detonating. However, there is no way that the S-mine or other mines of the same type could be accurate enough to castrate someone, except by chance. Sjö (talk) 05:45, 2 May 2013 (UTC)[reply]

I agree - damage to the enemies' legs was the real goal of these things - take people out of action as effective combatants - but without killing them, so they clog up the medical facilities, consume resources behind the lines, slow any advance, cause uninjured fellow soldiers to take extra risks to save them...that kind of thing. Castrating someone won't necessarily stop them from fighting again in the future - but ripping their legs to shreds most certainly will. SteveBaker (talk) 13:15, 2 May 2013 (UTC)[reply]

On a pedantic note shouldn't it be "debollocker". Richard Avery (talk) 13:23, 2 May 2013 (UTC)[reply]

'Ballocks' is a known alternative spelling. Chaucer uses it in (I think) the Pardoner's Prologue, and there's a type of Renaissance fighting-knife known as a ballock-dagger (apparently due to the handle-shape, rather than any potential target). AlexTiefling (talk) 15:36, 2 May 2013 (UTC)[reply]

The late Kingsley Amis steadfastly used the spelling 'ballocks', and I think his son inherited the affectation. AndrewWTaylor (talk) 21:08, 2 May 2013 (UTC)[reply]

There is a famous letter to Abraham Lincoln which (among other things) urges him to "call my Bolics[sic] your Uncle Dick." Interesting that the word now seems to have died out altogether in the States... Tevildo (talk) 23:18, 3 May 2013 (UTC)[reply]

Richard, best to be frank about this. It's clear that pedantry is not your true calling (which may make you very happy). Your pseudopedanticism is betrayed by the absence of a comma after 'note' and a question mark at the end of your question. There are still openings at my Winter Pedantry School; special rates for Wikipedia editors. :) -- Jack of Oz ^[Talk] 20:53, 2 May 2013 (UTC) [reply]

We should also consider the effect such a weapon would have on enemy morale. While a dead soldier returned home and buried with honor can actual inspire patriotism, a live soldier returned home, but missing a few key bits and pieces, can rather have the opposite effect. Not many men will rush to the recruiting office after seeing that. StuRat (talk) 07:05, 3 May 2013 (UTC)[reply]

D&D vets know the S-Mine as "the ultimate balls check" - when it pops up, it'll determine with scientific rigor whether its or your balls are harder. With extreme prejudice.

Oz: Do I get discount if I take ten? - ¡Ouch! (^{hurt me} / _{more pain}) 08:21, 3 May 2013 (UTC)[reply]

By all means. You understand, of course, that you will be required to exhibit decilocation, and empower each of your decuplicate presences to have a separate one of your ten personalities. The auditors get a bit funny about me teaching an apparently almost empty classroom. -- Jack of Oz ^[Talk] 09:23, 3 May 2013 (UTC) [reply]

Of the S-Mine:- "...there were extraordinary escapes. The chaplain of the 5th Seaforth Highlanders trod on one which bounced up and knocked his glasses off; perhaps divine intervention prevented the main charge from exploding. Pfc Larry Treff of the US 26th Division was lucky enough to have one bounce up and hit him in the groin without exploding; he was thrown several feet but survived with minor injuries, though his groin area was 'so purple and swollen' that he was temporarily immobile."[13] Alansplodge (talk) 16:13, 3 May 2013 (UTC)[reply]

Deriving the momentum operator.

In quantum mechanics, given the canonical commutation relation $[{\hat {x}},{\hat {p}}_{x}]={\hat {x}}{\hat {p}}_{x}-{\hat {p}}_{x}{\hat {x}}=i\hbar$ and ${\hat {x}}=x$ , how can one derive the result ${\hat {p}}_{x}={\frac {\hbar }{i}}{\frac {\partial }{\partial x}}$ ? I was told that this can be done, though I am unsure how to go about doing this. -- — Trevor K. — 05:40, 2 May 2013 (UTC) — Preceding unsigned comment added by Yakeyglee (talk • contribs)

Well, you can show calculate the commutator

[{\hat {x}},{\hat {p}}_{x}]

as follows. If F is any function of x, then

[{\hat {x}},{\hat {p}}_{x}](F)=({\hat {x}}{\hat {p}}_{x}-{\hat {p}}_{x}{\hat {x}})(F)=x{\frac {\hbar }{i}}{\frac {\partial }{\partial x}}(F)-{\frac {\hbar }{i}}{\frac {\partial }{\partial x}}(xF)=x{\frac {\hbar }{i}}{\frac {\partial }{\partial x}}(F)-\left(x{\frac {\hbar }{i}}{\frac {\partial }{\partial x}}(F)+F{\frac {\hbar }{i}}{\frac {\partial }{\partial x}}(x)\right)=-{\frac {\hbar }{i}}F=i\hbar F

Strictly speaking, this is a confirmation rather than a derivation. It shows that

{\hat {p}}_{x}={\frac {\hbar }{i}}{\frac {\partial }{\partial x}}

satisfies the canonical commutation relation, but does not show that it is the only solution. Gandalf61 (talk) 08:02, 2 May 2013 (UTC)[reply]

It is a consequence of the De Broglie relations as can be seen in the article titled - of all things - Momentum operator. Dauto (talk) 19:20, 2 May 2013 (UTC)[reply]

this page has a derivation of the reverse relation where x = i hbar d/dp in the momentum representation (start at equation 198). The proof you're looking for is completely analogous. Dauto (talk) 19:36, 2 May 2013 (UTC)[reply]

There is a subtle issue involving the theory of distributions involved here. I tried to look this up, but I can't find it in my old university lecture notes. Basically, the argument should go as follows. We want express <x|p|psi> in terms of <x|psi> by invoking only the commutation relation and nothing else. Let's write:

<x|p|psi> = Integral dx' <x|p|x'><x'|psi>

Then we can evaluate the matrix element <x|p|x'> as follows:

x <x|p|x'> = <x|x p|x'> = <x|i hbar + p x|x'> = i hbar delta(x-x') + x' <x|p|x'>

Therefore, we have:

(x-x') <x|p|x'> = i hbar delta(x-x')

The problem is then that can't just divide both sides by x-x', as that doesn't yield a bona fide distribution. If have the equation x T = delta, where we use the official math notation T(f) for a a distribution T acting on a test function f, the solution is not T = 1/x delta, as that doesn't define a distribution, rather it is T = -delta' + A delta. We have

x T(f) = T (x f) = -delta'(x f) + A delta (x f) = delta[(xf)'] = f(0) = delta(f)

We thus have:

<x|p|x'> = -i hbar delta'(x-x') + A delta(x-x')

This then gives:

<x|p|psi> = Integral dx' <x|p|x'>psi(x') = -i hbar d psi(x)/dx + A psi(x)

The constant A must be put to zero by hand. You can always add a contant times the identity to an operator without that affecting the commutation relations. Count Iblis (talk) 15:45, 3 May 2013 (UTC)[reply]

Inert gas

I was reading Nitrogen asphyxiation and it says suicide using inert gas doesn't cause pain. Is that really true? Btw I'm NOT thinking about suicide just wondering cuz it's pretty crazy that you can die without feeling pain. Money is tight (talk) 08:04, 2 May 2013 (UTC)[reply]

Yes, it is logical that it is painless. Otherwise, how else could there be accidental death by this manner? Plasmic Physics (talk) 08:09, 2 May 2013 (UTC)[reply]

Since four fifths of the air we breathe all the time is nitrogen, our bodies are quite tolerant of it. What you're really talking about here is removing the oxygen from the air so that you're breathing 5/5ths nitrogen. If you tried to breath only carbon-dioxide (for example) you'd start hyperventilating and all sorts of nasty things would result because your body is aware that CO2 needs to be dealt with - but we breathe mostly nitrogen all the time - so our bodies don't notice anything except the lack of oxygen.

So we might as well forget about the nitrogen here. What we're talking about is lack of oxygen - Hypoxic hypoxia - and anoxia...some of the symptoms of this are:

Cyanosis -- you "turn blue" as your blood loses oxygen.
Headache -- not completely painless.
Visual impairment, decreased reaction time and impaired judgment time -- hardly painful.
Numbness, drowsiness and euphoria -- At least that's going to make the headache seem less bad.
Lightheaded or dizzy sensation, tingling in fingers and toes -- not exactly painful but unpleasant.
Nausea -- also not great.

But it depends on how fast it happens. The above symptoms are likely if the reduction in oxygen is relatively slow...but if it's fast then seizures (painful - and might cause you physical injury through falling or something) and if you're a man, priapism (which can be very painful) are also possible.

The amount of time it takes (and therefore how long you suffer these symptoms) is also tricky to get a grip on. If you still have oxygen rich air in your lungs, you can hold your breath for maybe 30 seconds - but if you fill them with pure nitrogen by hyperventilating, you could lose "useful" consciousness in under 10 seconds.

I think the devil is in the details here.

SteveBaker (talk) 13:07, 2 May 2013 (UTC)[reply]

In almost any situation where someone takes a sedative toxic substance like drugs or alcohol in sufficient amounts to kill them their death will be preceded by drowsiness leading to sleep and then coma, during which it is extremely unlikely that they will be conscious of pain. Indeed, enormous numbers of people die every day under the influence of pain relieving medication in hospitals and hospices all over the world. Richard Avery (talk) 13:19, 2 May 2013 (UTC)[reply]

Any exertion speeds up the effect. If firemen come running into a computer room where some inert gas has been released to suppress flames they may immediately fall down, if this happens just hold your breath, go in and drag them out straight away into the fresh air. Dmcq (talk) 13:45, 2 May 2013 (UTC)[reply]

I presume that there are two reasons for that though: Firstly (as I said above), when you're breathing hard (as you would be if you were running with all of that fireman's gear on) then whatever oxygen would otherwise remain in your lungs gets flushed out and replaced with the inert gas more rapidly than if you were breathing at a steady pace. Secondly, the exertion increases your oxygen needs so whatever oxygen is left in there gets used up more quickly. However, if holding your breath and then exerting yourself to get a victim out of there actually works - then the former is clearly of much more importance than the latter. SteveBaker (talk) 14:21, 2 May 2013 (UTC)[reply]

Certainly does as four firemen neatly laid out on the grass can testify ;-) I must admit the gas seemed a bit excessive as it caused damage to the ceiling with the extra pressure when it was released. Dmcq (talk) 14:49, 2 May 2013 (UTC)[reply]

Sorry guys, your answers here are a little messed up. Some bullshit has been posted above. Several things to note:-

Presumably SteveBaker meant increased reaction time.
Fireman are NOT in the least bit likely to immediately fall down, or even fall down after a while, if they enter a computer room where gas suppression was used, for several reasons, the most important of which are:-
The most common system used is Inergen and similar - in these systems the amount of inert gas is selected such that the oxygen content is brought down (by displacement) to about half normal, i.e, roughly 10% by vol. Almost all computer room fires go out at this level, and the ones that don't (such as burning paper) slow down dramatically. The gas mix also contains carbon dioxide, so that the room carbon dioxide is raised from the natural level of 0.03% to about 0.1%. This raises the blood CO2 level similarly, which stimulates humans to breath harder, so that the blood oxygen remains close to saturation. Less commonly, CO2 suppression has been used, as it has been in aircraft flightdecks. If you enter a room where it has been used, you will know it from the sensation (assuming you didn't notice the mandatory flashing lights and bell), and you will get out.
Take a typical fit male human: say 80 kg weight. By standard medical calculations, such a male will have 7% of body mass as blood, i.e., 5.6 kg of blood. Each kg of blood has very nearly 210 mL of oxygen bound to haemogobin per kg of blood, i.e., our 80 kg fireman has 1.2 litres of oxygen stored in his haemoglobin, not counting oxygen dissolved in plasma. Male lung total capacity is aprox 0.53 litres per 10 kg of body mass, so our 80 kg fireman will have about 0.2 x 0.53 x 8 = 0.85 litres of oxygen in his lungs on entry to the room. Total oxygen he has is then 0.85 + 1.2 = 2.05 litres. No problem will occur until blood oxygen drops below 90% saturation. A fit male exercising (assume fireman has run up the stairs) will consume oxygen (the VO2 rate) at approx 50 ml/kg/min. So he will consume 4 litres per min and it will take him about 30 seconds before any distress or problems occur at all. It will take a full minute in a zero oxygen atmosphere before he has much distress - plenty of time to either get out or put on oxygen mask, even if there is NO oxygen in the room. However, as there will be about half the normal level, he would have TWO full minutes to take action. Even if he has all oxygen flushed out of his lungs upon entry, he can still fully function on blood oxygen for one minute.
Firemen are trained - they won't be that silly.
Those of us who have flown in an airliner will recall the safety instructions - where they tell you that should a sudden decompression occur (which roughly halves the oxygen partial pressure), put on your oxygen mask, but it isn't urgent - you have plenty of time to ensure your childen have their masks on and working before you put yours on.

I have witnessed a test dump of Inergen gas in a large computer room - I was in the room at the time. I noticed no ill effects.

I once took a tour of a hospital on an open day. We were taken into an out of service operating theater. An aneasthetist explained all his equipment, one item of which was a blood oxygen saturation monitor, which works by shining light of two wavelengths thru your finger. While he was talking, I put it on - it said my O2 sat was 100%, as it should be for a conscious healthy human not doing anything but stand. I then pinched my nose closed and kept my mouth shut, to see what my blood O2 level would be when I could not hold my breath any longer. The aneasthetist seeing me do this said "There's one in every group!". I kept holding my breath while listening to his talk and glancing often at my watch. After 6 minutes, I could not hold my breath any longer, it was most uncomfortable. My blood O2 saturation was still showing 99%! And, no, the machine was not faulty. When I took a breath, the aneasthetist said "About average, mate."

Ratbone 60.230.212.134 (talk) 15:36, 2 May 2013 (UTC)[reply]

I assure you that what I said was true. There was a fifth fireman who called for help from a bunch of students to get his colleagues out and they revived quite quickly when brought out but whatever you say they did collapse going into the room. And yes they did look a bit sheepish about having ignored the warning light outside the room. We gave them some tea and a biscuit and they went away again. The point about some gases is you don't notice the effect until you conk out and if you have not been running you probably are okay for much longer. Dmcq (talk) 16:04, 2 May 2013 (UTC)[reply]

Well, my calculations are verifiable from relevant wikipedia articles and elsewhere. What you describe is simply not possible with Inergen systems and it's competitors. It could happen with a CO2 system, but only if a) the gas dump was excessive, meaning it had not been tested per code requirements, and b) they were really stupid firemen. While you can conck out without distress with inert gasses, entering a room with CO2 produces immediate distress, even with plenty of oxygen. Remember that: normal oxygen content of air is 20%; you need much less than that to function; air CO2 is less than 0.03% and any increase triggers heavier breathing. CO2 has been a standard system to put out aircraft cockpit fires. Not much point if it puts the pilots out as well. What country was this in?

The only other possiblity I can think of - if lead acid batteries had been used in a UPS system, people can be overcome by stybine gas - stybine is a product of antimony in the pressence of water and other substances. However, a) the fire woukd have to have reached the batteries, b) the batteries would have to be the type using lead antimony plates, not normal in UPS service, and c) you cannot recover spontaneously from stybine inhalation. Ratbone 60.230.212.134 (talk) 16:12, 2 May 2013 (UTC)[reply]

Believe what you wish but I have not said a single untruth in all the time I have been on Wikipedia. Dmcq (talk) 16:25, 2 May 2013 (UTC)[reply]

I'm going to call bullshit on Ratbone item #6 recollection of the Emergency oxygen system instructions. FAA advisory AC 121-24C (7/23/03) appendix 1, item #10 states "passengers should be advised to don their own oxygen masks before assisting children with their masks." and I can't recall ever hearing anything else. Time of useful consciousness suggests you really don't have much time to figure out what to do and then do it. DMacks (talk) 17:14, 2 May 2013 (UTC)[reply]

We have an article on Gaseous fire suppression which mentions several different kinds with different methods of action. Rmhermen (talk) 17:16, 2 May 2013 (UTC)[reply]

Yes, I got #6 round the wrong way. I note that the article cited by Rmermen gives times, designed so that an elderly passenger of unknown fitness or baby with mimimal oxygen capacity will be catered for, gives cruise altidue (and therfore roughly the same oxygen partial pressure) times of the same order I calculated for a fireman, who can be expected to be in excellent fitness and do much better. And, as I said, firemen should not need to spend much time figuring out what to do, because they are trained - which is why I find it hard to believe that 4 firemen would enter a CO2 supprosssed room ignoring the signs, lights, and sound, and if they did, they should recognise the meaning of their immdeiate distress (from their training) and leave. Ratbone 121.215.74.116 (talk) 01:06, 3 May 2013 (UTC)[reply]

I remember seeing a doco film about capital punishment where they illustrated the painlessness of nitrogen asphyxiation by placing a pig in a room with a feed trough in a fume cupboard that recirculated 100% nitrogen. The pig would stick its head in to eat, pass out from lack of oxygen and fall outside where it would resume normal breathing of air, then wake up and stick its head back in. As for Dmcq's suggestion that you hold your breath and try to save people that have been overwhelmed by some gas; both my confined space and certified atmosphere tester training are telling "don't do that". Most industrial fatalities involving gasses are from trying to save someone else. Only emergency response personnel should take part in rescues of this type. 202.155.85.18 (talk) 02:48, 3 May 2013 (UTC)t[reply]

Well the last one did come out for help rather than going in himself and I'd guess it was pretty safe as there were a number of people around, it wasn't as though they had to go far. Whatever about that one should wait for trained people I would have difficulty with sitting around letting people die, and I can swim 50 meters underwater quite easily so I really don't see there would be a problem. It was rather a few years ago so I'd guess their training is better now. Dmcq (talk) 08:02, 3 May 2013 (UTC)[reply]

The first thing they teach in first aid courses is "DRABC" - Danger, Response, Airway, Breathing, Circulation. Or the equivalent in other languages. "Danger" in this case means you first check that there is no danger to yourself, BEFORE attempting rescue. I agree with 202.155.85.18 - If I was aware that more than one person collapsed after entering a room, there's no way I would enter. You don't know just why they collapsed, and if you enter and collpse, that just might mean n+1 fatalities instead of just n falalities. You last bit about swimming is self contradictory - if you can function for x minutes without breathing, then so should firemen. The training of firemen won't have changed in this regard since the advent of computer rooms and the like back in the 1950's. What country did this supposed event occur in? Ratbone 121.215.32.211 (talk) 11:34, 3 May 2013 (UTC)[reply]

I have to agree with Dmcq on all points here as most of you appear to be confusing theory with real life situations. Now lets get back to the firemen. Yes they are trained and are not stupid. Yet training costs money. It is more than likely in Dmcq case that these firemen had never before rushed into a building full of Inergen. These call-outs are rare compared to putting out cars, set afire by vandals etc. All the adrenaline is pumping with a focus to get into the computer room as fast as possible and save human life. Yes, the theory that has no doubt been taught to them in the class room leads them to understand you can breath in this atmosphere. Yet. isn't this a little bit like going on holiday and on arrival playing ball with your kids only to find you have collapsed in a gaping heap - because of the altitude. The holiday brochure pointed out the altitude but still you played foot-ball! Yes, the higher CO2 invokes faster lung action but if the available oxygen is only half that as sea level when one's action causes a high biological oxygen demand then whoops. This can happen without warning because of oxygen debt. Put simply, once you have stop running do you immediately stop panting? Of course not. One has to keep a high rate of respiration to stop from passing out. So for a fireman in such a situation. Just suddenly realizing that respiration is distressed, one has left it too late -but hey, in the class room weren't we were told we can still breath in this atmosphere? I would be interested to know from a current fire-fighter if his/her training included the cost of going into an Intergen atmosphere. In the days of Halon, such exercises where prohibitively expensive because fire-services are paid for by local taxation and nobody like paying tax – so they didn't do it. It may be the same to day. Are there any fire-fighters here that have had gas suppression training for real?Aspro (talk) 18:32, 3 May 2013 (UTC)[reply]

I doubt that many firmen would have actually had experience of Inergen during training, because it is expensive. The test dump that I witnessed cost ~$80,000. However, firemen do get to experience oxygen starved and smoke filled chambers during their training - at least they do here in Australia. But they'll get to hear about Inergen in classroom talk.

At least here in Australia, when called out to commercial buildings large enough to have a computer room with fire suppression, their normal practice is to send at least one guy in with breathing apparatus anyway, as burning plastics and electronics in commercial buildings can emit a range hazardous fumes, and rooms can be filled with smoke.

In fire suppression systems, the oxygen content is only brought down to no less than half the normal value. This is well and truely enough to function normally on, providing breathing is stimulated, as the CO2 content of Inergen does. Yes, if you are exercising heavily it won't be enough - after all if we run at maximum speed, we get puffed out in a normal atmosphere. But if the firemen actually collapsed, their body oxygen demand would have drops to the rest value, so they should spontaneously recover - within less than a minute. That's not what Dcmq said. It is possible that if the computer room was in a third world country that Western standards of workmanship, commissioning tests, and firemen training were not up to the situation, which is why I asked twice "what country was it in?" However, as a) Dcmg's intial post reads a hypothethical, but later posts after a challenge changed that, b) it is quite unlikely for technical reasons as explained, and c) Dcmq has not come back and answered the question, I rather think his story is just a story. Maybe he heard about an incident he wasn't personally involved in, and the facts got muddled up in the gossip - chinese wisper effect.

Ratbone 121.215.32.202 (talk) 01:15, 4 May 2013 (UTC)[reply]

As I said above 'they revived quite quickly when brought out'. I am glad standards for construction and firemens' training are so good nowadays that none of the incidents in [14] can happen. Yes I do refuse to say where or when. Dmcq (talk) 09:09, 4 May 2013 (UTC)[reply]

Several things to note about the EPA report you linked to:-

You said the incident occured in a computer room. None of the incidents listed occurred in a computer room.
You said firemen were affected - but the reported incidents involved all sorts of other workers.
The most common gas system in computer rooms is Inergen, which will not aspixiate as expalined. All the reported incidents involved CO2 suppression.
Many of the incidents reported occurrerd in US Navy vessels. I don't know much about the USN, but I do know about the Australian Navy, which largely tries to emulate the USN and mostly buys the same ships and equipment. The Aust Navy is notorious for ignoring civilian safety standards, not training crews properly, and consequently have an accident rate to match.
Many of the incidents occurred in facilities totally unlike computer rooms and may well require a greater degree of oxygen dispalcement than do computer rooms.

Ratbone 124.178.43.47 (talk) 09:26, 4 May 2013 (UTC)[reply]

Inergen was only patented in 1989 and introduced in 1992. I don't know what gas was used. I think I have had enough of this so bye. Dmcq (talk) 09:50, 4 May 2013 (UTC)[reply]

ITC

I read Isothermal titration calorimetry and couldn't find answer to the following question: In context of protein complexs, in ITC expriment there is a "receptor" in the solution (in the cell), and a "lignad" that is titrated into the cell, is it possible to swap them? would we get the same affinity? (for example suppose we find affinty trypsin-BPTI: will we get the same results when (1) have trypsin in the cell and titrating BPTI, or (2) having BPTI in the cell and titrating trypsin). Thanks, 192.114.91.228 (talk) 10:01, 2 May 2013 (UTC)[reply]

In concept, I don't see why not, but in practice, I think you'd have an issue that you want to add a small aliquot to a larger amount of solution. Because it's hard to purify macromolecules to a very high concentration, you'd be better off using a concentrated chemical to add. Wnt (talk) 17:02, 3 May 2013 (UTC)[reply]

Origin of the belief that parrots eat crackers

Anyone know for sure? I heard that it was something to do with parrots on ships eating hardtack and saltine crackers. Yes, they will certainly eat crackers if available and apparently enjoy them, but it's not their main diet. --31.185.233.239 (talk) 20:52, 2 May 2013 (UTC)[reply]

Obviously. Cracker trees aren't native to the same range as parrots. Well, they do overlap a bit with the range of the Norwegian Blue parrot, but alas, I think that species is no more, ceased to be, expired and gone to meet his maker... --Jayron32 21:09, 2 May 2013 (UTC)[reply]

No, he's asleep!!! Richard Avery (talk) 07:18, 3 May 2013 (UTC)[reply]

And now for something completely different ... The phrase "Polly want a cracker" goes back at least as far as 1848.[15] As Cracker (food) notes, the cracker is said to have been invented in 1792. Clarityfiend (talk) 22:20, 2 May 2013 (UTC)[reply]

Or maybe parrots are all just mildly racist. --Jayron32 23:39, 2 May 2013 (UTC)[reply]

When people first started keeping parrots as pets, you couldn't just head to the nearest pet store and pick up a bag of Purina Parrot Chow. So, crackers were a food item which people had on hand, which parrots could also eat. If we think of the stereotypical parrot kept on a ship, then fresh food would have been only available during, and shortly after, stops at ports. In between, parrots would have to make do with things like crackers, as would the rest of the crew. If you had a parrot with an exceptional vocabulary, he might say "Polly wants some fresh food, but since that's not available, Polly will settle for a cracker". StuRat (talk) 05:44, 3 May 2013 (UTC)[reply]

Isn't this just a meme based on classic children's cartoons, the same as the notion that mice like cheese, which they really don't? μηδείς (talk) 08:18, 3 May 2013 (UTC)[reply]

Did they have classic children's cartoons back in 1848? Or only classic children? -- Jack of Oz ^[Talk] 09:12, 3 May 2013 (UTC)[reply]

Jack, if you first heard this from 19h century literature and not a cartoon or similar juvenile entertainment, I'll eat a roo raw. μηδείς (talk) 19:31, 3 May 2013 (UTC)[reply]

A cartoon or one's mother or whoever may well be where you or I first heard the expression. But the question is about the "origin" (see the header) of the belief behind the expression, hence the origin of the expression itself. It's an etymological matter as much as a scientific one. We've traced it to at least as far back as 1848 at this stage. -- Jack of Oz ^[Talk] 19:40, 3 May 2013 (UTC)[reply]

If you had read the post above then you will see this pre-dates children's cartoons: The Knickerbocker: Or, New-York Monthly Magazine, Volume 34 1849... page 544 [16]. Would you like salt and pepper on you're raw roo or would you rather eat it purely al fresco?--Aspro (talk) 20:01, 3 May 2013 (UTC)[reply]

Yes, and there's still not a single person reading this thread who didn't learn that little bit of obscure high culture from Looney Toons or the like. Next you'll say the first place anyone ever hears Wagner's in the opera house. μηδείς (talk) 02:02, 4 May 2013 (UTC)[reply]

Are you deliberately missing the point? The q is not about when, where and how you or I or Joe Bloggs first heard the expression "Polly want a cracker". It is about how that expression came into being in the first place, and more to the point, why anyone thought a parrot's favourite food would be man-made crackers. The belief and the expression weren't independently created millions of times over. They were created way back when, and ever since then people have been copying other people. -- Jack of Oz ^[Talk] 04:10, 4 May 2013 (UTC)[reply]

Maybe some of those kangaroo meringues that Noël Coward was talking about. -- Jack of Oz ^[Talk] 20:09, 3 May 2013 (UTC)[reply]

Stop it Oz. Your making me feel hungry and my doc has warned me that I'm already 'morbidly' obese. Mind you... a kangaroo meringues does sound like it's light and fluffy and Oh, what the hell, a little mouth-full or two (or three) will not not make any difference. Medeis... add a little garlic to my portion please.Aspro (talk) 20:26, 3 May 2013 (UTC)[reply]

As I recall the classic parrot, Captain Flint, it said "pieces of eight" and ate bark. I never heard "polly want a cracker" until much later, probably as spoken by Disney's parrot in a bit of role-reversal that was lost on me at the time. I suppose a person's predisposed notions about parrot behavior are highly conditioned by the order in which one is exposed to classic parrots in literature and film. Nimur (talk) 15:45, 3 May 2013 (UTC)[reply]

Here's something from 1948,[17] and you can safely assume that the "Polly want a cracker" thing is much old than that. ←Baseball Bugs ^{What's up, Doc?} carrots→ 04:04, 4 May 2013 (UTC)[reply]

We're all overlooking something... when did parrots get to be named Polly? how do we even assume they're female? or that a male would want to be called Polly? Gzuckier (talk) 21:03, 3 May 2013 (UTC)[reply]

Read the posts above!:“Polly is a diminutive of Poll "as a female name, and name for a parrot," and Poll, altered from Moll, familiar form of Mary, is the traditional name for any parrot. The earliest quotation the OED gives for Polly as a name or designation for a parrot is from Ben Jonson's "Epigrams," 1616. ” If it is a he, then you can you can call him Joe. Or, if he is your-pal you can call him Al. He, she or it, wont care as long as Polly gets a cracker. They might have bird brains but they know how to train the humans around them to give them what they want. Aspro (talk) 21:41, 3 May 2013 (UTC)[reply]

Actually, Molly and Polly come from Mary. Moll or Poll would be single-syllable nicknames for Molly and Polly, as Mare is for Mary. How that figures into parrot names is anybody's guess. Maybe it's from "Paul"? As with the male bird in "Little Poll Parrot". ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:46, 4 May 2013 (UTC)[reply]

Paulie wants a cracker? --Jayron32 03:50, 4 May 2013 (UTC)[reply]

Various google items indicate that Poll is indeed a variant of Paul. So why Paul Parrot? Maybe just because it's nice and alliterative. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:59, 4 May 2013 (UTC)[reply]

I know a lovely lady named Polly. Her legal name is Paulette. -- Jack of Oz ^[Talk] 04:12, 4 May 2013 (UTC)[reply]

You can ask the World Parrot Trust.

World Parrot Trust - Ask an Expert

—Wavelength (talk) 04:28, 4 May 2013 (UTC)[reply]

Electromagnetic induction thru an unclosed ring

in a changing magnetic field. It's easy to see how an EMF comes about in the case of a closed conducting ring fixed in place in a magnetic field, since the the change in flux is related to the change in the magnetic field, while the area enclosed by the ring is constant. But how can we explain what happens in an incomplete ring (suppose a piece of the ring is cut away, leaving it unclosed) on the basis of change in flux ? In other words, how can we define an area here, in the first place (an open ring develops an EMF, but no induced current of course) ? BentzyCo (talk) 21:02, 2 May 2013 (UTC)[reply]

Area enclosed has nothing to do with it. A straight wire moving across a magnetic field will have a voltage induced along it. The voltage is proportional to the length of the wire, the strength of the magnetic field, and the rate at which the field moves in respect to the wire. In the case of a uniform magnetic field moving through a conducting ring that is gapped at one point, there will be no voltage across the gap, as the voltage induced in each half-turn will be the same. In the case of a straight wire subject to a moving/changing magnetic field, which will have a voltage induced along it, current will flow if, and only if, the ends of the wire are connected to a circuit outside the magnetic field (or a part of the field that is of lower intensity). Ratbone 121.215.74.116 (talk) 00:51, 3 May 2013 (UTC)[reply]

Your 1st sentence is wrong. Completely wrong. Do you how AC voltage is produced ? And the rest of your reply is redundant, since you weren't tuned to what was asked and its context. It's an incomplete ring. Stationary. Nothing's moving. Only the intensity of the magnetic field is changing. BentzyCo (talk) 10:38, 3 May 2013 (UTC)[reply]

Well, if you know so much about it, why ask the question?

AC is commonly produced in an alternator, which, at its most basic, is a loop conductor rotating in a constant and evenly distributed magnetic field, the loop being interrupted at the slip rings and connected to a circuit external to the magnetic field. As rotation means, during each half rotation, half the loop is going one way thru the field and the other half is going in the reverse direction thru the field, the induced voltage in each half turn is such that they add around the loop, instead of opposing in the case of a loop moving bodily thru a field. At each half turn, the direction each half turn is moving wrt the field in the opposite direction, thus the voltage at the slip rings reverses. Practical alternators are of course more complex both in conductors and in magnetic arrangements, but all of us who studied electrical engineering have studied simple loop-in-field alternators in 1st year, and, usually, done tests on lab models.

Now back to your loop conductor in your question, any magnetic field not penetrating the conductor cannot be inducing any electric tension in it - so area can have nothing to with it. Or, looking at it another way, you could have a great number of parallelled conductors entirely within a magnetic field, so that a considerable amount of the field cuts a conductor. You still get the same voltage end to end and no current. A loop is just two parallel conductors. Ratbone 121.215.32.211 (talk) 11:02, 3 May 2013 (UTC)[reply]

a. Your reply should be relevant to the question asked, irrespective of backgroung of the person asking the question, either he knows something about physics or even more than that. It isn't. Your 1st row up there is patronizing, just because I'm on the side of the question. Isn't it legimate to ask and consult with colleagues ?

b. You repeat your previous mistake regarding the consistuent of the flux, the effective area traversed by the magnetic field lines. Again, the part of describing how AC current comes about is redundant too, and part of it even repeats what I said. In other words, AC current is a phenomenon originating from a periodically changing effective area. I hope you know what magnetic flux is, and what Faraday's law says.

c. I think it's evident from my both writings that I'm quite in the field. My question is thus of an irregular kind, very intriguing and interesting, and I wanted to share it with others. It was very cut and clear - what's the origin of EMF across a bent wire making an arc of, say, 270⁰ ?

d. A complete loop is exactly one round conductor. BentzyCo (talk) 12:37, 3 May 2013 (UTC)[reply]

The emf is equal to minus the integral of E dot ds from one end to the other, but this is path dependent. If you e.g. measure the voltage betwen the two ends using a volt meter, then what the voltmeter will indicate is given by minus the integral the closed path that includes the connecting wires to the voltmenter and the voltmeter itself. Then the Maxwell equation nabla times E = -1/c dB/dt, makes that equal to Faradays law where the area is the area enclosed by the integration path. Note that any fields generated by moving charges don't contribute, because their electric fields are conservative (integral of E dot ds along a closed contour vanishes), therefore you can compute the contour integral for the hypothetical cases where all conductors are relaced by insulators. Count Iblis (talk) 12:34, 3 May 2013 (UTC)[reply]

For an loop conductor, broken by an infintessimal gap in a uniform magnetic field, the EMF across the loop gap is always zero. It matters not whether the loop is moving through the field (or a moving field is moving across a stationary loop, or the magnetic field is increasing or decreasing in intensity. So long as the field is uniform, that is everywhere the same strength, the EMF across the small gap gap is zero. If the incomplete loop is in fact a straight wire at right angles to the field, and the field is changing, then there will be an EMF from end to end. A partial loop acts between these two extremes - you can consider it as a number of straight wire segments, and add up the EMF's to get the total, which must be between the two limiting cases of EMF. Note that any voltmeter and its connecting wires used to measure the emf must lie outside the field, or EMF's induced in the meter and wires will oppose the EMF in the loop conductor, resulting in reduced or zero reading. It is indeed odd to attack a person answering. Is BentzyCo a troll? Certainly he writes in an odd way - his 1st sentence in his question is a nonsense, for a start, and he seems to want an argument. Wickwack 121.215.147.92 (talk) 13:05, 3 May 2013 (UTC)[reply]

a. 1st: put your reply in its right place.

b. 2nd: Your last sentences are just offensive & insultive, and will be treated as such. "attacking an answering person" ?, "troll" ? "my 1st sentence is nonsence" ? "want an argument" ? Your claims are unsubstantiated, to say the least. Will you, please, remain on disciplinary ground ?

c. Concluding my claim in the 1st place: the previous replies weren't relevant to what was asked, deviating the discussion from its intended focus. I think you've to apologize, Thank you. BentzyCo (talk) 13:34, 3 May 2013 (UTC)[reply]

@BendtyCo: Please bear in mind that this is a volunteer service - and that even as a questioner here, you are bound by the Wikipedia guideline to Assume Good Faith. If you don't like the answers, don't use them. Ratbone is trying hard to be helpful and explain his thoughts on this matter - there is no need to insult him for doing so.

@Wickwack: Same deal...not cool: WP:AGF OK?

SteveBaker (talk) 14:17, 3 May 2013 (UTC)[reply]

It seems like the wire is a distraction. We're only interested in how the electrons move when exposed to a changing magnetic field - given that they can't usually leave the wire. Unless it's a superconductor the field lines enter it, so the electrons inside are exposed to a changing magnetic field and by Faraday's_law_of_induction#Maxwell.E2.80.93Faraday_equation experience electromotive force. I'm afraid I'm quite rusty with this topic but working through it you should be able to see how the EMF adds up even without a closed loop. I suppose it doesn't with a superconductor because the electrons at the outer edge can just madly move to compensate for any induced potential without any force ever ... needing to be applied??? Wnt (talk) 15:26, 3 May 2013 (UTC)[reply]

Let's take the magnetic field to be non-zero only withing the ring. We take it zero also at the location of all the conductors, e.g. put a long thin coil at the center of the loop, and let the current in that coil increase or decrease, the magnetic field is fully contained in the interior of the coil, which is well away from the boundary of the loop. The potential difference between the (infinitesimal) gap is simply the induced voltage as follows from Faraday's induction law. To see this, write the voltage difference as minus the integral of E dot ds from one end of the gap P to the other Q where E is the total electric field (both induced by the changing magnetic field and the build up of charges at the ends due to them responding to the induced electric field).

Then because the charges in the conductor will make the total electric field zero inside the conductor, we can add to the integral from P to Q across the gap, the integral from Q back to P taken over the conducting loop, as the latter integral is zero. We then have an integral along a closed path to which the Coulomb fields of the charges do not make a net contribution. So we can compute this integral by replacing the total electric field by the induced electric field that follows from nabla times E = -1/c dB/dt. Then Stokes theorem says that the integral along a closed path of E dot ds equals the integral of nabla times E over the area enclosed by the path, substituting -1/c dB/dt for nabla times E then yields the Faraday's law result. Count Iblis (talk) 14:13, 4 May 2013 (UTC)[reply]

May 3

Dietary Reference Intake/Percent Daily Value

What's the relationship between Dietary Reference Intake and Percent Daily Values? Did the US government simply rename PDVs, perhaps when they made the food pyramid three-dimensional? PCV currently redirects to DRI, which doesn't mention PDV at all except in a single citation to a webpage that's now producing a 404 error, and a Google search didn't help. Nyttend (talk) 01:04, 3 May 2013 (UTC)[reply]

I think RDI is just the recommended amount of any particular "stuff" you should eat in a day, and so PDV is simply how much of that RDI any labelled food product contains. Just as a point of clarification, it's not technically the government which decides these things, these reccomendations are made by the Institute of Medicine which has a Congressional charter, but it's not in any way actually part of the government. Vespine (talk) 04:00, 3 May 2013 (UTC)[reply]

interesting that sometimes the RDI is the max amount of something you should eat, and sometimes it's the minimum amount. Tripped up a few students in health class. Gzuckier (talk) 21:10, 3 May 2013 (UTC)[reply]

recognizing a beetle

was photographed in East Talpiot, Jerusalem, Israel. anyone recognize? --SuperJew (talk) 05:11, 3 May 2013 (UTC)[reply]

Can't help with identity, although it looks like a weevil, but this site, What's that bug? is a very helpful resource. Richard Avery (talk) 07:15, 3 May 2013 (UTC)[reply]

It's not a weevil, they have snouts. μηδείς (talk) 08:16, 3 May 2013 (UTC)[reply]

This is a species of darkling beetle, possibly Adesmia abbreviata. --Dr Dima (talk) 08:27, 3 May 2013 (UTC)[reply]

here [18] [19] are two pages about it. --Dr Dima (talk) 08:36, 3 May 2013 (UTC)[reply]

I contacted this site: http://www.nature-of-oz.com/ I quote Oz Rittner: "This is Adesmia (genus), not possible to identify the species from this photo. It belongs to the Tenebrionidae family." 196.214.78.114 (talk) 08:52, 3 May 2013 (UTC)[reply]

The Lord sure do love Him some His beetles, don't He? Gzuckier (talk) 21:13, 3 May 2013 (UTC)[reply]

"inordinate fondness for stars and beetles", indeed [20]. --Dr Dima (talk) 22:20, 3 May 2013 (UTC)[reply]

Your use of some and his as specifiers at the same time is ungrammatical in the dialect you are attempting. It's like saying "This my book is interesting." μηδείς (talk) 01:57, 4 May 2013 (UTC)[reply]

Accelerator required for high intensity x-ray?

The voltage differential between cathode and anode in an x-ray tube translates into the "keV" rating of the resulting x-rays. Is it a requirement to use linear accelerators to get photons with higher energy. Or can one build a 50 MeV x-ray tube ..? An example being the Australian Synchrotron which generates 90 keV electrons using an electron gun and then accelerate these to 100 MeV. Electron9 (talk) 20:38, 3 May 2013 (UTC)[reply]

This might be an over simplification but once an x-ray or any other electromagnetic radiation setts off (emitted) – that's it. You can't give it extra electron volts in the real world. If you need harder x-rays then you need a greater differential. In the old days, one could tape a metal paper clip to a bromide paper and place it near to the TV tube – and hey presto – on developing you had an x-ray photo. A Synchrotron accelerates just the 'electrons' (read: increase of potential) so the electromagnetic radiation they 'stimulate ' (is that right in this context?) peak at higher energies. --Aspro (talk) 21:10, 3 May 2013 (UTC)[reply]

The reason linear accelerators (which accelerate electrons magnetically) are used for high energy X-rays in the medical filed and for industrial X-rays is because they are much more compact and cheaper that a conventional X-ray tube (which accelerates electrostatically) would be to get the same energy. Machinery to raise 20 MV at a useful current (several mA or more) needs large insulators and would need to be the size of a house, where as linear accelerator coil assembly is only the size of a couple of shoe boxes. That makes it a lot cheaper. A linear accelerator can be switched on and off virtually instantly, where as a many-megavolt power supply would take a second or maybe a few seconds to build up and die down - not very desirable.

The business of using a TV set to take X-Ray pictures is very nearly an urban myth. Except for the very early colour sets sold in the USA (1950's), the voltage used (about 17 to 20 kV depending on size) is not sufficient to get X-rays of sufficient energy to penetrate the picture tube front glass. The earliest coluor sets used up to 25 kV or so and a shunt triode regulator to regulate the voltage. These regulator tubes did emit very weak X-rays, but set manufactuers limited the X-Ray emission form the set by enclosing the regulator in a metal cage, and by arranging it so the the remaining X-Rays were directed downward thru the bottom case of the set, so that two thick layers of plywood, a metal sheet, and the floor of the dwelling would have to be penetrated.

Ratbone 58.167.231.148 (talk) 00:16, 4 May 2013 (UTC)[reply]

There's something wrong with the answer above.

Magnetic fields don't accelerate electrons, so it makes no sense to say something is accelerated magnetically
Magnetic fields bend the path of an electron bean, so it makes no sense to call an accelerator that uses magnetic fields a "linear accelerator"

Dauto (talk) 12:32, 4 May 2013 (UTC).[reply]

The magnetic keeps the beam on track and the electric field accelerate the particles/photons ..? Electron9 (talk) 13:36, 4 May 2013 (UTC)[reply]

What are the natural controls on Eh and pH?

What are the natural controls on Eh and pH (in areas of water) besides the amount of electronegative elements (eh) and the amount of acids (ph)?--149.152.23.34 (talk) 21:43, 3 May 2013 (UTC)[reply]

Think that's a very good question. Soil, is a mixture of many mineral and organic components. The answer of this question is not going to be straight forward. --Aspro (talk) 21:59, 3 May 2013 (UTC)[reply]

In general, see Soil pH and buffer (chemistry). More specifically, presence of limestone or sodium carbonate in the soil tends to prevent pH from going too low, or drives it up; whereas presence of humic acid / fulvic acid tends to prevent pH from going too high, or drives it down. Forest or swamp soils that are relatively poor in mineral content can be fairly acidic, while volcanic or some desert soils that are relatively poor in organic matter can be fairly alkaline. --Dr Dima (talk) 01:04, 4 May 2013 (UTC)[reply]

Was there ever a real species of dinosaur that resembled Barney?

In so much that it had the body shape and posture of a carnivorous dinosaur, but also had the blunt, wide herbivorous teeth? It's been a long time since I watched Barney The Dinosaur, but I think it's stated that he's a herbivore (or mostly so).

Discussion of the colour is optional. Though I don't suppose that a purple dinosaur would be out of the question, when you look at some of the colours we see in nature these days. Would it be any less wrong then showing them as being bright green?

Thanks. --146.90.56.134 (talk) 23:19, 3 May 2013 (UTC)[reply]

In my opinion Barney looks more like hadrosaurus than tyrannosaurus, but of course this is all just random. Looie496 (talk) 23:31, 3 May 2013 (UTC)[reply]

Barney was a guy. Don't you mean Dino?--Shantavira|^{feed me} 06:53, 4 May 2013 (UTC)[reply]

See Barney & Friends. HiLo48 (talk) 07:07, 4 May 2013 (UTC)[reply]

The members of Hevisaurus were of specific species.--Shirt58 (talk) 07:22, 4 May 2013 (UTC)[reply]

May 4

Trimmer vs shaver

What is the difference between electric trimmer and electric shaver? --Yoglti (talk) 01:43, 4 May 2013 (UTC)[reply]

About $15. μηδείς (talk) 01:52, 4 May 2013 (UTC)[reply]

The shavers are intened to get down to bare skin. The trimmer is intended to leave a short length of hair protruding above the skin ("stubble"). The trimmer likely has an adjustable setting to vary the amount of hair left, whereas the shavers have one setting ("as close to the skin as comfortably possible"). -- 71.35.109.118 (talk) 02:50, 4 May 2013 (UTC)[reply]

How energetic x-rays (kV) is required to reach above background?

How energetic x-ray photons (kV) is required for them to be stronger than background radiation ..? Electron9 (talk) 02:40, 4 May 2013 (UTC)[reply]

Your question does not make sense in the English language. I assume you meant to ask How energetic do X-ray photons (described in terms of the equivalent electron acceleration voltage) need to be to be stronger than the background radiation on Earth's surface? This is not a valid question. Do you mean the natural background radiation, or the averaged exposure due to nuclear fallout from accidents and explosions, medical X-rays taken during your life, use of nuclear isotopes in medical diagnosis and treatment etc? I will assume that you meant the natural background.

The higher the X-ray energy, the more penetrating it is. And if X-ray photons have fully penetrated a substance, then no energy was transfered to the substance and it cannot have been affected. This is why X-ray images taken to show bone structure are a lot less harmfull than X-rays taken to show soft tissue structure, where similar exposure times are used. It means that the natural X-ray exposure we experience includes X-rays from very high energy sources remote in the universe. In terms of exposure effects, very low energy man-made Xray sources very much over ride natural exposure.

Also note that in terms of effects on life, X-rays are just another sort of ionising radiation. The exposure to just natural X-rays is not important, but the total exposure to all sorts of ionising radiation can be.

Ratbone 124.178.43.47 (talk) 09:47, 4 May 2013 (UTC)[reply]

Maybe, he's talking about the cosmic microwave background? Plasmic Physics (talk) 10:43, 4 May 2013 (UTC)[reply]

I meant if you have a x-ray tube. How high acceleration voltage is needed to measure a higher dose than from ground rock (1 mSv/year?). I heard that CRT-TV-sets with acceleration voltage below circa 10 kV didn't make it out of the TV-set. So that only sets with higher voltage had any measurable radiation. Electron9 (talk) 13:30, 4 May 2013 (UTC)[reply]

Only the smallest CRT TV sets had an acceleration voltage as low as 10 kV. 17 to 20 kV is more typical for black and white sets, early colour sets were up to 25 kV. However, at typical voltages the X-rays are so soft normal materials used in sets (glass, wood, etc) stopped them.

You are still asking the wrong question - you are confusing photon energy with beam power. You can get a high effective dose from the lowest acceleration voltage that will produce X-rays of sufficient energy to penetrate the tube window - about 18 to 20 kV or so. You need to understand that Xray tubes are designed to produce Xrays - so the tube windows are constructed appropriately. TV sets are designed NOT to emit Xrays. For instance, the glass at the front of the picture tube is a three-layer sandwitch up to 18 mm thick and often lead loaded. Internally, older colour sets with internal parts such as the regulator triodes were designed so that Xrays from the triode had to pass through (typically) 2 layers of 12 mm plywood and a steel sheet barrier.

What affects dossage is the electron beam current and the exposure time. It is similar to exposing black and white photographic film with light. You can use a low power white light (say a 0.5 W krypton torch glode running at 4000 K filament temperature) or a high power light red light (say a 60 W globe run on low voltage so that the filament is running at only 1600 K and light output is reddish-orange). The first is analogous to making Xrays with a high voltage but a low beam current; the second is analogous to making Xrays with a low voltage but a high beam current. In both cases the higher power will have the greatest effect.

Not to be neglected is the fact that Xrays are emitted from Xray tubes in a fan-shaped beam, somwhat like light fans out from a light globe. This means that the further you are away from the Xray tube, the lower the dose, as you intercept a smaller fraction of the fan-shaped beam.

As I recall, you previously asked a question about making a homemade Xray apparatus. DON'T DO IT. You have so little undersanding of Xrays, you would be certain to cause harm to yourself and your friends.

As I said before, it is the low energy (ie from low acceleration voltage) that cause tissue damage. Very high energy Xrays pass through without lossing energy. If energy is not lost to the material passed through, there cannot be any effect on the material.

Ratbone 120.145.203.168 (talk) 15:21, 4 May 2013 (UTC)[reply]

There is information missing from your question. No matter how high the Voltage is, if no electron's are beamed down, no X-rays are produced. To answer your question you need to tell us the total number of electrons used by the machine per year. In other words, you have to tell us the current used by the machine. Other things to consider: How far from the X-ray source are you? Is there any radiation shielding? Dauto (talk) 15:28, 4 May 2013 (UTC)[reply]

Spin

If the moon did not exist our earth would spin faster. How short would our days be on the equator? Pass a Method talk 06:26, 4 May 2013 (UTC)[reply]

Not necessarily. The most popular theory Giant impact hypothesis is that the moon was formed in an impact that span the earth up to have something like a five hour day. Otherwise it would probably have something in between the very long days of Mercury and Venus and the roughly equal day of Mars is my guess, maybe somebody has worked out a typical value to be expected. Dmcq (talk) 10:46, 4 May 2013 (UTC)[reply]

What makes you think the Earth would spin faster? Dauto (talk) 12:14, 4 May 2013 (UTC)[reply]

Using debatable assumptions: momentum must be conserved, ergo, lunar recesion decreases Earth's angular velocity. Plasmic Physics (talk) 12:31, 4 May 2013 (UTC)[reply]

Also tidal locking between the Earth and the Moon is transferring rotational energy from the Earth into orbital energy in the Moon, and friction in the tides converts some Earth rotational energy into heating the Earth. Both of these cause the length of Earthdays to gradually increase. BTW, the day/night cycle are the same length no matter where you are, (as long as you are not within the polar circles. Did you mean the length of daylight? CS Miller (talk) 14:11, 4 May 2013 (UTC)[reply]

Hassium

Emsley's Nature's Building Blocks (both editions) say that IUPAC did not feel Hesse merited having an element named after it as a reason for their changing the name to hahnium in 1994. Does anyone know why they felt this way? It's quite odd in light of all those elements named after places! The only reason I've found that they mentioned themselves is that they wanted elements named after Hahn and Meitner to stand side by side on the periodic table to honour their joint discovery of nuclear fission. (Yes, this is for an article.) Double sharp (talk) 08:51, 4 May 2013 (UTC)[reply]

How to get shine/glow in face like celebrities?

They have glow in face [21][22][23] How can I get this glow? Note it is not medical advice, just a health and beauty question. --Yoglti (talk) 09:03, 4 May 2013 (UTC)[reply]

That's just the choice of lighting on the part of the photographer, nothing more. Someguy1221 (talk) 09:08, 4 May 2013 (UTC)[reply]

It's makeup too. Yes, even for Tom. Looie496 (talk) 14:50, 4 May 2013 (UTC)[reply]

efficiency

what is more efficient, a dolphin or shark moving its tale, OR, a machine as strong as the dolphin's or shark's muscles with a rotor? — Preceding unsigned comment added by 123lmon (talk • contribs) 13:16, 4 May 2013 (UTC)[reply]

Dolphins and sharks are much more efficient, they move their bodies in response to the actual flow of water so that it gets altered in the most optimal way. Count Iblis (talk) 13:34, 4 May 2013 (UTC)[reply]

Cooling of smartphones

Smartphones have nowadays as much as processing power as laptop did a while ago. However, when laptops had that much processing power, they had a cooling fan. Why don't smartphones don't need a cooling fan? Why do laptops need them? 123lmon (talk) 13:21, 4 May 2013 (UTC)[reply]

When a computer chip processes, that means electrons are whizzing around through it. That heats it up; if it heats up too much, it can actually melt the processor core. Newer chips can run much cooler than the ones of a few decades back — much cooler. The most common smartphone processor is known as ARM, and it was specifically engineered to have very low heat output and relatively low power requirements. So something on par with a smart phone, or even an iPad, doesn't really require a specialized cooling source, because they've been engineered to dissipate what little waste heat they have pretty effectively. (They don't alway succeed — the iPad will basically shut down if its internal temperature exceeds 95ºF.) Older processors, or modern processors of the speed that would be found in a laptop or desktop computer, still usually require fans to keep from overheating. --Mr.98 (talk) 13:43, 4 May 2013 (UTC)[reply]

Also the operating voltage of CPUs is gradually dropping - it used to be 5 Volts, and is now 0.8V (I think). The capacitance of each transistor is gradually decreasing, these both reduce the amount of power needed for each gate to change state, and thus for each operation the CPU performs. CS Miller (talk) 14:05, 4 May 2013 (UTC)[reply]

The low-power technology that permits all of that computation with such long battery life is the same thing that minimizes heat production. In the end, the energy from the battery turns (almost 100%) into heat inside the case...so things with long battery life and small batteries run cooler than things with short battery lives and large batteries. SteveBaker (talk) 14:06, 4 May 2013 (UTC)[reply]

I think smaller die geometries also plays into this. Electron9 (talk) 14:16, 4 May 2013 (UTC)[reply]

Also, the procesors used in cellphones where designed quite recently from scratch with an optimised instruction set. Intel compatible processors used in PC's and laptops have always been designed with backward compatibility stretching back to the original 8086 CPU designed in the early 1980's. Maintaining compatibility means they are a lot more complex internally. Also, the actual processor core is only part of what consumes power. A PC processor has to interface with external devices, external memory, and other bus devices. Cellphone procssors are one-chip devices that aviod the need for much power dissipating interface circuitry. Ratbone 120.145.203.168 (talk) 15:32, 4 May 2013 (UTC)[reply]

Science of mating in humans

In humans, mating and relationships have evolved into being sophisticated but could it be argued that the process of "picking up" women in nightclubs actually takes this back to being more animalistic. — Preceding unsigned comment added by Clover345 (talk • contribs) 13:44, 4 May 2013 (UTC)[reply]

Well, it could certainly be argued - but I'm not sure that the argument would be a valid one. What exactly is your question here? If you're asking: "Is the process of picking up women in nightclubs 'animalistic' behavior?" then we first have to ask what is meant by "animalistic". We are, after all, animals. Animals have a huge range of mating behaviors - from female spiders that eat their mates immediately after copulation to love-birds that mate for life and die soon after their mate is killed. I'm sure you could find at least one other species that exhibits comparable behavior to the one you're referring.

However, I think you're somewhat missing the point here. You're probably seeing this behavior from only one side - the male. Sure, men go out to nightclubs with the specific goal of finding a woman to mate with...but women go to nightclubs in the knowledge that this is a common thing to happen. This in itself is a sophisticated, nuanced, layered behavioral pattern...it's not so different from composing sonnets and singing outside of a woman's bedroom window...or whatever else you'd consider "sophisticated". SteveBaker (talk) 14:03, 4 May 2013 (UTC)[reply]