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November 6

Lizard Salivation

It was a particularity cold day today, even though I live Southern Florida. I have a Cuban Anole, and knowing that lizards are cold blooded desided to warm him/her up by placinh him/her in some lukewarm water (I was sure the water was not to hot). I left him there to warm up a rock and on my return I found him, seemingly sleeping, but more important he was what seemed to be salivating a lot. My Concern: Is this normal? Should I be concerned? And how do I now if it is a boy or girl? 66.229.227.191 (talk) 01:30, 6 November 2010 (UTC)[reply]

First of all, as evident by our article on Cuban Anoles, there are two types. Assuming yours is a Brown anole, then see if it has a dewlap to determine the sex (males do, females do not). And was the anole touching the water? If so, when the water evaporated the anole would be cooled even more (in the same way sweat cools us). If it was asleep, then we know it was just getting cooler. I suppose the drooling occurred because when it was asleep, it didn't know it was drooling. And I don't think you should be concerned as long as the temperature was above 0° C; reptiles, as the temperature drops, simply get less and less conscious. However, below 0° C, they will die, because the ice will damage tissues. --The High Fin Sperm Whale 02:25, 6 November 2010 (UTC)[reply]

I've never heard a brown anole called a Cuban Anole... but yea, its a Cuban Knight Anole. I understand that water evaporates, but thats normally under wind shear and he was inside and completely emerged. That's strange, I've never seen him/her do that before, maybe

someone could elaborate more 66.229.227.191 (talk) 02:39, 6 November 2010 (UTC)[reply]

If it's a Cuban Knight Anole, then still see if it has a dewlap. Water evaporates all the time, albeit slowly. Even if not evaporated, the water would get as cold as the air around it after you took it out, so warming by using water is not a good idea, I would think. And if you're wondering about the drooling, then see my earlier comment. Hope this helps, --The High Fin Sperm Whale 03:27, 6 November 2010 (UTC)[reply]

IBold text thought this said "lizard salvation".

percent algae death

What percent of oceanic algae death is necessary to reduce the amount of atmospheric oxygen by 10% and is there a curve for the relationship between oxygen and algae? --96.252.213.127 (talk) 03:37, 6 November 2010 (UTC)[reply]

Question one: I am not sure, but the amount would be enough to kill most organisms in the ocean. And even if all of those instantly died for some reason, they would reproduce so fast that in no time the population would be back to normal. Right now, I think the problem is to many algae, not to few.

Question two: Yes. There is a strong relationship between algae and oxygen.

Hope this helps, --The High Fin Sperm Whale 05:35, 6 November 2010 (UTC)[reply]

NO. It does not answer my question for I know that there is also a strong relationship between how much oxygen is needed and how many humans are asleep in a coal mine from which they must be rescued. What I am looking for is the numerical correlation or curve between oxygen production and algae on the planet Earth. --96.252.213.127 (talk) 14:42, 6 November 2010 (UTC)[reply]

It would be very tricky to set up a model to estimate a curve for the relationship. As oxygen decreases, carbon dioxide increases (at least as long as there are animals breathing), and this encourages the growth of plants, but they probably wouldn't fully compensate for the loss of algae. There are just too many unknown variables to derive an accurate relationship or to predict where an equilibrium would be reached for a given percentage of algae death. The coal mine is much simpler to model if there are only miners there. Someone might be able to make a guess at an answer for you, but experts are unlikely to agree or to be accurate because they will have to make many assumptions about what factors to include in the model. Dbfirs 17:47, 6 November 2010 (UTC)[reply]

The curve is very complex and needs to account for all the positive and negative feedback that occurs. You could try modeling it as a differential equation, but I suspect it will be very complex. John Riemann Soong (talk) 21:14, 6 November 2010 (UTC)[reply]

This has not been indubitably proven, but measurements using a Secchi disk and satellite analysis showed that the oceanic plankton population declined by 40% in the last 60 years. Also look at oxygen synthesis. ~A H 1^(TCU) 17:26, 7 November 2010 (UTC)[reply]

jeans

why do jeans feel different on the outside than the on the inside? insint it the same fabric —Preceding unsigned comment added by Kj650 (talk • contribs) 04:37, 6 November 2010 (UTC)[reply]

Jeans are made from denim, which is a type of fabric called twill. According to our article "Twill fabrics technically have a front and a back side", presumably because of the way they are woven (described in the article). The two sides are different: the technical face side (front) is usually more durable and attractive. Presumably (the article doesn't actually say) the material feels different on each side because of the asymmetric weave. Mitch Ames (talk) 05:51, 6 November 2010 (UTC)[reply]

how do u weave something that has 2 sides —Preceding unsigned comment added by Kj650 (talk • contribs) 08:01, 6 November 2010 (UTC)[reply]

How do you weave something that hasn't? --Stephan Schulz (talk) 09:56, 6 November 2010 (UTC)[reply]

Like this. Mitch Ames (talk) 12:15, 6 November 2010 (UTC)[reply]

I think he means "two sides that are different". Judging by the twill article, I'd say you'd need to know a bit about weaving to understand it... Vimescarrot (talk) 10:30, 6 November 2010 (UTC)[reply]

The front side of a twill should feel the same as the reverse side at right angles.

However jeans material will have had it's front surface treated differently, (confirmed by the presence of dye) - this will also affect the feel.94.72.205.11 (talk) 12:00, 6 November 2010 (UTC)[reply]

They won't necessarily be the same after rotation, even without differential treatment. On something like a 3/1 twill, one side will have mostly the warp fibers exposed, whereas the reverse will have the weft exposed. During weaving, there's different tension on the two, so the "lay" will be different. Examining a pair of jeans, this appears to be the case. The ones I looked at appeared to be a 2/1 twill, with the high-tension warp threads being prominent on the inside, and the lower tension weft threads prominent on the outside. This gives the inside a smoother, flatter texture than the outside. Although it doesn't appear to be the case on the pair I examined, different thread types can be used on the warp versus weft, accentuating the difference. -- 174.31.204.207 (talk) 17:25, 6 November 2010 (UTC)[reply]

that's right, good point. 94.72.205.11 (talk) 18:21, 6 November 2010 (UTC)[reply]

Geology abbreviation

While looking at a job application, in the job description it is mentioned: Grade: Band 7/6 (SO/HSO). What does this mean, and what other grades are there? Thanks in advance. --Danninja (talk) 07:11, 6 November 2010 (UTC)[reply]

These are (probably UK government) officer grades: Scientific Officer (SO); Higher Scientific Officer (HSO); Senior Scientific Officer (SSO)

According to this site, point 6 was £29,412 in August 2009, but you should check with the employer before assuming that this is correct because Medical Scientific Officers at point 6 on the Liverpool University scale are currently paid only £23,366, and if the employment is in India where there are also "SO" scales, then the remuneration will be completely different. Dbfirs 08:12, 6 November 2010 (UTC)[reply]

Coccyx decay rumor

If anyone searches about Coccyx decay, they will get many results on the top confirming this rumor without any scientific evidence:

"Coccyx tail is the only bone that doesn't decay after one's death"

Is there any scientific citation?--Email4mobile (talk) 09:57, 6 November 2010 (UTC)[reply]

What exactly does "after one's death" mean? 10 years? 100 years? 1000 years? There are many, many places all over the world where human remains from several thousand years ago have been excavated and many of them have most of their bones intact.[1] Just think logically about the "rumour", why is the coccyx any different from any other bones in the body? Correct, it's not. So why should it decay in a different way. Richard Avery (talk) 13:34, 6 November 2010 (UTC)[reply]

Schiller's experiment

Few days ago I posted this question on the Michelson–Morley experiment talk page, but will repeat here: I've read a related article and it was astonishing but I don't know if it were reliable to add it to Michelson–Morley_experiment article?--Email4mobile (talk) 10:08, 6 November 2010 (UTC)[reply]

What I find astonishing is not the article or the confirmation of the results, but the crank comments that follow. I'm not sure that this experiment really adds anything to a long-established result, but we could add a note that the result has recently been confirmed to at least one part in 10¹³. Dbfirs 11:38, 6 November 2010 (UTC)[reply]

health care

This request to diagnose the cause of a health problem has been removed. -- kainaw ™ 14:35, 6 November 2010 (UTC)[reply]

sorry, but I just had to edit your comment. --Chemicalinterest (talk) 15:02, 6 November 2010 (UTC)[reply]

How is combustion sustained in a jet engine?

Compressed air flows through the combustion chamber of a jet engine at several hundred miles per hour or more. How do they keep the flame from being extinguished? Do they have some kind of heating element that continuously tries to restart combustion? --173.49.14.225 (talk) 11:59, 6 November 2010 (UTC)[reply]

No, they don't. It's not necessary. 76.123.74.93 (talk) 13:58, 6 November 2010 (UTC)[reply]

Once a jet engine (where "jet engine" means turbojet or turbofan, for now) is started, the ignition system is turned off (it's mostly just a ring of spark plugs). From then on the flame is sustained by (subsonic) deflagration - the burning mass of fuel/air in the rearward end of the combustion chamber thermally heats the non-burning fuel-air mixture forward of it, causing it to burn. This means there's a stable flame-front in the engine, which self-sustains the flame without an external source of ignition. This process is aided by the incoming stream already being highly pressurised (overall pressure ratio gives ratios > 10:1). A flame holder in the air path provides an eddy environment to maintain the position of the flame front. Note that if things go wrong with the supply of air, fuel, or with the pressurisation provided by the turbine spools, the engine can flame out and must be reignited. This all works because the speed of air through a turbine engine is subsonic (even on supersonic aircraft); deflagration only works in a subsonic medium. If you tried to inject supersonic air, the deflagration couldn't keep up, the fire-front wouldn't maintain its position at the flame holder, would be blown backward, and the engine would flame out. As for truly supersonic engines, I think a scramjet needs a continuous ignition source, and a pulse detonation engine relies on supersonic detonation rather than subsonic deflagration, and I think may need a reignition for each pulse. -- Finlay McWalter ☻ Talk 15:34, 6 November 2010 (UTC)[reply]

Finlay's description above about the flame-front is excellent. The flame is constantly in a state of dynamic equilibrium as the chemical combustion takes place, using up fuel and air, and new air and fuel rushes in. Part of the difficulty of designing a safe jet engine is to make sure that the flame front is stable - that is, it must not be "wiggling around" a whole lot, or changing the magnitude or rate of combustion. This requires careful design (and possibly using active control of fan speed, turbo speed, fuel injection rate, and so on). If the combustion starts to oscillate (i.e., the "wind blows the flame front back and forth"), the thrust, fuel consumption, and other parameters may also oscillate - and this is a bad thing for safe flight.

Regarding scramjets: my understanding (and I admit my understanding of supersonic flows is limited) is that in a scramjet, the velocities and pressures are sufficient for autoignition - in a sense, you're creating hypergolic propellant out of jet fuel and highly supersonic atmospheric air. This can only happen stably in special high-velocity conditions; that is why scramjets typically require some other propulsion system to boost the vehicle up to speed. Our scramjet theory section outlines some math for comparing the aerothermal heating due to kinetic energy of the airstream; in sufficiently fast streams, the kinetic energy is greater than the ignition activation energy (hence, autoignition). Nimur (talk) 03:39, 8 November 2010 (UTC)[reply]

Air traffic control system

I may be getting the details of the story wrong. I think it was in the early '90s when I read about a problem with the ATC system in the US. The system was described as antiquated and running on obsolete hardware. Procuring repair parts for the hardware was getting difficult. And attempt to develop a modern software replacement was unsuccessful. I don't know if and how the situation have improved since, or whether the problem persisted.

My questions:

Has the US replaced the old ATC system with something modern and maintainable?
Why couldn't the US just buy whatever ATC system the Europeans were using at the time? Did the Europeans have the same problem?

--173.49.14.225 (talk) 12:15, 6 November 2010 (UTC)[reply]

The Wikipedia article Air traffic control will be of interest. It notes that In 1999, U.S.A. controllers began use of the Standard Terminal Automation Replacement System, which included new displays and capabilities for approach control facilities. Cuddlyable3 (talk) 23:10, 6 November 2010 (UTC)[reply]

ATC is extremely conservative. Over here, in Germany, the replacement system for the Upper Airspace Control in Karlsruhe has been under development for 15 years, but is still not operational. Of course, when they designed it they made it future-proof by using a client/server structure. And by picking the most advanced microprocessor for all components - the Alpha 21264, instead of one of the obviously dying Intel CISC kludges. For the US, buying an ATC system in Europe is hard to imagine. The US aviation market is very much cornered by the big US suppliers - Boeing, Raytheon, Lockheed Martin, and a few others - due to pork barrel spending policies and protectionism. See Next Generation Air Transportation System for the current transition in the US and Single European Sky for the next generation ATC systems in Europe. --Stephan Schulz (talk) 00:41, 7 November 2010 (UTC)[reply]

Possible future need for faster than light communication

What is to be done theoretically if the mankind becomes advanced to such an extent, when communication between remote human colonies and objects would require a faster-than-light transmission of signals? 85.222.86.130 (talk) 12:38, 6 November 2010 (UTC)[reply]

Why would it be required? If humans have extensive colonies thorough many different stars, they will learn to live with the fact contact between these will take several or many years (depend on how far we're talking about) and you may be dead before people in the other colony receive your message (although if you personally don't want that and with the tech required to set up these colonies who knows what may be possible, e.g. uploading yourself to a computer, some sort of stasis, a relativity time dilation device or even simple advances allowing near biological immortality). Considering it will take even more years for people to travel between these, I see no reason why people will somehow be unable to live with that. It may be nicer to have the soft SF near instant communication but there's no reason we can't live with the real physical universe we live in. Nil Einne (talk) 13:01, 6 November 2010 (UTC)[reply]

As far as we know, FTL communication is impossible — just plain ruled out by physics which seems pretty solid at this point. So I think we can throw that one out the window unless physics is totally wrong in some important respects. (Which is always possible — the quantum world certainly is weird and often surprising, even at this late date — but you can't bet on it.) What you more likely get are colonies that aren't in communication. That either means you have a limit on colonization, or you have a limit on how "connected" you expect your colonies to be, when they can only communicate with each other every few decades or so. The world of Ender's Game (or, rather, the sequels to that particular book) discusses these issues in great detail, although largely in reference to what FTL communication (through the fictional ansible) lets you do that otherwise you would not be able to do, politically, especially if you have it in absence of FTL travel (which means, in one book, that if you want to send a military fleet to an unruly colony, it'll arrive only after a number of decades from the colony's reference frame). In any case it seems pretty clear that you couldn't have the same degree of power projection between colonized worlds that you do between different nations on Earth, and your political models would have to reflect this reality. --Mr.98 (talk) 13:08, 6 November 2010 (UTC)[reply]

Browsed the superluminal communication, thanks. 85.222.86.130 (talk) 15:30, 6 November 2010 (UTC)[reply]

This form of communication may necessitate the controlled use of wormholes, strings or via the folding of space. ~A H 1^(TCU) 17:21, 7 November 2010 (UTC)[reply]

Or quantum devices that can violate causality. Googlemeister (talk) 15:15, 8 November 2010 (UTC)[reply]

Solstice dates change from one year to another

I am wondering why the dates for solstices and equinoxes change from one year to another.

I'm assuming it is because the sidereal year is 365.256363004 days and our calendar is 365 days (3 out of every 4 years).

If it is because of that discrepancy, could a calendar be devised in which it did not change? What elements would be required to accomplish this? -Joel (talk) 14:26, 6 November 2010 (UTC)[reply]

One small point. Because the Earth's orbit precesses, the sidereal year is irrelevant; the relevant version of the year is the tropical year. This is a bit under 365.25 days, which is why leap days are sometimes skipped rather than extra ones being added. --Anonymous, 20:13 UTC, November 7, 2010.

The relationship between the earth orbit's and the earth's rotation would need to change to integers. It's a good plan. --jpgordon^{::==( o )} 14:34, 6 November 2010 (UTC)[reply]

By that do you mean that our numeric system would need to change? There's got to be an easier way... I'm not thinking we'd need to ditch leap-years or anything. Could not the placement of the leap-year prevent the solstice date change? Like, with leap years in 2000, 2004, 2008 it will change, but with leap years in 2003, 2007, 2011 it won't. (It's really hard for me to think about trying to answer my question.) -Joel (talk) 14:48, 6 November 2010 (UTC)[reply]

I think jpgordon's reply was a subtle of saying that nature is messy and won't conform to our strictures or constructions. In other words, the plan would only work if the time for the Earth to go around the sun was an exact whole number multiple of the length of a day. Since it's not, that 'perfect' calendar simply can't exist. Changing when leap years are would make no practical difference. --jjron (talk) 16:04, 6 November 2010 (UTC)[reply]

... and even the year is not a fixed length. Our calendar is based on the average time between vernal equinoxes, and this is not quite the same as the mean tropical year. If we ignored sunlight and all lived according to an artificial "day" (not exactly 24 hours, and varying slightly from season to season) then we could have a fixed calendar, but I don't think it would be popular except possibly with those who work night shifts. Dbfirs 16:56, 6 November 2010 (UTC)[reply]

A further point is that solstices and equinoxes are instantaneous events. This means that the date when they occur depends on your local time zone. Since time zones exist that are over 24 hours apart (in Kiribati and Samoa, for example), any particular moment in real time always occurs on at least two different dates by zone time, sometimes three. For example, the upcoming solstice is at 23:28 UTC; it will be on December 21 in the UK and the Americas, but on December 22 in Australia, Asia, and most of Europe and Africa.

It would be possible to arrange a calendar so that a particular solstice was always on the same date, say December 21, in a particular part of the world. Whereas now we have leap years mostly at 4-year intervals with a fixed sequence of 8-year intervals, it could be decreed that leap years would be assigned dynamically at mostly 4-year and occasionally 5-year intervals, whichever would produce the desired effect. (The Islamic calendar is attached to the phases of the moon in a similar dynamic fashion.) But because of the time zones, this would not work for other parts of the world. There would be always be countries where it was not always on December 21.

Likewise, the time between one solstice or equinox and the event is not an integer number of days. For example, from the solstice this past September to the equinox coming in December is 89.853 days (the exact amount varies from year to year). This means that even if the equinox was fixed on December 21, the solstice might vary between September 22 and 23, for example.

--Anonymous, 04:57 UTC, November 7, 2010.

Leap year, leap second and nuclear clock may also be relevant. ~A H 1^(TCU) 17:14, 7 November 2010 (UTC)[reply]

I can't find any reliable information on it on-line, but Duncan Steel has published a theory that the hidden purpose of the Roanoke colony was to establish a settlement at longitude 77°W, where if the 33-year Jalali calendar were adopted, the equinox would always fall on the same date. This is a summary, but not a RS. --ColinFine (talk) 19:18, 7 November 2010 (UTC)[reply]

Work done by Ice-Skater

A skater's body has rotational inertia 4.2 kg m^2 with his fists held to his chest and 5.7 kg m^2 with his arms outstretched. The skater is twirling at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. Fine the work done by the skater pulling the weights to his chest.

I plan to find the work by subtracting initial kinetic energy (when his arms are outstretched) from the final KE (arms at chest). But I'm having trouble understanding the various rules regarding rotational/translational KE. Any help?199.94.68.201 (talk) 15:01, 6 November 2010 (UTC)[reply]

0. Cuddlyable3 (talk) 22:39, 6 November 2010 (UTC)[reply]

It's not zero, Cuds. Work is done pulling the weights into the body, increasing the kinetic energy. You have the right approach, Mr. 199.94. You want to find the final rotational speed using Conservation of Angular Momentum. Formulas can be found in that article, and Rotational energy. Buddy431 (talk) 02:40, 7 November 2010 (UTC)[reply]

What Happens if the gravitational force decreases?

Bold textItalic text I am very eager to know that what happen when the gravitational force is reduces? i know that the gravitational force is acting indirectly to keep the percentage of oxygen in air. if it start reducing will the percentage of oxygen will remain same or not? If that force is reduced what will be the reason for that? Waiting for the reply. —Preceding unsigned comment added by Yuvraja (talk • contribs) 15:05, 6 November 2010 (UTC)[reply]

Fear not. The gravitational force is not decreasing. See Gravity of Earth and Newton's law of universal gravitation. --jjron (talk) 15:35, 6 November 2010 (UTC)[reply]

Gravitational attraction is lowest at the outermost layer of Earth's atmosphere, the Exosphere which is mainly composed of hydrogen and helium. That might become the composition of the whole atmosphere if the Earth's gravity were somehow reduced, though that would not be the first abnormality we would notice. Cuddlyable3 (talk) 17:00, 6 November 2010 (UTC)[reply]

(edit conflict)No, it is not decreasing, the only way to do that would be to get rid of a huge amount of mass. One much more common (in space) way to loose oxygen is when a planet looses, or never has, a magnetic field. This causes the solar wind to blow away the atmosphere. --The High Fin Sperm Whale 17:04, 6 November 2010 (UTC)[reply]

It is decreasing because the mass of the planet is decreasing. Think of all those spaceships we released... --Chemicalinterest (talk) 18:59, 6 November 2010 (UTC)[reply]

True, but think of all the rocks that have hit us! --The High Fin Sperm Whale 20:35, 6 November 2010 (UTC) [reply]

But what about that matter we converted to pure energy through detonating a hydrogen bomb? Googlemeister (talk) 15:10, 8 November 2010 (UTC) [reply]

If the gravitational force were to "magically" decrease you would have a wide variety of effects. For example air pressure would be lower, more air would escape into space (but not oxygen specifically, it would be the regular mix of air). More interesting would be the effects on the orbits of the planets - the planets would suddenly be orbiting too fast, and will move to farther orbits. A more important effect would be on the sun. Unlike the Earth the sun is held together by gravity. Additionally the nuclear reactions are controlled by the density and pressure in the center of the sun. If gravity were lower the suns output would decrease, this would cool the sun, causing it to shrink, which increases the density and pressure. I'm not sure if the end result will be the same level of brightness or not. Ariel. (talk) 23:50, 6 November 2010 (UTC)[reply]

A reduction if gravitational force would require a reduction in mass of the Earth, the number of gravitons (theoretical particle) emitted or the gravitational constant. Gravity travels at the speed of light, so if the Earth and Sun both lost some gravity, the orbital effects would be felt some 8.5 minutes later. ~A H 1^(TCU) 17:11, 7 November 2010 (UTC)[reply]

Velocity management

The question appears quite weird in terms of special relativity, but anyway. Is it possible to extract the cause of light speed and manage it so that it could be multiplied? For example we take the light speed-producing conditions and multiply them just like we are adding more fuel to intensify the fire. Something like speed power of two photons cross-added, yielding roughly 600,000 m/sec. 85.222.86.130 (talk) 15:49, 6 November 2010 (UTC)[reply]

No. See article. Cuddlyable3 (talk) 16:42, 6 November 2010 (UTC)[reply]

(ec) You are correct in saying that the question sounds weird. Even if we knew the "cause of light speed", it is unlikely that we could "manage it" in any multiplicative way. From a mathematical and scientific point of view, the whole concept sounds like gobbledegook. Are you writing a science fantasy story? Dbfirs 16:44, 6 November 2010 (UTC)[reply]

To a certain crude approximation, the "cause of light speed" is the absence of invariant mass. (The "speed of light" is a fundamental property of the universe, and isn't special to light in any way.) It's not meaningful to say that something is "twice as massless". Two times zero is still zero. -- 174.31.204.207 (talk) 17:07, 6 November 2010 (UTC)[reply]

Heating house - turn off when I leave the house?

My husband and I debate over what we should do when we leave the house for a couple of hours. Should we turn off the heat, lower it, or leave it as is? Obviously, if we leave the house for a week it would make sence to turn the heat way down. Likewise, if we leave the house for 5 minutes, we would leave it alone. There must be a physics formula to answer my question. I've asked my friends, who just give me their opinions. —Preceding unsigned comment added by 74.90.197.121 (talk) 20:27, 6 November 2010 (UTC)[reply]

Much depends on the house and the climate. There's no one-size-fits-all answer about the time interval that would be suitable, but most programmable thermostats use four blocks of time: morning (warm), work (cool), evening (warm) and night (cool). I'd say around four to six hours of absence would be a good guide, based on that precedent, unless your house gets cold really quickly (and is quick to warm up again). Acroterion _(talk) 20:36, 6 November 2010 (UTC)[reply]

(edit conflict) Lowering the temperature always reduces heat loss, but whether this is a saving depends on whether you then turn the heat higher when you return. If you don't do this, then there will always be a saving in turning the heat off, but obviously for five minutes the few cents or pennies of saving will seldom be worth the effort. The graph never crosses zero, so there is no "break-even" point unless you factor in the wear on the switch. You will always save money by turning the heat off (though beware of frost damage in extreme cold). Dbfirs 20:38, 6 November 2010 (UTC)[reply]

To rephrase what Dbfirs said, it's always worth it to turn off the heat - unless, when you turn it back on, the furnace runs at a higher temperature. Usually the furnace does not run at a higher temperature, except if you have a modulating boiler or a heat pump. Most furnaces are either on or off, with no in between, but modulating boilers can change fire, and heat pumps will turn on auxiliary heat. Ariel. (talk) 23:46, 6 November 2010 (UTC)[reply]

Even if the furnace runs at a higher temperature for a while, it will not cost more unless the thermostat fails to turn it off at the set room temperature. The extra cost would only occur if there is some over-run, raising the room temperature higher than normal. I cannot see circumstances in which this would occur only after switching off, but I'm not an expert on heating systems, so I'm just looking at the whole-house boundary with the environment to give a simple answer. Dbfirs 00:24, 7 November 2010 (UTC)[reply]

In a modulating boiler the water temperature is adjusted based on the heat necessary. Usually it measure the outdoor temperature and sets it that way, but if the boiler sees that the call for heat lasts for more then X minutes it raises the temperature. The thing is, the higher the water temperature the lower the efficiency. The range is about 85% (180 degree water) to 98% (110 degree). In my house I only need 180 degree water when it's -20f outside. Interestingly with a modulating boiler, the typical recommendation to install a digital setback thermostat can actually cost you more, because the boiler has to raise the water temperature in order to heat the house quickly enough in the morning. The poorer efficiency when this happen can eat away any savings. I would love hard data on this subject (so I know what to do with mine), but I have not found any. Ariel. (talk) 01:19, 7 November 2010 (UTC)[reply]

Ah, yes, I hadn't considered the lowering of boiler efficiency where more heat or wasted fuel pours out through the boiler exhaust. Do no manufacturers provide an optional temperature-limiting control? Dbfirs 08:17, 7 November 2010 (UTC)[reply]

It's complicated. Cast iron boilers must run at 180 or above or they will rust from the water condensing from the exhaust. It's the same with the chimney - if water condenses in it it will dissolve the bricks. And finally if the temperature is not high enough the exhaust will not rise out of the chimney. All boilers (and furnaces) that can run at lower temperatures have stainless steel or aluminium parts, and exhaust using a fan. If you ever come across the terms Class 1, 2, 3 or 4 chimney, that's what the difference is. Ariel. (talk) 19:14, 7 November 2010 (UTC)[reply]

You should turn it off. The boiler (or whatever you call it where you live) will stop using fuel. And, as heat loss is proportional to the difference in temperature, less heat will be lost when the temperature difference between your house and the environment is lower, as it cools. Imagine a leaking bath. Rather than keeping water running into it all the time to keep it topped up, you use less water if you just fill it when you need it. I'm curious as to what the folk-model of physics would be that would give a different answer. (Hmmmn, maybe measuring rate rather than quantity). 92.24.189.145 (talk) 11:29, 7 November 2010 (UTC)[reply]

Yes, that was my analysis, but Ariel pointed out certain (very limited) circumstances in which the inefficiency of some boilers causes them to produce less heat for a given cost when running at high temperatures. If they do this for a while at switch-on, then there might be some circumstances in which money (not heat) might be wasted by switching off. I suspect that this effect is small. Dbfirs 18:00, 7 November 2010 (UTC)[reply]

They are wasting heat - it's going up the chimney, and it's not just at switch on either, the water is running at a higher temperature during most of heat cycle, while if you just left it it could run for longer, but at a lower i.e. more efficient temperature. But, the practical effect may be small like you say, I wish I had hard data. Ariel. (talk) 19:14, 7 November 2010 (UTC)[reply]

Another way to reduce energy use would be to have the heater turned off throughout a certain time period during the night, so that the temperature in your house varies similarly to the daily outdoor temperature fluctuations, and less energy is needed to heat during the day. ~A H 1^(TCU) 17:08, 7 November 2010 (UTC)[reply]

If you have a heat pump, you need to understand that it will take longer for it to get warm, as heat pumps operate on a low temperature differential compared to other heating systems; the air coming out of the register will be cooler than air from a gas furnace, so heat pump systems can end up "chasing" a set point if the set point is changed too often. Also, heat pumps and air conditioning systems generally incorporate delay timers to prevent damage to the compressor from frequent start-stop operations, which can cause high back-pressures in the refrigerant loop. Acroterion _(talk) 21:19, 7 November 2010 (UTC)[reply]

Moles

QI said that all pictures of animal moles are of dead moles, posed in the ground and fluffed up. Is this true? Is there not pictures of live moles? —Preceding unsigned comment added by 78.113.39.122 (talk) 20:45, 6 November 2010 (UTC)[reply]

The photographer needs an enormous amount of patience and luck to get a picture of a live mole. Moles spend at least 99% of their time underground. I occasionally see one above ground, but often moving fast, so difficult to photograph. I've never tried to take a picture of one, but I think I would be tempted to cheat. Dbfirs 21:43, 6 November 2010 (UTC)[reply]

People catch live moles on video [2] [3] [4]. I changed the title from "question" which is not informative. Cuddlyable3 (talk) 22:55, 6 November 2010 (UTC)[reply]

I got a video of me holding one once. It came out of the ground, so a grabbed it and kept it in my cage for a few days, during which time I got pictures and videos. --The High Fin Sperm Whale 23:31, 6 November 2010 (UTC)[reply]

Yes, it would be easier to take video footage than to obtain a high-quality picture with a manually adjusted camera. I suppose it would be much easier to catch a mole in action (the classic pose is just pushing up a molehill) using modern fast autofocus. The problem with moles is that one never quite knows where they are going to appear, so it is difficult to have the camera ready focused. I'm sure there are many pictures of live moles in existence, but I understand the psychology of using a dead mole to get a good photo of a "mole in action"! Dbfirs 00:16, 7 November 2010 (UTC)[reply]

Moles don't hang around for people to do this.Cuddlyable3 (talk) 19:51, 7 November 2010 (UTC)[reply]

Touching neutron star matter

The core of a neutron star is not well understood from what I heard, but it's mostly just neutrons right? If you were to take a mass the size of a baseball from the core and put it in outerspace, would it still maintain its immense density or would it kinda dissipate into a cloud of neutrons? I guess what I want to know is, is pressure causing it to be dense, or is it just an intrinsic property that the stuff has? Would touching it kill you? If so, why? Also since just a small amount of it is so massive, is it possible to arrange a bunch of them in a line in outerspace and climb them like a ladder even though they aren't touching each other? ScienceApe (talk) 21:05, 6 November 2010 (UTC)[reply]

I think it's kind of moot; you'd be dead before you could get close enough to a neutron star to touch the stuff, for I imagine many many reasons, but tidal forces will suffice if nothing else gets you first. Sci-fi writers sometimes like to imagine that you could hack off a piece of neutron degenerate matter separately from the gravitational field that squishes it together, but I believe that is not true. --Trovatore (talk) 21:08, 6 November 2010 (UTC)[reply]

Why not? ScienceApe (talk) 21:34, 6 November 2010 (UTC)[reply]

I'll leave it to a physicist to answer in more detail. But see the linked article, especially at the top, where it talks about the dominant contribution to the pressure coming from the Pauli Exclusion Principle. Stands to reason, though I certainly haven't worked out the equations, that if you take off that pressure, it all blows apart. --Trovatore (talk) 21:39, 6 November 2010 (UTC)[reply]

Neutron_star#Structure has a diagram of an expected structure and make up of a neutron star - also the text next to it has a description. In general pressure is required to maintain the structure.94.72.205.11 (talk) 22:40, 6 November 2010 (UTC)[reply]

Free neutrons have a half-life of about 15 minutes before they decay into hydrogen, or a proton and a neutron. The only reason that in a neutron the stuff stays neutrons is that in the crushing pressures there is simply not enough room to have atoms (remember, the nucleus is about 1/100,000 the diameter of an atom). They are squeezed until they merge, forming neutrons. And you couldn't touch the stuff; if you magically got a piece of it outer space it would instantly explode. --The High Fin Sperm Whale 23:29, 6 November 2010 (UTC)[reply]

I thought the neutron-star-ladder was kind of a neat idea, though, ScienceApe. In principle. It would probably be difficult to arrange everything in order to get the sequential gravity assist to work just right, though. WikiDao ☯ (talk) 01:18, 7 November 2010 (UTC)[reply]

A marshmellow dropped on a neutron star would release roughly the same amount of energy as 50 Hiroshima bombs. So you'd explode. ~A H 1^(TCU) 17:05, 7 November 2010 (UTC)[reply]

A baseball has a radius of about 37 mm, and thus a volume of about 212 ml. Then, according to the density from the article, it contains about 1.39*10⁴¹ neutrons. A neutron has a mean lifetime of about 886 seconds (the half-life is that times log(2)), and therefore the neutron baseball will have 1.57*10³⁸ neutron decays per second. The decay energy is 0.78 MeV = 1.25*10^-13 J, resulting in 1.96*10²⁵ W of power radiated from the baseball. Some of it would be in harmless neutrinos, but the rest would be still enough to make sure that you evaporate faster than you can move your arm in order to touch it... and the energy released in one second would be hundreds of billions times as large as that from the nuclear weapon that detonated in Hiroshima. Icek (talk) 03:52, 8 November 2010 (UTC)[reply]

Of course, that's why you need to put it in a Slaver stasis field. --Trovatore (talk) 03:56, 8 November 2010 (UTC)[reply]

If the neutrons are packed cheek to jowl, how can the ones on the inside decay at all? The electrons would be packed in at nuclear density which would be a very very high energy state for them. Wouldn't just the neutrons at the surface (or very near it) decay? Hcobb (talk) 04:29, 8 November 2010 (UTC)[reply]

What, no Larry Niven fans here? Never mind heat, think gravity. A neutron weighs 1.67*10⁻²⁹ kg, so 1.39*10⁴¹ of them would weigh 2.32*10¹² kg. The Earth weighs 5.974*10²⁴ kg, or 2.57*10¹² times as much. Therefore things will be attracted with a force (more precisely, an acceleration) of 1 gee to the neutronium ball if they are at a distance of sqrt(1/2.57*10¹²) Earth radius from its center. That is 1/1,600,000 Earth radius, or (6,370,000/1,600,000) meters: or just under 4 m. Come within 1 m and it'll be 16 gees. At 1 foot, 170 gees. At the surface, well over 10,000 gees. Get close enough to the thing to touch it and it'll wrap you around itself in a thin layer, moments after its tidal force (the difference in gravity between a nearer and a more distant position) rips your body apart. --Anonymous, 08:15 UTC, November 8, 2010.

Good thing your opponent cheated in the coin-throwing game, then. --Trovatore (talk) 08:32, 8 November 2010 (UTC)[reply]

That's the one I had in mind! Still can't remember the title—not "Neutron Star", of course. Well remembered. --Anon, 21:10 UTC, November 8, 2010.

There is a Tide. --Trovatore (talk) 21:16, 8 November 2010 (UTC)[reply]

Thanks! --Anon, 23:34 UTC, November 8, 2010.

I think you'd be vaporized before that. By the way, your number for the neutron mass is wrong by 2 orders of magnitude, it's 1.67*10⁻²⁷ kg, and therefore the distance at which the gravitational acceleration is 1 g is about 40 meters, and at the surface it's more than 1 million g. Another process that competes with a nice explosion is - if you are on a planetary surface - that it would fall towards the center of the planet, there is no rock that is hard enough to prevent that. Icek (talk) 09:08, 8 November 2010 (UTC)[reply]

Dang! All that calculating and I copied the number wrong. Sorry about that. But at least it was an error in Niven's tradition. :-) --Anon, 21:10 UTC, November 8, 2010.

On the decay within the neutron ball: I am not totally sure, but I would think that due to the release of the pressure the neutrons would already move further apart: If the density from the article is correct, then the distance between neutrons on a cubic face-centered lattice (which has the highest density possible for a lattice) would be 1.29 fm. Taking the diameter of the ²³⁸U nucleus (from Atomic nucleus), dividing the volume by the number of nucleons and taking the third root suggests an average distance of about 1.95 fm between neighboring nucleons.
But if it is true that the decay happens only on the surface, then we still have 1.03*10²⁸ neutrons at the surface, or 1.45*10¹² W of power, or 348 tons of TNT equivalent per second. Maybe it would evaporate a nice hole in the ground while falling towards the center of the Earth (if the neutrons in the inside decay, I would expect the ball to come apart pretty quickly). Icek (talk) 09:08, 8 November 2010 (UTC)[reply]

Suicide

What causes people to consider suicide other than depression? The Utahraptor ^Talk/_Contribs 21:51, 6 November 2010 (UTC)[reply]

Suicide#Causes - specifically see the social causes, but also read the whole article for idea, also ~~Assisted suicide~~ Voluntary euthanasia has some ideas too. —Preceding unsigned comment added by 94.72.205.11 (talk) 22:23, 6 November 2010 (UTC)[reply]

See suicidal ideation and parasuicide. Often it's the sense of hitting a "dead end" in someone's life, even if that dead end is illusory. ~A H 1^(TCU) 17:04, 7 November 2010 (UTC)[reply]

In some traditional societies it was expected as a matter of honour after certain failures. See Seppuku -- Q Chris (talk) 14:19, 8 November 2010 (UTC)[reply]