Wikipedia:Reference desk/Archives/Science/2013 January 29

= January 29 =

Question about the Sunrise and Sunset
It is nuff-said that the spherical earth not only rotates about its own axis but also orbits the sun in an elliptical orbit. If this is true then why the sun rises and sets almost at the same time on surface area of globe which is on or close to equator exempli gratia Sri Lanka when there is a geometrical shift change of sunrise and sunset on antipodal points of earth during its advancement from summer solstice to winter solstice in its yearly orbit for its four orbital events around the sun?

Places on earth experience day when facing sun while night if not at any given instant during the advancement of earth in its elliptical orbit around the sun. For the geometrical shift change of sunrise and sunset on antipodal points, let’s imagine earth positions at following 1- Summer solstice: Sun rises at “A” and sets at “B” while at

2- Winter solstice: Sun rises at “B” and sets at “A” similarly 3- Vernal equinox:  Sun rises at “C” and sets at “D” while at 4- Autumnal equinox: Sun rises at “D” and sets at “C” Since nights lag behind by *12hrs when earth moves around the sun in its yearly orbit from summer solstice (or vernal equinox) to winter solstice (or autumnal equinox) OR on any antipodal orbital (elliptical) points and similarly 24hrs in one complete cycle around the sun, ergo, shouldn’t the timing of sunrise on globe at its summer solstice be varied to sunset time instead of regular sunrise time for its winter solstice? 108.173.128.208 (talk) 03:47, 29 January 2013 (UTC)Eclectic Eccentric Khattak No.1
 * I didn’t calculate the lagging time
 * I'm not entirely sure I understand your question here, but there's a discussion above which may be related, titled "Why don't the the sunrise and sunset times go the other way at the solstice?". The article equation of time may be useful for you as well.  -- Jayron  32  03:51, 29 January 2013 (UTC)
 * I may be way off the mark, but I think what you're looking for is that the sidereal day is different from the solar day. We set the 24-hour length of the day as the average time from highest sun to highest sun, not the time it takes the Earth to rotate in place!  So the stars rise at different times each day.  The solar day is only an average - as the equation of time states, there is some variation from month to month based on the Earth's imperfectly circular orbit, which makes the Sun appear to move faster relative to the stars at some times of the year rather than others. Wnt (talk) 05:05, 29 January 2013 (UTC)
 * This was definitely my own reading of the question. It makes sense of everything the OP said, however hard to follow. IBE (talk) 09:38, 29 January 2013 (UTC)

I am unable post a diagram but you can find the four important events of earth in the following link when it orbits the sun in its elliptical orbit.



For simplicity

Mark two antipodal points “A” and “B” on the earth at its position #1 in the link diagram where the sun rises at “A” and sets at “B”. Move the same globe to its position #3. You can find the shift change in sunrise and sunset after copmaring position #1 and #3 i.e.

Sun rises at “A” and sets at “B”in position #1 while sun rises at “B” and sets at “A” in position #3

Mark two antipodal point “C” and “D” on the earth at its position #2 in the link diagram where the sun rises at “C” and sets at “D”. Move the same globe to its position #4. You can find the shift change in sunrise and sunset after copmaring position #2 and #4 i.e.

Sun rises at “C” and sets at “D”in position #2 while sun rises at “D” and sets at “C” in position #4108.173.128.208 (talk)EEK
 * I'm not sure about your question but the picture is slightly wrong in that the sun is at one focus of the ellipse rather than at the centre of the ellipse. Also the earth moves fastest when it is closer to the sun so you can't split the ellipse into quarters of the time the way you do. See Kepler's laws of planetary motion for an illustration of this. Dmcq (talk) 09:19, 29 January 2013 (UTC)
 * I agree the question is probably about Sidereal time. The earth does not go around once on its axis in 24 hours but in slightly less time - if we used the stars as our guide rather than the sun and so had shorter days a year would take one extra day. Dmcq (talk) 10:15, 29 January 2013 (UTC)


 * I believe the OP's question is this: If at the summer solstice the sun rises at 5AM and sets at 7PM, then at the winter solstice why doesn't it rise at 7PM and set at 5AM, since we've gone halfway around the sun? And more generally, if at one point in Earth's orbit it rises at time A and sets at time B, then at the opposite point in its orbit (6 months later) why doesn't the sun rise at time B and set at time A?


 * To elaborate on Dmcq's answer, I think we can say that the questioner would be more or less right if 24 hours was how long it takes the Earth to rotate completely once relative to the stars, so if you held the time of day constant (by stopping Earth from spinning relative to the stars) and moved Earth halfway around its solar orbit, you'd find yourself at the opposite point in the day/night cycle (but not exactly so since the orbit around the sun is not quite perfectly circular). But actually that one rotation relative to the stars takes only 23 hours, 56 minutes, and 4 seconds, whereas 24 hours is how long it takes on average for us to go from noon one day to noon the next day (i.e., one rotation relative to the sun), so that kind of shifting of the day/night clock-time ranges is prevented from occurring. Duoduoduo (talk) 18:15, 29 January 2013 (UTC)

It will just take a few seconds to understand the the crux of the problem if one start looking at all aforesaid positions of earth from sun angle/ top view of diagram for its all mornings and evenings. (please forget about the star or observing from other angle for a while)

All aforementioned pegs/ points posted at A, B, C and D on earth are on or close to its equator. 12 hrs (almost) days and nights on or close to the equator is ok if earth doesn't moves in its elliptical orbit around the sun but since it does move around the sun besides its rotating about its own axis therefore next sunrise will be on point A+1 not on A, similarly A+2, A+3 ……. and so on till the sunrise reached to sunset on point B (position#3) and finally to its original point A (position#1) after completion of its one cycle (one year) around the sun. I may repost my question withh rephrasing if difficult to understand 108.173.128.208 (talk) 22:33, 29 January 2013 (UTC)EEK
 * Did you read the bit in the answers about that the earth rotates in 23 hours 56 minutes and 4 seconds approximately, not 24 hours? Dmcq (talk) 00:49, 30 January 2013 (UTC)

That 24hrs(approx) was just for simplicity! Since planets move fatser near the sun therefore such shift change can be observed more rapidely when earth moves close to the fuci /sun from one endpoint of latus rectum to other thru vertex in its ellipitical orbit. So in whch part of the year such rapid shift change (approx 12hrs) occur even if it completes its rotation about its axis earlier than 24hrs. — Preceding unsigned comment added by 108.173.128.208 (talk) 21:27, 30 January 2013 (UTC)
 * Try multiplying the difference between 23 hours 56 minutes and 4 seconds, that is 3 minutes and 56 seconds, by half a year's worth of days and see how much it add up to. Dmcq (talk) 21:31, 30 January 2013 (UTC)

why isn't drinking yourself to death a more popular suicide option?
It seems kind of a peaceful way to die if you get your BAC over 0.50% ...there's also injecting oneself with alcohol too. I however don't hear a lot of people taking this route, but overdosing on pills seems quite popular instead, which generally isn't lethal. (This is for curiosity only.) — Preceding unsigned comment added by 71.207.151.227 (talk) 07:03, 29 January 2013 (UTC)


 * Answering the "why" question is likely impossible. We can answer other questions, such as what methods are most common.  Suicide has some numbers.  -- Jayron  32  07:15, 29 January 2013 (UTC)


 * At least in my case, drinking myself to death wouldn't work. If I drink too much, I barf it back out.  I've never had a hangover, as a result.  This seems like a rather sensible reaction to alcohol, and, based on the number of people who die from the chronic effects of alcohol, I'm surprised this evolutionary adaptation isn't more widespread.  Perhaps the few thousand years we've been distilling alcohol isn't long enough.


 * As for injecting alcohol, it would have to be fairly dilute, or it would be extremely painful. And, being dilute, you'd have to inject quite a bit. StuRat (talk) 07:31, 29 January 2013 (UTC)


 * I think most suicide victims are looking for a technique with which they can quickly pass the point of no return. You could down a whole bottle of sleeping pills in seconds. Shooting yourself, or jumping would also be over in seconds, but drinking yourself to death would take a little while, and you could stop at any point.
 * Besides, You'd hate to accidentally survive with permanent brain damage. APL (talk) 07:55, 29 January 2013 (UTC)
 * Craig Ferguson actually talks a little bit about this in this clip from his show: . Alcohol actually, according to him, prevented him from committing suicide. Ks0stm  (T•C•G•E) 08:08, 29 January 2013 (UTC)


 * According to List of preventable causes of death, 1.9 million deaths a year are attributable to alcohol.--Shantavira|feed me 08:22, 29 January 2013 (UTC)
 * I don't think the majority of those are due to blood alcohol toxicity. Cirrhosis of the liver is not an effective method of suicide, and choking on one's own tongue or vomit is far too hit-and-miss. AlexTiefling (talk) 09:28, 29 January 2013 (UTC)
 * I can think of two good reasons, both because alcohol is like a very sloppy poison: before someone gets to the point of death from alcohol, they would either feel *much* better about things & decide not to go through with it (i'm sort of oversimplifying, of course); or they would pass out because the brain's survival mode would have kicked in -- unfortunatelyasphyxiation would then be possible but as AlexT says it's not guaranteed, especially since one's choking reflexes might be enough to temporarily wake up or turn to the side. El duderino (abides) 09:49, 29 January 2013 (UTC)
 * I think the usual direction if any is from alcohol to suicide rather than the other way round. Dmcq (talk) 10:47, 29 January 2013 (UTC)


 * I suspect there is a very good chance that most people wouldn't think of alcohol in that way...either they wouldn't realize that this is a possibility - or when thinking of the possible ways to terminate their lives, that one simply doesn't pop into their heads. SteveBaker (talk) 22:03, 29 January 2013 (UTC)


 * Most healthy people not on other drugs who die from alcohol poisoning die from consuming it much too quickly, often from some or total lack of experience. There are plenty of cases of teens and twenty-one somethings in the US who do too many shots or jello shooters too quickly to feel the full effects before they get to the point of no return.  The Callahan hazing in NJ in the 80's was a famous case, with him doing two dozen "kamikazes" in under a half hour.  Of course Amy Winehouse manaegd it.μηδείς (talk) 22:22, 29 January 2013 (UTC)

Genus outbreeding
Can animals of the same genus breed with any of each other? Could, for example, an arctic fox and red fox successfully breed with each other? (I am not very familiar with biology.) --66.190.69.246 (talk) 08:36, 29 January 2013 (UTC)
 * In the case of the arctic- and red foxes, they can interbreed, but their offspring is infertile (see this article, page 5 under the heading "Genetics"). Note that the arctic fox was previously placed in a different genus (Alopex), but is now included in Vulpes. There is no general rule about animals of the same genus. Sometimes it is possible, sometimes it isn't. In some cases the offspring is even fertile, and in that case the line between the species may become blurred. Keep in mind that biologists do not test all possible combinations of cross-breeding before classifying species. In the case of the dog and the wolf, they were formerly considered different species, but have since been renamed because have been shown not to be genetically isolated from each other (and they can interbreed without any limitations, see wolfdog). - Lindert (talk) 09:17, 29 January 2013 (UTC)


 * See Hybrid (biology). Duoduoduo (talk) 17:47, 29 January 2013 (UTC)

A species is defined as a group of animals that can breed with each other successfully -- "successfully" meaning that the offspring are fertile themselves. So if animals are not in the same species, they cannot breed with each other successfully. They may be able to produce offspring, but the offspring will not be fertile. The caveat to this, as Lindert says above, is that biologists usually don't actually test for interbreeding before assigning a species designation


 * Actually defining a species is a lot more complicated than that -- see Species problem. Duoduoduo (talk) 19:55, 29 January 2013 (UTC)
 * As Duoduoduo alludes, that definition of species is traditional, not current. It does point to the right answer to the original question, though, which is "mostly no." In general, animals of the same genus but different species won't interbreed, and if they do the offspring will usually not be fertile themselves.--50.196.55.165 (talk) 18:48, 1 February 2013 (UTC)

Aakash Institute
please show me result of Aakash Institute (ANTHE-2012) roll no. 36001133 — Preceding unsigned comment added by 164.100.194.115 (talk) 09:27, 29 January 2013 (UTC)
 * Question heading added. AndrewWTaylor (talk) 09:38, 29 January 2013 (UTC)
 * Go here, then enter the roll number and the letters shown to view the results. (I can't link to the results themselves). - Cucumber Mike (talk) 15:13, 29 January 2013 (UTC)

Planet or satellites with the direction revolution opposite to that of its rotation
Viewed from the (so called) 'top', Earth rotates anticlockwise on its axis and also orbits round the sun in the anticlockwise direction. Are there any planets or satellites in our solar system that rotate in one direction but orbit their primary in the opposite direction ? Is there any name for such motion? I am not talking about retrograde motion - WikiCheng | Talk 14:29, 29 January 2013 (UTC)
 * Venus does exactly that. Uranus' axis is at a 90 degree angle to the ecliptic. Several of the gas giant moons have retrograde rotation too. Fgf10 (talk) 14:44, 29 January 2013 (UTC)


 * Uranus axial tilt is actually close to 98 degrees which means that - like Venus - it has a retrograde rotation. Dauto (talk) 16:38, 29 January 2013 (UTC)
 * Well, if you measure the axial tilt as 82 degrees, the planet has retrograde rotation; and if you define by sign-convention that the planet always has positive rotation, then it must have an axial tilt of 98 degrees... but this is really purely semantics about where you should place the negative-sign. We know which direction the planet rotates; we just usually happen to define the reference-direction to be oriented towards the "top" of the solar system.  Here's my standard reference to de Pater and Lissauer, for the interested planetary scientist.  Nimur (talk) 20:05, 29 January 2013 (UTC)
 * What in the world are you talking about? The axial tilt (as usually defined) is about 98 degrees. Since that angle is larger than 90 degrees, to rotation is considered retrograde (as usually defined). There is only one negative sign and only one place for it to go (as things are usually defined). Dauto (talk) 21:18, 29 January 2013 (UTC)
 * The angle between two non-directional lines has an ambiguity of ±180°. If both lines have an orientation, we can represent them as vectors and calculate a dot-product.  In this case, we define the axial tilt as the angle between the planet's axis of revolution and the axis of rotation.  The orientation of each axis is defined such that it satisfies a right-hand rule for the angular momentum; and therefore the axial tilt is greater than 90 degrees.  The negative sign can be found in the equation of the dot-product by noting that A⋅B = AB cosθ = AB cos(-θ) = -ABcos(θ-180°).  So, if you have a rotation dΦ/dt at an axial tilt θ, that is equivalent to a rotation -dΦ/dt around a flipped axis with tilt θ-180°.  Nimur (talk) 22:02, 29 January 2013 (UTC)

Thanks! After reading your answer, I found that this has indeed been mentioned in the article retrograde motion !. Sorry for not reading it fully - WikiCheng | Talk 17:14, 29 January 2013 (UTC)


 * Interesting. I also thought that the phrase retrograde motion implied apparent retrograde motion.  Live and learn. -- ToE 17:57, 29 January 2013 (UTC)

Manometery equation
In a manometer, i have the reservoir pressure as 0.5 Bar, the height between the pipe and the datum line as 0.5m, an angle of 20 degrees, the relative density of the fluid in the pipe as 13.6.

From my understanding the pipe pressure when the fluid is at its datum line, should be the pressure in the reservoir minus the pressure in the liquid minus the pressure if the liquid.

I.e. 50000 - (800*9.81*0.5) which gives 46076 pascals but apparently the answer is 46.1 pascals. Where have I gone wrong? Clover345 (talk) 17:53, 29 January 2013 (UTC)


 * Are you sure that the answer wasn't given in kilopascals? -- ToE 18:00, 29 January 2013 (UTC)
 * I'm sure. Could it be a typo? Because I've checked all the units and calculations numerous times. Clover345 (talk) 18:42, 29 January 2013 (UTC)
 * Possibly. A pascal is a tiny unit of pressure, equivalent to one newton of force spread over an entire square meter of area.  46.1 pascals would be nearly imperceptible.  46.1 kilopascals is a scoche less than half an atmosphere.  -- Jayron  32  19:08, 29 January 2013 (UTC)


 * My suggestion was based solely on the similarity of the numbers; I can't generate either their or your values from your statement of the problem. Someone else here might see what is going on, but if you'd like me to check your work, you'll need to explain the problem more fully.
 * Do I understand correctly that your manometer's pipe is tilted at an angle of 20 degrees from the horizontal? Is the reservoir filled to the datum line?  Is your "height between the pipe and the datum line as 0.5m" referring to the height of the fluid level in the pipe above the datum line, and by height do you mean vertical height, in which case the 20 degree angle is irrelevant, or is it the distance up the pipe along the 20 degree angle from the height of the datum line?  Working backwards from your numbers, I couldn't figure out how you got the 800 from the density and the angle (the two values stated in the problem which do not appear elsewhere in your equation).
 * Without understanding the problem further I can only say that it is either a typo or a poorly posed problem, because if the answer truly is 50,000 - 49,953.9 = 46.1 then the values in the statement of the problem were given with far too few significant figures. Given that the mantissa of your answer matches that in the answer key, you are probably correct and I just don't understand the full statement of the problem. -- ToE 19:17, 29 January 2013 (UTC)
 * The other possible typo would be if the height of the liquid was 0.5mm rather than 0.5m. But I agree that it's most likely a typo in the answer that should have been in kilopascals...which is by far the more common unit for problems like this. SteveBaker (talk) 23:45, 29 January 2013 (UTC)
 * Without a picture or more context, it's impossible to figure out. Could be an inclined tube manometer, the 13.6 suggests mercury, but where the 20 degree angle comes in, or the 800, I don't know. Like ToE, I can't make 800 from 13600 and 20 degrees. Ssscienccce (talk) 14:55, 30 January 2013 (UTC)
 * Clover345 responded on their talk page that the relative density of the fluid should have been 0.8, not 13.6. Based on that, and if 0.5 m represents the vertical height of the top of the fluid in their tube over that in the reservoir, then their calculated 46.1 kPa is correct. -- ToE 05:42, 5 February 2013 (UTC)

No sound
What are the values of density and compressibility so that dispersal of the sound is worthless?--YanikB (talk) 19:11, 29 January 2013 (UTC)


 * The best way for you to understand any answer to this question is to read about common equations that are used to model sound propagation, like the ones in our article on sound-speed for ideal gas. See how many parameters there are in there?  See the ones that affect speed?  And the ones that create anisotropic effects?  And the ones that attenuate the signal?  There is no specific set of parameters that makes sound "worthless;" that's not even a well-formed description of sound; but for example, you could contrive a parameter space and draw out the ranges of parameters that you care about, and (for example) graphically represent the region of interest where you would expect a sound-wave could propagate over a unit-distance at a reference sound intensity.  You could look at the set of all input-parameters that satisfy that criteria.  This is what a physicist would call a configuration space.  This style of thinking, and the mathematical and visualization tools that accompany it,  helps physicists work with multidimensional problems (like the way various gas parameters affect sound-propagation in 3 dimensions).  Nimur (talk) 19:53, 29 January 2013 (UTC)
 * OK, but in practice. There is no sound on the moon. Right? Then at what altitude this appen? --YanikB (talk) 21:35, 29 January 2013 (UTC) — Preceding unsigned comment added by YanikB (talk • contribs) 21:34, 29 January 2013 (UTC)
 * It wouldn't be at any definite altitude - it would just gradually get less and less. SteveBaker (talk) 21:58, 29 January 2013 (UTC)
 * It's very approximately where the wavelength of the sound wave is the same as the mean free path of the gas molecules. Already when the mean free path is still somewhat smaller than the wavelength, the waves will be distorted. By the way, the same applies to high-frequency (> 1 GHz) sound at sea level. Icek (talk) 01:31, 30 January 2013 (UTC)
 * That can't be it either - it's the mean free path. Meaning the distance that a molecule can travel before hitting another one on average.  Below that pressure, there will be plenty of molecules that do hit another one within one wavelength - and above that pressure, there will be plenty that don't.  So this isn't any kind of a limit.  My previous answer holds.  There is no specific pressure/altitude at which sound disappears. SteveBaker (talk) 14:26, 30 January 2013 (UTC)
 * How can it be approximative. No body mesured it ?--YanikB (talk) 03:39, 30 January 2013 (UTC)
 * Would you prefer if we said "at 7 km altitude, sound becomes worthless"? We could equally well say that at 4 km altitude, sound is worthless, and at 400,000 km away from Earth's surface, sound is even more worthless.  Much of the sound produced at sea-level is also worthless.  This just isn't a useful way to describe that the effect varies continuously.  For any given altitude, there is attenuation and other non-ideal effects on sound-wave propagation.  There is not a specific altitude or pressure beyond which sound ceases to propagate.
 * But, because you want a specific answer, and because I am an enthusiast for engineering approximations, especially ones passed down from old fogeys whose experience pre-dates modern digital calculators, I use the easy rule of thumb that we lose one inch of pressure for each 1000 feet above sea-level. And because Earth's atmosphere at sea-level is conveniently at a pressure of ... one atmosphere, or approximately exactly thirty inches of mercury, that means that the "top of the atmosphere" is approximately exactly 30,000 feet, and if we could find some way to stand at that altitude, we'd expect to measure exactly zero inches pressure there.  Above this, "there is no air."  Of course, this is a ridiculous over-simplification of the way things really are; we can't apply the rule-of-thumb with any accuracy beyond a few thousand feet altitude; if anything, this is a perfect example of why we can't use overly-simplistic models, because they break down in the limit case.  Nonetheless, if you've ever popped open the door or window on an airliner as it cruised along at 30,000 feet, you might conclude that there's "very little" air outside; and chances are very good that if you screamed as you fell out of the jet, nobody would hear you.  So the engineering approximation works anyway, insofar as it predicts the first-order effect of the non-propagation of sound at 30,000 feet.
 * Now, as any atmospheric physicist worth their salt will tell you, there are a few distinct boundary altitudes - the tropopause and stratopause and mesopause - whose altitude varies on an hour-by-hour basis - corresponding to some abrupt change in one or more properties of the atmosphere. But, it would be disingenuous to imply that these boundaries represent dividing lines between regions where sound does- and does-not propagate.  Nimur (talk) 06:18, 30 January 2013 (UTC)
 * The actual pressure at 30,000 feet is about 30 kPa, 30% of that at sea level. The Extravehicular Mobility Unit spaces suits used on the ISS are pressurized to that same level during space walks, and sound seems to carry well enough to their microphones that the astronauts' voices don't sound even distorted to my ear.  This would suggest that the above rule of thumb breaks down early enough that it does not inform the original question. -- ToE 09:53, 30 January 2013 (UTC)
 * What Nimur said about the question not having much meaning, because in this use the word "worthless" is indefinable, is never the less, true. I suggest that the question cannot be answered precisely, and so has been over-answered with detail the OP almost certainly does not want.  In terms of a human utilising sound, that peters out at relatively low altitude for various reasons.  Sound clearly does propagate well at 30,000 feet, because you can hear the engines of cruising airliners at ground level.  Physicists theorise that acoustic shock waves can travel in deep space (it is not a perfect vacuum), but clearly that is of no practical use.  Personally, I think the sound emitted by punk rock bands is worthless - others may disagree.  It is up to the OP to define "worthless".  Conversation not possible?? The height at which a rocket engine become inaudible to observers on the ground??  Wickwack 120.145.32.123 (talk) 13:51, 30 January 2013 (UTC)
 * Yes, exactly. For example, this article talks about measuring the shape of the universe by measuring sound waves propagating from the Big Bang in near (but not total) vacuum. The "mean free path" argument is circumvented by the extremely low frequency of the sound waves. SteveBaker (talk) 16:27, 30 January 2013 (UTC)

music and heat
I notice  when Im listening to very loud music that I particularly enjoy my  body temperature seems to rise  to the point of sweat. this even happens during the colder winter months. Could the sound energy be potentially creating the heat energy. or is it more of a biological nature. --86.45.153.90 (talk) 21:05, 29 January 2013 (UTC)


 * It's biological. The amount of actual energy carried by sound waves is very, very small unless the sound is beyond deafening.  Dragons flight (talk) 21:17, 29 January 2013 (UTC)


 * Agreed, and we can put some numbers on that. A speaker system might have maybe 20 watts per channel, for 40 watts total, while a typical space heater is about 1500 watts. StuRat (talk) 23:01, 29 January 2013 (UTC)


 * Also, a "40 watt" speaker system typically can't go above 20 watts before distorting, and that's the reputable ones. The makers of cheaper systems tell bigger lies. When playing music, even a true 40 watt system typically doesn't send more than a watt of average power to the speaker. (compare how hot a 40 watt speaker gets compared to a 40 watt light bulb in the same enclosure) Far less than that makes it to your body. --Guy Macon (talk) 07:22, 30 January 2013 (UTC)


 * Good point, although the waste heat will eventually warm the room somewhat and thus heat your body, if you remain there long enough, although some of it will also escape the room, and might affect the thermostat, if either a furnace or A/C is on. But these are all quite minor effects, to be sure. StuRat (talk) 19:25, 30 January 2013 (UTC)


 * If it's only music that you enjoy that does that, we can certainly eliminate all sources of heat except for the psychologically driven sources within your body. Much of the body's heat comes from organs like the heart and brain - so if really good music gets your mind racing and your heart beating faster - that might maybe do it.  I doubt that your body temperature is actually rising though - the whole point of the sweating is to prevent that from happening. SteveBaker (talk) 23:42, 29 January 2013 (UTC)


 * I don't doubt that you enjoy the music per se. But maybe you actually intensely dislike the volume at which you're playing it, but you've become so used to playing and hearing music at such absurd volumes that it never occurs to your conscious mind to turn it down a notch or ten, but your subsconscious mind knows exactly what's ideal for you and it is rebelling for all its worth and is sending you a message.  Maybe.  --   Jack of Oz   [Talk]  23:55, 29 January 2013 (UTC)


 * Well, unless the OP is playing enjoyable music at a much louder volume than less enjoyable stuff - this doesn't explain the symptoms described. SteveBaker (talk) 14:20, 30 January 2013 (UTC)


 * I reckon he's really getting into the groove, increasing adrenalin levels, and unconsciously giving it the 'ole dad-rock vigorous head nodding - such movements would generate heat in the muscles and making him sweat? That's what i reckon nonsense  ferret  15:15, 30 January 2013 (UTC)


 * Thanks for posing an interesting physiological question. There have been deaths at raves from elevated body temperature, and the obvious culprit is MDMA, but could the body's physical/reflex reaction to the sound have an influence on that?  Hmmmm.... Wnt (talk) 17:55, 31 January 2013 (UTC)  ... wasn't sure how I'd search for this - but at the first poke I stumbled on a smart kid's grade school report  which points at a reference Campbell 71 (alas, no expansion) for music elevating body temperature.  Which put me on the track of Don Campbell, The Mozart Effect and  which cites Savan, A. (1999). The effect of background music on learning. Psychology of Music, 27, 138-146.  for music decreasing body temperature.  They wave their hands in the general direction of the limbic system - go figure.  Well, finally I did the smart thing and searched NCBI - turns out, despite my initial skepticism, there's serious research about this indexed there!  I get  which supports a very small decrease in association with pleasurable excerpts.   also says something about it (any Norwegian speaking editors near a subscribing library?)  I suspect there's some distinction between the effect of very loud music and other music, which accounts for the degrees of increase versus 0.087 degree decrease thing, but I'm not quite on the mark yet... Wnt (talk) 18:08, 31 January 2013 (UTC)