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June 6

To charge a phone on a bicycle

Hi, is there any gadget I could get anywhere online (or preferably off) that will charge my phone while I pedal my bike? I remember bike odometers that hooked a little wheel to one of the tires (therefore, called "flywheels?") in order to spin the numbers. Could that same small wheel utilize the spinning of the tires to recharge my phone?

If so, where is a device that'll do exactly that? I would hope to find one before a long bike-ride. Thanks. --70.179.165.67 (talk) 04:29, 6 June 2011 (UTC)[reply]

Though someone smarter than I am on this matter will be able to give you the specifics, I recall that the amount of pedaling on a bike to produce any worthwhile amount of electricity makes the exercise impractical. So even if you were going to ride a marathon's length, there probably isn't a market for selling devices that charge cellphones this way, so you're out of luck unless you want to build your own contraption.--el Aprel (^facta-_facienda) 05:37, 6 June 2011 (UTC)[reply]

While the amount of electricity generated by a bicycle is low, and thus insufficient to heat, air condition, or light your home, or run a major appliance, it is sufficient for devices with minimal power requirements, like cell phones. However, the cell phone charger will need to be improved, as current chargers are rather inefficient (you can tell because they get hot). StuRat (talk) 05:59, 6 June 2011 (UTC)[reply]

Are you talking about the AC adapter? If so how old is your phone? All phones I've seen in the past 5 or so years, even very cheap ones come with SMPS power adapters (sometimes labelled as travel adapters). You can easily tell because of the weight (and to a lesser extent size and shape). Cheap electronics from China from eBay, DealExtreme etc also always come with (sometimes poorly made) SMPS. Various sources including our article say this has happened world wide, perhaps because of the decreasing cost of the electronics needed for an SMPS and the rising cost of iron (although the iron bit seems to have been removed from our article). Regulation or government pressure on power supply efficiency may have also played a part. Regardless, SMPS power adapter doesn't tend to get very hot. In any case it's a moot point since it's unlikely you're going to output 110 or 220V AC from your dynamo. Nil Einne (talk) 06:27, 6 June 2011 (UTC)[reply]

Yes, but inefficiency is possible with any charging system. And, unlike with the wall plug, there isn't much energy to spare in this setup, so efficiency would be key. StuRat (talk) 06:39, 6 June 2011 (UTC)[reply]

So? None of that relates to your original (generally false nowadays) claim about how inefficient current chargers are based apparently on some ancient phone with a linear/transformer based AC adapter. I suspect many dynamos output DC in the 4-12V range anyway since modern lighting systems tend to use that and ones with batteries often even use li-ion nowadays. And any slightly decent DC-DC adapter would generally be at least 50% efficient with something like 70-80% more likely and even 80-95% possible (at least from a more constant source, not sure how well they handle a more variable output you may get from a dynamo although again since dynamos for bikes tend to be targeted towards lighting systems it seems likely this is already something handled resonably well). The phone itself and the actually battery also has some loses but there's not much you can do about that.

Really you're approaching this from totally the wrong angle, sure the efficiency of the charging system and power adapter matters, but you don't seem to be considering the dynamo itself. For example [1] shows 60% but only for the best hub dynamos (which was where I found the link). 30% is more likely particularly for bottle dynamos and we're still talking about name brand dynamos. [2] from hub dynamos (where I found all the links) I think shows shows something similar. [3] ditto I think although you have to manually work it out. And this is at 15-20kmh, at high speeds you efficiency generally drops particularly in the crappier dynamos. Unless you plan to be charging your phone all the time, the 'daytime resistance' would likely matter (although on the flipside for a bottle dynamo you can just disengage it).

In other words, unless you're using a good hub dynamo like a Schmidt Original Nabendynamo, power loses in the DC conversion and charging are perhaps not your biggest concern. (Not saying they are not a concern but you obviously should be considering whether you want a better dynamo first.)

P.S. Worth remembering that given the 4.2V most li-ion batteries chargers or older mobile phones use and the 6V that our hub dynamo article and most of the refs above seem to say a typical dynamo for light systems outputs even a simple linear regulator just burning off the excess would achieve 70% efficiency. Of course nowadays phones are moving to microUSB for charging meaning they accept or require 5V so you get 83.3% with a linear regulator (the phone would have to do some conversion internally but it's likely most phones are fairly efficient).

Nil Einne (talk) 10:36, 6 June 2011 (UTC)[reply]

By "charging system", I meant to include the efficiency of the dynamo, as well. If this isn't the term you use, how do you collectively refer to it all ? StuRat (talk) 22:45, 7 June 2011 (UTC)[reply]

Many bicycle lights are powered by either rim or within-hub dynamos. You would need to build or commission a device that allowed this power source to charge or power a phone. This will increase the friction and thus effort needed to cycle. Fifelfoo (talk) 05:48, 6 June 2011 (UTC)[reply]

Another option might be regenerative braking, where you would pull more power from the wheel for charging purposes, but only when braking. This would be a big seller, if it worked. Obviously, such a system would provide more electricity in "stop and go" driving. One simple way to do this might be with the same dynamo, but where it had a user-operated switch to engage it or disengage (perhaps tied to the gear setting). This might work well in hilly areas, where excess speed could be bled off going downhill, but without the unwanted extra drag going uphill. StuRat (talk) 06:03, 6 June 2011 (UTC)[reply]

A simple search for 'bike dynamo phone charger' easily finds options both DIY and commercial [4] [5] [6] [7] [8] [9] [10]. So does 'bike phone charger' (mostly finding the Nokia device but also a few others) [11] [12] [13] [14] [15] [16] [17] [18] [19]. Other then specialist or biking companies evidentially Nokia and Motorola make devices too. Nil Einne (talk) 06:15, 6 June 2011 (UTC)[reply]

As an alternative, buy one of these traditional mobile phones (as opposed to these fancy new smartphones with large touchscreen), because they run much longer without recharging (despite that they typically have smaller batteries). – b_jonas 09:03, 6 June 2011 (UTC)[reply]

This answer is about as good as saying "Install Linux" when someone asks how to do something in Windows! APL (talk) 09:51, 6 June 2011 (UTC)[reply]

Another alternative would be simply to get yourself a solar charger..--Shantavira|^{feed me} 09:13, 6 June 2011 (UTC)[reply]

Unfortunately, Very few of those are worth the effort. In most cases you'd be better off economically and perhaps even ecologically just buying ones of those AA-to-phone adapters and popping in some AAs whenever you're away from line power. Problem is that solar panels with any sort of useful efficiency are very expensive, so the ones most of those "Solar battery chargers" use are pretty terrible. (Typically these devices are just rechargeable batteries that you charge from line-current, and the solar panel is mostly for show.)

You'll certainly never get enough juice from the sun with one of those devices to justify the cost, or the ecological impact of its own manufacture.

I've heard good things about the Voltaic backpacks, but even at it's theoretical peak output it's not as powerful as my phone charger. And you figure that unless you're pointing it right at the sun you won't get half that. (You don't normally keep your back to the sun while biking. You have to go where the roads go.) So it's still not really worth the effort unless you're honestly going to be away from line current for a while and for some reason can't carry AA batteries. APL (talk) 09:51, 6 June 2011 (UTC)[reply]

So, what would happen if you were to wire your bike lights dynamo to a spare car cigarette lighter socket (I think you can buy these from places like Halfords). You could then charge your phone from that using a car adapter/charger suitable for your phone? Astronaut (talk) 13:57, 6 June 2011 (UTC)[reply]

This is nitpicking, but the question says "on a bicycle", not "powered by pedalling a bicycle". When I go camping at music festivals I take a small 12V motorcycle battery and use that with the car cigarette socket charger to charge my phone. You could easily fit such a battery on a bike carrier rack - and charge the motorcycle battery up at home every few weeks. -- SGBailey (talk) 15:00, 6 June 2011 (UTC)[reply]

Earthquake tilting Earth axis

Originally asked over on the help desk [20] I've copied it here Chzz ► 12:14, 6 June 2011 (UTC)[reply]

I have been trying to find out more about the Earthquake in Japan. I read that it caused the earth to move 10 inches on it's axis. Is this true and are we tilted more or less on our axis. I have ask our TV stations abiut this and can't get an answer. I'm 71 years old and I took Astronomy Magizine for about 50 years but had to give it up because of my eyesight, it was just too hard to read. So if you can answer my question or tell me who to ask iI would appriciate it. (Redacted) — Preceding unsigned comment added by 75.221.228.71 (talk) 06:01, 5 June 2011 (UTC)[reply]

It was actually 10 cms. Please see our article on 2011 Tōhoku earthquake and tsunami.--Shantavira|^{feed me} 12:20, 6 June 2011 (UTC)[reply]

Hm, but in 2011 Tōhoku earthquake and tsunami#Geophysical impacts it says 25 cm (9.8 in), with this ref. And this one says 6.5". Chzz ► 12:28, 6 June 2011 (UTC)[reply]

Indeed this type of thing highlights the flaws with supposedly referenced information. The New York Times article that is supposedly a reference for the 10 cm figure in fact says "Dr. Gross said his calculations indicated a shift of 6.5 inches in where the figure axis intersects the surface of the planet". The other ref for that sentence gives the 10 cm figure. Regardless, all these articles are from fairly soon after the quake, but it seems that the closer the date of the article to the actual quake, the bigger the figure given, suggesting a bit of early sensationalism or hyberbole. Wonder if anyone can find a genuinely good source from a month or more after the quake that would likely be more reliable. --jjron (talk) 13:30, 6 June 2011 (UTC)[reply]

Also worth pointin out that the tilt didn't change. What changed was the position of the axis on earth's surface. Dauto (talk) 17:35, 6 June 2011 (UTC)[reply]

Is it more correct that the earth's surface changed relative to the axis? Getting back to the original issue though, it would really be useful to see this as a before/after two sets of axes on the same globe. Even if the actual number could be verified, it still isn't clear to me whether the earth tilted more/less, or the solid vs axis precessed a bit. Spheres have lots of "axis" directions. DMacks (talk) 18:27, 6 June 2011 (UTC)[reply]

Changing the tilt of the earth's axis would imply a change in angular momentum which is not possible without an external torque. Dauto (talk) 19:44, 6 June 2011 (UTC)[reply]

Cats can rotate themselves in mid-air with no externally applied torque (Falling cat problem). DMacks (talk) 22:43, 6 June 2011 (UTC)[reply]

Yes, and they do that by changing the position of their bodies around a rotation axis which remains fixed because of the already pointed out conservation of angular momentum, which is what I was pointing out. Dauto (talk) 15:39, 7 June 2011 (UTC)[reply]

I'm not so sure about that. Picture you have a rotating cylinder with heavy iron balls clamped to each of four sides, which are also attached by dangling chains. You spin up the cylinder and suddenly one of the balls gets loose and starts swinging around several feet from it. Well that cylinder is going to start rotating around a new axis. (Something similar can be done with an ultracentrifuge when one or more of the sample tubes break, but there the rotating mass changes) In this case, part of Japan moved down, and so the axis should be a little further "down" from that spot also. The angular momentum and thus the axis of rotation may be fixed, but only relative to the total rotation, not relative to the planet surface. Wnt (talk) 18:02, 7 June 2011 (UTC)[reply]

No it doesn't. The cylinder/ball assembly would continue to rotate around the same axis, since all you did was change the moment of inertia of the body. The ball swings out due to the lack of a centripetal force, and then air resistance begins applying an external torque due to a moved center of pressure. In this case, parts of Japan moved down, so the moment of inertia shrunk, producing a faster rate of rotation (similar to the often-used spinning ballerina) due to conservation of angular momentum. Titoxd^{(?!? - cool stuff)} 08:55, 10 June 2011 (UTC)[reply]

Hmmm, I wasn't clear there. I mean, the axis of rotation is fixed relative to the fixed stars, but it is not fixed relative to the Earth's surface. If a big enough chunk of crust around Japan drops by three feet, it's possible that the North Pole could move by a few inches. Wnt (talk) 17:20, 10 June 2011 (UTC)[reply]

So did it? How much? Sources seem to vary. Chzz ► 01:28, 11 June 2011 (UTC)[reply]

What are the materials of Polyethylene terephthalate

Does it come from Petroleum?--Inspector (talk) 12:55, 6 June 2011 (UTC)[reply]

Yes. If you read down to the Recycling section of the article you linked (and the information should perhaps appear more prominently), you will find the statement

". . . recycling back to the initial raw materials purified terephthalic acid (PTA) or dimethyl terephthalate (DMT) and ethylene glycol (EG) where the polymer structure is destroyed completely . . . ."

If you then read the articles about those ingredients, linking through where necessary to articles about their production, you will find that they derive from petroleum. {The poster formerly known as 87.81.230.195} 90.201.110.217 (talk) 13:14, 6 June 2011 (UTC)[reply]

Most PET is probably made from oil and/or natural gas, as are other thermoplastics, but as our polyethylene article notes, it is possible to make such materials from sugar cane and similar cultivated products. In addition, the level of recycling in PET is relatively high compared to other materials. AndyTheGrump (talk) 13:17, 6 June 2011 (UTC)[reply]

Checking on the PET article talk page, I see that PepsiCo are apparently working on making PET bottles solely from plant materials: [21]. How 'green' this actually is may be open to question though. AndyTheGrump (talk) 13:28, 6 June 2011 (UTC)[reply]

So, How many steps are there between petroleum and PET?--Inspector (talk) 13:41, 6 June 2011 (UTC)[reply]

[22] might be a good read. --Stone (talk) 21:12, 6 June 2011 (UTC)[reply]

Global Warming – Why Predominantly Negative Effects?

It seems that global warming produces predominantly negative climatic effects, i.e. ones that are inconvenient to humans. Why is this? Would a cooler climate make for milder weather? I find it hard to believe that we've found ourselves in such a climatic sweet-spot — especially when you consider that average temperature has been continuously drifting one way or another throughout history. I was also wondering if there is any geographic location that might actually benefit from global warming.

Thanks!

Alfonse Stompanato (talk) 14:59, 6 June 2011 (UTC)[reply]

We have not "found ourselves in such a climatic sweet spot" by accident, we have adapted to it. The climate has actually been remarkably stable for the last 10000 or so years - i.e. for the whole time of human civilisation. We have adapted our agriculture to work well with the combination of climate, soil, and seeds known to us. We have grown large populations where a reliable water supply is available year round by rivers that are buffered by glaciers fed by winter snow that melt in the summer. We build our cities on the current shoreline for both commerce and fishing. The primary problem is not that different global temperatures are inherently better or worse (though they may well be, depending on what criteria you use), the problem is that we experience changes to the status quo at record speeds. On a larger scale, the same applies not only to humans, but to whole ecosystems. Given enough time, both humans and ecosystems will be able to adapt to a new steady state. But we are a long way from reaching a new steady state, and the disruption will be predominantly negative, simply because our lifestyle is optimised to current conditions. There are some other concerns, primarily that a higher temperature means more energy in the climate system, which likely leads to more or stronger extreme weather events. As for benefits, we may be able to reliably use the Northern Sea Route or the Northwest Passage for shipping. --Stephan Schulz (talk) 15:25, 6 June 2011 (UTC)[reply]

(after ec)

To answer your last, "benefit" is a subjective term. No doubt the colder regions would benefit from being made warmer, if you're a human being who likes warm conditions. But if you're a capercaillie who has evolved to fit one particular niche, warmer conditions would spell the end for your species in that area. --TammyMoet (talk) 15:28, 6 June 2011 (UTC)[reply]

Another point is that we aren't necessarily in a "sweet spot". A few thousand years ago much of the Sahara desert was grassland, and in other parts of the world many areas supported people that are now flooded by rising sea level. Looie496 (talk) 16:09, 6 June 2011 (UTC)[reply]

The life around us now has evolved for the last couple of million years to live at about the current temperature or a bit lower as during the ice ages. Dmcq (talk) 16:57, 6 June 2011 (UTC)[reply]

I have to admit, I can't shake off the feeling that the Ice Ages that mark our current geologic era are in some way "pathological". If there is anything to the Gaia hypothesis then it would appear that Gaia noticed that the environment was constantly changing for some reason and decided to put some thought into how to dig up the old carbon and get it back out into the atmosphere... While I recognize, of course, that the very rapid warming caused by humans has harmful effects on the ecology, the world has been through such drastic changes quite a bit over the past few million years. So I'm thinking that while controlling carbon emissions is necessary, I wouldn't want to see atmospheric CO2 rolled back to 1600 or even to 1950. I think this position is implicit in climate plans that speak of partial (even 80%) reduction in CO2 emissions, but I haven't really seen the ideal endpoint seriously discussed. Wnt (talk) 21:42, 7 June 2011 (UTC)[reply]

See Holocene, Anthropocene, Holocene Climatic Optimum, Little Ice Age, Climate change and agriculture and current sea level rise. Several degrees Celcius of temperature rise would offset any benefits from CO₂ increases, while the greatest human benefits could be seen in previously subarctic areas that become arable land, but only after the permafrost problematicly melts. ~AH1 ^(discuss!) 17:07, 11 June 2011 (UTC)[reply]

Is it right that.. (Talking birds)

All bird kinds in the world, could imitate human Speech Just like the Parrots, Myna birds, and others do so?...

is it right that the only "barricade" from all other bird kinds to do so, is just Neural? (Their pro-Larynx Areas in their nervous system).

have i understand correct?

Best blessings. — Preceding unsigned comment added by 109.67.42.106 (talk) 15:35, 6 June 2011 (UTC)[reply]

No, not all birds. Only birds from three groups, the Myna and the relatives of the Parrots, the Crows are known to mimic human speech. These birds are relatively intelligent, and have vocal tracts capable of imitating speech. See talking birds μηδείς (talk) 15:51, 6 June 2011 (UTC)[reply]

The question is whether more birds than just these varieties have vocal tracts capable of imitating speech, even if the birds themselves don't actually do so. The article does not address that (or much else). --Mr.98 (talk) 16:29, 6 June 2011 (UTC)[reply]

I now understand that it's also because of this 3 groups of Birds - unique Larynxal Anatomy (off course, in addition to their neural capability).

Just to ensure i understand correct,

blessings 109.67.42.106 (talk) 17:00, 6 June 2011 (UTC)[reply]

Haven't you asked this already up there at #A_beatiful_question_about_Parrots_and_their_voice_producing_system..? – b_jonas 18:17, 6 June 2011 (UTC)[reply]

He'd better be careful, or he'll be charged with contributing to the deliquency of a Myna. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:43, 6 June 2011 (UTC)[reply]

Just for the record the adjective from larynx is usually laryngeal Richard Avery (talk) 07:14, 7 June 2011 (UTC)[reply]

Note also birds use a Syrinx (bird anatomy). Wnt (talk) 21:34, 7 June 2011 (UTC)[reply]

And all these years I've been saying "larynctive"! μηδείς (talk) 17:56, 8 June 2011 (UTC)[reply]

radiation from magnets

Since photons are the force carrier for electromagnetism, do magnets give of any form of radiation from the electromagnetic spectrum? Bugboy52.4 ¦ =-= 17:25, 6 June 2011 (UTC)[reply]

If you hold a magnet in your hand and make fast movements, it will radiate electromagnetic radiation, albeit it not very much. Count Iblis (talk) 17:38, 6 June 2011 (UTC)[reply]

(ec) A static magnetic field does not produce any radiation, but a changing magnetic field does -- this is a consequence of Faraday's law of induction. Looie496 (talk) 17:41, 6 June 2011 (UTC)[reply]

Is the frequency constant, or is does it change? Bugboy52.4 ¦ =-= 18:15, 6 June 2011 (UTC)[reply]

The frequency radiated will be the same frequency of the shaking movements you apply on the magnet. Dauto (talk) 19:39, 6 June 2011 (UTC)[reply]

That means 4.3 x 10^13 (43,000,000,000,000) vibrations per second for red light. μηδείς (talk) 20:05, 6 June 2011 (UTC)[reply]

So it is possible to emit light in the visible spectrum? Bugboy52.4 ¦ =-= 19:47, 6 June 2011 (UTC)[reply]

Not with your handheld magnet, your hand does not shake that fast. What you ahve here are virtual photons. Graeme Bartlett (talk) 20:52, 6 June 2011 (UTC)[reply]

I'm not sure what you mean. Radiated photons are real photons, regardless of frequency. -- BenRG (talk) 22:52, 6 June 2011 (UTC)[reply]

EM radiation from a rapidly vibrating permanent magnet should be indistinguishable from that generated by , say an Alexanderson alternator(early 20th century radio transmitter via rotating generator) or from a solid state radio transmitter coupled to an antenna. except for the difficulty of achieving physical vibration of a permanent magnet at frequencies beyond a few tens of Hz. Edison (talk) 02:31, 7 June 2011 (UTC)[reply]

(Edited to add) I had intended to say "10's of kilohertz," imagining a small permanent magnet bolted to a loudspeaker or an ultrasonic transducer. A permanent magnet could also be bolted to a vibrating rod or wire in tension, for instance, which could easily be tuned to vibrate at frequencies above the audible range, which would in fact be in the low frequency radio frequency range. A magnet held in the hand cold be vibrated manually at 5 Hz or so. This low frequency EM wave could certainly be picked up by a suitable nearby antenna or pickup coil. I've tried the experiment, and viewed the somewhat distorted waveform on an oscilloscope. I agree that the transmitted energy would be negligible. Remind me how EM waves work: would a rapidly vibrating magnet moving left and right produce a transverse electrical field moving up and down, both fields moving away from the origin? If I took an electrically charged object and caused it to vibrate left and right similarly, would it produce a transverse magnetic wave moving up and down? Surely electromagnetic waves are not restricted to electronic oscillators.Edison (talk) 05:16, 8 June 2011 (UTC)[reply]

Our piezoelectricity article talks about speed and size of deformation, but not general commentary on frequency limits to its oscillation. However, one of the noted applications mentions 100 kHz–1 MHz. Put a lightweight magnet on it, and you're well into the AM radio range. DMacks (talk) 03:15, 7 June 2011 (UTC)[reply]

With a handheld magnet, you can only move it at frequencies up to 10Hz. Radiation efficiency from this would be extremely low with less than picowatts of power transmitted as traveling electromagnetic radiation. Almost all the energy would be in the near field which would extend over 1000 km. This would be dissipated by electric currents flowing in the earth or the ionosphere and so just about nothing would escape into a radiowave far field. Even at the AM frequencies mentioned by DMacks, the antennas are dozens of meters high to get any significant power out. So a small vibrating magnet would still not do much. I am probably addressing Count Iblis q rather than bugboy52.40. For the original question the virtual photons answer still stands. The size of these photons is enormous, thousands of kilometers, so they are not like your normal particle! Graeme Bartlett (talk) 11:26, 7 June 2011 (UTC)[reply]

What? Radiated photons are always real, and photons are always point particles, and these photons are most certainly 'ordinary photons' in the sense that no other kind of photons exist. The wavelength of a photon is not a measure of it's physical size! — Preceding unsigned comment added by 129.67.37.190 (talk) 08:28, 8 June 2011 (UTC)[reply]

Yes, but I think Graeme's point is that an oscillating magnetic field which then leads to currents being induced should not be thought of as involving real photons. Suppose I have a magnet mounted on a spring that has an angular frequency of omega and next to it, I have a superconducting LC-circuit also with angular frequency of omega. Then both the spring and the LC-circuit have energy levels of

(n+1/2) hbar omega. The spring can jump to a lower energy eigenstate while the LC-circuit moves to a higher level. This process then doesn't involve excited states of the electromagnetic field (i.e. real photons). That process also can occur, but with much smaller probabilty. But in that case, most likely the spring would lose one quantum, while the photon simply moves away, it most likely won't interact with the coil to excite the LC-circuit. Count Iblis (talk) 16:14, 8 June 2011 (UTC)[reply]

Brightness of R136a1 when it goes hypernova

The article says it will have the luminosity of 100 typical supernovas. The brightness of supernova 1987A was magnitude +3, and its luminosity was about 1/10 of that of a typical supernova. So, does this mean that R136a1 when it explodes will become magnitude -4.5, making it barely visible at daytime? Count Iblis (talk) 17:44, 6 June 2011 (UTC)[reply]

They are about the same distance away from us, so yes. If those numbers are correct, then it would be about -4.5. --Tango (talk) 20:53, 6 June 2011 (UTC)[reply]

excited states / energy level diagram of nitrogen

If I look at the grotrian diagram of nitrogen in my book, I don't see a state 2²S_1/2, and the book says that it is not possible because of the Pauli principle. But if I assign the 3 valence electrons as follows, I don't see why this is wrong, where is my error:

n₁=2, l₁=1, m_l₁=-1, m_s₁=1/2
n₁=2, l₁=1, m_l₁=0, m_s₁=1/2
n₁=2, l₁=1, m_l₁=1, m_s₁=-1/2

So the resulting atom should be S=1/2, L=0, J=1/2. ??

best thanks --helohe (talk) 21:01, 6 June 2011 (UTC)[reply]

Wow. It's been many years since I studied this stuff myself, but there are ideas about Selection rules like the Laporte rule knocking around in my head, though I am not fully sure how to make it work for you. Remember that monatomic nitrogen is a radical, which may introduce some weird symmetry problems and degeneracies, akin to what you see in odd-electron molecules and ligand complexes, vis-a-vis the Jahn–Teller effect. I've forgotten most of this stuff, and how it works, but that may give you some leads... --Jayron32 05:48, 7 June 2011 (UTC)[reply]

Laporte rule might work but I still don't see it. Jahn–Teller effect seems a bit too advanced, as I'm currently only studying single atoms and this is for molecules. I guess I'm doing something wrong with the LS-coupling (Russell Saunders coupling scheme). But still thanks a lot for your answer :) --helohe (talk) 22:51, 8 June 2011 (UTC)[reply]

Desert basins

I was wondering what is distinguishable about desert basins which set them apart from deserts themselves. The article Kalahari Basin doesn't really help to explain what physical features separate it from the Kalahari itself.— Preceding unsigned comment added by Flaming Ferrari (talk • contribs) 21:08, 6 June 2011

I think the name Kalahari Basin refers to a drainage basin. The Kalahari Basin is an endorheic basin, which means that water flowing there does not reach the ocean. The same is true for the Great Basin in North America. In other contexts related to landforms, the word "basin" may mean something else; see Basin#Landforms. —Bkell (talk) 22:42, 6 June 2011 (UTC)[reply]

(Edit Conflict) So far as I understand it, there is no inherent connection between basins - large landform depressions of various sorts, and deserts - large areas with low precipitation: you can have either one without the other. However, if a basin happens to be in a desert, it's called a desert basin, and is likely to have certain features caused by the combination of shape and climate, namely seasonal water courses, salt lakes or salt flats (dried-up lakes), and perhaps subterranean water if the underlying geology is right. The Kalahari is a slightly different case in that the basin is larger than the desert, and the desert happens to be entirely contained within it; the Great Basin Desert in the USA is similar in this respect. Conversely, the Gobi Desert contains several basins. {The poster formerly known as 87.81.230.195} 90.201.110.217 (talk) 22:43, 6 June 2011 (UTC)[reply]

Well, as said in endorheic basin, they tend to be deserts because otherwise the water would break a channel flowing out. A desert is part of a system of classification of land that has a net moisture deficit. There's probably some slop here, I think, in that there are arid regions not classified as deserts that have a net moisture deficit, and lands with a net moisture deficit can still periodically overflow and erode a path to the ocean. Wnt (talk) 18:09, 7 June 2011 (UTC)[reply]

Not all desert endorheic basins are dry, however. The Caspian Depression is below sea level and has a huge inland sea. ~AH1 ^(discuss!) 17:01, 11 June 2011 (UTC)[reply]

Rotational physics

Hello all. This is a problem that I made up but can't solve :( Imagine an unevenly weighted barbell, say with 20 kg on one end and 10 kg on the other, dropped from an airplane. Intuitively I know that if it is allowed to fall for long enough, the heavier end will be pointed towards the ground. However how do I prove this? I know that I have to use torques but I'm not sure how. Thanks for any help. 72.128.95.0 (talk) 21:08, 6 June 2011 (UTC)[reply]

The center of mass will be approximately one third of the bar length away from the 20 kg object. This is where the weight of the entire barbell can be considered to be acting. As the speed of the barbell increases there will be a significant drag force acting on it. The drag on the 20 kg object will be greater than the drag on the 10 kg object, but probably not twice as much so the drag force will not act at the center of mass. As a result, there will be a torque on the barbell caused by the weight (acting downwards) and the drag (acting upwards) and the fact that the two forces are not acting at the same point. This torque will cause the barbell to rotate so that the end with the 20 kg object is pointing downwards, and the end with the 10 kg object is pointing upwards. Dolphin (t) 22:26, 6 June 2011 (UTC)[reply]

(ec) There is also an unstable equilibrium if the barbell is exactly vertically oriented, with the heavier weight on the upper side. You can formally solve for the stable orientations by construction torque as a function of rotation angles τ = τ(φ,θ), and solve for zeros of the torque function by setting its gradient equal to zero; and to determine stability, solve for the concavity of the function. Nimur (talk) 22:35, 6 June 2011 (UTC)[reply]

Here's a related question: I expect that when you drop it, there is liable to be a little bit of spin, which is going to cause it to rotate or oscillate as it falls (connected with the principle you've just stated), and eventually the heavier end will be pointing downward. My question is: If it were dropped lighter-end first, and with such machine-like precision that it had no spin, wouldn't the light end of it hit the ground first? ←Baseball Bugs ^{What's up, Doc?} carrots→ 22:33, 6 June 2011 (UTC)[reply]

If it were dropped with the lighter-end first the weight and the drag would act through a common point so there would be no torque. However, this arrangement would be unstable and practical experience shows that even with machine-like precision, unstable arrangements don't persist for very long and they quickly move towards a more stable arrangement - in this case the barbell would invert and end up with the heavier end pointing downwards. Unstable arrangements can only be maintained with some sort of an active feedback control system - way beyond the scope of the simple barbell experiment. Dolphin (t) 22:40, 6 June 2011 (UTC)[reply]

Sure. The slightest variation in the air it's falling through would start it tumbling, I'm sure. I'm just talking theory. Yet another question: The barbell will oscillate and the heavier end will rotate to the downward. But it won't just stop there, right? It will kind of act like a pendulum, oscillating for some time, with less arc each time until it stabilizes. Right? ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:02, 7 June 2011 (UTC)[reply]

It is dangerous to rely on intuition for things like this. The answer actually depends on how the weights are shaped. If you use two disc weights, a 20 kg on one side and a 10 kg on the other, then drag will be relatively greater on the lighter side, and you will end up heavy-side downward. But suppose you use three equal disc weights of 10 kg, with two on one side and one on the other. In that case the drag forces will balance, at least at the start, and there will be no intrinsic tendency to spin. However if there is any initial instability, things can get very complicated. Looie496 (talk) 22:54, 6 June 2011 (UTC)[reply]

@Looie496: Please explain drag will be relatively greater on the lighter side. Dolphin (t) 23:04, 6 June 2011 (UTC)[reply]

The surface-area-to-mass ratio will be larger for the lighter side. Looie496 (talk) 23:18, 6 June 2011 (UTC)[reply]

I agree that the surface-area-to-mass ratio will be larger for the lighter side. The larger object has greater surface area than that of the smaller so I would say the drag on the 20 kg object will be greater than on the 10 kg object, but by a factor less than 2 for the reason you have given. The factor is less than 2 so a torque acts on the barbell to cause it to fall heavy-side down. If the factor was exactly 2 there would be no torque and the barbell would fall with the bar horizontal. (If the factor was greater than 2 the torque would cause the barbell to fall light-side down.) Dolphin (t) 23:53, 6 June 2011 (UTC)[reply]

Any equations we specify to estimate drag based on the geometry and mass distribution will be just that - estimates. As every high-school physics student knows, force due to air resistance is a very difficult parameter to quantify and model. But, for any specified model of air resistance, we can construct the equations of motion and estimate the dynamics of the object as it falls and tumbles. As I described above, one such approach is to seek the steady-state, where net torque is zero. This is not the only way we could model the dynamics; but it's simple and allows us to find a "preferred" orientation. In reality, with turbulence, velocity-dependent drag, and so on, the actual model should account for all kinds of higher-order effects. When I need more sophisticated models, I toss out the elementary physics and use MDATCOM for modeling airflow and estimating turbulence. Nimur (talk) 00:23, 7 June 2011 (UTC)[reply]

Why are you all saying that the drag depends on the mass? Drag depends on surface area, velocity and, in very complicated ways, shape and texture, but it shouldn't depend on mass. The resultant force will depend on the mass, since it's the sum of the weight and the drag, but those are two separate forces. If we assume the weights on either end are identical except for their mass, then the drag will be the same on both, so can be considered to act through the geometric centre. That is not the same as the centre of gravity, which will be nearer the heavier end, so you have a torque and the object rotates. It will eventually reach a stable equilibrium when the heavier weight is at the bottom (since then the drag acts along the length of the object, so goes through the centre of gravity) and stay there. How long it takes to get to that equilibrium will depend on how big the drag is compared to the mass and how far from the geometric centre the centre of gravity is. A barbell would take a while to reach equilibrium, while a piece of paper with a blob of blutac on one edge would reach it very quickly. --Tango (talk) 00:31, 7 June 2011 (UTC)[reply]

We are all making the obvious assumption that the objects on either end of the bar are made of the same material so have the same density. Therefore their volumes are directly proportional to their masses. Dolphin (t) 00:37, 7 June 2011 (UTC)[reply]

Yes. Now here's a different question: What if the two ends of the barbell were identical in size, but one was made of iron and one made of aluminum. Does "drag" still figure into it? ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:01, 7 June 2011 (UTC)[reply]

See Tango's answer just above that you're indenting from. The forces will act through the geometric center because the shapes are symmetric, but will not act through the center of mass because of the density difference. This creates torque that will eventually leave the heavier end at the bottom. — Lomn 12:19, 7 June 2011 (UTC)[reply]

Another way to approach this problem is to look at the location of the center of mass (where "weight" forces balance) versus the location of the center of pressure (where "drag" forces balance). The object will be stable when the center of pressure is behind the center of mass (which is why arrows have feathers--the feathers move the center of pressure backwards without appreciably affecting the center of mass). In essence, that's no different than what everyone else is saying, but I think it might be easier to visualize. Andrew Jameson (talk) 12:40, 7 June 2011 (UTC)[reply]

Surely at terminal velocity we are way above the critical reynolds number for steady flow, and the barbell will tumble as it sheds vortices. Supposing that the lyapunov exponents are greater than one (which for such a flow they almost certainly are) then there is simply nothing that can be known about the orientation of the system on impact, based on its orientation at release. — Preceding unsigned comment added by 129.67.37.190 (talk) 15:00, 7 June 2011 (UTC)[reply]

Ionic liquids

Why don't the intermolecular forces between the positive and negative ions crystallize them into a solid? --75.40.204.106 (talk) 23:13, 6 June 2011 (UTC)[reply]

Our fairly detailed article entitled "ionic liquid" says that it's just an ionic compound with a low melting point ("liquid at room temp" for example), not an intrinsically "liquid" chemical. DMacks (talk) 23:29, 6 June 2011 (UTC)[reply]

Sleep and endurance athletes

This 2007 NYTimes article explores why endurance athletes sleep so much. I tried searching the scientific literature about this, but came up empty. Although I'm a bit groggy from a workout this morning. Can someone explain why long, intense workouts seem to cause sleepiness (preferably cited to review articles)? Yet short workout seem to have the opposite effect? I looked at our sleep article, but was unimpressed. Thanks. -Atmoz (talk) 23:46, 6 June 2011 (UTC)[reply]

For short workouts, you get the benefit of the adrenaline, which wakes you up. For longer workouts, this wears off and fatigue starts to take it's place. This means that damage, like tiny tears in muscles and cracks in bones, builds up and needs to be repaired during sleep. StuRat (talk) 22:37, 7 June 2011 (UTC)[reply]

The short answer is that there isn't really much to add to the NYT article. (Gina Kolata is very good, and doesn't miss much.) The literature confirms that prolonged intense exercise leads to increased sleep (e.g., PMID 9140908), but doesn't explain why. It isn't that hard, of course, to come up with more or less plausible speculations. Looie496 (talk) 23:31, 7 June 2011 (UTC)[reply]

Thanks to you both. ...but doesn't explain why. It isn't that hard, of course, to come up with more or less plausible speculations. Yup. The most common I've seen is actually StuRat's example that sleep is needed to repair tissue. But why would you need to sleep for that and not just rest? What are the biological processes that occur during sleep that facilitate this healing? I don't expect the ref desk to answer that, because based on my searching the last couple days, it doesn't appear that anyone knows the answer, or everyone is satisfied with the plausible speculations. I guess I'll just have to be satisfied with that for now. :) -Atmoz (talk) 02:05, 8 June 2011 (UTC)[reply]

Well, there actually is some information about how sleep promotes recovery -- our article Sleep#Restoration summarizes the evidence. Basically the sleep state induces hormonal and immune system changes that don't happen during a simple resting state. What is not clearly understood is how tissue damage produces an increase in sleep -- which I saw as the question that was being asked. Looie496 (talk) 03:09, 8 June 2011 (UTC)[reply]

I was originally asking what causes that increase need for sleep. But the lack of answer got me to wondering why we need sleep and not just rest. It would seem from an evolutionary biology point of view that needing to sleep after strenuous exertion would be a disadvantage over those that could get the same restorative properties from simply resting. The sleep article has a paragraph that speculates on this, and actually cites doi:10.1016/0304-3940(91)90276-Y where arctic ground squirrels actually arouse from hibernation to sleep! -Atmoz (talk) 14:58, 8 June 2011 (UTC)[reply]

That's an interesting article! I always thought that since almost all animals need to sleep, this outcome of evolution must be someting universal, so the fundamental explanation needs to come from mathematics (optimization/game theory) and not biology. If you have machines competing with each other, then a machine that operates in a single mode in which it also repars itself is going to be outcompeted by machines that don't carry out repairs in a "normal mode" and switch to a specially designed "sleep mode" that is optimized for carrying out repairs. Count Iblis (talk) 15:28, 8 June 2011 (UTC)[reply]

In addition to the need to perform repairs, the spinning of the Earth also makes day and night, with light levels so different that plants and animals must be specialized for each. Thus, it makes sense for them to sleep (and thus perform repairs) during the period which they aren't adapted to handle. StuRat (talk) 17:57, 8 June 2011 (UTC)[reply]

Another issue here is the increased sleepiness after exercise. I think this is really due to not getting enough sleep generally. If you always sleep 8 hours or more then you don't feel more sleepy after exercise, not even if the night before you happened to get less sleep. But if you usually sleep 7 hours or less, then you may notice increased sleepiness after doing hard exercise, even if you happened to get 8 hours sleep the night before. Count Iblis (talk) 20:10, 8 June 2011 (UTC)[reply]