Wikipedia:Reference desk/Archives/Science/2012 July 19

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July 19

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Image artifact of purplish and greenish/yellowish stripes

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Take a look at this scan of the actress Shivani Ghai: http://thedressdiscerner.files.wordpress.com/2011/07/asiana-mag-inner.jpg. There are alternating diagonal purplish and greenish/yellowish stripes running across the image. What type of image artifact is this, and what causes it? —SeekingAnswers (reply) 01:41, 19 July 2012 (UTC)[reply]

It's a Moiré pattern, and luckily the article somewhat explains the cause, since I can't. Dpreview, who should know, say "If a scene contains areas with repetitive detail which exceeds the resolution of the camera (1), a wavy moiré pattern (2) can appear" and provide onwards links. --Tagishsimon (talk) 01:43, 19 July 2012 (UTC)[reply]
For a rigorous mathematical overview of moiré patterns, there are articles on aliasing and Nyquist rate. Nimur (talk) 02:30, 19 July 2012 (UTC)[reply]
But for a simple answer, it happens when the image is made of tiny dots, and the dot spacing for the image does not match the dot spacing for the scanner. Looie496 (talk) 17:37, 19 July 2012 (UTC)[reply]

To what extent are astronomical observations automated ?

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For example, if one of our large telescopes is pointed at some spot in space, and there's a point of light there not in any previous observations, does a human have to spot it, or would it be spotted by software comparing with previous observations ? StuRat (talk) 03:47, 19 July 2012 (UTC)[reply]

It depends on what the goal of the observations is. There are projects that look for transient or variable objects (e.g. Supernova Cosmology Project, OGLE), and those apply automatic algorithms to their data to detect variability or "new" objects. In general, survey telescopes such as Large Synoptic Survey Telescope, that revisit the same part of the sky repeatedly, state the search for transient objects as one of their scientific goals and will run those algorithms on their data. On the other hand, data from "visitor telescopes" (like the Very Large Telescope) where astronomers propose observations for a specific scientific purpose are not routinely screened for transient objects (most of the time there will be no other observations of the same field of comparable quality), and it will be luck if an astronomer spots something they haven't looked for. --Wrongfilter (talk) 08:57, 19 July 2012 (UTC)[reply]
(ec) Astronomers usually agree to share their observations with the observatory where they make them. Many of the large observatories have standardized processing steps that they perform when they add new observations to their archive libraries. However, a new point of light isn't typically going to flag an image for review unless spectral characteristics are being measured and it doesn't look like something common such as a asteroid or artificial satellite. A missing object or significantly changed brightness in an expected point of light is another matter entirely, and is more likely to flag the observation for review. 75.166.200.250 (talk) 09:03, 19 July 2012 (UTC)[reply]
Considering that those new points of light could be a meteor headed for Earth or something else important, it's a shame they can be tracked from 1st discovery. You'd think they could be compared against an artificial satellite database to eliminate that possibility. StuRat (talk) 09:35, 19 July 2012 (UTC)[reply]
There are automated, specialized meteor-scanning telescopes plugged into computers. But not every telescope result is checked for meteors — the odds are super low and the number of false-positives would make such a system prohibitive, and many of them are not looking at the sky in ways that would even spot a meteor. I think you are underestimating the amount of astronomical data that is generated, as well. --Mr.98 (talk) 11:49, 19 July 2012 (UTC)[reply]
Is the amount of data collected from large telescopes just too large for all our computers to process ? If so, it sounds like we need to build up our computing capacity, maybe in the same way as SETI@home does (using volunteers' home computers). StuRat (talk) 18:34, 19 July 2012 (UTC)[reply]

Gluconeogenesis image review

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Could someone please check commons:image:Gluconeogenesis_pathway.png please? The final product (top) purports to be glucose (in its pyran form), but is missing the methylol(?) group. This make me question if there are other mistakes in it. One of the previous steps is a furan ring with two methylol groups, so its unlikely the target is a pentose. TIA, CS Miller (talk) 05:39, 19 July 2012 (UTC)[reply]

What worries me most about it aren't structural errors, but what's the source of the sequence? 75.166.200.250 (talk) 21:08, 19 July 2012 (UTC)[reply]

Galactic wind-up problem

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From our Milky Way article:

Another interesting aspect is the so-called "wind-up problem" of the spiral arms. If the inner parts of the arms rotate faster than the outer part, then the galaxy will wind up so much that the spiral structure will be thinned out. But this is not what is observed in spiral galaxies; instead, astronomers propose that the spiral pattern is a density wave emanating from the Galactic Center. This can be likened to a moving traffic jam on a highway—the cars are all moving, but there is always a region of slow-moving cars. This model also agrees with enhanced star formation in or near spiral arms; the compressional waves increase the density of molecular hydrogen and protostars form as a result.

So, how do these density waves manifest themselves ?

1) Do some arms fade away while other arms appear ?

2) Or is it that when they give the variable rotation speed of the inside and outside of the arms, what they really mean is the stars within the arms, but that, as new stars are created and old stars die, the actual location of the arms move relative to the stars, so that the arms really have constant rotation ?

Any clarification would be much appreciated. StuRat (talk) 06:55, 19 July 2012 (UTC)[reply]

The stars move at (roughly) constant speed (in km/s), the waves move at constant angular speed; their motion is essentially independent. Individual stars pass through the waves in the same way that water molecules pass through water waves (or the other way round - water molecules oscillate around a fixed position while the wave passes through). Unfortunately, we can only measure the motion of stars and gas clouds in galaxies, not the motion of the arms/density waves. --Wrongfilter (talk) 08:47, 19 July 2012 (UTC)[reply]
But the part I don't understand is how the stars and waves can move independently, if the waves are composed of stars. Are the stars inside a "crest" just pushed closer together as the wave passes ? Are the stars brighter while it passes ? StuRat (talk) 09:30, 19 July 2012 (UTC)[reply]
That's not specific to stars and galaxies at all, and a partial explanation is contained in Wrongfilter's reference to water waves. Generally speaking, when a wave propagates through a medium, it does not do so by moving all the medium with it, but by vibrating the medium. To take a more obvious example: vibrations move through the strings of a stringed instrument without the entire string moving. If I grab a loose garden hose and waggle one end, the entire hose can adopt a wave-shape without the end running away from me. So with water, the body of water doesn't move as a whole when the waves move - or the tide would come in vastly faster than it does. And with stars - although the forces holding them together are gravitational rather than electrostatic - the same principle applies.
(after ec) So, to answer your question - the stars don't get brighter, they get closer together. AlexTiefling (talk) 09:35, 19 July 2012 (UTC)[reply]
So, in that case, the wave must slow the stars at the leading edge and speed up the stars at the trailing edge, to force them closer together ? StuRat (talk) 09:38, 19 July 2012 (UTC)[reply]

FOLLOW_UP Q'S:

1) Why aren't pressure waves originating from the galactic core manifested as circular waves moving out from the core ? What gives them their spiral form ?

2) What happens to the pressure waves at the bar in barred-spiral galaxies, like the Milky Way ? StuRat (talk) 09:43, 19 July 2012 (UTC)[reply]

I really need to read up on this (it's not trivial), but I just point out that the waves in galactic discs are density waves, not pressure waves. The spiral pattern, i.e. the density pattern arises from the superposition of the stars' orbits, although the gravitational attraction also plays a role. It's a collective phenomenon in a non-collisional self-graviatating system, as is the bar. --Wrongfilter (talk) 10:55, 19 July 2012 (UTC)[reply]
Here's my casual explanation, no doubt wrong:
  • Galaxies have a core, which contains sites where stars are formed, at least some of which for some reason then travel away from the core.
  • They travel at different speeds, causing clusters. The clusters still move away from the core. This is like a series of small, non-stationary traffic jams.
  • Galaxies rotate.
  • Consider a long straight single-lane north-south road, with clusters of traffic on it. The traffic all emanates from a building in the middle point of the road, half going north, half going south. Rotate the entire road around this building, and the cars will form a spiral pattern (as the slower moving ones, and the ones stuck behind them, are left behind, falling off the rotating road, to form trails).
That analogy became increasing bizarre as I edited and improved it. Oh well. Hope it helped.  Card Zero  (talk) 12:41, 19 July 2012 (UTC)[reply]
Yeah, I'm pretty sure that is wrong. Stars orbit the centre - they tend to stay at a roughly constant distance. There will be some stars moving outwards or inwards (mostly due to close interactions with other stars), but I doubt there are enough of those for it to be a significant effect. I'm almost certain it won't be a large enough effect to cause the spiral arms. --Tango (talk) 16:36, 19 July 2012 (UTC)[reply]
Ooh, so rotated concentric ellipses cause the arms. How nice.  Card Zero  (talk) 17:43, 19 July 2012 (UTC)[reply]
 
Explanation of spiral galaxy arms.
Yes, that makes some sense, but can those spirals create more than 2 arms ? I'd like to create some animations of this illustration and one with multiple arms, if I can figure out how the multiple arms case works in this model. StuRat (talk) 18:29, 19 July 2012 (UTC)[reply]
Well, you can get that by superposing two (or more) independent waves. Dauto (talk) 19:06, 19 July 2012 (UTC)[reply]

Density wave theory is only a part of the set of theoretical explanations there are about galactic spirals; see Spiral galaxy#Origin of the spiral structure. In general, the dynamics of stars within galaxies is not as well understood as one might expect, with theory at times matching poorly with observation; see Galaxy rotation curve. Red Act (talk) 21:47, 19 July 2012 (UTC)[reply]

There is little discrepancy between theory and observations as far as galaxy rotation curve is concerned as long as that theory includes dark matter (which it should). Dauto (talk) 02:46, 20 July 2012 (UTC)[reply]
But dark matter fits solidly within the category of "not as well understood as one might expect". Without a clear-cut understanding of what exactly dark matter is, direct (as opposed to gravitational) evidence for its existence, or experimental evidence that would clearly rule out the alternative theories that compete with dark matter, dark matter currently isn't much more than a fudge factor that describes the discrepancy between observation and what the theoretical results would be without dark matter thrown in. That may well change if some good experimental evidence about dark matter starts coming in over the next decades, but at the moment, dark matter isn't a very satisfying explanation. Red Act (talk) 04:21, 20 July 2012 (UTC)[reply]
StuRat, perhaps you have figured it out already from the various links, but the density wave is principally a property of the galactic gas. We think of the interstellar medium as empty, but it is filled everywhere with diffuse gas. Though almost empty, waves can propagate through that gas provided that the waves are large enough (in this case on the scale of a whole galaxy). As the wave compresses the gas, that perturbation provides just enough impulse to trigger star formation in gas clouds that were already on the brink of collapsing. This means that the passage of the density wave is marked by a surge of young stars. It is those new stars, many bright and large, that make the spiral arms stand out visually. The stars, once formed, follow elliptical paths and are hardly affected by the density wave at all. However, many of the brightest stars are so big that they live short lives (only ten million years, for example), so they die out quickly compared to the travel time of the density wave. Hence the spaces far from the front of star formation appear darker. So yes, the arm positions (really a front of star formation) is moving relative to the actual stars. Dragons flight (talk) 19:23, 20 July 2012 (UTC)[reply]
OK, thanks all, but what actually generates the density waves ? StuRat (talk) 20:26, 20 July 2012 (UTC)[reply]

Li-ion current price per kWh

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I was wondering how much rechargeable li-ion battery cost per kWh. To grab very rough estimate I gone ebay and searched for different types of li-ion batteries there. To my huge surprise, it seems that 18650 form factor cells are most cost effective... And 10 batteries 3800mAh each 3.7V cost $14.94 with free shipping...

My question, if advertised characteristics reflect real performance, that translates for 140.6 Wh for $14.95, or ~$107 per kWh. My calculations correct? Or have I made a mistake?

And second question, how likely that such lots(not only on ebay, but alibaba etc) reflect the cost of production of 1 kWh rechargeable li-ion battery selling for around $100 and still making profits? Or such batteries are left overs from bigger orders or even flawed cells that haven't passed quality control, and thus price of them do not really reflect cost of kWh...

Example ebay slot 70.49.169.86 (talk) 16:32, 19 July 2012 (UTC)[reply]

Um, don't ever trust mAh ratings from eBay. Given the price, you'd probably be lucky if the cells even have half the capacity [1]. I believe the real high end 18650 cells from Panasonic etc tend to be around 3100mAh [2] [3] [4]. From what I've read, cells with Xfire labels can vary from reclaimed battery pack cells to those that didn't make the grade for a battery pack. (You can get okay ones but you need to do a bit of research and ensure it's a supplier you can trust.) In any case, I wouldn't read too much in to the price, but it sounds like you should be more interested in the wholesale price anyway not the price from random eBay sellers who stick random labels on them. P.S. I'm pretty sure your calculation can at best be called simplified since you didn't take in to account the actual discharge curve for your cell which will depend on discharge current and to a smaller extent what voltage you stop at. Nil Einne (talk) 17:48, 19 July 2012 (UTC)[reply]
''it sounds like you should be more interested in the wholesale price anyway'' Yes, this is what I'm really interested to figure out. But not GM/Nissan/Ford claimed price of li-ion battery packs, but cheap modern li-ion chemistries produced out there(in China?) and wholesale price per kWh they are selling for... 70.49.169.86 (talk) 18:09, 19 July 2012 (UTC)[reply]
My point is the correlation between the price you pay retail for individual cells and the wholesale price is weak at best. Particularly in a case like this where the cheap retail options are of unknown heritage with random labels and the wholesale price will depend greatly on things like what sort of supplier you're willing to trust (given the way things can go wrong with li-ion cells, although of course one cheapish seller may give you total crap which blow up regularly, another selling at the same price may give you decent quality cells most of the time) and capacity (choosing the highest capacity cells is probably not the cheapest option), not to mention the typical stuff like quantity. Your best bet is probably looking at something like Alibaba or one of the other China B2B sites out there for the sort of quantity ranges you're thinking out, but you'd also have to have some idea of what you're looking at which to be honest it sounds like you don't. You can also try asking suppliers since they don't always bother to properly specify. Of course it does depend somewhat on what level you're actually looking at, I would guess at a low level range like 10000 cells a year you may get an okay idea from such sites, but if you're thinking of 100k a month, probably not. It's not that this info is super secret, it's just that it's not out there on some internet site because that's not how people who actually make the decisions are going to look for it. And navigating the China market can be difficult, if you don't much experience, given the variety of problems there is like unreliable suppliers. I don't really get the relevence of GM/Nissan/Ford as I didn't mention them, despite the recent interest in electric cars and the fact some of them now use li-ion cells, I'm pretty sure they still only make up a tiny percentage of the market. Nil Einne (talk) 04:44, 20 July 2012 (UTC)[reply]
First of all thank you Nil for pointing out that 3700 mAh was totally unrealistic to expect from 18650 cell. That helped me a lot:) Yes, correlation between retail and wholesale prices is hard to estimate precisely. But it still better then nothing and lowerest retail price would give some idea about wholesale price level... I have mentioned GM/Nissan cuz price of battery packs of Leaf and Volt has been published ($375 and $550 per kWh 'IIRC'). And that theoretically should help estimate wholesale price of batteries... But automotive packs are something different then what I'm looking for(at least the way GM define it). I have got a two question for you, Nil.
1) Wouldn't you think that LiFePO4 chemistry should be cheaper then bunch of 18650 cells with similar capacity?
2) Could you make an educated guess for me about a price of average capacity (2200 mAh?) 18650 cell. What would you consider cheap, but realistic and a good price for ~100k cells? Would you say that $1.50-$1.80 price range per cell is achievable? I know Tesla Motors was using hundred thirty one 18650 cells per kWh in their battery packs. So that translates into 7.6Wh per 18650 cell, should be a good estimate for price effective cell capacity I guess....
Thank you in advance, 70.49.169.86 (talk) 05:59, 20 July 2012 (UTC) PS. I have tried to use Alibaba even before I have asked question here, with very little luck. And really do not feel like approaching companies to get quotes...[reply]

energy drink

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why we add salt,sugar in water to give energy to a person — Preceding unsigned comment added by 59.177.161.174 (talk) 17:08, 19 July 2012 (UTC)[reply]

Such drinks, like Gatorade, have three purposes:
  • To provide a quick source of food energy, which comes from the sugar
  • To rehydrate someone, due to sweating during exercise, this comes from the water
  • Sweating also contains a lot of salt, which needs to be replaced. If you drink too much water and not enough salt, you can get Hyponatremia, so the salt is there to keep the electrolytes in the body at the right concentration.
Does that help? --Jayron32 17:25, 19 July 2012 (UTC)[reply]

For why they put salt in energy drinks, see water intoxication. Whoop whoop pull up Bitching Betty | Averted crashes 17:33, 19 July 2012 (UTC)[reply]

Water intoxication and hyponatremia are basically two sides to the same coin. Not exactly, but they are very closely related conditions: if you have too much water, it also means you don't have enough sodium. If the two are in correct relative balance, your kidneys should be working efficiently enough to keep things working well; that is for a properly balanced solution with the right amount of salt, it is hard to over-consume it: you'll just tend to piss a lot more. If you don't get enough salt, you'll experience the symptoms of water intoxication/hyponatremia. Too much salt or not enough water leads to dehydration and Hypernatremia. --Jayron32 17:39, 19 July 2012 (UTC)[reply]
That's all correct, but my understanding is that there is also another factor: it is easier for the body to absorb fluids if they are approximately in osmotic balance with the bloodstream -- meaning that they have a comparable level of dissolved molecules, of whatever sort. Looie496 (talk) 18:00, 19 July 2012 (UTC)[reply]
The answers above are correct if you're talking about liquids consumed by professional athletes during strenuous training sessions. If you're talking about supermarket "energy drinks" consumed by non-athletes, the sugar and salt are there to make them taste good. -- BenRG (talk) 22:06, 19 July 2012 (UTC)[reply]
Yes, the answers above are for sports drinks, rather than energy drinks. Sports drinks are used by amateurs for the same reasons as professionals. Energy drinks are a completely separate market (do they usually contain salt? They're mostly caffeine and sugar, I think). --Tango (talk) 00:37, 20 July 2012 (UTC)[reply]
This may be an ultra-dumb question even for me, but one thing about Gatorade is that it tastes a lot better when you've been exercising strenuously than if you just take it like you would soda pop. So, does that have to do with a stressed body "craving" what Gatorade contains? ←Baseball Bugs What's up, Doc? carrots04:08, 20 July 2012 (UTC)[reply]
I don't know what is in Gatorade, but I do know you lose the ability to taste salt when severely dehydrated, which makes oral rehydration solution taste a lot better (it tastes like sugar water, rather than sugar and salt water). If Gatorade contains quite a lot of salt, then it could be the same effect. --Tango (talk) 06:04, 20 July 2012 (UTC)[reply]
That's exactly it: the salt. But it isn't that severe dehydration makes you unable to taste salt, it's that moderate loss of salt via sweating makes salt taste better than it usually does. Looie496 (talk) 06:39, 20 July 2012 (UTC)[reply]

I suspect that the real reason for companies adding anything to water, is so that they can convince people to buy it and thus take your money. Of course other companies manage to extract money from the consumers without even adding anything to the water. Mitch Ames (talk) 03:13, 21 July 2012 (UTC)[reply]

Earth and Jupiter

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At what point would Earth have to be to Jupiter before it is sucked into its orbit? Reticuli88 (talk) 17:17, 19 July 2012 (UTC)[reply]

See Gravitational binding energy. If the kinetic energy of a body exceeds the gravitational energy holding it to another, it will be able to escape. If the kinetic energy of the object is less than the gravitational energy, it will eventually be "sucked in" to the other object. To answer your question meaningfully, which I take it to mean "how close must an Earth-sized object be to a Jupiter-sized object before it will be gravitationally bound to it" the answer is "it depends"; mostly on the relative motion of the two objects. The dynamics are very complex, and can't be answered without more information regarding the specific orientation and relative velocities of all objects involved. However, I wouldn't worry about it happening any day soon. The distance would be a LOT closer than they are now; Jupiter is 5 times as far from the Earth as the sun is at the closest it ever gets to the Earth, and it is only 0.001 times the mass of the Sun. Roughly speaking, that means that Jupiter should exert a gravitational influence over objects only 0.001 times as well as the Sun does, and the Earth falls well outside of that sphere. The earth does experience a tiny effect from Jupiter's gravity, but it isn't going to send us crashing into Jupiter. --Jayron32 17:35, 19 July 2012 (UTC)[reply]
Binding energy isn't really the best concept to look at. What you're really interested in is escape velocity. --Tango (talk) 00:39, 20 July 2012 (UTC)[reply]
See Hill sphere. manya (talk) 03:52, 20 July 2012 (UTC)[reply]
And also see Sphere of influence (astrodynamics). manya (talk) 03:56, 20 July 2012 (UTC)[reply]
Sorry, these links are not relevant to answer the question. manya (talk) 04:00, 20 July 2012 (UTC)[reply]

Cell Signalling in Development

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Hello. Two-cell (comprised of P1 and AB) embryos were incubated either in a translation inhibitor or in a transcription inhibitor. The AB cells were then isolated, washed, and grown in culture. AB cells of embryos treated with the translation inhibitor produced only neurons and skin, while AB cells of embryos treated with the transcription inhibitor also produced muscle–their normal fate. The direction of signalling is from P1 to AB. Why would the signalling interaction at the two-cell stage most likely involve proteins? If proteins are the product of transcription and translation, wouldn't the P1 cells of embryos treated with transcription inhibitor not produce mRNA and, as a result, proteins? This would block the signalling pathway, right? Thanks in advance. --Mayfare (talk) 19:56, 19 July 2012 (UTC)[reply]

Because the two cells are adjacent. Low molecular weight hormonal signaling paths are unnecessary under those conditions. Yes to your second and third questions. 75.166.200.250 (talk) 20:46, 19 July 2012 (UTC)[reply]
In most organisms, transcription is marginal or completely absent for the first several rounds of cell division after fertilization. In Drosophila melanogaster, for instance, this stage lasts for ~2 hours and 12-14 rounds of cell division. Virtually every mRNA found in very early embryos has actually been produced maternally. And so the transcription inhibitors would have no effect because transcription isn't even on, or is nearly irrelevant if it is. Translation, on the other hand, is quite active, working from the maternally provided mRNA. Someguy1221 (talk) 22:14, 19 July 2012 (UTC)[reply]

Equivalent Animal Processing Power of Modern Computers

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Out of some recent boredom, I have been reading up on the computer vs. human brain speed dealie. Though I knew that there is no easy way to measure and compare the two, I was interested in some of the (questionable?) methods used to approximate the processing speed of human brains and that they were considered to be vastly superior to current technologies. I was wondering, then: what animal's mental "processing power" is closest to the fastest computer processor available to consumers today? Have we even gotten past ant? Thanks, Sazea (talk) 19:57, 19 July 2012 (UTC)[reply]

Well, my opinion is somewhat heretical here, but since I have a Ph.D. in neuroscience I can present myself as enough of an expert to express it: I believe that the computational power of brains is way overstated, and that at a practical level a human brain has capabilities comparable to a powerful modern desktop PC -- except in terms of memory access bandwidth, where the brain really stands out. The thing that really makes brains superior, I believe, is the vast amount of information about the world they take in during the process of development. Looie496 (talk) 20:17, 19 July 2012 (UTC)[reply]
Aren't neurons arranged in more than 2 dimensions like computers are?--Canoe1967 (talk) 20:22, 19 July 2012 (UTC)[reply]
I'm not sure why you ask that, but there is no doubt that digital computers differ from brains in a variety of important ways. They differ so much that it is hard to find a basis for comparison. In my view, the only proper way to do it is to ask how powerful a computer would be needed to do the same tasks that the human brain can do -- for example, to pass the Total Turing test. I think it is likely that modern PCs, augmented by a few special chips, are powerful enough to do it -- that the real difficulty is in programming them, not in making computers that are strong enough. Looie496 (talk) 20:40, 19 July 2012 (UTC)[reply]
I just heard somewhere that they are thinking of biological memory to get some of the advantages of animal brain material and the configuration was one of them. I also remember that if we can make a wire small enough to connect we may try programming animal neuron groups.--Canoe1967 (talk) 21:22, 19 July 2012 (UTC)[reply]
Looie, what would be your estimation of the human memory access bandwidth? And yes, hard disks suck. Pitiful bandwidth. Sagittarian Milky Way (talk) 23:34, 19 July 2012 (UTC)[reply]
It's been quite some time since I worked through the calculation (calculating the rate at which information in the synaptic weight distribution generates information in the population firing pattern), but my recollection is that I could get a value on the order of 1 terabyte per second without making any assumptions that seemed unreasonable. Looie496 (talk) 03:04, 20 July 2012 (UTC)[reply]
Which even ultra high end GPUs are a while away from matching Comparison of AMD graphics processing units, Comparison of Nvidia graphics processing units. Nil Einne (talk) 04:35, 20 July 2012 (UTC)[reply]
I dunno, I see 264 GB/s and 2 x 192.256 GB/s for the desktop in those lists. What's the Moore's law for GPU bandwidth? That's only 2.6 times to 1 TB/s (wow, a TB/s, that's fast). Some people put 2 or 3 (4?) graphics cards in the same PC, right? We've already matched it on some PCs. With coolers for each card that can play top games at max settings on a 26, 36 inch or something screen that's like 4 HDTVs worth of pixels. There are supercomputers with 20 TB of RAM, they probably blow away human brains in terabytes/s. Sagittarian Milky Way (talk) 18:07, 20 July 2012 (UTC)[reply]
But it doesn't make any sense to multiply the bandwidth, because a single GPU only has whatever amount of bandwidth. Otherwise you might as well just talk about how we've matched it with a Beowulf cluster. You can see the history of GPU bandwidth in the articles. Nil Einne (talk) 17:02, 22 July 2012 (UTC)[reply]
Looie, what are your views on whether brains can even be considered computable. I have read penrose's thoughts on this, but for some reason have completely forgotten the gist of his argument. Perhaps it unsettled me. Egg Centric 17:55, 20 July 2012 (UTC)[reply]
I think it was that it's not (fully) computable due to quantum mechanics. I hope that's true, that's where free will comes from. Sagittarian Milky Way (talk) 18:10, 20 July 2012 (UTC)[reply]
Well I rather (personally) hope quantum mechanics explains my self-evident consciousness (as I define it) although I can't possibly see how it could. Free will I see no reason to belive in tbh. Egg Centric 21:35, 20 July 2012 (UTC)[reply]

Energy requirements for reverse osmosis

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How much energy does it take to desalinate 10,000 cubic meters of seawater by reverse osmosis? 75.166.200.250 (talk) 20:38, 19 July 2012 (UTC)[reply]

You could try websites that sell them. They may have those specs. I assume the larger the more efficient.--Canoe1967 (talk) 21:24, 19 July 2012 (UTC)[reply]
Using numbers from http://urila.tripod.com/desalination.htm/ (for which I do not vouch), and assuming perfect efficiency, it comes to around 1000 kilowatt hours. Looie496 (talk) 21:55, 19 July 2012 (UTC)[reply]
You should buy one of these though, which I endorse. http://www.ansl.ca/candesal/article.html --Canoe1967 (talk) 23:40, 19 July 2012 (UTC)[reply]

"Mr. De Villiers says if CANDESAL can produce water for less than $1 a cubic metre, the technology has great potential. "If their business plan is right, then they're onto something really big," he says." --Canoe1967 (talk) 23:49, 19 July 2012 (UTC)[reply]

You can easily compute this from first principles. It follows from the change in the Gibbs energy. Suppose you have seawater with a concentration of n ions per unit volume in it. You take a volume V from this seawater, which will contain N = V n ions. The Gibbs energy of N ions in a volume V of water is approximately given as g = N k T Log[N/(V Z)] where Z is the single ion partition function. Then the initial state is the water of volume V with ions at concentration n, and the sea which has some huge volume V' with ions at concentration n. The Gibbs energy of the initial state is then

g1 = N k T Log[N/(V Z)] + M k T Log[M/(V'Z)]

We then remove all the ions from the water by dumping into the sea. The limit of N to zero of the first term is zero. The Gibbs energy is thus given by:

g2 = (M + N) k T Log{[M + N]/(V'Z)}

The maximum amount of work that can be extracted from this process is the drop in the Gibbs energy, g1 - g2. This will be negative, so it will require a minimum amount of work of g2 - g1 to get to the final state. We have:

g2 - g1 = (M + N) k T Log{[M + N]/(V'Z)} - N k T Log[N/(V Z)]

      -  M k T Log[M/(V'Z)]

We then want to take the limit of V' and M to infinity such that M/V' = n. We can simplify the first term:

(M + N) k T Log{[M + N]/(V'Z)} =

(M + N) k T Log(n/Z) + (M + N) k T Log(1 + N/M)

The first term of this cancels against the last two terms in the expression for g2-g1, we thus have:

g2 - g1 = (M + N) k T Log(1 + N/M)

For small x we have Log(1+x) = x - x^2/2 + ..., so the limit for M to infinity becomes:

g2 - g1 = N k T

Seawater contains 35 grams per liter of salt. Sodium chloride has a molecular weight of about 58.45 u, so 35 grams of salt contains 3.606*10^23 ions of sodium and 3.606*10^23 ions of chlorine. One liter of seawater thus contains 7.21*10^23 ions, 10^4 cubic metres contains 7.21*10^30 ions. If the temperature of seawater is 15°C, then g2-g1 = 2.87*10^10 Joules. Count Iblis (talk) 03:58, 20 July 2012 (UTC)[reply]

That's nearly 8 megawatt hours. I wonder why the discrepancy with the 1 MWh figure above. 75.166.200.250 (talk) 07:02, 20 July 2012 (UTC)[reply]
Looie probably forgot a factor of ten somewhere, the source he uses says "0.66 kcal / liter is the minimum energy required to desalination of one liter of seawater, regardless of the technology applied to the process.". If you use this then you get almost the same figure as I obtained (they use 33 g/L for the salt concentration and I took 35 g/L, if you correct for that, then the agreement become even better). Count Iblis (talk) 15:35, 20 July 2012 (UTC)[reply]

Hi. Recently, I became interested in the concept that some animals other than humans can "understand" music and appropriately respond to it. We all know that humans have a range of musical perception, interest and appreciation, and that other animals use music or music-like sounds as a form of communication. Also, there are videos on Internet of dog and cat dancing to music. Birds, on the other hand, use song that humans interpret as musical. Meanwhile, in the Chinese language there is a proverb that states "playing piano to an ox", suggesting perhaps that oxen do not understand music. In the extreme case, insect buzzing can also be music-like, but their brains are small. Finally, the clade of dinosauria may have used music[citation needed]. Dolphins and other cetaceans use human-understandble music, have the second-largest encephalization quotient in all of Animalia, and some evidence of human-cetacean communication is potentially documented. All of these are examples of anecdotal evidence. However, could there be an overlap between human and non-human music? More specifically:

  • What brain structures and patterns allow for the comprehension of music?

Which lead to a more interesting question:

  • Is there any corresponding resonance pattern between the types of music that a certain species "understands" and the more subconscious brainwave patterns such as delta, mu and theta?

Furthermore, this leads to an even more interesting question:

I recently came across the above-linked articles, and they may be relevant in answering the question, as I knew nothing about this topic beforehand. Thanks. ~AH1 (discuss!) 22:40, 19 July 2012 (UTC)[reply]

Music appreciation is probably a learned social behavior. Please see PMID 22732561 (figures, tables, and supplementary material.) 75.166.200.250 (talk) 22:44, 19 July 2012 (UTC)[reply]
Let me just recommend the book Musicophilia by Oliver Sacks, if you want to know more. I haven't actually read it, by everything by Sacks is very readable. Looie496 (talk) 01:37, 20 July 2012 (UTC)[reply]
It is largely a collection of case histories, not a theoretical book. But yes, almost everything he writes is excellent, even his account of falling down a mountain. μηδείς (talk) 04:34, 20 July 2012 (UTC)[reply]
I was somewhat underwhelmed by it, I must say. I loved some of his early work (Awakenings was by far the best book I read in 1976, probably for the entire second half of the '70s), but Musicophilia didn't do it for me, and it received very mixed reviews. What I would recommend, though, is Robert Jourdain's book Music, the Brain and Ecstasy. -- ♬ Jack of Oz[your turn] 11:22, 20 July 2012 (UTC)[reply]
"Undertanding" music and enjoying it requires at fundamental level 3 things (1) a sense of timing, (2) a sense of pitch (ie frequency as it is called scientifically), and (3) able to sense and enjoy repetition. The are a number of subtleties, such as what pop music composers have called "ear grabber" notes (notes that are harmonically slightly out of place, or melody lines that sound exciting), but that's what it comes down to. A quite good explanation of music in terms of expliting these 3 aspects was published in a book on how to write pop songs by, of all people, Rolf Harris, about 40 years ago. You may be able to obtain a copy via your local library. Aspect (3) is of course short term memory. Aspect (2) is of general use to animals as it helps to recognise the sounds of prey and predators, and is vital to communication. Aspect (1) is of general use in communication and in understanding the behavior of prey and predators. The brain structures responsible for aspects 2 & 3 have not been pinned down precisely, but a multitude of texts have been published on it. Aspect 1 has until recently been pretty mysterious, however special "clock" neurons have been discovered through the brain. Scientific American and Scientific American Mind has carried articles about it in the last year or so. Given that enjoyment of music is based on such fundamental requirements of intelligence, it should not be a surprise that animals have some sense of music. Having said that, my experience in owning several dogs, from dumb breeds like spaniels, to highly intelligent breeds like german sheherds, is that our music is just noise to them. Wickwack124.178.34.188 (talk) 02:21, 20 July 2012 (UTC)[reply]

Medical Term

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What is sphincter pharyngoplasty? Does it have anything to do with tonsils? Daonguyen95 (talk) 23:28, 19 July 2012 (UTC)[reply]

See Hypernasal speech#Sphincter pharyngoplasty. Red Act (talk) 23:38, 19 July 2012 (UTC)[reply]