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April 28

hydration of methane

Why dsn't it happen? —The preceding unsigned comment was added by Bastard Soap (talk • contribs) 00:34, 28 April 2007 (UTC).[reply]

It does happen, see Methane clathrate GB 01:22, 28 April 2007 (UTC)[reply]

Hydration is the addition of H₂O across a carbon-carbon double bond. There are no C-C bonds in methane, and definately no C=Cs. Aaadddaaammm 01:37, 28 April 2007 (UTC)[reply]

Methane clathrate is a crystal, not exactly what I was looking for, but thanks anyway it was an interesting read. I made a mistake, it's the hydration of carbon tetrachloride. At school it was said that it has to do with the d orbital but in the article orbitals are not mentioned, it is said that it is steric hindrance. Can anyone clear things up a bit?Bastard Soap 10:43, 28 April 2007 (UTC)[reply]

As soon as the phrase "d-orbitals" is uttered when discussing main group chemistry, stoplistening because you are talking to a geezer. Hydration means many things, including the addition of water to make conventional bonds to OH and H (organic chemists use term this way), formation of water of crystallization (inorganic chemists like this application of the term), and even the curing of concrete. In general, terminology involving water is very rich and complex because H2O is so pervasive. Regarding carbon tetrachloride, there is no significant reaction with water although at high temperatures, such as in fires, it may give phosgene and HCl.--130.126.228.60 17:03, 28 April 2007 (UTC)[reply]

Carbon doesn't have any d orbitals anyway. I agree with the "there are no double bonds" answer. --Bennybp 05:40, 29 April 2007 (UTC)[reply]

Propose a mechanism and then see what looks unlikely about it:) Maybe nucleophilic attack by water (or hydroxide) in S_N2 fashion is sterically blocked (large halogen atoms prevent access to back-side attack at carbon) and net zero dipole of CCl₄ makes it less likely to interact with a polar molecule? Under high-energy conditions, maybe could knock off a chlorine radical or ion, allowing further reaction to trichloromethanol (decomposes to phosgene). Cytochrome P-450 can do CCl₄→COCl₂ too (PMID 6700577). DMacks 06:58, 30 April 2007 (UTC)[reply]

Torque horsepower revisited

on the artical about topfuel dragsters it states

"Power output of these engines is most likely somewhere between 6000 and 8000 horsepower (approximately 4500-6000 kilowatts). This is calculated from performance as a dynamometer capable of measuring power output of these magnitudes has yet to be built. This would suggest a torque output of 5100-6750 Nm (3760-4980 lb-ft) and also a brake mean effective pressure of 80-100 bar."

I asume horsepower in that case was calculated by how much work the car did over a period of time. w=fxd P=w/t That being the case, how was the torque figure calculated?

furthermore it has come to my attention that at an rpm of 5252 the torque and horsepower of an engine are equal, if there is such a thing.....

Power is measured "instantaneously" via a derivative of work with respect to time. This is a common tool for analysis of dynamic systems. As far as your "equal" assessment, it is more correct to say that "in the units of horsepower, foot-pounds, and rpm, 5252 rpm yields a horsepower and torque which both equal ___." Torque and horsepower are never "equal," because though they are related, they are different quantities which describe a different physical attribute of the moving system. Nimur 05:10, 28 April 2007 (UTC)[reply]

Thanks for responding, it cleared up some confusion I had.

Solvay Process

Hi there, I got seven questions.

1. What are Intermediate products?

2. Raw materials are the materails that are consumed in the net reaction. What are the raw materials for the Solvay process? Where are these raw materials obtained? What makes these materials suitable for a large-scale chemical process?

3. The primary products and the byproducts of a chemicalprocess depend on how marketable the products are. What are the primary products and byproducts of the Solvay process?

4. What intermediate product in the Solvay processis highly marketable? What are some consequences of removing this intermediate from the system of reactions?

5. Resources other than chemical and technological resources are required for most chemical processes. What additional natural resources are needed for the Solvay process?

6. Large quantities are used in newer processing plants that are built on what is called a world scale to produce quantities for international distribution. Why do you think chemical plants are being built on an increasingly large scale?

If you going to tell me to read Solvay process article, well, you better show me which line says what.

Thanks. —The preceding unsigned comment was added by 76.64.132.195 (talk) 03:56, 28 April 2007 (UTC).[reply]

Due to the intricacies of the modern web browser rendering engine, your line numbers may vary based on font size, system display resolution, window toolkit use, maximization or windowed viewing, and any number of other factors. I would really recommend lines 5-100 of the Solvay process article. Nimur 05:13, 28 April 2007 (UTC)[reply]

In which part of the article? Byproducts or what?

I think you will find that the answers to all of these questions are very straightforward from the article. You might also want to read background about marketing, globalization, and chemical engineering, which will help you appreciate the full context. Nimur 05:17, 28 April 2007 (UTC)[reply]

Balancing chemical equations

Is there website where you can put a chemical equation and it will show you the balance? Thanks —The preceding unsigned comment was added by 76.64.132.195 (talk) 03:58, 28 April 2007 (UTC).[reply]

I'll look, but like all skills, it gets easier the more you practice. --Russoc4 04:15, 28 April 2007 (UTC)[reply]

Edit: I found two of them. One here and another here. --Russoc4 04:16, 28 April 2007 (UTC)[reply]

Virus or bacteria phases

Are there any viruses or bacteria that exist in a liquid or gaseous state (not meaning in a liquid or gaseous medium) or do they only exist in a solid state? Nebraska Bob 06:20, 28 April 2007 (UTC)[reply]

They only exist in solid state. − Twas Now ( talk • contribs • e-mail ) 06:34, 28 April 2007 (UTC)[reply]

Er, solid and liquid, you mean...? (Never liquid alone, mind.) TenOfAllTrades(talk) 15:39, 28 April 2007 (UTC)[reply]

A Bacteria or course, being a cell could never exist as a gas, although most of its cytoplasm and its plasma membrane(s) exist in a quasi-liquid/gel state. A virus is mostly a mix of protein and nucleic acids, which could be converted into a gas phase through extraordinary measures, but both the protein and nucleic acid portions would be completely denatured, making the virus inactive or "dead" if you prefer--VectorPotential^Talk 15:44, 28 April 2007 (UTC)[reply]

Why limit ourselves to states of matter and not waves? As of 2002, buckyballs have been diffracted and work was under way to do viruses. DMacks 07:27, 30 April 2007 (UTC)[reply]

Overarm throw analysis

Hi

I need to analyse the sequential muscle contraction of an overarm throwing action, including joints. It has to include the function of the muscles, concentric or eccentric contraction, which muscles act as stabilizing muscles etc. I have no idea where to start and couldn't find any info... Please help.

Mike196.23.232.64 09:16, 28 April 2007 (UTC)[reply]

You probably want to get a list of all the muscles in the arm, to start with. The hand is probably ignored, right? Then there's the trapezius, in the back, Deltoideus on the shoulder, Pectoralis minor and major in the chest, Brachialis flexing the elbow joint, and Extensor carpi radialis muscles and friends, but if you're ignoring all the ones in the hand then you can't ignore those really can you? Somebody else can give a better answer than me. ;)[Mαc Δαvιs] ❖ 10:14, 28 April 2007 (UTC)[reply]

Before burying people

Is it true that a funeral home or morgue has to wait 72 hours before a body can be buried after a person is pronounced dead and during the 72 hours the body has to be monitored by sensors to determine if there are any remaing signs of life? 71.100.8.252 11:34, 28 April 2007 (UTC)[reply]

Not true. --Zeizmic 12:14, 28 April 2007 (UTC)[reply]

The embalming process removes all doubt. --TotoBaggins 13:21, 28 April 2007 (UTC)[reply]

Being buried alive was more of a legitimate concern in the 19th century, per [1] and the fear was enhanced by Edgar Allen Poe's story The Premature Burial which is available free online at [2]. Embalming or an autopsy removes all doubt. Edison 16:08, 28 April 2007 (UTC)[reply]

Organ donation, followed by an autopsy and then embalming removes all doubt. --TotoBaggins 01:51, 29 April 2007 (UTC)[reply]

It's not true, and it's not true whether or not the body is embalmed. Many unembalmed people are buried within 24 hours for religious reasons. - Nunh-huh 16:32, 28 April 2007 (UTC)[reply]

Burial within 24 hours means that rigor mortis has set in and departed, but putrification has not yet set in. There is the slight possibility of the "deceased" not actually being dead, and there is a window of time within which they might awaken after being buried alive, and before the oxygen supply is exhausted, however unlikely that might be. In the 19th century, there were patents for devices consisting of an alarm bell with a control in the casket, so the buried person could summon help if he awoke after burial and before death from hypoxia.[3][4][5][6][7] My favorite method was a device whereby the towns most patient and easily amused individual operated a machine having a rotary crank which was clamped to the deceased tongue and which yanked on same repeatedly for several hours. [8] Today, a disposable cell phone might be a better choice. (Damn! No service, and the battery is about dead. Oh well, have to claw my out like a vampire.) A modern physician should be able to determine whether the person is deceased. Electrical monitoring of heart and brain activity should be more diagnostic than 19th century methods such as placing a silver object in front of the nose to detect respiration and feeling for a pulse. Edison 01:25, 29 April 2007 (UTC)[reply]

To make sure people can be shot between the eyes, or cremated. Or you could drive a stake through their heart and scatter rose seeds around the grave.

lol Rfwoolf 20:48, 1 May 2007 (UTC)[reply]

Equation problem

This was in a past paper, and I don't know how to do it. There is a diagram of a titration with a burette containing sulphuric acid, 0.20 mol/l and a conical flask with 20cm³ of potassium hydroxide solution and indicator; and a table of results:


_	Rough titre	1st titre	2nd titre
Initial burette reading/cm³	0.5	21.7	0.3
Final burette reading/cm³	21.7	42.4	20.8
Volume used/cm³	21.2	20.7	20.5

The first question I can get easily (what everage volume should be used to calculate the number of moles of sulphuric acid needed to neutralise the potassium hydroxide solution): 20.6cm³.
The second part I can get fine, which is calculate the number of moles of sulphuric acid in this average volume: n = CV = 0.20 * (20.6*1000) = 0.00412 mol.
Finally, and this is the part I cannot get. The equation for the titraion is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Calculate the number of moles of potassium hydroxide in 20cm³ of potassium hydroxide solution. Any ideas are much appreciated. Thank you. 81.154.211.80 12:39, 28 April 2007 (UTC)[reply]

In the equation shown, how many moles of potassium hydroxide react with each mole of sulfuric acid? TenOfAllTrades(talk) 15:35, 28 April 2007 (UTC)[reply]

One mole of sulphuric acid reacts with two moles of potassium hydroxide. 81.154.211.80 15:58, 28 April 2007 (UTC)[reply]

So 0.00412 moles of sulfuric acid reacts with...? TenOfAllTrades(talk) 16:54, 28 April 2007 (UTC)[reply]

Applications of Linear Algebra

Could you, please, cite some direct applications of linear algebra in the natural sciences? Thanks. --Taraborn 14:38, 28 April 2007 (UTC)[reply]

Quantum mechanics, Molecular orbitals, Computational chemistry, NMR, Group theory, just to name a few. All of these require a basic understanding of Linear algebra--VectorPotential^Talk 15:15, 28 April 2007 (UTC)[reply]

I would suspect systems of linear equations to be the most common application of linear algebra. They arise everywhere and if you want, I and others could give you lots of specific examples, but perhaps you already know some. —Bromskloss 16:43, 28 April 2007 (UTC)[reply]

Classical mechanics, especially anything involving resonance, the moment of inertia tensor, etc. All this even more so in engineering applications. Even more important: As foundation for the idea of orthonormal function system, which leads to Fourier transformation, multipole expansion, etc and has applications in virtually any kind of physics. And much more. I think there is no part of physics that you can really understand in depth without linear algebra. So: Hang on and study the stuff. No first-year student of physics understands what it is good for, but after your Masters, it will have become so embedded in your thinking that you use linear algebra as naturally, frequently and intuitively as you now do use elementary arithmetics. Simon A. 17:39, 28 April 2007 (UTC)[reply]

reg hall effect non contact type current sensors

Insert non-formatted text here respected madam/sir

like i would like to know exactly is ther any formula to caluclate the turns ratio of the secondary winding it the current sensors .the details given below

originally we use an current sensor with input current of 5000A with an output of 1705 milli Amps with the turns of 6000 in the secondary,so i want to modify it with decreasing the input current to the required level and making the output constant(1705) so please guide me in this and i would like to know the relation how to caluclate the secondary windings when we are using an hall effect non contact type current sensors

lokking for ward for your reply.

thanks & regards, —The preceding unsigned comment was added by 202.63.119.180 (talk) 16:31, 28 April 2007 (UTC).[reply]

The question as stated is not clear enough to allow a good answer. You say originally you used 5000 amperes of input current and 1.705 amperes of output current, which implies a turns ratio of 5000amperes/1.705amperes=2932.5513 to one, but then you say 6000 turns in the secondary, wihich is very unclear. Then you say it is a Hall Effect sensor, which is not a current transformer, but which could be connected to the secondary of a current transformer. Please provide a diagram of the circuit showing all components. Current transfoemers used in utilities might have 5000 or 6000 amperes as the primary rating with perhaps 5 amperes as the secondary rating. Edison 01:18, 29 April 2007 (UTC)[reply]

NACA 632

The wing on the Koenigsegg CCGT is said to have the NACA 632 profile [9]. However, as far as I can see, the article does not mention any three-digit profiles. Does "NACA 632" exist or is it a typo? —Bromskloss 16:36, 28 April 2007 (UTC)[reply]

Gaussian curves: peak height vs area

When comparing two Gausian peaks, do the peak heights correlate with the areas under the two curves? Where would I go to find this answer?--130.126.228.60 16:53, 28 April 2007 (UTC)[reply]

The area under the curve of a gaussian distribution is always one for every curve. The height of a peek corresponds to the width of the peek at half height. It is either narrow and high or stretched and low. The obvious place to look is Gaussian function. —The preceding unsigned comment was added by 84.187.18.197 (talk) 17:13, 28 April 2007 (UTC).[reply]

Well, strictly speaking a Gaussian function does not have to integrate to one. Only when said Gaussian function is the probability distribution function (PDF) of the normal distribution must it fulfill the requisite integrate-to-one condition for valid PDFs. (P.S. - Questions like this should go on the mathematics reference desk) -- mattb 18:16, 28 April 2007 (UTC)[reply]

If one has fit a pair of peaks that were each fit to a Gaussian, how does one determine their relative areas? I think that is the question being posed. Does one need only compare peak heights or what calculation exactly is required to get the two areas?--Smokefoot 18:22, 28 April 2007 (UTC)[reply]

No, comparing the peak heights isn't enough (look at the aforementioned normal distribution page for examples of Gaussian functions with different peak heights but the same area). You can determine the area under the curve by integrating it over the real line. Looking at the solution to the Gaussian integral, you can see the simple relationship between peak height and area. -- mattb 18:38, 28 April 2007 (UTC)[reply]

I think you could multiply the height of your peak with the "width" of it to get something that is proportional to the area and can be compared with the corresponding value of other curves. By "width", I mean for example what is illustrated in the figure, i.e., the distance between the two points where the peak have fallen off to half its maximum value. —Bromskloss 19:41, 28 April 2007 (UTC)[reply]

Abundance of land slightly above see level

There has been a fact abouth the geology of earth, that always puzzled me, but I neved noticed it enough to ask the obvious question. Take a look at this diagram: http://en.wikipedia.org/wiki/Image:Earth_elevation_histogram.svg.

You can see that few areas are mountains, few are deep see trenches and almost all of the surface area on earth is either flat land from -200 to 1000 meters or deep see from -2000 to -6000 meters. There is a small costal shelf followed by a steep cliff and a deep ocean.

Why? What is the reason for the existence of the little bump, that cliff in this diagram that gives us so much more habitable land than we should expect?

That's an interesting question. One thought is that it's based on relative erosion rates. Mountains erode quite quickly because of the water pouring down in torrents during rains and winds slamming into them. Relatively flat, low land, on the other hand, has a slower erosion rate. There is also one case where it forms that way, from volcanic lava. This lava tends to flow until it hits water, then it hardens and cools. This leads to a long flat lave field just above the water line. What I would have a harder time explaining is the low amount of relatively shallow water. StuRat 17:20, 28 April 2007 (UTC)[reply]

I would expect the land formed in this way to be moved up or down by continental drift soon. And how much land is formed by lava flowing into the see? It is so little per day, can it accumulate to a whole continent over time?

This would also explain why sea level is nearly exactely at the cliff. ~~It is remarkable that this way a rise or fall in sea levels will not much change the area covered with water.~~ (Not my day today) 84.160.242.138 17:42, 28 April 2007 (UTC)[reply]

That was one point why I found this so interesting. This is exactly the situation in which a small rise or fall of the see level will make a huge change in the area covered by water.

All the Hawaiian Islands are volcanic, as are many more. I suspect that large portions of the continents are also very old volcanic shelves, having since become metamorphic rock or weathered into particles and become sedimentary rock. See our supervolcano article to get an idea for some of the huge volcanoes we've had in the past. StuRat 22:06, 28 April 2007 (UTC)[reply]

Spacecraft chart

Does anyone know where I can find a chart of all the current spacecraft in the solar system outside of Earth orbit with their location and vectors noted ? It would be similar to this one [10], but would include all spacecraft, including non-NASA ships. StuRat 17:29, 28 April 2007 (UTC)[reply]

It's not a chart, but User:Michaelbusch showed me this fun site [11]. Someguy1221 22:55, 28 April 2007 (UTC)[reply]

Are you sure there are any non-American spacecraft that qualify? Your reference seems to imply that there are only those five. After reading Soviet space program, I suspect that the Soviets were more concerned about projects with a more immediate payoff and propaganda value. According to this (past and current missions), those five seem to be it. Clarityfiend 03:52, 29 April 2007 (UTC)[reply]

Those are just the 5 which have left the solar system, I'd expect far more to be inside the solar system, yet outside of Earth's orbit. Some may be lost, I suppose, if they've stopped transmitting, but we could still make a reasonable guess as to their current location. StuRat 08:48, 29 April 2007 (UTC)[reply]

I don't know what's wrong with me lately. That's the second question I've misinterpreted. D'oh. Clarityfiend 10:27, 29 April 2007 (UTC)[reply]

experiment to induce a mild form of (Synesthesia?)

A subject is placed in an environment where all ambient noise of a sufficient loudness is automatically recorded and stored in a sampler. At frequent intervals, the sampler plays the recordings back as music. After the subject has spent some time in that environment, he is then removed to a "normal" environment with no such sound manipulation. The subject then reports a subjective perception that "the music is still there" and he claims to hear everything as including some kind of musicality. Eventually this fades.

what is the formal name for this kind of "subjective change in perception"?
is the result described above plausible?
if yes, are there any other circumstances where similar results might be produced through other kinds of manipulation of sensory input?

TIA. dr.ef.tymac 17:59, 28 April 2007 (UTC)[reply]

I don't understand what you mean. How can the "ambient noise" be replayed "as music"? Do you mean that music is mixed with the recording somehow? Nimur 21:54, 28 April 2007 (UTC)[reply]

No, consider for example someone dropping a wine glass on the floor, or a dog barking. That sound would trigger a recording device, the recording device would split the sound into 700ms samples, and those samples would be mapped to individual keys on an electronic keyboard. The keyboard would then randomly play MIDI files using the "ambient noises" as the timbres (instead of recorded musical instruments as is usually done with a sampler). The term I am looking for something akin to the auditory equivalent of "phantom limb" syndrome or synesthesia. dr.ef.tymac 05:52, 29 April 2007 (UTC)[reply]

This is a little weird and I'm not at all sure it's really a reference desk question, but here's my take, insofar as I understand what you're getting at. It's not clear to me whether you're suggesting that the samples be played back at different speeds for the different notes. If not, there's nothing melodic going on- it won't really sound much like music to us. If you do want to gets notes out of the samples, then sure, music could be played using environmental sounds as the instrument (many examples exist of people playing melodies using weird samples). But, these same environmental sounds as they naturally occur aren't going to sound musical, just like a little kid banging randomly on a piano doesn't sound musical. Friday (talk) 06:05, 29 April 2007 (UTC)[reply]

I am a big fan of experimental, ambient, electronic and industrial music, using every day sounds sampled as instruments is nothing new. One of the long standing jokes about people like me, collectively sometimes referred to as rivet heads was: "You know you are a rivet head if you miss the concert because you got caught up dancing to the construction site you passed on the way." I hear musicality in very many every day objects, things like my hard disk clicking, the train going by, the pedestrian crossing clicking, they are all like a symphony to my ears ;) Vespine 03:19, 30 April 2007 (UTC)[reply]

I can understand where Vespine is coming from regarding "experimental music," but I disagree with the word-choice. I think the defining characteristics of music, which distinguish it from noise, are rhythm and tonality. Without these two elements, sounds do not fit the definition of music; regardless of how pleasing the sounds may be (because "pleasing" is a purely aesthetic and subjective definition). Nimur 20:14, 30 April 2007 (UTC)[reply]

Have you never heard rhythm and tonality in, say, a washing machine cycle? And hum a tune over a train going past if you think it's lacking tonality, because it certainly has rhythm;). Vespine 22:55, 30 April 2007 (UTC)[reply]

That might have been what inspired Honegger to write Pacific 231. JackofOz 01:39, 1 May 2007 (UTC)[reply]

Speaker power

My audio system merely uses 100 watts of electric energy but the labels on the speakers show 1500 watts.Where does such power come from? —The preceding unsigned comment was added by 202.70.64.15 (talk) 18:19, 28 April 2007 (UTC).[reply]

Marketing.

Atlant 23:14, 29 April 2007 (UTC)[reply]

An audio amplifier, presumably. The power rating on your speakers defines a maximum; you can certainly deliver less power to the speakers (and you usually will). -- mattb 18:21, 28 April 2007 (UTC)[reply]

That may also be the maximum allowable power for those speakers; you could probably hook it up to a larger amplifier to deliver a louder sound. Nimur 18:43, 28 April 2007 (UTC)[reply]

What power is it really the speakers are labeled with? Is it the maximum electric power you shuld feed them with or the power they will let out into the air (as a sound wave) when you feed them maximally? —Bromskloss 19:25, 28 April 2007 (UTC)[reply]

I think the 1500W rating is just PMPO which is less than useless IMHO. --antilived^{T | C | G} 01:07, 29 April 2007 (UTC)[reply]

Also, power for audio amplifiers is usually specified with Total Harmonic Distortion percentage. Depending on the output stage of the amplifier (Class A preferably, Class AB probably more common), the distortion is reduced with percentage of total power. Therefore, for a given loudness, a 200 Watt amplifier will generally have less distortion than a 100 Watt amplifier. When comparing amplifier's the power number matters when the THD is the same. This is the reason why amplifier power matters and also why THD is also relevant when comparing two amplifieres. Beware of the 150W 0.1% THD compared to the 100W 0.05% THD. They could very well be the exact same amplifier. Likewise, the difference between a 200W 0.05%THD amplifier and a 100W 0.05% THD might be very noticeable not for volume but because the THD of the 200W amplifier running at 100W should be significantly better than the 100W amplifier running at 100W. It's not louder, it's cleaner. --Tbeatty 03:32, 29 April 2007 (UTC)[reply]

Capillary threads for power production

Hi, I have two questions. The first regards the capillary threads/tubes that exist in trees. These are super-smart, super-thin threads that don't actually spend any energy at all on carrying water from the soil, up the trunk of the tree. Since this is making use of surface tension, I was thinking if this could be used in a closed system to produce electricity.

Basically the idea is to have a standard dynamo connected to a watermill, on which water falls from a pool above. The water then flows into a large pool that surrounds this watermill, where millions, maybe billions of tiny such capillary threads made of glass or similar are. The water gets sucked extreeeemely slowly up this way, before it reaches the top pool, and falls down on the watermill again. Even if the watermill was made from super-light material, you'd no doubt need a wast quantity of such threads. My question, if I haven't asked it yet, is if there's any future in this? I realize it would have to take a blasted lot of room... :)

The second question is, does Wikipedia have an article on inventions that are meant to last forever, perpetuum inventione or something like that? Thanks for all help! :) 81.93.102.185 18:58, 28 April 2007 (UTC)[reply]

Without even addressing the specifics of capillary action, the flaw with this idea is that it's just another take on the classic perpetual motion machine. Getting energy from "nowhere" violates the first law of thermodynamics, and thus renders this construction impossible. Perhaps someone a little more familiar with fluid dynamics can point out the specific reason this wouldn't work. -- mattb 19:11, 28 April 2007 (UTC)[reply]

That's simple. The water does not fall out at the top. It sticks in due to (doh!) capillary force. —The preceding unsigned comment was added by 84.187.18.197 (talk) 19:18, 28 April 2007 (UTC).[reply]

That's right, c-action sucks water into small tubes, whether that means sucking upwards or downwards. StuRat 21:59, 28 April 2007 (UTC)[reply]

Sounds right... It might also be noted that at a certain height, the weight of the column of water will equal the capillary force holding it up. If memory serves, this is one of the mechanisms that restricts the maximum height of plants. -- mattb 19:24, 28 April 2007 (UTC)[reply]

If you did this outside in the sun, I think it would work. Just like with plant xylem, the water would evaporate at the top of the tubes, and more water would be pulled up. You'd need something above the tubes to re-condense the water vapor so it would fall back down. But in this case you don't have a closed system, so you'd be talking about a method of harvesting solar power, not a perpetual motion machine. --Allen 19:31, 28 April 2007 (UTC)[reply]

A much bigger system is called "hydroelectric power" :). --Tbeatty 03:51, 30 April 2007 (UTC)[reply]

Though that rather relies on the water already having plenty of readily usable gravitational potential due to geographical features... -- mattb 13:51, 30 April 2007 (UTC)[reply]

Cosmic Inflation

I have read in some popular science publication that inflation ended when the size of the universe was about the size of a baseball. Can you give me a technical publication reference which discusses this matter?

Thank You,

Alfred Bussian Bussian 19:52, 28 April 2007 (UTC) <email address redacted>[reply]

Have you started with our article on cosmic inflation? There are a great many references available there. — Lomn 20:23, 28 April 2007 (UTC)[reply]

Note that there is a difference between "inflation" and "expansion" - they mean two totally different things in cosmology. --bmk

As an aside, does it mean anything to talk about the size of the universe at such an early stage? What does "Size of a baseball" mean when matter/energy will warp space time to such an extent? what about time in the early universe? (i.e. talking about the universe in the first bilionth of a second - does it mean anything? ). --Tbeatty 02:32, 29 April 2007 (UTC)[reply]

There is no evidence that any baseball was played at that time, when the universe was a billionth of a second old. Having to watch a baseball game would have made time seem to pass much more slowly, had there been any observers. Edison 06:41, 29 April 2007 (UTC)[reply]

Distances don't mean much at this early time, no. I asked a cosmologist for a straight answer about this once, but he kept evading the question - sometimes deliberately, sometimes not even realising he was doing it. In the end we gave up, realising both the question and the answer had little meaning. Cosmologists measure this early phase purely in terms of the time since the big bang, and the universe's scale factor, not 'real' distances. Spiral Wave 18:41, 29 April 2007 (UTC)[reply]