Wikipedia:Reference desk/Archives/Science/2008 October 25

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October 25

Looking at Earth

What does Earth look like in the multiple dimensions (say 7th dimension) of M theory ?

Would the answer to such a question be useful to someone without seven-dimensional eyes? —Tamfang (talk) 07:54, 25 October 2008 (UTC)

Ordinary physics is confined to 3+1 dimensions in these models, so there's nothing to see in the extra dimensions (no light) and nothing to see it with (no way to construct an eye). -- BenRG (talk) 16:21, 25 October 2008 (UTC)

So are everyday objects 0-sized in the extra dimensions or just very very small? --Tango (talk) 16:32, 25 October 2008 (UTC)

Objects will retain their size if you add extra dimensions, it is just if you measure in the extra dimension, there will be no thickness, of course this will depend on your fabricated physics if there is or is not any extent onto the extra dimension(s). Graeme Bartlett (talk) 20:13, 25 October 2008 (UTC)

Of course it depends on the theory, I'm asking whether there is any spacial extent to everyday objects in the extra dimensions of String/M theory. --Tango (talk) 21:21, 25 October 2008 (UTC)

I don't know, sorry. I gather than in brane cosmology the Standard Model particles are supposed to be open strings whose ends are confined to a 3+1 dimensional subspace, but the middles aren't confined, so I think the answer is "sort of". -- BenRG (talk) 22:22, 26 October 2008 (UTC)

a 2D projection of a 4D object

There are some misconceptions here. One is that we can't possibly answer this question...when in fact we can...and it's rather easy in fact. First we have to think carefully about how our visual system works. Our eyes don't see in three dimensions - each eye is like a camera - it sees a two-dimensional projection of the 3D world onto the backs of our eyes. Using two eyes - separated horizontally in space, we can infer some information about that third dimension by a process of triangulation. So - if there were additional dimensions - we'd see projections of those onto our same 2D eyes - and because the separation between our eyes only adds ONE new piece of information (the lateral displacement of the image in our two eyes), we would be unable to deduce anything 'new' about the 4th and subsequent dimensions. When you see pictures of things like tesseracts (four-dimensional hyper-cubes) - they are generally represented as 2D projections.

Some people are going to argue about this (they did the last time we answered this kind of question). I'm a computer graphics guy. I can quite easily program my computer to generate 2D, 3D, 4D, 5D...or any other number of dimensions inside the computer's memory - the computer really doesn't care how many dimensions there are - the software is very similar no matter how many dimensions there are. But the computers' display is only a 2D surface - so I have to write software - even for three dimensional objects that explicitly reduces them to two dimensions so I can display them onto that 2D screen. We do this in various ways - but the one that's simplest to envisage here is called "Ray Tracing". The idea is that you calculate an imaginary 'ray' that comes out of your eye - passes through a pixel on the computer's screen and goes off into the 'virtual world'. We calculate what part of what object that ray first hits. We take the color of that object and place that color onto the pixel that the ray passed through. We do this thing for every single pixel on the screen - and then we have an image. This process is identical to what happens with our real 2D eyes - and that process of tracing rays is just the process of figuring out where a ray of light came from on it's journey onto our eyes. That process is EXACTLY the same in 3D as it is in 4D, 5D, and so on.

So it's really easy to have the computer draw things in more dimensions - in a manner that PRECISELY shows what it would be like if we lived in a 4D, 5D or more-D world. So what does it look like? We might hope that some amazing revelation would appear - but really, it doesn't. It's pretty disappointing actually. Even when we do ray tracing separately for each eye and feed the two views into our eyes using a virtual reality helmet or something - the 4D world doesn't look any different from our normal 3D world. There are plenty of places on the web with animations of 4D objects which are spinning or something. They look much like ordinary 3D objects (which we're seeing in 2D because that's how our eyes work) - except that they seem to change shape in strange ways.

Hence - we know exactly what a 7D world would look like with our normal 2D retinas. What we don't know (and can't easily imagine) is how it would look if we were in a 7D world with 6D retinas...but it's a somewhat meaningless question because if we had bodies (and especially brains) with more dimensions, we wouldn't be anything like we really are - so asking how we would see things if we were not ourselves is a really meaningless question.

SteveBaker (talk) 21:55, 25 October 2008 (UTC)

So if we lived in a 7D world but were surrounded by 3D objects and we only moved in the standard 3 dimensions then everything would look the same - we could not tell that extra dimensions existed. But if we moved along one of the extra dimensions - well, it's not clear that we would "see" anything, since (I think) the inverse-square law tells us that photons only travel in 3 dimensions (on a macroscopic scale). I suppose we might conceivably wander into "parallel" 3D worlds filled with their own 3D objects and 3D photons. Gandalf61 (talk) 10:58, 26 October 2008 (UTC)

I think you're basically describing brane cosmology. --Tango (talk) 14:02, 26 October 2008 (UTC)

When you move around a 3D object in our 3D world - viewing it with our 2D eyes the shape of the projection of the object changes. A cube morphs from a square (when it's exactly lined up in front of us - to something that's almost a rectangle if we move off to one side or a hexagon if we move up or down and off to one side. The only reason we know it's a cube is because we have two eyes - and each gets a slightly different view - and because the effect of light striking the cube and reflecting different amounts at different angles makes the faces appear in different colors. The same thing happens if you move around 4D objects in a 4D world with our 2D eyes. The 3D projection of a tesseract looks like a cube from some angles (like the view in the picture above) and like four cubes that are slanted together at other angles. Which means that when you project it into 2D - to us it's like a square or a rectangle or a hexagon or an octagon - and just like the cube in 3D, it changes shape as you move around it. The effects of light reflection in the 4th dimension will allow us to recognise more of the shapes of things - but it's really not going to look all that strange. But you don't have to take my word for it - there are plenty of web sites that show 4D (and more) objects projected into 2D. Everyone is acting as if there is some deep mystery - but there really isn't.

It's possible that the physics of light transmission in the presence of additional dimensions might be wildly different. But I suspect that can't be true because if it was, we'd have concrete proof that the 26 dimensions of string theory were not a physical truth. Since we have not yet debunked string theory - I don't see how there can be solid proof that light acts differently in universes with more than 3 spatial dimensions. If you want 'weird' - start thinking about cosmologies with multiple temporal dimensions! SteveBaker (talk) 19:04, 26 October 2008 (UTC)

You say "That process is EXACTLY the same in 3D as it is in 4D, 5D, and so on", but it's not the same: if you cast rays from a point (the eye) through a 2D surface (the computer screen), the locus of points that those rays intersect is three dimensional. You would only see a slice of the 4D object if you ray-traced that way. Pictures like that one of the tesseract are made by a two-step process, first projecting the tesseract onto a three-dimensional "screen", then reinterpreting that screen as a new 3D scene and projecting that onto the real screen. The human eye won't do that intermediate step, and I'm having trouble imagining any physical object that could. Even if you can concoct a 4D physics in which all four spatial dimensions are on equal footing (so that shapes like the tesseract make sense) and in which a 3-dimensional eye can somehow exist, there's no way the 3D eye will see pictures like that one of the tesseract, unless maybe with the help of some magical material that's transparent to "3D light" but emits 3D light when it's struck by "4D light". If there's only one kind of light and it travels through all four dimensions then the 3D eye will just see a uniform blur, because the interior of the eye will be flooded with light that gets in via the extra dimension. I can't figure out what you mean by "different" in your last paragraph. On the one hand light in these theories behaves like light in the real world: it's confined to three spatial dimensions. On the other hand that makes the other dimensions different from the three usual dimensions with regard to how they transmit light. You seem to be pulling a definitional switcheroo, saying that since these models haven't been falsified, light transmission must be confined to three dimensions (anything else would be "different" from what we observe) and then saying that the tesseract picture makes sense (because otherwise the extra dimensions would be "different" from the usual three). -- BenRG (talk) 22:22, 26 October 2008 (UTC)

Well, you'd stand a chance of being right if I hadn't actually done this with computer software. There is absolutely no problem in tracing the path of a ray in four dimensions instead of three. 4D geometry works just fine. I can simulate the motion of our 3D eye with its 2D retina inside that 4D geometry - have it translate and rotate about all four axes, place 4D objects into that simulated world - and so long as we can assume that 4D photons would somehow transfer energy into our retinal cells and that 4D photons travel in straight lines in a 4D world, everything works out just fine. If those assumptions are not true - then the idea of 'being in a 4D world' becomes rather meaningless anyway - but taking the premise that "IF we could be in a 4D (7D) world - what would it look like?" - then this is a reasonable supposition. With such software, I can use a joystick with some extra axes to manouver around in the 4D world and watch pictures of the results at close to realtime rates (for a small image area) - and there is no special difficulty. I'm emphatically NOT converting a 4D image into 3D and then into 2D - I'm tracing the path of photons directly in 4D and seeing where they impact the 2D screen (which might as well be a "retina" except that I don't want to model a lens and deal with flipping the image up the right way at the end). SteveBaker (talk) 14:28, 27 October 2008 (UTC)

But that's not true. I don't care if you've written software that does it, it's not true. An eye is a camera. A camera is a dark room. It must be dark, or it will not work at all. All light must pass through the aperture, or else no image will be formed. Where do you cast your rays from, and in what direction? Do they originate from the retina? If so, do they all pass through the pupil (which would fix the direction), or do they go "out the side", and in the latter case, which way do they go? Or do they originate from the pupil? In that case, how do you map a direction from the pupil to a location on the retina? Or let me put it this way: in appropriately chosen coordinates, the 3D-to-2D projection maps (x,y,z) to (x/z, y/z). You say the 4D-to-2D projection is exactly analogous. So, in whatever coordinates you like, what is it? Is it (x/zw, y/zw), or (x/(z+w), y/(z+w)), or what? I'd love to see the source code to your ray tracer so I can understand what the heck you're doing, but I can tell you right now that it isn't optics, whatever it is. -- BenRG (talk) 22:21, 31 October 2008 (UTC)

A small niggle. You show a picture of a projection of a tesseract. But it would have to be transparent like a transparent cube to see all those cubes and lines in it. If one was 4 dimensional and had an analogue of our eyes you'd see only the four nearest cubes of the tesseract at most - like you only see the nearest 3 squares of an opaque cube at most. Dmcq (talk) 15:33, 26 October 2008 (UTC)

Yes - I agree. Images of "solid" tesseracts look pretty much just like normal 3D polyhedra that change shape as they move. SteveBaker (talk) 19:04, 26 October 2008 (UTC)

I've sen those too, I prefer just to remove the extra lines in the projection picture. Just because the 4d figure is opaque doesn't mean the image has to be. Really one would have to see the 3D in depth like a dolphin with sonar to have the equivalent of an eye for 4D. Dmcq (talk) 20:56, 26 October 2008 (UTC)

ISS flyby and binoculars

In a few hours, the ISS is going to make a visible pass almost directly above where I live. heavens-above.com says that it will be magnitude -2.2 which should be about the brightest I've ever seen it. It will fly almost directly overhead. Anyway, I've heard of some people viewing the ISS through binoculars and supposedly you can see solar panels and other doohickeys (couldn't think of a better word) with them. I was just wondering if the glare from the ISS would be too much to see anything in detail if it's so bright or if I should wait for a dimmer flyby to see if I can't see shapes. What would be the ideal brightness to see shapes?

If I can get a good view of it through binoculars in the limited time I have during the flyover, I might try to place my digital camera behind one lens of my binoculars to try to get a good picture of the station, so stay tuned for updates! (if the video is good, I'll post it on youtube) 63.245.144.77 (talk) 06:38, 25 October 2008 (UTC)

Does the pass enter the Earth's shadow while still above the horizon? If so, then even if it's too bright to see close up for most of the pass you should be able to see it for the last couple of seconds. --Tango (talk) 11:58, 25 October 2008 (UTC)

Hi. I've never been able to see the solar panels of the ISS on binoculars on my 8x30, but I think I have been able to see a bit of detail with my telescope operating at 36x. It's really difficult to catch one on camera through the telescope because it moves so quickly (and you'd need a special camera adapter). Nevertheless, there are videos on youtube showing the ISS. Thanks. ~AH1^(TCU) 19:06, 26 October 2008 (UTC)

An Electronic Planet

Is it possible to have an electronic planet. By that I mean, imagine lots and lots of electrons in space with no protons or neutrons. Imagine the mass of electrons is such that gravity takes hold and forms a spherical gravitational body like a planet. How big would the planet be and what colour is it? Can a spaceman with a spaceship land on the electronic planet?

122.107.147.49 (talk) 09:48, 25 October 2008 (UTC)

Since Coulomb repulsion is going to be several orders of magnitude larger than gravity, it will require some other kind of external force to stabilize it. On top of everything, you will have the Fermi pressure destabilizing the sphere even more. In other words, no, it is not possible. MaNiAdIs-Talk-GuestBook 09:56, 25 October 2008 (UTC)

• You can have neutron stars though. I guess if I fired an enormous number of electrons at a black hole I could get an immense electrical field.I'd like to do this in near earth orbit, the funding agency just won't give me the money though. Peasants, they just don't understand the importance of science. :) Dmcq (talk) 14:17, 25 October 2008 (UTC)

  Highly charged black holes have some strange properties - I think one made up entirely of electrons would contain a naked singularity. --Tango (talk) 15:11, 25 October 2008 (UTC)

The current thinking is that it's impossible to have a black hole with a charge/mass ratio larger than 1 (in natural units where G = 1/4πε₀). The electron's charge/mass ratio is about 10²¹. Simply firing electrons at a charged black hole won't work to increase the charge because electric repulsion will prevent their getting anywhere near the event horizon. -- BenRG (talk) 16:47, 25 October 2008 (UTC)

Ben - could you calculate that for a couple if different sizes of black hole? On a previous question there was consideration of the energy to have one mole of electrons in a confined space with out other protons. Graeme Bartlett (talk) 20:19, 25 October 2008 (UTC)

Calculate what? The largest possible charge? That's just a matter of plugging numbers into formulae. If my derivation is correct, the formula you need is $Q=2M{\sqrt {G\pi \epsilon _{0}}}$. Give me a minute and I'll convert that into something more usable. --Tango (talk) 21:10, 25 October 2008 (UTC)

Ok, taking constants to 1 sig fig, that comes out in SI units as $Q={\frac {\sqrt {7}}{3}}\cdot 10^{-10}\cdot M$. --Tango (talk) 21:14, 25 October 2008 (UTC)

So that means a proton or electron with mass 1.6x10^-27 or 8.1x10^{-31 kg}and with charge 1.6x10^-19 Coulombs could not be compressed to a black hole, In fact only if 1 in 10^-18 of particles in an object are charged you still could not make a black hole out of it, is that roughly correct? Graeme Bartlett (talk) 21:38, 25 October 2008 (UTC)

I haven't checked your arithmetic, but basically yes. If a charged sub-atomic particle (or collection of them) were to collapse into a black hole it would result in naked singularity and it is hypothesised that they cannot exist. --Tango (talk) 22:04, 25 October 2008 (UTC)

I suspect that the energy in the electric field when converted to mass would exceed the mass in the black hole if the charge were too high. This means all the earlier speculation about micro black holes we had before is probably irrelevant as the tiny black hole could not swallow a charged particle. Graeme Bartlett (talk) 10:44, 26 October 2008 (UTC)

My guess then is that a highly charged black body would emit an enhanced amount of Hawking radiation which had a preponderance of negatively charged particles in it. Does that sound about right? Dmcq (talk) 20:18, 25 October 2008 (UTC)

A highly charged black hole (by which I mean one with greater charge than mass in the appropriate units) just leaves the laws of physics in shreds. If such things are possible, our theories can't handle them. A black hole with a lower, but non-zero, charge is called a Reissner-Nordstrom black hole - I have no idea what the Hawking radiation is like for a R-N hole, but someone here might. --Tango (talk) 21:10, 25 October 2008 (UTC)

The Hawking temperature is ${\frac {\kappa }{2\pi }}$ where $\kappa$ is the surface gravity. For a charged unrotating black hole that's ${\frac {1}{2\pi }}{\frac {\sqrt {m^{2}-q^{2}}}{2m^{2}-q^{2}+2m{\sqrt {m^{2}-q^{2}}}}}$. So a charged black hole radiates less than an uncharged black hole at a given mass, and an extremal black hole (|q| = m) has zero temperature and doesn't radiate at all. Large black holes radiate mostly photons, so a large charged black hole in a vacuum will tend to lose mass until |q|/m reaches one and then stop radiating. I think a small black hole will tend to lose its charge by radiating electrons or positrons. -- BenRG (talk) 21:19, 26 October 2008 (UTC)

That's interesting, thanks. How are you defining "large" and "small"? --Tango (talk) 11:37, 27 October 2008 (UTC)

Instead of a electron planet, look out for a cold neutrino planet. Perhaps these will form later in the life of the universe. Graeme Bartlett (talk) 10:44, 26 October 2008 (UTC)

Grounding & Bonding - static electricity safe guards

Moved question from VPT Franamax (talk) 09:30, 25 October 2008 (UTC)

According to the NFPA all flammable containers must be grounded and bonded to ensure no static electricity build-up resulting in an explosion. The question: is bonding necessary when the transfer container is plastic?

Ira Hayes —Preceding unsigned comment added by IraHayes46 (talk • contribs) 08:50, 25 October 2008 (UTC)

You said it yourself. "All" containers. --Russoc4 (talk) 13:40, 25 October 2008 (UTC)

This is a question that can't be met by a simple yes-or-no. A proper answer will depend on the type of container, the size of the container, the environment, and the type of flammable liquid being transferred, not to mention local health and safety regulations. You need to consult an expert to get a reliable answer to this type of question; some random guys on the Internet just aren't going to cut it. Your local fire department may be able to offer your advice, or at least be able to point you to appropriate resources. TenOfAllTrades(talk) 16:33, 25 October 2008 (UTC)

Static electricity builds up more efficiently on plastic containers than on metal ones - so yes - they do need to be grounded. SteveBaker (talk) 21:32, 25 October 2008 (UTC)

How do you ground a plastic container? Do you need to cover it in some kind of metal mesh? Just connecting a grounded wire to one part of it wouldn't seem to have much effect. --Tango (talk) 22:07, 25 October 2008 (UTC)

Exactly. For this to make sense, the plastic would need to be at least a little bit conductive, like and antistatic bag. Perhaps plastic gasoline containers are made that way? I can't find my DMM, or I would go check.-Arch dude (talk) 00:58, 26 October 2008 (UTC)

human eye.

when we come from intense light to dim light why human eye take time to see thing clearly? —Preceding unsigned comment added by 119.154.22.96 (talk) 12:06, 25 October 2008 (UTC)

See the article Adaptation (eye), which explains the phenomenon. The article doesn't really answer the question as to why it takes so long (20-30 minutes), however. --NorwegianBlue ^talk 12:36, 25 October 2008 (UTC)

20-30 minutes seems excessive. I'm typically able to see well in a changed lighting environment in 5 minutes maximum. —Cyclonenim (talk · contribs · email) 12:42, 25 October 2008 (UTC)

From this article, full light adaptation takes from tens of seconds to several minutes. There are two phases. Important factors are changes in calcium levels and cGMP. Axl ¤ [Talk] 12:58, 25 October 2008 (UTC)

The longest times are for full adaptation to night vision, and are correct. It takes up to 30 minutes for rhodopsin (a retinal pigment sensitive to very low levels of light) to recover after being bleached by exposure to bright ambient light. Rhodopsin is essentially insensitive to long-wavelength (red) visible light, which is why amateur astronomers are often seen carrying red-filtered or red-LED flashlights at observing sites. TenOfAllTrades(talk) 13:12, 25 October 2008 (UTC)

There are two completely separate things that happen. Firstly the iris in your eye shrinks down to let less light in - this takes less than a second. Secondly - IF you've been sitting in the dark for long enough to get dark-adapted, the chemical rhodopsin has formed to make your retina more sensitive. When the sunlight hits your light-adapted eye, the extra sensitivity you've built up causes that extra jolt of pain - but the rhodopsin is bleached out and ceases to function within just a few seconds. Going back into the dark - the process reverses - your iris grows large to let in more light and the rhodopsin starts to form again to make your retina's more sensitive. The 10 to 30 minutes time that some people here are thinking of is the time it takes the rhodopsin to build back up to its maximum level again. We can get un-dark-adapted very much quicker than we can become adapted again. SteveBaker (talk) 21:30, 25 October 2008 (UTC)

chemicals

how can we prepare low temperature colours which can stick on ceramic tile? —Preceding unsigned comment added by 119.154.22.96 (talk) 12:18, 25 October 2008 (UTC)

Public Health Question:Handwashing & faucets w/sensors-transfeered frWater Conservation

water-saving mechanisms for washing hands--how have public health experts reacted to them, does anyone know about this? I've experienced in restaurants how frustrating they can occasionally be as the mechanism ages and it takes longer to provide less water, so I wonder if boards of health have voiced concern about impatient food workers etc leaving the bathroom without washing their hands. As experts have gotten more worried about possible epidemics, perhaps literature critical of the new faucets has appeared? Thanks, Rich (talk) 03:24, 25 July 2008 (UTC)

not that i know of. However, Elementary School Washroom Faucets in Ontario (not saying anything about Canada as a whole) have become frustrating to use, according to a friend of mine's son (and many others that I have asked around our community). He says: "I wish the water would come out faster! I spend, on average, about 10 minutes in the washroom, 7 of which are spent waiting for the water to come out fast enough to clean my hands!"

-GameLoRDz (talk) 21:18, 22 October 2008 (UTC)

---

Thanks for any and all responsesRich (talk) 13:24, 25 October 2008 (UTC)

I can't answer the specific question but I don't know whether water is the biggest issue when it comes to hand washing. The biggest issue is most people don't wash their hands properly. Proper hand washing with soap, as described in Hand washing and Hand washing with soap generally requires about 20 seconds or so of rubbing wet soapy hands together and this is outside the stream of water. Only then should the hands be (thoroughly) rinsed. They then have to be dried thoroughly (usually for around 20 seconds as well). Many people don't rub their hands for close to long enough. Many people don't even use soap at all. And many people don't dry their hands properly (ironically if you don't dry your hands at all it may actually be worse then not washing your hands). I would say if people would wash their hands properly it would make a far bigger difference then whatever negative effect from water conservation measures. P.S. Yes I'm one of the many poor hand washers. Nil Einne (talk) 14:51, 25 October 2008 (UTC)

tsunami

What is the name given to the time interval when the sea recedes and the tsunami arrives? —Preceding unsigned comment added by Paulpearn (talk • contribs) 14:05, 25 October 2008 (UTC)

The article tsunami indicates that there is not necessarily a draw back prior to the wave itself hitting the shore. The article also gives the effect of wave shoaling as "the wave length diminishing to < 20 km" travelling at below 80 km /h. Thus the time for the tsunami travelling 1/2 of a wavelength can be roughly calculated. In ocean surface waves the time of a full wave is termed the wave period, so the time between the trough and the crest may be called "half a wave period". Of course, different topologies of the sea floor rising to the shore line will make this a variable. If at all, there may be a Japanese name for it, but I could find no reference. --Cookatoo.ergo.ZooM (talk) 16:35, 25 October 2008 (UTC)

How do violin strings make sound?

How do strings on a violin make music? I know that you use the bow and draw it across the strings. What I want to know is how does it make sound?

Clare age 7 —Preceding unsigned comment added by 72.81.184.136 (talk) 16:55, 25 October 2008 (UTC)

Christyn2 (talk) 17:02, 25 October 2008 (UTC)

See String_instrument#Sound_production. Axl ¤ [Talk] 17:38, 25 October 2008 (UTC)

You've got to be kidding, she's 7. Theresa Knott | The otter sank 20:12, 25 October 2008 (UTC)

When the bow is moved across the string it's scratchy surface makes the violin string jiggle to and fro. This is called "vibration" and all sound is made by something vibrating. The vibrating string makes the air jiggle as well and it is this vibration of the air that gets picked up by our ears and we hear it as a sound.

The faster the air vibrates the higher the note we hear. There are a few ways of making the string vibrate faster. One is to make the string shorter, and this is what you do when you put your finger on it to play it. Strange as it may seem bowing the bow faster will not make the string vibrate faster. It will make it vibrate more and this makes the air vibrate more. We hear this as a louder sound.

Our ears can hear many different sounds but note that are too low or too high cannot be heard even though the air is vibrating. Interestingly children can hear high sound better than grown ups can, but I bet even you will not be able to hear sounds as high as a dog can. Try looking up dog whistles! Theresa Knott | The otter sank 20:12, 25 October 2008 (UTC)

(Edit conflict) Clare, your question is a good one. I was about to answer something like this: When the bow first makes contact with the string, the bow will take the string along in its motion. The displacement of the string makes it longer, and creates a force which within a fraction of a second will become greater than the force between the bow and string. At that point, the built-up energy will make the string move back towards its starting position. Because of inertia, it will move a little past the starting position, then back again a couple of times, but the vibrations will be dampened, and when its energy of motion is smaller than the frictional forces between bow and string, the bow will again take the string along in its motion. The result is that the string vibrates, the vibration is transmitted to the body of the violin, through the air and to your eardrum. "Sound" is what you experience when your eardrum vibrates.

Then I thought, "why is the frequency of a violin string played pizzicato style the same as when it is played with a bow?", and really couldn't come up with a convincing answer. --NorwegianBlue ^talk 20:52, 25 October 2008 (UTC)

Because the frequency depends on the natural frequency of the string and that is a property of the string only not how it is caused to vibrate. The bow has multiple strands and I would think that the bow is likely to cause vibrations at multiple frequencies, However only those that are resonant with the string will build up any kind of serious vibration in it. Theresa Knott | The otter sank 21:04, 25 October 2008 (UTC)

The vibrating string would not be very loud, except that the body of the violin is made to be a sounding board, to send the sounds out to the listener. Edison (talk) 00:30, 26 October 2008 (UTC)

Initially, when the bow is moved across the string, static friction causes the string to momentarily move with the bow. As the string moves, its tension increases. A point is reached at which string tension is great enough to overcome the static friction between the string and bow. The string is then released. The coefficient of sliding friction allows the string to return almost to its starting point. When tension becomes low enough, the entire action is repeated. The process continues, alternating between static friction and sliding friction. The vibration of the string produces the sound. Rosin on the bow increases its coefficient of static friction with respect to the string, allowing the bow to momentarily pull on the string (if string tension is not too great) instead of sliding over the string. The form of the violin, its wood thickness, varnish etc., modifies the sound produced. —Preceding unsigned comment added by 98.17.36.129 (talk) 11:11, 27 October 2008 (UTC)

On second thoghts, I think the violin string returns fully to its starting point on the return vibration (while under sliding friction). It then overshoots its starting point and continues until increased string tension brings it to a halt or very nearly so. Static friction with the bow then takes over from sliding friction. The whole process repeats for the duration of the bow stroke. Stop-motion photography would show what actually happens.

Spectacles deglazing

Is it possible to deglaze plastic scratched lenses in the home? —Preceding unsigned comment added by Stargazermum (talk • contribs) 17:34, 25 October 2008 (UTC)

Deglazing says it means removing a shiny surface. Do you want to change the lens surface to look like Frosted glass , or are you trying to remove scratches and leave the surface smooth and clear? Edison (talk) 00:27, 26 October 2008 (UTC)

You can use acetone to turn clear glasses into frosted glasses (if they're made of polycarbonate). 96.242.34.226 (talk) 01:38, 26 October 2008 (UTC)

From experience, you cannot use heat to remove scratches, the lenses are not made from a thermoplastic. They just crack. Polypipe Wrangler (talk) 20:52, 26 October 2008 (UTC)

something we dug up in our back yard!!

Hi, I hope someone can help with this question/explanation of my circumstances. Today, my husband was going to plant some things in the backyard of our 1923 home and while digging, he hit concrete. He followed it as he dug and it goes down 4 feet and is 6 foot in cir. It is made of concrete and has one 4" hole on one side at the bottom of the thing. There is a round indentation on the bottom but is concreted over. We don't know what it could be. It seems too small for a septic tank, too small for a cistern (besides, it is underground), and too shallow for a well tank. There are no pipes but the one near the bottom on the wall that goes into the earth. If there is anyone out there who could tell me what this is, it would be appreciated. This area was owned for over a hundred years by one family and it was all farmland. It is located on the central west coast of Florida.

Thanks, JSRny1JSRny1 (talk) 18:52, 25 October 2008 (UTC)

Maybe it's the remains of a dried up borehole, with the water pump on the surface removed and sold once the water stopped. Does anyone from the previous owners' family still live nearby who you could ask? — FIRE!^{in a crowded theatre...} 21:10, 25 October 2008 (UTC)

The description is unclear. Is it a solid concrete cylinder 6 ft diameter and 4 ft high? That would make no sense. Is it a hollow cylinder of those dimensions? Does it have a lid or hatch or manhole cover. or metal rings for lifting the top off? Is (or was) there inlet and outlet pipes at opposite ends? It could be part of a well, part of a cistern, or a septic tank. Or perhaps Amelia Earhart, Jimmy Hoffa, Judge Crater , Peking Man and the Ark of the Covenant are inside it. Edison (talk) 00:25, 26 October 2008 (UTC)

I think the OP meant 6' in circumference, not diameter (at least, I can think of anything else "cir." could mean). --Tango (talk) 00:43, 26 October 2008 (UTC)

Whatever it is, it was probably meant to store water, either fresh or waste. I'm sure you know, living in Florida, no one digs more than 5 feet in the state and builds a structure that can stay dry. There are no basements or cellars in houses. --Moni3 (talk) 02:42, 26 October 2008 (UTC)

A photo would be helpful! --S.dedalus (talk) 07:34, 26 October 2008 (UTC)

Actually, it is hollow with no lid or hatch. It might have had one at one time, but there seems to be no hinges or anything. The only previous owners passed away and we got it as an estate sale 15 years ago and am always finding obstacles in our way when we do any work. The old man who built it didn't appear to have any logic in some areas. They have no family left to ask. As for a basement, this old house has a cellar..unheard of in Florida, but we have it. It's only 10' x 10' with a 7' ceiling but hey, it's there. At any rate, nothing was inside it but dirt, which we dug out. It is now empty and even clean. We are contemplating turning it into a concrete pond but just wanted to know what the heck it was to begin with. It just seems too small to be a cistern or a septic tank due to it's dimensions of 4' deep by 6' in circumference. I have a pic, will upload later today. Thanks, JSRny1JSRny1 (talk) 10:21, 26 October 2008 (UTC)

Hole In Yard

I uploaded a photo of this thing in the common upload area. The file is named holeinyard.jpgJSRny1 (talk) 12:22, 26 October 2008 (UTC)

Maybe I haven't got the perspective right, but it doesn't look like it's four feet deep. Axl ¤ [Talk] 15:29, 26 October 2008 (UTC)

Agreed. That looks wider than it is deep. 6ft circumference would be about 2ft diameter, so it should be about half as wide as it is deep... Was the photo just taken at a strange angle or are your measurements off? --Tango (talk) 16:46, 26 October 2008 (UTC)

Perhaps the bottom is a total of four feet below ground level? In any case, it looks like it was once taller and has been partially demolished. APL (talk) 19:36, 26 October 2008 (UTC)

Thanks for the picture! It looks like the sides are sloping, rather than cylindrical, and like it has rather thin sides. I wonder if the builder dug the hole, then lined it with concrete, perhaps with wire mesh reinforcement, to seal it. The sides look fairly thin, and it looks like the surface was crudely formed. Could it be a small "cement pond?" [1]. In Florida it would not be destroyed by the ground freezing as might be a problem up north. The drain could have once had a plug, or it could have been connected to the inlet of a pump. The hole or the bottom indentation might have been part of a filter system/aerator system (fountain) connected to the discharge line from the pump. "Cement pond" reminds me of the Beverly Hillbillies, where they used the term for their swimming pool. Try digging by the visible exit hole to see if there is a pipe and where it goes. Perhaps there was once a pump and a little waterfall above the pond, to make a "water feature." There would have been a buried electric line from the fuse box/breaker box in the house to power the pump. Edison (talk) 22:07, 26 October 2008 (UTC)

My parents have something similar to this at their house in South Carolina--a big hole with one exit pipe near the side of their house. The pipe runs to the ditch in the front yard to help drain water away from the side yard. Laenir (talk) 14:31, 27 October 2008 (UTC)