The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.
Purple is perceived when the blue and red cone cells are stimulated but the green ones aren't. Can we perceive another colour in an equivalent way, with the blue and green cones stimulated but not the red ones? Or would it look the same as a spectral blue-green?
Magenta is unique in that it doesn't appear in a rainbow. Because our three color sensors are arranged in order of frequency: Red, Green, Blue and their regions of sensitivity to frequency overlap - colors like yellow and cyan can either be a mixture of two primaries - or a frequency midway between two primaries. Our eyes can't tell the difference. But the color that's midway between red and blue is green - but because we have a green sensor - we see that color as green...red+blue doesn't look anything like the frequency that's midway between red and blue. However, the frequency between red and green - and the frequency that's between green and blue look exactly the same as mixtures of those primaries.
I believe the spectrum is continuous - but I could be wrong about that. Our eyes are not infinitely sensitive though - so there is a limit on the number of colors we can truly distinguish. SteveBaker (talk) 02:13, 13 April 2009 (UTC) SteveBaker (talk) 02:13, 13 April 2009 (UTC)[reply]
The colors that we perceive are really only indirectly related to the receptors we have. We have three types of receptors (red-green-blue), but we perceive colors as combinations of four primitive hues, red-green-blue-yellow. These are arranged in opponent pairs, red-vs-green and blue-vs-yellow. The relationship between physiology and color perception is a fascinating topic, with endless surprises. Looie496 (talk) 02:44, 13 April 2009 (UTC)[reply]
In what sense "as combinations of four..." ? Speak for yourself, perhaps I perceive colours (after interpretation in my mind) as combinations of seven hues, or as pantone swatches. I mean it seems a rather arbitrary thing to say, what do you mean? Where do you spring this yellow from? It's complementary to blue, but so? Are you talking about something that actually happens in the retina involving, in some sense, yellow? (Uh, that last question is probably the best one to answer.)
Edit: come to think of it, orange is usually shown as complementary to blue, and yellow to purple.
OK, I found the article on Opponent_process, and I see that, yes, there are such things as P cells which turn the three signals (R, G, B) into two signals: red-green, and (red+green)-blue, which is to say yellow verses blue. Magenta, then, would be blue with some amount of yellow (which is to say, lots of red) on the yellow-blue channel, and red with no green on the red-green channel. My proposed colour would be the same, only with the red-green channel reversed. I see that this is either similar or identical to a spectral cyan (depending on whether that would activate the red cones a bit or not) - not really analogous to purple in terms of being a non-spectral colour.
Interestingly, I see potential for yet more unexpected signal combinations - the yellow-blue channel can say that there isn't any yellow, meaning that there isn't any red or green, while the red-green channel can contradict that. I suppose there's no way to make that happen just by looking at colours. 81.131.19.13 (talk) 05:07, 13 April 2009 (UTC)[reply]
I was supposing that spectral blue-green would activate the red cones a bit. My reasoning is that the green frequency we are sensitive to is relatively close to the red, and the green cones react slightly to light which is nearly blue (allowing us to see it as greenish-blue), therefore the red cones ought to react a similar amount to light a similar distance from red, which would be I suppose more of a blueish green. Removing that small amount of red signal ought to create a non-spectral colour, that's my reasoning. It ought to be a different experience from seeing spectral blueish green. Do the red cones not in fact react to frequencies on the blue side of green?
I don't understand what Looie is talking about. For what reason is purple and orange excluded from the list? To answer the OP: Adding a little bit of red to the blue-green (or cyan) will not give you a new color. It will simply make it look a little bit more washed (white-ish). Dauto (talk) 05:04, 13 April 2009 (UTC)[reply]
No, I'm talking about removing a little bit of red - which I assume must exist, I mean in terms of signal in the retina - from the blueish green. My assumption may be wrong. 81.131.19.13 (talk) 05:10, 13 April 2009 (UTC)[reply]
Removing a little bit of red (if possible at all) should have the oposite effect of adding a little bit of red. It won't give you a new hue. Your brain automatically normalizes the perception to a bi-dimensional chromatic space. For instance lets use the symbols r, g and b to represent how much the red, green and blue cones are being stimulated. Lets say your monochromatic blue-green gave r=2, g=10 and b=10. your brain sees that as (w)hite=2, g=8, and b=8. If you remove the extra red you get r=0, g=10, and b=10. That is more blue-green and less white. A less washed blue-green is still blue-green.read that. Dauto (talk) 05:37, 13 April 2009 (UTC)[reply]
Does this mean that (regarding the opponent process) if you can somehow contrive to have a red or green signal along with a completely blue signal (i.e., not yellow, therefore indicating that there is light, but that it doesn't contain any red or green), that wouldn't create a new colour either, because the brain would just normalize the contradicting signal away? What would you get? 81.131.19.13 (talk) 05:51, 13 April 2009 (UTC)[reply]
If it contains no red and no green, it is percieved as a deep blue sometimes called violet. See the bottom of the picture I posted. Dauto (talk) 06:08, 13 April 2009 (UTC)[reply]
It wouldn't correspond to any actual wavelength or combination of wavelengths: it would be a set of contradictory nerve signals - in the retina, I suppose, if that's where the P cells are - that say the red and green frequencies both do and don't exist in the colour. 81.131.19.13 (talk) 06:28, 13 April 2009 (UTC)[reply]
Experiments show that across cultures, there are four hues that people perceive as "pure" -- red, green, blue, and yellow. People perceive other colors as mixtures of these, for example orange as red-plus-yellow or purple as red-plus-blue. See opponent process for more information (although it's a pretty sketchy article). Looie496 (talk) 05:23, 13 April 2009 (UTC)[reply]
Looie, I beg to differ. How many 'pure' hues do you see in this picture?
In that diagram, why is the gamut horseshoe-shaped? Do theoretical colours with greater saturation lie beyond the edge of the horseshoe? Also: "given three real sources, these sources cannot cover the gamut of human vision," because ... something about triangles. I don't understand. We have R, G, and B cones: surely they can be stimulated with R, G and B light sources to make them do anything they are capable of (with the exception that their sensitivities overlap a bit). How is that naive? 81.131.19.13 (talk) 06:20, 13 April 2009 (UTC)[reply]
I'm not sure I should get involved in another color vision thread after what happened in the last one, but here goes. The sensitivities of the three cones overlap a lot, not just a little. The "red" and "green" cones, in particular, have very similar sensitivity curves that are only shifted slightly with respect to each other. They both peak in the yellow-green region and respond noticeably across the whole visible spectrum. Because of this, it isn't possible to stimulate them independently; you always get a mixture. The horseshoe shape reflects that. If the cones could be independently stimulated then the horseshoe would be a triangle instead. That would leave us with inferior color discrimination, though, because for each cone there would be some range of frequencies over which just that cone was significantly stimulated, and over those ranges we couldn't distinguish different frequencies. -- BenRG (talk) 12:41, 13 April 2009 (UTC)[reply]
The problem arises in trying to replicate fully (or near fully) saturated colours that have a wavelength in-between the wavelengths of the three chosen sources, for instance, a strong yellow (assuming RGB sources are used - others are possible). The hue can be reproduced exactly by a suitable mix of red and green, but not the saturation. The green source will also stimulate a lot of blue receptor. It is true that monochromatic yellow would also stimulate some blue, but not as much as the green source does because the green is closer to the blue receptor peak that yellow. The net result is that this is interpreted as yellow plus some white light (ie all three colours) and the colour appears de-saturated. A similar result occurs with blue-green mixtures with red this time being overstimulated. Most colour space diagrams (including this one) are arranged so that linear mixtures of any two sources lie on straight lines between those points. Since such a mixture is de-saturated, it follows that a monochromatic source of light (by definition fully saturated) must lie outside the triangle - hence the curve. SpinningSpark13:48, 13 April 2009 (UTC)[reply]
Here are a couple of illustrations. The first shows the responses of the three cone types at different wavelengths (normalized to have the same peak value). If you treat the height of the three curves at a given wavelength as coordinates in an abstract 3D space, and plot those points across all wavelengths, you get the green curve shown in the second image. One of the dimensions of the 3D space is just brightness (twice as much light with the same spectral distribution doubles all the coordinates). If you're only interested in color and not brightness, you can divide out the brightness, which is equivalent to projecting onto a 2D surface like the one shown by the triangle in the second image. That projected shape is the horseshoe. If you have a computer monitor with three phosphors which can be made to glow in various proportions up to a maximum brightness for each one, then the points in the 3D space that you can "reach" with various combinations of the phosphors form a linearly distorted cube (a parallelepiped). When you project that onto the 2D plane, you get a triangle whose vertices are the colors of the individual phosphors (usually red, green, and blue). You can see by inspection that no triangle with vertices inside the horseshoe can cover the whole horseshoe. The approximate locations of the phosphors used in most monitors are shown here. The colors outside that triangle are incorrect since your monitor can't reproduce them. The three prominent radial lines on the horseshoe image (colored yellow, cyan, and magenta) correspond to three of the edges of the RGB color cube (HTML #xxFFFF, #FFxxFF and #FFFFxx). They only show up because the RGB values in the image were normalized to be as bright as possible within the separate 0–255 limits of red, green and blue, which leads to a sharp peak of brightness at the corners. If they had been normalized to be of constant brightness instead, those colors wouldn't look as special. The other edges of the cube project to the boundary of the triangle. -- BenRG (talk) 13:56, 13 April 2009 (UTC)[reply]
The proof of the validity of the tri-color stimulus theory is sitting right in front of you as you read this. Your computer monitor only produces fairly pure red, green and blue light. There are no yellow emitters in there (take a magnifying glass and look closely at some white area of the screen and you'll see the little clusters of red/green/blue emitters). Yet, it does a pretty good job of producing every color you'll ever see. You never take a digital photograph and say "Oh no! All of the orangey-yellow is missing!". There are limits to the intensities of color that the computer can display and the 'gamut' of colors it can produce is a little less than (say) printer's ink in some regions - but there are no hues that it cannot reach. The business of "opponent color theory" lies in that after the rods and cones have captured that red, green and blue light - subsequent processing turns that into 'color difference' signals. That's similar to how broadcast television works (PAL, NTSC, SECAM, etc) - where it is desirable to transmit a monochrome signal that's compatible with old-fashioned black-and-white TV's - and TWO color difference signals - the resulting three signals being called 'YUV'. These are then converted into RGB inside the TV. The cells behind the retina also do various color difference calculations before passing on the results to higher brain functions. They also do things like 'shape-from-shading', 'edge enhancement' and 'motion compensation'...what goes to your higher brain functions is a vastly different description than just a two-dimensional array of pixels. However, all of that complexity doesn't change the fact that all of the color we see can be described in terms of red, green and blue alone - hence the whole digression above is irrelevent to the question and my first answer remains. Fortunately - you don't have to take my word for it - you can use a magnifying glass and a digital camera and find this out for yourself. SteveBaker (talk) 14:00, 13 April 2009 (UTC)[reply]
I come across the mail recently regarding the Tsunami will going to happen in July 2009.
I wonder this is the rumour or truth.Can anyone help me to find out? —Preceding unsigned comment added by 58.27.115.86 (talk) 03:50, 13 April 2009 (UTC)[reply]
At least no one can predict them that far in advance. Once an earthquake or other geologic event which causes a tsunami occurs, the waves of the tsunami can be (and routinely are) predicted. But that is on the order of hours, not months. No one can predict when an earthquake, landslide, or other geological event will occur, so no one knows when a tsunami (caused by those events) will occur. --TeaDrinker (talk) 05:32, 13 April 2009 (UTC)[reply]
OP, your version of the rumour may have originated here. The writer himself calls it "pseudoscience" and says it's based on the idea that the eclipse of the sun (which is genuinely predicted for that date) could cause an earthquake, which could cause a tsunami. He mentions the "original guy who came up with the theory", but doesn't give his name. You can find some links to such theorists in the comments below the blog post. It's amazing how many people come up with correct predictions that they forget to mention until after the event. ;-)
I can think of two errors with this theory, but there are probably more. First, there are many occasions when the sun and moon are almost in line, not close enough to cause an eclipse but close enough to cause almost the same gravitational effect, so we should get earthquakes whenever this happens. Does anybody predict this? I doubt it, because eclipses are newsworthy but near-eclipses aren't. Second, the sun and moon don't just suck up the ground on the part of the Earth that's facing them. It's more complicated than that. The gravitational field causes a tidal effect that deforms the entire Earth. You would be just as likely to get an earthquake on the opposite side of the planet. Try telling that to the lunar gardening people. --Heron (talk) 10:33, 13 April 2009 (UTC)[reply]
Tsunami's are mostly caused by earthquakes - and a few by landslides or meteor impacts - we can't predict any of those things reliably - so we can't predict tsunamis. QED. SteveBaker (talk) 13:46, 13 April 2009 (UTC)[reply]
So it's economical problem rather than technical problem. Well I think someone would pay a lot to be able to fly using his own power. I might consider if this thing is for rent, maybe $5 for a short period of time. roscoe_x (talk) 11:05, 13 April 2009 (UTC)[reply]
It's also a flimsy and delicate piece of equipment that needed repairs (sometimes extensive) after every single flight it ever made. Insurance would be impossible - and the upkeep of the thing would be ruinously expensive. SteveBaker (talk) 13:41, 13 April 2009 (UTC)[reply]
Also, due to its flimsy nature, it can only be flown under ideal weather conditions, with little or no wind, and so is quite impractical for anything but setting a record or proving what can be done. -AndrewDressel (talk) 14:08, 13 April 2009 (UTC)[reply]
As far as I can tell, most Ultralights have engines about 20-50hp. Humans struggle to maintain a quarter of an hp. I mention this to point out how different the Gossamer Albatross is from normal ultralights, It has a huge wingspan and has a very light, delicate construction. Even ultralights that appear to be less substantial than the G.A. require much more engine power than a human can provide.
You could easily pay 5-20 thousand dollars on a commercially produced ultralight, I imagine that a G.A. work-alike would probably be a lot more. APL (talk) 16:31, 13 April 2009 (UTC)[reply]
physics/Crystal growth
respected sir
i am doing Phd in physics. and i am interested to work in crystal growth.
My question is that is there any crystal ,in which not too many studies are not yet done? —Preceding unsigned comment added by Rmgirish1 (talk • contribs) 06:18, 13 April 2009 (UTC)[reply]
People are again talking about raising the legal drinking age in Australia to 21 (from 18). I doubt it'll happen, but it lead to me wondering whether there was any good reason for drinking ages to be set where they are. The page linked above was little help - it is but a list. So can anyone here point me to any rational arguments over the age? Is 18 much worse than 21? Should it *really* be 25.. or 16, or different between males and females?
--Polysylabic Pseudonym (talk) 11:58, 13 April 2009 (UTC)[reply]
I think the strongest argument is that teenager's brains have not yet finished maturing at age 18. Indeed there have been studies that show that as each source of various hormones matures at a different rate, the brain of an 18 year old is actually less able to cope with self-control and behavior moderation than (say) a typical 16 year old. Since we know that alcohol further dis-inhibits the brain, this can result in kids who have little or no ability to avoid getting into stupid and dangerous situations. I can appreciate that as an 18 year old, you might think this is impossible - but that's precisely because your brain chemistry is all screwed up! I'm tempted to make remarks about my own 18 year old and how he backs up this theory perfectly...but I don't think he'd thank me for that. So - between ages ~16 to ~20 you need to be doubly careful about any decision you make and always err on the side of caution.
Having said that - I don't advocate a 21 year old drinking age. We make too much of the role of alcohol. My wife (being French) has an entirely different cultural attitude - which is that if you bring up your kids thinking that alcohol is really nothing all that special - then when they can legally go out and get drunk as a skunk - they'll be less likely to see that as a novel and exciting thing to do. She claims that French high-schools used to allow kids to drink 'small wine' (wine diluted with water) at school lunchtimes! So we've always allowed our kid to have a small glass of wine or champagne on special occasions. I always offer to grab a beer for him (even though it's not legal in Texas)...because I think it's important that it be "no big deal"...and 99% of the time, he says "No thanks". I never liked the idea that suddenly, you reach age X (where X is an arbitary number chosen by the government) and you can go from never having tasted alcohol to buying it in large quantities and getting smashed out of your skull at your X'th birthday party. If X is 18 - that's a recipe for disaster - so perhaps 21 is a better choice. However, a more gradual approach seems more sane to me - and in that case, it probably doesn't hurt to do it more gradually.
In most European countries I know, there is a staggered approach - you can buy beer (and maybe wine) at 16, but spirits only at 18. Southern European states, with wine being a much more integrated into the culture, are even more relaxed, certainly in practice. --Stephan Schulz (talk) 23:03, 13 April 2009 (UTC)[reply]
I agree with Steve. In the UK, the legal drinking age in a public place is 18, but at home, it's just recently been raised to 5, as opposed to there not being a legal drinking age. People drink at home when they are younger, say at Christmas or other special occasions, and I am sure this would help to alleviate people going out and getting mullered when they get older. However, we still have the minority that go out binge-drinking. Plus, we are now seeing teenagers (and younger!) out on the streets and in the parks with bottles of super-strength cider and vodka. Our government is taking a different step, though. Rather than changing the age, they are raising the prices to ridiculous levels to make it harder for people to buy the alcohol.--KageTora (talk) 08:35, 15 April 2009 (UTC)[reply]
Also, I think the most important thing to note is that the legal age for drinking away from home is, in the UK, a year after the legal age for driving (17).--KageTora (talk) 08:38, 15 April 2009 (UTC)[reply]
Well, some of the references that our article point to state that the day-name (like "Tuesday") of last day in February is known as "Doomsday" for that year (so for this year, Doomsday is Saturday, for next year, it'll be Sunday) - but that just pushes back the question to "Why is the last day in February called 'Doomsday'?"...for which I have no answer. Some sources kinda suggest that Conway (who discovered the algorithm) just needed a name for the day of the week on which "March 0th" fell - and he chose that name arbitrarily. If that's the case then probably someone should ask him why - because otherwise we may never know! SteveBaker (talk) 13:22, 13 April 2009 (UTC)[reply]
Conway loves to give silly names to things. I don't think there's any deeper reason than that. Read Winning Ways and you'll see what I mean. Evil and odious numbers, the Uppitiness Exchange Principle, the atomic weight of lollipops... -- BenRG (talk) 14:43, 13 April 2009 (UTC)[reply]
How to get a women to cum
My girlfriend will not cum during intercourse or oral or masturbation. Is there anything wrong. How can I help her to acheive orgasm? —Preceding unsigned comment added by 97.65.234.3 (talk) 12:36, 13 April 2009 (UTC)[reply]
Have you tried searching this reference desk? Try doing a search using the box directly above here at the top of the page (not the one on the left, but the one directly above) and terms like "orgasm girlfriend" and "orgasm female". You'll find that this question has been asked and answered many times, in a remarkable variety of forms, over the years. Hope that helps. --Scray (talk) 12:46, 13 April 2009 (UTC)[reply]
Guessing that Reticuli88 is referring to a fourth spatial dimension, one definition would be that you have your normal three dimensions, up/down, left/right, forward/backward, each at right angles to the others. The fourth dimension is like those and at right angles to all the others. You can't imagine it. --Polysylabic Pseudonym (talk) 13:49, 13 April 2009 (UTC)[reply]
I disagree. Not only can we imagine it - with computer graphics we can actually DO it...for real...without any approximations or assumptions. People like to make this more mysterious than it really is. The truth is we can easily simulate EXACTLY what it would be like to be in a 4D world...the image at right of a spinning hypercube is exactly right in that regard.
There have been many efforts to try to help people visualise a fourth spatial dimension - one is to map it onto time. Thus you could imagine a hyper-sphere as a regular 3D sphere that grows and shrinks. However, I have a better answer. Our eyes do not see things in three dimensions. We have two eyes - each of which sees in only two dimensions. We infer the third dimension (actually, just the distance at right angles to our face) by seeing how the images from our two 2D retinas differ due to parallax - and we note how much we have to distort the lens in order to get the image into focus - and we can get shape information from the way light falls onto flat and curved surfaces. This is why a 2D photograph or a movie or TV screen looks fairly realistic. In a 4D world, things would not be all that different. We'd still only have two eyes and still only 2D retinas. So just as the three dimensions are 'projected' down into two when we do things like 3D computer graphics - we can project four dimensions in the exact same way. We still only have two eyes and 'depth perception' doesn't work any differently. Consequently - all we'd see would be 2D projections of 4D objects - which would look exactly like the picture to the right here...assuming the hypercube were made of a clear substance with opaque edges. The four-dimensionality would perhaps be more noticable when our heads could be rotated around that fourth axis of rotation - so that distances at right angles to our faces would measure something different. All of this can be precisely simulated inside a computer and presented to us as 2D images - and in truth, it's not really all that remarkable.
The problem here is that what you REALLY mean to ask (because it's a tough question) is "What would a 4D world look like if I had 3D retinas?" - and that's something we can't answer. We can't answer it because we have no idea what it would be like to truly see in 3D in a 3D world...let alone 4D! Aside from anything else, we'd probably need a third eye - stuck out into the fourth dimension - in order to get better range information. But once you start to speculate about true 3D vision - you have a non-human brain attached to it and truly all bets are off.
I remember there was a previous thread in which you claimed you could put a 3D eye (with a 2D retina) in a 4D world and everything would just work. But that isn't true—you must be misremembering the program you wrote because the idea of a direct 4D-to-2D projection isn't mathematically or optically sensible. The animated tesseract is projected in two stages, first from 4D to 3D and then from 3D to 2D, with different viewer locations in each stage. (If you disagree then tell me what the direct projection formula is. A 3D-to-2D projection takes (x,y,z) to (x/z, y/z) for appropriately chosen coordinates; what's the 4D-to-2D equivalent?)
I don't think you would need three 4D eyes to get depth perception in four dimensions. A 3D retina gives you three constraints on the object's position and two-eye parallax gives you one; taken together that's enough to locate it in 4D space. -- BenRG (talk) 15:20, 13 April 2009 (UTC)[reply]
I don't think you know enough about computer graphics. You are thinking of the interactive kind of 3D graphics based on projecting triangles onto the screen such as one would use for computer games. However, that's not the only graphics algorithm. You can also do ray-traced graphics - which work just fine in 4D...or 5D or a million dimensions. The standard "reverse-ray-tracing" algorithm traces the history of a light ray that ends at your eye - through one pixel on the screen. It repeats this process for each pixel on the screen in turn. For each of those 'reversed' rays, it traces back until the ray hits something solid out there in the virtual world. Typically - you'd then go on to figure out how that ray of light was reflected from that surface towards the eye and deduce how incoming rays must have travelled in order to produce that reflection (or refraction or scattering). For each incoming ray, you repeat the process, following the incoming rays of light back to their sources - until you either reach a light source or the "infinite depths of space" (or the sky or whatever you decide is the edge of your virtual universe). When you know how all of the light rays that contributed to that particular point on the screen originated and were reflected/refracted/scattered - you know what color to assign to that pixel. This process is extremely routine - just about every piece of computer graphics you see on TV and in the movies is done like that. You can do the analogous thing for tracing rays from points on your retina - through the pupil of your eye and out into the world...but the result is exactly the same - and thinking about a 2D "video-screen" with a zero sized eye is easier. That 'tracing' process works just fine in 4D - you can easily compute the 4D intersection between a ray and a hyper-cube or whatever - and this results in an entirely reasonable 4D to 2D projection - which (as far as we could reasonably say) would represent how the photons in a 4D world would appear in a 2D retina. I see no problems with that...and indeed I've done 4D rendering before (although not precisely like that) and it works perfectly. I don't know how the animated hypercube was rendered - but when you do a 4D raytracing - that's exactly how it looks - so whatever they did clearly works OK. SteveBaker (talk) 17:23, 13 April 2009 (UTC)[reply]
I've written a ray tracer and four is my favorite number of dimensions. Ray tracing won't give you an image like the one above, it will give you a slice through the tesseract. Subtract one dimension from everything to make visualization easier. Now we have an opaque circle with a 1D retina and an aperture (pupil), and it's looking at a cube. The rays from the retina through the pupil all lie in the plane of the eye. Only the part of the cube that intersects that plane will be hit by the rays. Other parts of the cube might be seen by reflection, but they won't be seen directly. Also, this kind of ray tracing makes little sense from a physical perspective. The reason you cast rays this way ordinarily is because the only way to the retina is through the pupil. When you have another dimension and the eye is "open from the sides" then that's no longer true. Light will reach every point on the retina from every direction, so you ought to cast rays in every direction from every pixel, which just gives you a uniform blur. That's what the eye would actually see (to the extent that this thought experiment makes any sense in the first place)—a uniform blur, like a camera with a severe light leak problem, which is exactly what it is. -- BenRG (talk) 12:19, 14 April 2009 (UTC)[reply]
No! As I carefully explained - the hypercube imaged above is translucent with opaque edges...and that's how a 4D raytrace comes out. An unshaded opaque hypercube would look like a smoothly colored polygon - much as an opaque spinning 3D cube looks like a hexagon in utterly uniform lighting. The argument about light leaking around the eyeball in the extra dimension is certainly a practical issue...but again, we have to come back to the total impracticality of the experiment in every other way...your blood would simply squirt out of your veins through the "open" 4th dimension. But if you stipulate that you could somehow survive all of those practical problems - then the animation above is an excellent representation of a shiney translucent hypercube imaged on a 2D retina. Maybe you have to wear some opaque 4D headgear to keep the light from escaping around the sides of your eyes and some kind of very skinny 4D spacesuit to keep your insides...inside. (I'm also super-skeptical of BenRG's claim to have tried a 4D raytracer. If he had - he wouldn't have thrown up the entirely bogus claim of mathematical impossiblity of doing a 4D to 2D projection in his post a couple further up this thread - only to completely ignore that objection on the next post after I explain how it is in fact easily possible.) SteveBaker (talk) 23:09, 14 April 2009 (UTC)[reply]
No, you are wrong. Raytracing from a 2D screen spans a 3D space - 2 dimensions of the screen and 1 dimension of a ray makes 3 dimensions. The animation above is not done by simple raytracing. If you are going to have something that works like a human eye in 4D you need a 3D retina. --Tango (talk) 11:43, 15 April 2009 (UTC)[reply]
I agree that the image above may well not have been raytraced in 4D...but it could have been. A conventional, off-the-shelf raytracer doesn't work in 4D - but if you write your own, you can certainly make it do that (I've actually done this - so I know). You position the eyepoint in 4D - you position the corners of the screen in 4D and you do 4D math to compute vector/hyper-solid intersections, reflections and refraction - and it all works just fine. And the results of 4D raytracing a spinning shiney-translucent hyper-cube look pretty much just like that image (mine didn't have the cylindrical rods highlighting the edges of the geometry and I filled the interior of the hyper-cube with a low-opacity "gas" to attenuate the light to give you a better idea of depth). My actual application was totally off-the-wall though...not aimed at 4D visualisation at all - it was to do with data compression of volumetric descriptions of 3D smoke plumes as they evolve over time - 4D raytracing was just one small step in the process. Rendering hypercubes with it was just a part of the testing the actual results of the program were data files - not pretty pictures. But if you follow the 4D light rays through the 4D world - reconstructing their history - then what happens, happens. That's what it would look like - I don't see how you can argue with that. SteveBaker (talk) 18:11, 15 April 2009 (UTC)[reply]
Oh - and as for the rays from the 2D screen only spanning a 3D volume - again, you don't understand what's going on here. The first bunch of rays do indeed stay confined to a 3D cross-section of the 4D world - but once they hit something that's sloping in the 4th dimension, the reflections go off in all directions. Think of looking through a thin, 2D slit into our 3D world. Sure - you only see a 2D slice of the 3D universe...unless, there is a mirror that's tilted at some angle to that 2D 'slice' - now your reflected rays can be anywhere in the 3D world. Add a dimension and the same thing is true. But no matter the limitations...those are the true limitations of a 2D retina in a 4D world...yes - you don't see everything at once unless you turn your head or something. But that doesn't take away the fact that this is a representative view of what (under extremely extenuating circumstances) you "would see". SteveBaker (talk) 18:18, 15 April 2009 (UTC)[reply]
A mirror wouldn't increase the number of dimensions, it just changes which 3D subspace of the 4D space you can see. Raytracing works by projecting a ray (1D) to a point (0D), that's a reduction of one dimension. Since you end up with a 2D image that means you must have started with a 3D space. I don't know what programs you've written, but I do know that 2+1=3. Either your program wasn't actually a raytracer and was doing something more complicated, or it only showed you a 3D subspace of the 4D space (projected to 2D). --Tango (talk) 18:56, 15 April 2009 (UTC)[reply]
No - because the retina has to be opaque in order to absorb the light and therefore react to it. If you had a "retina" that was (say) a solid 1" sphere of light-sensitive cells - with some (!?unknown?!) means to project the 3D world onto it - only the cells on the surface would see anything. You'd need to come up with some completely different way to perceive things...and I don't think there is anything like that out there. SteveBaker (talk) 17:27, 13 April 2009 (UTC)[reply]
One fairly comprehensible -- or at least illustrative -- way of doing it would be to grab a piece of paper and a pen. First just draw a dot: that's represents a zero-dimensional space. If you were in there, you wouldn't really have anywhere to go.
Next you extend the dot into a line. That's a one-dimensional space. If you were in there, you could go in either of the two directions, but only in those directions -- it's like a narrow passage that's just wide enough for you to move in. If you encountered an obstacle, you couldn't pass it. (You should understand that we're not talking about an actual corridor, of course! You couldn't climb over something, because that direction -- "up" -- doesn't exist, nor could you give someone room to pass, or even slide a sheet of paper past your body. Your body would be just as one-dimensional as the corridor is.)
Next you take up the entire piece of paper and declare that to be the two-dimensional space. Now you can move not only along the line, but you can also move around the line. It's as if the corridor just became a wide room; now people can pass each other. The ceiling is still awfully low, though, and nothing can be on top of anything else. Everything is like a game of air hockey; things can slide around, but everything is the same height -- or more to the point, nothing has any height, because that dimension doesn't exist. You couldn't even put anything in your pocket, because the very concept of pockets is three-dimensional, and you only have two. The best you could do is push and pull things around.
Third dimension: you can now abandon the piece of paper and look around. You can move in any direction -- up, down, left, right, backward, forward. Before, in two-dimensional space, you couldn't even look up, because there was no up to look at, and such a direction would've been inconceivable to you. Now, though, you can play leapfrog with your friends, climb trees, look up to the stars and avoid open manhole covers. Welcome to real life!
You notice, of course, that every time you add a new dimension, your perceptions change radically, and your freedom of movement increases by such a magnitude that anyone or anything stuck in a space one dimension less would be ridiculously hindered compared to you. That would also be true for the fourth dimension: a two-dimensional being in a two-dimensional world can't see behind a wall, but a third-dimensional being viewing a two-dimensional world can easily do so, because he's looking at the whole thing from above. In the same vein, a four-dimensional being could look at a three-dimensional world and see not only us, but also what's behind and even inside of us.
That's the thing with the fourth dimension: it's pretty freaky. You cannot perceive it, and you cannot even properly comprehend it. If I were to stand next to you, and you suddenly moved in a direction that only exists in the fourth dimension, I couldn't even understand what I'm seeing. The classic novella Flatland explores these concepts in a fun and informative way, and the trailer for the movie version probably helps illustrate these things. Alan Moore also wrote a Hypernaut story in his awesome retro comic book limited series, 1963, in which Hypernaut has to fight a four-dimensional being and, being severely outclassed, has to defeat the enemy with non-physical means. You can find that story online. It's a fun conceptual romp. -- Captain Disdain (talk) 15:04, 13 April 2009 (UTC)[reply]
I don't know...why would we? We have 2D retinas in our 3D world. 1D retinas "work" in our 3D world. If you are asking what the 4D world would look like to a normal person - then you have to ask what 2D retinas would do...because that's what we have. SteveBaker (talk) 17:23, 13 April 2009 (UTC)[reply]
He doesn't really says more than what has already been said above. A particular point to infer from what he mentions is that the above animation is what an image of a hypercube viewed from certain different angles looks like in a 3 dimension further projected onto 2 dimensions. - DSachan (talk) 15:53, 13 April 2009 (UTC)[reply]
True, but his demonstration with the apple (3D as seen in a 2D world) as well as the shadow of the cube in a 2D world helped me understand the 4D problem. David D.(Talk)16:09, 13 April 2009 (UTC)[reply]
@The Hand That Feeds You. In a cube the face that is furthest from you becomes larger as the cube is rotated around. Observe in the hypercube how the cube that is furthest away (in dimension 4) from you becomes larger as the hyperecube is rotated. The baffling "turning inside-out" behaviour is explained as the the cube that was "behind" in the fourth dimension is now "in front". This is entirely analogous to the 3D version where the rear face somehow gets in front by going through a third dimension which would be equally baffling to a 2D being. SpinningSpark17:09, 13 April 2009 (UTC)[reply]
You really need a 3d retina to do the job properly, the image of a tesseract above is a projection of a wire frame rather than a view of a solid tesseract. With a solid tesseract and using the analogue of our normal vision you'd only see 4 cubes and the centre point where the 4 cubes join is the point nearest to you. If you shade the cubes you wouldn't be able to see the closest point with the projection way of doing things. basically the image above hasn't had the equivalent of hidden line elimination done. The way I like to think of it is to form a 3d image and then have different properies apply to the 3d image where the ra[idly changing areas are solid and the slow changing areas are transparent. This alows the image to be see wih the nearest point surrounded by light coloured transparent solids. Dmcq (talk) 22:01, 13 April 2009 (UTC)[reply]
Yeah - I just grabbed the first picture to come to hand from the "Hypercube" article. But I have used 4D raytracing to do similar things...although the application wasn't exactly that kind of thing. SteveBaker (talk) 02:38, 14 April 2009 (UTC)[reply]
That doesn't seem to be mentioned in spirit, but then again I don't know of any reliable evidence saying spirit aren't four dimensional. Dmcq (talk) 18:33, 14 April 2009 (UTC)[reply]
No!! The definition of a ghost or spirit is a figment of some people's imagination. The universe doesn't have four spatial dimensions...it has three (duh!) - so nothing "real" can be a 4th dimensional entity. Please - don't fall into the ridiculous trap of saying "Wow - the 4th dimension! That's weird! Ghosts are weird too! Maybe they're the same thing?"...that's just fuzzy thinking. The sane, rational approach is: "We have no evidence whatever that the universe has a 4th spatial dimension. We have no evidence whatever that there are ghosts. Both are 'unfalsifiable' hypotheses - so we must treat them both as false - and there is absolutely no reason to 'connect' these two ideas." SteveBaker (talk) 22:59, 14 April 2009 (UTC)[reply]
I just finished a lab on light diffraction and interference. For my calculations, I'm supposed to use my data from the double-slit diffraction to determine the slit width and slit spacing. I already know how to calculate the slit spacing (using dsinθ=mλ), but how can I calculate the slit width?--Edge3 (talk) 15:39, 13 April 2009 (UTC)[reply]
I'd suggest looking at Fraunhofer diffraction. See if you can write down an integral representation of the intensity pattern in terms of a transmission function. If you then have a look at how the intensity changes with different transmission functions (I'd recommend looking at perfectly thin double slits and then at a single finite width slit) you might see a way of calculating the double finite-width slits problem (convolution theorem might help there). Once you've got an expression for what you want, you'll be able to see the parts you'd need to measure to experimentally determine the slit width. 163.1.176.253 (talk) 22:18, 13 April 2009 (UTC)[reply]
I never realized this could get so complicated! Could you possibly give me a simpler equation? I'm in a physics class that doesn't use calculus at all (but I still know how to do calculus if it's absolutely necessary). In the double-slit part of the lab, I measured the spacing between bright fringes and the distance of dark fringes from a pre-selected point. For the second part, I picked a bright point on the fringe pattern, and I calculated the distance from that point to three dark fringes on each side of the point. Does that make sense?--Edge3 (talk) 23:54, 13 April 2009 (UTC)[reply]
(This is 163.1... at a different IP) Well that's the beauty of physics, even the simplest of things can be as complex as you want to make it. I wouldn't worry about what I said if you haven't studied much calculus but essentially the same method is used at the end of diffraction formalism to derive the N-slit intensity pattern for slits of finite width. Now if you set N=2 in this expression, we can get an equation relevant to your experiment (It simplifies if you note that sin(2x)=2cos(x)sin(x) in the last factor). I would strongly advise looking at the maths and asking if you have any questions because it's not too complicated and I'd say it is the important part of the experiment; infact, it also justifies the d*sin(θ) = mλ you used earlier.
Now, I didn't really follow what you've measured: lots of maxima and minima and distances between them. If you look at the equation at the end of diffraction formalism, you'll be able to work out when the function is at a maximum and when it is at a minimum (you could try getting a computer to plot it to make that bit easier). If you work out how the distances between these points will change as you vary the slit width, I'm sure you'll find a distance to measure which will give you the slit width. 129.67.116.90 (talk) 11:53, 14 April 2009 (UTC)[reply]
Great! Thanks for the help! So I'm now looking at the equation at the bottom, and I'm wondering whether I can eliminate I0 for the purposes of my calculations. Also, the variable I'm looking for is a, right?--Edge3 (talk) 01:33, 15 April 2009 (UTC)[reply]
Well, I0 is only going to be important if you measured the intensity at each point; from what I gathered, you ony measured the position of the maxima and minima. Now, you can see the general shape of the intensity function: it's a sinusoid, with a sinc(x) envelope function. Somehow you're going to have to identify the envelope function's period as this is what is related to the slit width. Were some maxima dimmer than others? 163.1.176.253 (talk) 11:23, 15 April 2009 (UTC)[reply]
Thank you so much!!! That was very helpful! I think I won't worry about the intensity too much, since all of the minima have intensities of zero. With a few organized manipulations of the equation, I think I can get the function to work out. Thanks again!--Edge3 (talk) 00:19, 16 April 2009 (UTC)[reply]
cars
I've always wonder about the new cars which they say its fuel saving
and as i understod they said it work by fuel (petroliuom) and while its moving it charges some
kind of battrey which its going to use later.
so the question is how could this method save fuel like some car can do 40 km consuming
1 liter of fuel thats ( 800 km / 20 liters ) , where korean cars do (300 km / 20 liters)
mabey they manage to cut the loses like wasted heat and such but it could never save that mush.
and if we use the fuel motor to generate elecricity which means turn the kinatic energy from
the tires to charge the battrey then this electricity will generate the same amount of energy
What I think you are asking is whether you could use the engine to charge a battery - and then use the battery to drive the wheels using electric motors. Well, that's exactly what a hybrid car like the Toyota Prius does. The reason is saves energy is a three-way thing:
Gasoline powered engines work most efficiently at one particular speed - perhaps 3,000 rpm. If you push them harder than that - or less hard - then they need more gasoline per unit of energy they deliver. With a normal car, the number of rpm's you need depends on what gear you are in and on what speed you are going - but it's only rarely turning the engine at the best speed. In the Prius, the engine only ever runs at this perfect speed - and when the battery is fully charged, the engine shuts off and stays shut off until the battery drains down and needs a recharge.
When you push on the brake in a normal car, you are wasting the kinetic energy in the motion of the car to wear down the brake pads and heat up the disks. With a hybrid, the electric motors that normally power the wheels can be used 'backwards' as generators - so you slow the car down by (effectively) using the battery charger to extract energy from the car's motion. Of course you need conventional brakes too - and this process doesn't work when the battery is already fully charged...but still, it makes some significant savings. This is called 'regenerative braking'.
Because you don't need that peak power from the gasoline engine when you do a (rare) hard acceleration - you can have a smaller engine. The engine only has to be large enough to provide the AVERAGE amount of power the car needs - not the MAXIMUM amount. Since you (mostly) don't go around accelerating hard all the time - this means that you have a smaller, lighter engine, more fuel-efficient engine - and let the battery provide the power for short bursts of speed.
Having said that, hybrid cars are not the perfect thing some would tell you. Most of the reason the Prius gets such good gas mileage is because it's super-streamlined, it's actually not a very fast car and it has relatively poor air-conditioning and such. If you did all of those things to a conventional car - and DIDN'T have to carry around all of those heavy batteries - you can do just as good as the Prius. The Prius actually gets rather poor miles per gallon on long freeway trips because in that case, the regenerative braking and the average-versus-peak thing doesn't work out too well - and pretty much any decent car, when driven in "overdrive" or topmost gear will have the engine running at it's most efficient rpm's. Hence the Prius has no special advantages in that case. However, for in-town stop/start driving, it works amazingly well.
I test drove Honda's new Insight, their new hybrid, last week. Part of the reason why it is so economical on fuel is that if you are stationary (e.g. in traffic, or at traffic lights) the engine cuts out and cuts back in when you take your foot off the brake to start moving. The battery cuts in to power the drive when you are going slowly in queueing traffic, which again makes it economical. --TammyMoet (talk) 18:22, 13 April 2009 (UTC)[reply]
That's true - but LOTS of non-hybrid cars do that. Most BMW's do it in their european versions. However, there are some weird laws in the US that (I believe) make cars that do that illegal. However, in hybrids, the engine isn't driving the wheels - it's only charging the battery - so under US law, it's not an "engine" so much as an "on-board-generator". However, if the US lawmakers would only get into the 21st century - we'd find that almost all modern cars could do it. The British version of the MINI Cooper (a non-hybrid) can use the cars starter motor to start the car rolling in stop-start traffic - so the engine doesn't even need to start...but again - not legal in the US. SteveBaker (talk) 22:31, 13 April 2009 (UTC)[reply]
What causes the stabbing/crampy pain you get sometimes due to exercise. I used to think (and was told) it had to do with the spleen but I and several friends can get it on both sides (not only the left). Is there a singular cause? PvT (talk) 16:46, 13 April 2009 (UTC)[reply]
Though only an entire hair, including the follicle, meaning it has to have fallen out or been pulled out. Cut hair, which includes only the shaft, is not sufficient. Rockpocket23:15, 13 April 2009 (UTC)[reply]
I think they generally like to have a sample from the mother as well, and all the possible fathers. It makes it easier to work out what DNA in the child comes from where. --Tango (talk) 18:30, 14 April 2009 (UTC)[reply]
Constructing 3D Sight
How would you biologically create sight` that is capable of viewing things in 3D? I know the brain will be significantly altered to understand 3D images, however, biologically, what will need to be constructed (even if imagined)?. Do we have AI's that can "view" things in 3D?--Reticuli88 (talk) 17:47, 13 April 2009 (UTC)[reply]
You already have a 3D image of the world. The image in your brain contains depth perception of the objects you see, derived from stereoscopic vision and focus information from the eyes. AI can do this just as well, if not better, than humans. However, the image in your brain is a model of the world quite different from the image projected on to your retina. I think perhaps you are referring to a Steve Baker post in a previous question, which I will go and copy here before answering further. SpinningSpark18:01, 13 April 2009 (UTC)[reply]
On second thoughts, I won't copy it, the discussion got too long and involved, Steve can post something here himself if he wants. To directly produce a 3D image in the eye (rather than a model in the brain) requires a 3D "screen" in which to form the image. There are many possible technologies for doing this, but that is not the problem. The big problem you have is that the eye must be out of the space of the thing it is trying to see, just as our 2D eyes have to be out the plane of the thing they are trying to see (you cannot see flat things edge on). In other words, your need an eye in the direction of the 4th dimension. This is simply not possible, a 4D body is now required which we do not have. SpinningSpark18:11, 13 April 2009 (UTC)[reply]
Dolphins and bats probably do this already using their sonar. A dolphin can see inside other creatures, not just the surface. Dmcq (talk) 21:48, 13 April 2009 (UTC)[reply]
Full 3D "sight" would require a lot of brain reorganization. Each part of the early visual system is organized as a 2D layer, matching the topography of the retina. All of this neural circuitry would have to be completely reorganized. Basically you'd be starting from scratch. Looie496 (talk) 22:21, 13 April 2009 (UTC)[reply]
We probably already have some mechanism for imagining it ok though. When I think of something 3d like a building where I know what the rooms are like I feel the whole structure inside myself rather than imagining looking at it. Dmcq (talk) 23:07, 13 April 2009 (UTC)[reply]
If dolphins and bats can see inside objects it's simply because those objects are (semi)transparent to the sound waves they use. It would be like the way we see the inside of a jellyfish, not a fundamentally different form of perception. Our bodies are transparent to light too at many frequencies (like X-rays).
We have various technological means of taking volumetric photographs, like magnetic resonance imaging (awesome animated GIF in the lede), but they require the subject to stay still for long periods of time. Although, come to think of it, so did early photography. -- BenRG (talk) 23:14, 13 April 2009 (UTC)[reply]
Seeing through transparent things is I think a poor man's 3d version of what dolphins can do. With sonar you get a click back for each of the various distances to objects and there is very little masking. The difference is great enough that it probably qualifies as a completely different type of experience. Dmcq (talk) 23:45, 13 April 2009 (UTC)[reply]
Some blind people have done an amazingly good job of teaching themselves to echolocate. We even have an article on it: Human echolocation. Having seen a TV documentary about Ben Underwood accomplishing this and being tested by scientists to measure his abilities - it is simply incredible what he has found himself to be able to do. It wouldn't surprise me at all to discover that he has a rudimentary "3D acoustic vision" as a result of being able to echolocate through softer objects. SteveBaker (talk) 02:30, 14 April 2009 (UTC)[reply]
Hmm, could we view small things in better 3D simply by having a 3rd eye which extended from our foreheads on a stalk? If we held things in front of ourselves, so that our stalk-eye saw the other side of the object to our current eyes, we could see these objects in 3D. There wouldn't be much point though, since in that situation we can 'see' these objects in 3D by feeling them, adding visual detail by turning them to look at different sides. I agree that echolocating seems the best way to 'see' the wider world in 3D. 217.43.141.59 (talk) 22:04, 14 April 2009 (UTC)[reply]
According to this article, are they saying that there is a possiblilty of universes within universes? If so, does that mean that there can be another universe existing in the exact space, time of where I am right now? --Reticuli88 (talk) 18:25, 13 April 2009 (UTC)[reply]
Not exactly, what they are postulating is that the 3D universe we can see is just a slice of the actual 4D universe. Just like a slice through a 3D object will result in a flat 2D object. This braneworld cosmology as it is known is saying there is a lot more to the universe than we can see, rather than any sort of many-worlds interpretation. It remains to be explained why we have been utterly unable to detect the existance of all this extra universe so far. SpinningSpark18:40, 13 April 2009 (UTC)[reply]
You may have also been confused by them saying that our observable universe is part of the whole universe, but there's nothing mysterious about that. The observable universe is just the region around an observer where they could, in principle, see light from. Anything outside that region is moving away from the observer (due to the universe expanding) faster than the speed of light, so the light could never arrive at the observer. It's unfortunate that there's not a better term for it, since it confuses so many people. --Sean19:10, 13 April 2009 (UTC)[reply]
I have re-read my post quite carefully and do not see the phrase "observable universe" anywhere in it. This indeed is something quite different and has nothing to do with the question being asked. SpinningSpark19:38, 13 April 2009 (UTC)[reply]
I think Sean meant this sentence which is in the second paragraph in the link:
"The theory holds that the visible universe is a membrane (hence "braneworld") embedded within a larger universe"
Right. SpinningSpark, your original question was about the possibility of a "universe within a universe". The observable universe is exactly that, and is the only such thing mentioned in the article. --Sean21:16, 13 April 2009 (UTC)[reply]
Well it wasn't me who asked the question, but my apologies, I now see you were replying to the OP, not to me. I ought to know what the indent means by now! SpinningSpark21:59, 13 April 2009 (UTC)[reply]
I'm skeptical that music could really make plants grow faster. I know a lot of people insist otherwise, but people insist on all sorts of crazy stuff. That said, an episode of Mythbusters seemed to indicate that plants exposed to music and speech grew faster than their control plants, which were kept in a quiet spot. I find the results kind of dubious, and suspect that a more controlled experiment would bring about different results, but them's the breaks. In the episode, the plants that got a constant blast of death metal did best. (As for seeds, talking to them strikes me as even more nonsensical.) -- Captain Disdain (talk) 23:53, 13 April 2009 (UTC)[reply]
It is a known effect that many plants grow stronger in response to varying wind, while if grown in a bell jar will grow spindly and weak. Is it possible that loud death metal music is stimulating this response through mechanical vibration? SpinningSpark06:33, 14 April 2009 (UTC)[reply]
That would make sense, sure. I really don't see how that would have any effect on the seeds, though -- one of the most interesting characteristics of seeds is that they are kind of dormant, and can remain that way for a long time, until they end up in an environment where they can start growing. -- Captain Disdain (talk) 11:28, 14 April 2009 (UTC)[reply]
Sure, though not necessarily all instances of vomit. The vomit itself -- that is to say, that charming combination of stomach acid, half-digested food and whatnot -- probably doesn't contain any, but it's not uncommon for the vomit to have bits of stomach lining in it, for example. Likewise, people generally do a lot of spitting in the process. (That said, Nigel Tufnel famously opined that you can't dust for vomit, but I grant that he may not be the greatest authority on this subject. Or any subject.) -- Captain Disdain (talk) 01:57, 14 April 2009 (UTC)[reply]
Vomit may represent the worst sort of specimen for DNA testing, because a large proportion of the DNA in it may represent a source (food) other than the person you mean to be testing. Ideally, of course, if you can ensure your testing only detects human DNA then you'll only be confused if you're testing a cannibal. --Scray (talk) 02:35, 14 April 2009 (UTC)[reply]
stars outside the boundaries of their constellation
is there a name for a star that is named as being a member of one constellation, but actually exists inside the boundaries of another? does such a thing even exist? i had a dream that a star like that was called a "tapuo" or something, so now i'm curious if it actually exists. 99.245.16.164 (talk) 00:35, 14 April 2009 (UTC)[reply]
I have a copy of A Field Guide to the Stars and Planets, one of the books mentioned in the constellation article. I'm pretty sure that its maps don't show any such thing as a star named as part of a different constellation. --Anonymous, 00:53 UTC, April 14, 2009.
The star Sirrah might be considered an example of what OP refers to: It forms the north-eastern corner of the big square that visually defines the constellation Pegasus (constellation), and historically, it was called δ Pegasi. According to the modern definition of the constellations it belongs to Andromeda, and is even known as α Andromedae. Another example might be β Tauri/γ Aurigae. --Wrongfilter (talk) 07:54, 14 April 2009 (UTC)[reply]
Diamonds have some properties of ionic compunds (they are hard, have high melting and boiling points and can conduct electricity) but other properties (such as diamonds not being able to dissolve in water and not being brittle) are not properties of ionic compounds. —Preceding unsigned comment added by 174.6.144.211 (talk) 01:52, 14 April 2009 (UTC)[reply]
But if diamonds are covalent compoounds then why are they hard, have high melting and boiling points are able to conduct electricity? —Preceding unsigned comment added by 174.6.144.211 (talk) 02:18, 14 April 2009 (UTC)[reply]
Usually covalent compounds will have covalent bonds (which are very strong) within a molecule but those molecules will bind to each other by week Van der Waals atraction. I diamond has strong covalent bonds all over the whole crystaline structure. In simple terms, it is one single macroscopic molecule. Dauto (talk) 03:11, 14 April 2009 (UTC)[reply]
Look up diamond. most diamonds are insulators and they more burn rather than melt. The reason they are so hard is how well they electronic/bonding structure matches their packing crystal lattice. Diamonds also have a lot in common with ionic materials. Diamond aren't so much a molecule or polymer as a material made of repeating unit cells like a ionic solids.--OMCV (talk) 02:24, 14 April 2009 (UTC)[reply]
More generally, lots of things you learn in school are general guidelines or observed trends, not the actual reasons. "Ionic has property X, non-ionic has property not-X" is only true if being ionic is the exact and only cause of property X. But that's not true for this case: the actual cause of "high-melting" isn't "ionic", but rather it's related to how the molecules are held together. Ionic molecules are merely a common type that are held together in a way to give high-mp. For that reason, "ionic gives high mp" doesn't mean you can say "therefore non-ionic will not be high mp": being ionic is not the only type of molecule that is held together in this certain way. Same goes for any property and any characteristic: need to see how/why the trend works, not use the observed trend as the actual cause and black'n'white separation (correlation vs causation). DMacks (talk) 14:18, 14 April 2009 (UTC)[reply]
To expand a little bit, the binary "covalent vs. ionic compounds" thing they teach you in basic high school chemistry is a bit oversimplified, and it does not apply to many substances including diamonds. Reorganize your thinking this way. Compounds either exist as a collection of discrete molecules or they exist as a network of roughly equally bonded particles.
Molecules are the "special case" for us to consider. In the case of molecular substances, you have a small number of atoms which are strongly tied to each other, making a tiny packet which itself is electrically neutral; meaning it does NOT much attracted to other substances, and only subject to smallish forces like dispersion forces and dipole interactions. Therefore molecular substances (calling them "covalent" compounds is misleading, see explanation below) will tend to have low melting and boiling points, be mostly gasseous or low boiling liquids, etc. etc. The bonding inside of the molecule is covalent bonding, but the bonding between the molecules are relatively weak, and THAT is what makes molecular substances behave like they do.
All other compounds consist of billions of atoms all bonded to each other in a highly stable crystal. The strength of these bonds is comparible to the strength of the internal bonds in a molecule, thus ripping apart a salt crystal or a diamond is roughly on the order of physically pulling the hydrogen atoms off of a water molecule; it requires huge forces, which is why "non-molecular substances" are universally strong, hard, and have high boiling and melting points. Now, these non-molecular substances can be further classified according to the type of bonding holding their atoms together, but these distinctions are not as large as one might imagine. The type of bonding in non-molecular substances falls into three types: covalent bonding, ionic bonding, and metallic bonding.
For covalent and ionic bonding, look at the crappy MS Paint diagrams I uploaded. The both consist of the force of attraction between a positive atomic center and an negatively charged electron cloud. It's only the location of that cloud (either between the atoms or around one of them) that determines the difference between the two. Since BOTH consist of the same maginitude of electrostatic attractions, BOTH are going to be on the same order of strength. Thus so-called "network covalent" solids, like diamond and zinc blende are about as strong as so-called "ionic" solids like sodium chloride and potassium nitrate.
For metallic bonding, the electron cloud, instead of being localized either between or around some of the atoms, is instead shared roughly equally between all atoms, creating a so-called "sea of electrons" which explains metals electrical conductive properties. The material is hard because ALL of the electrons in bulk are holding the whole mess together, but the individual electrons are not tied to any individual nucleii, so they are free to move around easily given a little "push".
I hope that clears things up a bit for you. So, you see that diamond is infact a covalently bonded substance, but it is not a molecular substance, which is why it does not behave like a molecular substance. --Jayron32.talk.contribs01:24, 15 April 2009 (UTC)[reply]
Atropine as anaesthesia pre-med
Atropine is historically used as an anaesthetic, and apparently is used as a premedication for general anaesthesia today, I can see why you would want to have a lower heart rate, bronchodilation and the like, but as atropine is a CNS stimulant, my question is, why would you use it as an pre-anaesthesia. I know hyoscine would be a better choice. MedicRoo (talk) 11:14, 14 April 2009 (UTC)[reply]
You may find your answer to why hyoscine (scopolamine) has limited usage compared to atropine at Muscarinic_antagonist#Effects, which states that scopolamine crosses the blood-brain barrier, whereas atropine does not. Therefore, scopolamine can cause increased CNS effects such as amnesia, which can in some cases be beneficial, but not all. Also, atropine given intramuscularly initially results in bradycardia. —Cyclonenim | Chat 11:43, 14 April 2009 (UTC)[reply]
I understand most of the effects and differences, more I'm not clear on why atropine would be used for anaesthesia when it causes CNS excitation instead of depression.MedicRoo (talk) 13:15, 14 April 2009 (UTC)[reply]
I'm not too sure, all I know is that there is a weak agonist effect in low doses of atropine, which lowers the heart rate. Presumably, this agonist effect spreads throughout the entire parasympathetic nervous system, not just the heart. —Cyclonenim | Chat 13:42, 14 April 2009 (UTC)[reply]
Other possible benefits are relatively unimportant. The sedative effect of hyoscine is not usually regarded as beneficial when giving general anaesthesia. Indeed the sedative effect of hyoscine may be tricky to predict, especially in the elderly. Other anaesthetic drugs give a more reliable response. Axl ¤ [Talk]17:34, 14 April 2009 (UTC)[reply]
Bugger, somewhere along the line me and atropine went seperate ways and I thought it was right behind me. Thanks for the help.MedicRoo (talk) 18:05, 14 April 2009 (UTC)[reply]
I know that atropine is frequently used when doing surgery on rats, because it dries them out and makes it less likely that they will choke on their saliva. It isn't an anesthetic as far as I know. Looie496 (talk) 01:35, 15 April 2009 (UTC)[reply]
Separating Water(H2O) and Hydrogen Peroxide(H2O2)
The hydrogen Peroxide(H2O2) you buy at pharmacies are so low concentrated and are dissolved in water(H2O), and I want more pure hydrogen Peroxide, but don't want to use so much money to buy the expensive high concentration hydrogen Peroxide(H2O2) from large companies.Can I remove the water by evaporation(H2O2's boiling point is 150.2 °C but water's boiling point is 100 °C)? Please tell me other methods too!The Successor of Physics15:19, 14 April 2009 (UTC)[reply]
Have you read our excellent article on hydrogen peroxide ? Especially the part that says "Above roughly 70% concentrations, hydrogen peroxide can give off vapor that can detonate above 70 °C (158 °F) at normal atmospheric pressure. This can then cause a boiling liquid expanding vapor explosion (BLEVE) of the remaining liquid. Distillation of hydrogen peroxide at normal pressures is thus highly dangerous." Maybe there's a good reason it's only generally available in low concentrations .... Gandalf61 (talk) 16:42, 14 April 2009 (UTC)[reply]
The main reason it's only available at low concentrations is because high concentrations are very dangerous. It looks exactly like water, but if you take a swig of highly concentrated H2O2 and swallow it before you know what you're doing, say goodbye. Looie496 (talk) 01:38, 15 April 2009 (UTC)[reply]
I know what Looie496 and 65.121.141.34 said but forgot what Gandalf61 said! Well if I can't seperate them by evaporation, then how about cooling(I'm not even sure is there such a separation method)(H2O2's melting point is -0.41 °C but water's melting point is 0 °C)?The Successor of Physics13:05, 15 April 2009 (UTC)[reply]
If we were (somehow) a 3D object in a 4D world, we'd look kinda "flat" - much like a drawing on a 2D sheet of paper looks to us in the 3D world. But this is basically a silly question. If the universe had 4 spatial dimensions (and not "curled up" dimensions as string theory supposes) then we would be 4D entities with perceptual systems to match. With 4D brains, we'd have evolved to be entirely different creatures than we really are...in all likelyhood, there is some deep, dark, physics-ish reason why things like people, DNA, planets, stars and galaxies might simply not be able to exist in a 4D world. Certainly, such things are impossible in a 2D world. So this "what would it be like if..." trail our OP is on is entirely fruitless. All we can say (and even then, not entirely uncontroversially) is what a 4D world might look like to us...as we are right now with our 2D eyes...assuming that the phenomenon of light is possible in a 4D universe, assuming we don't explode into fundamental particles the moment we pop into existence in the 4D world, etc, etc. SteveBaker (talk) 19:41, 14 April 2009 (UTC)[reply]
The reason I brought this up was because of the link below:
The author describes what a 4D entity would appear to us: "....resemble flesh-colored balloons constantly changing in size." Also, as scene in the movie Contact, Jodie Foster's character experiences a sort of 4D experience, I think. I mean the scene when she exclaims "They're alive!" the first time; the scene where it appears she splits in two. Does anyone know this reference? I'm just trying to get some handle what 4D would appear to us lowly humans. --Reticuli88 (talk) 20:40, 14 April 2009 (UTC)[reply]
You might like to read Flatland; it's a good way to start thinking about these things. It uses a common approach to these things; what would 3D objects look like to 2D entities? Then extrapolate. For example, (and this is probably something like what the author of that link is getting at) a 2D entity experiencing a 3D entity would itself really be a 3D entity which only experienced 2 dimensions (like an ant that never looks up or down). If a sphere passed through the 2D entity's plane, the 2D entity would experience it as a circle which got bigger and bigger, then smaller and smaller. Equally, a 4D hypersphere passing through our 3D space would appear as a 3D sphere that got bigger and bigger, then smaller and smaller. More complicated shapes would change in more complicated ways.
I don't understand why people keep recommending Flatland - I can't believe that anyone who has actually read the book would ever wish it on anyone else! It's a simply awful book! It's turgid, sexist, racist, class-ist and everything-else-ist. It's really painful to read. It tells you nothing meaningful about different numbers of dimensions and the author is clearly quite confused about his explanations of the 2D world (his description of houses is particularly ill-thought-out. So - forget that piece of junk and read Planiverse by A.K. Dewdney - this is an entirely self-consistent world in two dimensions - it has plot, it has amazing diagrams of 2D gadgets (if you have any interest in mechanical gadgets, you'll spend ages staring at the wonderful 2D steam engine - and thinking SURELY there is a way to have 2D wheels that work usefully (there isn't)...the 2D house and 2D space-station are also pretty interesting) - and it has a comprehensive appendix that explains in detail the problems with there being an actual 2D universe. It's written by a mathematician so you can be reasonably sure that it's properly thought through. It remains one of very few works of fiction that survives on my "top 20 books" bookshelf! Read it! NOW!!!! SteveBaker (talk) 22:42, 14 April 2009 (UTC)[reply]
One aspect of the steam engine diagram that may not be evident without reading the book is the question of how 2D beings could assemble such a machine? When you look at it at first sight, it seems impossible. However, what you may not appreciate is that the parts that look like the artist shaded them with a bunch of thin parallel lines - actually represent a glued laminated material that the people of the 2D world have developed to solve this exact problem. By building up the steam engine from layers of thin material, glued together, they are able to fabricate the machine perfectly well. When you understand THAT - you've gotta go back and stare at the diagram for ANOTHER 10 minutes to convince yourself that's really true. I just flipped through the book again - the stuff about the 2D computer is really quite clever...how do you make a computer when wires can't cross each other without shorting out? Also how does a 2D being eat and excrete without falling in half? All of these things are quite nicely explained. Like I said - it's an amazingly well put-together book. SteveBaker (talk) 22:40, 15 April 2009 (UTC)[reply]
Flatland is a satire. You're not supposed to like the sexist and classist society it describes. As a satire I think it's pretty clever. But I don't understand either why people keep recommending it as a book about life in two dimensions. Abbott obviously didn't much care about that side of the story. There's nothing in the book about two-dimensional physics or biology, and what little it says about geometry could fit on one page. For that you have to slog through 150 pages of the narrator complaining about the baseness of women and the isosceles class. Think of it as part five of Gulliver's Travels, though, and it's a good read. -- BenRG (talk) 12:55, 15 April 2009 (UTC)[reply]
Yeah - I've heard it described as a satire - but it doesn't seem in any way to be poking fun at these ideas. If you told me it was a stinging diatribe about the fickle nature of women and the uselessness of the working classes - I'd have been more inclined to believe it. SteveBaker (talk) 17:57, 15 April 2009 (UTC)[reply]
Well, consider that as the dimensionality decreases, so does one's respect for the inhabitants. The King of Lineland isn't much when compared to A Square, and the sole inhabitant of Pointland is a horrid egomaniac. By analogy then, the implication is that the figures in Flatland are not only physically, but morally, inferior to we Spacelanders. And, BTW, the Sphere, itself, is a bit on the pompous side. (Perhaps the Hyperspacers have higher ethics.) B00P (talk) 23:10, 15 April 2009 (UTC)[reply]
A friend of mine told me that you can see the constellation Orion in Australia, but that it looks upsidedown then what is normal in the northern hemisphere. Was he just joking with me, or is there truth to this? 65.121.141.34 (talk) 16:12, 14 April 2009 (UTC)[reply]
You can currently catch Orion for about three hours after sunset, looking westwards, tilted about 90° clockwise compared to the northern hemisphere (with Sirius above) after which it disappears below the horizon to reappear later during the day lost in the sun glare (and "upside down" for a while). EquendilTalk17:03, 14 April 2009 (UTC)[reply]
It's true - from the point of view of someone looking at Earth from the outside, someone in the Southern hemisphere is "standing on their head" relative to someone in the Northern hemisphere. That means that, from their perspective, anything outside the Earth is upside-down compared to what someone in the North sees. --Tango (talk) 18:24, 14 April 2009 (UTC)[reply]
Obviously, all of the constellations, the sun, the moon and the planets are "upside down" - but not many of the constellations are visible from both the far northern and southern hemispheres...but I guess Orion is one of them. The MUCH more noticeable thing for me is that the moon is "upside down". A new moon in the northern hemisphere looks like ')' but in the southern hemisphere it looks like '('. On the equator, it is an upward curving arc. I was born and lived most of my life in the UK - which is pretty far north. A waning moon looks like '(' - and I'd read many times that some cultures believed the waning moon was a boat that carried the gods around (or some such twaddle). I could never understand why they thought that...until I moved to Texas where the waning moon is rolled over at 45 degrees...and on the equator...it looks just like a boat (well, a canoe, perhaps). Even now - after living 16 years in Texas - the moon always looks "wrong" to me. SteveBaker (talk) 19:22, 14 April 2009 (UTC)[reply]
I recall looking at the full moon in New Zealand and trying to figure what was wrong with it. Not sure why but I was totally unable to recognize the "patterns" on the surface at all (I wasn't aware that it was simply upside-down at the time), as if it had been randomized and now I was forced to find another character or face to associate it with. 124.154.253.25 (talk) 03:43, 15 April 2009 (UTC)[reply]
Orion's head and feet are at 10 degrees north and 10 degrees south respectively, so he's partly visible everywhere on Earth and fully visible (not counting some dim stars) within 80 degrees of the equator. —Tamfang (talk) 06:32, 15 April 2009 (UTC)[reply]
Note that it's not as simple as "right-side-up or upside-down".
Even at the same location, the same constellation will be seen in
different orientations at different times of day when it is in
different parts of the sky. Similarly, it is not true that "down"
is exactly the opposite direction for people in different hemispheres
unless they happen to be in antipodal locations. I'm in Toronto;
compared to someone in Buenos Aires I'm not upside-down, I'm sideways.
For a simple example that doesn't require you to wait for nightfall,
look at the moon at different times on the same day. The center of
the lighted side always points toward the sun -- therefore it points
above the horizon when the sun is up, but below the horizon when the
sun is down. But the same features on the moon's face are visible,
so you can see it's a different way up with respect to you. Well, for
someone in a different part of the Earth it's different again.
--Anonymous, expanded 04:55 UTC, April 15, 2009.
A stock solution of HCl(aq) has a density of 1.19 g/mL and a percentage purity of 39.1%. A lab requires 7.25 L of 0.500 mol/L solution. What volume of stock solution do I need before dilution?
I divided the amount of hydrochloric acid needed (3.625 mol) by [HCl(aq)]39.1% (12.76165661 mol/L), reaching the conclusion of 284 mL. The answer key says 303 mL. This is not homework counting for marks. Otherwise, my teacher would not have given me the answers. Did I do something wrong? Thanks in advance. --Mayfare (talk) 20:58, 14 April 2009 (UTC)[reply]
If there is an obvious mistake out there, you can count on me to make it, I get the same answer as Mayfare also (well 284.7 to be exact). The book answer requires a molecular weight of 38.8. The value I am using is 36.45. SpinningSpark22:18, 14 April 2009 (UTC)[reply]
One thing to double check is what is meant by "39.1% purity" - is it w/w, w/v, v/v, v/w, mol/mol (mole percent) or something else? I would usually expect it to be w/w in this context, but you never can tell. I tried to run the calculation several other ways but none gave me 303 mL, though. -- 128.104.112.117 (talk) 23:14, 14 April 2009 (UTC)[reply]
No. In order to catch fire, the object has to be able to combine with oxygen and give off energy. Wood has energy to give off - and the carbon in the wood can combine with the oxygen in the air to make carbon dioxide. Rocks have no spare energy to give away - the chemicals they contain don't easily combine with oxygen - so with rare exceptions (Coal, for example), rocks don't catch fire. SteveBaker (talk) 22:27, 14 April 2009 (UTC)[reply]
Does something that burns (rapidly oxidizes) always have to be exothermic ? I've noticed that aluminum foil seems to "burn" in a hot enough flame, yet doesn't appear to give off energy, since the flame is not self-sustaining (if you try this experiment, please do it outside with a torch, as aluminum fumes can be toxic). As for the rocks, the two alternatives would seem to be melting and burning. So, do all other rocks melt at high temps (or perhaps some sublimate) ? StuRat (talk) 15:25, 15 April 2009 (UTC)[reply]
It will definitely depend on what is eaten before the fasting starts. The obvious possible negative effects would be dehydration, hypoglycaemia and possibly electrolyte imbalance. See: Fasting#Health effects for a list of some benefits of fasting. Also, the ramadan article states that "Fasting is meant to teach the person patience, sacrifice and humility", so that could be a good effect of fasting, although whether it will teach those things is quite hard to test. --Mark PEA (talk) 22:22, 14 April 2009 (UTC)[reply]
How would someone be able to eat their whole own self? Since eating one's whole self requires eating their mouth, how would they be able to eat afterwards? And since it requires eating their digestive system, how would they be digested? 58.165.25.29 (talk) 23:26, 14 April 2009 (UTC)[reply]
No, fava beans and a nice chianti are only for the liver, not for the whole thing. Still, your approach is perfectly valid. I suggest a simpler, alternative approach. Step 1: spew generous amount of digestive juices. Step 2: digest self. Or, a slightly fancier alternative approach. Step 1: make a cocoon. Step 2: undergo histolysis. Step 3: remember that you are not a butterfly, stop. Step 4: Have a nice day :) . --Dr Dima (talk) 00:46, 15 April 2009 (UTC)[reply]
Most human cells are not particularly long-lived, so if you were to consume all of your waste products, hair, nail trimmings, and sloughed skin cells, you will have eventually eaten most of yourself, in a manner. --Sean16:36, 15 April 2009 (UTC)[reply]
Put on your will that your body be sent back in time to the present, then, assuming time travel exists and people are willing to fulfill your will, you can eat it. — DanielLC—Preceding undated comment added 01:22, 16 April 2009 (UTC).[reply]
Mailbox Key question.
In the neighbourhood where I live, all of the houses share a single mail pick-up point - it's like an apartment complex - there are no mailboxes outside each house. Instead, over by the park, there are rows of tiny locked mailboxes - one for each house - for which you have a key. There is also a row of much larger mailboxes for packages and such. When you get a large parcel, the mailman puts it into one of the large boxes, locks it and places the key into your individual mailbox. Now - here's the thing. The large mailboxes are set up so that you put the key into the lock - turn it to get your package out - but the key won't turn back again - and it won't come out of the keyhole so nobody can steal it. There is a second keyhole that the postman puts a special key into that releases the first key.
I've been trying to figure out how this fiendish mechanism works - the mailbox key looks like a perfectly ordinary key. It's obviously a ratchet or something - but curiosity demands that I know how it ACTUALLY works. Does anyone know of a diagram or something?
I also recall seeing that same mechanism in lockers in bus/train depots and airports. You would pay a quarter and get to use the locker/key once. Of course, as a result of recent terrorism, such bomb-friendly lockers have probably disappeared. StuRat (talk) 15:17, 15 April 2009 (UTC)[reply]
The authors of the paper have not explicitly stated the reason for the patient's symptoms (i.e. the diagnosis). One could debate whether this patient has "Congenital absence of appendix" or "Rudimentary appendix". If the latter, perhaps the diagnosis is indeed appendicitis of the rudimentary appendix? The authors do not describe their further management of the patient. Did they remove the rudimentary appendix? Did the patient improve?
The main problem here is with the definitions of "congenital absence of the appendix" and "rudimentary appendix". If they were defined more clearly, we could give a more helpful answer. Axl ¤ [Talk]08:28, 15 April 2009 (UTC)[reply]
(ec)You can start by looking at Isotope#Occurrence in nature and Nucleosynthesis. In general, the various processes that transform the elements yield nuclei with various combinations of protons and neutrons. Only certain combinations are stable. Others will almost immediately decay into other elements. Still others will be relatively stable and may take years millennia, or longer to decay. An element is defined by the number of protons contained in the nucleus, and a given element may only be stable or have a measurable life with a certain number of neutrons. For example, Carbon (see also Isotopes of carbon) have three isotopes with a half life of more than a few seconds, but those three isotopes – with six protons, and six, seven, or eight neutrons – have very long lives (in human terms). Those are the isotopes general identified as naturally occurring. -- Tcncv (talk) 03:26, 15 April 2009 (UTC)[reply]
(Sorry, I guess I got off track) To generalize, there is no reason why an element cannot have more than one stable isotope. hydrogen has two, oxygen has three, and as you found antimony has two. As to why they have different abundances, I expect that is due to the different isotopes being formed through different mechanisms, or if formed through the same method, that method (such a supernovas) just happens to produce more of one isotope than another. -- Tcncv (talk) 03:56, 15 April 2009 (UTC)[reply]
It does, to some extent. Loosely, radioactive decay can be thought of as a Quantum tunelling phenomena and the tunneling probability is dependent on the released energy. However, the transition rate is more heavily influenced by various selection rules governing the decay mode. For example, in beta decay there are selection rules which arise out of conservation of angular momentum and parity. The influence of these selection rules can be huge - it is energetically possible for Calcium-48 to undergo single beta decay but it has never been observed to decay via. single beta decay (it has been observed to undergo double beta decay though).
Going back to the original question, it is possible to have two stable (or effectively stable) isotopes of an element as long as there is no significant decay mode to another element or isotope. A nuclear decay chain can be thought of as a waterfall, where nuclei keep "falling" into lower states until there's nothing to fall into. It just so happens that both Antimony-121 and Antimony-123 are at the bottom. Someone42 (talk) 09:09, 15 April 2009 (UTC)[reply]
Precambrian craton formation
Hi, I've been studying a little about the geological history of the Earth preceding the ordovician period, and I've come across some information on the cratons that are known to have existed, as well as an animation of how they were believed to have been positioned at the point when the crust first formed into plates. There is an animation on Wikipedia here (warning, may enduce seizures), but I find the morphing process is much easier to see in this video clip (youtube, starts at about 2:15), though I can only assume they are working from similar data.
I'm specifically concerned with the peculiar string of "islands" linking the Siberia craton and the Baltica craton. By the Carboniferous period they seem to have all been gobbled up by the larger cratons as pangea is formed. I assume that the reason they have been separated from Siberia and Baltica in the first place is because of some difference in the mineral/fossil record that indicates separation before Carboniferous, but I'm curious about the positioning of these mini-cratons, and the size. Are they being placed in a string simply because it is impossible to tell exactly where they were located, or is there some reason to believe that they were arranged in such a manner? And if the land fossil record for that period is as sparse as I think it is, how can they rule out that these mini-cratons were not in fact much larger, or merged together into a larger mass?
It sounds like you are talking about a volcanic arc. I don't know why geologist believe that there was an arc link the cratons you are talking about. Volcanic arcs, like the one at the caribean or aleutian islands for instance, are actually quite common and they do indeed align along a string of islands as you described. Dauto (talk) 04:36, 15 April 2009 (UTC)[reply]
Yes, it does look very like a volcanic arc, and if it is, the real question is what kind of evidence led scientists to their conclusion? I thought perhaps that they may have evidence for a certain number of small cratons separate from Siberia and Baltica, likely created by a similar event (such as a volcanic hot spot under a floating plate) assuming they are geologically similar. Fossil evidence could then further show that they were separated by water, or alternately the cratons could have been discovered very far from each other. I'm not educated enough in any field to make wild guesses like that though! 124.154.253.25 (talk) 06:35, 15 April 2009 (UTC)[reply]
The evidence nowadays would be a line of acid volcanics such as dacite or andesite from that period, possibly embedded in marine sediments. There is likely to be folding in marine (or back arc) sediments, as the volcanic arc was fused onto a craton. In this particular case I don't know the specific evidence. Graeme Bartlett (talk) 06:46, 15 April 2009 (UTC)[reply]
This paper by Robin Cocks and Trond Torsvik [1] includes a fairly detailed discussion of all the 'bit and pieces' or terranes caught up between Baltica and Siberia. I'm assuming that you're mainly talking about 'Kazakhstania'. I think there is some faunal evidence, i.e. an analysis of high v. low latitude types of fossil and possibly palaeomagnetic data allowing the construction of an apparent polar wander path for at least some of the terranes. Remember that geoscientists have difficulty reconstructing continents where there are no 'magnetic stripes' preserved to help close up intervening oceans (that's pretty much anything before the middle Jurassic or about 200 million years ago); several distinct reconstructions may be possible from the same data. Mikenorton (talk) 10:22, 16 April 2009 (UTC)[reply]
Pharmaceutical naming
How do drugs get their common or generic names? I'm thinking here of lorazepam, minoxidil, sildenafil, etc. They're somewhat random syllables, although I'm pretty sure that anything ending in -mab is a monoclonal antibody, so there's some correlation sometimes. Is there a convening body to dispense these names, is there some centralized random generator everyone can go to, or what?
Related question, when the actual corporations do their branding, are they wide open, or do they obey some rules also? For the examples above, how do we arrive at Ativan/Temesta; Rogaine/Regaine; Viagra/Revatio?
a generic name, which is the active ingredient of the medicine
a brand name, which is the trade name the manufacturer gives to the medicine.
The generic name is the official medical name for the active ingredient of the medicine.
The brand name is chosen by the manufacturer, usually on the basis that it can be recognised, pronounced and remembered by health professionals and members of the public. An example would be Viagra - this is the well-known brand name given by Pfizer to the generic medicine sildenafil. (Brand names are capitalised; generic names are not.) More here.Cuddlyable3 (talk) 09:56, 15 April 2009 (UTC)[reply]
We've discussed these issues before on the ref desk, e.g. generic names and trade names. These are just 2 of the first few hits I got after I searched the RefDesk archive (above) for "drug names". BTW, I was told 5 or 6 years ago that focus group research had indicated that "v", "x", and "z" were "strong" letters to include, and since then we've seen "Vioxx", "Zyvox", and the list goes on... --Scray (talk) 11:28, 15 April 2009 (UTC)[reply]
Hey, you noticed! :) The marketing committee was up all night on that one. It's a combo of my given name and "max", connoting large or major. If the Pope can be pontifex maximus I figured I could follow the same strategy. Sadly, I am not yet considered infallible. ;) Franamax (talk) 18:31, 15 April 2009 (UTC)[reply]
Ooohh, an absolute superlative in singular masculine form - that's good! The other strategy is to go for self-deprecation. When I first saw User:Mild Bill Hiccup, I laughed for quite a while. (For those not familar with American culture, the contrast is here). Franamax (talk) 20:03, 15 April 2009 (UTC)[reply]
Thanks for the pointers! The World Health Organization Expert Panel on the International Pharmacopoeia and Pharmaceutical Preparations handles the generic names - why didn't I think of that. Boy, the United Nations has an article on everything! Franamax (talk) 18:31, 15 April 2009 (UTC)[reply]
No, it's infinite. Since ρ is defined as electric field divided by current density, and there is no current in a vacuum (since there are no charge carriers), ρ is infinite. Note that this is only hypothetical, since nobody has ever created or discovered a perfect vacuum. --Heron (talk) 13:07, 15 April 2009 (UTC)[reply]
Are you sure? If the electric field was strong enough I would expect electrons to be ripped off the negatively charged object and be attracted to the positively charged object, so a current would flow. (Although, I guess technically it wouldn't be a vacuum then.) --Tango (talk) 14:46, 15 April 2009 (UTC)[reply]
A "perfect vacuum" is merely" a philosophical construct that is never observed in practice" per the article. Outer space is 1×10-6 to <3×10-17 torr, compared to zero torr for a perfect vacuum. Intergalactic space could have as little as 1 molecule per cubic meter. But a heated filament in that region would emit electron toward a positively charged plate and the current would be as great as in a common vacuum tube(valve) or cathode ray tube, since these devices do not depend on residual gases to carry the current. That current, known as the "Edison effect" or thermionic emission, can be characterized by a curve relating current and voltage. It does not have a constant resistance, and the space does not have a constant restivity. It does not follow Ohm's Law any more than a semiconductor does. The tubes are evacuated to the greatest extent possible. Do electrons travelling through a region of otherwise pure vacuum count as "particles" and how much pressure would each electron in, say, one cubic centimeter create? Edison (talk) 16:08, 15 April 2009 (UTC)[reply]
Electrons have mass, so when an electron strikes a surface it exerts a force – a pressure. The amount of force depends on the speed of the electron. In the case of a radio vacuum tube the situation is special because the electrons are not moving at random. They are all moving in one direction – toward the anode. – GlowWorm. —Preceding unsigned comment added by 98.17.38.45 (talk) 22:01, 15 April 2009 (UTC)[reply]
A perfect vacuum is not merely philosophical, since I can define some arbitrarily small volume of space which contains no particles at the time I measure it. But at the next instant, those damn physicists insist it will get filled up again - nature abhors a vacuum and Casimir keeps putting in new particles.
Nevertheless, if the definition of a perfect vacuum is zero particles and you send a particle (or wave that may also be a particle, whatever) into that space - it is definitely not a perfect vacuum anymore!
And "pressure" is a statistical concept, so I'm not sure how valid it is to talk about a single particle exerting a "pressure" - wouldn't we instead term that an "impact force"? Franamax (talk) 22:25, 15 April 2009 (UTC)[reply]
To mend the gaping flaw in my reasoning, for "particle" substitute "particles possessing mass". EM waves pervade the universe and are indeterminate in location. Erk! Franamax (talk) 22:41, 15 April 2009 (UTC)[reply]
I spoke of a single electron, but I had in mind the statistical effect of many electrons. – GlowWorm.
While it's true that it is not possible to create a perfect vacuum, that does not make a merely philosophical construct. It is a physical idealisation. There is nothing wrong with talking about the properties of physical idealisations. They are very usefull for thought experiments, for instance.
The OP asks about the vacuum resistivity. Paradoxically, the answer to this question can be both zero and 'infinity' depending on how we chose to understand the question. If you shoot electrons creating a streem, these electrons would continue to move unoposed even in the absence of an external field, zero resistivity! I seriously suspect that this was the setup that the op had in mind. But that's not what people usually understand by measument of resistivity. The normal understanding is to apply an external field and see what happens. If you do that, you get no current because there are no carriers, infinite resistivity! —Preceding unsigned comment added by Dauto (talk • contribs) 03:45, 16 April 2009 (UTC)[reply]
It's zero. The more fundamental understanding of resistance is not the ratio of potential difference to current (for one thing, that works only for DC), but rather a measure of the energy dissipated when a given current flows through the medium (power dissipation is I2R, where I is the current). Since vacuum dissipates no energy, it has no resistance. --Trovatore (talk) 04:01, 16 April 2009 (UTC)[reply]
I disagree with the "zero resistance" claim. A superconductor will carry a certain amount of current with no voltage across it and no energy dissipate, so zero resistance applies. A perfect vacuum (no electrons or ions) can have substantial voltage across it with no current. Don't tell me a vacuum tube diode with the filament cold has zero resistance when I apply 12 volts across the cathode to anode gap and no current flows. Edison (talk) 19:05, 16 April 2009 (UTC)[reply]
I stand by what I said. The vacuum has no resistance at all. That does not, of course, make it a superconductor, or even a conductor; it means only that no energy is lost to scattering when charge carriers flow through it. --Trovatore (talk) 18:36, 17 April 2009 (UTC)[reply]
Regarding your difficulty in making the current actually flow from the cold filament, well, the fact that it depends on the temperature of the filament should give you a clue that this is not a property of the vacuum. If the issue were related to the vacuum, then heating the filament wouldn't change it.
I don't think the resistivity will be infinite even in the absence of carriers, because beyond a certain field strength the vacuum itself will become a carrier: electron-positron pairs will be created out of the void and split apart, going in opposite directions and producing a net current. This phenomenon is a consequence of quantum electrodynamics. Looie496 (talk) 05:11, 16 April 2009 (UTC)[reply]
Looie, that which you talk about is called the dielectric strength. It's not the samething as resistivity. You are right. Vacuum's dielectric stregth is not infinite. A simple back of envelop calculation (Or better yet, a dimentional analysis) gives the vacuum dielectric strength which is 9 to 10 orders of magnitude higher than the dielectric strength of any material. Dauto (talk) 17:57, 16 April 2009 (UTC)[reply]
There is a voltage drop across a resistor if another resistor is in series with it to form a voltage divider. (Without the second resistor, the full voltage of the power supply will appear across the resistor, and this would not be called a voltage drop.) A voltmeter will show a voltage drop between the filament and plate of a radio vacuum tube. This indicates that the vacuum in a vacuum tube is a resistor, even though it dissipates no power.(There is a resistor in the plate circuit to form a voltage divider.) – GlowWorm. —Preceding unsigned comment added by 98.17.41.25 (talk) 10:35, 16 April 2009 (UTC)[reply]
If you have two electrodes in a perfect vacuum with some constant potential difference then there will be a nonzero tunnel current. Count Iblis (talk) 13:25, 16 April 2009 (UTC)[reply]
formation of pure alkyl nitrile
Hi, I need to find a way to synthesise the following product: it's a n-hexane with a CN group attached to the 2-carbon. My "base" reactant has to start with a hydrocarbon, though presumably I can employ a whole host of special reagents after that. I can't find a suitable process? I must obtain a single pure product, no isonitrile stuff (unless you can tell me an easy way to purify it without jumping through hoops). I don't think Kolbe nitrile synthesis will give me that, and all the other processes I've looked up either seem to apply to aromatic compounds or they leave oxygen groups on the carbon chain. John Riemann Soong (talk) 11:28, 15 April 2009 (UTC)[reply]
Working forwards, what reactions do you know that convert an alkane (is it correct to assume "hydrocarbon" means alkane?) into something-else (i.e., what kinds of functional groups can you make in Step 1)? Working backwards, what are ways you can make a nitrile (what would be a precursor) and what are ways you can add a "prebuilt" nitrile or nitrile-containing structure onto some other structure (i.e., what could be the precursor for the final step)? Do you have to start with a 6-carbon hydrocarbon? DMacks (talk) 04:23, 16 April 2009 (UTC)[reply]
Invasive species
(Not homework.) Dog (something)bine. Southern Ontario. Grows in underbrush of dead pine forest. A short human-height plant, wood-like or stalk-like (don't remember which). Nearby red pine dying from fungual infection. Any ideas? Thanks. ~AH1(TCU)11:33, 15 April 2009 (UTC)[reply]
Do scientists have to pay to have their work published in a peer-reviewed journal?
I was wondering if scientists have to pay to have a paper published in a peer-reviewed journal? The Open Chemical Physics Journal says that they "aim at providing the most complete and reliable source of information on current developments in the field" but charge anywhere from $450 to $900 to have a paper published. This sounds like a vanity press. The fees are listed towards the bottom of the page. [2]A Quest For Knowledge (talk) 12:18, 15 April 2009 (UTC)[reply]
Yes, and that's partially why I'm asking. In particular, Bentham's Open Chemical Physics Journal published an article [3] authored by 9/11 conspiracy theorists. The most prominent is Steven E. Jones who was relieved of his teaching duties and placed on paid leave at Brigham Young University after he published an article on BYU's web site promoting 9/11 conspiracy theories.[4] They claim that the destruction of the World Trade Center was not the result of terrorist attacks. Instead, they claim that the US government had planted explosives inside the WTC prior to 9/11 and that the towers were brought down by controlled demolition. The article Bentham published claims that active thermitic material was discovered in the dust from the 9/11 World Trade Center remains which 9/11 conspiracy theorists argue are the result of a controlled demolition. If Bentham is a vanity press, that would explain how this paper was published. If, however, this is a legitimate peer-reviewed scientific journal, how did it get past a peer review? A Quest For Knowledge (talk) 13:07, 15 April 2009 (UTC)[reply]
While I know nothing in this case it is rather an open secret that peer review depends very heavily on the opinions of the editor, who is in charge of assigning reviewers, shepherding the article through to publication, etc. Peer review is not without its problems. (And not all important articles are peer reviewed.) --98.217.14.211 (talk) 01:15, 16 April 2009 (UTC)[reply]
I don't know whether this journal is vanity or not, but with open access journals often the costs are put upon the authors of the articles (more usually their institutions), and the scholars in question usually have money for this written into their grant proposals for this purpose if they are planning to go in this direction. For example, the Public Library of Science, which is very reputable, works this way. --98.217.14.211 (talk) 12:31, 15 April 2009 (UTC)[reply]
Journals – open access and otherwise – often defray some of their costs through page and color charges (extra fees for color figures). Page charges are higher for open access journals which enjoy little to no other income from subscription fees. As the anon above notes, the very reputable PLoS family of journals charges substantial publication fees: [5].
That said, it's worth looking closely at any journal which charges page or publication fees — some of them are vanity presses. Reputable journals will usually have an indication that they waive fees for authors who can't afford them; the PLoS statement is in the second paragraph of that link, above the prices:
We offer a complete or partial fee waiver for authors who do not have funds to cover publication fees. Editors and reviewers have no access to author payment information, and hence inability to pay will not influence the decision to publish a paper.
The journal(s) in question (Bentham) began in 2007. To quickly fill up about 300 "open" journals, Bentham spammed universities looking for papers. Anything that was thrown at them that appeared remotely interesting was "peer reviewed" and published online as quickly as possible. The main goal was to get a lot of hits. The second goal was to fill the journals with articles as quickly as possible. The third (perhaps fourth, fifth, or sixth) goal was to ensure the articles had a respectable peer review. -- kainaw™13:29, 15 April 2009 (UTC)[reply]
I think there are plenty of legitimate journals with page charges. Not the most high profile journals, but nonetheless respectable places for researchers in specialty fields to publish their work. I believe my colleagues who work in leather chemistry pay page charges. Some fields are very small, and their journals just don't sell a ton of subscriptions. I don't think this necessarily says anything about the quality of the work or the journal. ike9898 (talk) 13:39, 15 April 2009 (UTC)[reply]
There seems to be an inherent conflict of interest there. That is, if the journal is short on funds and can get them by publishing articles, they may be tempted to publish total crap just so they can meet their payroll. StuRat (talk) 15:01, 15 April 2009 (UTC)[reply]
If its a peer reviewed journal, then that shouldn't happen as total crap should not pass review. Besides, journals don't chase money that way, because its ultimately self defeating. What they really chase is impact factor. The higher your IF, the more people want to publish in your journal and the more people that want to read it, which means more money for you. If you publish crap, your IF will decrease which means less people will want to publish in your journal which means less money for you. There is possible COI in a peer reviewed journal, though, as publishers can manipulate IF's in devious ways. Rockpocket19:15, 15 April 2009 (UTC)[reply]
What you say makes sense for the long-term future of the journal, but short-term concerns like avoiding immediate liquidation tend to outweigh lofty long-term goals. StuRat (talk) 16:14, 19 April 2009 (UTC)[reply]
I found a second Bentham article promoting 9/11 conspiracy theories in a discussion from the archives of the Reliable Sources Noticeboard regarding another Bentham journal, The Bentham Open Civil Engineering Journal. [6] Unfortunately, the discussion of the journal's reliability doesn't go very far. A Quest For Knowledge (talk) 15:52, 15 April 2009 (UTC)[reply]
The existence of page charges does not provide any evidence of a journal being a less than reliable source. Googling "journal page charge" showed "article processing charges" of $1400-$1550 (US) for BiomedCentral, which in turn tabulates the page charges for a number of journals, including Cambridge University Press ($2700), Blackwell ($3000), American Physiological Society ($3000), Elsevier ($3000), Wiley ($3000), and Springer ($3000). Some journals have lower page fees. If you do not like Jones and do not believe his analysis of the building collapses on 9/11 (notably the only modern highrises which ever collapsed from a fire or impact), do not try to discredit him or his analysis by claiming that page charges or "processing charges" make a journal "vanity press." Edison (talk) 15:53, 15 April 2009 (UTC)[reply]
It should be noted that it costs more to publish if the figures are in colour. Agree with Edison that a journal should not be considered a vanity press due to page charges. David D.(Talk)15:59, 15 April 2009 (UTC)[reply]
Note that Nature (journal) (which is about as respectable as you can get) charges fees for colour pages (though not non-colour pages), but its rather difficult to publish a paper in Nature without colour. Scientific publishing has a strange business model. Most journals have a very limited circulation, so to cover costs (and make a profit) the publishes charge at both sides - they charge the scientists to publish and they charge the scientists to read. Amazingly, the scientists also provide most of the content (the manuscripts), yet the publishers often demand the scientists hand over the copyright of that work also!
For such smart people, the scientists appear complicit in getting screwed over every which way. Why? Well, its because the scientific publishing system is so important to progressing in a scientific career, that scientists are willing to do whatever it takes to get their papers published. One could take a principled stand and refuse to comply, but ultimately your paper will not get published in a top journal and your career will not progress. There are efforts to change the system, such as the PLoS and BioMed Central journals, where you typically pay for publishing only. These are similar to Wikipedia is spirit, freely disseminating knowledge is the goal. What distinguishes them from vanity publishing is, as other point out above, they all have clauses saying you only have to pay if you can. If you genuinely don't have the grants funds to cover the costs of publishing, then you don't have to pay. Rockpocket16:27, 15 April 2009 (UTC)[reply]
If I were trying to get a job in academia or to get tenure or to get a research grant renewed, $3000 for a publication would seem like chump change, besides the fact that the grant is likely to pay it. How many hours and dollars went into the research and writing, do you suppose? Anything to pad the resumé. Edison (talk) 18:52, 15 April 2009 (UTC)[reply]
Many would argue is that publication is just one of the many costs of doing research. Just like presenting your work at conferences - some people get invited and have their way paid, but others are on their own to find funds they can use for this. ike9898 (talk) 19:50, 15 April 2009 (UTC)[reply]
Just to clarify, I've never heard an employed scientist spending their own personal money to publish. Sometimes you can use grant money or department funds or something else. It's just that you'd rather save that money if you could for supplies, equipment and salaries. ike9898 (talk) 20:03, 15 April 2009 (UTC)[reply]
When I was an unemployed scientist, I would certainly have paid page charges to get a publication in a good journal. Maybe you cash in some U.S. Savings bonds, or hit up the parents for some cash. It could lead to becoming an employed scientist. The alternative might be writing off many years of college and graduate school and accepting a less desirable career path. Do not underestimate the value of a publication if the resumè is a bit anemic. Edison (talk) 18:59, 16 April 2009 (UTC)[reply]
Hi all I after having read the article on rate constant I am still a little confused over what k is. Is it the rate of the reaction when the reactants are at a concentration of 1 mol dm-3 I've also heard that k isn't even a constant anyway as it if affected by the likes of temperature. How bizarre to call it a constant then! Another thing which particularly confuses me is writing the rate equation for a zero order reaction. I believe it would be written: Rate = k. Which means rate = rate constant. What does this mean? I hope what I've written makes sense - I find chemistry very confusing. Thanks in advance to all who help. —Preceding unsigned comment added by 92.10.162.104 (talk) 12:29, 15 April 2009 (UTC)[reply]
The arrhenius equation states what k is. Yes k is not really a constant, rather a coefficient. If the order of reaction is 0 with respect to all reactants, then yes, rate = k (mol dm-3 s-1).
Clearly, if the total order of reaction (m + n) is 0, then changing the concentration of reactants will have no effect on how fast/slow the reaction is (because any number to the power of 0 equals 1). --82.21.28.65 (talk) 12:58, 15 April 2009 (UTC)[reply]
They may call it a "constant" since, in this particular equation, it is constant with respect to reaction progress. When you use the rate equations, you pick a value of k and stick with it. But yes, it's not really constant - I think maybe "rate coefficient" would probably have been a better name, but I think we're stuck now :) --Bennybp (talk) 13:34, 15 April 2009 (UTC)[reply]
Armour made from tantalum-niobium alloy - how effective?
Any metallurgists here? In Terminator: The Sarah Connor Chronicles, it has been stated that the armoured parts of the more recent Terminator T-8xx models are constructed from an alloy of tantalum and niobium, as it was discovered that this was more effective on the battlefield of the future than the previously-used titanium, due to its greater heat-resistance (in response to the development of laser cannons and particle beam weapons, I suppose, as the war has been going on for decades in the story), comparible resistance to bullets and lighter weight (I think).
So, in the world of Real Life, how would armour plating made from an alloy of tantalum and niobium really measure up in a war situation? Not nescessarily when attached to a robotic soldier, just in general, I mean. Thanks. --81.77.155.81 (talk) 15:13, 15 April 2009 (UTC)[reply]
My first thought is that the armour would be vastly more expensive than titanium, since there appears to be three orders of magnitude difference in crustal abundance.
Looking at the various properties: Ta/Nb is much more dense than Ti, so a given thickness would be much heavier; the specific heat capacities are comparable, but Ta/Nb has greater conductance, so it would be better at absorbing laser input; also Ti becomes brittle on heating in presence of oxygen, if you can get a good laser shot in, that bit will turn blue and fracture on mechanical impact; Ta/Nb would probably be better with particle beams, due to its higher density; the hardnesses are comparable but I have a suspicion that Ti might be more susceptible to brittle fracture than Ta/Nb, not sure there - recall that the SG-1 team beat the replicators by using bullets, which shattered them.
If you're giving me unlimited funds, I'd probably pick the Ta/Nb solution. In fact, I'm not sure why you made us use Ti in the first place. In the end, it depends on what you want to use the armour against, and how fast you want the thing you're protecting to move. The best way to withstand an enemy shot is to move out of the way, heavy armour restricts that ability. If you're going to stand and take fire, there's a lot to be said for the sandwich technique using depleted uranium.
The high density of tantalum and uranium as well makes it a problematic alloy to use in large scale, for a humanoid cyborg, but if you do not care if one terminator is only capable to run ultra hard surface and simply would get stuck in a normal madow it is good. Normally a superalloy similar to the stuff used in turbines, a alloy based on nikel or cobalt with aditions of other refractory metals like tantalum and rhenium gives the best performance when heated. But a woven fabric of carbon nanotubes as interior of a composit material would be my favorite. For a particle beam weapon the exact sort of particle is important, protons, neutrons, electrons, alpha particles are positrons would all need a material capable to with stand the negative effects of nuclear conversions of the material. For Protons the answer can be found in the normal satelite business, because solar wind is made mostly from protons.--Stone (talk) 21:01, 15 April 2009 (UTC)[reply]
The OP is asking about the present-day performance of the armour, so perhaps I've led us astray discussing particle-beam weapons. Laser and microwave weapons are reality or close to it, so that's where we should focus, along with projectile and heat munitions (and vacuum bombs, which render the issue moot if you don't have a secondary air source). I think we agree on the tradeoff between weight and mobility.
As to a nanotube inner lining, would this be part of the breathable atmosphere inside the armour? There is growing evidence that carbon nanotubes are as toxic as asbestos, so I think maybe I'd rather have a chrysotile blanket instead (not amphibolic!) At least then I could quantify the risk. Franamax (talk) 21:50, 15 April 2009 (UTC)[reply]
Hair & Facial Hair Question
Hello, if this has been asked before, I apologize, I couldn't find where it had been asked. Anyway on to the question:
How is it possible for some body hair to be a completely different color than all the other hair on someone? For example I am blonde most of my arm and leg hair is blonde (dirty blonde to be exact). But my facial hair is RED. I'm not really upset about it and I'm not looking for answers on how to dye or change the color of my hair, I'm just curious about the origins and reasons behind this.
Us blondes have hair of many colours, on our heads and elsewhere on our bodies. You will probably have red hair on your head as well as blue (honestly), black and red. It's just that the blonde is more prominent on your head, whereas the red is probably more prominent on your face. Nothing to worry about - it's a quirk of nature. For the record, the only blonde hair on my body is on my head!--TammyMoet (talk) 15:48, 15 April 2009 (UTC)[reply]
Yes I've noticed the array of colors I seem to have! I wasn't too sure if it was kosher to talk about but my pubic hair isn't blonde..that's for sure. It's good to hear this isn't uncommon, but I'm still wondering why this is? Aside from artificial coloring you don't seem many dark haired people with such a wide variety of colors. 12.204.178.35 (talk) 16:19, 15 April 2009 (UTC)[reply]
I'm afraid TammyMoet's response isn't quite true. The gene responsible for red hair in most people is MC1R. When this was first being studied, the scientists found a strange correlation. They found that if you have two null alleles (two copies of a gene variant that doesn't function fully) then you have an increased likelihood of having red hair. However they also found that if you have just one null allele (what we call heterozygous for that gene), then you may have a different hair colour on your head, but that you have an an increased likelihood of having red body hair. I was one of the scientists working on that project, and having a reddish beard myself, I decided to sequence my own MC1R gene to see if I was a heterozygote myself. Turns out I am.
Quite why being a heterozygote results in this effect isn't quite clear. However it may be something to do with different effects of haploinsufficiency on different hair types. More simply, that the hairs on your head manage to make enough eumelanin with just a single copy of MC1R, but the hairs on the rest of your body need more MC1R. If you want to read the scientific data behind this, let me know and I'll find it for you. Rockpocket17:25, 15 April 2009 (UTC)[reply]
I don't see how what you said contradicts what I said: indeed, it seems to be the science behind the phenomenon. I was sure there was a genetic component to it, however I was also sure there'd be a geneticist along in a while - and I was right! Thanks.--TammyMoet (talk) 20:45, 15 April 2009 (UTC)[reply]
The thing I took issue with was the suggestion people have blue hair. There are no blue pigments naturally occurring in mammals, so (healthy) humans don't make blue hairs. Rockpocket21:35, 15 April 2009 (UTC)[reply]
However, healthy humans might look at a hair and perceive it as being blue. I run into that terminology problem all the time. Franamax (talk) 21:58, 15 April 2009 (UTC)[reply]
There are plenty of dog breeds called "Blue somethingorother" - they aren't literally blue either - but the problem (again) is that blue is just a word - and it means whatever people want it to mean in some specific context. SteveBaker (talk) 22:25, 15 April 2009 (UTC)[reply]
Just to be clear, the preceding reflects two very different things. Blue jays have no blue pigment, but are coloured blue due to a process referred to as structural colour. Grue refers to the different ways languages classify colours. See here for the actual article on the topic. Matt Deres (talk) 13:33, 16 April 2009 (UTC)[reply]
Calculating Kc equilibrium constant
Hi I am really stuck with the following problem. The reaction needed for the question is: Cu2+ + H2A (a fictional substance)(equilibrium arrows - don't know how to insert them) CuA + 2H+. The question states that at the beginning there are 0.005 moles of Copper ions and 0.005 moles of H2A. At equilibrium there is 0.0049 moles of CuA. Therefore:
Kc = [CuA] [H+]2/ [Cu2+] [H2A]
The question asks me to calculate the number of moles of all substances at equilibrium. So I already know that equilbrium the concentration of CuA is 0.0049 moles. Therefore at equilbrium there is (0.005-0.0049) 0.0001 moles of Copper ions left. Now comes the part I'm not too sure on. Looking at the answer the concentration of H+ = 0.00492 = 2.401x10-5 mol. My question is how can it be known for sure that the concentration of H+ ions is the square of the concentration of CuA? Is it not possible that the concentration of H+ ions would not be the square of 0.0049 moles at equilibrium? How do we know this is the case? Sorry for asking silly questions and thanks in advance to anyone who can help. —Preceding unsigned comment added by 92.8.198.137 (talk) 18:51, 15 April 2009 (UTC)[reply]
I don't think squaring the concentration of CuA will give you the concentration of protons. Check the stoichiometry of the reaction (and note the link I provided), specifically the ratio of CuA to H+ in the equation, and ask yourself: for every mole of CuA, how many moles of H+ will be generated? Looks like AFTER you have answered this question and determined the concentration of protons, you may want to square it in calculating Kc. I hope this helps. --Scray (talk) 22:10, 15 April 2009 (UTC)[reply]
As Scray said. Rethink the meaning of the formula: [H+]2 does not mean you have to square [H+] to figure out the concentration, but the other way round: [H+] is the concentration and it needs to be squared to obtain the equilibrium constant. Boris (talk) 03:34, 16 April 2009 (UTC)[reply]
I can't answer your question, but I just wanted to help you understand the relationship between volume and intensity. Volume is a human phenomemon and the correct word is called loudness and is measured in the decibel system while you are asking a question that will probably get answered using the terminology of sound intensity which is power per square meter and is related to the strength of a sound wave. A very intense sound wave is, for example, has a very high value of watts per square meter and vice versa. Even if you can't hear the sound wave you speak of in your question, it will still have intensity, although it will not have loudness because if you can't hear a sound, that means it has zero decibels, but not necessarily zero watts per square meter and if I can further request to whoever answers the OP's question: is my understanding of the relationship between intensity and loudness correct? is loudness, or intensity--specifically, what causes hearing damage? The article goes into generalities such as "caused by a wide range of biological and environmental factors... etc...". I suggest read the Hearing impairment article because it gives some good info, but doesn't specifically answer what you are asking--which is a good question, because I imagine those who read the article would like to know as well. 71.55.186.77 (talk) 20:28, 15 April 2009 (UTC)[reply]
It is incorrectly said above that if you can't hear a sound, that means it has zero decibels. Without a specified reference sound pressure, a value expressed in decibels cannot represent a sound pressure level. With any specified reference sound pressure, a sound pressure that is equal to the reference has zero decibels relation to it. The "loudness of inaudible sound", regardless of reference sound pressure, is actually minus infinity decibels. Cuddlyable3 (talk) 21:29, 15 April 2009 (UTC)[reply]
There is evidence from studies of the effects of ultrasound scans on unborn children that this does not harm their hearing - even though the intensities (watt/sq.meter) are quite high and the amniotic fluid (which is mostly water) transmits the sound very efficiently. I assume that this is because the eardrum has inertia and elasticity and at sufficiently high frequencies is simply not moving measurably in the tiny amount of time before the sound wave reverses direction. This (admittedly sketchy) information suggests that the transmission of sound energy from outside of your ears to inside (where high energies could certainly wreck your hearing) must be a function of frequency. That doesn't mean that a super-loud burst of 20kHz wouldn't harm you though...it just says that we can't simply measure the raw intensity and nothing else. SteveBaker (talk) 20:47, 15 April 2009 (UTC)[reply]
Agree. The possibility of damage to hearing depends on the efficiency of the amplification mechanism in the human ear. Quite simply, if the eardrum and the little bones can't transmit the vibrations to the cochlea because they don't respond fast enough, those little hairs (cilia) won't get damaged. Where the exact cutoff line is, I dunno. It's very likely that the human ear has evolved to efficiently filter only the frequencies that matter most. I'd speculate that ~20KHz might still be in the danger zone, but I don't know of any detailed mechanical studies on the amplification and coupling factors in that region. I do know that I cover my ears as soon as I hear feedback rising in frequency, since it quickly becomes very painful. Actually, as a former soundman, I would leap for the volume knobs first. Intense high-frequency sound is double-plus-ungood. Franamax (talk) 21:10, 15 April 2009 (UTC)[reply]
Yes on 20KHz not being "imperceptibely high frequencies"--many people can hear in that range quite well. Of course not all loudspeakers are capable of producing sound in that range. Pfly (talk) 05:25, 17 April 2009 (UTC)[reply]
I don't know what has the analgesic effect, but the antiseptic effect is due to the way the high concentration of sugars absorbs water, killing the bacteria. --Tango (talk) 20:43, 15 April 2009 (UTC)[reply]
D'Oh! I overlooked the word ingredients [don't misguide old men by a "poorly formulated title" of your request ;-)]. Coming from the media you would call them something like non-defined (or poorly defined) protein complements. If you search for them in the Sigma-Aldrich Catalogue (they should have an interest to label this "Class" correctly, right?), they call them "Microbiology, Basic Ingredients, Protein Sources", or "Peptones" (which has another more specific definition). In my view (as suggested above): No generally accepted "Class" designation. Better? ;-) --Grey Geezer 07:30, 16 April 2009 (UTC) —Preceding unsigned comment added by Grey Geezer (talk • contribs)
Yeah. The reason I ask is because not too long ago a microbiologist got all bothered when I referred to this sort of thing as a "nitrogen source". I never got the chance to ask her exactly why she thought that was wrong or what term she would prefer. It is possibly because these ingredients do more than just supply nitrogen, they also contain growth factors and minerals, and provide some energy as well. However, I haven't been able find to good single word to describe these things. Maybe I should come up with one (remember Sniglets?). ike9898 (talk) 13:28, 16 April 2009 (UTC)[reply]
Very small creatures in my fish tank?
I have something very small living (uninvited) in my freshwater tropical aquarium. It is circular/oval, white/grey in colour and about 1mm-2mm across in size. It moves very slowly across the surface of the glass or the leaves of the plants, as if it were a tiny mollusc or similar. There are several of them in the tank and the numbers seem to be steady. They don't appear to be damaging the plants although it's possible they are attacking the fish. Every so often a fish appears to have a twitchy fit for about a day and then dies, but this is rare. Is it likely they are harmless or could they be some kind of parasite?Popcorn II (talk) 20:51, 15 April 2009 (UTC)[reply]
Hmmm. Possibly but I don't think so. These things are about 2mm at their largest. Many of them are less than 1mm. They also seem to move as a whole, solid form without stretching, as a Limpet might.Popcorn II (talk) 22:55, 15 April 2009 (UTC)[reply]
Ostracods are enclosed in a little shell, and that is all you usually see: they look like little beans, usually whitish. The crustacean-looking parts (jointed legs and mouth parts) are enclosed inside.--Eriastrum (talk) 19:38, 16 April 2009 (UTC)[reply]
Feelings (affection, humor, fear, even belief etc.) can be associated with "biological functionality". However, what is the "functionality" of grief (e.g. after the loss of a child, partner, parent)? A functional reaction should be "feeling strong" or even "euphoria" to start up again (to find a new partner, have another child etc.). Griefing for a loss (which will not be returned) does not really make sense, right? If the loss is by accident, predator, or disease, the learning effect (how to avoid) is low: the griefing person is depressed, inactive, sometimes suicidal, has no appetite, which should increase the chance for an accident, falling prey to a predator or even getting weak and sick: So exactly the opposite = grief makes things worse. The question is not: How to overcome grief but "what is its function in the big game"? Any suggestions? --Grey Geezer 22:08, 15 April 2009 (UTC) —Preceding unsigned comment added by Grey Geezer (talk • contribs)
It probably has something to do with the fact that we've evolved in tribal settings with small groups of a dozen or two people. We have evolved a sense of community. Just as we've evolved responses such as pain so that we know that when we've been bitten in the leg by a rampaging marmosette we should avoid letting that happen again - so taking a 'wound' to the community is something we need to try to avoid - so we have a different 'pain' response to that. If there were no emotional consequences to losing tribe members ("friends" and "relations") perhaps we'd be less cohesive as a group? SteveBaker (talk) 22:16, 15 April 2009 (UTC)[reply]
Its very beneficial to be part of a group, from a group of two people (couple relationship) to a family/tribe group, to a polical/national group. The emotions try to make you stay part of that group no matter what. These emotions fire even if the group doesn't physically exist anymore. Remember that the whole idea of being in a group is a mental idea - so its sort of an idea organism that isn't going to know if the group physically exists - for most of your life the group will exist and these emotions will keep you in-line with group which is beneficial biologically. Also the reason can be non-biological. There is no biological reason for a person to kill themselves for the group (that doesnt include their offspring), but the idea organism that permeates all minds in the group has alot to gain from someone doing this - its strenghtens the idea of group, and thus the group is likly to expand and beat other group ideas. So basically you need to understand that once we had evolved a mind with certain capabilities, we also had 'ideas' that could evolve, and these virtual ideas can very often beat out the biological requirements of their hosts (us). It is beneficial for the rest of the group (which is really just an idea of what the group is) if you are seen to endure great hardship from leaving or someone else leaving the group. Likewise people are typicalyl very moved if they see someone else who has lost their partner, often it motivates them to form even strong group connections with there still existing partner, this is how an idea (like marraige) can be a duplicate of groups, an idea that copies itself as small groups everywhere. So while its pointless for us to greive, it helps everyone else with the idea of marraiage, and it likely results in the idea of marraige being further conveyed.--Dacium (talk) 03:28, 16 April 2009 (UTC)[reply]
Like other aversive things, its main function is probably preventative: the knowledge (obtained by observation of other people) of how terrible grief is makes you work very hard to avoid it. Looie496 (talk) 05:21, 16 April 2009 (UTC)[reply]
Thanks for the comments! So grief, which is often used as a "personal problem" could be a "group signal" ("Look how valuable members of this group are to me!". And like "Look, I suffer to have all those piercings / brandings done to show that I am really part of your group!") and this signal would be "more worth" than the "temporary inefficiency" of the grieving group member. Makes kind of sense. However, grieving for pets then seems to be a "side effect". It seems that Inuit or other native groups are very pragmatic (more realistic?) with the dogs who guarantee part of their survival (also very important!), e.g. killing off "weaker" female pups or disposing of the corpses of long-time sleigh dogs, but skinning them and using their fur. --Grey Geezer 07:51, 16 April 2009 (UTC) —Preceding unsigned comment added by Grey Geezer (talk • contribs)
Yes, I think that's true. We've "adopted" pets - they are a part of the "tribe" - they are treated like children and grieved for in the same way. When animals are "working animals" they are more like tools than members of the family - we can be more hard-hearted about the whole thing. When your old car is beat up and rusty - you get a new one - you throw the old one out...unless you can salvage something from it. SteveBaker (talk) 12:56, 16 April 2009 (UTC)[reply]
Grieving is a retreat into a passive state when one finds that previously held assumption(s) about one's safety, posessions or relationships have been broken. Its survival value lies in 1) the retreat from action to allow time for, hopefully, the sufferer to evaluate their changed situation
and 2) the person's grief signals can bring aid from other(s). Babies use 2) as feeding strategy. Cuddlyable3 (talk) 23:20, 16 April 2009 (UTC)[reply]
Tendons (and all soft tissue in general) has elasticity. If you keep the tendon tight, it will stretch a bit when you apply weight to it and release some of that energy to propel you up to the next step. Of course, this only works if you go up the stairs on the balls of your feet, not the heels. -- kainaw™12:14, 16 April 2009 (UTC)[reply]
Well no. Im looking for the actual transfer function of the human ear canal (including the pinna), but excluding the ear drum, choclea etc —Preceding unsigned comment added by 79.67.15.85 (talk) 22:53, 15 April 2009 (UTC)[reply]
This book has what you want. You will also find a lot of papers on the subject if you search for "head related impulse response" or "HRIR" on google scholar. Here are some from the Acoustical Society of America [8][9][10]. There is also a lot published by the Audio Engineering Society and the IEEE, but sorry, I don't know if they are exactly what you want as I can mostly only read the abstracts. SpinningSpark13:48, 18 April 2009 (UTC)[reply]
April 16
Ionic detox
Recently, a co-worker met with an alternative health practioneer and participated in an "ionic foot bath." He described all of the ickey things coming out of his feet (including what appeared to be "live" worms. Ever the skeptical one, I did some investigating. Here on Wikipedia, the closest thing I could find was an article on detoxification foot pads. Using Google, I found a plethora of info (all promoting ionic foot baths!). So, here is my question: Since there is no scientific evidence to support the claims of ionic foot baths, what is really happening in the wather that makes it look like things (including worms) are being drawn from the body via the soles of the feet? —Preceding unsigned comment added by 209.161.212.185 (talk) 03:35, 16 April 2009 (UTC)[reply]
With a little googling you can find accounts of people who run the "ionic baths" without putting their feat in and the water still turns murky. It's a trick. It's been mentioned on The Amazing Randi's site several times. here's the most recent mention.
The live worms are an interesting touch, but once you acknowledge that it's a hoax, then it's easy to use slight of hand to add some of those tiny worms that you can buy to feed your Betta fish.
In general the idea that there are numerous common "toxins" build up in your body and must be cleansed is new age nonsense. If you've got a healthy liver and a pair of healthy kidneys, most of your toxins are being taken care of. (And, if not, all the foot baths in the world won't save you.) APL (talk) 04:49, 16 April 2009 (UTC)[reply]
Here are some YouTube videos. Here. On closer examination, this seems to be similar technology that is used in saltwater etching(Scroll down), except instead of removing metal from an electrode to create an artistic pattern, you're removing metal from an electrode so that the waste metal will darken the water and create a "toxic"-looking effect.
I'm also less convinced that there were actual live worms, now that I've read about it more. Perhaps tiny particles of metal held together by magnatism or electric charge would move around in a lifelike way? APL (talk) 05:06, 16 April 2009 (UTC)[reply]
There is absolutely no doubt that this (and those damned stupid foot-pads) are a total scam. The adverts for the foot pads actually make me laugh: "Just as the leaves of a tree draw toxins out of the air and expel them through their roots - so the Kinoke foot-pads...yadda, yadda, yadda"...EXCUSE ME?! Since when do trees absorb toxins through their leaves? Trees don't expel ANYTHING through their roots. Actually, they soak up stuff through their roots and expel it (well, water and CO2) through their leaves! And just because a tree does it (or not) why would that make a difference to humans...we don't even have leaves or roots! Gah! Evil bastard scammers with annoying TV ads. Argh! Anyway - once you accept that it's a scam - you have to ask how those things got into the water. I presume that if you were observant and skeptical during the "treatment" you could spot the moment when these things were dropped into the water by the practitioner - perhaps they were inside the 'healing crystals' or whatever else was placed into the water? Perhaps they were concealed inside the water recirculating mechanism (or whatever it uses). Perhaps the scammer 'palms' them (as a magician palms a playing card or a coin) and slips them into the water at some point. For sure there are no worms in your feet! Ask your co-worker to look at his/her feet and see if the holes from which the worms emerged are visible? No? No holes? Did the worms teleport through the soles of the feet? If the government isn't going to enforce trading standards and medical practitioner licensing laws against these people then it's up to everyone to be super-skeptical of any of these kinds of claims. Mostly they don't stand up to any inspection at all. I really wish I had the time to go to some of these places - take the treatment - spot the scam and sue the evildoers for fraud and practicing medicine without a license. We're gradually sliding backwards into the Dark Ages - and with all we know and understand - that's just ridiculous. SteveBaker (talk) 12:51, 16 April 2009 (UTC)[reply]
I don't think that you could "spot the moment when these things were dropped into the water by the practitioner". I'm pretty sure that the murkiness of the water is actually rust, or other material from the electrodes. They are consumed by the "ionic bath" process and need to be replaced on a regular basis. (This adds a nice 'don't sell the razor, sell the blades' aspect to the scam.)
They sell these machines (at about $1000!) for home use, and they produce the same effect when operated by the end user.
Of course, live worms aren't going to come out of an electrode. So if there really were live worms, they must have been added by slight of hand. A bit over the top, if you ask me. If I were working this con I would have just stuck with "toxins". APL (talk) 15:40, 16 April 2009 (UTC)[reply]
Hmmm - perhaps these "worms" are something like dried out Sea-Monkeys that could be inside whatever stuff they sell you to put into the water? Suppose they sold you a large tablet - like the kind you put into your dishwasher - with a capsule in the center that would dissolve and re-animate the dried worms. They'd probably scare the bejeezus out of the gullible 'mark'! But you could probably make sufficiently convincing "worms" out of all manner of inert materials that would expand and look suitably yukky. SteveBaker (talk) 20:36, 16 April 2009 (UTC)produce[reply]
Well, I suppose you could do that, but there'd be some setup involved. SeaMonkey "instant life" packets take about 24 hours to come to life. I can't find any evidence that it's ordinarily done. None of the videos of the devices (Either the credulous ones, or the debunking ones) show anything that looks like a live worm, though the orange muck seems to eventually coalesce into a few 'floaters' that might look alive if you didn't look too close. These machines involve dunking a pair of electrodes into salt water. The electrodes apparently corrode away, so that explains the gunk in the water. If there were live worms involved, then I'll bet they were put in there by the salesman to try to punch-up the effect of the machine and not part of machine's usual scam. If that makes sense. Like a vacuum-cleaner salesman who isn't content with ordinary dust, so he hides some graphite in his vacuum. APL (talk) 23:07, 16 April 2009 (UTC)[reply]
A good rule of thumb: If someone talks about "toxins" but fails to mention what specific substances they mean, you can be 99 and 44/100 % sure they're full of shit. Friday(talk)15:52, 16 April 2009 (UTC)[reply]
Sadly, while that's certainly what's going on...there was no mention of the worms. These scammers are on to a good deal then...they take one of these foot-spa's (Cost: $15.14 at Walmart) - hook up a couple of electrodes (6" nails ought to do it - that's another $1) and a suitable battery a switch and some wires (maybe another $10) - and sell the thing for $1000. They'd be making a spectacular profit at $100 - at $1000 it's like printing money! I could build 50 of these a day by myself with just the tools in my garage. Then they have the nerve to sell the "refill kits" for $44 plus postage...$55 to you! Pulling this kind of scam on the American public is like shooting fish in a barrel. There are times when I really wish I was not such an honest person! SteveBaker (talk) 00:41, 17 April 2009 (UTC)[reply]
Minor and off-topic correction for SteveBaker: trees do expel lots of stuff through their roots, primarily sugars destined for the commensal organisms helping them to thrive. It's a fascinating topic, especially when you think about how easy it should be to cheat a tree. It's even vaguely possible that the tree would acquire and pump toxins into the cheating fungi, though I've never seen a single word to that effect. Totally agree though on the other commentary on this scam, my only regret is that I haven't dreamed one of these puppies up! Franamax (talk) 00:51, 17 April 2009 (UTC)[reply]
Ideal physical attributes for boxing
In boxing, are smaller or larger fists better (assuming they're on the end of the same arm throwing the same punch)? And what would be the ideal build in terms of height, chest size, reach, muscle mass etc and the relationships between those characteristics? Thanks. 86.8.176.85 (talk) 04:49, 16 April 2009 (UTC)[reply]
The fist part of your question is easy, the second one is not. When someone (constant weight = same punch power) steps on your hand with high heels (small fists, high relative impact) or Gramps slippers (large fists, low relative impact)... what hurts more? --Grey Geezer 08:01, 16 April 2009 (UTC) —Preceding unsigned comment added by Grey Geezer (talk • contribs)
Force equals mass times acceleration - so heavier fists (not necessarily larger ones) with the same exact speed of swing will produce a larger amount of force when they hit something. However, Force equals mass times acceleration - so when you throw that punch, a heavier fist requires more force to get it going fast...or to put it another way...for the same amount of musculature, etc - a heavier fist will end up travelling more slowly than a lighter one - and the benefit of that heavier fist will end up cancelling out. Now, admittedly - as User:Grey Geezer says - the same amount of force over a smaller area produces more pressure - and therefore (presumably) more pain...hence the reason one stabs people with a sharp knife and not (say) a brick! As to muscles...the force a muscle can produce is proportional to the cross-sectional area - so if you have two people with the exact same body proportions - but one is twice as tall/wide as the other - then the larger person has EIGHT TIMES the amount of weight to move around - but only FOUR TIMES the amount of muscle force to move it with. This is why elephants move so slowly compared to (say) dogs...and insects so much faster still. So we're back to that same Force=Mass times Acceleration problem - if you are smaller, you can move (relatively) faster - but with lighter fists, you need to be faster in order to produce the same amount of force. But this is all very superficial reasoning - can a larger person train harder? Can they take more punishment? Can a smaller person simply dodge all of the big guy's punches so it's irrelevent how hard the big guy can hit? All in all, it's a very complicated question. SteveBaker (talk) 12:38, 16 April 2009 (UTC)[reply]
Yeah - but they're more of a disadvantage when it comes to the running-away-screaming...which is probably higher on my list! :-) SteveBaker (talk) 20:28, 16 April 2009 (UTC)[reply]
I noticed that in some cities, trolley-bus routes may be circular or non-symmetrical (e.g., A-B-C and C-D-A even if there are no one-way streets).
Let's say you own a bus company in a small town having a bus terminal (A) and three stations (B, C and D). Let's say your buses are all diesel powered. There can be two routes (there are no one-way streets):
A-B
| |
D-C
Route Clockwise: A-B-C-D-A
Route Counterclockwise: A-D-C-B-A
Now if your buses are all trolley-buses. Running these two routes requires that you install TWO SETS OF OVERHEAD WIRES for your entire bus routes.
So you may decide to run one route only and double the frequency (e.g., 10 mins a bus -5 mins a bus).
I guess that if the routes are light capacity (e.g., low population density), trolley-bus companies may have their routes covering more roads one way or circular to increase the number of stations or reduce the length of wires. For example:
A
|
B
|\
C D
|/
E
|
F
Route A->F: A-B-C-E-F
Route F->A: F-E-D-B-A
On second thought, this logic may not be applicable to railroads. You cannot afford to run a railroad system on a single set of tracks design. If a train is derailed, you can do nothing before the derailed cars are removed.
If you have a subway system, you may want to have at least two sets of tracks because it's much more expensive to dig two smaller tunnels than a big one. One bidirectional underground station is also less expensive than two one-way stations -- Toytoy (talk) 05:26, 16 April 2009 (UTC)[reply]
No matter whether you're talking about a trolleybus line, a streetcar line, a subway, or a railway, it's always cheaper to
build a single line than a double one. But it also severely constrains your operational flexibility and capacity if you want
to operate vehicles in both directions. Loop routes are indeed a possibility for avoiding this issue while building only a
single line, but the trouble is that they produce indirect routes, which aren't as good for attracting riders. Passengers
want to go more or less directly to their destination, not detour around the sides of a loop. So when transit systems do
use one-directional loops, they're typically only small ones, just a few blocks long or wide, usually at the end of a route.
Here's an example on a subway line (trains run anticlockwise around the single-track loop),
here's one on a streetcar route, and
here's one on a trolleybus line, which also includes some other
one-way sections due to one-way streets. These examples are in Paris, New Orleans, and Geneva respectively.
Single-track lines are much more common on railways, where distances are long and frequencies may be relatively low, than on
urban transit systems such as trolleybuses and streetcars. It is true that a derailment may block the line completely,
but derailments are rare (and for that matter, if one happens it may block both tracks of a double-track line).
Capacity limits and dealing with ordinary delays are much more of an issue.
It's not always correct that on a subway system "it's much more expensive to dig two smaller tunnels than a big one".
It depends on the construction methods. In London, for example, all of the deep-level tube
lines do use two separate tunnels for their tracks, and often even for
the larger tunnels at stations.
But generally in urban transit, if the frequency of operation on a route is low enough that a single direction is
acceptable, it will be operated by buses, which need no tracks or overhead wires. --Anonymous, 08:44 UTC, April 16, 2009.
In terms of digging tunnels - the amount of dirt you have to dig and the amount of tunnel-walls you have to erect to make two identical one-track tunnels is (obviously) twice the amount for a single one-track tunnel. But because tunnels have to be circular, making one two-track tunnel requires removing FOUR TIMES the amount of dirt and erecting twice the amount of wall compared to a single one-track tunnel. So if the cost of removing dirt is high - then making two small tunnels could easily be around half as expensive as one larger tunnel. Of course there are many other issues involved - such as the size and running costs of your tunnelling machines and the amount of surface disruption caused by your activities. SteveBaker (talk) 12:26, 16 April 2009 (UTC)[reply]
Factors that favour dual- over single- tunnels for two tracks include 1) the possibility to close a tunnel temporarily for maintenance while keeping a reduced single-track service, and 2) in case of an emergency such as a derailment, fire or chemical spill, the parallel tunnel can be used for access and escape if cross connections are provided. That arrangement has been tested by a number of fire incidents in the England-France Channel Tunnel. Cuddlyable3 (talk) 22:24, 16 April 2009 (UTC)[reply]
Tunnels certainly do not have to be circular. Some are oval, some have vertical sides and and an arched roof, some are rectangular or close to it, and so on. Again, it depends on the technology (not necessarily a tunnel-boring machine; less automated techniques were common in the past and are still used for short tunnels, and many transit systems use cut-and-cover tunnels) and on what you're tunneling through.
On a subway system, one advantage of double-track tunnels is that since the tracks are close together, they can lead directly into a station that has separate platforms outside the tracks for the two directions of travel, which gives a convenient separation of passenger flows compared to a center "island" platform. Another is that if crossovers between the two tracks (for turnbacks) do not require additional tunneling. Another is that a single train does not restrict the flow of air through the tunnel significantly, which improves ventilation and also reduces air resistance on the trains. On the other hand, when circular tunnels are in use, a double-track tunnel requires more height clearance around other things that may be in the ground, and then there are the issues mentioned by the last poster. --00:00 UTC, April 17, 2009.
Oval tunnels have one major problem though - you can't use a simple tunnel boring machine to make them because the cutter heads go around in a circle and the tunnel walls have to be lined as the TBM inches forwards. Agreed that oval tunnels are fine when you're doing cut-and-cover though...but then the amount of material you have to remove and replace is large compared to the volume of the tunnel anyway. As for the air flow argument - I'm pretty sure I read somewhere that the progress of trains though the London underground system is designed precisely so that it DOES move air through the tunnels - and that saves them from having to pump fresh air around the system. Those trains go so slowly that air resistance probably isn't a large fraction of their running costs.
Streamlining trains is less important than (say) cars and trucks anyway because they only pay the price once for the entire length of the train - compared to each car or truck on a freeway having to pay the energy price again and again. Anyway - we're getting quite a way from the OP's question here. SteveBaker (talk) 00:27, 17 April 2009 (UTC)[reply]
Again with the boring machines! (Pun intended.) Cut-and-cover tunnels are usually rectangular. Oval ones were probably excavated using a shield or by blasting.
It's not correct that trains pushing air through the tunnels provides useful ventilation. The builders of the early tube lines in London HOPED it would, but it turned out they were wrong. It is correct that streamlining is generally less important for trains than for cars and trucks moving at similar speed, and is generally less important at low speeds, but the situation is different when the train is blocking a large fraction of a tunnel's area. The newer tube tunnels in London (Victoria and Jubilee Lines) are generally at least 6 inches larger in diameter than the older ones, and air friction is the main reason.
--Anon, 07:25 UTC, April 17, 2009.
concurrent queues
at the bank I got a 369, there was someone at 365 but also 500's and 900's. However the bank teller next to me surprised me by offering to tell me when I would come up -- she told me I would be 8th in line. So if her computer shows all three queues as mixed into a single queue order, why keep them separated from each other? Thank you! 94.27.231.11 (talk) 10:17, 16 April 2009 (UTC)[reply]
I presume that there are multiple ticket-issuing machines with rolls of tickets pre-printed with consecutive numbers. So if you go to the first machine, you maybe get a ticket that starts with a '3' - at the second machine they start with a '5' and at the third machine, they start with a '9'. The ticket machines tell the computer system when each ticket was issued - irrespective of the number printed on it. Alternatively - sometimes these machines are set up so that the number tells what kind of service you want - so perhaps if you needed a foreign currency transaction, you'd get a ticket starting with a '9' - and then only the desks that have tellers who are qualified to do foreign currency work will accept those tickets - where every teller can take '3' tickets because those are for general check cashing, etc. But without knowing how your bank issues those tickets, it's hard to be certain. SteveBaker (talk) 12:19, 16 April 2009 (UTC)[reply]
I've seen systems that intentionally keep the ticket-numbers non-consecutive (one fast-food place was totally random, some skip around) for no reason other than to keep customers from knowing that they have a long wait ahead for a noticeably higher number than a single linear "now serving #..." queue. DMacks (talk) 14:55, 16 April 2009 (UTC)[reply]
It is probably done to prevent anyone knowing that they will be served next, allowing them to hover ready to pounce immediately the preceeding customer leaves. Cuddlyable3 (talk) 22:05, 16 April 2009 (UTC)[reply]
My question here is: why wasn't that helpful teller helping the first customer in the queue? Do they plan this stuff during coffee break? :) Franamax (talk) 00:30, 17 April 2009 (UTC)[reply]
Paradox in current through vacuum
please visit the following post in the refdesk : [[11]] .
Here, some of them are saying that vacuum has zero resistivity. So, if the resistance is zero, for any finite voltage V,
I = V/R
so for any value of V, the current would be infinite. It means that infinite number of electrons would be passing through the vacuum at a given time. Please note that the vacuum will cease to be a vacuum if electrons are there but nevertheless, the resistivity will not change as the electrons are not moving at random. But how will the universe contain an infinite number of electrons?
--harish (talk) 12:43, 16 April 2009 (UTC)[reply]
No paradox. If you'll noted, the discussion that gives rise to "resistance = 0" is actually saying energy dissipation is zero. No energy lost can be equated to no resistance. In practice, this means that any given voltage can be transmitted at any given current, but this is not a case where division by zero equals infinity. 64.238.255.250 (talk) 13:12, 16 April 2009 (UTC)[reply]
What does any number divided by 0 equal? If there is no current, there is no defined resistance. Ohms law applies to a conductor with a voltage across it and current flowing through it, nothing more.--58.111.134.75 (talk) 13:16, 16 April 2009 (UTC)[reply]
Division by zero means that your mathematical model broke down, and you need a better one if you want to know what happens then. — DanielLC00:58, 17 April 2009 (UTC)[reply]
No, R would be infinite, not zero. Another thing is that there will still be a tunnel current, and Ohm's law is not in generally valid in the first place. Count Iblis (talk) 14:34, 16 April 2009 (UTC)[reply]
One can "shoot" a charge carrier, such as an electron, through a vacuum. That means a voltage (the shooting force) has caused a certain amount of charge to move from A to B. The charge flow per time is the effective current. The vacuum has neither helped nor hindered the transfer. Because of its non-hindrance to current one may declare the vacuum to have zero resistance but because it does not itself provide any current carrying path for an applied voltage one may declare the vacuum to have infinite resistance. Since the vacuum has not participated, no property of the vacuum, such as resistance, can be deduced from the experiment. Cuddlyable3 (talk) 19:27, 16 April 2009 (UTC)[reply]
When an electron strikes the anode in a radio vacuum tube (or similar device), it loses its kinetic energy. The energy will be converted to heat. So there is a power loss.(Large transmitting tubes have heat dissipating fins or may even be water cooled.) Another effect may be secondary emission in which one or more electrons are knocked loose from the anode by the arriving electron. The energy imnparted to the secondary emission electrons will be another power loss. - GlowWoprm. —Preceding unsigned comment added by 98.16.66.104 (talk) 20:57, 16 April 2009 (UTC)[reply]
At the risk of sounding alittle dense, I have to ask how is the vacuum permittivity related to the vacuum resistivity? I don't see the connection. Dauto (talk) 18:20, 17 April 2009 (UTC)[reply]
CO2 production vs Absorbtion
I know there are lists of how much CO2 countries emit, but is there a list of how much CO2 is absorbed (mainly by vegetation) per country? The 'land use' seems to be something completely different. I am trying to make a list of which countries are net CO2 emitters and which are net CO2 absorbers and by how much per capita.--58.111.134.75 (talk) 13:00, 16 April 2009 (UTC)[reply]
This is poorly understood at present. Lots of CO2 absorption is into soil, and it's quite difficult to calculate the rate at which that happens across all the different kinds of soil in a region. Looie496 (talk) 17:21, 16 April 2009 (UTC)[reply]
And it will stay poorly understood for a while longer. The OCO satellite mission couldn't get the door open and returned to Earth in a non-optimal fashion. Franamax (talk) 00:20, 17 April 2009 (UTC)[reply]
Isn't it Algae, i.e. the oceans, which do the really big job? All is one. --Grey Geezer 07:34, 17 April 2009 (UTC) —Preceding unsigned comment added by Grey Geezer (talk • contribs)
A stable and healthy eco-system is usually CO2 neutral, i.e. there is no significant net absorption. Most of the CO2 that is consumed by vegetation is released within a short term (weeks to years) when the vegetation rots or is consumed. Only a few geologic formations permanently sequester CO2 by (eventually) turning it into fossil fuels or carbonate rocks over long periods of time. You can draw some CO2 out of the air by planting previously fallow ground, or, to a certain extend, by harvesting plant matter for long-lived products like books and furniture. You might want to take a look at Carbon cycle. --Stephan Schulz (talk) 12:51, 17 April 2009 (UTC)[reply]
GG, you may be thinking of the oceans physically absorbing the CO2, partial pressure dictates that if you put more of a substance into the gas phase of a system, you will also get more in the liquid phase. Thus the problem of ocean acidification.
Certainly though, while SS is correct in stable natural cycles being generally CO2 neutral, there is a net sequestration of carbon to the ocean floor, mostly I believe from diatoms. I'm not sure exactly how large that flux is though. Franamax (talk) 20:15, 17 April 2009 (UTC)[reply]
CDC doesn't keep records for location of contraction of Lyme disease, only location of diagnosis. Is there any research between the correlation of the two? With the incubation period averaging 10 days, do many people contract the illness (for example, while hiking on vacation) and then travel a significant distance before being diagnosed? Or do most people contract it at or around home? 71.62.175.234 (talk) 18:12, 16 April 2009 (UTC)[reply]
Well, firstly, Lymes disease isn't tough to diagnose - so I would imagine most people are going to their doctors or some very local hospital emergency room to get their diagnoses - so there probably isn't much error in using the location of diagnosis. Secondly - how would the victim KNOW where (or when) they contracted it? The incubation period is anything from days to YEARS! How the heck would the average patient be able to know which of all the places they visited in the last six months was the one where they were infected?
Video games? Movies? eBooks? If you weren't worried about copyright, you could pack a large variety of all of those into a rather small, lightweight package. (A solid state hard-drive, for example.) You could also bring an iPod, of course. APL (talk) 18:31, 16 April 2009 (UTC)[reply]
If you're headed to Mars, I'm not sure you need to worry much about the copyright-holders coming after you:) Are Martians even party to the Berne Convention? Though the MPAM/RIAM might have a thing or two to say about it... DMacks (talk) 19:45, 16 April 2009 (UTC)[reply]
Even with todays' technology, it would be quite possible to bring along a snapshot of the entire Internet and every piece of music and every episode of every recent TV show and every movie made in English over (say) the last 20 years...let's toss in every digital book and every videogame ever written too. But even if not - those things could be downloaded via telemetry as-needed - so long as you were prepared to order movies and as you would from a subscription DVD service like NetFlicks. So there is really no need to fill the ship with hard drives just to do that. As '65 says - playing online games with anyone other than your crewmates would be impossible even after just a day or two of travel...the latency would make that impossible...and I think that probably rules out the otherwise excellent teledildonics idea from Edison too. The speed of light is a harsh mistress.† Personally, I think the thing the crew would get most benefit from would be access to email and Internet forums. Using those kinds of inherently non-realtime service - you can have close friendships with people even when you are halfway across the Solar System...and I think the biggest stressor for such long duration flights would be having to see the same half dozen faces day after day. But finding things to do that DON'T involve staring at a screen is by far the hardest problem. SteveBaker (talk) 20:19, 16 April 2009 (UTC)[reply]
Last phrase from the article you linked: "The Local Interstellar Cloud's potential effects on Earth are prevented by the solar wind and the Sun's magnetic field". The cloud is too thin. It pays to read an article all the way to the end. Dauto (talk) 18:53, 16 April 2009 (UTC)[reply]
Temperature is almost meaningless in such a near-vacuum situation. At reasonable gas densities it's a measure of how fast the molecules are rushing around. But when there are hardly any molecules there, the distinction between a large body of very thin gas that's merely travelling along and an actual hot gas, becomes almost non-existant. What you have to consider is the amount of energy contained in that gas - that's a function of it's density and it's temperature - but because it's practically a vaccuum - it's density is all but negligable - so it's not able to transfer significant energy into nearby planets and such. If it did transfer energy into the earth - then the parts of the cloud nearest us would instantly cool down to the same temperature as the earth and because thermal conductivity of such a thin gas is terrible - it would take millennia for any significant amount of additional heat to make it through to us. So the total energy transferred from the cloud to us is far FAR too small to measure. SteveBaker (talk) 20:06, 16 April 2009 (UTC)[reply]
Temperature is a measure of the average kinetic energy of the molecules of a substance. Basically, its the speed at which those molecules move. One substance can transfer thermal energy to another substance only when molecules of that substance bump into other molecules and make them go faster. So, there are two important things here: the speed the molecules are moving, AND how often those molecules bump into you. A substance which is so diffuse that a molecule only hits you like every few seconds will not efficiently transfer much energy to you. A single molecule striking you at a temperature of 6000 degrees has such a tiny effect on heating you up that you wouldn't even notice.
So, even though those molecules are whizzing around at a super fast speed, there are not enough of them around to effectively raise your temperature much at all. In fact, if you were in this environment, uninsulated, you would radiate heat energy much faster than it could be replaced by the occasional 6000 degree molecule bumping into you. You wouldn't burn up, you'd freeze to death, not because the temperature of the surrounding molecules isn't higher than your base temperature, but because there just aren't enough of them around to keep you warm, regardless of how fast each molecule is moving. --Jayron32.talk.contribs01:16, 17 April 2009 (UTC)[reply]
Well, I think that density X temperature (X volume) gives you the total thermal energy of the system, which is a quantity. Density X temperature per unit volume would be specific energy, since volume is not an absolute term (it depends on the relative scale you measure it with, unlike density and temperature which are absolute scales). Or possibly energy density, unlike several others here I'm not an expert!
But you asked why we're not burning up - that's not a quantity, it's a function, namely the heat transfer function. In this case, there are so few of those really hot molecules hitting us to make a difference. We absorb and emit far more energy through other processes, it's there, but it doesn't make us burn up. Generally, I think Jayron said it best. Franamax (talk) 00:34, 19 April 2009 (UTC)[reply]
Ecology book, online source?
My girlfriend is trying to refrence a refrence from an article, and her instructor won't let her use it. I was wondering if anyone could help me find an online or downloadable copy, preferably a PDF, of the journal. The journal is "Perspectives in Plant Ecology, Evolution and Systematics," the author she's referancing is S.J. Wright. Though, this may present a problem, in that, the author, of the journal, may be referencing another author, I just don't know.
Would the title of the article be The myriad consequences of hunting for vertebrates and plants in tropical forests ? That's the only article I can see by SJ Wright in that journal. It looks like you can only get the full article by buying a downaloable copy, which costs $31.50. If your girlfriend is attending a college or similar institution, she may be able to access it for free, if they have a subscription. The college librarian(s) should be able to point her in the right direction.--Kateshortforbob22:38, 16 April 2009 (UTC)[reply]
Google Scholar is great for stuff like this, because often you can find downloadable versions of things that aren't easily available from the journal web site. This is such a case: here is a pdf version. Looie496 (talk) 23:45, 16 April 2009 (UTC)[reply]
When I studied for my last degree, this sort of thing was counted as academic misconduct: quoting a reference that was not consulted. The idea was to note down the actual document used, not necessarily where that author got their original idea from. Then you are covered if there is misinterpretation or alteration of the original author's writings. And you are not claiming to have read all those references you did not see. Graeme Bartlett (talk) 22:20, 17 April 2009 (UTC)[reply]
Yes, but reviewers often complain if the original source of an idea is not cited -- especially if the reviewer is the original source. I believe it is pretty common to cite papers that have only a peripheral importance based on a general understanding of their findings rather than a close reading. It's a writer's responsibility to cite papers correctly, but this is not something that people frequently obsess over. Looie496 (talk) 16:41, 19 April 2009 (UTC)[reply]
Just help me, dangit. I've got C6H806 + NaHCO3 -> H2O + the formula for sodium ascorbate + unknown. I'm thinking carbon dioxide, but I'm not sure. Reywas92Talk23:32, 16 April 2009 (UTC)[reply]
If it's homework, then making you do it yourself is helping you. We can give you help if you're stuck on specific concepts, but we can't just give you the answer. Why are you assuming a 1:1 ratio for the reactants? What atoms do you have left over? Franamax (talk) 00:01, 17 April 2009 (UTC)[reply]
Do the metathesis (or double replacement reaction) just as you would expect to, and you will get two products. Just swap the positive ions between the two compounds, and write the new formulas, balanced for charge of course, and show what you EXPECT to get.
Read this article or, look at the formulas of the two compounds you got doing this the standard way, and compare to what your homework problem SAYS you can get. It will require you to do a little subtraction problem.
If you do these things, you will get the correct answer. If you just berate people who are trying to help you figure out the right answer, you won't get much more help from us... --Jayron32.talk.contribs01:07, 17 April 2009 (UTC)[reply]
Does anyone know of the best free dicom viewer out there? I want one with all of the features of Osirix, but that runs on windows. Specifically I want the following features:
-A GUI interface.
-Installed from an .exe file, or other autoinstaller file.
-not web based (although if you find a web based one and you can help me set it up then that's fine)
-Viewer supports:
-Regular tools: cine, zoom, magnifier, pan, windowing
-Measuring
-ability to compare two or more images
-ability to colour enhance the images
- 3D mpr and post processing like OSIRIX.
It doesn't have to be open source. It just has to be free of charge. I do not need a pacs server, just a full featured radiology workstation. Please help me. I've searched google for months and come up with nothing except osirix. If there is a windows version of that i'll take it!!!!!
WTF are you screaming at us? I can't help but someone who can help out may decide not to by your ALL CAPS approach. Feel free to edit your question. -hydnjo (talk) 00:57, 17 April 2009 (UTC)[reply]
Is there another explanation for Black Holes besides being dense matter from which light cannot escape?
Has anyone considered the possibility that a black hole is simply a phenomenon produced by 2 bodies, such as galaxies, moving away from each other at a combined speed that is greater than the speed of light. For instance, our galaxy may "appear" to be a black hole to inhabitants of a neibouring galaxy which is moving in the opposite direction. Acknowledging that nothing can travel faster than the speed of light it is still possible for 2 bodies to be moving away from each other at a almost twice the speed of light. The light from either body would never reach the other thereby making them invisible to each other. Would this not produce a black hole effect?
Actually, even if two galaxies (on different sides of us) are moving away from us at near the speed of light, an observer in either one would still be able to see the other and would not see the other galaxy as moving faster than the speed of light (see this section on relativistic addition of velocities). Additionally, even if there were a galaxy we couldn't see, it wouldn't explain black holes or look like a black hole at all. It would just be something you couldn't see. It wouldn't explain any of the Black hole related phenomena that have been observed. Someguy1221 (talk) 03:56, 17 April 2009 (UTC)[reply]
We do not "see" distant galaxies because the light is red shifted. But the fact that far off galaxies are moving away from us is observed phenomenon, not just theoretical. We can observe galaxies moving away faster than light provided that they are no so far off that their light has not reached us yet. Galaxies which are too far off are beyond our observable universe. But yes, regardless of speed of our and far off galaxy, the speed of light coming from the distant galaxy will be the same. Black Hole is region of space that is within our range of observation and we cannot see light from it because light cannot escape from Black Hole due to very high gravity. It is nothing to do with speed of observer or that of Black Hole. If light not reaching us is criteria for Black Hole, then we can say we are in enclosed inside of a black hole ( which region of non observable universe) manya (talk) 05:03, 17 April 2009 (UTC)[reply]
From special relativity we know that the speed of light is the same in all frames. Starting with two planets moving away from some central point at near light speed in opposite directions, we can change to a frame where one of the planets is stationary. In this frame, any light emitted towards an observer on the stationary planet from the moving planet will still approach the observer at 'c'. So it's still going to get there, you can't out-run light. (Also, as mentioned by Someguy, the moving planet in this frame would not exceed the speed of light itself.) 129.67.117.21 (talk) 12:27, 17 April 2009 (UTC)[reply]
It is not the case that there were many observations of black holes and a theory needed to be developed to explain them. Actually, it is entirely the other way round. Black holes were predicted to exist (or at least the possibility) from the theory of gravity, and then astronomers went out looking for them. Thus there is no question of there being any alternative explanations. Although many candidate black holes have now been identified (Cygnus X-1 being the first) it would be a very rash astrophysicist who would make the claim that their existence is established beyond all doubt. SpinningSpark13:04, 17 April 2009 (UTC)[reply]
You don't need to change reference frames to understand why your reasoning is flawed. Say you're going to the "right" at 3/4 c and an object to your "left", that's going left at 3/4 c, emits some light toward you. The light goes to the right at c, which is faster than you're going, so it will eventually reach you. In cosmology, though, what you describe is possible, and the effect is something like a black hole turned inside out—the event horizon is still a sphere, but instead of the inside being invisible from the outside, it's the outside that's invisible from the inside (and we're inside). That's still not the same as a black hole, though—there's no way to confuse the two. -- BenRG (talk) 15:00, 17 April 2009 (UTC)[reply]
Coriolis effect and sniping
In the game Call of Duty 4, there is a part where we have to snipe a guy 896 meters away. Our captain says that we should take 3 things into consideration:
1 Variable Humidity
2 Wind speed
3 The Coriolis effect
My question is how does variable humidity and the Coriolis effect effect the bullet? You can read it here http://en.wikiquote.org/wiki/COD4#Cpt._MacMillan under Capt. MacMillan. —Preceding unsigned comment added by 116.71.49.171 (talk) 05:32, 17 April 2009 (UTC)[reply]
Wind is obious - it exerts a force on the bullet pushing it away. The coriolis effect is when the earth spins as the bullet is travelling, so by the time it hits the target, the bullet is now offline, even though it travelled perfectly straight, because the earth rotated relative to the bullet. Humidity reduces the density of the air, which will effect the total force of the wind and humidity.--155.144.40.31 (talk) 05:53, 17 April 2009 (UTC)[reply]
I haven't played the game, but I have had military training in long-distance shooting. Everyone has to do up to 500 meters and in special schools we go beyond that. Wind and humidity play a factor. The coriolis effect is brought up, but doesn't matter. The sights are set for distance to the target, taking the actual arc of the bullet into consideration. So, as you "click click click" the distance to the target, you are adjusting for the actual arc of the bullet. The main thing you do is adjust for windage (left and right) and then distance (up and down). If it is extremely humid, you add distance. Then, breath, relax, aim, steady, squeeze. In Call of Duty, do you have to adjust your sight left/right and up/down or do they have you just winging it by aiming a little up and a little off to the side? -- kainaw™06:33, 17 April 2009 (UTC)[reply]
You don't adjust the sights in CoD4, you just aim a little off to the side. (Also, they don't really simulate the coriolis effect there at all, that's just patter. Ditto for the humidity. What matters is the wind (which is indicated by a flag nearby, so you can tell what direction it's blowing in, and how hard); the rest of it is just there to bring atmosphere to the scene. It does the trick, too.) -- Captain Disdain (talk) 10:23, 17 April 2009 (UTC)[reply]
Setting the distance isn't enough to account for the coriolis effect - you need to know what direction you are firing in as well. The maximum acceleration due to coriolis effect is velocity*angular momentum (but rotated 90 degrees). If we take a muzzle velocity of 1000m/s (and assume it doesn't slow down during travel, which isn't realistic, but nevermind) and the mean angular velocity of the Earth (from Earth's rotation) of 7.2921150 × 10−5 radians per second, we get a coriolis acceleration of about 0.07m/s2. If the target was 1000m away, the bullet would take 1s to get there, resulting in a deflection of 35mm. That's a noticeable amount (although it's an overestimate in most cases), but probably far less significant than the wind. --Tango (talk) 11:15, 17 April 2009 (UTC)[reply]
The Coriolis effect does not exist when firing due east or due west. It has maximum effect when firing due north or due south. It has intermediate effect when firing at directions in between. – GlowWorm.
GlowWorm, that's not right. To eliminate the coriolis effect you would have to shoot paralel to the earth's axis. That would be firing along the north-south direction with an upward (or downward) inclination identical to the local latitude. If you forget about the vertical component of the coriolis force (which gets masked by gravity anyways), the coriolis effect is identical in all horizontal directions. Dauto (talk) 19:02, 17 April 2009 (UTC)[reply]
I don't see that, Dauto. If a gun is north of the equator and fires due north at a target due north of it, the gun and the target are both on the same line of longitude. During the flight of the shell the target moves to the east. So does the line of longitude. So the shell will land west of the target. This is tricky to think about. Am I wrong? – GlowWorm. —Preceding unsigned comment added by 98.16.66.104 (talk) 01:23, 18 April 2009 (UTC)[reply]
Yes, you are wrong. In the nothern hemisphere the coriolis force will deflect the bullet towards its right. If you shoot towards the north the deflection will be eastward. That happens because as it moves towards the north it is actually approachingthe earth's axis. from the equation for the speed of an object in circular motion we see that the bullet is actually moving eastward faster than the target (larger ) and that's why it appears to be deflected eastward. Dauto (talk) 05:11, 18 April 2009 (UTC)[reply]
Wouldn't it also have an effect on an east-west shot? Not because it's going off target to one side or the other but because the bullet is traveling slower or faster relative to the target. So if you're shooting eastward, the target is coming at the bullet as the bullet is moving towards the target. Therefore, you don't need to aim as high. Right? Dismas|(talk)05:19, 18 April 2009 (UTC)[reply]
Yes, there's a second effect but your description of it is also incorrect. Due to the earth's rotation, in a referential attached to the earth's surface there is a centrifugal force. We don't notice that force because it gets included into our perception of the gravitational force as the earth itself deforms under the influence of that force becoming a little flat (Yes a small fraction of what we perceive as gravity is actually centrifugal force). If you now shoot a bullet eastward, that bullet will be rotating around the earth's axis faster than objects attached to the earth's surface, creating an extra centrifugal force. That extra centrifugal force is not included in the local gravity force and must be perceived as a new force pushing the bullet away from the earth's axis. This force has both a vertical component upwards (that is an effect in the opposite direction to the one you incorrectly described) as well as a horizontal component southward. If you shoot the bullet westwards, you get the opposite effect and end up having an extra force with components downward and northward. This effect I describe now together with the one I described earlier are included under one umbrella as the coriolis effect. If you ignore the vertical component of the coriolis effect, you will see that between the two effects there is always a force towards the right of the direction of the movement of the bullet (in the nothern hemisphere). Moreover, this horizontal component of the coriolis force turn out to have the same magnitude independently of which direction you shoot the bullet. Neat uh? Dauto (talk) 06:40, 18 April 2009 (UTC)[reply]
The behavious of things like for example electrons are described by complicated maths formulas, that require a high intelligence and a good education to understand. So how come a little electron knows how to behave? They cannot have computers hidden away inside them. 78.146.249.32 (talk) 11:12, 17 April 2009 (UTC)[reply]
They don't have to think how to behave. Their behavior is guided automatically according to the laws of physics, which we are still trying to decipher. It is like we have to calculate with what velocity and initial angle an object has to be launched to reach a certain distance according to the laws of physics, but when the object is actually launched, it (the object itself) doesn't calculate in the beginning of launch how far it has to go. It has no other choice but to reach that distance because it has no control over itself, it cannot think and react. It has to obey the laws of physics. For the projectile motion, those laws are simple. For describing electrons, they become complicated and require sophisticated mathematics and high intelligence. But the analogy still holds and electron having no other choice and no control over itself has to behave the way the laws of nature demand it to. - DSachan (talk) 11:22, 17 April 2009 (UTC)[reply]
And what's the problem with saying that? Your question is like asking 'when I let go of a stone, how does it know it's supposed to fall?' It just do. Dauto (talk) 14:05, 17 April 2009 (UTC)[reply]
Yes, that's exactly what he meant. You've got two alternative explanations here. 1) They just do 2) They were designed that way by someone or something that created them. But whichever it is, electrons don't need to be intelligent to do this, any more than a rock needs to be intelligent to know that it should roll downhill. DJ Clayworth (talk) 14:01, 17 April 2009 (UTC)[reply]
Actually, the behavior of an individual electron is not all that complicated, the problem with the calculation of their movement comes in when you have many subatomic particles acting together and influencing each other in a non-linear fashion. Truthforitsownsake (talk) 14:08, 17 April 2009 (UTC)[reply]
Some very complex-appearing behavior can arise from very simple rules. For example, Conway's Game of Life consists solely of rules simple enough that an 8-year-old can follow them. Yet from these rules you can get crazily complex behavior which would be difficult for people to explain. That's not unique to the GoL. Other simple rulesets show similarly complicated results (see Cellular automaton). Now, I'm not claiming that the behavior of electrons is the result of emergent behavior: what I'm saying is that electrons are following what are to them very simple rules (e.g. "move to a lower energy state"), but because of how these rules are constructed and how they interact with the environment, the overall behavior of the electrons appears very complex to us and requires complex math to represent because we don't "see" the world the same way an electron does. -- 128.104.112.117 (talk) 23:47, 17 April 2009 (UTC)[reply]
Also, the electron doesn't "know" how to behave. The OP is putting the cart before the horse here. We invent mathematical equations to model the behavior of electrons. But the equations don't tell the electron how to behave, they tell US how the electron will behave. The electron would continue to do what it did even if we didn't try to figure it out... --Jayron32.talk.contribs02:19, 18 April 2009 (UTC)[reply]
78.146, how do you behave at any given moment? You have a set of forces pulling you through life and at each instant you respond to those forces. Sometimes certain forces are stronger than others. Sometimes you do dumb things (so there goes your argument of having intelligence) - but you do those dumb things because local forces caused you to act that way. Other times you do smart things - but only because other forces drew you that way.
You are an electron, and an electron is you. You both respond to a complex set of inputs and neither of you have a "computer" (well, you have a brain, but it uses electrons to act). On a semi-related topic, see The Tao of Physics. Franamax (talk) 23:46, 18 April 2009 (UTC)[reply]
bodybuilding
hi everybody and thanks for reading this .. my question is :
what are the negative results I can get from bodybuilding ?
Does it have any negative effects on hormones and sperms ? —Preceding unsigned comment added by Sha9law0 (talk • contribs) 11:43, 17 April 2009 (UTC)[reply]
You should talk to an expert about that, rather than random people on the internet. If you are a member of a gym, they will probably have personal trainers that can advise you. We have an article, Overtraining, which might answer some of your questions, but don't rely on us when your health is at risk. --Tango (talk) 12:01, 17 April 2009 (UTC)[reply]
A normal and healthy human can do heavy exercise -- including bodybuilding work -- with no adverse effect on hormones or sperms. Unfortunately, many professional-level bodybuilders have cheated with anabolic steroids, which can have serious negative effects on both hormones and sperms. --Sean14:22, 17 April 2009 (UTC)[reply]
I'd like to see some sources if you're going to make such claims. I don't deny of course that anabolic steroids seriously screw up your body and are one of the reasons many body builders suffer serious adverse effects. However while the human body is very good at compensating, it seems a fairly extreme claim to me that there will definitely no adverse effects on hormones or sperm from body building. Presuming we're talking about quite hefty body building, the exercise and diet can itself have some fairly major effects and it seems unlikely to me that these will have no effect (positive or negative) on your hormones or sperm (or for that matter other bodily functions). This doesn't mean I believe your going to go infertile from body building, simply that there's a difference between saying it will have no adverse effects and saying that it's unlikely to have extremely detrimental effects Nil Einne (talk) 22:23, 17 April 2009 (UTC)[reply]
I agree that I should have qualified what I said. I was thinking of a sensible amount of bodybuilding, but I agree that an absurd amount of bodybuilding could very well affect the sperms, for some definitions of "sensible" and "absurd". --Sean22:47, 18 April 2009 (UTC)[reply]
If you take steroids, obviously there are tremendous negative effects. If you don't, the major negative effects probably come from the huge amounts of food you have to eat to do top-level bodybuilding, and the heavy weighting of the food toward protein. There's also a problem in that bodybuilders are sometimes so musclebound that they have difficulty doing ordinary activities. Looie496 (talk) 18:19, 17 April 2009 (UTC)[reply]
You should also consider what else you would be doing with all your time that you would be spending in the gym. Will it stop you working or ruin your social life, or stop you having time to edit Wikipedia? There are also injuries resulting from improper use of equipment. Graeme Bartlett (talk) 22:01, 17 April 2009 (UTC)[reply]
Questions on cell size and multicellularity
What limits the size of a living cell? Why don't we see (for e.g.) cells the size of a dog?
Do prokaryotic cells show multicellularity? I know they form colonies, but do individual cells show division of labor/specialization? If not, why not? —Preceding unsigned comment added by 121.241.167.100 (talk) 12:56, 17 April 2009 (UTC)[reply]
One thing that limits cell size is the way that cells function. Cells require a large surface area to volume ratio. Now while technically, you could still have very massive cells with the same surface area to volume ratio, the structure would be such that the cell would pull itself apart from gravity. 65.121.141.34 (talk) 13:25, 17 April 2009 (UTC)[reply]
Cells can get bigger than a dog! A Hypselosaurusegg (which, assuming it's like modern eggs, is a single cell) could reach a foot in length, larger than the dog at right. You could probably find an ostrich egg bigger than some dogs. And even the biggest dog has a nerve cell that runs from the tip of its tail to its brain, which is pretty long, if not "big". Also, see the discussion regarding division of labor here. --Sean14:38, 17 April 2009 (UTC)[reply]
I do not believe that large eggs are just one cell. I think thats a myth. For one thing, they would have terrible problems with the celluar equivalent of respiration, as their volume to surface area ratio would be very much different from a normal cell. 78.149.194.28 (talk) 18:26, 17 April 2009 (UTC)[reply]
To clarify the common misconception. Ova, that is the female gamete colloquially known as "eggs" are a single cell. Actual eggs, certainly are NOT. No ova I know of are larger than microscopic for any species. The largest cells are still likely nerve cells, as mentioned, there have been some documented to be several feet long; even given their thinness, that still makes them quite sizable. --Jayron32.talk.contribs19:34, 17 April 2009 (UTC)[reply]
Also, from our Egg (biology) article, "The 1.5 kg ostrich egg contains the largest existing single cell currently known...". Seems we really have a contradiction here. --Scray (talk) 20:44, 17 April 2009 (UTC)[reply]
Yeah, OK, so the human ova is a bit larger than microscopic. But really, only a bit. And the canard that an egg is a single cell misrepresnts what a cell is. It doesn't have organelles, it doesn't contain a cell nucleus or anything like that. Its basically a big sac of amniotic fluid and surrounding a smaller sac of nutrients to provide fuel for the growing embryo. Our article is actually probably wrong here. It calls avian and reptilian eggs "zygotes" That makes no sense. If the zygote develops into the embryo, then why is the egg still here?!? So yes, I will buy that the ovum is the largest cell in the body. But a chicken egg is still not a "cell", unless you redefine what a cell is to mean "everything that a cell is, and eggs too". --Jayron32.talk.contribs02:15, 18 April 2009 (UTC)[reply]
I just re-read your comment, and realize that you're specifically referring to the avian/reptilian yolk and albumin as lacking a nucleus and organelles. All right, I'll agree with that as a generalization. BTW, "ova" is plural, "ovum" is singular. --Scray (talk) 03:57, 18 April 2009 (UTC)[reply]
One more note: the Squid giant axon (not specific to giant squid - it's just a giant axon) can be 1 mm in diameter - a diameter easily visible to the naked eye; with a length that can be measured in meters, this is a huge cell. --Scray (talk) 16:40, 18 April 2009 (UTC)[reply]
The article correctly states not that an egg is a single cell, but instead it states that the ostrich egg contains the largest known single cell. So, tell me Jayron, If the yolk of the egg is not that cell, where's the cell? Dauto (talk) 04:55, 18 April 2009 (UTC)[reply]
As to division of labour, there is the awesome dictyostelium - but it's a eukaryote. :( As far as the "why not" bit goes, I'm not sure that lack of a nuclear membrane is the defining factor in how cell specialization works, it comes down in the end to appropriate communication. Perhaps prokaryotes don't adopt this strategy because they've never had to? There are many paths to evolutionary success. I can't think of any definitive (published) reason why a prokaryote couldn't do the same thing. Interesting question though - there's my whole afternoon shot now! :) Franamax (talk) 20:38, 17 April 2009 (UTC)[reply]
Suppose I shoot a bullet into a block of wood, knowing the velocity and mass of that bullet. How could I estimate the final resting position of the bullet? What other information would I need to know? 58.6.129.205 (talk) 13:42, 17 April 2009 (UTC)[reply]
I don't know the exact answer to your question, I'll leave that to our resident physicists, but I think you'd need to know the distance of the wood from the initial position of the bullet (in the gun), and the strength of the wood involved. Both these factors would have a large role to play in how far the bullet can penetrate the wood. Regards, --—Cyclonenim | Chat 13:54, 17 April 2009 (UTC)[reply]
(ec)Our article on Penetration (weapons) sadly doesn't provide any information. This webpage [13] gives an equation by the 19th century French engineer Jean-Victor Poncelet, and a calculator to plug your own numbers into, which may be of use (Note that one of the users of the page spotted an error in the Java script for the calculator, hard to tell if it's been fixed). I presume that the various constants quoted would vary depending on the type of wood, orientation of the grain etc. Mikenorton (talk) 14:05, 17 April 2009 (UTC)[reply]
One questionably-useful data point from personal experience. A plain old lead round-nose .22LR round will penetrate a 2x4 of plain old lumber (I think probably pine, but I'm not sure.) However a 2x4 of weather treated wood will stop the same bullet. Clearly, the density and strength of the wood is a factor. Also consider bullet shape and cross-section. A smaller, more pointed bullet will penetrate most materials better than a bigger, blunter bullet. This sounds like one for Mythbusters- I imagine trying to estimate this is more complicated than simply performing some tests. Friday(talk)14:09, 17 April 2009 (UTC)[reply]
PS. (I mean penetrate the short way, not the long way. I don't remember trying it the other direction.) And the poster above brings up a good point about grain orientation- if a bullet can slip in between things, this is going to take less energy out of it than having to blast through things. Friday(talk)14:11, 17 April 2009 (UTC)[reply]
My dad said (WP:OR) that a 22 long would go an inch and a half into the end of a 2x4, farther than I would have supposed. I know that a 22 LR will penetrate seasoned 1x lumber (something over 3/4 inch thick: very old boards) and richochet around inside the building a couple of times, from shooting at an unoccupied farm outbuilding I owned, so it did not have the velocity to penetrate the 1x board on the far side of the building. Milage may vary depending on muzzle velocity, distance to target, mass and jacketing of bullet, and type of wood, grain orientation,angle of impact, and seasoning of the wood. Snipers, assassins, CSIs and police must have some empirical data. Edison (talk) 16:06, 17 April 2009 (UTC)[reply]
Paradox
One of my friends asked me this, and i just can't figure it out. Somebody please help me!!
Proof that unicorns exist For me to prove that unicorns exist, it is enough if i prove that there exists at least a single existing unicorn(possibly a more stronger statement). so there are 2 cases
I don't think there's a paradox so much as a silly phrase. What is the distinction between "unicorn" and "existing unicorn"? If you strip that word, then the choices of "at least one unicorn exists" and "no unicorns exist" is quite clear. In this specific case, it looks like "existing" is used to mean living. Again, though, swap the word and there's no contradiction. — Lomn15:25, 17 April 2009 (UTC)[reply]
You almost convinced me! Nice. Isn't the joke based on a shift from "there exists no existing unicorn" to "there exists a non-existing unicorn", that is not exaclty the same statement. pma (talk) 15:42, 17 April 2009 (UTC)[reply]
Listen here (for your friend). A contradictory object does not exist, but also, any object that does not exist may well have a contradictory property: this can be harmlessly assumed, for it does not exist. A contradictory property for a non-existent object is, of course, existence. Therefore a non-existent object enjoys the property of existence, so it exists pma (talk) 16:23, 17 April 2009 (UTC)[reply]
Assuming you don't just have a fetish for horses with strange frontal growths, you chose to discuss unicorns because they are objects which by definition do not exist. Any proposition in which they exist is wrong before it leaves the unicorn paddock. --Sean16:42, 17 April 2009 (UTC)[reply]
A silly play with words does not a paradox make. Said that, who is to say that unicorns do not exist? Any animal with one horn fits the bill. Dauto (talk) 17:19, 17 April 2009 (UTC)[reply]
It's pretty nonsensical. "Existing" is just an adjective - it's like "Blue" or "Pointy" - if I put the phrase "existing unicorn" in quotes then either there is an "existing unicorn" or there is no "existing unicorn" - which is clearly a true statement - just like either there is a "blue unicorn" or there is no "blue unicorn". But in any case - even with your interpretation - the two sentences are only "obviously true" when you interpret "existing" like you'd say "blue". Otherwise I can make paradoxes all day long by saying things like:
Either:
There are pink unicorns ...or...
There are green dragons
That's not a paradox - it's a set of statements that are simply false. So if you insist on this rather crazy interpretation of "existing unicorn" then you simply invalidated the entire statement by turning it into "Either unicorns exist or unicorns exist"...which is false. SteveBaker (talk) 17:41, 17 April 2009 (UTC)[reply]
This sounds a little like a play on the Black Swan Problem? Though I could be wrong as i don't fully understand why their statement is supposed to be a paradox. I could sort of read it similar to the Rumsfeld statement about there being known knowns and known unknowns, and unknown unknowns? (http://www.brainyquote.com/quotes/quotes/d/donaldrums148142.html). But at least Rumfeld's made a bit of sense. 17:55, 17 April 2009 (UTC)
I would say this is clearly a play on the ontological argument for the existence of God. It runs: Define God as a being that is best in every way. Clearly it is better for God to exist than not to exist. Therefore God exists. Vast tracts of philosophy have been written to argue about the validity of this argument. Looie496 (talk) 18:14, 17 April 2009 (UTC)[reply]
Me too, I thought to the logic-ontologic trick, but in fact it's not exactly the same... It seems to me that this one is more linguistic than logic.--pma (talk) 20:02, 17 April 2009 (UTC) As to Rumsfeld I thi(self censored).[reply]
Rumsfeld was a monst(self censored) but I think he made an very perceptive point and got screwed over for it. There really are known-unkowns and unknown-unknowns! Now, the context of his pontification, that was incredibly (self censored) and dismissive of (self-censored) and repugnant to every (self-censored). I do appreciate that contribution to world thought though, flawed as its presentation may have been. Franamax (talk) 20:55, 17 April 2009 (UTC)[reply]
This is discussed in one of the Smullyan books but I'm too lazy to look up which one even though I have them at home. Sorry. – b_jonas18:30, 18 April 2009 (UTC)[reply]
positron behavior
dear sir
what will happen when we put a positron in an electrostatic field?
does positron behave like an electron or as proton?
if we put an electron and positron together. do they attract each other or repel?
In an electrostatic field, a positron will not move in the same way as either an electron or a proton. It will feel the same force as a proton, but because it's a lot lighter than a proton, it will move along a different path. Looie496 (talk) 18:59, 17 April 2009 (UTC)[reply]
File:SmartseLeg hair.jpgA close up shot of the hair in question with normal hairs for comparison
I have one 10cm long leg hair (not quite a world record, see here) why has this one hair grown so long? It's a lot thinner and whiter than my other leg hairs. Any ideas? Smartse (talk) 19:26, 17 April 2009 (UTC)[reply]
I've got a very fine, long, white hair on the side of my forehead, which I doubt was ingrown at any time. It grows back after plucking. --Kjoonlee17:13, 18 April 2009 (UTC)[reply]
Redox reactions and Lewis Acids & Bases
If an oxidizing agent gains electrons, does that make it a Lewis Acid? Similarly, if a Lewis Base donates an electrons, is it also a reducing agent?130.127.99.54 (talk) 20:16, 17 April 2009 (UTC)[reply]
Not really. A Lewis Acid-Base reaction is a covalent bond formation reaction. And its a specific kind of covalent bond formation reaction. Remember that a chemical bond consists of two electrons. Under most reactions, one usually thinks of each atom on either side of the bond donating one electron to make a two-electron bond. Kind like this:
X• + •Y → X—Y
However, in a Lewis Acid-Base reaction, in the formation of the new bond, BOTH electrons come from one of the reactants. Thus, a Lewis Acid-Base reaction always looks like this:
A + :B → A—B
In this case, the A is the "acid" and the B is the "base". A Bronsted-Lowery acid-base reaction is thus a special case of the Lewis reaction, where A is ALWAYS the H+ ion. But the more general Lewis Acid-Base reaction encompases a whole lot of other scenarios, such as the following reaction:
F3B + :NCl3 → F3B—NCl3
In this case, the boron compound is "electron deficient", that is it has an empty orbital, or if you prefer, less than a full octet, which makes it an ideal "lewis acid". The nitrogen compound has a spare, unbonded pair of electrons, which makes it an ideal "lewis base". The thing about this is, unlike other acid-base theories, Lewis theory doesn't even deal with "H+" ions at all.
Now, redox reactions involve the transfer of electrons from one element to another, without necessarily forming any new bonds. For example, if you add say a piece of magnesium metal to a solution containing Copper (II) ions, you will get the following redox reaction to occur spontaneously, forming magnesium ions and copper metal:
Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
No new covalent bonds have formed. What happened here is that two electrons were transfered from the magnesium to the copper. So you see, though both types of reactions involve the rearangement of electrons in some way, they are very different sorts of reactions. Actually, EVERY chemical reaction involves the rearrangement of electrons in some way, so these two types of reactions are not all that unique in that regard. I hope this makes a little more sense. You may want to read Lewis acids and bases and Redox for more on these processes in general. --Jayron32.talk.contribs02:07, 18 April 2009 (UTC)[reply]
And a poorly done one too. Look at where the light is falling on the skull - only the front part is bright, the top and sides are in shadow. Now look at the shadow of the guy right next to it - the sun is falling on his back - so the skull should be brightly lit from it's right side...and it's not. SteveBaker (talk) 20:04, 19 April 2009 (UTC)[reply]
April 18
Coriolis force on a river in the northern hemisphere
"A straight river flows with speed v in a direction α degrees East of North. Show that the effect of the Coriolis force is to undermine the right bank. Does the magnitude of the effect depend on α?"
My brain has shut off, I don't even know where to start! Thanks for any help,
This seem's to be a homework. Is that the case? I hope it isn't because the statement "the effect of the Coriolis force is to undermine the right bank." is actually incorrect. Dauto (talk) 01:01, 18 April 2009 (UTC)[reply]
Oh dear, it is part of my set work, yes! Perhaps that explains why I've spent so long trying to understand why and getting nowhere - what does happen then? I've read through the wiki article on the effect but I'm still not completely certain, gah! Mathmos6 (talk) 01:41, 18 April 2009 (UTC)[reply]
I think it should be true in the northern hemisphere. The rotation of the earth about its axis gives the water angular momentum. As the water moves north, it gets closer to the axis, so conservation of angular momentum means it tries to rotate more rapidly about the axis, like an ice skater drawing in her arms to spin -- this pushes it against the right bank. Looie496 (talk) 01:51, 18 April 2009 (UTC)[reply]
Okay, so assuming this is talking about the northern hemisphere, does the angle east of north affect the magnitude? Or equivalently, I suppose, will you have a different effect if the magnitude of α is 0/90 degrees? You should have no effect along the hemisphere and 100% of the effect northwards, right? Mathmos6 (talk) 01:58, 18 April 2009 (UTC)[reply]
The effect doesn't depend on the angle . The coriolis force will indeed push the water rightward. The problem with the statement is that this rightward push will very lightly raise the water level on the rightside but that does not undermine the riverbank. To undermine the riverbank you would need an effect that speeds up the flow of the river on its right side. Coriolis effect doesn't do that. Dauto (talk) 02:10, 18 April 2009 (UTC)[reply]
Actually, a flowing river, being a fluid, would also be deflected rightwards as it flows. I am not certain the effect would be great enough to cause undermining of the river bank, but it may do things in the large scale like cause subtle differences in the overall way rivers flow across open plains and the like... --Jayron32.talk.contribs21:31, 18 April 2009 (UTC)[reply]
I'll try a qualitative answer here: steel is more efficient to recycle overall. I base this statement on the fact that steel is more widely recycled.
One factor in this is that it's more easy to collect and sort the steel. Crushing a car is easier than sorting six million pieces of plastic, and a bridge or a ship can pretty much be just thrown into a mini-mill. Plastics come in a wide variety of types (look at the number inside the recycling symbol on plastics, it goes up to at least nine) and are difficult to segregate since they need humans or spectroscopic/laser methods to sort. Steels by contrast can be sorted by weight and magnetic properties.
Another factor is that steel is much more dense than plastic. This comes into play when considering the cost of transporting the material back to the recycling facility, since steel "weighs out" before it "cubes out". This means that the trucks/trains/ships are carrying their maximum load, which is usually more fuel-efficient. Put another way, you probably need fewer trucks to ship the equivalent amounts of steel and plastic for their strength-weight-volume ratios.
And steel has the big advantage in thermodynamics that you just have to remelt it, you don't have to reform the molecular structure. With steel, you just make a melt, check its properties, then throw in a little more of what you need to get the exact metal alloy you're trying to produce. With plastics, you have to deal with thermal dissociation of the plastic molecules and more severe negative properties of the contaminants. Steelmaking is a metallurgical process, plastic is a chemical process.
As I said, those are qualitative impressions. If you had a million steel computer cases sitting beside a million plastic cases just outside your recycling facility, I'm not sure which one would win. I'd put my money on the steel though. The economics of recycling are very complex. But since people will come over to your house and tow your old car away for free, whereas they won't come over to get your plastic bags for free - again I'm betting on steel. Franamax (talk) 23:16, 18 April 2009 (UTC)[reply]
Recycling economics is complex; the deciding factor however in determining if it is economically feasible to recycle a material is in having uses FOR the recycled material. Take rubber tires as an example. They are a complete nuisance as a waste product (take up a lot of space, never degrade, etc. etc.) and so should be an ideal candidate for recycling them. However, they are completely worthless in that regard, as there are so few uses for recycled tire rubber. There are a few weird esoteric uses, such as paving running tracks and things like that, but really you can't even use it to make new tires out of, its a worthless material, so there is really no good reason to recycle it. The chemistry of plastics is similar. With steel cans, you melt down the steel cans and make more steel cans. Its simple and easy and means that you will always have enough demand to make steel recycling economically feasible. With plastics, it just doesn't work that way. You can't melt down plastic milk jugs and make more plastic milk jugs; the chemistry just does not support that sort of application. Sometimes, you can use the recycled plastics for alternate applications, but like with the rubber tires example, the demand created by these alternate applications is just not enough to make it as econimical as metal recycling. --Jayron32.talk.contribs04:56, 19 April 2009 (UTC)[reply]
I thought they fed milk jugs back to the cows who made them! And I'm pretty sure we ship all our old tires to Springfield, that fire's been burning for 20 years now. :) Franamax (talk) 20:13, 19 April 2009 (UTC)[reply]
Power production from sea water by electrolysis process.
Would it be efficient to seperate hydrogen from sea-water by electrolysis process and then to produce electrocity by using the hydrogen in a gas turbine? —Preceding unsigned comment added by Shamiul (talk • contribs) 07:59, 18 April 2009 (UTC)[reply]
No. See First law of thermodynamics for why you cannot win, and Second law of thermodynamics for why you cannot break even. Splitting a water molecule requires exactly the same amount of energy that is returned when you recombine it. But none of the processes are 100% efficient - i.e. you lose energy. As a rough ballpark figure, in you configuration I would expect the loss to be around 60%. --Stephan Schulz (talk) 08:46, 18 April 2009 (UTC)[reply]
Certainly that isn't going to work in terms of getting free energy (but then there is no possible way to get free energy - period - because of the laws of thermodynamics). But there may be reasons to do it anyway. For example - it's proving very difficult to design batteries to make electric cars a particularly viable option - so using the electricity to make hydrogen and putting the hydrogen into the car makes some sort of sense. Also, if you were to supply power from wind turbines or nuclear power exclusively - then there would be times of day when you'd be producing more electricity than you need - and turning that energy into hydrogen, which you'd store and burn later when the need for electricity is higher. So there are reasons to do it - but none of them relate to getting energy for free - which is completely impossible by any means. I think Stephan's estimate of a 60% loss is an under-estimate. There aren't many gas turbines that are more than 60% efficient - and electrolysis is at best only about 80% - there are bound to be other losses. I'd be surprised if you got back 40% of what you put in. Probably, a gas turbine would be a bad way to get electricity from hydrogen - hydrogen fuel cells seem like a better bet. SteveBaker (talk) 20:02, 19 April 2009 (UTC)[reply]
Sleep
Last night, I had approximately five hours of sleep and not suprisingly, I am unable to think about mathematics today. Similarly, with regards to learning vocabulary (from another language), my short term memory did not seem to be up to its uaual standard. I also seem to be consistently thinking about sleep the whole day, unable to do much. I have thoroughly read the article on sleep but could not find the answer to this question - thinking in which areas of mathematics is affected most by a lack of sleep (and which areas are affected the least) and which aspects of thinking in general, are affected by a lack of sleep? I think this reference desk is the best place to post but hopefully I will get answers from the Wikipedians at the other reference desks. Also forgive any errors in my writing. :) --PST10:55, 18 April 2009 (UTC)[reply]
Have you read Sleep deprivation? I wouldn't expect getting only 5 hours sleep for one night to have a significant effect, though. Plenty of people sleep that little routinely (studies vary in their conclusions about whether that is a good idea or not). I expect your problem is not that you are too tired, but rather that you are too sleepy. You need to do something to wake yourself up - I find sugar and light exercise to be the best way to do that. Once you've woken up and got going, you should be fine. --Tango (talk) 12:49, 18 April 2009 (UTC)[reply]
Thanks for your advice and comments. The other question I have is about sleeping too much. If I were to sleep in, although having slept for 10-12 hours, I feel sleepy 2 hours after having woken up. Describing it as sleepy might be inaccurate - rather it is a temporary drowsy sensation. Is it better for thinking, to sleep for around 10-12 hours or more, or a moderate amount? For me, it is usually around 8-10 hours of sleep that is the best for logical thinking. --PST14:27, 18 April 2009 (UTC)[reply]
I know what you mean - I find sleeping too long leaves me kind of fuzzy. Again, exercise seems to help (just 10 minutes brisk walk in the fresh air is often enough). I think a typical amount of time to sleep is 6-8 hours, although opinions vary on what is best (I remember a study that showed people that sleep an average of 6 hours a night live longer than those than sleep 8, which is the opposite of conventional wisdom). --Tango (talk) 14:59, 18 April 2009 (UTC)[reply]
Having a short (20 minute) midday nap can help immensely if you aren't sleeping enough at night. Our article Sleep is full of useful advice and data. SteveBaker (talk) 19:52, 19 April 2009 (UTC)[reply]
I agree - it's amazing how much better I feel after a short nap when I haven't been sleeping properly. However, it seems from the OP's question that the lack of sleep was a one-off, so hopefully they slept properly the next night and all is well now. --Tango (talk) 21:58, 19 April 2009 (UTC)[reply]
hiv transmission
can hiv be transmitted through kissing?
and
which is more likey a way of hiv transmission, swallowing cum or bareback sex?
I believe HIV can be transmitted through kissing, but it is quite unlikely. Cuts, sores, bleeding gums, etc. can increase the risk, though. Unprotected vaginal or anal sex is almost certainly a greater risk than oral sex, although cuts, sores and bleeding gums do, again, increase the risk. I remember an excellent table being linked to in a previous ref desk question showing the risks of various sexually transmitted infections being transmitted is different ways. I'll try and find it. --Tango (talk) 15:04, 18 April 2009 (UTC)[reply]
The only was HIV could be transmitted through kissing if the receiver has some form of cut in their digestive tract before the stomach. If not, the virus will be broken down in the stomach. It's pretty rare to catch HIV this way.
As for your second question, the former could only result in transmission if the above provision still applies. The latter, however, is much more likely as anal sex is, quite frankly, done in an area not designed for such activities and can result in more cuts within the anal canal. If semen enters a cut, it can transmit the virus to the other person. Regards, --—Cyclonenim | Chat 16:06, 18 April 2009 (UTC)[reply]
There have not been any documented cases of HIV transmission by French kissing. This paper discusses reasons for the low HIV content in saliva. Despite this, the CDC cautions against French kissing with HIV positive people due to theoretical risk of transmission. Axl ¤ [Talk]16:56, 18 April 2009 (UTC)[reply]
Figuring out when a star does heliacal rising
Hello Science Deskers!
(a) If I know the right acension and declination of any star, how do I do the calculation to figure out when that star has its heliacal rising? (I'm hoping there is a way to do some math and get an answer such as "19 days after the summer solstice", that kind of thing.)
(b) Once I figure out the answer, how would I adjust it to figure out when that star would have had its helical rising for any year in the past (such as 5 or 10 or 15 thousand years ago)?
Thank you, WikiJedits (talk) 16:00, 18 April 2009 (UTC)[reply]
What you need is a planisphere. Make sure you get one that is for the latitude where you live. You can also get free ones off the internet you can print yourself or software you can run, just try googling for it. If you want to do the trigonometry yourself, you will need a book, this one iss very old (1834) but because it is out of copyright you can view it online or print it out. I don't think trigonometry has changed much since then. SpinningSpark17:12, 18 April 2009 (UTC)[reply]
Thank you SpinningSpark. I have found a planisphere but can't figure out how to read a heliacal rising from it. For example, I put Spica over where it says Eastern Horizon, but the earliest time it gives is 5 am and the month over there is November. I know sunrise is way after 5 am in November. Is there a way to work it out by doing a calculation instead? I have looked at the book but it is way beyond me. Is there a method a beginner could use? (Also, do you or anyone else know the answer to my question b?) Thanks again, WikiJedits (talk) 01:52, 19 April 2009 (UTC)[reply]
My Planisphere is the Philips' one like the picture in the article so if you look at that you will understand what I am talking about. You are right that the first step is to put Spica on the Eastern horizon. The moving part of the planisphere has a time scale, and the fixed part has a calendar. At any particular date, this will tell you the time Spica will rise. If you look for 19th April, you will find that Spica is rising at about 7pm in the Northern hemisphere. This depends on your latitude which is why it is important to get a planisphere for your own location. Ok, so 7pm tells you that Spica is rising in the night sky (or before) at the moment and is nowhere near its heliacal rising. Next you look around the scale till you find a date at which the timescale is showing the time of sunrise for that date (again dependant on your location). 6am is at 30th October on the planisphere, so if sunrise is at 6am on that date, then that is the also the heliacal rising of Spica.
If you cannot deal with the trigonometry in that book, then you are not going to be able to do the calcualtion by hand. I suggest you get some software to do it for you, or else go and learn the maths first.
For 5 or 10 years in the past, there is virtually no difference to the date of the heliacal rising. For thousands of years in the past, you must take into account the precession of the earth. To do this you must first locate the position of the North Celestial Pole in the epoch which you are interested in. You can get that from this diagram. Next, you must move the pivot point of your planisphere to the location of the ancient North Celestial Pole (don't destroy your good one, print a throwaway one off the internet). Then you just proceed as before.
Thank you, I really appreciate the time you've taken. Think I get it now - I need a table of sunrise times and to compare it to the times and dates shown on the planisphere. Got one off timeanddate.com and doing that for Spica again (at 30N) gives me a best match with October 17, 5:58 am. Additional question 1 - the heliacal rising isn't exactly at sunrise, is it? Do I need to allow extra time? Would it be better to pick July 3 when sunrise is at 4:58 am? Cos that's quite a difference.
Additional question 2 - About the past calculation, I must be doing something wrong? I drew the north celestial pole circle on the rotating part of the planisphere and put the pin in a different place along the circle. But this didn't seem to make much difference (other than the planisphere not fitting together so well). For example, I tried it for halfway around the circle, or about 10,000 years ago, and putting Spica on the eastern horizon still has the fall months aligning with the morning times, when I would expect it to be the spring months. Do you know what I'm missing? Much thanks for all your help, WikiJedits (talk) 19:42, 19 April 2009 (UTC)[reply]
Turning radius of a vehicle
The turning radii of all four or more wheels of a vehicle are always different.What does the mean turning radius given in their specification mean? —Preceding unsigned comment added by 202.70.64.16 (talk) 17:10, 18 April 2009 (UTC)[reply]
Mean turning radius is the radius through which the centre-line of the vehicle turns. It is usually defined at the front wheels, the mean radius is different along the length of the vehicle in general. SpinningSpark17:38, 18 April 2009 (UTC)[reply]
Car manufacturers sometimes talk about the "turning circle" which is the smallest circle within which the car can do a complete 360 turn...that's a more useful measure - and refers not to the wheels but to the parts of the vehicle that stick out the most. For a car that has lots of "stuff" sticking out behind the rear wheels (like a US-style school bus), that can be a significantly larger number than the radii of the circles made by the wheels. For my car, the number in the handbook specifically excludes the door mirrors! SteveBaker (talk) 19:46, 19 April 2009 (UTC)[reply]
Sedimentary rock, evaporites: How does gypsum end forming a stratum .50m thick?
Where can I learn about the processes by which gypsum -probably originally sedimented in the form of thin little crystals- end up forming strata half a meter thick and spanning kms?
This question applies (in my curiosity) to all kind of evaporites.
In the articles about evaporites and sedimentary rocks, and in the books I've read, I don't see anything explaining the processes that undergo the evaporites once sedimented.
Jorgemelis (talk) 17:17, 18 April 2009 (UTC)[reply]
See also Sabkha (a supratidal salt flat), which is an environment in which thick gypsum deposits are often formed, sometimes in the form of desert roses. Evaporation at the land surface causes the rise of brines by capillary action and as these evaporate they deposit salt at the surface and gypsum and aragonite in the subsurface. In deeper water areas gypsum layers may be formed by crystallization at the air-water interface followed by sinking of the crystals to the bottom. Mikenorton (talk) 19:31, 18 April 2009 (UTC)[reply]
Lazy smokers
Which of these would improve the overall health of an inactive (no exercise) smoker with a typically bad, high-fat diet more?
I doubt there is an answer that applies in all cases. People have lived to be over 100 while smoking 20 a day, so stopping smoking won't necessary improve anything at all (although it most likely will). Exercise and diet are complicated things, there aren't simply "good diet" and "bad diet", "exercise" and "no exercise", so more information would be required. Also, it's necessary to know in what way the person is unhealthy - if they are suffering from Emphysema, stopping smoking will almost certainly improve their health more than the other two, for example. --Tango (talk) 20:26, 18 April 2009 (UTC)[reply]
I wondered if I might get that. Since I'm just curious in general, not looking at anyone or any particular set of circumstances in particular, I probably can't get a real answer as such - though anyone is welcome to come in with more input and opinions. Vimescarrot (talk) 20:37, 18 April 2009 (UTC)[reply]
Of course, health is highly individualized. One can speak in general terms for what works in averages, but you will always find anecdotes of 110 year old men who sit on their ass all day chain smoking unfiltered Camels and eating 5 pounds of bacon a day and washing it all down with Thunderbird wine. And you can not smoke, not drink, run 10 miles a day, and live by eating nothing but tofu, broccoli, and wicker and die of a heart attack at 45. Behavioral effects on health are measurable across large populations, and you should probably do whatever you can to improve your own health, but you can do nothing at all to guarantee that you will live longer only because you do everything you are "supposed" to do. --Jayron32.talk.contribs21:27, 18 April 2009 (UTC)[reply]
Yes - but it's a matter of probability. One person in a thousand who does all of those terrible things to their body will survive - but the other 999 will die an early death. Sure, there are anecdotes of individuals who smoked their entire lives and through sheer, blind luck survived. On the other hand, you have people like my father who tragically died in his early 50's from smoking just 10 cigarettes a day - or like my mother who suffered from most of the bllod pressure and heart disease symptoms of a 10-a-day smoker despite never having smoked in her life because of the second-hand smoke. That my father died so young is probably the only thing that saved my mother from doing the same. Picking the one person in a thousand who survived and hoping that you'll also be that lucky one person in a thousand and not one of the 999 more typical cases is exactly like saying "I'm going to give up working and assume that I'll win the lottery" - nobody in their right mind does that yet it's the exact same ridiculous thinking pattern. SteveBaker (talk) 19:29, 19 April 2009 (UTC)[reply]
It kind of depends on your exit-strategy. Do you want to live with maximal indulgence and go out with a bang? Suffer through lingering health problems? Exercise will improve your cardiac state, help your lung capacity to a certain extent and improve your feeling of general wellbeing - but there's all that dicomfort and sweating. Improved diet will lessen your risk of atherosclerosis, stroke and metabolic syndrome/diabetes - but there's the factor of being hungry all the time. Quitting smoking improves almost every risk factor there is (not diabetes) - but that could just mean you spend a long life wishing every minute of the day that you could have another smoke.
If quitting the evil weed is off the table, then a little of both of the other two. A half-hour brisk walk every day is a good start and eatcher veggies. But really, anything other than continuing to sit in a chair smoking and eating pork rinds is probably going to be an improvement. In the end though, everyone chooses their own way out the door.
Maybe you need to engage with your mother more - offer to cook some neat recipes and ask her to go out on regular walks with you. That way you get an equal benefit. Franamax (talk) 22:24, 18 April 2009 (UTC)[reply]
Exercise should be enjoyable if done correctly (endorphins and all that), and a healthy diet shouldn't leave you feeling hungry unless you are trying to lose weight (which should only by a short term issue). --Tango (talk) 01:02, 19 April 2009 (UTC)[reply]
For a heavy smoker, the answer is very likely stopping smoking. The other things make a difference, but unless the person is an inert obese blob, not as huge a difference as smoking. Looie496 (talk) 02:22, 19 April 2009 (UTC)[reply]
A mixture of OR, unsourced and read from sources but not cited:
@Tango, granted on the endorphin reward, but they take time to develop hence the initial discomfort. There's also the social factor, which is overall important. I tried working out at a gym once and didn't return. OTOH I will go for multihour walks with my friends or family anywhere and always. And despite the obvious handicap of being a smoker, I love playing squash, just because I'm competing against myself as much as the other player. Plus it's among the most exertive of sports, so you get the good stuff that much faster I suppose. :)
And the unsourced part is to feeling hunger pangs. Perhaps I should call them cravings. Long-term exposure to high-sugar and/or high-fat diets set up feedback rhythyms which can result in habitual behaviour. Our reward systems are treated independently in our wiki articles - since I'm in the "unsourced/OR" bit here, I'll point out the large number of neurons comprising the "gut brain" concerned exclusively with feeding.
@Looie - oh yes, undoubtedly quitting smoking is much more than the "likely" answer. The way the OP posed it as a simple choice, it wins hands down. I'm pretty sure the various physiologists who attend this page will agree with me on the risk factors. Trouble is that life is a little more complicated than that. You have to address the risk factors that are easiest to change (IMO) first. To use your terminology, the OP was in a way describing an "inert obese blob" they are very concerned about. They wish to modify someone else's lifestyle and outlook. There's a lot of factors there beyond just presenting "what the people on Wikipedia said". But yes of course, quit smoking! Easier said than done. :( Franamax (talk) 03:24, 19 April 2009 (UTC)[reply]
These activities complement each other nicely. When a heavy-smoking friend of mine started exercising, he found his desire for cigarettes decreasing. Conversely, if one stops smoking, then one breathes better, and might feel more like getting up and exercising! -GTBacchus(talk)02:44, 19 April 2009 (UTC)[reply]
The classic studies are MRFIT (Multiple Risk Factor Intervention Trial) and the Framingham Heart Study, although I don't have access to those studies at the moment. This study included 84,129 women. Relative risks of coronary events:-
Relative risk of coronary events over 14 years in the Nurses' Health Study
Risk factor
Increase in relative risk
Poor diet (quintile 1) vs. good diet (quintile 5)
1.90
Smoking ≥ 15 cigarettes per day vs. ex-smoker
3.5
Exercise < 1 hour per week vs. > 5.5 hours per week
Give up smoking - by a very large margin. Our article Smoking cessation gives a TON of useful information - what techniques for giving up work best, etc. I'll quote this for you because I think it's inspirational for the person doing the quitting:
The immediate effects of smoking cessation include:
Within 20 minutes blood pressure returns to its normal level
After 8 hours oxygen levels return to normal
After 24 hours carbon monoxide levels in the lungs return to those of a non-smoker and the mucus begins to clear
After 48 hours nicotine leaves the body and tastebuds are improved
After 72 hours breathing becomes easier
After 2–12 weeks, circulation improves
After 5 years, the risk of heart attack falls to about half that of a smoker
After 10 years, the chance of lung cancer is almost the same as a non-smoker.
If the person doing the quitting can understand on a day by day basis what's getting better - they'll be more inclined to stick with it.
I don't think you'll get such immediate or profound benefits from either diet or exercise. However, it's very hard to be specific without a lot more information - and in any case, we're not allowed to give that kind of specific medical advice. Probably the very best advice we can give is "Go see a doctor" - they can weigh the pro's and con's with full knowledge of the state of the person involved. SteveBaker (talk) 19:19, 19 April 2009 (UTC)[reply]
Having quit once for 8 years, I can testify that the effects are immediate and profound. Not all of them beneficial, like searching the ashtrays in the basement to find stale old butts and light them up for two more drags, right down to the cellulose acetate in the filter. Overall, yes of course it's good to quit smoking (although I didn't have one-half a heart attack in those years, so I can't confirm your statistic ;). But the little trap in my brain circuitry was still poised, it never went away - 8 years later when a friend offered me a $200 Cohiba at a wedding (I just had to destroy that value, c'mon), the trap was sprung.
My point is that all risk factors exist, but some can be mitigated more easily than others. It's not a trinary choice, it's a matter of what can most easily be achieved in the circumstances. GTB puts it well, an overall approach yields synergy. That's not what the OP asked originally, but they later revealed their overall concern - so I suggest an overall approach. It's what I do with my own mum, she's made her own decision to keep smoking - but she smokes nothing when we go hiking. :) PS: I understand that this is not a scientific response Franamax (talk) 19:50, 19 April 2009 (UTC)[reply]
Actually, my overall concern is not with my mother. I was genuinely just curious. If I try any sort of logical approach with my mother, depending on her mood, she'll either look at me like I'm a joke, look at me like I'm stupid, or look at me like she's never seen me before. Oh - and thank you, everyone, for your epic input. Uhm...I apologise if you were only doing it because you thought it would be in some way functional. :( Vimescarrot (talk) 23:11, 19 April 2009 (UTC)[reply]
O yes, I've googled it first. But many of the links are selling products. I found this one [15].What I want to know is how common it is? And its affect to health. roscoe_x (talk) 08:10, 19 April 2009 (UTC)[reply]
It depends. It will affect your health in two and only two ways. First, it will consume as much of your available income as you are willing to give the peddlers. If you otherwise spend this on Hamburgers and cigarettes, it will improve your health. If the alternative is to pay for and go to a gym, it will usually be detrimental. For the second way it may affect you, see Placebo effect. In other words, it's complete bunk, but should be fairly harmless by itself. --Stephan Schulz (talk) 09:15, 19 April 2009 (UTC)[reply]
There are a bunch of different type of electric field therapy and little evidence for effectiveness for any of them. It's not scientifically implausible that something like this could work, but there's hardly any actual positive evidence at this time. Looie496 (talk) 16:52, 19 April 2009 (UTC)[reply]
(I've got to stop looking at the websites for these things...it only makes me angry). [17] claims to explain how this works. Aside from the fact that this page is one large image - so you can't cut and paste the text (grrrr) and it's riddled with typo's - the actual "science" there is just a bunch of random jargon stringed end-to-end to look impressive. In the section "How the Electric Field Therapy works", we have:
"EFT is a simple, non-invasive technique that uses static electricity...". OK - fact #1 - it uses static electricity.
"The theory is that when an electric filed with high voltage AC is applied on a human body..." - well, firstly it's a "FIELD", not a "FILED" - but fact #1 said they use STATIC electricity. You can't have AC static electricity! Plus, if you apply high voltage AC to the human body (eg by sticking your fingers into an electrical outlet, it tends to DIE - not get better!
"...cells metabolism is stimulated by supplementing ions and the Acid-Base Balance of Electrolyte in the blood results adjusted." ...that's not even a grammatically correct sentence. But what makes you think the pH of your blood is likely to be wrong? There is no science saying that any of the conditions it claims to fix is caused by an error in the pH of your blood - and even if it were - passing AC current at high voltages isn't going to change the pH. Do you really want to take a machine made by people who can't put together a single coherent sentence and have it try to electrocute you? I really don't think you do.
"The H+ flow of ions in the organism can optimally be directed."...er what? Positively charged hydrogen atoms...like in water..."can optimally be directed"...Where? Why? What happens to all of the OH- ions that would be left behind? If you try to adjust the pH by moving the H+'s in one direction than you'll be adjusting the pH wherever the OH-'s would be in the opposite direction. But this is crazy! This is doing electrolysis on your blood! If it actually did this, it would kill you amazingly quickly!
"This is important because beside others, the co-ordinates of enzyme, the Ca++ household, the proteins of transportation in the blood and the general acid-base-homeostasis depend on these flows of protons and ions." - OK - the only way to write a sentence with so little meaning is to open a dictionary of scientific terms and write them down at random. There is no way to even try to analyse what this sentence means!
It gets worse, the company claims to have "ISO9001" certification. I've actually been through the certification processes for those standards and they are pretty hard to get - they relate to how products are designed and documented and represent the result of frequent inspections of your processes, etc. However, when you click on the image of the "certificate" - it's not an ISO document - but something awarded by some unheard-of Chinese organization. It's certainly not a proper ISO-9001 certificate.
This is so very bad...it's beyond ridiculous. PLEASE - don't pay them a cent of your money.
Do you mean by the body or in a laboratory? In the body it would be some kind of white blood cell which would then present a viral protein to another type of white blood cell which would then produce antibodies against it.
Thanks. The question is about how viral detection/identification can be performed in a lab context. What if the virus is unknown, or new? Would a clinical lab, say one at a hospital, be equipped to detect and confirm a new form of viral infection? Or is that only possible at a very well-equipped research lab? 173.49.18.189 (talk) 14:52, 19 April 2009 (UTC)[reply]
Hospital labs are equipped to detect known pathogens. Novel viruses would not routinely be screened for. The detection of novel viruses is a huge topic - there is no simple way to do it. Some viruses have been detected by looking for cytopathic effect in tissue culture. The hepatitis C virus was known to exist (as "non A, non B viral hepatitis") for more than a decade before Michael Houghton's group at Chiron discovered it by screening over a million cDNA clones in a phage library using serum from infected people to detect viral proteins. Today, techniques like random-primed PCR (with random-primed reverse transcription to detect RNA viruses) are available, but far from perfect. Of course once you've detected a virus, showing that it causes a specific disease is another challenge - we'll all loaded with viruses. --Scray (talk) 18:39, 19 April 2009 (UTC)[reply]
Photovoltaics and thermodynamics; ie, efficiency vs 'wasted' energy
Some considerable pondering hasn't cleared this up, and neither have I been able to find reasonable info in any of the articles I looked into, nor any of the archived reference desk questions some searches turned up.
Imagine a photo cell exposed to monochromatic photons are at the precise energy (ie, wavelength) which exactly matches the optimum energy for absorption in the cell; this is, I understand the band gap energy in traditional photo cells (eg, silicon). We will thus, be generating no phonons carrying 'excess' energy from the absorbed photons, there being an exact 'fit'. Phonon generation (energy therein will be wasted w/regard to electricity output) will be at least minimized. As nearly as I can make out, absorption and emission of photons being quantum mechanical events, there is no energy "leak" which eventually ends up in increased temperature of the photocell or attached structure.
It seems to me to be an exception to the usual energy conversion stricture which inherently includes some losses to the energy uber despot, entropy.
One thing that you've missed is recombination. You might absorb every incident photon with no losses, but the resulting charge carrier pair might not make it to the external circuit. See Quantum efficiency of a solar cell. I don't know if this is the main method by which entropy gets its revenge, but it's certainly one of them. There is also the fact that "no real source of electromagnetic radiation is purely monochromatic" (according to Monochrome). --Heron (talk) 11:30, 19 April 2009 (UTC)[reply]
(ec) I am not sure that the second law of thermodynamics prevents 100% conversion of energy from one form to another. What it does do is prevent 100% conversion to useful work. This will certainly come into play once you start trying to extract current from your cell; if through nothing else the joule heating of the circuit. In any case 100% conversion of photons is not practically possible. Some of them will hit the crystal lattice and become phonons, some will be re-emitted before the electron can be made to do anything useful, all real crystals have flaws which will lose some of the photons etc etc. SpinningSpark11:41, 19 April 2009 (UTC)[reply]
It does not violate thermodynamics because a pure plane wave of zero frequency width has zero radiation entropy, thus, there is no maximum possible conversion efficiency below 100%. The opposite extreme would be blackbody radiation at the same temperature as the cell itself, having maximum radiation entropy, none of the energy could be used to create work. Radiation obeys the laws of thermodynamics just as matter does. 129.2.43.42 (talk) 16:06, 19 April 2009 (UTC)Nightvid[reply]
My thanks all. It may be that the blinders of ignorance have loosened their hold slightly. My thought experiment was deliberately constructed to evade most practical issues and focus (ahemm..) on what seemed to be the key question. So the objection that there can't be a monochrome source is skew to the hypo; and likewise inefficiencies (ohmic and otherwise) during use of the electricity generated to something helpful (like running my automated beer fetching and opening robot, say). I was attempting to exclude such issues, mostly because they're as obiter dicta and so obscurantist. Recombination does indeed seem be something I'd not considered. Easy to miss, not being one at home in the labyrinthine complications of doped and dopey micro lattice structures, so I don't feel as terminally thick as I probably should. The anon contribution by 129.2... may be the very answer I was seeking, but alas I fell off the wagon as it made the turn into pure plane wave territory... In English this historian type can cope with, please...? ww (talk) 18:06, 19 April 2009 (UTC)[reply]
It would be great if they really were "energy saving" but I think that would contradict the second law of thermodynamics. They are compact fluorescent lamps - I think this should help. Incidentally DYK that it's best to leave them on if you leave the room for less than 15 minutes because switching them on/off wears them out and could end up making them less efficient that incandescent light bulbs. Smartse (talk) 11:04, 19 April 2009 (UTC)[reply]
They are energy saving as you get the equivalent of 60w worth of light from an incandescent bulb for only 11w. The packaging of the 11w (230-240v) bulb I have says it gives 600 lumen, and is supposed to last 10000 hours. 89.242.147.172 (talk) 11:23, 19 April 2009 (UTC)[reply]
I don't believe CFLs are likely to be less efficient then incandescent bulbs if turned off in less 15 minutes. They do wear out a lot that way however and so in overall energy consumption and pollution terms as well as cost to the consumer terms, leaving them on is sometimes better then turning them off. See [18] for example Nil Einne (talk) 13:22, 19 April 2009 (UTC)[reply]
It's important to be clear on this: CFL's have a MUCH longer lifespan than incandescents when used in "normal" ways. They are vastly more efficient because they generate much less heat - so the laws of thermodynamics are not violated! They are more expensive - but pay for themselves in a very short amount of time because of the energy savings and the increased lifespan. The only significant drawback is the mercury they contain - which is a nasty pollutant. However, they use a lot less mercury than old-fashioned 'striplight' flourescents - and right now, properly recycling the darned things - or cleaning up mercury pollution afterwards are a LOT easier than dealing with global warming caused by the wasted energy from incandescent lights. SteveBaker (talk) 18:24, 19 April 2009 (UTC)[reply]
Steve, we must disagree on what "normal" use is. To me, it's switch it on when you need it, then switch the damn thing off when you don't. I put a CFL in my bathroom and it was toast within about four months. The incandescent in the fixture beside it is still working two years later. It operates maybe 30 minutes a day, but gets cycled regularly (I work at home and drink a lot of water). Do the math on that and consider that it takes 9 cubic metres of landfill soil to neutralize the mercury emitted if you throw the bulb in the trash (which I don't, but many people do). Furthermore, during heating season, six months of the year, I save NO energy by having a more efficient light. All that happens is that the other energy source comes on. If that happens to be fuel oil in a furnace vs. hydroelectricity, I'm net-subtracting from combatting global warming. (Mercury emitted from new dam reservoirs is a wash vs. CFL bulbs).
If your definition of "normal" is leaving a light on all day in an air-conditioned house, unquestionably a CFL is better. Whereas I call that aberrant behaviour! :) Franamax (talk) 20:46, 19 April 2009 (UTC)[reply]
No matter ho much water you drink, that CFL should have been ok. You cannot generalize from a single fluke. Mine are warranted for 3 years minimum, and I have so far (3.5 years since I switched) had one dud after 3 weeks - and that was replaced without any discussion. As most electronics, CFL failure will tend to follow a bathtub curve, so a few early failures are to be expected. --Stephan Schulz (talk) 21:06, 19 April 2009 (UTC)[reply]
In contrast to the CFL sitting beside my couch, which has operated reliably for long durations for the last two years? I don't think it was a dud, although it's certainly possible that when you open a two-pack of bulbs one will fall into the bathtub :) I disagree, current switching is a known factor which shortens the life of CFLs compared to incandescents. My habit, ingrained in childhood, is to switch off a light when you don't need it. I find the suggestion that I should leave the light on, consuming energy, in order to save energy - paradoxical.
You're right in a way, I should have kept my receipt, gone back to the store, complained. They would have replaced the bulb and thrown the old one into their garbage. I've researched this a lot - switching and enclosed fixtures are failure modes for CFLs, or at least were. The various commentary available covers several generations of bulb, so I'm not sure. What I do know is that I use CFLs everywhere I find them appropriate. In the bathroom, where they are switched 10-15 times a day - no, I don't see the math that tells me I should spend more for no net energy/CO2 saving and use mercury in the process. The one in the living room, definitely. The bathroom, nope! Franamax (talk) 21:33, 19 April 2009 (UTC)[reply]
Well, our article claims (with sources) that a 5 minute on/off cycle reduces CFL lifetime to about that of an incandescent lamp, or alternatively, that one on/off cycle is equivalent to about 6 minutes of lifetime. With a rated lifetime of 6000h (the lower limit) they should be good for 60000 cycles, or about 10 years - half that if you also actually leave them burning 6 minutes (which also factors into their lifetime, of course). Even if the numbers are somewhat optimistic, it looks very much as if you hit the front side of the bath tub.... --Stephan Schulz (talk) 22:20, 19 April 2009 (UTC)[reply]
I have the feeling that they do flicker, even if I can only detect it subconciously. Although I use them they do not seem as nice as incandescent bulbs, but do feel like flourescent lighting which I do not like. I'd like it if someone could provide the evidence to prove me wrong. 89.242.147.172 (talk) 11:48, 19 April 2009 (UTC)[reply]
There are many different kinds of CFCs, with widely varying properties and prices. I use the Phillips Softtone, which after warming up looks much like an incandescent to me. It did cost me around EUR 6 or 7 per bulb, IIRC. Cheaper bulbs will quite possibly have worse performance, both in light colour and in flickering. --Stephan Schulz (talk) 12:39, 19 April 2009 (UTC)[reply]
Note it is important to consider what you mean by 'worse performance'. For example, everything else being equal, cool coloured fluorescents (cool daylight for example) generally have better colour rendition (see colour rendition index) then warm ones. (The phosphors uses makes a big difference however whatever the colour temperatures). However in Western cultures, warm coloured lights which are closer to incandescent lights are generally preferred particularly for living spaces and are sometimes considered more natural (even though cooler lights are far closer to the epitome of natural light, the sun and bright daylight). From sources I've read and anecdotal evidence these preferences don't hold for Asian cultures however where preferences are more varied and often tend to the cooler end of the spectrum. If you can only detect something 'subconsciously' it could easily be simply that you want flourescents to be bad so they are. Nil Einne (talk) 13:42, 19 April 2009 (UTC)[reply]
That's all very true - but since you can buy CFL's that match either incandescent light's color or natural daylight color - there is no way that 'color' is a good excuse for not buying them. The fact that you get the choice is a huge plus. SteveBaker (talk) 18:24, 19 April 2009 (UTC)[reply]
I seem to get a peculiar feeling on my skin when in a room illuminated by flourescent light - as if it were something to do with static electricity perhaps. Perhaps its just my imagination. 89.242.147.172 (talk) 14:06, 19 April 2009 (UTC)[reply]
Well if you get a peculiar feeling on your skin, then that's easy to test in a double blind fashion. You leave the room and someone throws a coin, or uses a better random method to decide whether to put an incandescent or CFL in the room. The person then leaves the room. 5 minutes later (well a preset interval so there is no need for contact between the two of you), you come in to the room well blindfolded so you can't possible see what light is present. Repeat multiple times (I would say at least 20.) If you are able to detect reliably whether CFLs or incandescents are present, then we know you are somehow detecting CFLs, perhaps because of a peculiar feeling on your skin. Nil Einne (talk) 14:11, 19 April 2009 (UTC)[reply]
The trouble is that it may be the color rather than flicker or (the extremely unlikely) skin tingle that would clue the person in to which is which. CFL's exist that are a pretty good match for incandescent color - but they aren't 100% perfect. Controlling for color in your experiment is much harder. SteveBaker (talk) 18:24, 19 April 2009 (UTC)[reply]
One last comment here - I regard CFL's as an emergency measure. Incandescents are such a huge problem for global warming that we simply MUST get rid of them as quickly as we can. In many places around the world, laws are being passed to outlaw these ridiculously inefficient devices...and incandescent light bulb factories are gradually switching production to CFL's. They simply have to go - whether you like it or not. However, LED lighting is the true way of the future. LED lights have an amazingly long lifetime - they are yet more efficient than CFL's - you can dim them nicely (tricky with CFL's) and you can even get them where you can adjust the color they produce to whatever your heart desires (I have one of these). Because they last so long, whatever pollution they create when they eventualy DO die is relatively unimportant because we won't be throwing them away so often. The only limitation on them right now is that the really bright ones are pretty expensive. We're seeing them appear in things like car tail-lights and traffic signals where the consequences of failure and the cost of replacement are high. This year, I think there were more LED Xmas tree lights on sale than incandescents in local stores here in Texas. I have a really fancy tri-color LED "bulb" in my home-office area (you can actually turn it on and off and set the color remotely from your computer using X10 protocol over the power lines!) - it was really just a hideously expensive gimmick - but it does work. It's plenty bright enough - and you can adjust it to produce "warm" lighting or "cold" lighting depending on your mood (you can also make it flash in various colors in time with music on your PC if you're into 1970's disco lights!) But as soon as they get cheap enough (and they undoubtedly will), I plan on switching over my home to slightly saner LED lights completely. SteveBaker (talk) 18:24, 19 April 2009 (UTC)[reply]
I look forward to getting something similar to sunlight rather than the yellow gloom we are used to - although on the other hand it does signal the evening and hence probably helps you prepare to sleep. 78.146.75.232 (talk) 19:54, 19 April 2009 (UTC)[reply]
The great flaw of CFLs is that it costs about $1 (U.S) to recycle a CFL. An incandescent bulb can be purchased for less than 50 cents (US). A CFL can be purchased for about $3 or less with occasional rebates. Most consumers will not pay $1 to recycle a CFL, so they go into landfills and the mercury pollutes the environment. This is a neglected externality of the "bulb of the future." Edison (talk) 21:23, 19 April 2009 (UTC)[reply]
You're repeating some common, misleading arguments. There are organizations in many areas that recycle CFLs free of charge, including The Home Depot. You can also get mercury-free ones if that's for some reason not an option. If you get your power from a coal plant, the reduced power requirement saves more mercury from entering the environment than is actually contained in the light. Decent CFLs last longer than incandescents and obviously take less power, so your point on price is weak. -- Consumed Crustacean (talk) 21:30, 19 April 2009 (UTC)[reply]
How long do American electric kettles take to boil?
I'm used to 220volts, but in the US you only have 110volts. Does that mean that American electric kettles take an extremely long time to boil water? And are electric room heaters unknown in the US? I was reading the BS 1363 article and its discussion. Edit: I've just tried boiling one litre of tap water in a 240v electric kettle, and it took 2m 45s to boil, and exactly three minutes before it turned itself off automatically. 89.242.147.172 (talk) 11:30, 19 April 2009 (UTC)[reply]
In principle, you can get the same amount of energy independent of the voltage - you just draw a higher current. I would expect US equipment to be made to the relevant specifications. In other words, a simple heating device designed for 230 volts will work in the US, but will be much less effective. A device designed for 110 volts will likely fail, possibly in spectacular and dangerous ways, if connected to 230 volts. More complex devices will run into trouble either way, although many modern electronic devices are built to accept 100-240 volts, 50-60 Hertz, to be usable off nearly any mains supply in the world. --Stephan Schulz (talk) 11:58, 19 April 2009 (UTC)[reply]
American plugs are rated at 15 amps so the maximum power that can be extracted is 1650 watts. Poking around on US eBay their kettles seem to be generally in the 1000 to 1500 watt range, although I did find one that claimed 1750 watts and can boil seven cups in 5 minutes. By contrast, British plugs are rated at 13 amp and kettles are usually in the 2000 to 3000 watt range which is considerably more. Before anyone points it out, I know that 220 volts x 13 amps is less than 3kW but the British supply is not really 220 volts. It used ot be 240 volts, when we signed up to European harmonization at 230 volts (edited) the only thing that changed was the specified allowed variation in voltage! Don't believe everything you read on labels. So in summary, American kettles will boil slower, but not extremely slower as you state. SpinningSpark12:16, 19 April 2009 (UTC)[reply]
In NZ, the maximum current allowed by our normal plugs is 10 amps. Looking at my kettle which is the cordfree kind, it's rated at 2200-2300W. Nil Einne (talk) 13:18, 19 April 2009 (UTC)[reply]
I'd also point out that electric kettles are relatively rare in the US. When we wanted to buy one, we had quite a hard time of it. They aren't things that are ubiquitous in every household...and indeed the one we have has a low capacity and is incredibly slow compared to the ones we've owned in the UK. Most of the time, if we want boiling water, we use either the microwave or a pan on the stove - the former is faster for small quantities - the latter is faster for large quantities because the kettle is so small. There are similar issues with many other appliances. Mains-electric lawnmowers, for example, are relatively pathetic affairs compared to the ones we had in the UK - so relatively few people have them. Whether 110v or 240v was the better choice is a tough one. 110v is undoubtedly much safer around the home than 240v - but the hassle of having to have special multi-phase 220v outlets for running things like washing machines and hot-tubs is a real pain. I've recently been looking to buy an air compressor for running pneumatic tools in my garage - the 110v ones are crap but having to wire a special 220v outlet just to run the darned things is a real pain. I suspect that when electric cars and plug-in hybrids become common - the UK will use three-phase 480v charging outlets - and here in the US, we'll be stuck with 220v...which is going to be painfully slow. SteveBaker (talk) 18:01, 19 April 2009 (UTC)[reply]
That explains (partly) why they dont drink tea in the US - they cannot boil the water. They must just have coffee percolators. Makes me wonder how they "invented" instant coffee. Using an electric stove for cooking must be difficult too. 78.146.75.232 (talk) 19:47, 19 April 2009 (UTC)[reply]
I doubt that 100V is much safer than 240V (or 230V these days, +/- enough that it makes no difference). It's probably a bit safer, but is it really "much safer"? Especially since UK appliances will generally draw less current than their US equivalents (most appliances don't use anywhere near the maximum power available, so the UK and US ones will presumably use the same power) - current can be just as important as voltage when determining risk from electrocution. As for electric cars - UK charge points in public car parks, etc., might use 3-phase, but ones in private garages at people's homes won't - individual houses aren't generally connected to a 3-phase supply. It would probably be easier to increase the current than the voltage - the 13A limit is determined by the wiring inside the house, a separate high-current connection to the mains supply could be made (I don't know what the maximum current that can be drawn by an entire house is, but it is clearly much greater than 13A, although perhaps less than 13A times the number of sockets in the house). --Tango (talk) 19:05, 19 April 2009 (UTC)[reply]
First, the correct nominal North American voltage is 120, not 110 or 100. As to safety, 110-120 V is slightly better than 220-240 V in terms of shock hazard but slightly greater in terms of fire hazard since higher currents are required. However, a factor of about 2 is only going to matter if a shock or overheating incident is right on the boundary between serious and not-serious.
To match the original poster's original research, I just measured a liter of water into a Canadian 120 V electric kettle and boiled it. This kettle is old enough that it doesn't have an automatic shutoff and if it ever had a wattage rating displayed it's illegible now.
But it took between 5 and 6 minutes to reach a full boil, and I think you'll find that'd be typical today. Although the circuits are rated for 15 A, nobody makes appliances that draw a full 15 A because one other thing on the circuit would trip the breaker. The tap water going in would have been between 5° and 10°C, so, taking the mid-range of those numbers, the current draw would be 92.5/(120×5.5) kilocalories/volt-minute = just about 10 A. I'd be surprised if you can get a significantly more powerful one in North America.
--Anonymous, 19:40 UTC, edited 19:45, April 19, 2009.
Electric space heaters in the U.S. designed to operate on 120 volts only draw about 12.5 amps. so they only put out about 1500 watts of heat, enough for a small room. 240 volt heaters are also available, which can heat a larger room. 120 volt electric kettles in the U.S are sadly underpowered and do an inadequate job of boiling water compared to those running on 220 or 240 volts in the rest of the world. The inadequacy of 120 volts is the result of being the pioneer country in having central station electricity for heating, lighting and power. Poor insulation on the available wiring in the 1880's and a concern for safety from electrocution resulted in a distribution voltage lower than desirable in the light of 2020 hindsight. Edison (talk) 21:18, 19 April 2009 (UTC)[reply]
HIV
I found out recently that in many countries, men who have sex with men are nto allowed to donate blood because they are in a higher risk group for HIV. But I don't see what the problem is; surely as long as you're tested and you're negative then you're fine to donate blood. What else is there to it? Thanks 92.7.20.157 (talk) 11:50, 19 April 2009 (UTC)[reply]
The problem is simply that the tests aren't 100% reliable (especially in the early stages of infection). There is a risk with any donor, but the risk with men that have had sex with men is greater and someone has determined that the former risk is acceptable and the latter isn't (probably because you can exclude men that have had sex with men without significantly reducing blood stocks, so there is very little cost and a significant benefit). --Tango (talk) 13:23, 19 April 2009 (UTC)[reply]
We have an article MSM blood donor controversy. As has been mentioned no test is 100% effective. In particular many of the older tests had large windows where a HIV would not be detected, however the new tests common in developed countries are better and have a shorter window. The bans vary from country to country but as our article notes, for a number of reasons the bans are sometimes controversal, particularly when it's a lifetime ban even more so when other high risk groups such as sex workers and intravenous drug users don't receive a similar ban. There are also questions raised about whether all MSM should be considered in the high risk group or whether some may be lower risk then people who are currently allowed to donate. [19][20][21]Nil Einne (talk) 14:06, 19 April 2009 (UTC)[reply]
Am I likely to have any of the water molecules from this iceberg in me now? How many? In the same way that my next breath is said to contain molecules breathed by Julius Caeser, the dinosaurs, and so on. I've been wondering what volume six billion water molecules woould have, if everyone in the world has one molecule from the iceberg in them. I wantched a tv programme last night (Channel 4, UK) that said the iceberg was ten times the size of the Titanic. Incidently, some of the details given in the programme were different from those given in the Wikipedia articles. 89.242.147.172 (talk) 13:55, 19 April 2009 (UTC)[reply]
If one assumes that the Titanic iceberg was only 10 metres on a side (and it was certainly larger than this), it contains 103 metric tonnes of water whereas the total water in the oceans is 1.3 x 1018 tonnes. Assuming this is all now thoroughly mixed we can expect this proportion of Titanic iceberg water to be in an average (say 100kg) human. Dividing by the molecular weight of water (34 u) gets about 1012 (one trillion) molecules per person. SpinningSpark14:59, 19 April 2009 (UTC)[reply]
Actually, there are even more than that, Stephan Schulz has pointed out that the molecular weight of water is 18, not 34, but was kind enough not to write my mistake on the page. SpinningSpark17:21, 19 April 2009 (UTC)[reply]
Ok, I did misinterpret you. I thought the 1012 was the number of water molecules in a typical person, rather than the number of water molecules from the iceberg in a typical person. (Had I thought about it, I would have realised that was far too small a number...) --Tango (talk) 16:39, 19 April 2009 (UTC)[reply]
(ec)Well, 1 g of hydrogen has 6E23 atoms, or 6E14 (600000000000000) per person. For H2O, or water, 18g holds as many molecules. Titanic displaced about 50000 tons. Depending on what "10 times means", the iceberg should have between 500000 tons and 50000000 tons. You do the math ;-). The other way round, 6 billion (US reading, i.e. 6E9 ) water molecules is about 1E-17 mol, or 0.000000000000000018 g (give or take a few zeros if I miscounted). --Stephan Schulz (talk) 15:03, 19 April 2009 (UTC)[reply]
I'm not sure this is a meaningful question - in a body of liquid water the molecules of H2O and constantly dissociating into H+ and OH- ions and recombining (but not necessarily in the same order), so the same molecules don't exist for long. I'm not sure how long a typical water molecule lasts for, but it's probably much less than the time since the Titanic was hit - although actually, you need to measure it from when the iceberg melted, which could have been at least a few years later. --Tango (talk) 15:12, 19 April 2009 (UTC)[reply]
Yes, you could, but that is a different question and could have a different answer. It's also a harder one to answer - I'm not sure how to determine how much oxygen or hydrogen there is in circulation - lots of it is locked up in the crust, mantle and core of the Earth, but the upper layers of the crust are involved in the mixing going on, so it becomes very difficult to work out where to draw the line. --Tango (talk) 16:39, 19 April 2009 (UTC)[reply]
Well, actually, the answer is easy, it's "yes". The difficult question to answer is "how many", that there is at least 1 is easy enough to show by taking extreme estimates. --Tango (talk) 16:52, 19 April 2009 (UTC)[reply]
The really difficult question here (aside from all of the annoying nit-picking) is whether sufficient time has elapsed for the atoms from the iceberg to have mixed sufficiently with those in the rest of the planet to make it reasonable that one or more would have made it into your body. The more recent the event, the less likely that is. So when you pick the more common choices like water that Julius Caesar peed - then the odds are much higher than for someone who was around in the 1990's. The Titanic iceberg only melted something like 100 years ago - so I doubt that it has been completely mixed into the rest of the water on the planet to the degree that Julius Caesars excretions have. But the efficiency of that mixing is undoubtedly the key here - and I'm not aware of any way to measure or even estimate the degree of that. If all the water in the world were not being swirled around by tides and such - I suppose you could use the Maxwell–Boltzmann distribution and the Random walk math to make some kind of measure of the rate at which the molecules would propagate outwards...but that would neglect other water transport mechanisms like tides and clouds being pushed around by the wind...it's a tough thing to attack convincingly. SteveBaker (talk) 17:45, 19 April 2009 (UTC)[reply]
Well, we can make a rough guess. [22] lists the self-diffusion constant of water as 2.2x10-9 m2/s. Using Fick's_law#Example_solution_in_one_dimension:_diffusion_length, with the concentration at position x=0 of 55.5 mol/L (the concentration of pure water), we find that at a range of 20,000 km (half the circumference of the earth) and at 97 years the concentration would be 55.5 mol/L * erfc(3853500). Now erfc(10) is about 2x10-45 (by [23]), and a larger parameter only makes the value less. (Note that extending the time to 2 thousand or 65 million years doesn't appreciably change things. Even at the age of the universe at 14 billion years we're still only at erfc(320).) Working in the other direction, in 97 years diffusion results in a concentration of 1 molecule/L at a distance of 38.9 meters (65 million years is 32 km). - I'd say that diffusion would play only a minor role, and other transport mechanism are much more important. -- 128.104.112.117 (talk) 23:28, 19 April 2009 (UTC)[reply]
Is there a cartographic projection such that if you cut it in half along the equator, rotated one half 180 degrees, and overlaid it on the other half, all points would be overlaying those directly opposite them on the globe? I've got a kid here who wants to dig holes through the earth and would appreciate this sort of visual reference. ;-)
I am no expert on this subject but I very much doubt that there is a named projection for this. I certainly cannot find anything even remotely like that in my textbook (Steers) on map projections, nor does our article map projection. Of course, such a projection is possible, just about any surface is possible to project, it just hasn't been done. Can I suggest you download Google Earth which may help your youngster to visualise this by manipulating the globe? SpinningSpark15:39, 19 April 2009 (UTC)[reply]
No, certainly not without some bizarre discontinuity or pathology. This is forbidden on account of parity within the surface. A pure rotation keeps left turns left turns and right turns right turns. Yet if you travel on the surface of the Earth and turn right, a person remaining opposite you will make a left turn. 129.2.43.42 (talk) 15:58, 19 April 2009 (UTC)Nightvid[reply]
Ah, just beaten to it! I don't think there's a specific projection name for a map that shows antipodes. It's not really a projection per se, but a way of overlaying other projections. --98.217.14.211 (talk) 17:07, 19 April 2009 (UTC)[reply]
After looking on the article about antipodes, I found it amusing that there is almost no overlap of land on land in those superposed maps. Dauto (talk) 22:19, 19 April 2009 (UTC)[reply]
Decay chain
Hi, what are the equations used to calculate how much of each substance there will be at a certain point in time in a decay chain? Thanks, 96.255.93.227 (talk) 19:49, 19 April 2009 (UTC)[reply]
Why does the universe seem to be older than it is?
If I look in one direction using a powerful telescope I can see objects that are 13 billion years old. Now if I look in the opposite direction, I can also see 13 billion years old objects, which means that these objects are separated by 26 billion light years. However the age of the universe is only 13.7 billion years old, so how come the distance between these objects appears to be bigger than the universe itself? Laurent (talk) 21:54, 19 April 2009 (UTC)[reply]
If an explosion travels at lightspeed, its radius will be (lightspeed x time). Its diameter (which is what you're measuring) will be twice that. Vimescarrot (talk) 21:56, 19 April 2009 (UTC)[reply]
That doesn't answer your question. Never mind. It's something to do with relativity and the different perceptions of time and...and...and someone else will come along and correct my mistakes and explain it better, so I'd be better off not trying. Vimescarrot (talk) 21:57, 19 April 2009 (UTC)[reply]
The real reason is the metric expansion of space. Things cannot move faster than the (always constant) vacuum speed of light through space. But it's space itself that is expanding. Consider, as an example, a rubber balloon that doubles in diameter every second. Even if ants living on it are constrained to 5 cm/second, the distance between two given points on the surface will eventually increase much faster, and two ants at these points will move apart with the same speed even if they don't move with respect to the surface at all. For concrete numbers, assume the balloon has a diameter of 10 cm, and the ants are 1 cm (measured along the surface) apart. After 1 second, the balloon will have a diameter of 20 cm and the ants will be 2 cm apart. After 3 seconds, the balloon will have 80 cm, and the ants will be 8 cm apart. After 4 seconds, the balloon will have 160 cm, and the ants will be 16 cm apart - oops! The distance between the ants now has increased 8 cm in one second (and even weirder, they think they have not moved at all).--Stephan Schulz (talk) 22:34, 19 April 2009 (UTC)[reply]
As the last poster has pointed out the distance between objacts in the universe is expanding almost exponentially, so that distant galaxy that emitted some light 13.7 billion years ago that you can see now has been moving away from us all that time at ever increasing rates and is much further away now than 13.7 lightyears. (more like 40 lightyears if memory serves), and twice that distance if you compare galaxies in oposite sides of the sky. Dauto (talk) 22:42, 19 April 2009 (UTC)[reply]
![{\displaystyle r\;=\;k\left(T\right)\left[A\right]^{m}\left[B\right]^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b224483895f79ac334d0887e6a3e71b51c262850)
