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September 10

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identical location

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How will astronomers be able to take pictures from the exact same location 100 years from now for comparison purposes? -- Taxa (talk) 00:45, 10 September 2009 (UTC)[reply]

Clarification: In the documentary "Extreme Ice" the way the exact location of the camera was maintained was to fasten it to rock above the ice flow. This was probably okay for showing what happened each hour for over a year and maybe even in the event of an earthquake because the field of view is so wide. Doubtful even an institution would have use for lengthier time lapse but then we still spend a lot of money to drill holes in the ice to get data that goes back lots of years. I'm speaking theoretically in the event someone wanted data that required the exact location from the object being recorded. -- Taxa (talk) 01:11, 10 September 2009 (UTC)[reply]
I'm sorry, your question is not very clear. What do you mean by "location"? And do you mean "How will" or "How do"? "Will" implies the future, and Wikipedia is not a crystal ball. However, if you mean how astronomers are able to take pictures of celestial bodies as they were 100 years ago, it is because even light does not move instantly. Light moves at a constant rate. The reflected light off of a celestial body may have been reflected off of the body 100 years ago, but is only reaching earth now. Therefore, astronomers looking through telescopes see the celestial body as it was when the light reflection they are viewing was reflected off of its surface. For example, if the sun suddenly disappeared, we, on Earth, would still see the sun for about eight minutes after it actually happened. Intelligentsium 00:58, 10 September 2009 (UTC)[reply]
At a distance of 100 light years an astronomer almost certainly sees light emitted not reflected from a celestial bodies, and that is definitely so in the case of viewing our Sun. Cuddlyable3 (talk) 11:44, 10 September 2009 (UTC)[reply]
I think the objects they are comparing are so far away that an area in the general vicinity will suffice. DRosenbach (Talk | Contribs) 01:00, 10 September 2009 (UTC)[reply]
(ec)I'm thinking this has more to do with a question of whether it is possible to take a comparison picture in 2109 at exactly the same location in space, relative to the sun, given the somewhat eccentric and very slightly chaotic earth orbit. I'm wondering of how much importance to comparative astronomy is the change of location of the camera base due to earth orbit. --Tagishsimon (talk) 01:01, 10 September 2009 (UTC)[reply]
The problem is really half a million years from now, what with the sun's orbit around the galactic center and whatnot. John Riemann Soong (talk) 02:21, 10 September 2009 (UTC)[reply]
Obviously, without a lot of expensive and impractical spacecraft, it's going to be impossible to get perfectly repeatable positioning. However, the size of the error due to the earth going around the sun - and the sun moving around the perimeter of our galaxy are very small indeed compared to the distance between stars. Stars in our local group are moving at almost the exact same speed as us around the edge of the galaxy - so the closest stars are moving the most slowly relative to us. Their relative positions will change very little over very long timescales. For stars at much longer distances, the amount of change compared to the distance between us means that those stars move VERY little relative to us. Take some hard numbers: The sun is moving at about 20 km/second relative to our neighboring stars. Over 100 years, the sun (and therefore the earth) will have moved 100x356x24x60x60x20 kilometers - that's 63 billion km - about six lightyears relative to the other stars in a region perhaps 20,000 lightyears across. So most of those stars won't move more than one part in several thousand compared to us...that's measurable - but not huge. The entire galaxy is of course spinning - we're going to have moved about 60 light years over the next 100 years compared to more distant galaxies...our galaxy is moving at about the same kind of speed relative to the other galaxies closest to us - but now we're talking about looking at objects like other galaxies...those are at distances measured in the millions of lightyears. Truly, our motion over 100 years is negligable to their distances. SteveBaker (talk) 02:26, 10 September 2009 (UTC)[reply]
You may want to check those numbers - a light year is just shy of ten trillion kilometers, so we'd move less than a hundredth of a light year in a century. If we take the diameter of the solar system as twice the semimajor axis of Neptune's orbit (screw you, Pluto, and your Kuiper belt buddies, too!) 63 billion km is only seven times the width of the solar system. That's enough that you could see slight shifts in the apparent positions of a few of the nearer stars, but even the nearest stars are almost a hundred times that distance from us. TenOfAllTrades(talk) 02:38, 10 September 2009 (UTC)[reply]
Yeah - that didn't seem right...oh - I see what happened. I slipped between meters and kilometers halfway through the mental arithmetic! Never trust a software guy without a calculator! Thanks for the double-check! Anyway - the point is made - we move a TINY distance compared to the distances to the objects we're observing. SteveBaker (talk) 02:49, 10 September 2009 (UTC)[reply]
Just to be pedantic, "local group" usually refers to our local group of galaxies. The nearby stars are in the Local Interstellar Cloud or Local Bubble or Local Spur (depending on the scale you want). (I know you were using the words descriptively, but this is ref desk - pedantry is what we do!) --Tango (talk) 13:36, 10 September 2009 (UTC)[reply]
If by location you mean taking a picture of the night sky from the same latitude and longitude, then precession and perhaps crustal movement will likely have more of an effect on the placement of the stars than the solar system's movement through the galaxy. There's also the proper motion of the stars themselves. ~AH1(TCU) 23:01, 11 September 2009 (UTC)[reply]

Preventing/Dissuading housecats from slaughtering houseplants...? Hizzelp!

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So I've got 4 wonderful little housecats and a ~100sqm apartment (a bit crowded, yes, but they are all street-rescues from urban China). Until recently, I've had houseplants scattered around on high shelves and the occasional wall-mounted planter or hanging basket, but what I'd really like to do is green the place up to the max... which would require plants be placed on counters, tables, and the like.

To date, whenever my cats have been presented with an opportunity to reach one of my plants (say, by careless placement of a chair or box) they have proceeded to go all Paul Bunyan / Mike Mulligan & Mary Anne.

Has anyone had success deterring this behavior? The cats already know it's not acceptable, so only do it when I'm sleeping or at work. Is this driven by the rarity of the occurrence? In other words, if they had a wide selection of targets would the urges eventually be satisfied? Or do I need to be more proactive?

Thank you! 218.25.32.210 (talk) 01:33, 10 September 2009 (UTC)[reply]

Start with cacti? --Tagishsimon (talk) 01:38, 10 September 2009 (UTC)[reply]
I'd be thinking boredom as the first problem, too many cats, small place. What other interesting things are there for cats to do at your place? If they can't pursue their traditional activities (chasing/hiding/jumping), they will do other and weirder things.
As for deterrence, cayenne pepper sprinkled onto the soil? Water an ashtray and pour a bit of the foul water onto the plant (bet they've never attacked an ashtray!) I think you need to solve the boredom problem first though. Franamax (talk) 01:53, 10 September 2009 (UTC)[reply]
Oh yes, if the cats really do know that wrecking your plants is unacceptable - and think about it, did you really scold the right one at the right time, so that they learned the right lesson? They don't always have very good memory and understanding of causality. The cats themselves form a peer group, so they will be watching each other to see who gets awsy with what. Anyway, when a cat really does think it has got your measure, then specific destructive acts are usually commentary on what the cat sees as the latest thing you did to piss it off. In other words, the cats are engaged in a power struggle with you, so it's best to appease them rather than punish them. ;) Franamax (talk) 02:14, 10 September 2009 (UTC)[reply]
If they're chewing them up, there are sprays made from bitter melon that might work. These should be non-toxic unlike some of the other bitter sprays, so it should be safe for both the cats and the plants. I've never used them myself, but was actually thinking about trying it on electrical cords as a precaution. If they're urinating in the pots, then I've read in many books that covering the dirt with marbles work well. --Wirbelwindヴィルヴェルヴィント (talk) 04:28, 10 September 2009 (UTC)[reply]
Cats actually need to eat grass. They are obligate carnivores, but because they clean themselves, they need to induce vomiting and so they need to eat grass in order to do this. My guess is that your cats are self-medicating. We found that our cats will eat plants shaped like grass (spider plants and the like), but not plants with rounded leaves. So maybe stick to plants like African Violets (Saintpaulia), Begonia rex or Anthurium. Oh and the tip about marbles works: we use largish stones from the garden, you might be able to get some shingle.--TammyMoet (talk) 12:11, 10 September 2009 (UTC)[reply]
Another thing, you might want to check that any new plants you might add are not toxic to cats. 65.121.141.34 (talk) 13:04, 10 September 2009 (UTC)[reply]
The usual trick is to provide the cats with cat grass, which is also sometimes relabled oat grass or wheat grass. You can usually find pots of the stuff in pet stores. -- 128.104.112.179 (talk) 16:12, 10 September 2009 (UTC)[reply]

Strange haze around the ISS

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I just finished watching a visible pass of the International Space Station and Space Shuttle Endeavour. I noticed that there was a strange glowing haze around the ISS to one side, in the shape of a crescent moon. My guess is that it's some kind of out-gassing or perhaps the remnants of a burn they just did (wouldn't they do some kind of burn shortly after the departure of a shuttle?). I was worried that the station has depressurized and what I was seeing was the atmosphere leaking into space, but NASA would know about that and there is no mention of such a catastrophe having occurred on their website or on NASA TV.

I took photographs of the ISS, but they're long exposure so the station shows up as a bright streak. You can still see a bit of a haze around the station, but you can't see the shape that the haze took, so these pictures don't really show what I observed very well at all.

http://nickwarren.200u.com/pics/9-9-9-ISS/

Whatever it was, it was following the station in its orbit because it kept pace with it as it moved across the sky. It made it appear as though the station had a comet-like "tail". I've seen dozens of ISS passes and I've never seen anything like that.

What did I see? 63.245.144.68 (talk) 02:11, 10 September 2009 (UTC)[reply]

You might want to ask NASA directly. Quoting the time at which the photos were taken (in UTC) would probably help. --Tagishsimon (talk) 02:17, 10 September 2009 (UTC)[reply]
Was this a naked eye observation? Binoculars? Telescope? SteveBaker (talk) 02:27, 10 September 2009 (UTC)[reply]
Just naked eye. I regret I hadn't brought out the binocs now. 63.245.144.68 (talk) 02:31, 10 September 2009 (UTC)[reply]

Correction: it seems I got the wrong shuttle. It was Discovery, not Endeavour. I should have known that. 63.245.144.68 (talk) 05:56, 10 September 2009 (UTC)[reply]

Alright, seems I have an answer now: it was Urine! http://spaceweather.com/ 63.245.144.68 (talk) 07:05, 10 September 2009 (UTC)[reply]

Ah! I'm too late. Yes, most often, if you see a large gas plume coming out of a manned space vehicle, it is... urine. This is a disaster for scientific instrumentation. A professor of mine was principle investigator on a Space Shuttle mission to study plasma ion effects in the high ionosphere; in at least one experiment, the trailed antenna became so coated in urine ions that it directly affected the scientific measurements. So much for the vacuum of space... Nimur (talk) 08:19, 10 September 2009 (UTC)[reply]
In the Mercury space program an astronaut referred to the little sparkly drops of frozen liquid outside the spacecraft as "the constellation Urion." Edison (talk)
Wasn't that a line from the movie Apollo 13? 146.74.230.106 (talk) 21:48, 10 September 2009 (UTC)[reply]


How much urine could there be? 98.14.222.248 (talk) 22:05, 10 September 2009 (UTC)[reply]
About 68 kg (150 pounds), according to Space.com, an unusually large amount. ~AH1(TCU) 22:52, 11 September 2009 (UTC)[reply]

history of the octet rule ... and proving that something is an element

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It is the 19th century. You don't have any NMR, or mass spectroscopy. How do you prove that something is an element? Couldn't you say that the substances in your hypothetical compound are too bound together to release them by most known means? For example, in Davy's conclusion that HCl did not in fact contain oxygen, how did he defeat the idea that the oxygen was just really really strongly bonded to the other compounds.

The other thing is that I'm really confused at how people came to realise the octet rule. I mean, not long ago, Descartes was thinking that elements bonded to each other via hooks. How did people discover the chemical bond and the octet rule before the discovery of the electron? How did people propose enantiomers, Kekule structures and structural isomers and acetylation reactions before they had even discovered the electron?

It is clear that in the latter 19th century, people recognised the principles of unsaturation, electrophilic and nucleophilic attack ... but I can't really see how people discovered them if they didn't know the electron existed. How did people discover carbonyl reactions if they had no idea of partial charge and bond polarity? John Riemann Soong (talk) 03:09, 10 September 2009 (UTC)[reply]

The "Octet rule" is a historical artifact from early versions of the Periodic Table of the Elements, especially John Newlands's "Law of Octaves" which observed that, when arranged in order of atomic weight (protons and "atomic number" would not be elucidated for many years) chemical elements would repeat their properties every 8 elements. If you arrange the elements in order of atomic weight, then the 3rd, 11th, and 19th elements (lithium, sodium, and potassium) all display strikingly similar properties, as do other elements spaced similarly. Furthermore, Gilbert N. Lewis and Irving Langmuir devised important atomic models which basically connect this empirical observation with the electronic structure in the atom.
As to your first question, reliable methods of calculating the molar mass of a substance predates understanding of actual atomic structure by almost a century. Since elements cannot be broken down into smaller substances, you just keep doing things like electrolysis or other similar processes until you don't produce any new subtances with smaller molar masses. Once you rigorously show that something cannot break down anymore, you likely have an element. However, some early models of atomic structure were based on the theory that there was exactly one element, hydrogen, and that ALL other "elements" were merely increasingly more complex conglomerations of hydrogen. Its an easy mistake to make, since the molar masses of nearly all elements are simply interger multiples of the molar mass of hydrogen. See Prout's hypothesis for more on this reasonable, but ultimately dead-end theory. --Jayron32 04:29, 10 September 2009 (UTC)[reply]
Another general comment is that you don't need to know how something works to know that it does work. The actual nature of a bond ("covalent electrons") isn't needed to recognize that "something" holds molecules together in certain ways, and using certain chemicals, those things can change, and then eventually one sees patterns in what kinds of chemicals cause certain changes. For example, you don't have to know that "unsaturation" means two shared covalent pairs, you just have to see that one "unsaturation" in a molecule absorbs one hydrogen molecule or dihalogen molecule. The names of many old concepts are based on observed behavior, not on the now-known chemical structure: if something is "saturated" it cannot absorb any further. DMacks (talk) 04:36, 10 September 2009 (UTC)[reply]
If you are interested in understanding the nature of the chemical bond, one could do no better than to go to The Source, which in this case is the book The Nature of the Chemical Bond by Linus Pauling. The section on covalent and ionic bonding in EVERY single general chemistry textbook for the past 50 years is basically cribbed directly from Pauling's work. --Jayron32 04:39, 10 September 2009 (UTC)[reply]
Did people realise that alkenes had stronger "double bonds" (cf. to single bonds), smaller bond lengths, etc. or only unsaturation? How did they rationalise why C=O was preferred over C=C, for instance? (How did they interpret keto-enol tautomerism)? John Riemann Soong (talk) 13:44, 10 September 2009 (UTC)[reply]
At the time concepts like "saturation" WRT hydrocarbons was developed, people had almost no idea what was actually going on, things like "double bonds" and "keto-enol tautomerization" were unknown at the time. All that was known was that some compounds tended to react with excess hydrogen in the precence of a metallic catalyst, and others did not. This sort of work was going on in the late 1700's and early 1800's. At that time, people had means of deducing the molecular formula of a compound, indeed the basic concept of the isomer was known by 1827-1828 or so. People knew that isomers had to result from difference in molecular structure, however they had no idea what those structures looked like. However, the basics of organic chemistry structure didn't even come into place until 1858, when Kekule proposed the tetravalent carbon atom. Remember, before Kekule was born, chemists knew what isomers were, and could identify the molecular formula of a compound, but they did not even know enough about the structure of these molecules to know that carbon formed 4 bonds. It wasn't until Gilbert N. Lewis in the 1910's when people even had an idea what a double bond was. True experimental evidence the structures of compounds probably only came about in the 20th century, with modern analytical techniques like Mass spectrometry and Nuclear magnetic resonance. --Jayron32 05:49, 11 September 2009 (UTC)[reply]

heat of fusion of aluminum oxide...

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Need this to solve my thermodynamics problem set, but google is being obnoxious. In fact, the scientific community as a whole is being obnoxious. How the hell are we missing the heat of fusion for such a common substance? And why isn't there an easily-available database somewhere? Why does "aluminum oxide" + "heat of fusion" turn up no useful results? John Riemann Soong (talk) 03:42, 10 September 2009 (UTC)[reply]

This source from google books: [1] does not give the value directly, but notes that liquid Aluminum Oxide is a well studied material, which may mean that enthalpy of fusion data may be availible. I would check the footnotes of that reference for more details. However, it may also be that such data has never been reliably measured given Aluminum Oxides rediculously high melting point. It could just be that its hard to do a reliable calorimetry experiment to find the heat of fusion of a material that melts at a higher temperature than your measuring devices do! --Jayron32 04:04, 10 September 2009 (UTC)[reply]
Per CRC handbook, the enthalpy of fusion of Aluminum oxide (α) (MP 2054°C) is 111.1 kJ/mol. Tim Song (talk) 04:29, 10 September 2009 (UTC)[reply]
Good old Chemical Rubber Company. I should have plucked my CRC off of the shelf rather than tried to search the internet. --Jayron32 04:34, 10 September 2009 (UTC)[reply]

michelson and morley experiment

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210.212.239.181 (talk) 04:27, 10 September 2009 (UTC)sandeep[reply]

q: how were michelson and morley able to get the initial interference pattern if their did not exist any path difference initially??

I don't think it matters whether the two paths are perfectly identical. Interference fringes will be seen in the field of view and the object of the experiment is to see whether the fringes move when the apparatus is rotated. They don't. Cuddlyable3 (talk) 11:27, 10 September 2009 (UTC)[reply]
I think the question is, if the path distances are equal and the beams are perfectly aligned, wouldn't the phase difference on the detector be essentially uniform across the surface of the detector, and therefore show no fringe pattern? Rckrone (talk) 16:43, 10 September 2009 (UTC)[reply]
And to answer that, the two distances won't be exactly equal, and the wavelengths involved are small enough (~500 nm) that even small discrepencies will cause fringes. In particular, if d is the approximate path length and a is the discrepancy between the two paths then the change in path difference between the center of the detector and a point at radius r is approximately ar2/d2. Just to pick some numbers, with d = 10m and r = 5cm, there would be 1 fringe showing up for each 2cm that the path distances were off by. You could probably set it up so that initially there were no fringes if you wanted, but I assume it's easier to see a change in the pattern if there are some. Rckrone (talk) 17:18, 10 September 2009 (UTC)[reply]

temperature-dependent heat capacity of aluminum

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I note that aluminum exhibits a relatively significant temperature-dependence ... a lot of books and internet pages mention this, but won't give me a function. Can someone give me a function? If I extrapolated experimental data onto excel, what minimum degree polynomial would be useful? John Riemann Soong (talk) 06:34, 10 September 2009 (UTC)[reply]

Look at [2] for some graphs. But no formula. Did you see Resistivity#Temperature dependence? It has a formula: Bloch–Grüneisen formula. : Graeme Bartlett (talk) 07:08, 10 September 2009 (UTC)[reply]
Wait, but I need heat capacity, not resistivity. John Riemann Soong (talk) 07:10, 10 September 2009 (UTC)[reply]
As far as I remember, the reason is as follows. Alumin(i)um has a relatively high Debye temperature (about 400 K), that is, some of its phonon modes have energies well above the room temperature (about 295 K). As a result, at room temperature the heat capacity of aluminium is below the 3Nk value of Dulong-Petit law, and growing with temperature. Please see our article on Debye model, or Ashcroft & Mermin textbook. You can use the formulae derived in the framework of the Debye model. You should also be aware that the heat capacity of a solid may actually exceed 3Nk close to the melting point, and that there is also a (small) electron contribution to the aluminium heat capacity that is not accounted for by the Debye model. --Dr Dima (talk) 08:33, 10 September 2009 (UTC)[reply]

How to calculate the density of a gas

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How do you calculate the density of hydrogen at 50 Celsius and 0.5 MPa? This is not a homework question, I need it for work. I suppose you perhaps could use the ideal gas law pV=nRT but dont know how. Thank you! Wikifantast (talk) 06:47, 10 September 2009 (UTC)[reply]

This kind of question can be answered at Wolfram Alpha, with this link: http://www.wolframalpha.com/input/?i=Density+of+hydrogen+at+50+degrees+Celcius+and+0.5+MPa I get 374.1 grams per cubic meter. Graeme Bartlett (talk) 07:12, 10 September 2009 (UTC)[reply]
Or if you want to do it by hand, just substitute in the mass as in the ideal gas law article and rearrange:
From molar mass, M(H
2
) = 2.015 88(14)x10-3 kg/mol. AlmostReadytoFly (talk) 07:34, 10 September 2009 (UTC)[reply]

Mixtures and phases

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do mixtures have always more phases? —Preceding unsigned comment added by 121.54.107.254 (talk) 07:56, 10 September 2009 (UTC)[reply]

More phases than what? DRosenbach (Talk | Contribs) 16:51, 10 September 2009 (UTC)[reply]
Yes, mixtures tend to have more phases compared to the pure substances, see Gibbs' phase rule. For a more general discussion see Phase diagram. All the best, --Dr Dima (talk) 17:57, 10 September 2009 (UTC)[reply]

Predicting lottery numbers

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From a scientific point of view, how does Derren Brown predicts the lottery numbers? My personal guess was that the numbers were written on the balls after it was announced on TV? Any takers? 122.107.207.98 (talk) 11:46, 10 September 2009 (UTC)[reply]

There is no prediction involved, it's a trick - the idea is to try and work out how the trick is done - predicting the numbers is not one of the options. my own take is that it involves mirrors and the balls are simply showing what is reflected onto them from the floor. That's why he does not lift the balls as it would spoil the effect. it's also why we get the misdirection about him not being able to reveal the numbers before the drawn occurs because of legal rights. No legal rights would stop him making a prediction, it's simply because the trick needs him to know the numbers before they 'appear' on the balls - however he does it. --Cameron Scott (talk) 11:52, 10 September 2009 (UTC)[reply]

It's not science, it's 'magic' and from my experience the answer is usually something dull and 'obvious'. One thing to note about Derren Brown is that he often 'reveals' how a trick is done. Don't be fooled into believing that this is necessarily any more the truth than the first trick you saw. (i.e. the explanation of how the trick was done is a lie itself, the 'real' way is not revealed). 194.221.133.226 (talk) 14:21, 10 September 2009 (UTC)[reply]

One technique he is said to use, which might be applied here (I have not seen the specific trick) is to tediously film all the possible outcomes of a trick (say, drawing a specific card), each time without any camera cut, and then show the appropriate "take". Obviously this would not be possible for the 14 million-plus possible outcomes of the Lottery, so some combinatory cutting might be employed, but I also thought that he usually avoided such simple cutting "cheats." 87.81.230.195 (talk) 15:13, 10 September 2009 (UTC)[reply]
I saw this and he didn't predict anything, he only revealed the numbers after they'd been drawn, so even as a magic trick it was quite disappointing. Anyway, he himself will be revealing how he did it tomorrow evening, so you won't have long to wait.--Shantavira|feed me 16:44, 10 September 2009 (UTC)[reply]
His programme is subtitled "How to win the Lottery", if he could do that we could all win the lottery and probably only get our stake back. —Preceding unsigned comment added by 86.4.181.14 (talk) 14:50, 11 September 2009 (UTC)[reply]
...and winning the lottery is easy - you just have to buy 14-million-and-something tickets. Winning the lottery and making a profit - that's the hard one ! Gandalf61 (talk) 15:01, 11 September 2009 (UTC)[reply]
Have you tried buying 14 million tickets? I don't think there's any way to buy them in bulk. — DanielLC 00:29, 12 September 2009 (UTC)[reply]

And what a disappointment that show turned out to be - he constructed a pseudo-science narrative so he did not have to simply say "it was spilt screen". --Cameron Scott (talk) 21:13, 11 September 2009 (UTC)[reply]

Yeah, very disappointing. The coin tossing thing is well known and has nothing to do with the law of large numbers, which is what the rest of it focused on (albeit a misapplication - wisdom of crowds is one thing, psychic power of crowds is quite another). --Tango (talk) 00:33, 12 September 2009 (UTC)[reply]
Split screen is most likely, or mirrors, or an ink-jet spray to the bottom of the balls, which are then imperceptibly rotated! What I thought was interesting about his "explanation" was that it was what he does best - misdirection. It was purportedly about probability and prediction, but it was nothing of the kind. It was really about how he can manipulate people. A woman scared of mice will put her hand in a box with a 50:50 chance, she thinks, of a mouse being in there, simply because Derren Brown asks her to. There is no mouse. A man will stamp on an upturned paper cup with a 50:50 chance of stamping on a knife, so he thinks. There is no knife. And a group of people will believe that they can collectively predict lottery numbers using "automatic writing" because Derren Brown persuades them that it works, despite precognition being impossible and everyone knowing that Derren Brown is a wily trickster with no psychic powers. And, most impressively, he can make an entire nation talk about him "predicting" the lottery and watch his TV show about it, when all he did was some petty trick that stage magicians have been pulling off for decades. Clever bugger. Fences&Windows 00:07, 13 September 2009 (UTC)[reply]

heart

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how many hearts does a octopus have? —Preceding unsigned comment added by 117.196.176.228 (talk) 12:23, 10 September 2009 (UTC)[reply]

3. Algebraist 12:24, 10 September 2009 (UTC)[reply]
It depends on your definition of heart. In the colloquial sense, a heart is an internal organ that serves to pump blood (or other circulatory fluid, as in hemolymph), thereby maintaining circulation. The human heart, along with other mammals, birds and crocodilians is divided into two halves that are synchronized in their beating -- one pumps the systemic circulation while the other pumps the pulmonary circulation. Annelids may have enlarged vessels that contract, thereby pushing blood in the closed tube system, or a number of vessels connecting the dorsal and ventral main vessels that contract -- some would refer to each of these as a heart. So does an earthworm have 5 hearts? Not nearly in the sense of the human heart. There's probably a similar thing going on in cephalopods like octopi. DRosenbach (Talk | Contribs) 15:36, 10 September 2009 (UTC)[reply]
"The circulatory system of cephalopods is an extensive system of vessels and capillaries . . . Blood within the vessels follows more or less the same route through the body as in other molluscs, but in addition to a systemic heart, the circuit includes two branchial hearts, which pump blood through the gills." --Edward Ruppert and Robert Barnes, Invertebrate Zoology (quote taken from this web page). Looie496 (talk) 17:26, 10 September 2009 (UTC)[reply]

Extremely fast Blu-Ray recorder required

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I want to record rock concerts (with the permission of the artists) and almost instantly burn blu-ray disks for sale to fans after the concert-whilst I am doing this just to promote my favorite acts, completely free of charge right now, I later plan to turn it into a commercial business. I have heard about 8x blu-ray recording, but I need MUCH more speed for this proposition, even if I used an array of burners the crowd would have gone home before the disks were burnt. Is there any technology that could get around this problem on the horizon, such as a burner with multiple lasers for example? Supposing a powered vacuum suction plate clamped the disk tightly to a platter-could a much higher speed (presumably using a higher power laser as well) be achieved without turning the disk into a safety hazard?80.2.195.218 (talk) 12:56, 10 September 2009 (UTC) —Preceding unsigned comment added by 80.2.195.218 (talk) 12:53, 10 September 2009 (UTC)[reply]

You might need to consider a parallel burning scheme if you want to make a business out of this. Unfortunately, fast burning is not scalable. You can invest in several slow burners and have them running simultaneously. Keep in mind the important difference between throughput and latency. These terms apply equally well to the low-level "bits" burning on to the Blu-Ray; and also the high-level, conceptual throughput for your business model. You want to generate the maximum number of discs in the shortest amount of time - so while a faster burner decreases delay per disc, it may be less effective at increasing total discs per unit time than investing in several dozen cheap burners. Nimur (talk) 13:52, 10 September 2009 (UTC)[reply]
I understand that you want to catch the fans before they leave but consider that they won't actually get to play the music until they get home. (Can't your favourite acts pre-record for you?) An idea is that you could give away CDs that give users access to a website where they can download a recording of the concert. CD players are commoner than Blu-ray players and I suppose they all have Internet. For you the advantages would be 1) Easy, cheap no-rush technology. 2) You have time to listen to and edit the material. 3) You can use the website for any kind of promotion, knowing who are the visitors. Cuddlyable3 (talk) 13:55, 10 September 2009 (UTC)[reply]
Bands usually already do sell their CDs at concerts. --98.217.14.211 (talk) 14:14, 10 September 2009 (UTC)[reply]
But not CDs of the very concert at (the end of) which they're being sold, surely? I thought Midge Ure had once done so during an April 2002 solo tour, but in fact the double-live album in question had been compiled during the earlier gigs of the tour. What had me going was that a very similar incident to what occurred at the gig I attended (Midge telling off a fan for singing along at the beginning of "Dancing with Tears (In My Eyes)") appears on the recording, but the recorded event's details differ slightly from my (admittedly unsober) recollection.
Sorry, I misunderstood what the previous poster was suggesting, and thought he/she was just suggesting they sell CDs. Instead it is more like the "voucher" model that I mention below. --98.217.14.211 (talk) 16:50, 10 September 2009 (UTC)[reply]
I think that if the technical problems could be overcome, the OP would find a small but dependable market (given typical post-gig euphoria and the desire for personal souvenirs), but making it cost effective might prove difficult. 87.81.230.195 (talk) 14:50, 10 September 2009 (UTC)[reply]
This sounds... not very feasible. Technically, or as a business model. (What happens if there is a glitch with the machine, or the concert gets out early? Then you've just ended up with 100 BluRays that won't offload any time soon.) You could, of course, sell vouchers -- they buy the disk, you ship it to them the next day. But I imagine that isn't as fun. It doesn't really seem feasible to me unless you lock the crowd up for a little while after the show ends. --98.217.14.211 (talk) 14:14, 10 September 2009 (UTC)[reply]
There are already people who are doing this exact thing with CD's. Certainly my son went to a concert over 2 years ago and bought a live, recorded-on-the-spot CD containing the musical highlights as he left the concert which included still photos of the crowd - which we know had to have been taken during the concert. So the only thing that's novel here is to record a Blu-Ray rather than a CD. The problem is indeed the speed of recording. But a standard drive cannot simply be made to spin faster to record faster! That's an exceedingly naive expectation! There is the matter of the laser delivering enough energy to one tiny spot on the disk to do whatever it does to record a '1' or a '0' there. If you make the disk spin 10x faster, the laser has to deliver 10x the energy per second to be able to burn the disk. It's possible that the chemical processes that cause the disk to change state require a certain minimum time to happen. That means that you quite simply cannot use the same laser diode that an off-the-shelf BluRay recorder uses...you'd need some really fancy industrial laser. That laser has to be at exactly the right frequency - which might mean that you need something rather exotic. Can you even buy such a laser with a sufficiently small spot-size? It would also have to be focussed the exact same way as the original LED in order to deliver the laser light to a sufficiently precise spot on the disk and also to put the energy into the layer of dye that you want to record onto without melting the plastic on the outside of the disk. Would the cheap lenses they use stand up to that amount of energy delivered that quickly? Worse still, you probably have to cool the disk because as it absorbs all of that energy, it's going to get hot. Then, the data rate that you have to supply to the laser would require entirely new electronics - and probably new software. All-in-all, this is an exceedingly tricky technical problem - requiring an understanding of lasers, chemistry, physics, electronics, software and mechanics. I would be EXCEEDINGLY surprised if our OP were to be able to pull off such a system without a large team of experts working for him/her. SteveBaker (talk) 20:23, 10 September 2009 (UTC)[reply]

Supposing there was a way to quickly create a photo-mask http://en.wikipedia.org/wiki/Photomask like the ones used to create silicon chips and then use it to burn a blu-ray disk using a blue laser with a beam the size of the disk? Or could a high powered digital laser projector achieve the same effect with the right focusing lenses? Any heat could be dissipated by a cooling plate with liquid nitrogen. User:Trevor Loughlin —Preceding unsigned comment added by 80.2.200.204 (talk) 05:30, 11 September 2009 (UTC)[reply]

This discussion has now become a question about what the current limiting factors are for modern data transfer to optical storage media (rather than strictly about faster Blu-Ray burners). I suspect that, per Steve's answer, there are so many bottlenecks that you need technological advances on all fronts - mechanical, chemical, electrical, digital/bandwidth, that making a dramatic revolution in one area is still insufficient to speed up the entire process. Trevor's suggestion of a photomechanical transfer seems unlikely to speed the process - having spun photomask polymers a few times myself, I would definitely not call the process "quick and easy". To do it on a one-off basis requires a research laboratory and a lot of expertise; to do it in a controlled, mass-production sort of context requires a clean room. Assuming the technicalities of turning one of those facilities into a standalone box were solved, we haven't even addressed the issue - because in order to photomask, you need a mask - so you'll need to transfer the data to the mask first. Presumably once you have that mask manufactured, you can re-use it; but all told, at least using present technology, that idea would be much much more complicated than simply burning the discs individually. The ultimate goal, which is to parallelize the data transfer, is fundamentally the right direction; but I don't think we're going to see this scale of "all-at-once", gigabytes-of-parallelism data copying until a few more miles down the road of technological progress. By that point, optical storage media will probably be less desirable, less economically competitive, and a whole new array of parallelization options will become available. Nimur (talk) 11:23, 11 September 2009 (UTC)[reply]

Have an array of burners recording at 1x, basically in real time while the concert is happening. When the music stops, close the sessions, and all discs are finished in a few moments. 70.90.174.101 (talk) 08:11, 12 September 2009 (UTC)[reply]

Car tire wear

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I have a car that is rear wheel drive, but for some reason, the right front tire seems to wear the fastest. Why would this be? We drive on the right hand side. Googlemeister (talk) 14:37, 10 September 2009 (UTC)[reply]

Are you properly rotating, balancing, and aligning the tires/wheels? I'd put any of those three well above which side of the road you drive on. Also consider damage to the wheel bearings or other components that may cause the wheel to lose alignment even if you're following the basic maintenance schedule. — Lomn 14:56, 10 September 2009 (UTC)[reply]
(EC) The most likely cause is that the wheel is slightly misaligned (see Wheel alignment); the exact pattern of wear would indicate to a mechanic the nature of the misalignment. Less likely is that, overall, you make significantly more left turns than right - possible if you habitually follow different "out" and "return" routes, as many people do. 87.81.230.195 (talk) 14:59, 10 September 2009 (UTC)[reply]

If the wear isn't even it could be due to incorrectly inflated tyres. Take it to a local garage and they'll do an alignment/balancing check and get it sorted - also remember to check your tyre pressure regularly (i'm terrible for doing it only before long-ass trips). 194.221.133.226 (talk) 15:14, 10 September 2009 (UTC)[reply]

You could be taking far more left hand turns and if the car is improperly balanced, will put a great deal of pressure on the right front tire. Livewireo (talk) 19:19, 10 September 2009 (UTC)[reply]
This answer seems to imply that the driver is going in circles. :) Statistically speaking, you really should be turning both ways with equal frequency. When you go back home on the same route you came, all your turns are reversed. Franamax (talk) 10:48, 11 September 2009 (UTC)[reply]
True, I don't drive for NASCAR. Googlemeister (talk) 13:20, 11 September 2009 (UTC)[reply]
Too bad, if you did you could just tell your pit boss to get the g-dam RF wheel into shape. Or else you're trying to sucker us Wikipedians into helping your NASCAR team for free. I get confused on this stuff... :) Franamax (talk) 14:12, 11 September 2009 (UTC)[reply]
You are an ambi-turner right? Not to put you on the spot... TastyCakes (talk) 14:21, 11 September 2009 (UTC)[reply]

Otitis

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Does Otitis lead to high frequency loss in Adults? Otitis? after treatment is it recommend a audiometer exam be done for adults who had otitus? Finally, can otitus be associated with hearing loss and or tinnitus?

Thanks for your timely response,

Jackk Devlin USN (Ret.) —Preceding unsigned comment added by DevlinJohnA (talkcontribs) 15:04, 10 September 2009 (UTC)[reply]

This question appears to be a request for medical advice. It is against our guidelines to provide medical advice. You might like to clarify your question. You may also find it helpful to read the article: Otitis, and form your own opinion from the information there. Thank you.

Responses containing prescriptive information or medical advice should be removed and an explanatory note posted on the discussion page. If you feel a response has been removed in error, please discuss it before restoring it.

. Nimur (talk) 15:13, 10 September 2009 (UTC)[reply]

Oh, I think it can be answered without offering any advice. There are three kinds of otitis: otitis externa, affecting the outer ear, otitis media affecting the middle ear, and otitis interna, involving the inner ear. The most common of these is otitis media, which causes a (usually transient) hearing loss, but the poor hearing is caused by a loss of conduction of sound to the inner ear. But high-frequency loss is more suggestive of sensory-neural injury, involving the inner ear, rather than conductive loss. As for the advisability of audiometry: if you are having hearing problems or tinnitus, this is something you should ask your doctor. It's not routinely done after every bout of otitis. - Nunh-huh 18:59, 10 September 2009 (UTC)[reply]
Medical information is not the same as medical advice. Please people, try not to pounce on every question that even alludes to medicine and/or the law... -Elmer Clark (talk) 21:51, 13 September 2009 (UTC)[reply]

displacement

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is displacement is vector quantity? how is it ? —Preceding unsigned comment added by Rucha6 (talkcontribs) 15:25, 10 September 2009 (UTC)[reply]

Yes. Algebraist 15:32, 10 September 2009 (UTC)[reply]
Unless you are asking about the fluid displacement :) --Dr Dima (talk) 18:05, 10 September 2009 (UTC)[reply]
Displacement is a vector because in order to describe WHERE something has moved to, you need to say BOTH how far it moved AND in which direction. That's pretty much exactly what a vector is. --Jayron32 20:42, 10 September 2009 (UTC)[reply]

Reflex enclosure design before TS

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How were reflex cabs designed by hi fi loudspeaker designers before Thiele Small parameters were discovered in the 70's? was it mainly rule of thumb/guesswork?--79.75.118.147 (talk) 15:29, 10 September 2009 (UTC)[reply]

Sound has a wavelength, and resonance was known about, so it would have been possible to make back of the envolope calculations to calculate the resonant frequency of a ported loudspeaker with out too many tears.
More relevently sealed box, mass damped and of course horn type loudspeaker where much more likely to be encountered then.
Specifically Helmholtz resonance would be one concept available to them.
This site may be of interest [3] - note a reflex design from 1930 which may be worth investigating. I'm sorry I haven't really answered your question, but hopefully it will serve as an introduction to a better answer below:::83.100.250.79 (talk) 16:32, 10 September 2009 (UTC)[reply]
An early version of the now widely-used bass-reflex cabinet design was patented by Albert L. Thuras of Bell Laboratories in 1932. The patent shows that the mathematics was already understood at that date. Cuddlyable3 (talk) 20:47, 10 September 2009 (UTC)[reply]
There were excellent engineers designing audio equipment, ranging from microphones to amplifiers to loudspeaker enclosures, at such places as Bell Labs, back in the 1920's. They went far beyond guessing and rule of thumb. For some early Google Book hits before 1950, see [4]. Unfortunately, the books and journals are not old enough to be more than snippet view, but the journals could be accessed at a good engineering library. The research publications from Bell Labs apparently are not in the Google Book search index, but could also be found in an engineering library. The Society of Motion Picture and Television Engineers (back before TV) also worked on well engineered sound systems for movie theaters, and again the journals are in engineering libraries. The Journal of the Acoustical Society of America also has papers on loudspeaker enclosure design dating back to the 1920's [5]. The Proceedings of the Institute of Radio Engineers (predecessor of IEEE) published studies on the design of speaker enclosures. Here are some from the 1930's:[6]. It must be noted that lots of radios and record players back then had beautiful wood outside and crummy speaker enclosures. Edison (talk) 04:12, 11 September 2009 (UTC)[reply]
Using transistors it is now possible to make a radio that has both a crummy box and a crummy loudspeaker. Cuddlyable3 (talk) 13:50, 11 September 2009 (UTC)[reply]

aliens & ufo

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it have seen that if an ufo is looked anywhere , then it is seen that aliens use zigzag motion to come at the earth, so why can not use this type of maotion to go to any other planet? —Preceding unsigned comment added by Aatmic (talkcontribs) 15:45, 10 September 2009 (UTC)[reply]

If your question is why don't humans travel to other planets in a zigzag motion, then the answer is that it requires a large amount of energy (in the case of our spacecraft, rocket fuel) to change the direction of a vehicle traveling at such speeds. Googlemeister (talk) 15:50, 10 September 2009 (UTC)[reply]
ZigZag is for coming towards the Earth, so we would have to go ZagZig to go away from Earth.Cuddlyable3 (talk) 20:53, 10 September 2009 (UTC)[reply]
I don't know about that...from our perspective, their approaching ZigZag looks like бɒZб!Z. DMacks (talk) 21:09, 10 September 2009 (UTC)[reply]
Your question reads to me like you're asking why aliens don't zigzag over other planets as well as Earth. In which case - they probably do. If you're asking why we don't do it to travel to other planets; zigzags are not efficient for travelling from one location to another, as pointed ut by Googlemeister. If they are zigzagging, it's not to reach the planet - it's for some other reason (photography, malfunction - take your pick, the possibilities are limited only by your imagination). 90.208.66.97 (talk) 21:07, 10 September 2009 (UTC)[reply]
I don't know what UFO's you're thinking of, but keep in mind that coming to Earth is a very different deal than going from Earth to another planet. The former requires you to slow down from an orbital speed to a speed suitable for flight, whereas for the latter you'd want to accelerate as quickly as possible to get out of Earth's atmosphere. As it turns out, the space shuttle does zigzag across the sky upon re-entry: it traces S patterns in the sky to lengthen the distance it travels and bring it down to landing speed. --99.237.234.104 (talk) 00:01, 11 September 2009 (UTC)[reply]
That would not help in space though as there is no atmosphere providing friction. Googlemeister (talk) 13:19, 11 September 2009 (UTC)[reply]

I can't remember what it's called

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What is the name of the machines or prcosses that perform simple function (turning on a light) by using convoluted tasks to perform the action (a ball drops into a cup which tips over causing a basket to rise on a pulley, withsomething attached to flick on the light switch) —Preceding unsigned comment added by 138.64.2.77 (talk) 16:31, 10 September 2009 (UTC)[reply]

A Rube Goldberg machine. Recury (talk) 16:38, 10 September 2009 (UTC)[reply]
Yes and also Heath Robinson contraption in the UK.83.100.250.79 (talk) 16:39, 10 September 2009 (UTC)[reply]
And 'Storm-P' machines in Denmark thanks to Robert Storm Petersen. Heath Robinson was the first - pretty much everyone else 'borrowed' his idea. SteveBaker (talk) 20:02, 10 September 2009 (UTC)[reply]

physics(electricity)

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On what factors does the resistance of a conductor depend? Will current flow more easily through a thick wire or a thin wire of the same material when connected to the same source? Why does the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value? What change will occur in the current through it? —Preceding unsigned comment added by 59.96.142.2 (talk) 17:01, 10 September 2009 (UTC)[reply]

I have taken the liberty of fixing the grammar and spelling in this question, and adding punctuation. Regards, Looie496 (talk) 17:16, 10 September 2009 (UTC)[reply]
Please do your own homework.
Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Refer Resistor for starters. Rkr1991 (Wanna chat?) 17:55, 10 September 2009 (UTC)[reply]
Also Electrical resistance and Resistivity might be helpful. Rckrone (talk) 00:58, 11 September 2009 (UTC)[reply]

dark energy

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Can dark energy be the other "pole" of gravity or anti-gravity? -- Taxa (talk) 18:57, 10 September 2009 (UTC)[reply]

It sort of behaves as a kind anti-gravity, in that it causes objects (that are far apart) to accelerate away from each other, while gravity causes them to accelerate towards each other. I don't think it is actually related to gravity in any way, though. There is a theory that it is a result gravity itself (not any kind of anti-gravity) behaving differently are large distances, but I don't think it has much support. --Tango (talk) 19:18, 10 September 2009 (UTC)[reply]

Lux and sunlight

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i've never hear the term lux. All it is is is measurment of sunlight fraction of earth. how much dimmer than light is nighttime or midnight on Earth? Total blackness would be 450,000 times dimmer than light? Will 100 times dimmer be sunset or sunrise?--209.129.85.4 (talk) 20:46, 10 September 2009 (UTC)[reply]

Lux has an list of examples you may find useful. --Tango (talk) 20:59, 10 September 2009 (UTC)[reply]
(after ec)You'd struggle to find "total blackness" anywhere on the surface of the Earth, since it's all illuminated by the Sun during the daytime, and by the Moon and stars at night. "Total blackness" (zero light) cannot be expressed as a fraction of another light on a linear scale. Time to check the Lux article...You also asked "how much dimmer than light is nighttime or midnight" - do you mean how much dimmer than daytime? If so, that will depend on weather (cloud cover would reduce light both at night and during the day), the time of the month (full moon will illuminate more). A location and general weather you're expecting for these answers would help. 90.208.66.97 (talk) 21:02, 10 September 2009 (UTC)[reply]
On reading Lux, it turns out every answer is nicely tabulated. Way to waste your time, 90.208.66.97. 90.208.66.97 (talk) 21:04, 10 September 2009 (UTC)[reply]
I was involved in a "struggle to find total blackness" at a company making a test card for TV cameras. We could find no black paint that was truly black so the solution was to build a little box behind the card, viewed through a hole in the card. The inside of the box was lined with black velvet. Cuddlyable3 (talk) 13:37, 11 September 2009 (UTC)[reply]
A hole in a box is a very common way of simulating a black body, which is basically what you wanted. --Tango (talk) 14:43, 11 September 2009 (UTC)[reply]

What causes older people to have trouble rising?

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I noticed with my grandparents and other older people that they sometimes need help getting out of regular chairs. Is this something with the back muscles, such as the spine? Is it generally arthritis - I don't recall my grandparents, at least, having complaints about that, though. Or, is it just a combination of factors? Notice I'm trying to be vague so it doens't sound like a request for medical advice or diagnosis; I'm jsut asking about a general problem. Maybe you can just link an article or two. The article on sitting really didn't answer it. (Maybe I should have looked under "rising" :-) )4.68.248.130 (talk) 23:21, 10 September 2009 (UTC)[reply]

As somebody who's approaching that zone myself, I can tell you that the problem is that your joints tend to stiffen when you sit for a while. Rising from a sitting position requires a substantial amount of leg muscle strength and a rapid change in joint angles. Looie496 (talk) 23:40, 10 September 2009 (UTC)[reply]
Especially if it's a deep, soft chair that you used to be fine with getting in and out of. Hence the market for chairs that slowly push you up and let you straighten your legs out. A great invention. I'll get one when I'm old and crotchety. >:) Baseball Bugs What's up, Doc? carrots 01:57, 11 September 2009 (UTC)[reply]
I hear there's a few months' lead time on them, Bugs. Better get your order in now :-). 87.81.230.195 (talk) 02:27, 11 September 2009 (UTC)[reply]