Wikipedia:Reference desk/Archives/Science/2011 December 2

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December 2[edit]

Unidentified moss[edit]

Unidentified moss

Can anyone identify the type of moss shown in File:Moss closeup.jpg? Apparently, the photo was taken in Beacon Hill Park, Victoria, Canada in March 2007.  Chzz  ►  00:37, 2 December 2011 (UTC)

Looks like Sphagnum to me, but as to which of the 200 or so species, it may be tough to narrow it down. The picture of Sphagnum flexuosum in the infobox of the Sphagnum article looks like a halfway decent match, but there could be dozens of other sphagnums which look similar. --Jayron32 05:02, 2 December 2011 (UTC)
Need a moss specialist. I don't think you can identify it as Sphagnum with any confidence whatsoever though. Shame it doesn't have sporophytes yet. Some species found in B.C. which resemble it (the most likely candidate is Grimmia trichophylla):
Also see Moss photos, Bryophytes of Stanley Park, Bryophytes of British Columbia -- Obsidin Soul 09:45, 2 December 2011 (UTC)
Thanks very much, both. Good info!  Chzz  ►  20:56, 3 December 2011 (UTC)

Classified US space rocket launches during the Cold War[edit]

So, my dad and I watched Apollo 18 (film) tonight and got into a discussion about the feasibility of secret space missions afterwards. My dad asserted that there was no way that NASA (in real life) could have sent up a Saturn V without the whole world being aware of it, on account of it being a huge, very distinctive, very conspicuous rocket that anyone living within 300miles of launch could've seen going up - to which I responded that he'd missed the bit in the film where it specifically stated that it wasn't a 'secret' Saturn V launch, rather that the US govt. had put it about that it was a Saturn V carrying a classified military satellite to earth orbit (i.e. a normal launch with a 'secret' cargo).

Anyway - my question is thus. When the Cold War was on, how did the US government actually deal with media and local interest into rocket launches containing classified payloads, when it was impossible to hide the launch itself from the public? Was it a case of saying 'Yes, we're launching a rocket - but we can't tell you any more about in in the interests of national security'? --Kurt Shaped Box (talk) 01:15, 2 December 2011 (UTC)

That's pretty much how they did it. I dunno what they'd have to do in order to carry out a classified manned launch, though... 67.169.177.176 (talk) 01:21, 2 December 2011 (UTC)
A few small points:
  1. Saturn Vs are huuuuge and were used for veeeerrry heavy payloads. Which is to say, people. If you want to put up a satellite, or test an ICBM warhead, generally speaking you don't need that much payload, and you can do much more nimbler operations. If you were just putting up a satellite or something you'd use a much smaller rocket, and you can launch those from much more isolated areas (e.g. Johnston Atoll, where the US sent up the rockets with its spy satellites early on).
  2. The government is actually in a much better situation when it can admit to doing classified things for the reason you describe. It's easy to say, "yes, there's a secret program involved here," than it is to try for a complete cover-up. So the Manhattan Project was a huge pain for security people, because the fact that the US was working at all on atomic weapons was a secret, and even the fact that there was a secret of it was a secret. In the postwar period, it was a lot easier, because the government could say, "well, we're doing secret stuff over there," and leave it at that — a lot less work to do. When the US gov't detonated the first atomic bomb in the summer of 1945, they had to put out fake press releases claiming an ammunition dump had blown up, and then check around the local towns to see if anybody suspected otherwise. When nuclear testing was a known thing, they could just say, "yeah, it was another nuclear test, but we can't tell you anything about it." Much easier.
I'll see if I can dig up actual press coverage of the first spy satellite launches — it would be interesting to see how much detail was given out, if any. --Mr.98 (talk) 01:51, 2 December 2011 (UTC)
You mean like mission USA-226, launched in March?  Card Zero  (talk) 01:59, 2 December 2011 (UTC)
There's really no such thing as a "secret launch" of a satellite, regardless of where it's launched from or whatever orbit it's heading for. Firstly you have to tell anyone who might fear you're launching an ICBM at them that you're launching, lest they get the wrong idea and panic. When the detect a launch, Soviet warning systems can't immediately distinguish between a Titan launcher putting a satellite in orbit and a Titan missile (which is just about the same thing) chucking a suite of thermonuclear MIRVs at Moscow. So a few days before a launch the US would post a notice that it intended to launch, which had the launch window(s) and some of the characteristics of the launch trajectory. Other launcher and ICBM countries would do the same. Secondly, as any launch may fail and scatter debris downrange, the launching agency has to issue a NOTAM to make sure no civil aircraft is in the credible debris footprint. You might think the public information about the launch trajectory would be enough to determine the final orbit of the satellite (which is vital for tracking it), but they don't advertise much information, and nothing about the settings of the last (usually third) stage. So you don't know the elements of the final orbit. And many military satellites carry quite a lot of fuel, so they can adjust their own orbit repeatedly, and try hard to be difficult to spot optically or with radar. As Mr.98 notes, Saturn V was only used for Apollo and Apollo/Soyuz; US military launches variously used Atlas/Centaur, Delta, and Titan - these are smaller rockets which launch lighter payloads. I guess it's not impossible someone could cobble together a lunar mission out of multiple Atlas/Centaur launches, but with a bunch more docking in orbit and certainly not with the same equipment used in Apollo. As for military payloads in general: the US and Russia continue to launch military payloads (recon, mapping, communication, launch detect, etc.) and they just announce that it's a military payload and leave it at that, and won't be moved as to what's what and what went where. The US didn't really admit the NRO existed for decades, and still says essentially nothing about which launches are for it or what they might be doing. You might think that a look at the rocket, prior to launch, might be informative; but probably not much. The payload is integrated with the rocket in an indoor building, and is covered by an aerodynamic faring, so you can't see it. We only know details of classified payloads decades later, when they're declassified, or we find out odd things by association, such as KH-11 Kennan's apparent similarly to the Hubble Space Telescope. 2.122.75.79 (talk) 17:59, 2 December 2011 (UTC)
Ah, I forgot - there were also a few classified Space Shuttle launches, such as STS-51-C. The role of these missions is also classified. 2.122.75.79 (talk) 18:07, 2 December 2011 (UTC)

Atoms[edit]

Scanning tunneling microscope image showing the individual atoms making up this gold (100) surface

What do atoms really look like? I mean a real picture with electron microscope, not the usual diagram showing nucleus etc. And is it possible to see what protons and quarks look like? Money is tight (talk) 01:18, 2 December 2011 (UTC)

Regarding your second question: No, it's not possible to see individual subatomic particles, let alone quarks. 67.169.177.176 (talk) 01:38, 2 December 2011 (UTC)
When you start talking about things that are in the same order of magnitude as photons themselves, you have to start asking what you mean by look like. We tend to think about vision as a straightforward thing, but it's the reflection of photons off of things. (Colors are the way our brains interpret the frequency of the photon in question.) Things smaller than photons cannot really be visualized. What you can do is find ways to make data that you can convert into visualizations — so a scanning tunneling microscope can register very small changes in voltages that coincide with electron rings, and you end up with visualizations of things that we can say are "atoms" but really what we mean is "here is the change in voltage on a very small scale and it looks like a little ball." Is that what one means by look like? We don't have any way of visualizing bare protons or quarks (we can see their tracks in a bubble chamber but that's not really seeing the particle, just evidence of its passage). --Mr.98 (talk) 01:40, 2 December 2011 (UTC)

The above response is much better than I could manage; I just wanted to add the picture really. I think the best we can say is, it's a "fuzzy blob" - the bits whiz around, mostly within a certain area. Colour has no meaning at that level (the pic colour is 'fake'). We can't "zoom in", because of the problems mentioned above; it's just... fuzzy, uncertain.  Chzz  ►  02:18, 2 December 2011 (UTC)

You are seeing atoms and exactly what they really look like right now. You just can't distinguish them (technically the photons they emit) from each other. In effect, no different from asking what grass looks like, except that with grass we can get close enough to distinguish individual blades. But you wouldn't, sitting up in the stands, tell the guy next to you that he wasn't actually seeing what grass really looks like. The problem is assuming that things have an inherent look. But appearances are always relative to the perceiver, the perceived, and the conditions, including scale. Hurriquake (talk) 02:33, 2 December 2011 (UTC)
Oooh, I must disagree. The OP is asking about seeing individual atoms, not aggregates. The equivalent is trying to see blades of grass from outer space. If someone was able to tell, from outer space, that grass was green, and had theoretical reasons to believe it was made of blades, well, that's great. But that's not the same thing as being able to see individual blades, and the kind of information you get from the different levels of scale is significant. Even more significant is the fact that quantum effects kick in, which do, in fact, bring into question what one would even mean by "seeing", because you simply cannot "zoom in" on a proton in the that you can on a very small, but still macroscopic, object. --Mr.98 (talk) 19:04, 2 December 2011 (UTC)
Guess who saw Frank Wilczek give a presentation on this very subject in 2006?! End result (after thousands of hours of computation time based on iffy measurements): a quark, the most basic building block of matter we know, looks like an oblate spheroid. The reason it's "oblate" (that is, bulges out) is because of its spin, which is analogous enough to actual rotation that it "looks" the same. Of course, what the "image" actually means or if its worthwhile scientifically is hard to say -- as a lay-audience talk, the image of a quark was more for a "wow" effect.
As for an atom, that's easy: we can image individual atoms using a scanning tunneling microscope - IBM loves to show off how good it's got at this. As you can see, they're essentially spheres or spheroids. Boring.
But these are big, heavy atoms that are easy to see, which means that their outer electron shells (which is the thing we interact with when we touch any matter) are thick and round. If you look at our article on atomic orbitals, however, you'll see that the outer electron shell can make fun, cool shapes (scroll down to the table). These are only visible on the lightest atoms, such as hydrogen, however. Except now we can even image individual hydrogen atoms on graphene! Which means maybe soon we'll be able to see atoms that aren't just spheres! Cool! SamuelRiv (talk) 03:28, 2 December 2011 (UTC)
p-wave AFM lets you detect and visualize the shapes and phase/node patterns of the HOMO and LUMO molecular orbitals specifically (unlike normal s-wave AFM that only gives overall electron density). So some of the π and π* (composed of p atomic orbitals) of pentacene becomes visible. A few years ago someone had published direct evidence for the shape of dz2 and others, but I can't find the reference. DMacks (talk) 17:12, 2 December 2011 (UTC)
Though I would suggest that being able to infer the shape of a quark is not quite the same thing as being able to actually "see" one. --Mr.98 (talk) 19:04, 2 December 2011 (UTC)
Mmm, a strange world, in which colour and even shape, has no real meaning. "Anyone who says that they understand Quantum Mechanics does not understand Quantum Mechanics"-Richard Feynman.  Chzz  ►  20:59, 3 December 2011 (UTC)

Sea Level Rising[edit]

Hi.

I'm a 6th grader working on a science project about sea level rising. I have one question about it. If the sea level gets to high what would happen? My teacher asked that I get information from an expert source so I hope to get the answer to my question. — Preceding unsigned comment added by SentryBravo (talkcontribs) 02:14, 2 December 2011 (UTC)

I'm glad you're working on an important topic like this. One of the most important things for a scientist to do is to make sure that the terms he or she uses are very clear. Think about what "too high" might mean: a sea level that is "too high" for one purpose might be just fine for some other purpose. It might help if you can think of a more specific term than "too high" (unless that is what your teacher told you to do). Short Brigade Harvester Boris (talk) 02:18, 2 December 2011 (UTC)
Start by reading the article Current sea level rise it's quite thorough and even has a section on effects of sea level rise. Depending on what your teacher means by "expert source" that might not include wikipedia, in which case you should look at the sources which are referenced by our article. Vespine (talk) 03:08, 2 December 2011 (UTC)
"Too high for whom?" is a good question. If you were living on a boat, the sea level would never get too high. If you were living on a very low area near the ocean, too high might, over ten years or a century, mean that area would soon be permanently flooded with water. Another article to look at is mean sea level. Answering the question "what is the sea level?" is a lot harder than it seems. The water is constantly moving, and the ocean can be very different heights in different parts of the world. Imagine asking what the height of land is? It's actually many different heights, and it changes over time. Just because water appears flat, doesn't mean it's all the same height. It is also a fact that land itself and the continents can rise and fall according to their temperature and weight. When the earth goes through Ice Ages, the amount of ice on land is strong enough to press the land downward with extra weight, and when the ice melts, the land will actually rise up, out of the ocean. This is called isostasy, eustasy, and post-glacial rebound. Mac Davis (talk) 15:42, 2 December 2011 (UTC)
Low-lying islands, coastal cities, and megadeltas (regions where multiple mouths of rivers empty out into the sea, like the Ganges Delta and the Nile Delta) would be engulfed by the sea. Also see Climate change in Tuvalu, Regional effects of global warming, Effects of global warming#Sea level rise, and Current sea level rise.-- Obsidin Soul 17:04, 2 December 2011 (UTC)

Well... people in formerly landlocked countries or relatively elevated cities would get to go fishing... Those whose dwellings were flooded would quickly learn just how effective their swimming lessons were.Heck froze over (talk) 19:14, 5 December 2011 (UTC)

Unidentified Cricket[edit]

click for bigger version

Can anyone identify this insect? Family members in New Jersey have had their house overrun by them, but the species has never been seen with decades at the same location. The length of the body is about a centimeter. (The cricket is photographed on a standard sized paper table napkin.) They do not "sing". The stripes are brown and orange. At least half a dozen have been found in all levels of the house including in a bath tub.

Also, the "crickets" appear to have multiple "fingers" forming hands at the end of their limbs. (This is visible in the full size picture.) I have never heard of or seen such a thing. Which insects have this feature? What might it be called?

Thanks. Hurriquake (talk) 02:19, 2 December 2011 (UTC)

Well i'm not much of a bug guy and I don't live in the states, but my google fu was strong today. First I thought it had more then a passing resemblance to weta, but after a bit more digging, I found this pest control site specific to New Jersey, I present Rhaphidophoridae, which does actually include "cave wetas" so I wasn't far off at all:) But this one is specifically a Camel cricket. Vespine (talk) 04:18, 2 December 2011 (UTC)
Camelback crickets seem to be common in the mid-Atlantic area this year; I've got a few in my basement and had never seen them there before. Acroterion (talk) 05:13, 2 December 2011 (UTC)
If you're interested in reading more about the critters, you might try chapter 13 of Sue Hubbell's Broadsides from the Other Orders, in which she describes her observations of them. Deor (talk) 11:54, 2 December 2011 (UTC)

Catalase[edit]

What food has the most catalase? More than chicken liver that is.--KAVEBEAR (talk) 04:00, 2 December 2011 (UTC)

Second question: Does soybean have much catalase?
You can't get catalase from food. You have to make your own. Any catalase you eat is destroyed during digestion and cannot be absorbed in useful form by your body. Dominus Vobisdu (talk) 04:18, 2 December 2011 (UTC)
That's not exactly the question he asked. He asked which food has the most catalase, not which food can your body obtain it from. Liver is a good bet, but something like raw blood sausage or other blood-based food may be another good one. --Jayron32 04:56, 2 December 2011 (UTC)

orbitals...[edit]

I tried to find where is the place around the nucleus that has the most amount of probability density and what the amount of is in that point. now, since for 2px and 3px orbitals we know that this point is in xy plane, and on the x axis, I replaced y and z in the wave function equations with 0, and found the derivative for psi^2, and found the root for d(psi^2)/dx=0 and placed the root in the psi^2 equation.Now the strange thing is that the amount of psi^2 in the anti-node of 3p is 0.135 while it's 0.005 for 2p. but shouldn't it be more for 2p? what is my mistake?and please give me answers that are clearer than "it doesn't work that way" and stuff like that.Thanks in advance.--Irrational number (talk) 12:30, 2 December 2011 (UTC)

What makes you say it should be more for 2p? Dauto (talk) 14:31, 2 December 2011 (UTC)
since it is nearer to the nucleus, it must have a more probability density near the nucleus?--Irrational number (talk) 17:24, 2 December 2011 (UTC)
Just because the mean of one probability density function is greater than the mean of a second PDF, doesn't mean the mode is. Red Act (talk) 17:50, 2 December 2011 (UTC)
Ha, I get it.... but a little more explanation makes it better...
I still don't understand what's the problem. The 3p is further out but more concentrated. Dauto (talk) 18:48, 2 December 2011 (UTC)
now I get it.--Irrational number (talk) 19:00, 2 December 2011 (UTC)

Mitosis[edit]

Hi there. Not a biologist here, just an interested onlooker. During anaphase of mitosis, when the two (identical?) sister chromatids of a chromosome split, one chromatid ends up in each daughter cell. I've looked up different references (including Wikipedia), but none seem to go into the details of how each chromatid becomes a genuine chromosome in each daughter cell. As I see it, each daughter cell inherits one sister chromatid from each chromosome in the parent, leaving it with half the number of chromatids it needs to have a full complement of chromosomes. What's going on? Do they make copies of themselves somehow? It's been years since I studied biology, so be gentle! Thanks, Icthyos (talk) 12:59, 2 December 2011 (UTC)

The DNA is copied before the division, not after it. Dauto (talk) 14:27, 2 December 2011 (UTC)
Ah-hah, I see. Does this happen during interphase? So during the entirety of mitosis, the cell already has double the number of chromosomes it normally does? I remember asking my high school biology teacher this, and she said the new cells reproduced the new required chromatids. I never quite bought that... Icthyos (talk) 15:09, 2 December 2011 (UTC)
Yes, this happens during Interphase, during the S-subphase, between the G-phases. The classical scheme of mitosis you learn in school textbooks is a bit outdated and simplified for pupils. A real cell, after the S-subphase of the interphase, is already preparing for a cell division. A cell that has doubled its number of chromosomes only has one way to go: into mitosis. If it doesn't, there are controls in place, in a healthy human cell for example, to ensure that such a defect cell dies. So from a modern, conceptual view, it probably would be better to group the S-phase already into the "mitosis-part" of the cell cycle. The original naming of phases was all done by light-microscopy, where the cell mostly looks the same during the whole interphase. --TheMaster17 (talk) 15:28, 2 December 2011 (UTC)
I think there is a misunderstanding here. After cytokinesis, the two daughter cells have each the exact same chromosomes that the parent cell had had before entering mitosis. A "normal" chromosome, outside of mitosis, consists only of a single strand of DNA. The double structure (looking like an X) you see often depicted when someone talks about chromosomes is in reality only present in a cell during the metaphase of mitosis, after the doubling of its chromosomes into two sister chromatids. Have a look at the first picture in mitosis and follow each chromosome from the parent to the daughters, perhaps this already clears up the misconception. --TheMaster17 (talk) 15:18, 2 December 2011 (UTC)
Okay. That's perfectly clear now, thanks. The doubled-up chromosome (with the two chromatids) picture had stuck with me for some reason, and I thought that's what a 'normal' chromosome looks like. Perhaps because most of the pictorial flow diagrams I'd seen missed out that first picture, and there was never any mention of anything like interphase. Thanks! Icthyos (talk) 15:23, 2 December 2011 (UTC)
Note that you don't really see chromosomes under a microscope during the bulk of the cell cycle - chromatin condensation is a process which is done pretty much specifically for the purpose of dragging them apart (caveatting that "purpose" is something we assign, not the biology, and probably messing with the process would foul up something else). So the X-form chromosome is not really uncommon in the sense that when you see a chromosome it often looks like that. But hunting around for cells with any visible chromosomes on a mitotic spread is not so easy; mitosis is a short phase of the cell cycle and so such cells are not common even in rapidly dividing populations. Wnt (talk) 00:45, 3 December 2011 (UTC)

Late C19—early C20 British diplomat and entomologist named Alfred E. Wileman[edit]

Hi all,
One Alfred E. Wileman appears to have been a late C19—early C20 British diplomat and entomologist, and the binomial authority for Abaciscus costimacula, Taicallimorpha albipuncta, and so on. Possibly a fellow of the Royal Entomological Society of London.
Looking for refs to start article. Any clues about when he was born, died, and so on?
Thank you!--Shirt58 (talk) 15:03, 2 December 2011 (UTC)

Hm... I figured out his middle name - Ernest. Searching for "Alfred Ernest Wileman" and "AE Wileman" gives you more results. He was consul-general to Hawaii, the Philippines, Taiwan, and Vice Consul/Consul for Japan (and [20]).
Wikispecies gives his year of birth and death as 1860 to 1929.
From a preview of The Diaries of Sir Ernest Satow, British Minister in Tokyo (1895-1900): A Diplomat Returns to Japan, I have:
Alfred Ernest Wileman (1860-1929) was appointed a student interpreter in Japan in 1882. Employed as Assistant in the Japanese Secretary’s office from October 30, 1894 to March 31, 1896. H. M. Vice Consul for Hiogo (i.e. Kōbe) and Osaka from December 28, 1896. Transferred to Hakodate on April 1, 1901. (Foreign Office List, 1930)
From The Correspondence of Sir Ernest Satow, British Minister in Japan, 1895-1900:
Alfred Ernest Wileman (1860-1929). Collector and breeder of butterflies and moths. Fellow of the Entomological Society of London. Appointed student interpreter in Japan, 1882. Assistant Japanese Secretary and British Vice-Consul at Tokyo, 1892. Acting Japanese Secretary 1892-94. Assistant in the Japanese Secretary's Office from October 30, 1894 to March 31, 1896. In 1896 promoted Vice-Consul for Hiogo and Osaka. Acting Consul at Hiogo, 1898. Vice-Consul at Hakodate, 1901. Consul for Tainan, Formosa, 1903. Acting Consul-General, Yokohama, 1907. Consul-General for the Philippines, 1909. Retired 1914. (See Kuwata, 2003, pp. 481-482)
A Japanese author also seems to have written biographies of him (but they're probably both in Japanese):
  • Esaki, Teiso (1953). "Foreigners' collecting in Kyushu" [in Japanese]. Shin Konchu, 6y(3):2-7. Itineraries of J. J. Rein, G. Lewis, J. H. Leech, A. E. Wileman, H. Fruhstorfer, & F. Silvestri recorded. Chronology for 1690-1939 supplemented.
  • Esaki, Teiso (1956). "Alfred Ernest WILEMAN (1860-1929)". TINEA 3(1/2):127-128.
There's also this Lepidoptera blog which mentions Wileman thrice, in what seems to be a biography. It's in Japanese though and Google Translate results in gibberish. The Natural History Museum has a picture of him and a short description in this file. That's pretty much all I can glean from the internet.-- Obsidin Soul 16:45, 2 December 2011 (UTC)
That's magnificent work! Thank you so much! --Shirt58 (talk) 03:03, 4 December 2011 (UTC)

Melting plastic[edit]

Odd question- suppose I bought this nice tool I saw today, a heat gun for stripping paint off stuff, 2000W, 350-600°C, then suppose I wanted to use it to melt the surfaces of a couple of pieces of plastic and press them together so they stick as they cool, would it work? What sort of effect would it have on the plastic, would they melt properly, and would they actually stay stuck if I tried something like this? If not, what other sorts of tools are there around that could do the job?

148.197.81.179 (talk) 15:53, 2 December 2011 (UTC)

It depends on what type of plastic is involved. Also there is a high probability that you will make a mess, since the air from a heat gun is hard to control. Looie496 (talk) 16:13, 2 December 2011 (UTC)

OK, is there any way of finding out which plastics this would or would not work with? 148.197.81.179 (talk) 16:18, 2 December 2011 (UTC)

Soft plastics are usually thermoplastics, which should behave as you want. Hard plastics are generally thermoset plastics which will just smoke if you heat them. In any case make sure you do this outside, as heated plastics give off toxic fumes, and be careful as getting melted plastic on your skin will severely burn you. StuRat (talk) 16:31, 2 December 2011 (UTC)
Start in Category:Plastics and read up on plastics' melting points. I know from friends' MakerBots that Acrylonitrile butadiene styrene, polylactic acid, and high-density polyethylene are thermoplastics with relatively low melting points. ~Alison C. (Crazytales) 16:39, 2 December 2011 (UTC)
(e/c)And both pieces of plastic need to have the same melting point for this to work. The manual welding of plastics is a skilled job and even more tricky than welding metalwork. Use of a solvent adhesive is a much safer bet, otherwise be prepared to lose a few pieces of plastic before you succeed in welding them.--Shantavira|feed me 16:44, 2 December 2011 (UTC)


(ec) Be careful with heating and burning plastics - some common, household plastics give off very noxious gases when heated.
If you buy stock plastic material from a supplier, like McMaster-Carr, you can get detailed technical information (including safety information) on the exact chemical compound. "Plastic" is such a generic term that it covers tens of thousands of different commonly-used chemical compounds. If you're using "home-supplied" plastics, you can often check the recycling code to determine the chemical makeup of a particular plastic (at least to get a ballpark idea about what "general class of materials" the object was made from). Unfortunately, though, most plastic you find around the house is pretty "anonymous." Here's Identifying Plastics by Burn Test - also from McMaster. By browsing their website, you can find a lot of other useful information - which glues/epoxies/bonders are compatibile with different plastics/substrates/metals, and so on. Nimur (talk) 16:46, 2 December 2011 (UTC)

So, if I have the right sort of plastic, enough to practice on a few times, and a good place to work, what do I then use to heat the things up? The paint stripping gun was the only heat based tool I could find around the shops, would there be anything better I could look for or ask for that wouldn't cost a fortune? 148.197.81.179 (talk) 19:46, 2 December 2011 (UTC)

Something that hasn't been mentioned yet is that plastic items often still has traces of mould-release compound adhering to it. This is a very tenacious lubricant that can make it difficult to achieve a good homogeneous weld between the two plastic surfaces. Therefore, remove this with some fine glass paper. The roughened surface will also act as a guide to when it has reach melting point because it will start looking wet. Quickly Press the two bits together hard.--Aspro (talk) 20:19, 2 December 2011 (UTC)
There are of course Hama Beads (what? No article?) that are welded together with a clothes iron (through a piece of greaseproof paper). 109.148.238.255 (talk) 00:20, 3 December 2011 (UTC)
For connecting polyethylene pipes welding works. An amateur version was a block of metal heated up, the pipes held on it, and the the pipes put together. I suspect that applying heat direct to the pipes risks overheating (burning and smoking). You can see a picture at commercial site http://www.dixonind.com.au/#/fusion/butt-fusion and smaller ones elsewhere on the site. Polypipe Wrangler (talk) 00:03, 7 December 2011 (UTC)

Heating with an electric heater[edit]

Is it possible to be efficient here or do all the electric heaters convert all electricity to heat? — Preceding unsigned comment added by 80.58.205.34 (talk) 18:09, 2 December 2011 (UTC)

No. Some electric heaters produce light. Light is not heat. -- kainaw 18:15, 2 December 2011 (UTC)
But, won´t this light be converted to heat when it hits (heats?) a curtain or wall? — Preceding unsigned comment added by 80.58.205.34 (talk) 18:26, 2 December 2011 (UTC)
Yes. :All of the electricity is eventually converted to heat, though. The light energy wll eventually be absorbed by a surface and be converted into heat. Dominus Vobisdu (talk) 18:30, 2 December 2011 (UTC)
But, light converted to heat by another object is not electricity converted to heat by the heater. It is electricity converted to light by the heater. -- kainaw 18:36, 2 December 2011 (UTC)
The question was about the relative efficiencies of electric heaters for the purpose of heating a room. Whether the heater does this directly or indirectly is immaterial. Actually, any electrical device is exactly 100% efficient in heating a room: a refrigerator, and electric shaver, a blender, a cell phone, a vibrator. All of the electricity consumed will eventually end up as heat. Dominus Vobisdu (talk) 18:47, 2 December 2011 (UTC)
You're being pedantic, which isn't very helpful. Light gets absorbed by the atmosphere or other objects and becomes heat. ScienceApe (talk) 19:12, 2 December 2011 (UTC)
I'm actually trying to be practical. If I want an "efficient" electric heater, I want to turn it on, sit near it, and get warmed up quickly. But, the answer here is that I could turn on a vibrator and sit near it and somehow I'll get warmed up because all that electricity will, eventually, be turned into heat. When running a heater, it isn't important that you will eventually get the paint on your walls to warm up a tad. You want to get immediate heat transferred to your body. -- kainaw 19:19, 2 December 2011 (UTC)
I wasn't implying that heating with vibrators was practical, just that they are as energy efficient as heaters as any other electrical device. And if you want to get your legs warm as fast as possible, you do NOT want a heater that releases heat, but one that releases light, specifically infra-red light, which gets converted into heat when it hits your legs. You forgot that light can be delivered in a focused beam much more precisely and more comfortably than hot air can in a directed stream. Dominus Vobisdu (talk) 19:32, 2 December 2011 (UTC)
You might want to read what Dauto posted here about light being heat here http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Science/2011_October_22#More_redox_questions
But you're still being pedantic anyway. What's the point of a heater if it only makes you warm when it's blowing directly on you, but the rest of the room is ice cold? I want the whole room I'm in (at least) to be warm. ScienceApe (talk) 19:28, 2 December 2011 (UTC)
And yet, it is possible to do better than that by using a heat pump instead of an electrical heater. Dauto (talk) 18:35, 2 December 2011 (UTC)
Err, aren't those mostly used in air conditioners? I'm pretty sure generating new heat is more efficient than pumping heat out of an already cold environment. ScienceApe (talk) 19:16, 2 December 2011 (UTC)
Heat pumps can be more efficient, but only at certain times of year. To heat efficiently, you need the ground or outside air from which you steal the heat to be warm. StuRat (talk) 20:54, 2 December 2011 (UTC)
No, a heat pump is more efficient than an electric heater. It might be more efficient to use a stove and burn fuel instead of using electricity because electric power is produced by a power plant which usually has an efficiency in the range of 35% to 40%, so that gives the stove a head start relative to any electric device. But if you're comparing an electric heater with an electric heat pump than the heat pump is better every time. Dauto (talk) 20:57, 2 December 2011 (UTC)
Not according to our heat pump article:
"When there is a high temperature differential on a cold day, e.g., when an air-source heat pump is used to heat a house on a very cold winter day of say 0 °C, it takes more work to move the same amount of heat indoors than on a mild day. Ultimately, due to Carnot efficiency limits, the heat pump's performance will approach 1.0 as the outdoor-to-indoor temperature difference increases for colder climates (temperature gets colder). This typically occurs around −18 °C (0 °F) outdoor temperature for air source heat pumps. Also, as the heat pump takes heat out of the air, some moisture in the outdoor air may condense and possibly freeze on the outdoor heat exchanger. The system must periodically melt this ice. In other words, when it is extremely cold outside, it is simpler, and wears the machine less, to heat using an electric-resistance heater than to strain an air-source heat pump.:
So, it would be the same efficiency as an electric heater, if you didn't have to melt the ice that condenses on it. This brings it below the efficiency of an electric heater at outdoor temps under −18 °C (0 °F). A geothermal heat pump solves this, since below ground temps rarely got that low on Earth. StuRat (talk) 21:19, 2 December 2011 (UTC)
The efficiency approaches 1 from above, so the efficiency is always higher than the electric heater efficiency. There is no need to melt the ice outside, except that the ice might cause some problems but that is a matter of poor design that could be solved. No engineer bothers solving this problem since that's not the intended use for the heat pump. Dauto (talk) 21:29, 2 December 2011 (UTC)
You have the cause and effect backwards. The reason why heat pumps aren't intended to be used to heat homes when the outdoor temp is that low is that they would be inefficient. If you have a solution that makes them efficient at low temps, I suggest you patent it and get rich. StuRat (talk) 21:36, 2 December 2011 (UTC)
Modern heat pumps are, in effect, 300 to 400 "% efficient" at mild exterior temperatures, with the "% efficiency" gradually dropping closer to 100 as the exterior temperature falls. "% efficient" is not really the appropriate term, but reflects the fact that the amount of heat delivered to the room is always more than the amount of electricity used, except when the system fails to circulate at extremely low temperatures. I thinks that's actually what you both said above! The reason that heat pumps are "inefficient" in another sense at low external temperatures is that the motor will be working very hard for very little benefit, and will thus cost much more to maintain than the extra heat is worth. Dbfirs 00:18, 3 December 2011 (UTC)
StuRat, I wouldn't get rich because it is cheaper to burn natural gas or heating oil because electricity is produced by power plants that have an efficiency of about 35% to 40% (didn't I say that already?). That's why it's only worth to use a heat pump if you can get efficiencies of 300% or more. There is no money to be made by solving the engineering difficulties you describe and that's why they have not been solved. Dauto (talk) 06:06, 3 December 2011 (UTC)
But if you can rig a heat pump to work like a window air conditioner, and only heat one room, then you have the huge savings of not having to heat the rest of the house. This is why electrical resistance heaters are used, after all, and if heat pumps could be made more efficient at any outdoor temp, then they would be the superior choice. StuRat (talk) 17:39, 3 December 2011 (UTC)
And they are the superior choice. Dauto (talk) 18:15, 3 December 2011 (UTC)
Except for icing up. So we're back to you patenting your solution and getting rich. StuRat (talk) 18:44, 3 December 2011 (UTC)
The icing can be avoided. The only reason it's not done is because it's not worth the trouble. No body is going to get rich that way. Dauto (talk) 21:10, 3 December 2011 (UTC)
Avoided how ? StuRat (talk) 00:56, 4 December 2011 (UTC)
Just keep the temperature of the heat exchanger above the frost point, but still below the outdoor's temperature. That's always possible except when the relative humidity is 100%. Dauto (talk) 04:56, 4 December 2011 (UTC)
In other words "not always possible". The exchangers would also have to be huge, due to the slight temp difference they could support without freezing up, at low outdoor temps. StuRat (talk) 16:18, 5 December 2011 (UTC)
It is always possible, but it's not always practical or affordable and that's why it is not used at such low temperatures. The best solution at low temperatures is to burn fuel locally instead of burning it at an electric power plant. Dauto (talk) 18:35, 5 December 2011 (UTC)
Except that it's often impractical to heat a single room that way. There are kerosene and propane heaters, but the fumes must either be endured, or vented, which then lowers efficiency dramatically. Using a central heating system to heat a single room is also quite inefficient. Thus, electrical resistance heaters are the best choice, to heat a single room in cold weather. StuRat (talk) 18:09, 6 December 2011 (UTC)
Btw, Dauto mentioned that light IS heat. I recalled him saying that, and I found the section here, http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Science/2011_October_22#More_redox_questions
So if Dauto is right, then a heater DOES convert all electricity into heat. ScienceApe (talk) 19:23, 2 December 2011 (UTC)
Dauto is wrong. The OP used the term heat meaning infrared radiation, light is heat only when heat means entropy. Obviously, a heater does not make energy disappear, it just transform it into other forms of energy (IR and sometimes light). — Preceding unsigned comment added by 88.14.196.65 (talk) 00:39, 3 December 2011 (UTC)
Infrared radiation IS a frequency of light. ScienceApe (talk) 02:05, 3 December 2011 (UTC)
And you end up rehashing a decades old argument. "Heat" originally had two meaning. Average idiots walking down the street called anything "hot" to have "heat". Scientists used "heat" as a measurement of entropy. Then, not long ago. an influx of physicists came in and started using "heat" to mean "energy" as if all energy in the universe was synonymous with entropy. The old argument... Is a high-energy particle floating through a complete vacuum considered heat? The smartass young physicists said it was. Then, the Internet was invented. It wasn't long before a bunch of self-trained scientists came around and decided that if heat is energy and light is energy, then light is heat. Thanks to all these Internet scientists, we no longer have to pay for heaters to keep our houses warm. Just shine a flashlight at your toes and you'll be plenty warm. If you aren't, then you're one of those old curmudgeons who apparently differentiate between entropy and energy and should be executed for making life complicated. -- kainaw 21:17, 3 December 2011 (UTC)
Shining a flashlight at your feet does make it warmer, it's just not a practical heater. ScienceApe (talk) 00:17, 4 December 2011 (UTC)
Practical? Wasn't I just called out for claiming that light isn't a practical heater? Wasn't I just lectured that using light to heat up the paint on the walls is more practical than heating the cold person? Please make up your mind. -- kainaw 17:49, 4 December 2011 (UTC)
Well, radiant heaters use infrared light as a heat source. I don't really like them, though, as the side away from it is always cold. Some ambient heat is also needed. StuRat (talk) 03:58, 5 December 2011 (UTC)
Kainaw, what you said is completely wrong. Physicists don't use the word heat to mean entropy. Heat and entropy are not even commensurable. Heat, being a form of energy, is measured in Joules, while entropy is measured in Joules/Kelvin. Light is a clearly a form of heat and that's why wearing dark closes out in the sun makes you feel warmer. — Preceding unsigned comment added by Dauto (talkcontribs)
Study your history, or at least read heat. Even the Wikipedia article states that "heat" (which was introduced by Joseph Black long long long before this whole internet thing) was what we now distinctly refer to as entropy. So, either I am wrong, Wikipedia is wrong, and every book on the history of Physics and Chemistry is wrong - or - you are ignorantly attempting to make a claim that is wrong. -- kainaw 17:49, 4 December 2011 (UTC)
Historically heat and entropy were indeed confused with each other but the modern definition of heat is that it's a form of energy while entropy isn't. Dauto (talk) 21:17, 4 December 2011 (UTC)
I just read the second comment you made. Wearing dark clothes makes you warmer, so light is heat. You are claiming that if I walk outside wearing all black, the LIGHT will somehow pass right though my clothes and the light will warm my body. So, if I hang black curtains in my window, the light will pass right though and the light will warm my room. All of this must be true because some guy on the Internet is the lord of physics and has declared that light is heat and heat is light and nothing between shall ever exist? I suppose that there are no atoms in my black clothing. I suppose that none of those atoms have electrons. I suppose that none of those electrons are affected by electromagnetic radiation (light). I suppose that those electrons do not increase energy when absorbing light. I suppose that increasing the energy of an atom does not make it move, wiggle, or spin a little more. I suppose that increased energy and movement in a closed system does not increase entropy. I suppose that when a system of higher entropy is joined with a system of lower entropy that entropy does not move to create a more uniform mean entropy in the joined system. I suppose that when my black clothes have more entropy, contacting my body which has less entropy will not cause the entropy of my body to increase. I suppose that everything on my bookshelf is absolutely wrong and the entire physics department at my university should be fired and replaced with some random guy on the Internet. -- kainaw 17:59, 4 December 2011 (UTC)
Kainaw, you're also just a random person on the internet so stop complaining about that. I happen to have a Phd in Physics. What have you? I don't know where you got the idea that I was saying that the light passes right through your dark closes. Obviously heat is being transferred from the closes to your skin through the process of conduction. Heat was also transferred from the sun to the dark closes through the process of radiation. A sizable fraction of that radiation was visible radiation and it heated up your closes. Dauto (talk) 21:16, 4 December 2011 (UTC)
  • Kainaw and Dauto are just random guys on the Internet, but only one can be right. Dauto seems to be the classical random guy on the Internet who, when proven wrong, claims to have a PhD My vote goes clearly to Kainaw.
Unfortunately, this is not a popularity contest so your vote counts for nothing. I wasn't proven wrong. Just read the article titled (of all things) heat and you will see that radiation is a form of heat. Here's a quote from my copy of Thermodynamics by Enrico Fermi which happens to be at arms length. Page 17, Line 6: "Q can be interpreted physically as the amount of energy that is received by the system in forms other than work". Q is the the heat and it includes everything but work. That means that radiant energy, including visible light, is a form of heat. Usually I don't bring up the fact that I have a PhD in Physics but Kainaw's bad attitude declaring that the whole physics department of his university should be fired because I'm the Lord of physics was dripping with a bit too much sarcasm. I don't know who Kainaw is or what university he's talking about but I'm sure he doesn't speak for the University's department of physics. Dauto (talk) 04:03, 5 December 2011 (UTC)
A couple points:
1) The light generated from electric heaters, whether visible or infrared, is a very small portion of the electricity used.
2) Of that light, only a tiny portion escapes the room, through windows.
So, the efficiency of an electric heater at changing electricity into usable heat is probably over 99%. Of course, this doesn't mean it's always the best choice, since electricity costs more than many other energy sources, especially natural gas. It works out that heating the entire house with electricity is usually prohibitively expensive (unless you live someplace where it only gets a little bit cool in winter). However, if you only heat the room you're in at the time, and let the rest of the house remain cold (but not so cold that the pipes freeze), then an electric space heater can be more economical. As for type, the oil-filled electric convection heaters that look like radiators don't have an annoying fan or make a burning smell when you turn them on, and, if you leave them on low, probably won't get hot enough to burn you or cause a fire. They do tend to heat the ceiling, though, but that's not bad if you don't mind heating the room above you, too. StuRat (talk) 21:10, 2 December 2011 (UTC)
With heaters all this talk about theoretical efficiency is interesting, but doesn't help with buying decisions. With heaters "efficiency" is more about how the heat is distributed in your house, and using the smallest possible heater to warm the things you want warm and not warming things you don't care about. An "efficient" heater for keeping your toes warm under your desk is different than an "efficient" heater for warming a large living room with a twenty foot ceiling. (Even though both heaters will convert >99% of their electricity to heat.)
You could argue that I'm abusing the definition of "efficient", but I'm not, I'm just pointing our that the technical scientific definition of the term does not always perfectly align with the 'everyday' definition of the term. APL (talk) 06:13, 3 December 2011 (UTC)

Diameter of an electron[edit]

What is the diameter of an electron? Whoop whoop pull up Bitching Betty | Averted crashes 21:26, 2 December 2011 (UTC)

did you read the article Classical electron radius? I have created a redirect for those that ask the same question. As a number the diameter from this point of view would be 5.6 attometers. Graeme Bartlett (talk) 21:37, 2 December 2011 (UTC)
The "classical" part is important here, as it refers to size estimates that predate the modern understanding of quantum mechanics. While it is useful for some calculations (such as the scattering of photons against electrons), this is not actually a measure of the size of an electron. As far as we know, electrons are point particles with no intrinsic size at all. The current limit is that an electron has no internal structure larger than 10-18 m, consistent with zero size and already 1000 times smaller than the classical electron radius. Dragons flight (talk) 00:15, 3 December 2011 (UTC)

Galilean moons and longitude[edit]

In Greenwich Time and the Discovery of the Longitude, Derek Howse discusses Galileo's discovery that the Galilean moons could be used to demonstrate one's longitude; as he notes, Galileo observed that moons appear to be eclipsed behind the planet at the exact same time, regardless of where on Earth the observer is located. How did Galileo discover that their eclipses appear worldwide to be simultaneous? Or did he simply calculate it and find no reason to disbelieve the calculations? I've checked the article on the moons; you'll see that I added a citation to Howse for the longitude sentence, which was previously uncited. Nyttend (talk) 22:15, 2 December 2011 (UTC)

Hmm, that's perhaps not a sufficiently clear question. I'm essentially trying to ask the following: how did Galileo discover that observers anywhere in the world would see the eclipse at the same time? Nyttend (talk) 22:20, 2 December 2011 (UTC)
Maybe the parallax of Jupiter (against the background of stars) indicated that this planet was so far away, that he reasoned that these eclipses would appear the same from wherever there are viewed from earth. The earth revolves, so he had a base line, roughly the diameter of Earth from which he could try and ascertain the parallax. --Aspro (talk) 22:33, 2 December 2011 (UTC)
Whoops... Or should we include the Earth's orbital diameter as a base line - for distance estimation as well?--Aspro (talk) 22:42, 2 December 2011 (UTC)
Well, we need the distance from the Earth to Jupiter, which, at the minimum, is Jupiter's minimal orbital diameter minus the Earth's maximum. I get 740 million km minus 152 million km, or 588 million km for the min distance. The Earth's diameter is around 12,740 km. I get under half a degree when I take the arctangent of 12,740/588,000,000. So, not much difference will be visible with only half a degree in parallax. I imagine the moons will cover up a slightly different spot on Jupiter, depending on your longitude on Earth, but you'd need a rather powerful telescope to be able to distinguish that. StuRat (talk) 23:10, 2 December 2011 (UTC
Since Galileo was a smarty-pants he wouldn't have any difficulty in working out its 11 or 12 year orbital period based on several years of observation and then sussed out the the angular orbital change of position around the sun (he was, I believe, famous for this rearrangement of the firmament) in six months (a diameter). From that he could work how far away Jupiter was in units of 10 inch pizzas or what ever units of measurement they used in those days.--Aspro (talk) 23:35, 2 December 2011 (UTC)
Half a degree of parallax is huge! That's the angular diameter (not radius) of the Sun and Moon, and about half the FOV of a telescope on low power. It's easily perceptible to even the naked eye. What matters is not the parallax of Jupiter itself, but the difference in parallax between Jupiter and its moons. --140.180.15.97 (talk) 02:33, 3 December 2011 (UTC)
Interesting... Would this be a practical method for determining longitude (assuming of course that your GPS receiver got fried by a lightning strike)? 67.169.177.176 (talk) 22:59, 2 December 2011 (UTC)
No easy answer to that question. It could be, but other factors have to be taken into account and if one was on say a sailing ship, then the delay whilst one did all the corrections could end up with one's ship on the rocks. One problem that springs to mind is that due to the distances involved one would have to allow for the the time it takes light to travel from Jupiter to Earth (the moons obit the same period regardless). The Earth's orbit alone is about 17 light minutes in diameter. Then one has to consider the constantly changing distance to Jupiter. Lastly, as the Royal Navy used to issue a tot of good Jamaican rum each day to the crew. I wouldn't trust any of them to punch the right keys on their slide-rules (given the rums proof -which was something like 60). Personally, I would rather have my rum on the rocks than my vessel. --Aspro (talk) 23:56, 2 December 2011 (UTC)
I believe the navigator would be an officer, and, as such, would be expected to remain sober while on duty. A more serious problem would be trying to locate Jupiter in a telescope on a swaying ship. A stable platform is a must for such work. StuRat (talk) 00:00, 3 December 2011 (UTC)
What about using the Gatty bubble sextant for this job? 67.169.177.176 (talk) 00:04, 3 December 2011 (UTC)
The bubble only provides and artificial horizon. It magnification is chosen for solar/lunar sightings. On a heaving ship it would be difficult to view Jupiter with a hand-held telescope but for Galilean moon sightings I think I could just do it with a low power scope. However, to get a good fix one need to take several sightings and take the average. Whilst this is easy with a celestial body like the Sun, the eclipses of Galilean moons are a momentary events. Then there is human reaction time and the slowness of the moon appearing/occulting. It would be easy to end up with several several nautical miles of error. Not much use if one is sailing between the Islands of Penzance. --Aspro (talk) 00:31, 3 December 2011 (UTC)
Using Galilean moons is one of several astronomical methods of measuring longitude proposed before the invention of the marine chronometer. It was used successfully on land during a few occasions, but the technical difficulties made it too hard to actually use on a moving ship. Dragons flight (talk) 00:22, 3 December 2011 (UTC)
It was known before Galileo that Solar parallax is tiny and the distance between the Earth and the Sun is more than 1,000 times the Earth's radius. (The actual value of the ratio is 23,455, but ancient astronomers were using wrong values in 1,100 to 1,200 range.) Copernicus worked out that the distance between the Sun and Jupiter was five times the distance between the Earth and the Sun. From these two numbers, using the false lower value in the first case, Galileo could have worked out that the parallax of Jupiter would be around 1.5 arcminutes. Then observers on two opposite ends of the Earth would see the same eclipse separated (time-wise) by about 2 minutes for the outermost moon (Callisto), and 10 seconds for the innermost moon (Io).
So, having a table of eclipses computed for, say, Paris or London, a telescope, and a good clock that is set to local solar time, you can compute your longitude with the accuracy of (10 seconds) / (86400 seconds) * 360º = 2.5 arcminutes. Which is, I think, vastly better than the precision of existing navigational instruments like the astrolabe at the time of Galileo.--Itinerant1 (talk) 02:14, 3 December 2011 (UTC)
But if you have a good clock set to local solar time, why would you need the moons to determine longitude? There are much easier ways, which is why the problem of determining longitude at sea was solved by the invention of the chronometer. --140.180.15.97 (talk) 02:33, 3 December 2011 (UTC)
As far as I understand, to use the easier methods, you need a clock that is set to a standard time, such as Greenwich time. If you have a marine chronometer, you'd set it to Greenwich time before leaving and it has to remain accurate for weeks or months on end. Galilean moons play the role of such a chronometer. The clock that is set to local time should only run for a day or two, to let you measure the interval of time between the eclipse and the nearest solar noon. (Or you could dispense with the clock altogether, if you know how to calculate time from positions of stars, but that is more complicated.)--Itinerant1 (talk) 02:56, 3 December 2011 (UTC)
The problem is that good chronometers hadn't yet been invented in Galileo's day: they couldn't keep going for long periods of time without significant inaccuracy in the rapidly changing weather and frequent turbulence of a ship at sea. Galileo's method worked well on land (a king of France sent surveyors to map the country; when they produced a new map of France that was vastly different from the very-badly-distorted former official maps of the country, he remarked that they had lost him more land than his armies had gained in a long time!), but it pre-dated many of the inventions that have been suggested here. Nyttend (talk) 04:45, 3 December 2011 (UTC)
During the era of the first circumnavigation of Vancouver Island, by the Spaniards Galiano and Valdés, along with George Vancouver, 1792, marine chronometers were not reliable after sailing halfway around the world—they required calibration. All three captains calibrated their chronometers whenever practical by going ashore and setting up a little "observatory", taking readings of the moons of Jupiter. From what I've read, it was usually easier for them to use lunar distance observations, because transits of the moons of Jupiter are not an everyday thing. During their circumnavigation, Galiano and Valdés had four chances to observe a Galilean transit, but on two of the nights it was cloudy. Pfly (talk) 04:47, 3 December 2011 (UTC)