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November 27

Does blood really thicken in cold weather?

I have heard people say that their blood thickens in colder climates, and that this is why one can be more accustomed to the cold if one has lived their whole lives in, say, Canada versus visiting after spending their whole lives in Mexico.

Is this true? Or, is it just a joke, and something else causes one to become more accustomed to the cold? I suppose that it *could* be true, in that cold contracts things and heat expands it, but there are so many other things, such as nerve cells, that impact how we feel heat and cold. 209.244.187.155 (talk) 02:23, 27 November 2009 (UTC)[reply]

See Acclimatization. There is a real biological/psychological process whereby someone becomes accustomed to (acclimated to) their environment, to the point where they become more comfortable with the conditions where they live. Blood itself doesn't become "thicker", per se, but such phrases are merely a "figure of speech" which is understood to refer to acclimatization to temperatures. --Jayron32 03:41, 27 November 2009 (UTC)[reply]

Dehydration sets in if the amount of water in the blood is just 2% below normal. At 5% below normal, one may become groggy, dizzy, tingly and get serious headaches and such. At 10% to 15% below normal - you're dead. So if there were any real "thickening" - it would have to be by a very small amount. I agree with Jayron32 - this is just a figure of speech that describes some more complex acclimatization. SteveBaker (talk) 04:07, 27 November 2009 (UTC)[reply]

Short-term exposure to cold can significantly "thicken the blood" (as measured by blood viscosity): PMID 17929604; however, longer term acclimatization to cold weather results in a compensatory thinning of the blood: PMID 10627870. Such studies aren't conclusive, but they do seem to make intuitive sense. -- Scray (talk) 04:14, 27 November 2009 (UTC)[reply]

Water doesn't thicken until it freezes, while oils and fats do thicken gradually when they cool. Since blood contains both water and (a small amount of) fat, I'd expect it to thicken some when it cools. However, note that the core body temperature doesn't drop much in cold weather (unless you go into hypothermia, that is). However, the blood temp in the extremities, like fingers and toes, could drop considerably during cold weather, if you aren't dressed properly. So, blood in those areas could become significantly thicker. StuRat (talk) 05:47, 27 November 2009 (UTC)[reply]

Water's viscosity does almost double when its temperature drops from 37 degrees C (body temp) to 20 degrees C, though the viscosity is still low compared to oils for example. -- Scray (talk) 06:58, 27 November 2009 (UTC)[reply]

Blood viscosity is much higher than that of water, even though blood is mostly water, because proteins (e.g. fibrinogen) and other constituents of blood have a big effect on viscosity, as noted in the blood viscosity article and Pubmed references I cited in my prior edit. Hence, blood really is thicker than water, even though it's mostly water. -- Scray (talk) 16:55, 27 November 2009 (UTC)[reply]

Blood is a non-Newtonian fluid - it behaves in very strange ways and conventional measures of viscosity are not entirely meaningful. SteveBaker (talk) 19:50, 28 November 2009 (UTC)[reply]

Certain situations can make a person more adaptive to temperatures than otherwise. For example, getting out of the shower or harnessing energy in martial arts. ~A H 1^(TCU) 01:58, 2 December 2009 (UTC)[reply]

quantum mechanis

How much quantum mechanical is experimental or a theoritycal subject thats means scope for thinking?Supriyochowdhury (talk) 03:24, 27 November 2009 (UTC)[reply]

The computer you are sitting in front of wouldn't work if it were not for Quantum tunnelling. So-called "NAND flash memory" - such as you find in USB memory sticks use a technique called tunnel injection for writing and tunnel release for erasing. Neither of those things would be possible if quantum theory were not true. No - it's safe to say that quantum theory is very real - and quite applicable to all sorts of real-world applications. SteveBaker (talk) 04:15, 27 November 2009 (UTC)[reply]

Thae article Quantum mechanics is your friend. Cuddlyable3 (talk) 09:23, 27 November 2009 (UTC)[reply]

Even some of the parts that appear on the face of it to be pure philosophy—like the debates over the EPR paradox—can turn out to have not only testable results, but practical importance (see quantum cryptography). --Mr.98 (talk) 13:47, 27 November 2009 (UTC)[reply]

theoretical chemistry

will u give me information about theoritical chemistry.Supriyochowdhury (talk) 03:30, 27 November 2009 (UTC)[reply]

We can give you the information in the article Theoretical chemistry. There is also the professional journal Theoretical Chemistry Accounts which presumably discusses cutting-edge work in the field. --Jayron32 03:37, 27 November 2009 (UTC)[reply]

I have worked with theoretical chemists; their discipline runs across a loosely-defined boundary. You might also want to read about applied physics, specifically molecular physics; also, chemical physics, computational chemistry, molecular dynamics are commonly synonymous. In some cases, protein folding or biophysics can all be classified as "theoretical chemistry" (I have known self-described theoretical chemists in all of these disciplines). Most often, the theoretical chemists I know are using mathematical and computer-based models to predict chemical properties and the aggregate behaviors that result from microscopic chemistry. In some cases, there are no applications. My personal exposure is more applied; so the theoretical chemists I have worked with have also had applications in biochemistry, petrochemistry, atmospheric chemistry, and photochemistry. "Theoretical" can mean a lot of things to different people, ranging from thought-experiments and chalkboards, all the way to lab-work for next-generation materials that don't exist yet. For comparison, see applied chemistry or chemical engineering, which are categorically the opposite of theoretical chemistry. Nimur (talk) 16:55, 27 November 2009 (UTC)[reply]

which are the compounds in which boron has -3 oxidation state?

which are the compounds in which boron has -3 oxidation state? —Preceding unsigned comment added by Danishmanzar (talk • contribs) 05:03, 27 November 2009 (UTC)[reply]

I am not sure there are any; there are a class of compounds called Borides, which have nominally negative oxidation states for boron. However, these are network solids for which concepts like "oxidation state" has little meaning. You could probably find some "boride"s with a nominal -3 oxidation state, but this has little meaning when discussing compounds of this type. --Jayron32 05:13, 27 November 2009 (UTC)[reply]

Are there any compounds in which boron has -4 or even -5 oxidation state? --84.61.167.221 (talk) 12:13, 27 November 2009 (UTC)[reply]

Again, it may be possible to find compounds where you could calculate a nominal oxidation state of Boron at just about any negative number; however the nature of such compounds is such that those are meaningless oxidation numbers. Many boron-containing compounds are network solids, for which "oxidation state" is a meaningless statistic. Many borides, in fact, have fractional oxidation numbers, which again is a meaningless reality, since you cannot have a part of an electron. There may be, or there may not be, boron-containing compounds which could be calculated to have -3 or -4 or -5 oxidation states, but the chemistry of these compounds is such that such calculations aren't going to be useful. --Jayron32 04:48, 28 November 2009 (UTC)[reply]

Are there any compounds in which boron has +6 oxidation state? --84.62.199.19 (talk) 22:02, 28 November 2009 (UTC)[reply]

No, Boron only has 5 protons. Most commonly, if one really wants to count it, it has a +3 oxidation state, as in Boric Acid. I'm not sure that, under normal room conditions, one would find boron in any positive ion except +3, since those two core electrons are bonded rather tightly. --Jayron32 02:13, 29 November 2009 (UTC)[reply]

I assume each boron atom in bis(pinacolato)diboron is formally +2. DMacks (talk) 15:59, 29 November 2009 (UTC)[reply]

Deep sea fish: do we know how the buggers cope with the insane pressures?

Well I'm curious if the mechanisms by which a small puny fish survives while machanical behemoths may fail have been understood. Also do we know how do they breath? They must have special adaptations in there too right?Bastard Soap (talk) 07:58, 27 November 2009 (UTC)[reply]

"Mechanical behemoths" have to survive both standard atmosphere and deep-sea pressures, and everything in between. Deep-sea fish only have to survive one, which is a lot easier. (Similarly, I've heard that designing something that can go at Mach 2 isn't difficult - the difficulty is making something that can reliably do Mach 2 and Mach 0.2.) Vimescarrot (talk) 09:10, 27 November 2009 (UTC)[reply]

Deep sea fish (see article) are aclimatised to their environment by having their internal pressure matched to the external pressure so there is no differential stress on the body structure. That is unlike a submarine in which the hull must resist external pressure to maintain comfortable air pressure for humans. A deep sea fish would experience difficulty if it were to ascend rapidly to shallow water, comparable to The Bends known to divers. We have no reliable sources to confirm or deny the occurence of buggery or insanity at deep sea depths. Cuddlyable3 (talk) 09:20, 27 November 2009 (UTC)[reply]

Or bastardy, for that matter. ;-) --Heron (talk) 20:32, 27 November 2009 (UTC)[reply]

Please do not make fun of an OP's name from Heron. Cuddlyable3 (talk) 09:19, 28 November 2009 (UTC)[reply]

We know more about the possibility of life on other planets than we do for the depths of our own oceans. So perhaps there is a food source such as brine pools, dead corals and plankton, or sulfur from sources such as hydrothermal vents. ~A H 1^(TCU) 01:57, 2 December 2009 (UTC)[reply]

Unmanned submarine design

This brings up a design question I thought of before. In an unmanned submarine (one with just cameras and such), could the pressure be kept equal inside and out, since there's no human the pressure changes would kill ? I doubt if you'd want seawater inside the submarine, so you could use pressurized air tanks (or some other gas), instead. Of course, the electronic and mechanical components inside the sub would need to be able to withstand pressure changes, but the hull would no longer need to be thick. Has anyone tried such a design ? StuRat (talk) 07:52, 28 November 2009 (UTC)[reply]

In principle yes but the submersible would be loaded with an extra pressurised gas tank, a pressure regulator and a compressor pump. Omit the compressor pump and you have to vent the gas on returning to the surface, which means you need a compressor to refil the tank anyway. However why should a submersible need a controlled pressure compartment at all? Cuddlyable3 (talk) 09:26, 28 November 2009 (UTC)[reply]

But that compressor needn't be aboard the sub, which makes all the difference. StuRat (talk) 13:41, 29 November 2009 (UTC)[reply]

Some electronics don't do well when wet. There are options, such as coating everything in epoxy or waterproofing everything, but this is not always possible; it's especially notable that deep oceans are pressurized and saline - so you really need to be careful about where seawater can be allowed to flow. Most often, the electronics cabin is inside a pressure hull, which is filled with air at or above atmospheric pressure. It's not totally unheard of to have wet electronics, e.g. no "interior", but it's less common. Nimur (talk) 21:06, 28 November 2009 (UTC)[reply]

You don't need tanks, or a pressure regulator or anything like that. A simple flexible bladder filled with air (a big bag o' air) outside the submarine would work just fine. As the pressure went up, the bladder would get smaller, but the pressure inside and out would be the same, and no saline would enter. Ariel. (talk) 03:35, 29 November 2009 (UTC)[reply]

I like that idea. One negative, though, is that such a large bag of air might cause currents to carry the sub away. Also, such an expanding and contracting (rubber?) bag would be likely to rupture eventually. StuRat (talk) 13:41, 29 November 2009 (UTC)[reply]

Another design might be to have a hose reaching down from the surface to pump in pressurized air as needed. At the top could either be a manned ship or perhaps an unmanned service vehicle which is dragged along on the surface behind the sub. A variation on this theme might allow the submersible vehicle to disconnect from the hose once it reaches proper depth and interior pressure. It could then go into sunken ships, caves, etc., and vent the excess air as it returns to the surface.

Or, instead of a hose from the surface, there could be a (manned or unmanned) "base" at proper depth underwater that would deliver pressurized air to the sub. The sub could then return to that base to drop off samples, recharge it's batteries, etc., and the air could therefore be retained. This approach might work well to do very detailed studies of one particular area of interest, like a region of deep sea smokers. StuRat (talk) 13:47, 29 November 2009 (UTC)[reply]

I seem to recall that underwater electrical plugs used by divers are in things like bottomless bottles filled with oil. If oil does not transmit electricity then perhaps a robot-submarine could be filled with it, or some other incompressable non-conducting fluid with an opening underneath to the outside. 78.146.171.75 (talk) 12:13, 30 November 2009 (UTC)[reply]

is it hard to make vacuums?

Is it hard to make a large area of vacuum on Earth, for example the size of a football stadium etc? I would think it would be really easy, because you could just have a bunch of layers, for example 100, then each one would only have to be able to support the pressure difference of 1/100th of an atmosphere... but maybe I'm not thinking clearly... so anyway: would it be hard to cover a football stadium, and make a vacuum out of it, using many layers or any other technique? Thanks. 92.230.68.102 (talk) 13:02, 27 November 2009 (UTC)[reply]

There are several problems here. But the main one is coping with atmospheric pressure, which is about 101 kN/m², or the equivalent of 10 tons per square meter. So, roughly, every part of your construction must be strong enough to carry that weight - unless I missed a zero somewhere, that means that if you pack your construction with cars, one next to the other, with so space in between, it would need to be able to withstand the weight of about 60 layers of (empty) Toyota Corollas. --Stephan Schulz (talk) 13:11, 27 November 2009 (UTC)[reply]

This is a moot point, and it's pretty much categorically wrong (sorry Stephan Schulz). DOT-rated tanks (the sort you see every single day) are rated to 3000 psi gauge pressure. A theoretical perfect vacuum only has 14.7 psi gauge pressure. It'd be a piece of cake to make a tank or structure of aluminum that can withstand this - and in fact, they are commercially available. Anver or Fisher sell all the parts you need. If you felt like building a vacuum system the size of a football stadium, some unique engineering challenges would arise; but the total material stresses would probably be less severe than, say, a large crane or cantilever bridge. Sorry Stephan Schulz, your calculations are very wrong. For comparison, think of an average tire - it's inflated to a ballpark of around 30 psi - which means that the contact force of a Toyota Corrolla is ... 30 psi. This is actually twice atmospheric pressure, on the road surface, everywhere the tire contacts the road. Nimur (talk) 17:05, 27 November 2009 (UTC)[reply]

Actually, it's three times atmospheric pressure, as the gauge only registers pressure difference to the outside. A Corolla's weight is 1300 kg, divided by 4 that's roughly 400 kg per tire. Assuming a contact patch of 200 cm², that is 400*50kg per square m, or 20 tons - with 2 times overpressure that is exactly compatible with my computation. The other way round: The Corolla covers 4.3 times 1.7 m, so the atmospheric load on a Corolla-sized patch is 7.3 times 10 tons, or 56 Corollas. What you have shown is that we can make containment vessels that can withstand this force (see below for the submarine examples), not that my calculation is wrong somewhere. --Stephan Schulz (talk) 17:32, 27 November 2009 (UTC)[reply]

There is one thing you didn't address - my question about layers. If you have a 1 meter-per-side cube, then if you empty it you are right, you create 10 tons (101 kN) of force on each side. Okay, 10 tons is too big, I want to reduce it to 500 pounds. So, I only create a 2% vacuum, so that the pressure differential is 0.02 atmospheres. Now, my question is, why can't you create that pressure differential even if immediately inside the large cube there is a SLIGHTLY smaller cube, that also supports 500 pounds of force on each side -- and, you guessed it, inside of it you put another 0.02 atmospheres of pressure difference. So the difference between the outside air and the inside of the small cube is now 0.04 atmospheres, but each one only has to support 0.02 atmospheres. The difference in size between the two cubes could be negligible: a millimeter is plenty. Now you repeat it for 50 cubes (like Matryoshka dolls), getting down to a perfect vacuum. If there is a millimeter of buffering each one (2 millimeters per side total), then the 50th cube should be 100mm, ie 10 centimeters smaller than the largest one. So it will be 0.9 meters per side, instead of 1 meter. But inside it it will have a perfect vacuum, and none of the intermediate cubes have to support more than 500 pounds per face. 500 pounds is not so hard to support, I bet styrofoam can do it (if there's no sudden jerk but the pressure is slowly laid on). My question is why this couldn't be done on a larger scale -- why couldn't you have a football stadium inside 100, 1thousand, 100thousand, however many layers it took to get the pressure differencial so slight that it is very easy to build nested structures that each support it. What is wrong with my reasoning here? 92.230.68.102 (talk) 13:53, 27 November 2009 (UTC)[reply]

For one, you neglect the thickness and the weight of your layers. 500 pounds per square meter still is significant loading. Styrofoam is not airtight, and to hold 500 pounds on it you would need a very significant thickness - certainly more than 10 cm. And whatever you build, each layer must have completely independent structural integrity. You cannot rest the outer layer on the inner layer, as that would also transfer the force. So you are talking about building 100 complete enclosures around your stadium, each one a little bit bigger than the one before. I don't think that is even technically feasible - it's certainly not feasible economically. --Stephan Schulz (talk) 14:32, 27 November 2009 (UTC)[reply]

Thank you, but I was not asking not in a technical, or economical sense, but a physical sense. Nice catch that you can't in any way rest the outside ones on the inside ones, as that would transfer force. Unless of course you somehow had "pylons" go through seals (it's easy to seal 0.02 atm) directly to the center, if you can imagine what I'm talking about: I mean like a Kline bottle. So, again, in a physical sense (not really an economic one), imagine if cost wasn't an issue. So, I want a football-stadium size vacuum, so the technique is:

like this (I don't know how to embed this so it appears inline -- if someone else can, please replace this with an embedded version! Thank you.)

where you imagine these are spheres, the pressure difference between each one is small, and when you see one going "through" another it means a seal lets it through without equalizing the pressure. This is only 4 layers, but I don't see why a thousand spheres couldn't be put one inside the other, again the only constraint for me would appear to be the total thickness of each sphere. You are right that styrofoam is probably a bad idea -- way too thick. But, again if cost were not an issue, would this general idea work to set up a vacuum in a very large (say football-stadium sized) cavity without bearing the enormous atmospheric pressure on a single surface? Thanks for any further analysis! and if you can put my image inline somehow. 92.230.68.102 (talk) 15:19, 27 November 2009 (UTC)[reply]

Now you talk about 100 (or even 1000) sets of pylons. I don't know if that's impossible, but certainly very complex, and there wont be much space for the vacuum, with, say, 10000 pylons in the way. Why not just build one thick containment vessel instead? If you are interested only in the physical principle, and not in technical and economic plausibility, just create a sufficiently large and thick steel sphere. A Los Angeles class submarine at depth withstands 20 times the pressure, and it already is 110 m long/10m wide. The Russian Typhoon class submarines are 175 times 23 m, and can go down 400m, for 40 times atmospheric pressure (but they have 2+1 separate pressure hulls). Neither is quite stadium-sized, but they are not too far off. --Stephan Schulz (talk) 16:17, 27 November 2009 (UTC)[reply]

I second What Schulz said above. There is no reason not to build a single hull strong enough to hold the atmospheric pressure. Dauto (talk) 16:28, 27 November 2009 (UTC)[reply]

Is there ANY limit on the size of the cavity inside that could be produced in this way (outside technically/economically)? Is there a size that the steel sphere would collapse on itself or something due to the limit on how much compression it could take?

also: is there any difference then between one large sphere and the pylons solution above, in terms of the total amount of material you would end up using? Is it the same to have 10,000 or 100,000 matryoshka dolls getting the pressure down to zero by degrees, versus melting them all together and making one large one with the pressure going from atmosphere straight to zero? What is the mathematical/physics analysis of this equivalance? (if it guaranteed that the total structure would have as much mass, and it's not possible to save on weight by doing it this way). By the way thanks for your answers.

and finally: what would the approximate thickness be for a large hollow steel sphere with a vacuum inside, with a cavity of a volume equivalent to the size of a stadium. How much would it weigh? Is the sphere the best shape for this? (as opposed to, say, a dodecahedron).92.230.68.102 (talk) 16:35, 27 November 2009 (UTC)[reply]

The problem here isn't going to be material properties. We have very strong metals and we know how to engineer good trusses to support large semi-hollow spaces. The hard part will be pumping. It will take a lot of very powerful pumps to get this much air out of a big chamber. Even small (10-cm diameter cylinder) vacuum chambers suffer from a variety of interesting phenomena - first, of course, is leakage and backflow through the pump system; so you switch off the roughing pump and turn on the bigger roughing pump or the turbo pump (which you can't even operate until you're at pretty low pressure!). Next, leakage through the real, non-ideal seals, flanges, and gaskets; and then when you get down to ultra high vacuum, weird things begin to happen. Your pumps have set up a steady state, you're accounting for leakage; but the pressure doesn't go down any more! Weird! Where did all these gas atoms come from! Every time you pump them out, something still registers. By now, your pressure is so low that hydrogen atoms which used to be embedded into the metal lattice of your structure start to outgas. So you turn off the turbo and turn on the ion pump; and you get most of the hydrogen out. But now some really weird things happen. Helium and Neon start percolating through the two inches of solid, hermetically sealed steel. Because these noble gases are sort of small and they don't really interact with the electrons of transition-metal atoms, they just "squeeze" through the pore spaces. And, because they aren't ions, you can't get rid of them with the ion-pump! So, you turn down the cryo and liquefy everything that is gas; and you can get rid of most of the neon... but the Helium won't go away! Well, your cryo is made of helium, so you can't get it much colder than that; and something about the Second Law of Thermodynamics is kicking in. No worries - turn on the optical laser system and start plucking any of the remaining atoms out of the vacuum chamber - ... you say you want this for a foo tball-sized vacuum-chamber arena? The logistics problems will be in the pumping, not the material strength. Nimur (talk) 17:15, 27 November 2009 (UTC)[reply]

Wow, thank you for the really interesting read! I really didn't mean to imply anything about needing anything near a perfect vacuum though... what percent of a vacuum can you reach with just the initial pumps? Will it just take a really long time, or do they need to be very special to pump out air even when the pressure differential is close to 1 atmosphere? (which doesn't seem THAT much, especially since you can pump out over as small an area as you want...) —Preceding unsigned comment added by 92.230.68.207 (talk) 19:30, 27 November 2009 (UTC)[reply]

Different pumps have different operating regimes. A roughing pump can operate from atmospheric pressure down to pressures on the order of millibars. Some are even more powerful than that. Our article says 1x10^-3 torr which is about 1 millionth of atmospheric pressure. (That's still a lot of atoms per cubic centimeter). The other types of pump I mentioned require a pretty good vacuum already (e.g. require a roughing pump behind them); these ultra high vacuum pumps can not operate from room-pressure and would be damaged if you tried. It's been a while since I played with vacuum systems, and I can't remember who made our roughing pumps, but you can buy simple pumps for under $1000 (here's a nice pump table from University of Alberta). Here's some product literature on top of the line rotary-vane roughing pumps. It takes a different amount of time to pump down, depending on a lot of factors - size of your chamber, size of your pump, material in the chamber, any residues or oils left over in the chamber, etc. If you are actually building a vacuum system, I would suggest finding an expert at your local university and getting some experience working with them before you try on your own. Be aware that there are a variety of unique hazards when dealing with pressurized (and unpressurized) gases. Watch out for moving parts and electrocution, make sure your system is pressure rated for what you want to do; don't mix oil-lubricated roughing pumps with ox-clean plumbing or experiments. Nimur (talk) 21:02, 27 November 2009 (UTC)[reply]

I think the answer to the original question is that the idea would work just fine. However, after you have gone to the trouble of designing 100 thin, strong, concentric cubes or spheres, each of which can withstand a pressure difference of 0.01 atm without collapsing under the pressure or its own weight, you will find that you have just designed, the hard way, a 1 atm pressure hull. So, philosophically speaking, you will not have gained anything. Just as Schultz and Dauto said. --Heron (talk) 20:30, 27 November 2009 (UTC)[reply]

Can you tell me in physics or math terms why 100 hulls with a pressure difference of 0.01 atmosphere are equivalent to one big hull that can stand 1 atm? I just don't understand what of necessity makes the two possibilities need to have the same sum mass or in any other way represent no preference for one over the other (except for the preference for 1 large hull because of the engineering challenge of making a hundred separate parts). I'm just not grasping the key thing that makes them equivalent. —Preceding unsigned comment added by 92.230.68.207 (talk) 21:07, 27 November 2009 (UTC)[reply]

They would not be equivalent. Monolithic pieces of metal have different properties than individual pieces of metal, of the same size, that are "stuck together". This is a crucial bit of material science. You might want to read about metal lattice structure if you're into the gory details. Heron is making a simple approximation, suitable for a limited range of properties. I am highly suspicious of this approximation - especially since estimating pressure tolerances is very important for establishing safety margins. Even metal welded together is not the same as a single piece of metal. Even polycrystalline metal that is forged together is not the same as monocrystalline metal that is slowly cooled. There's no reason to assume that a metal sphere suitable for 1 atm gauge pressure would be equivalent to 100 concentric spheres each suitable for 1/100th atmosphere gauge pressure. I would expect the single hull to be made of much less material. The only real way to know is to consult an empirical table of material pressure strengths; e.g. Metal Chamber Pressure Standards from the AIAA. Don't mess around with approximations when it comes to pressure-tolerance. Consult a reliable reference. NASA, DTIC, and AIAA all publish similar standards handbooks. Nimur (talk) 21:15, 27 November 2009 (UTC)[reply]

thanks, but this is theoretical. I actually don't have anyplace to put the football-stadium-sized structure at the moment, so it's just sitting out back waiting to be put together. 92.230.68.207 (talk) 21:34, 27 November 2009 (UTC)[reply]

I might suggest Sandusky, Ohio. You might have some competition. Nimur (talk) 03:13, 28 November 2009 (UTC)[reply]

can you make it out of anything? (non-porous)

here is another question: could you make a structure to house a football-stadium sized vacuum cavity out of ANY material (that was non-porous), including different types of steel, plastic, fiberglass, even treated wood? (so as not to be porous), with the only difference being how thick you would need to make it? If you can't make it out of ANYTHING (even plastic) then what must the material's properties be, and which materials would and wouldn't qualify? Would some kind non-porous styrofoam be possible, so long as you had ENOUGH of it?

I mean this question purely on a theoretical-level; maybe you would need a half-mile-thick sphere to get a cubic-meter vacuum ... if the sphere is made of chocolate. Is it possible? Can you make it out of chocolate? (so long as you ahve enough, again only theoretically). If not, why not?92.230.68.207 (talk) 21:37, 27 November 2009 (UTC)[reply]

No, you can't make it out of chocolate. Sturdiness notwithstanding, chocolate walls would erode and oily residues would gassify. In fact you should not put chocolate in a vacuum system, let alone build the walls from it. I don't even have any idea how you might hermetically seal chocolate; it might actually be pretty decent if it is thick enough. Again, "sealing" from outside to inside is usesless if the material is going to erode and gassify into the vacuum chamber. Nimur (talk) 03:10, 28 November 2009 (UTC)[reply]

A good thing to remember whenever you are thinking about the structural requirements of containing a vacuum is that the pressure difference between sea level atmosphere and a perfect vacuum is approximately the same as the pressure difference between sea level atmosphere and 10 metres underwater. If we ignore the weird outgassing and erosion that Nimur talks about (which you can't for chocolate, but probably can for metal or plastic) then anything which can be sealed and taken 10m underwater without being crushed can also have all the air sucked out of it while at sea level - it is exactly the same problem (in terms of material strength - you may need different types of seals and things to keep water out vs keeping air out). --Tango (talk) 10:15, 28 November 2009 (UTC)[reply]

I'm not sure but I don't think so. The main problem, is to stop the weight of the material itself crushing the material. The way I'd go about it is to use a number of domes within each other. Each would be supported by lots of thin struts forming triangles and the outside of each dome covered with a thin sheet of material. The question then boils down I believe to whether one can make a dome framework of plastic or even steel of any size say 20 miles round. I'm not sure what the precise problem would be though. Dmcq (talk) 15:22, 28 November 2009 (UTC)[reply]

As long as you are allowed to include internal supports or some kind, it shouldn't be too difficult. If you aren't allowed internal supports then the problem is very similar to building a single-span bridge. --Tango (talk) 15:43, 28 November 2009 (UTC)[reply]

I totally don't understand - you guys seem to be in disagreement, as you're talking about a series of domes but above someone else said there is NO advantage to have a series of domes one inside the other versus having one large dome with a thickness equal to the individual domes combined - in fact, they said the monolithic (single-domed) structure would probably end up THINNER! So who's right? If there is no advantage to a series of concentric domes, why are we even talking about it anymore? 85.181.144.200 (talk) 15:52, 28 November 2009 (UTC)[reply]

I'm not talking about concentric domes, just some kind of pillars or girders to hold up the structure. --Tango (talk) 15:56, 28 November 2009 (UTC)[reply]

The part about concentric domes vs. single dome was strictly with regard to pressure tolerance. I mentioned earlier about the need for trusses or internal support structure if the dome got sufficiently large - as Tango said, this would be kind of like girders for a bridge. Nimur (talk) 16:19, 28 November 2009 (UTC)[reply]

Moon Landing

Who was the youngest person, at time of landing, to set foot on the moon and who (currently) is the youngest person to have ever done so? What about oldest for both questions? TheFutureAwaits (talk) 13:20, 27 November 2009 (UTC)[reply]

List of Apollo astronauts#Apollo astronauts who walked on the Moon has everything you want. Charles Duke was the youngest when he stepped, and Alan Shepard the oldest. Both would have retained their titles, but since Shepard died, Buzz Aldrin is the oldest moonwalker. Shepard was the only one not born in the 30s. ~ Amory (u • t • c) 14:14, 27 November 2009 (UTC)[reply]

Cool thanks! TheFutureAwaits (talk) 16:19, 27 November 2009 (UTC)[reply]

quantum mechanics

I want to research on theoretical chemistry specially on quantum part in future ? how do I start and go for it?Supriyochowdhury (talk) 15:18, 27 November 2009 (UTC)[reply]

The answer to that question depends on how much you already know about the subject. Why don't you start by telling us wheather you have a degree in chemistry or related subject? Dauto (talk) 16:34, 27 November 2009 (UTC)[reply]

If you want to work on quantum mechanics, I recommend boosting your math skills very significantly. Take several university-level courses on calculus. Follow this with several university-level courses in differential equations. Also take several courses in linear algebra. When you take your quantum class, be sure to take at least one quantum course from a physics department and one quantum course from a chemistry department. A standard chemistry curriculum is sometimes lax in the mathematical requirements, because most chemists do not need the gorey details of quantum mechanics, but if you intend to make it your research focus, you will need to get into the math-heavy parts. Nimur (talk) 17:19, 27 November 2009 (UTC)[reply]

To what Nimur said I would add at least one course in complex variables and analysis. Note that if you really want to do research, you will most likely be required to take some graduate level classes as well. Dauto (talk) 19:19, 27 November 2009 (UTC)[reply]

statistical testing for a feeling of being watched

all right, I've had it enough with this pseudoscience. I want to test whether I can feel being watched. I propose the following methodology. I will print a lot of random "1. watch" and "2. watch", "3. don't watch." etc and then, without looking at the person, have someone do the required action (either look at me or don't) while I call the appropriate number. I will then write down whether, after I called out 1, I felt watched (or didn't feel watched), and repeat on and on. at the end I will compare my results with the instructions, statistically. My problem is: how should I do this? How many numbers should I even have? The real problem isn't that the effect might be a weak one -- my problem is that there is a little thing called prior probability. And the prior probability that I can feel where someone's eyeballs are directed without any further sensory input is precisely 0. But if the prior probability is 0, then even if I can feel 100/100 of the gazes (ie 100% match the random instructions) it will still fail to show that there is a feeling of being watched, due to the prior probability of this which is zero. What do you all suggest? How many trials should I do, and what percent should I set the threshold at? I am at a loss, as prior probability to me implies that it is literally impossible to test for something like this, because of its probability 0 of being true. Thanks for any advice you might have. —Preceding unsigned comment added by 92.230.68.102 (talk) 16:09, 27 November 2009 (UTC)[reply]

Have you contacted Rupert Sheldrake, who has published research on this subject? --TammyMoet (talk) 16:46, 27 November 2009 (UTC)[reply]

I believe there's a story of Feynman being rejected for the army because they asked him if he thought he was being watched and he said yes. He had a look around and half of the people waiting for the test were actually looking at him. :) Dmcq (talk) 17:17, 27 November 2009 (UTC)[reply]

I've read that anecdote in his biography - and I think you're explaining the story in a misleading manner. He didn't say that because he 'felt' that he was being watched. He simply reasoned that with nothing else interesting going on during the long, boring wait in line to see the army recruiter - that a lot of people would indeed be watching him. So he replied logically. That being the case, Feynman's story has zero relevance to this question. SteveBaker (talk) 19:18, 28 November 2009 (UTC)[reply]

If you say the prior probability is zero, there is no reason to do the experiment. But you don't really think it is zero, you just think it's very small, say one in a trillion. Statistically, the exact number you assume doesn't have a huge effect on the number of trials needed. (To put it more precisely, the number of trials needed increases in proportion to the logarithm of the prior probability.) What matters a lot more is the effect size. To resolve a 10% difference between watched and unwatched, you need a couple of hundred trials, and the number of trials needed increases in proportion to the inverse square of the effect size—the article on Pearson's chi-square test will give you more information. It is also critical, if you really want to do an experiment like this, to make sure that the person being watched has no information whatsoever about what the watcher is doing. Looie496 (talk) 19:47, 27 November 2009 (UTC)[reply]

I believe (without proof) that this feeling comes about from subliminal cues we're picking up from the environment. If someone tiptoes up behind you - they'll still subtly block ambient light - and reflect light - and absorb or reflect ambient sounds and emit smells in ways that are perhaps so subtle as to be below our conscious ability to detect - but plenty enough for our subconscious mind to pick up on. Since no definite information is available - all your consciousness gets is a vague feeling of unease. This seems like an entirely reasonable thing - after all, our ancient ancestors would have had to be very attuned to predators sneaking up behind them - and any teeny-tiny cue that we could pick up would have evolutionary benefits - so it's quite possible that we've evolved this ability. So any experiment you did would have to very carefully test for this effect because it's presence would prove something - and it's absence would be hard to set up controls for because we don't know how subtle that effect is. SteveBaker (talk) 19:15, 28 November 2009 (UTC)[reply]

so basically, you want me to change my test from "can I feel being watched" to "is my wife any good at sneaking up behind me"? :) 92.230.69.34 (talk) 20:49, 28 November 2009 (UTC)[reply]

No - I just maintain that on those occasions when we think we're successful at intuitively feeling we're being watched, we're actually picking up on very subtle cues with perfectly normal senses. An overly tightly controlled experiment (eg where a "watcher" randomly either closes their eyes or does not) might come up with a negative result - when in reality there is something super-subtle going on that is worthy of note. It's hard to comment on the quality of your experiment because it's not clear what hypothesis it's trying to demonstrate. I think you need a much clearer statement of your hypothesis before you attempt to devise an experiment to prove or disprove it. SteveBaker (talk) 01:06, 29 November 2009 (UTC)[reply]

While most likely it is impossible to detect whether someone standing behind us is looking at us or looking away, there are plausible cases when we can detect the following: 1. someone standing a few meters behind us vs. no one standing there. 2. Someone standing just barely at the edge of our peripheral vision and looking at us vs. looking away. I've seen a paper about it, which I cannot find now, but the main idea was that our brain is very good at pattern recognition, and it uses subtle inputs (which we don't consciously know) like reflections on the wall, distribution of brightness in the room, heat, sound, slight shadows, and, in the second case, visual input with very low resolution. All these signals are too week to be consciously detected individually, but all of them together can trigger this sensation of being watched. --131.188.3.21 (talk) 14:53, 29 November 2009 (UTC)[reply]

Paramecium may be able to communicate by electromagnetic radiation. How this may be relevant I cannot say. But if humans or other forms of life can possibly possess this sort of capability to some degree maybe it could have bearing on an ability to know when the eyes of another organism have become intently focused within our vicinity however vicinity might have to be defined. Admittedly the distances between paramecium in the experiments conducted are minuscule relative to the distances necessary for this to be a factor between large animals such as ourselves. Bus stop (talk) 15:43, 29 November 2009 (UTC)[reply]

If somebody walks up near you to look at what you are doing, it is possible to detect the subtle sounds or vibrations of the footsteps. And then there is psi but that is a different matter altogether. ~A H 1^(TCU) 01:53, 2 December 2009 (UTC)[reply]

hydride donors and nitrogen

So if you use sodium borohydride (or forbid, LiAlH4) on say, an imine in aprotic solvent, you really get an anionic amide species until you expose the reagent to protons, right? Or how else does the amine get protonated (to neutral) form? I guess I'm surprised at the reactivity because RNH- is a considerably worse anion (conjugate acid pKa 35) than RO- in the hydride donation to carbonyls.

Also, does the electron-donating effect of nitrogen in some nitrogeneous aromatic rings make the rings more susceptible to non-catalytic hydride donors (LiAlH4?) -- cuz imidazole only has a resonance stabilisation of 14 kcal/mol, right? John Riemann Soong (talk) 16:31, 27 November 2009 (UTC)[reply]

After reactions are done, during the "work-up" phase, water or other proton sources are added to neutralize anionic positions, destroy and/or remove unreacted reagents, etc. I can't make any sense of thse cond part of the first paragraph...the relative reactivity of something in two separate reactions does not affect whether those reactions works, especially since there are other components involved as well. As long as there is enough instability in "all the reactants in a certain reaction", the reaction works.

To get good reaction with a hydride donor, you need an electron acceptor, not just a "not extra-stabilized π system". DMacks (talk) 17:32, 28 November 2009 (UTC)[reply]

There's a significant dipole moment in imidazole... so I think that would set up a sort of Michael-ish addition right there. Basically I'm wondering since the RNH- anion is less stable than RO-, whether this causes problems for some hydride donation schemes... John Riemann Soong (talk) 03:38, 29 November 2009 (UTC)[reply]

Hydride attack is almost always irreversible. But the whole reason one chooses a certain donor is to guarantee that situation (which is why you learn about differences of reactivity, etc.) for a reaction being studied. Some hydride donors are not strong enough to attack some carbonyl-like compounds. More reactive compounds are more easily attacked and a weaker donor is "good enough". Remember you have to consider "all reactants → all products" not just "stability of one product". There are definitely reactions involving nucleophilic attack on aromatic rings, but it's hard and not usually the ultimate product because loss of aromaticity is such a large energy cost. The product often re-aromatizes by tetrahedral collapse/ejection of a leaving group. This is one of the few situations where hydride actually can be a viable leaving group. DMacks (talk) 11:52, 29 November 2009 (UTC)[reply]

Seagull ID needed

Just watched this video:

http://www.youtube.com/watch?v=LjgmXxxM6zM

What sort of gull is that? Never seen one like it before. --84.64.96.34 (talk) 22:36, 27 November 2009 (UTC)[reply]

Have you asked this guy - he's our resident expert in these matters. hydnjo (talk) 06:26, 28 November 2009 (UTC)[reply]

Nice story, I'm intrigued with the presence of the tennis ball, did it go fetch it when you threw it ;-)). Come on Karl, we're all waiting. Richard Avery (talk) 08:27, 28 November 2009 (UTC)[reply]

I have read that ducklings will adopt the first creature they see as "mother" and follow him/her/it around. I wonder if that would have happened if the bird had been taken in as a hatchling and/or hand fed. Cuddlyable3 (talk) 09:16, 28 November 2009 (UTC)[reply]

I think that this may be an immature Northern Fulmar, actually a petrel, rather than a gull. Mikenorton (talk) 12:12, 28 November 2009 (UTC)[reply]

Looking a bit further, it could also be a dark morph adult. Mikenorton (talk) 12:36, 28 November 2009 (UTC)[reply]