Wikipedia:Reference desk/Archives/Science/2012 March 29

= March 29 =

Distance to ULAS J1120+0641 and the age of the universe
The article about ULAS J1120+0641 gives its comoving distance as 28.85 billion light years. The article on the age of the universe gives the age of the universe as 13.75 ± 0.11 billion years. This appears to mean that the light emited by ULAS J1120+0641 can't have reached us yet, and won't for at least another ~15 billion years. I originally thought that the solution was that the space between earth and ULAS J1120+0641 has expanded since it first emitted light in our direction, so the heavily redshifted light that we see from it was emited at a time when it was much closer. But the article on comoving distance says that "Comoving distance factors out the expansion of the universe, giving a distance that does not change in time due to the expansion of space". So how is it that ULAS J1120+0641 is visible given the age of our universe? 203.27.72.5 (talk) 00:42, 29 March 2012 (UTC)
 * I find this question really hard to get my head around, it's very counter intuitive when you're talking about BOTH distance AND age in terms of light years.. Here is a couple of articles which attempt to explain it... Vespine (talk) 01:15, 29 March 2012 (UTC)


 * Suppose that you could, at this very moment, freeze the expansion of the universe. If you measure the distance to ULAS J1120+0641, you would get 28.85 billion light years.  If you send a radio signal, it will reach the object after 28.85 billion years.  However, the light we receive now from that object was emitted billions of years ago.  For all of those billions of years, the universe was smaller than it is now, so light travelled, in one year, farther than 1 light year of comoving distance.  --140.180.39.146 (talk) 01:56, 29 March 2012 (UTC)

Proper unit and force
Which unit would suit to measure the force that seeks to stop the moving object (can't recall the proper name of that force, maybe that from Newton's 3rd law), specifically by pulling the accelerating object backwards or holding it from the front? And how to calculate that stopping force for any speed and/or acceleration given (such as, theoretically, for a superman who is pulling the accelerating car backwards?) Also in the case of a superman is a frontal stopping of, say, a moving car more advantageous due to better friction or there is no difference?--46.205.11.219 (talk) 00:50, 29 March 2012 (UTC)
 * That force would be measured in Newtons. Your second question: no difference (if you apply the same force in the same direction). F = m(vf - vi)/t. Plasmic Physics (talk) 01:03, 29 March 2012 (UTC)
 * That's only for the total force though. OP seems to imply there are two forces working against each other, in which case you need to add or subtract a magnitude (say $$F = F_a - F_b$$, then $$F_a = ma + F_b$$). To stop an accelerating car from accelerating you need to apply a force of equal magnitude and opposite direction. --145.94.77.43 (talk) 08:17, 29 March 2012 (UTC)
 * Will I need conversion of car's horse powers to newtons in that case or something like that?--46.204.18.135 (talk) 14:05, 29 March 2012 (UTC)
 * That 'stopping force' is friction, there is no other natural force that has the same effect. You may be confused with inertia, which is a different concept. Plasmic Physics (talk) 01:07, 29 March 2012 (UTC)
 * Horsepower is a measure of power nor force, and power is alternatively measured in Watts not Newtons. Plasmic Physics (talk) 20:37, 29 March 2012 (UTC)

The first question is really vague. Is the object just moving, or is it actually accelerating? Forces are usually measured in Newtons, but I have a feeling that's not what you're really asking for here. A force of a given magnitude has the same effect as long as it acts in the same direction, so it doesn't matter if it is a pull from the back or a push from the front. As for the question "how to calculate that stopping force for any speed and/or acceleration given "; to over come the acceleration, the stopping force must be greater than the accelerating force. If there is no acceleration, that force is any number greater than zero; FStopping > 0. If there is acceleration then stopping force must be equal to the accelerating force plus any number greater than zero; FStopping > FAcceleration. You can calculate the accelerating force by taking the mass of the object mObject and multiplying it by its acceleration aObject; FAcceleration = mObject x aObject. 203.27.72.5 (talk) 22:29, 30 March 2012 (UTC)

Genome testing
When submitting your DNA sample for genome testing, is there any way to verify that they actually did it, versus just cashing your check and sending you back randomized results ? For example, can they tell you your eye color, hair color, hair texture, etc. ? StuRat (talk) 06:48, 29 March 2012 (UTC)


 * In principle, yes. In practice, it will depend on what you are having tested.  Unless the purpose of the test is to obtain some sort of comprehensive profile, then they are unlikely to be looking for things like eye color and hair.  It would generally be considered an invasion of privacy for the lab to look at traits beyond those they were explicitly asked to examine.  Beyond that, there is also the issue that the function of many genes is still unknown and even for some common traits (e.g. eye color) there isn't enough knowledge of all the genes and possible alleles to reliably predict outward appearance with high confidence.  However, there are enough traits that we do understand that a testing company could provide the kind of verification that you are looking for, provided they performed the appropriate tests and you had given your permission to do that.  Dragons flight (talk) 08:39, 29 March 2012 (UTC)


 * Actually, if you tested with 23andme, which provides health details as well as genetic results, they will provide you with guesses as to your eye color, blood type, height and other traits. If you tested with FamilyTreeDNA, there will be no traits (but you will have access to their more extensive genetic database, if you are looking for relatives). And if you test at 23andme and transfer your results to FamilyTreeDNA, you will have all of the above. The best way, though certainly not the least expensive, to verify results is to have the test done at two different labs and compare results; testing relatives who should have similar results is another way to be sure actual testing is being done... - Nunh-huh 10:23, 29 March 2012 (UTC)


 * Some companies will tell you what they can directly, others will give you a file with all your SNPs (or, rather, with all SNPs that their machine can recognize), and you can try to interpret the results on your own.
 * At the very least, the lab should tell you your mtDNA haplogroup and, if you're a male, Y-DNA haplogroup (and you should have a pretty good idea what Y-DNA haplogroup to expect, if you know your ancestry.)
 * To give an example, 23andme.com tells me that I have "72% chance of blue eyes; 27% chance of green eyes; 1% chance of brown eyes" because I have GG at the marker rs12913832, and "greatly increased odds of having red hair" because I have TT at the marker rs1805007. But these are characteristics where the connection between genetics and appearance is well understood. They can't tell me anything meaningful about my height, because height is influenced by many genes, most with small effects, and not all of them are known.--Itinerant1 (talk) 10:35, 29 March 2012 (UTC)


 * Visible physical characteristics would make a very poor check, since if they company is a scam to begin with, obviously they could just look and see. Even health conditions would offer little reassurance in the U.S., where health information is anything but closely guarded.  By far the most important check is that the test or part thereof might be repeated for some reason (especially likely when large amounts of information are collected incidentally as part of a standard screen) - any inconsistency would potentially cause them grief.  Of course, real scam artists are very good at avoiding scrutiny and gabbing their way out of contradictions. Wnt (talk) 15:17, 29 March 2012 (UTC)


 * The idea is that you would mail in a DNA swab, anonymously, so they would have no idea what you look like, other than to the extent they derived that info from your DNA. Payment could be done by money order, to keep it all anonymous.  You'd want the info to be anonymous anyway, lest they sell info to potential employers on your genetic likelihood to have a foot fetish. :-) StuRat (talk) 17:10, 29 March 2012 (UTC)


 * Relevant literature for anyone considering sending samples for DNA testing: My Beautiful Genome by Danish author Lone Frank. I've been told it's a good read, and bought the book a week ago, but have only reached page 18, so I cannot really review it. Good reader reviews on Amazon.com. --NorwegianBluetalk 16:26, 29 March 2012 (UTC)

german raider "wolf", at the end of first world war
recently i read, a rather well written tale of the german raider "wolf". by the end of her odyssee, she apparantly visited trinidade isl. but fled at the sight of approaching vessels. the book in question is: "wolf" by guilliatt & hohnen, random house 2009,  isbn 9780593060759 an entry concerning this episode could be of interest. peter varenhorst — Preceding unsigned comment added by 81.205.241.214 (talk) 10:34, 29 March 2012 (UTC)


 * SMS Wolf (auxiliary cruiser), which cites the same book you mention. That article doesn't mention a trip to the Caribbean, however. -- Finlay McWalterჷTalk 11:07, 29 March 2012 (UTC)
 * The Transcript of Naval Staff Intelligence Division Report "Cruise of the German Auxiliary Cruiser WOLF", March 1918 in the UK National Archives might have more information. I haven't had a chance to read through it yet, but the Google search result suggests that the island in question is Trindade and Martim Vaz which is in the Atlantic, not Trinidad in the the Caribbean. Alansplodge (talk) 15:26, 29 March 2012 (UTC)

Unidentified flowers
Could anyone help me with the two flowers linked here? I'm not sure at all on the first, and the second is a species of guava (methinks). Both are fairly good pictures, so it would be a shame to not have them in an article.

Thanks. Crisco 1492 (talk) 12:54, 29 March 2012 (UTC)


 * Considering the first image, it appears to be an actinomorphic flower, which as best as I can see has four sepals, four small pink petals, and some absurd number, maybe over 100 stamens, hiding any gynoecium that might be present. This gives a floral formula of "Ca4Co4A∞G?" ... I think.  After that I just googled "guava indonesia flower" like everybody else and got, identified by the poster as a Malay apple flower.  Some day I mean to learn botany to the point of being able to actually use a key, really, but by the time I do I think they'll be obsolete. Wnt (talk) 15:45, 29 March 2012 (UTC)


 * Thanks, that appears correct. Crisco 1492 (talk) 15:57, 29 March 2012 (UTC)
 * The yellow one appears to be Cosmos caudatus. Crisco 1492 (talk) 16:03, 29 March 2012 (UTC)

Titration curve of pure water
What does the titration curve (pH vs volume of acid added) look like? Is it a straight line? (This isn't a homework question, I'm not in school) ike9898 (talk) 17:45, 29 March 2012 (UTC)


 * Titration curve. TenOfAllTrades(talk) 18:15, 29 March 2012 (UTC)
 * If you're not going to help, please consider not posting anything at all. The linked article doesn't explicitly answer the question, at least not for someone without a complete understanding of the subject. ike9898 (talk) 20:07, 29 March 2012 (UTC)
 * Almost all titration curves look approximately the same, the only key differences are the exact position and number of the (comparably) steeper rises along the graph compared to the more gentle slopes between them--these positions are dependent upon the pKa of the substance being titrated (once you account for the actual absolute amount of that substance). If there is any substance present that can react with the solution being added (your acid), the graph will certainly not be just a straight line, because by definition "some" of that substance would be consumed at first but eventually no more could react. Because the graph is a measurement of either the amount of unreacted substance being titrated or the amount of titrant being added (or some complex function of them), different slopes would be present at these two extremes. Water is not an inert solvent when it comes to adding H+ because it is both a weak acid and a weak base. DMacks (talk) 20:19, 29 March 2012 (UTC)
 * Thanks. The pKa of water is >15. Do you think the titration curve would be 'practically a straight line' in the range of, say, pH 3-10? ike9898 (talk) 20:47, 29 March 2012 (UTC)
 * Assuming you are adding an acid that dissociates completely in water (a strong acid), and you plot it in the usual manner (pH vs added acid) it won't be a straight line, but rather a logarithmic curve (because pH is logarithmic with regards to acid concentration, and was originally defined for being in water). There are also complications due to the fact that adding acid probably changes the volume perhaps significantly.  However, it really isn't appropriate to speak of a titration curve in this regime - a titration curve is typically used either when a buffer is being looked at (theoretically water could be used as a buffer, but I don't know any situations where it is), or where you are doing a titration, trying to fully react the species.
 * Put acid added on a log scale and you get your straight line. Alternately, you could remove the log from the pH and just plot [H+] vs Acid added, which if you think about it makes it quite clear why it should be a straight line. 203.27.72.5 (talk) 04:36, 30 March 2012 (UTC)

Thanks for the help y'all. ike9898 (talk) 13:15, 30 March 2012 (UTC)
 * to close this topic, consider this $$pK_W=pH+pOH $$. using $$pK_W \approx 14 $$ at room temperature, therefore $$pH=14+log(c(OH)) $$ --Biggerj1 (talk) 16:36, 26 January 2015 (UTC)

chaos
I'm soon gonna be giving a lecture about chaos (and many other things) and there's a point about chaos I want to understand. They generally say that chaotic systems are unpredictable because although they're deterministic, they're very sensitive to changes in the initial conditions and small changes rapidly amplify, making it too difficult to predict the evolution and final conditions of the system. My question is HOW sensitive? Is there a point to which if we know all the details in the conditions of the system, we won't need to know any further details in order to fully predict the behavior of the system (by fully, I mean at least like planetary motions) or does it go all the way to a point that it isn't theoretically possible to measure the conditions of the system? What I mean is, is it right to say "If I know the conditions of the system to a certain degree(not fully of course), I can fully predict the system."?--Irrational number (talk) 17:49, 29 March 2012 (UTC)


 * I'd say the degree to which you can predict the outcome depends on the degree of knowledge of the initial conditions, and the more time elapses, the more questionable the predictions become. For example, in weather forecasting, today's weather is right maybe 90% of the time, but next week's only 50%.  And to put it in terms of the butterfly effect, killing most butterflies won't much matter, but a small portion may be absolutely critical and have a huge effect.  StuRat (talk) 18:01, 29 March 2012 (UTC)


 * You are right, in that sometimes the sensitivity to changes in initial conditions does not pose a problem for prediction of chaotic systems, while sometimes it does. For a given system, the value that tells you how concerned you have to be is the Lyapunov exponent. Also, in my opinion, mentioning quantum mechanics is a red herring here. With respect to the underlying math and physics, the uncertainty principle is completely different from deterministic chaos. The only similarities are extremely hand-wavy and superficial. SemanticMantis (talk) 18:25, 29 March 2012 (UTC)
 * Thanks for the answers, I've already read the Lyapunov exponent article, my question is that will your predictions become questionable no matter how good your measurement of the initial conditions are? or is it like, if your error is less than, say 0.00000001 percent, you can fully rely on your prediction, and your predictions will become more or less time-independent?--Irrational number (talk) 18:45, 29 March 2012 (UTC)
 * The former. For any Lyapunov type system, arbitrarily small errors will eventually become arbitrarily large differences.  The Lyapunov exponent helps tell you how long that will take for a given level of uncertainty in the initial conditions, but eventually you get large errors regardless of how accurate the initial measurements.  This is fundamentally what it means to be sensitive to the initial conditions.  Dragons flight (talk) 18:54, 29 March 2012 (UTC)


 * Consider a system whose state is a real number between zero and one, and changes once per second to the fractional part of ten times the previous state. If the initial state is 0.231486... then the next state is 0.31486... and the next state is 0.1486... and so on. If you can measure the initial state to one part in 10n then you can predict the system's behavior pretty accurately for n seconds, and after that you can't predict it at all. That's typical of chaotic systems. No particular level of detail is enough to predict the behavior into the indefinite future. -- BenRG (talk) 18:39, 29 March 2012 (UTC)
 * Well then in some sense every system is more or less chaotic, is it the "aplification" of changes that makes the system more chaotic?--Irrational number (talk) 18:52, 29 March 2012 (UTC)
 * SemanticMantis, I didn't say that they're related, I just asked do we have to fully know the conditions of the system in order to predict its behavior, and fully measuring a system's conditions, to my limited knowledge, is somehow related to the uncertainty principle, I didn't say that chaotic systems are unpredictable because of the seemingly non-deterministic behavior of small particles, I understand that these are different subjects... (or at least I think I understand...) --Irrational number (talk) 19:02, 29 March 2012 (UTC)


 * In principle it's not always true that no level of detailed knowledge is enough: if you know the exact initial conditions, that is enough to predict infinitely far into the future if your computer can handle an infinite number of decimal places -- so computer storage capacity is the limiting factor when there is perfectly detailed knowledge.
 * For you to know the exact initial conditions, they must be rational, and in some cases this is consistent with a chaotic trajectory. For example, if $$x_{n+1}=4x_n(1-x_n)$$ there is an analytic solution $$x_n=\sin ^2 (2^n\theta \pi)$$ where $$\theta$$ is such that $$x_0 = \sin ^2 \theta \pi$$. It's chaotic if and only if theta is irrational. If for example $$x_0$$ = 1/3, or almost any rational number, then theta is irrational. Then the trajectory is chaotic, and since theta is irrational the initial condition for theta cannot be known with perfect precision, but the initial condition for x is known with precision so 4x(1-x) can be iterated arbitrarily far into the future until the numbers get too big for the computer to store them. See Chaotic map.


 * Of course, in physical systems there may be no circumstances in which you know the initial conditions, even if rational, perfectly. Duoduoduo (talk) 19:13, 29 March 2012 (UTC)
 * True, I didn't mean to put words in your mouth. But I stand my my claim: the relationships are superficial; I'd advise that you not go looking there for inspiration, understanding or intuition ;) As to your more specific question, Dragons flight makes a good point above; no amount of partial information gives you predictability forever, but the lyapunov exponent can give you a good indication of how your errors will propagate in time, and therefore on what time horizon you can hope to make decent predictions. SemanticMantis (talk) 00:36, 30 March 2012 (UTC)
 * Thanks, I kinda get it now, so no matter how much you know the details, given enough time, your predictions will become less and less reliable...--Irrational number (talk) 05:39, 30 March 2012 (UTC)


 * I think you have to be careful about what you mean by 'prediction'. For example, look at the picture of the Lorenz attractor at the top of the Lorenz system article.  If you know the initial condition of this system with a small but finite precision, your predictions of the state of the system (meaning the values of x, y, and z) will get exponentially worse with time, for small times.  However, after a long time, you can be sure that the state will be very close to the attractor - in other words, you can narrow the state down to quite a small region.  In a practical application, that might be all you need.  The concept of fractal dimension is relevant here - if the attractor has a large dimension, then it will tend to fill a large region of space.  There is a nice paper that links the dimension of an attractor to the predictability of the system - I think it is this one (you need a subscription to Physical Review Letters - which I don't have). 81.98.43.107 (talk) 17:43, 31 March 2012 (UTC)

Plugging into the sun with carbon nanotubes + what is the world economy's ATP (energy carrier, biology metaphor)
This is a two-part question.

I.

First, assuming that the Sun is a ball of electricity we can plug into and get AC voltage. How much would a one-atom (or however thick it needs to be) carbon nanotube in each direction (positive and negative polarity) from Earth to Sun (1 AU) weigh. If we had such a ball of carbon nanotube and it would unroll slowly and gracefully if we flung it toward the sun, and the two lines got there, then assuming their attraction to each other did not cause the two lines to fall into each other (they were sufficient distance apart) would there be practical problems with orbits that would cause planets to intercept these lines? With a few active movement points (rockets) could we move parts of the line out of planets' ways?

Second, look at our article on sun, we find that it is not in fact a ball of electricity we can plug into and get AC voltage. And that's not because it's DC. Instead, our article informs us that it is hot plasma interwoven with magnetic fields. What interests me is whether these magnetic fields, if they are stonger than Earth's could be harnessed by lassoing our line (the one going form Earth to Sun and back), around and around and around the sun (like someone doing whatever you do when you're jumping/skipping rope) then would we get electricity, or enough for it to be worth our trouble? If not, we will have to do something more drastic at the end of the leads. I'm not sure what, but for part 2, assume we just get AC back to Earth.

II.

Here I am making an analogy with biology. The 'energy carriers' of the Body are ATP which depletes to ADP when used up, and is replenished to ATP as an energy source. My question concerns the world economy. At the moment, instead of having litle carriers of energy that get used up and replenished, instead we have fuels we burn off. So, what would the energy carrier be that is a physical object that gets depleted and centrally replenished, were we to get leads with unlimited AC at some central location on Earth? Would we have to literally have giant transformers and strap this whole thing into the existing electricity grid, or is there some other great physical carrier (whether physical, mechanical [torque, pressure, whatever], electrical, magnetic, chemical, kinetic, biological, whatever) that for example could power a house for a while, a year say, and that you then can send along to be replenished from the AC leads coming from the sun. I am thinking of something small.

Basically, both of these questions are probably not of much interest, but the explanations might enlighten me. Thank you for your time. 188.156.115.197 (talk) 18:18, 29 March 2012 (UTC)


 * You don't need to run a line all the way to the Sun. The solar wind completes the circuit in some sense.  All you need is to tap into the ionosphere, which has a high charge difference from the ground which I believe is ultimately due to the solar wind, and tap into near-limitless energy by shorting out this immense atmospheric capacitor.  At least, that's the idea proposed by Nikola Tesla, and still viewed as theoretically possible, I think.  (Sort of a more ambitious version of the lightning rocket)  Though I would greatly miss the aesthetics of it, as the mythological wedding of Persephone and Prometheus draws nigh, it is about time for once-mortal men to steal Jove's trove of thunderbolts. ;) Wnt (talk) 18:54, 29 March 2012 (UTC)


 * OP here. Jesus Christ, man, don't do this!!!!  Our ionosphere is the ONLY thing protecting us from immediate annihilation.  If anyone is reading this - D O   N O T   D O   T H I S.  Our system of commerce is not set up to protect public goods such as the ionosphere.  No, better get it from the sun.  As for my original question: how much nano tubing do we need? 188.157.120.238 (talk) 19:29, 29 March 2012 (UTC)


 * The ionosphere is not what protects us. The magnetosphere is. Whoop whoop pull up Bitching Betty 19:58, 29 March 2012 (UTC)
 * I'm no physicist, but I'm reasonably sure that 8 1/3 light-minutes of wires would have enough resistance to nullify any gains we could hope to get. Unless you do superconductors, I guess.  Also, wouldn't work when the part of Earth that has the wire's endpoint rotates away from the sun. Writ Keeper &#9863;&#9812; 20:05, 29 March 2012 (UTC)


 * doesn't space automatically cool everything off to 0 kelvin and superconduct any wire? also aren't carbon nanotubes very good at transferring current. also if it's the sun who cares how much is lost, it could be 7 orders of magnitude and we'd still have enough.  if there were an ac jack on the sun to plug into.  Finally as to rotation I guess we have to do something about that. 94.27.177.124 (talk) 21:04, 29 March 2012 (UTC)
 * It is just overall crazy with so many flaws in your logic, I won't even try to explain. Plasmic Physics (talk) 21:08, 29 March 2012 (UTC)
 * all great ideas start as overall crazy with so many flaws in logic that they are not even worth explaining. Einstein had to go to a friend who paid attention to math he didn't learn properly before he could even work out special relativity.  Let alone, let alone, general relativity.  Also your name is plasmic physics, which this text field (firefox) redlines.  Google returns only your username here on Wikipedia when I put "plasmic physics" in quotes on the search engine.   That does not scream nobel prize to me.  But hey thanks for your help and insight. 94.27.177.124 (talk) 21:18, 29 March 2012 (UTC)


 * The article outer space says "The current black body temperature of the background radiation is about 3 K (−270 °C; −454 °F)." Not quite absolute zero, so I don't know if it's low enough to "superconduct any wire", but I doubt it. I think it is low enough to superconduct even the first (low temperature) generation of superconductors.


 * I personally like your thought experiment. Who knows where thought experiments lead? (To me, maybe this one leads to Earth's rotation being no problem -- just use the Sun's boundless energy to stop the rotation of the Earth. But that's probably just hypothetical.) Duoduoduo (talk) 21:45, 29 March 2012 (UTC)
 * If Earth stopped spinning it would either collapse (or everyone would be crushed) or explode spewing magma and with everyone flung off into space - only one of those things would happen and this is the type of thing where if you bother to learn it you would know, which I didn't. So: what happens if the Earth stops spinning?  Presumably not 'nothing' though maybe so, given that the Moon is rotation locked with us and seems to be doing just fine, though it doesn't have much of any life. 149.200.72.244 (talk) 22:12, 29 March 2012 (UTC)

also (op here) i did not get any answer to the SIMPLE question of how much 2 AU (a unit of length) of carbon nanotubing weighs (when it's sstill on Earth) nor any response to the second part of my qeustion. 94.27.177.124 (talk) 21:14, 29 March 2012 (UTC)
 * What is the diameter of the nanotube in terms of carbon atoms, and is the wire tensioned? You are essentially asking how long is a piece of string? We need a bit of information before we can calculate an answer. Plasmic Physics (talk) 21:23, 29 March 2012 (UTC)
 * I'm NOT asking "how long is a piece of string", but out of curiosity how long is it? As to my actual question, say the diameter of the nanotube is 1 carbon atom, 0 stress since in space things float and we get the tension to balance out somehow (naively, since it goes there and back again that cancels out).  in this primordial case, how much will 2 au of 1 carbon diameter nanotubing weigh while it's on Earth? Then we can multiply by some factor, Gee, in case any of this is realistic.  It's simple enough to go from 1 carbon diameter to 2 or 6, you just multiply whatever answer you're about to give me.  It's the 2 AU that is the bigger factor accounting for most of the order of magnitude...so...?  how much does a 1 carbon diameter nanotube that is 2 au (a 1 dimensional measure of length) long weigh? 21:36, 29 March 2012 (UTC)  — Preceding unsigned comment added by 149.200.72.244 (talk)
 * I get only .04 grams --Digrpat (talk) 21:59, 29 March 2012 (UTC)
 * OP here. Pardon my French but are you shitting me?  I *thought* it might be some tiny amount, like a few pounds or a ton or two.  You did the full 2 AU?  See, this is what I expected.  Now from 0.04 grams, you know if you want to go from 1 atom to 100 in diameter be by guest, that 4 grams is not going to be impossible to launch.  I wonder about the other parts of my question - orbits intersecting our nice little power cord... woudl we have to actively move it?  How much transmission loss could we expect?  (Again, I think 7 orders of magnitude is fine, given we're dealing with the sun).  Then we'll work out the sun and and the Earth end, but let's concentrate onthis.  How much current do you think a carbon nanotube can carry, and at what resistance?  (or what's the graph look like). 149.200.72.244 (talk) 22:33, 29 March 2012 (UTC)


 * A single nanotube would weigh about 15000 kg over 2 AU, and be able to carry a maximum hypothetical current of about about 6 microamps with a resistance of about 1 Ohms at room temperature. For a transmission loss of around 4 W, somewhat larger than the entire world's current energy consumption at present.  Dragons flight (talk) 22:44, 29 March 2012 (UTC)
 * A nanotube of one atom circumference is hardly a tube, more like nanostring. This means that a different typr of bonding may be in order - what is the bond length between the atoms in the carbon nanostring? Will the new bonding type allow for electronic conductivity or inslation? Plasmic Physics (talk) 22:52, 29 March 2012 (UTC)
 * "doesn't space automatically cool everything off to 0 kelvin and superconduct any wire?" Well that depends on your time horizon. Space contains only a sprinkling of matter since it is almost a total vacuum. The matter that is present is mostly high speed (i.e. high temperature) particles in the solar wind. But even if you ignore the solar wind, any object in a vacuum has nothing around it to act as a heat sink to transfer its heat to. So it doens't get instantly chilled. On the contrary, it can only cool down by emitting electromagnetic radiation (see black body radiation. So the end result is that it cools down at an increasingly slower rate forever. And all that is not taking into account the fact that it will be bathed in EM radiation from the Sun. 203.27.72.5 (talk) 07:52, 30 March 2012 (UTC)
 * I would think that even synthesizing a nanotube on an astronomical length scale in the first place would fall somewhere between chemically implausible and completely absurd. The real limiting factor isn't the amount of carbon, or the length of tube, but the ridiculous number of defects you'd wind up with.  My (somewhat limited) understanding is that a nanotube hundreds of microns long is so full of defects that it may as well be amorphous carbon, to say nothing of a nanotube that you'd have to measure in AUs.  If you want a conducting material, that's just not going to work.  (+)H3N-Protein\Chemist-CO2(-) 14:24, 30 March 2012 (UTC)
 * Note that linear acetylenic carbon, carbon nanowires, linear carbon, crystalline carbyne, as can be generated by additions into CnO which produce a resonance structure between C=C=C=C=C=O and -C*C-C*C-C*O+ etc. ... whatever you call it, it's a real thing.  Carbon nanowires are more stable than fullerenes, less so than carbon nanotubes.  However, they can become energetically unstable at lengths over 20 unless confined in carbon nanotubes ... probably an interesting story there.  They can also apparently transform a semiconducting SWNT into a metallic one. Wnt (talk) 20:19, 30 March 2012 (UTC)


 * In general, I'm not sure what I'm talking about with this subject, but some basics at least: we're talking about a circuit here.  What's being plugged into?  I suppose the emission of plasma from the coronosphere has some intrinsic bias that should charge up the Sun to a certain degree, but that same charge is being transmitted to the Earth via the solar wind.  I picture a perfect circuit from the Sun and what would it be?  A huge carbon nanotube full of solar wind particles moving straight down the center without hitting anything to carry the charge without resistance.
 * Now eventually these particles strike the Earth's ionosphere, and the Earth picks up a charge, but I'm still not sure how the circuit works. Why doesn't the whole Earth pick up the same charge?  I know the ionosphere varies by day and night, and surely the magnetic field of the Earth must have some differing influence on positive and negative per Van Allen belts - I suppose the charge we're picking up by day must be leaving somehow - but I don't really know at all.  Is it discharging back out to space by night, some kind of "terrestrial wind"?  We're not hot enough to spew plasma, but what about charge?  Hmmm, I suppose not... we must be completing the circuit by intermittently? feeding from those different Van Allen belts of protons and electrons.
 * My feeling is still that by tapping into the ionosphere we might as well be tapping into the Sun; that conduction from there to here, via solar wind, is not the issue. If we could ground that and tap into that huge power source, the question still remains, how does the current move on from there?  There must be some mechanism by which the ionospheric potential is preserved.  It might be that getting the current back out is just as important as getting it in for such a scheme.
 * For that matter, why aren't satellites powered by solar wind? Trail two (widely separated) very fine nanowires, one that passing protons tend to stick to, one that electrons reach more readily, harvest the current, and emit a small amount of atomic or molecular hydrogen gas.  Is that a kind of heat engine?  The article on the Van Allen belts has a scheme for "removal" - isn't there some less ambitious means to simply harvest such energy?
 * I'm afraid I'm far, far out to sea on this one. It was indeed a very good thought experiment for the OP to propose this. Wnt (talk) 20:19, 30 March 2012 (UTC)

Just thought I'd point out that all of those pi bonds linking the carbons together are prone to electrophilic attack. And you're sticking them in a proton plasma. 203.27.72.5 (talk) 21:27, 30 March 2012 (UTC)
 * Hey, I didn't think up this scheme, and I didn't say it was perfect. Maybe a big shade at the Sun end and a strong positive charge (p-doping a linear carbon chain??) would keep away protons... of course, the whole scenario of digging a hole in a conductive medium in order to install a single tiny wire that won't conduct except as a matter of rhetoric is wacky to start with, but... Wnt (talk) 14:11, 31 March 2012 (UTC)

VSEPR theory
What geometry does VSEPR theory predict for molecules with a steric number greater than 9? Whoop whoop pull up Bitching Betty 18:25, 29 March 2012 (UTC)
 * VSEPR theory is pure geometry: Take X number of objects distributed around a central point, and arange them so they are a maximum distance from each other. Hypothetically, any number of objects could be used, and indeed the shapes so described have almost nothing to do with chemistry, it is just a geometry exercise.  There isn't really any need, for a chemist, to describe any molecule which cannot actually exist, but geometrically, one can describe the optimum arangement for any arbitrary number of objects.  As a geometry exercise, I would imagine that the solutions for molecules with greater than 9 atoms around the central atom have been solved, but as a chemist I am unfamiliar with those shapes as I never need to use them for any reason.  -- Jayron  32  02:55, 30 March 2012 (UTC)
 * Apparently, no general solution exists, but you can write a computer program that will arrange any arbitrary number of points onto the surface of a sphere. One example that is easy to find is for 12 atoms around a central one. This will form an icosahedral pattern, equivalent to the hexagonal sphere packing. This paper has more info, though it looks at all n-spheres (i.e. all sphere-like objects in any number of dimensions), and barely touches on the classic 2-sphere (what we'd call a sphere in everyday language). Smurrayinchester 07:46, 30 March 2012 (UTC)

looking to side to yawn
if i look to the side (left or right) with my eyes closed, in the position i'm in, it sounds in my ear like yawning. why is this. what is that sound. does something open or close.

in this case i dont open my mouth at all but it sounds / feels the same. closed mouth yawn. 149.200.72.244 (talk) 21:29, 29 March 2012 (UTC)


 * At a guess, is its because your eye muscles ( the ones controlled by you're sixth cranial nerve) are tremoring. Yawning muscles of the jaw do the same thing at 'maximum' flexation. Bone and muscle/connective tissue  transmits the tremor   and back to the ear. This tension may  also distort the shape of the ear and changes the amplitude and shape of its dynamic response so that the ambient noise sounds different. Never been asked this question before -so thanks.--Aspro (talk) 21:48, 29 March 2012 (UTC)

Do planes cast full-sized shadows?
do planes cast full-sized shadows? if so/if not is this due to the fact that the sun is not a point source (like a very small candle) but in fact a huge ball of 'light' far larger than Earth? Is it the distance that does it? If so at what distance would planes start casting shadows that weren't/were (opposite of fact) their own size? 149.200.72.244 (talk) 21:54, 29 March 2012 (UTC)


 * The Sun is so incredibly far away that the light from it is, for all intents and purposes, parallel. So illuminated by the Sun overhead, the size of an aircraft's shadow (on a level surface) is the same as the aircraft. -- Finlay McWalterჷTalk 21:57, 29 March 2012 (UTC)


 * But even at the distance it is, the Sun isn't a point source. Looked at (with appropriate precautions of course) it's still a disk (because it's s incredibly huge). If it was a point source the aircraft would cast a crisp shadow; as it's not, the shadow has fuzzy edges (areas on the ground for which the plane blocks some, but not all, of the Sun's disc). -- Finlay McWalterჷTalk 22:00, 29 March 2012 (UTC)
 * This is correct, but unless the plane is quite low the fuzzy area extending from the edges meets in the middle and the whole shadow becomes faint and fuzzy. As Baseball Bugs notes below, to form a dark shadow with reasonably clear edges, the plane has to be low enough to totally eclipse the sun from the point of view of observers on the ground.--Srleffler (talk) 17:53, 2 April 2012 (UTC)


 * The moon, for example, doesn't cast a "full-sized" shadow, i.e. a cylinder - it's more of a cone. ←Baseball Bugs What's up, Doc? carrots→ 21:58, 29 March 2012 (UTC)


 * When an aircraft is close to the ground it casts a shadow that is clearly recognisable as the shadow of the aircraft, although it is a little fuzzy around the edges due to the penumbra. Higher above the ground an aircraft doesn't cast a shadow at all, but a Glory (optical phenomenon) which is a bright circular patch on the ground, a bit like there is a searchlight shining brightly at a point on the Earth's surface.  The bright patch (the glory) moves across the Earth's surface at the same speed as the aircraft.  Dolphin  ( t ) 22:03, 29 March 2012 (UTC)


 * Looking at it from the other direction, consider looking up at a plane crossing the sun's disk (let's supposed one is immune from being blinded). Unless the plane totally "eclipses" the sun, there's not going to much of a shadow discernible from the ground. ←Baseball Bugs What's up, Doc? carrots→ 22:15, 29 March 2012 (UTC)


 * If there's no shadow, what if (immune from being blinded) you look up and into the hole of the glory? What do you see? - is the sun occluded there or not? 149.200.72.244 (talk) 22:26, 29 March 2012 (UTC)


 * There is a shadow, you just can't see it. All you see is the plane, looking black, against the sun - as with, for example, an annular eclipse. ←Baseball Bugs What's up, Doc? carrots→ 22:29, 29 March 2012 (UTC)
 * I have been on planes where the shadow is clearly visible on the ground, but I believe at a suffieciently high altitude, like cruising altitude for a jet liner, the shadow becomes almost indescernible. The atmosphere also plays a non insignificant part in this question. Even though it "looks" transparent, the air does reflect some light. It was noted that on the moon, shadows appear MUCH darker and "crisper" then on the earth. This effect is ever greater if there are any clouds in the sky which act like big light diffusing boxes. Vespine (talk) 23:19, 29 March 2012 (UTC)
 * BS. Airplanes do not cast glories on the ground. The glory is an optical phenomenon that happens when light around a shadow falls on water droplets in the air (mist or clouds). The phenomenon is created by the water droplets, and the rainbow pattern is cast back toward the source of the light (which is why you only see it if your shadow is falling on the mist or clouds. You have to be on the plane to see a glory around the plane's shadow). --Srleffler (talk) 17:48, 2 April 2012 (UTC)


 * Google image [airplane shadow] and you'll see all kinds of interesting stuff. ←Baseball Bugs What's up, Doc? carrots→ 00:16, 30 March 2012 (UTC)
 * There is a couple of good images of optical effects from aircraft - see Glory (optical phenomenon). Dolphin  ( t ) 01:06, 30 March 2012 (UTC)


 * Ignoring atmospheric effects, we can pretty easily calculate all the values explicitly. First, I want to know the apparent angle of the diameter of the sun as viewed from Earth.  This is just the ratio of its actual diameter to its distance, so that's 1.392×106 km/1.496×108 km = 0.0093 radians.  Dark full shadow (umbra) occurs where the plane is fully obscuring the disk of the sun.  Supposing the plane is wide-body aircraft let's assume the diameter of the fuselage is about 6m.  For the body of the plane to fully obscure the sun at any point on the ground, the distance would have to be less than 6m/0.0093 ≈ 640m away.  Above that, no point on the ground will be in full shadow.  The "fuzz" on the shadow (penumbra) grows from the edges linearly in the plane's altitude, until eventually it engulfs the whole shadow.  Rckrone (talk) 02:45, 30 March 2012 (UTC)

wnhat would happen if the Earth stopped spinning
say over a 72 hour period it wound down and stopped spinning (you can make this slower, a year or two if it helps preserve life and sanity). we need this to harness infinite energy (let's say) and can use it to stop te earth. But what happens if we do? Nows in the interest of fairness (one half would always be dark) we could transfer the power station every 3 months and rotate the Earth by a quarter, to change who gets light. (seasons however are due to the angle with the sun i understand, and aren't due to spin - after all it's day and night in every season). This movement is a massive pain in the ass though so you would only rotate the earth every so often. at other times it becomes locked facing the Sun (s the moon is locked facing us).. in this case would this work, or how often would we have to give the old rotesserie a turn to keep things from burning? Finally are any of the planets (mars, venus, jupiter, etc) locked iwth the sun, and if so how are they dealing with it? 149.200.72.244 (talk) 22:16, 29 March 2012 (UTC)


 * No, none of the planets are tidally locked to the Sun. People used to think Mercury was (there's an Arthur C. Clarke story about pogo-sticked aliens living in the twilight band between the burning hot and the unliveably chilly), but it's not.   To answer your question: in such a short time, we'd all die. But if we have infinite energy, can't we can just live off that?  Given more time than your draconian deadline, we can all move a mile or so underground (hey, it's like The Matrix) where geothermal heat predominates, and screw the Sun. -- Finlay McWalterჷTalk 22:22, 29 March 2012 (UTC)
 * Wouldn't the infinite energy collapse into a black hole and suck us all in? --Trovatore (talk) 23:30, 29 March 2012 (UTC)
 * see my other question. apparently we only need to stop the earth if we want this infinite energy (which we would also use to stop the earth) but a bit of googling what happens if the earth stops shows that tthe whole thing might not actually be worth it. 149.200.72.244 (talk) 22:29, 29 March 2012 (UTC)


 * If I may quote myself, from a similar question in September of last year: (I have taken some liberties to modify the quote to increase relevance to the current question)
 * The answer to this type of physics question always depends: how would [the earth stop spinning]? If you can specify that, we can follow through with the consequences by solving the equations of motion for the Earth-Moon[-Sun] system.

For example, if you hypothesize that a giant comet large and fast enough to change the [Earth's rotation] impacted our planet, .... well, we would need to calculate the effect that such a large comet has on the orbits of Earth and everything else in the solar system, too. We could solve that problem by setting up an n-body problem to model the solar system, including Earth, Moon, Sun, and other planets; and we would use perturbation theory to study how sensitively the system reacts when we add in a new comet on a course to impact the moon. The results are difficult to compute, but this can be done in a reasonable amount of time with a reasonable amount of effort.

...

On the other hand, if you just want to make something up, "just imagine" that the [Earth] magically changes its [rotation], all bets are off. We can't meaningfully speculate what consequences follow when one law of physics breaks "because of magic." Anything could happen. Everything we know about the way the Moon's orbit couples into the Earth's rotation depends on the rules of physics as we currently understand them.
 * So - does that answer your question? Anything.  Anything could happen.  Nimur (talk) 22:44, 29 March 2012 (UTC)
 * I saw a documentary on exactly this topic, they descibed the weather system shutting down, the oceans receding from the equator, all life on earth dying, etc.. Plasmic Physics (talk) 22:57, 29 March 2012 (UTC)


 * The English language lacks a word to adequately describe such hyper-speculative programs, which are indeed a staple of science channels the world over. May I suggest "makestuffupumentary"? They've taken a model with hundreds of unrelated variables and woggled the dials around (usually to the most cinematically exciting values) and say "what would happen if". -- Finlay McWalterჷTalk 23:08, 29 March 2012 (UTC)
 * It would be certain death to anything on the "sun" side in a very short amount of time, far less then 3 months. Without a night time to cool things down, my completely uneducated speculative guess, between the tropics everything would be dead within a week. Vespine (talk) 23:26, 29 March 2012 (UTC)
 * On the Sun side, temperatures would go up by about the fourth root of 2 times the Kelvin temperature, so a daily average of 20 C would go to 70 C, which I agree is pretty unpleasant. Offhand I think it would take ~2-3 days before it is too hot to be tolerable to humans.  Dragons flight (talk) 00:04, 30 March 2012 (UTC)
 * Nay, it would take much less than one day. :) Really though, there are some factors that would be hard to figure out - for example, 70 C would put so much water vapor into the air - what would that do to the greenhouse effect and albedo?  But meanwhile water would be getting lost to the dark side where it would be locked up in ice.  There would also be huge salt deposits where I suppose briny water would lurk and defy (almost) all attempts at evaporation.  I'm not sure if you end up with Venus or the dark side of Mercury, or something truly weird and novel.  I doubt it would look like in the "crockumentary" though. Wnt (talk) 14:36, 30 March 2012 (UTC)


 * This is the television program.  No less inane than any other speculation on the topic, I suppose. Wnt (talk) 23:57, 29 March 2012 (UTC)


 * By the way, has no one thought to ask Alan Jackson? --Trovatore (talk) 00:39, 30 March 2012 (UTC)


 * To answer Finlay's question: the term "crockumentary" has been used to describe perceived junk documentaries. Fox News has used it before. 207.6.208.66 (talk) 01:14, 30 March 2012 (UTC)
 * Fox "News" questioning the academic stridency of anything is pure, unadulterated irony. -- Jayron  32  02:40, 30 March 2012 (UTC)
 * there will be wave ,  Water Nosfim  — Preceding unsigned comment added by 81.218.91.170 (talk) 07:58, 30 March 2012 (UTC)

If the Earth were tidally locked, presumably even with no rotation the atmosphere would redistribute heat to the night side. Given the thickness and composition of Earth's atmosphere, I wonder how the day side-night side temperature difference would be mitigated as compared to a situation with a very thin atmosphere. Duoduoduo (talk) 15:08, 30 March 2012 (UTC)
 * You would end up with very strong winds blowing from the day side to the night side because of convection currents (it's the same process as makes prevailing winds blow from the equator to the poles, just more extreme). That would be enough to transfer a lot of heat. It is very difficult to judge exactly how much, though. What kind of greenhouse effect you end up with would be significant - there isn't a great deal of difference between the day and night side of Venus despite a day lasting several months, and that's because of the greenhouse effect - the night side can't actually cool down much because the heat is trapped. I'm speculating here, but I expect the convection currents would be enough to keep the levels of water vapour reasonably constant on both sides of the Earth, so the night side would be kept reasonably warm. If we compare it to the moon, which has no atmosphere to regulate temperature at all, the day side gets up to +180C and the night side down to -150C. The Earth would show a significantly smaller variation than that, and presumably a fairly similar average to now (depending on exactly what happens to the greenhouse effect), so I think Dragon Flight's estimate of +70C on the day side is reasonably (hotter at the equator and cooler at the poles, the same as it is now). The night side might get down to -50C or so? Those are just educated guesses, though. --Tango (talk) 20:21, 30 March 2012 (UTC)