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August 6

can ethers act like alcohols in acid-catalyzed addition?

My lecturer never really seemed to cover this (or I must have forgotten) and a quick google search isn't helping. Basically I'm wondering since ethers also have lone pairs to donate, whether they can interact with carbocations ...

So let's say we have an (R2)+ carbocation after the addition of H+ to a C=C alkene bond, and a R1-O-R1 ether, and say R1 was a shorter alkyl chain than R2. Shouldn't a R2-(O+)-(R1)_2 intermediate develop, with the resulting ejection of one of the alkyl groups? I'm basically wondering if you can exchange alkyl groups in an ether this way. Shouldn't the lone pairs on an ether oxygen be even more reactive than an HOH or ROH, since the ether oxygen will have more electron density. I know alkyl groups aren't that labile, so would that just make it kinetically unfavourable to form the ether cation?

I assume one of the difficulties is that acids don't like to dissolve into ethers -- and since the ether doesn't have an H group, it can't "return" the H+, so H+ is getting consumed....and I suppose nature would prefer to have H+ ions than R+ ions? But to address the first problem, we could choose a mixture of methylene chloride and ether to try to solvate the acid, and choose a strong acid whose conjugate base makes a poor nucleophile... (magic acid?). John Riemann Soong (talk) 02:15, 6 August 2009 (UTC)[reply]

Alkyl groups are not great leaving groups in these sorts of reactions. And the lone pairs are LESS reactive on the ether, because the long alkyl chain acts as a "soft" electron sink, basically giving the electrons a place to disipate to, making them less concentrated on the oxygen. This is in contrast to the "hard" hydrogen, which cannot absorb extra electron density. These two effects (poor leaving group + less electron density around oxygen) probably make the reaction unlikely to happen to a great extent. --Jayron32 02:30, 6 August 2009 (UTC)[reply]

That's not quite right - the lone pairs on ethers are less reactive maybe due to steric effect - but they have a greater ability to stabilise a cation compare to water when R=alkyl,benzyl, ethenyl etc..83.100.250.79 (talk) 13:04, 6 August 2009 (UTC)[reply]

In other words: definitely not (at least under anything like "normal" conditions). 98.234.126.251 (talk) 02:46, 6 August 2009 (UTC)[reply]

Well yes, but carbocations aren't too great either -- would that in fact prevent the H+ from attacking the alkene? (As in, the intermediate will convert back to the alkene + H+ before an ether would attach?)

Why is that ether oxygen lone pairs are less reactive? Because alcohol oxygens are still reactive (the reaction proceeds merrily), and I note the pKa of an alcohol is higher than that of water ... and the pKa of a tert alcohol tends to be highest. My interpretation of this is that the electron-donating effects of the alkyl chain were greater than any electron-withdrawing effects -- that is, the alkyl chain destablised the lone pair conjugate base. John Riemann Soong (talk) 03:16, 6 August 2009 (UTC)[reply]

Carbocation formation is different cause there ain't nothing being actually ejected from the molecule in this case (i.e. no C-C or C-O bonds being broken, as they would be in your scenario). It's the strength of the bonds to be broken that accounts for the reaction not happening. FWiW 98.234.126.251 (talk) 03:25, 6 August 2009 (UTC)[reply]

So what would happen after the acid addition step? Would we have a "stabilised" carbocation solvated in solution? (Assuming the acid's conjugate base is a very poor nucleophile, e.g. SbF6-.) After we evaporate the ether and the methylene chloride off, would I get a strange crystalline salt of alkyl carbocations and SbF6- anions? (Granted, it would probably be hygroscopic and quite reactive to water, but it would be interesting....) John Riemann Soong (talk) 05:03, 6 August 2009 (UTC)[reply]

I think what will really happen is the anions would add to the carbocations forming a neutral covalent compound. Or if not, then the carbocations will instantly react with any moisture available (even the moisture in the air, no matter how slight) and form a sec / tert alcohol and regenerate the acid (and prob'ly give off a whole bunch of noxious vapors and maybe splatter the reaction mixture all over the place cause this final step is very exothermic). FWiW 98.234.126.251 (talk) 05:50, 6 August 2009 (UTC)[reply]

Mmmm, but diborane is extremely reactive with water as well, and yet we can still observe it meaningfully without too much hassle. And one could just view it in dehumidified air... John Riemann Soong (talk) 06:27, 6 August 2009 (UTC)[reply]

You said you formed the carbocation by protonation of an alkene. That's essentially an equilibrium process--loss of H⁺ adjacent to a carbocation forms an alkene as the standard E1 reaction mechanism. On the other hand, reaction of carbocation(-like species) and ethers is known, and leads to alkylation of the ether as you suggest (an O⁺ compound). These compounds can be isolated in some cases, for example, triethyloxonium tetrafluoroborate. DMacks (talk) 05:58, 6 August 2009 (UTC)[reply]

So where exactly would the carbocation add to the ether? And would the alkylation product be stable? 98.234.126.251 (talk) 06:12, 6 August 2009 (UTC)[reply]

It attatchs at the ether oxygen, and the resultant compound may or may not be stable, depending on what exactly it is, and the temperature, and other factors.83.100.250.79 (talk) 13:45, 6 August 2009 (UTC)[reply]

Well, I see several scenarios: the ether is alkylated and doesn't eject any of the alkyl groups, and forms a stable cation, assuming again that the conjugate base of the acid used is a poor nucleophile; an alkyl group is eventually ejected (would the ether oxygen prefer to retain the longer-chain, and if so, could we exchange alkyl groups on an ether this way?); though much less labile than H+, the alkyl groups act somewhat like H+, being constantly exchanged; the ether doesn't react, leaving just the carbocation (the strange salt described above); the ether doesn't react and the carbocation isn't stabilised enough to keep the H+; so the equilibrium favours the reactants and most of the alkene is left unreacted. John Riemann Soong (talk) 06:27, 6 August 2009 (UTC)[reply]

Comment In the above discussion - only DMacks (and to an extent the orignal questioner) above is really correct in what they say.83.100.250.79 (talk) 13:11, 6 August 2009 (UTC)[reply]

Assuming you have produced a carbocation, it will and can react with an ether - but you need to consider steric effects, which will tend to hinder their production in general.

Compounds R1R2R3O⁺ are called oxonium ions. eg the triethyl oxonium ion mentioned above, which can actually be bought as the tetrafluoroborate in solution (needs to be kept cold)

Yes you can/could exchange alkyl groups this way. As you expect the reaction will tend to favour the most stable carbocation being ejected (all other things being equal)

see Triethyloxonium tetrafluoroborate 83.100.250.79 (talk) 12:57, 6 August 2009 (UTC)[reply]

hovever bear in mind that if you try to protonate an alkene in ether solution with an acid, you'll end up protonating the ether instead83.100.250.79 (talk) 13:09, 6 August 2009 (UTC)[reply]

Wouldn't the protonated ether then protonate the alkene? Or would the H+ prefer to stick with the oxygen? I mean H3O+ protonates alkenes in the classic acid addition of water setup... John Riemann Soong (talk) 13:17, 6 August 2009 (UTC)[reply]

Yes, I was assuming you wanted pure solutions of carbocation - the reaction will still work.

By the way did you know that one of the common reactions of ethers is acid catalysed decomposition - eg

Et₂O + H+ >>>EtOH + ethene + H+

This can be a problem if the reaction is warm83.100.250.79 (talk) 13:49, 6 August 2009 (UTC)[reply]

How important is the consumption of H+? I guess it's not acid-catalysed addition anymore, is it, since H+ is being consumed? What determines whether the oxonium ion is stable or not? And even if alkyl exchange occurs, we'll still get an interesting crystalline species once we evaporate off the solvents... John Riemann Soong (talk) 13:15, 6 August 2009 (UTC)[reply]

The same things that stabilise carbocations, stabilise oxonium ions.

The concentration of H+ (from which the concentration of C+ derives via H+ addition to alkenes) will affect the rate of reaction. But it's not catalytic if oxonium cations are the product - since H+ is used up.

However if alkyl exchange is the only reaction and the products are different ethers and alkenes - then H+ is catalytic83.100.250.79 (talk) 13:42, 6 August 2009 (UTC)[reply]

The oxonium cation will be the product if the ether oxygen is the strongest base in the system - most oxonium cations will be stable at low enough temperatures - at high enough temperatures - decomposition occurs.

It's probably worth noting that ethers can interact (form bonds) with a wide variety of lewis acids such as BF3to form Diethyl ether boron trifluoride adduct (aka Boron trifluoride etherate), and in general can act as ligands (as lone pair donors) or lewis bases with a wide variety of compounds (not just carbocations, and H+) 83.100.250.79 (talk) 14:53, 6 August 2009 (UTC)[reply]

Interesting -- normally I assume high temperature favours carbocation formation .... but I guess this principle applies only as long as we want the carbocation to be an intermediate? I assume this is the same for oxonium too. Re: oxonium stabilisation: I suppose we can't do anything about trying out allylic-ish oxonium cations, since the pi bonds are going to be attacked. And if oxonium stability is analogous to carbocation stability, I guess actually alkylated ethers are more stable than H3O+?

This makes me very interested now -- well, if tri-alkyl oxonium cations are more stable than say, a protonated alcohol or a protonated ether, why don't most textbooks mention ethers as something you could play around with during acid addition? I assume the ether would be a very good competitor for the cation compared to other nucleophiles. Is the fact that H+ might not be returned, potentially making a reaction scheme infeasible?

Also, so many different possibilities -- I'm trying to work them out. I guess a useful reaction might be mixing a straight-chain alkene, an ether with tertiary alkyl groups and the strong acid with a poor nucleophile conjugate base; the ether solvent binds to the carbocation and expels one tertiary alkyl group. So if we run the reaction at high enough temperatures, this favours both the decomposition of oxonium and the alkyl group into a new ether and a new alkene...but then it occurs that some of the ether molecules might react twice with the protonated alkene -- I guess it depends on just how much alkene is there compared to ether, isn't it? (And what would be a general principle of separating two different ethers? I assume that if tert groups are being ejected in favour of primary groups, the boiling point of our products will be higher and we can evaporate off the original solvent -- but that's one case.)

To me, I'm seeing this as a game of which type of cation is more stable -- protonated ether, tri-alkyl oxonium, or the alkyl carbocation. What if we had straight chain alkenes and ethers? I assume in general the longer-chain primary carbocation would be more stable? And now I realise that tertiary carbocations are more likely to be ejected, but primary carbocations are more likely to convert back into alkenes. Aaaahh, so many different considerations! John Riemann Soong (talk) 11:43, 7 August 2009 (UTC)[reply]

Why is the Mandelbrot Set so complicated?

Infinitely complicated in a way that is awesome, distressing, beautiful, fearful, samey, completely obsessed and obsessive, suggestive of an infinitely talented but hermetically sealed divine lunatic of a geometer. Pray answer me in such a way as this. Suppose that I had asked: “Master, why is this river so quiet and this other river so wild and noisy.” And the Master replied: “Grasshopper, the first river has little water and flows over land only slightly tilted so it runs slowly and quietly, but this other river runs over steep ground with many rocks and crevasses, and there is much water so it is churned up as it runs, and thus the noise and the spectacle.” You see, Grasshopper can understand that. And I am such a one as that Grasshopper. I am Termite. And I would appreciate you answering me in just such a way as the Master answered Grasshopper. Why is the Mandelbrot Set so complex, and why is a triangle so simple? Myles325a (talk) 05:50, 6 August 2009 (UTC)[reply]

I think you should try the Mathematics desk, they may be of more help to you. John Riemann Soong (talk) 06:29, 6 August 2009 (UTC)[reply]

I don't think the mathematicians are going to help you. You may get an answer there - but I'd bet good money that you'd be unable to understand it. This is a case of observer bias. The fact is that if the Mandelbrot set wasn't all of the things you describe, nobody would have given it a second glance. There are an infinite number of mathematical objects - and only the interesting ones are ever investigated or written about. The Mandelbrot set (and it's close relatives) is well known BECAUSE it's complicated. The triangle is known because it's useful. Anything significantly simpler than the Mandelbrot set is simply too boring to have made it into your stream of thought. There are many other things that are as complex as the Mandelbrot set (a Julia set, a Sierpinski sponge, or a Koch snowflake for example) - but somehow we don't see them as beautiful - so most people never hear about them. Check out List of fractals by Hausdorff dimension for lots more examples. SteveBaker (talk) 13:04, 6 August 2009 (UTC)[reply]

Triangles are one of the abstractions of Euclidean geometry that Buckminster Fuller pointed out cannot exist[1]. The Mandelbrot set reveals the form in 2-D of an iterative oscillation that actually exists, in the same way that prime numbers actually exist. Triangles are useful for designing things and bracing them, while the other two are more intriguing than useful. No one promised that reality would not be complicated. Cuddlyable3 (talk) 20:56, 6 August 2009 (UTC)[reply]

Huh? Triangles and prime numbers are both ideal objects; I don't see any basis for saying one really exists and not the other. Of course triangles are completed infinite totalities, and it's possible that finite ideal objects exist but infinite ones don't. But empirically this doesn't seem to be a very productive assumption to make. --Trovatore (talk) 21:01, 6 August 2009 (UTC)[reply]

None of the Euclidean primitives point/line/plane can exist. Try to make one of them if you can. Cuddlyable3 (talk) 21:18, 7 August 2009 (UTC)[reply]

As soon as you make a prime number. --Trovatore (talk) 08:01, 8 August 2009 (UTC)[reply]

I consider the Julia set to be more beautiful than the Mandelbrot set. I have some home-generated prints produced on a BBC Archimedes computer many years ago. Most of the set is not particularly inspiring, but I spent many hours looking for the beautiful bits. I think the same is true to some extent of Mandelbrot: we are shown just the attractive bits. Dbfirs 01:59, 7 August 2009 (UTC)[reply]

Which Julia set? The M-set is a sort of catalog of an uncountable family of J-sets. —Tamfang (talk) 03:21, 11 August 2009 (UTC)[reply]

Yes, do ask at the mathematics desk. But the Mandelbrot set is complicated because of the phenomenon of sensitive dependence on initial conditions (SDIC). If you're really a grasshopper, then among your insect relations you have probably encountered a few butterflies, so you can understand the butterfly effect, another example of SDIC. A triangle is simple because moving from one point to another in one operation doesn't have SDIC. If you start from a slightly different place, you end at a slightly different place, and there you are. The SDIC in the Mandelbrot set comes from the repeated iteration of the squaring function, like in the butterfly effect it comes from evolving the chaotic system for a long period of time. Points in the Mandelbrot set can behave much differently from very nearby points under the iteration, because of the SDIC. The exact pattern of differences is the complexity that you see. 70.90.174.101 (talk) 07:09, 7 August 2009 (UTC)[reply]

Biology has its own version of such sets, in even more abstract form. Deuterostome gestation arguably follows a fractal pattern (radial cleavage) -- one cell's signals ends up determining the positions of other new cells, but the conditions for such patterns must all be stored within a single zygote cell. Mysterious, because there are no such things as "arm genes" or "elephant trunk genes" -- just a whole lot of signalling genes cooperating together to encourage division in the right direction and arrest division in the wrong direction (over and over). What's especially fascinating is how did such a gestation pattern evolve? Some dynamic patterns you might be interested in might include Conway's Game of Life. John Riemann Soong (talk) 08:35, 7 August 2009 (UTC)[reply]

amper-current and relativity

hello, this is a passage from article and i can't understand it,the context is that scientists didn't want to deal with theories that they couldn't measure.

"One can see an analogous sort of indeterminateness at the basis of A. Einstein's (1952, p. 37) complaint against pre-relativistic mechanics and electrodynamics. His criticism was that they lead "to asymmetries which do not appear to be inherent in the phenomena." Whether one assigns an absolute velocity of zero to a conductor and a non-zero velocity to a magnet, or vice versa, the measurable result (current) is the same. Hence, the absolute velocity is indeterminate: "The phenomena . . . possess no properties corresponding to the idea of absolute rest," as Einstein put it. Interpreted a la Glymour, the point is not the naive one that absolute velocity is not measurable-that would be alright-but that it cannot even be computed from measured quantities via any seriously proposed (let alone well-tested) hypotheses.

why does Einstein says there is no symmetry? why it is not the same to rotate object around another which stand still and then swap the rolls? why can't we measure current according to the text? if so, what is ammeter??

thank you —Preceding unsigned comment added by מני111 (talk • contribs) 07:10, 6 August 2009 (UTC)[reply]

"why does Einstein says there is no symmetry?" He means that Maxwell's equations are not invariant under the Galilean transformation. In simple terms : In Einsteins time, people thought they could transform from a "stationary" frame of reference to a moving frame of reference simply by adding velocities. The equations describing electromagnetism change when you do this. In reality, the observed phenomena stay the same for both observers. It is this observation that eventually led to the discovery of the Lorentz transformation, which does have the required symmetry : the Maxwell equations are invariant under the Lorentz transformation. http://en.wikipedia.org/wiki/Maxwell%27s_equations

http://en.wikipedia.org/wiki/Galilean_transformation

http://en.wikipedia.org/wiki/Lorentz_transformation —Preceding unsigned comment added by 81.11.170.162 (talk) 10:48, 6 August 2009 (UTC)[reply]

I'm not really too sure what you're asking, but to give you I think a suitable analogy, take the Earth-Sun system. If they were stationary relative to one-another, the Earth would be sucked into the Sun. And there is a suitable rotating reference frame in which they are stationary relative to one another, but even in this frame the Earth is not sucked into the Sun. This is an asymmetry; the laws of physics do depend on your frame of reference. Incidentally, even in Special Relativity, acceleration is still absolute, it's only in General relativity that there are no inertial frames. I'm not too sure if I've addressed your clearly, so please post again if I haven't.--Leon (talk) 10:51, 6 August 2009 (UTC)[reply]

There was an asymmetry in the physical theories of the time that did not correspond to any asymmetry in the real world (the phenomenon). We can measure current, but the theory at the time seemed to do a bad job of explaining why, and Einstein was criticizing it for that.

I would advise against reading this article, whatever it is, because it looks like bad philosophy. It seems to imply that "absolute velocity is not measurable" is different from "absolute velocity [...] cannot [...] be computed from measured quantities"—which is nonsense and certainly not something that Einstein believed—and it implies that "absolute velocity is not measurable" is an obviously true statement a priori; see Section 16–1 of the Feynman Lectures on Physics for a response to that. -- BenRG (talk) 11:17, 6 August 2009 (UTC)[reply]

thank you all. i can't understand why when you swaping the objects the result shuold be a-symmetric. why the frame of reference imoprtant when i (the observer) don't move and one time i take the counducter rotate it around the magent and the other time, the smae places but the magnet is rotating. each object was one time static and the second time mobile - soo it should'nt change the current result. it is the way you look at that, one can say both pbjects moving but one is faster (the frame reference is earth). if we will think about a person seeing a child throw an apple on a moving train (a magnet make electron move) is the equivalent for the child to see the man outside throwing an apple. [not connected to the fact the apple will move slower for the an outside) - i really hope i didn't make it much to confusing the speed of light is absolut, so there is an absolute speed or not? about the article itself is michael r. garder - realism and instrumentalism in 19th century atomism

if i  may,the lorentz transformation article is complicated, is there intuative explantion?

thank you —Preceding unsigned comment added by מני111 (talk • contribs) 12:34, 6 August 2009 (UTC)[reply]

Einstein was pointing out a problem : in reality, the result is symmetric. Science in Einstein's time suggested it should be asymmetric.

The Maxwell equations allow you to calculate the speed of light without specifying an observer who measures this speed. In classical mechanics on the other hand, all calculated speeds are relative to a (usually implicitly specified) observer : if you calculate the speed of an apple that falls from a tree, your result will be relative to an observer who saw the apple at rest in the tree.

Thus, consider a light in a train with speed v. According to the Maxwell equations, an observer in the station sees the light moving with a speed c. According to gallilean relativity, an observer in the train will see the light moving with a speed c-v, but according to the Maxwell equations, he will also see it moving with a speed c. This is a problem! In reality, both observers will observe the light moving with speed c, but that is not what Einstein's contemporaries thought.

Einstein (and Lorentz) discovered that you must not only transform velocities, but also lengths and durations, when calculating what a moving observer sees. This is known as the Lorentz transform.

The above example involves the speed of light. Light is an electromagnetic phenomenon, and comparable examples can be found for other electromagnetic phenomena, including moving magnets.

---

If you read the whole paragraph (the first one) in Einstein's paper it is clear what he meant by the asymmetries. The problem is that theory of the day gave a different physical explanation for what happened when you had a moving magnet and a stationary coil, and when you had a moving coil and a stationary magnet (this is a problem relating to a unipolar dynamo, if one is interested). Einstein says that this is silly and means something is fundamentally wrong: you can't distinguish between the two if you don't believe in absolute rest (which he doesn't). It's an introduction to a general attack on the theoretical framework of the day (a clever one, but not necessarily a convincing one in its time, more of a logical attack than an empirical attack), and gives some indication of Einstein's characteristic style in attacking problems (Machian, axiomatic, concerned with basic concepts and definitions, not specific experimental results). Asymmetries are a huge deal to Einstein—he says if you can't possibly distinguish between two different states, then you shouldn't have a separate set of physical explanations (or physics) for them. It's a philosophical critique, though one that led him to some rather fruitful revisions of the theories. The specific criticism is something of a throwaway introduction—here's one of many things that indicate something is wrong, so let me try and rework this from the ground up without assuming the existence of concepts (e.g. absolute rest) that we can't measure. --98.217.14.211 (talk) 16:39, 6 August 2009 (UTC)[reply]

thank you very much, that is very intersting, why the physicists at that time thought those are completly different phenomena? can you give me a link please to those different explanations? —Preceding unsigned comment added by מני111 (talk • contribs) 22:38, 6 August 2009 (UTC)[reply]

Click the link—he discusses it. It isn't very interesting in and of itself. I don't remember the details of it, but it's basically that in both cases you get an electric current. One explanation was used to say that this came from some sort of interaction with an electric field. The other said there was an electromotive force involved. These are two rather different types of answers and Einstein said, "you can't have both of them." this (from "Electric Dynamos" on) gives a good discussion of the technical aspects of what Einstein was talking about. --98.217.14.211 (talk) 14:12, 7 August 2009 (UTC)[reply]

thanks (meni) —Preceding unsigned comment added by מני111 (talk • contribs) 11:37, 8 August 2009 (UTC)[reply]

METHANE ON MARS

An article on BBC News Science page says there is methane being produced on Mars it says this could be geological in origin or biological, Geological would be Active volcano not heard of this before on Mars 1.is there known active Volcanoes? or by a process of Serpentinisation - which is a process involving water. 2.is water in suffcient quantity to produce this effect would it need to be free flowing or in a solid form Ice maybe? and last but not least Biological activity which i assume they mean life in some form? —Preceding unsigned comment added by Chromagnum (talk • contribs) 07:30, 6 August 2009 (UTC)[reply]

Our article Atmosphere_of_Mars has an extensive section on Methane that explains the current theories rather well. We also have a section in Life on Mars that talks about how Methane might imply the presence of life on that planet. SteveBaker (talk) 12:46, 6 August 2009 (UTC)[reply]

Maybe the methane is being produced because the Martians are drilling for natural gas! :-D 98.234.126.251 (talk) 23:12, 6 August 2009 (UTC)[reply]

Thanks SteveChromagnum (talk) 08:39, 8 August 2009 (UTC)[reply]

Ipod

(moved from Miscellaneous reference desk)

How can I connect my Ipod to my Sony (maybe Panasonic) Hi Fi? It has a usb port on the front, but when I put my Ipod, or a flsh drive in there it says Unsupport In fact what can I plug into this port to make it work. I have put about 3 months worth of music on my Ipod, but some times I want to listen to music through proper speakers, eg a party. Any help in this would be greatly appreciated. Thank you. ~~Zionist —Preceding unsigned comment added by 62.172.58.82 (talk) 06:58, 6 August 2009 (UTC)[reply]

I don't know about your specific hi-fi, but I would connect the two via audio cable. On the back of your Hi-fi there is likely an auxilliary input (probably an RCA connector, but check). You can buy a cable that plugs into the headphone jack on your iPod at one end and into the auxilliary input on your Hifi on the other. That way, your Hifi is acting as speakers for your iPod. — QuantumEleven 08:39, 6 August 2009 (UTC)[reply]

This is exactly what I did for an outdoor party at my house. Dismas|^(talk) 20:21, 6 August 2009 (UTC)[reply]

Your Ipod doesn't support the USB mass storage device class (MSC) interface the way that a pen drive or some other mp3 players do. You have to get an ipod docking adapter that connects to the docking socket on the bottom of the ipod, or else use an analog audio cable as Dismas suggested. There are also lots of boom boxes with dedicated ipod ports these days. 70.90.174.101 (talk) 07:18, 7 August 2009 (UTC)[reply]

Another option is to buy a transmitter for your iPod which will transmit what your iPod is playing over an FM radio frequency, and tune your hi-fi to this. These are predominantly used in cars without iPod inputs to the stereo, but I have used mine with my home stereo and it works fine. --jjron (talk) 08:34, 7 August 2009 (UTC)[reply]

Presuming your hifi still has a tape deck, buying a tape one may be a better option to reduce the chance of interference etc Nil Einne (talk) 09:28, 7 August 2009 (UTC)[reply]

But the solution suggested by QuantumEleven above is by far the simplest and cheapest. The connector (with a 3.5 mm stereo jack at one end and two phono plugs at the other end) costs only a few dollars/pounds, and most amplifier phono inputs are happy with an earphone output, though you can add an impedance matching circuit if you are concerned about highest quality of sound. The next cheapest is the tape adaptor (a mock cassette tape at one end and a 3.5 mm stereo jack at the other end). Dbfirs 19:33, 7 August 2009 (UTC)[reply]

If you store them as files (ie just copy the songs onto the hard drive when you plug it into your computer, don't use iTunes) does it work then? Your iPod wouldn't be able to play the files, but I think it would just work as a USB drive which your Sony could play from. TastyCakes (talk) 20:24, 11 August 2009 (UTC)[reply]

Fast poison

I've recently read a novel (Joe_Abercrombie-Best Served Cold) in which one of the characters poisons a number of his victims via fast acting poisons. Specifically;

Using a needle/pin dipped in a vial of poison to prick his victim.
Brushing a solution of poison onto books/metal and allowing the victim to touch the poison.

My questions;

Assuming that the pin prick allows direct transfer to the blood, do such poisons exist that work fast at such a small dosage (the amount that would be left after the pin is pushed through the skin)?
Do fast acting poisons exist that could be transmitted by touch?
Lastly, how many of these would be available to medieval level technology?

Thanks, Kellhus (talk) 11:22, 6 August 2009 (UTC)[reply]

A lot of poisons work at such small dosages (see LD50 for some good examples) -- botox seems like a good candidate. John Riemann Soong (talk) 12:28, 6 August 2009 (UTC)[reply]

There was that case of Georgi Markov - a Bulgarian dissident working for the BBC who was killed in London by a soviet agent using a poison-tipped umbrella. I know that sounds like the plot of a James Bond movie...but it's true. The poison use in that case was Ricin. The lethal dose is 500 micrograms...which could certainly be administered with a pinprick. The stuff is found in the beans of the Castor Oil plant - it's easy to extract and separate - so I presume it would have been possible to make it in medieval times. (Oh - but that's no good - you wanted "Fast acting". Markov died 3 days later.) SteveBaker (talk) 12:36, 6 August 2009 (UTC)[reply]

Not strictly a poison, but in medieval times, perhaps venom from a viper or something could fill the bill. Googlemeister (talk) 14:04, 6 August 2009 (UTC)[reply]

These are are excellent examples of injected poisons. As for touch, the only thing I can think of off the top of my head is some variation of organophosphate nerve agent, like sarin or VX, which will soak right through the skin. Most of the really nasty ones, though, are also pretty volatile since they were designed to be employed as gas weapons. – Clockwork Soul 17:55, 6 August 2009 (UTC)[reply]

This fits number 1, but not so much for 2 and 3. From our article on Suicide pill, The Central Intelligence Agency began experimenting with saxitoxin, an extremely powerful neurotoxin during the 1950s. It was rumored that they issued a tiny, saxitoxin-impregnated needle (hidden inside a fake silver dollar) to each American U-2 pilot, with instructions to stab themselves with it if shot down over the USSR.. Vespine (talk) 23:11, 6 August 2009 (UTC)[reply]

Maaan -- from the article it sounds like that's one of those things where you're completely conscious, but can't breathe. I wouldn't do it. I'd take my chances with the Reds. --Trovatore (talk) 23:27, 6 August 2009 (UTC)[reply]

A pin dipped in cyanide would work pretty quick too. As for medieval stuff -- well, Black widow venom might be a good candidate (better even than snake venom, which is not that fast-acting at any rate). FWiW 98.234.126.251 (talk) 23:17, 6 August 2009 (UTC)[reply]

I remember reading a story where an assassin used Dimethyl sulfoxide to carry a poison through the skin and into the target's bloodstream. It turns out that is actually possible. 152.16.59.102 (talk) 10:42, 8 August 2009 (UTC)[reply]

Frying an egg with a nuclear weapon.

I have found the figure 4.184×1017 J for a potential nuclear weapon yield, and 17500 Joules to fry an egg. I can't, however, find or understand any figures on the dissipation of energy per unit of distance from ground zero. Does anyone have these figures? I want to work out how far from a nuclear blast I need to be to fry an egg, my cooker is broken. SGGH ^ping! 11:52, 6 August 2009 (UTC)[reply]

Not sure, but you need to make sure you can actually touch it afterward. :-) I'd suggest using the rule of thumb Johnny Carson suggested once for turkeys, and extrapolate down for eggs. He noted that if you heat your oven to 10,000 degrees, you can cook your turkey in 2 seconds - but you won't be able to touch it for six years. :-)Somebody or his brother (talk) 12:05, 6 August 2009 (UTC)[reply]

That would be: 4x10¹⁷J, not 4.184x1017J...anyway: The big question here is how the energy dissipates over distance - and that's a real tough one because we don't know what's between you and the bomb. We also don't know what percentage of that energy is heat, radiation, the pressure wave, etc. But we also don't know what percentage of the energy impinging on the surface of the egg will be absorbed by it and turned into heat...an egg isn't like a 3" thick slab of solid lead...it doesn't stop all of the energy passing through it. Furthermore, it's very likely that if you throw enough radiant energy at an egg to cook it - but you do it all within a couple of milliseconds - then it's pretty safe to assume that a lot of that energy would actually go into splattering the egg all over surrounding landscape rather than cooking it. In short, I think the error bars on this thought experiment are so huge that the answer could be incorrect by several orders of magnitude. In other words, we might calculate the answer at 1 mile - when in fact it's something like 100 yards or perhaps 10 miles. Bottom line - we don't know. SteveBaker (talk) 12:31, 6 August 2009 (UTC)[reply]

... or indeed 93 million miles - see solar cooker. Gandalf61 (talk) 12:38, 6 August 2009 (UTC)[reply]

If you wanted to do it "safely," use a parabolic mirror from a safe distance to focus the heat energy. Ted Taylor did this to light a cigarette with a nuclear bomb, once, if I recall from The Curve of Binding Energy. --98.217.14.211 (talk) 15:23, 6 August 2009 (UTC)[reply]

Do NOT try this at home because cigarettes are unhealthy. Cuddlyable3 (talk) 20:03, 6 August 2009 (UTC)[reply]

That's officially the coolest thing I've ever heard. I bet he told that story to girls for years. – Clockwork Soul 17:51, 6 August 2009 (UTC)[reply]

If I recall correctly, there are places in the world where you can cook an egg just by dropping it on a rock. Probably somewhere with a lot of geothermal energy. Iceland? Jeremy Clarkson did it. Vimescarrot (talk) 18:23, 6 August 2009 (UTC)[reply]

I'm sure you can do this on some of the rocks in Rotorua at least some of the time. However your eggs may also start to smell rotten if you leave them too long Nil Einne (talk) 20:33, 6 August 2009 (UTC)[reply]

Cooking eggs in hot springs is a feature of Icelandic cuisine where the sulfur is thought to add to the flavor. Rmhermen (talk) 22:11, 6 August 2009 (UTC)[reply]

From accounts of the effects on people and structures at the Hiroshima and Nagasaki bombings, along with U.S. tests, it should be possible to determine what distance from a blast of what magnitude would produce the desired effect. If your skillet is too close, the blast effect would scramble the egg and send it flying. But farther from ground zero the thermal effects might provide enough heat energy to do the trick, without excessive radiation or blast. Edison (talk) 18:43, 6 August 2009 (UTC)[reply]

According to the article Effects of nuclear explosions, the thermal radiation would be enough to cause second-degree burns (and also to fry an egg) at about 3.2 kilometers from ground zero for a 20-kiloton bomb detonated at 1800 feet. Unfortunately, at that range the blast wave would be powerful enough to splatter the egg all over the place and also send the pan flying through the air at Mach speed. Also, the thermal radiation will have the effect of cooking the OP's skin, not just the egg, and it will hurt like you wouldn't believe. "Prompt" radiation will be negligible at that range, but radioactive fallout could be a concern if the nuclear explosion is upwind of the frying pan. FWiW 98.234.126.251 (talk) 00:18, 7 August 2009 (UTC)[reply]

Ah, but the radiation travels at the speed of light whereas the blast wave is only somewhat faster than ordinary sound. I'm not sure exactly how much faster, but supposing it is at Mach 1.1, say, what you need to do is cook the egg for 8 seconds and then you have 1 second to cover the pan. --Anonymous, 01:28 UTC, August 7, 2009.

But then you'll still have your (covered) pan flying through the air at Mach 1.1... 98.234.126.251 (talk) 02:27, 7 August 2009 (UTC)[reply]

The trick is to use a sufficiently heavy lid. --Anon, 22:06 UTC, August 7, 2009.

Wouldn't it squash the eggs and the pan as well? 98.234.126.251 (talk) 00:37, 8 August 2009 (UTC)[reply]

So you also need a sufficiently strong pan, that's all. --Anon, 23:41 UTC, August 8, 2009.

What you do is cook it inside a supersonic vehicle. Just outrun the blast wave. — DanielLC 05:02, 7 August 2009 (UTC)[reply]

You could also melt some Velveeta over it during your supersonic trip, allowing you to make a nice Mach & cheese. DMacks (talk) 05:37, 7 August 2009 (UTC)[reply]

Supersonic flight tends to heat up the skin of aircraft fairly well. When Concorde was fliying, it would be at 100C on the outside surface of the aircraft. Fry the egg on that. Googlemeister (talk) 13:34, 7 August 2009 (UTC)[reply]

Just make sure you hold on tight so the shockwaves don't carry you (or the pan) away. 98.234.126.251 (talk) 00:26, 8 August 2009 (UTC)[reply]

Please clarify the question: 1) Which egg? 2) Can the egg be still inside the bird ? 3) Is it a requirement that the OP survive to eat the egg ? 4) Is this method of frying to be used regularly until the OP's cooker can be fixed ? Cuddlyable3 (talk) 20:20, 6 August 2009 (UTC)[reply]

You forgot point 4. Will the egg be radioactive enough to kill you if you eat it. Maybe not the egg, but I bet the pan would be pretty nasty to be around. Googlemeister (talk) 20:36, 6 August 2009 (UTC)[reply]

Mmmm... probably not. Thermal radiation extends much further than the ionizing radiation does in most scenarios (neutron bombs excepted). Mildly radioactive eggs will probably not kill you. Trivia: During Operation Teapot, they did test the effects of nuclear tests on varying types of food products, including beer. Radioactive beer apparently is somewhat stale tasting but definitely potable. --98.217.14.211 (talk) 01:20, 7 August 2009 (UTC)[reply]

How do you know what it tastes like -- were you the brave soul who tasted it? ;-) 98.234.126.251 (talk) 02:24, 7 August 2009 (UTC)[reply]

The reports actually discuss the results of their atomic taste testing... --98.217.14.211 (talk) 14:06, 7 August 2009 (UTC)[reply]

The U.S. also exposed live pigs to nuclear blasts (in lieu of U.S. servicemen) at close enough distances to produce 1st, 2nd and 3rd degree thermal burns. So you could have ham or bacon with your hypothetical eggs. Wear oven mitt or welding gloves, and hold the skillet out the window for a few seconds until the ham and eggs are cooked to taste, then bring in and cover before the blast wave and fallout arrive. Edison (talk) 05:30, 7 August 2009 (UTC)[reply]

Would that be Operation Crossroads? ~A H 1^(TCU) 02:10, 8 August 2009 (UTC)[reply]

From Tsipis, K. "Arsenal" c. 1983 App. D: (brokenly translated by me to an ASCII approximation of the math):

Energy of a nuclear explosion = 4.2x10E15 joule / megaton
Neutron flux in rads = 5x10E13 * (Y/R**2)*exp(-R*rho/780)
- Y is yield in megatons henceforth, rho is density of air = 1.1 g/l, exp is ~~the natural logarithm~~the exponent of Euler's number, the inverse of the natural logarithm oops, I meant what a calculator does when you push the e**x key!, R is distance in feet (bear with me, the guy was writing for an American audience)
- And neutron flux and the forthcoming peak overpressure are important. Presumably you will be persuading someone else to hold the frypan out there, preferably swinging the pan face on into the blast at just the right time, otherwise I don't think the egg will get properly cooked. You don't want the egg to be overdone on one edge, right?
Thermal radiation for an airburst is Q = 1.8X10E3*(Y/R**2)*t
- Q is cal/cm**2, R is in km, t is the transmission factor through the atmosphere, ~= 0.8 at 1km
- Groundburst, use 1.15 instead of 1.8
- And the big question which sunk my ship is whether these are "big" calories or "little" calories. Comments later.
- Also from Tsipis: there are two thermal pulses, the "prompt" pulse is 0.1% of the total energy before the fireball becomes opaque (the double-flash) and the following emission when the fireball has cooled enough, for a 1-megaton device this lasts about 10 sec and emits approx. 1/3 of the total energy.
Airblast, as in peak overpressure: P-naught = 3300*(Y/R**3) + 192 * (Y/R**3)**-2
- P-naught is in psi, R is in "kilofeet"
- Again, this will be important for whoever you persuaded to cook your eggs. The peak overpressure at your location will determine whether their arm gets blown off. (Or whether the tank gets pushed away)
So I've calculated so far that by analogy to my stove burner, you could cook a good egg at 4.5 km distance from a 500 kiloton detonation, or 0.9 km from a 20 kiloton Hiroshima-style egg cooker. I need to look at those peak overpressures though, to find out if 1 psi overpressure exceeds the sonic velocity needed to push that tank off of the frypan. :) Franamax (talk) 05:01, 8 August 2009 (UTC)[reply]

At 0.9 km from a 20-kt "egg cooker", the blast wave is strong enough to throw a train off the tracks. 98.234.126.251 (talk) 11:01, 8 August 2009 (UTC)[reply]

Julia Set

Does the Julia Set happen in nature? If so, where? --Reticuli88 (talk) 12:48, 6 August 2009 (UTC)[reply]

In a literal sense, no, because a (typical) Julia set has complexity (lack of smoothness) on every scale, whereas the complexity of natural objects is limited ultimately by the size of atoms and molecules. However, it is possible to use Julia sets and other fractals to model or approximate natural objects - see fractal landscape for an example. Gandalf61 (talk) 13:04, 6 August 2009 (UTC)[reply]

Idealised, perfect, fractals are impossible in the real world - but things like clouds, landscapes, trees, coastlines, (broccoli!)...those are all fractal to a point...they can't be fractal down to the smallest scales of atoms because the laws of nature are very different down at those sizes. Coastlines can't be fractal above the scale of a continent because the earth isn't flat. Clouds can't be fractal on scales larger than a few miles because the weather patterns that form them aren't fractal - and around the size of a single water droplet, the cloud is dominated by the 'new' force of surface tension that's not relevant when it's hundreds of feet across. But within the range of a couple of inches to a mile or so, clouds are pretty decent fractals. You can even calculate things like their "fractal dimension".

So if, for example, you measure the angle between the trunk of a tree and the first two main branches - you'll find that the angles are pretty much the same as between the main branches and the secondary branches, and between the secondary branches and the tertiary branches - and so on down to the smallest twigs. The ratio of the diameter of the trunk and that of the main branches follows a similar rule. One consequence of this is that if you cut a main branch off of a tree and stick it upright into the ground...it looks more or less like a little tree. This 'self-similarity' is the same kind of thing you're seeing in the mathematical fractals like the Mandelbrot and Julia sets - and it's at the heart of fractal mathematics. It's interesting to note that humans find both natural and mathematical fractals beautiful. We've evolved to enjoy things like mountainscapes, trees and clouds - and that's probably at the heart of the reason that we enjoy mathematical fractals. SteveBaker (talk) 13:59, 6 August 2009 (UTC)[reply]

And we enjoy Romanesco broccoli because it's not only delicious and nutritious, but eating it makes you smarter and better at math. That's a scientific fact! ;-) —Scheinwerfermann ^T·_C17:56, 6 August 2009 (UTC)[reply]

^{[citation needed]} Nil Einne (talk) 20:22, 6 August 2009 (UTC)[reply]

Yeah - indeed, I find it very hard to believe that anyone actually eats broccoli without being forced to by their moms. We need references for any claim that someone actually does that! SteveBaker (talk) 00:31, 7 August 2009 (UTC)[reply]

Just steam them a little so then soften up! Or put them in the oven with a little bit of olive oil and something slightly spicy. Mother will be soooo happy with you. --98.217.14.211 (talk) 01:09, 7 August 2009 (UTC) [reply]

^{[citation needed]} SteveBaker (talk) 01:34, 7 August 2009 (UTC)[reply]

I like broccoli. This is unpublished work. --Trovatore (talk) 01:36, 7 August 2009 (UTC)[reply]

So do I. Broccoli is awesome (though easy to ruin if you don't know what you're doing). -- Captain Disdain (talk) 11:54, 7 August 2009 (UTC)[reply]

WP:NOR. SteveBaker (talk) 18:15, 7 August 2009 (UTC)[reply]

It's not my fault you're not bold enough to love broccoli, Steve. -- Captain Disdain (talk) 17:47, 9 August 2009 (UTC)[reply]

Omg! I thought that type of broccoli was fake. It's real!--Reticuli88 (talk) 18:32, 7 August 2009 (UTC)[reply]

Frightening, horrifyingly, gutwrenchingly real. SteveBaker (talk) 22:56, 7 August 2009 (UTC)[reply]

You forgot 'eldritch'. —Tamfang (talk) 03:35, 11 August 2009 (UTC)[reply]

Jubilee Clips?

Where did the name 'Jubilee Clip' come from? Why are they called this?57.86.133.98 (talk) 14:29, 6 August 2009 (UTC)[reply]

Jubilee Clip states that the producing companies were named "Jubilee Components Ltd and Jubilee Clips Ltd", having been invented by Royal Naval Commander, Lumley Robinson, in 1921. 1921 was not a Christian jubilee, nor was in a jubilee of George V of the United Kingdom. So, I am at a loss as to why. SGGH ^ping! 14:49, 6 August 2009 (UTC)[reply]

Perhaps Lumley Robinson celebrated his own golden jubilee in 1921. Cuddlyable3 (talk) 18:24, 6 August 2009 (UTC) (updated)[reply]

Possible - I found that he died in 1939 [2], if he was 50 in 1921, then he would have been 68 in 1939 - probably typical life span for that time.

Robinson was born in 1877. MilborneOne (talk) 22:18, 6 August 2009 (UTC)[reply]

Poisoning the sea

If you took a cupful of something like, say, Botulinum toxin, and poured it into the sea...how big would the effect be? Vimescarrot (talk) 18:27, 6 August 2009 (UTC)[reply]

It would be time related. At the moment of pouring the concentration would be high, but the toxin will start to diffuse away, and some time later, depending on the speed of the wave and tide actions, the concentration will fall below a "no effect level". Ronhjones ^(Talk) 18:56, 6 August 2009 (UTC)[reply]

What kind of radius are we talking before that happens? Vimescarrot (talk) 19:27, 6 August 2009 (UTC)[reply]

Probably not too big - no more than a thousand feet I'd say. A cup is 236 cm³, and that in a cubic kilometer of water is a concentration of 2 × 10^-13 by volume. Botulinum toxin has a toxicity of 1 ng/kg, but only when administered intravenously. Its effect would dissipate quite rapidly. On the other hand, if you spill a tanker full of oil into the sea, the entire area can be quite severely affected, not least because oil floats on the surface. Dcoetzee 19:40, 6 August 2009 (UTC)[reply]

The effects in order of increasing radius (dilution) would be: (i) kills everything, (ii) kills some things and paralyses others, (iii) paralyses some things only, (iv) temporarily hides wrinkles on a mermaid, (v) does nothing and can't be detected, and (vi) can be marketed as a homeopathic remedy for literally anything. Cuddlyable3 (talk) 19:58, 6 August 2009 (UTC)[reply]

One old saw that I always thought was unfortunate was, "The answer to pollution is dilution." Tempshill (talk) 20:05, 6 August 2009 (UTC)[reply]

I've heard that one as "The solution to pollution is dilution." More rhyming makes Mother Nature smile while we poison her. DMacks (talk) 22:20, 6 August 2009 (UTC)[reply]

Diluting a toxin until it has negligible effect has more basis in scientific reason than the concept of "mother nature". This is not to suggest that I enjoy polluting lakes, rivers, or oceans - but a thorough, unbiased analysis is necessary to determine if a particular action is environmentally harmful, or merely "repugnant". Nimur (talk) 23:44, 6 August 2009 (UTC)[reply]

Of note is that we (in which "we" can mean I think any of the five "nuclear weapons states," among others) have, in fact, dumped tons and tons of radioactive wastes into deep sea trenches. The idea here is that the buffer of the ocean would generally keep it in one place and if it got spread out a bit, no harm, no foul (unless you are some sort of deep sea creature who is unfortunate enough to live near it). --98.217.14.211 (talk) 01:05, 7 August 2009 (UTC)[reply]

By "we", do you mean the former Soviet Union? 98.234.126.251 (talk) 02:29, 7 August 2009 (UTC)[reply]

I specified who I meant. The US, the USSR, the UK, France—all of them did ocean dumping of radioactive wastes. Look it up! (I don't know about China, but I would be surprised if they didn't.) --98.217.14.211 (talk) 13:07, 7 August 2009 (UTC)[reply]

I'm absolutely sure the USSR and China did the most radioactive dumping of all (easy enough for them, cause they don't let anyone know what they're doing); France and the US did some "dumping" of radioactives (mainly through atmospheric nuclear weapons testing), but not nearly as much. I don't think the UK did any, though -- why don't you look it up? 98.234.126.251 (talk) 22:25, 7 August 2009 (UTC)[reply]

The Convention for the Protection of the Marine Environment of the North-East Atlantic says the UK has dumped 200kg of plutonium into the Irish Sea, per the Sellafield article. -- Finlay McWalter • Talk 22:29, 7 August 2009 (UTC)[reply]

Only 200 kg? The Soviets used to dump nuclear waste in the Arctic Ocean by the shipload. BTW, the claim in the Sellafield article that you're referring to has a "citation needed" tag. FWiW 98.234.126.251 (talk) 22:41, 7 August 2009 (UTC)[reply]

There is a significant difference between "nuclear waste" - which can be things as (relatively) harmless as the disposable paper coveralls used by workers in nuclear power plants...and "plutonium". The lethality of plutonium is not just due to the radioactivity - but it's also poisonous in a conventional sense...however, a lot depends on how big the chunks were that they dumped - what it was encased in - and (especially) what mix of isotopes it contained. If it were Plutonium 241 - encased in a good amount of lead and concrete and in small chunks, then after 50 years, there would only be 1/16th as much of the stuff as they dumped - most of it having decayed to something less lethal. SteveBaker (talk) 22:55, 7 August 2009 (UTC)[reply]

By "nuclear waste", I specifically meant high-level nuclear waste, like plutonium, fission products, etc. As for what the Soviets used to dump in the ocean: they used to dump many tons of spent nuclear fuel, which is even worse than plutonium. 98.234.126.251 (talk) 23:12, 7 August 2009 (UTC)[reply]

Gravitational waves

If they are able to carry and transmit energy, does that mean that they contribute to the stress energy tensor as Einstein's continuity equation $T^{\mu \nu }{}_{;\nu }=0$ seems to indicate? 70.24.37.78 (talk) 18:46, 6 August 2009 (UTC)[reply]

Yes, I believe so. The stress energy tensor includes everything in the universe, as far as I know. --Tango (talk) 21:07, 6 August 2009 (UTC)[reply]

Yes, absolutelly, and that contribution will also have to be included in the Einstein's field equations which will feed back as a gravity source. That's the main reason why Einstein's field equations are not linear. Dauto (talk) 21:49, 6 August 2009 (UTC)[reply]

Actually, gravitational stress-energy cannot be expressed as a nonzero tensor. It can, however, be expressed as a pseudotensor. See Stress-energy tensor#Gravitational stress-energy and Stress-energy-momentum pseudotensor. Red Act (talk) 01:54, 7 August 2009 (UTC)[reply]

Gravitational waves do not contribute to the stress-energy tensor. In weak-field gravity they clearly carry energy, as seen in PSR B1913+16, so

T^{\mu \nu }{}_{;\nu }=0

doesn't express conservation of energy even though it looks like it ought to. As Red Act said, you can define a "pseudotensor" that includes gravitational energy, but that breaks general covariance. In general spacetimes it's far from clear what energy even is. See this Physics FAQ entry and also §19.4 of MTW ("Mass and Angular Momentum of a Closed Universe"). -- BenRG (talk) 04:47, 7 August 2009 (UTC)[reply]

Jumping Americans cause earthquake in China

Is this possible or just a myth? If most of the population of the USA co-ordinated jumping in the air at the same time, would this cause an earthquake in China? Or vice-versa - jumping Chinese cause an earthquake in the US? 92.26.30.9 (talk) 21:21, 6 August 2009 (UTC)[reply]

Seems to me I read this in one of Cecil Adams books, or antoher like that. the short answer - it is a myth. If 300 million people exert enough force, coming down at the same time, and each is an average of a little over 100 pounds, that's only 30 billion pounds of force, and when cmopared witht he mass of the earth (not sure the exact number, but I'm sure more than what, 30 million tons?) it wouldn't be nearly enough. it's like saying "what's the effect if a few thousand people tried to push a WW II battleship." You also have the problem of the cushioning of the erth's crust, too.Somebody or his brother (talk) 21:57, 6 August 2009 (UTC)[reply]

As an aside, I'm pretty sure it's come up here before that one person can push a battleship (actually a destroyer IIRC, but same sort of thing) when it's in still water enough that it moves. 91.143.188.103 (talk) 17:29, 8 August 2009 (UTC)[reply]

Earth's mass is 5.9742 × 10²⁴ kg, according to Wikipedia. Rmhermen (talk) 22:12, 6 August 2009 (UTC)[reply]

There would also be the counteracting force of the gravitational pull of the 30,000 airborne people just above the surface of the planet Earth. That too should probably be factored into any total calculation. Bus stop (talk) 22:10, 6 August 2009 (UTC)[reply]

Uncle Cecil answered this question, sort of, in 1984, starting off with one of his better intros, "Believe it or not, I'm actually going to answer this ridiculous question." Tempshill (talk) 22:15, 6 August 2009 (UTC)[reply]

It certainly isn't going to throw "earth of its axis". Jumping or not the centre of gravity of "earth+humans" stays the same. 1 human (75kg) jumping to 0.5m high against gravity at 10m/s^2 has an energy of 375J. If 1.000.000.000 people do that its 375GJ or about 80t TNT. The Hiroshima bomb was about 15000t TNT. There was no earthquake in the USA after that. I always thought it is more interesting if by coordinated repeated jumping you could somehow trigger some resonance frequency of the earth/earth's crust. But I would guess the calculation above suggests that the energies involved are far too small - apart from the considerable synchronization problem195.128.250.173 (talk) 22:41, 6 August 2009 (UTC)[reply]

The part of this that most people forget is that the total momentum of the earth/people system never changes as they all jump. So when they push off - they make the earth move a little bit in one direction - but the force of gravity pulling them back down again accelerates the earth towards them. When they land, all of those forces exactly cancel out and the earth is exactly where it was before they jumped. That's not to say that there wouldn't be some vibration and such - but you can't knock the earth out of orbit that way. SteveBaker (talk) 01:33, 7 August 2009 (UTC)[reply]

If they have any effect at all, your jumping Americans are probably more likely to trigger an earthquake along the San Andreas fault (but only because it is going to happen soon anyway). Dbfirs 02:07, 7 August 2009 (UTC)[reply]

Ah. but what if Nikola Tesla had determined the sequencing and used radio to direct the repeated jumping, so as to establish a resonance which focussed the reflected disturbance at a particular spot in China? The amplitude of motion of the earth there should be huge. Edison (talk) 05:24, 7 August 2009 (UTC)[reply]

Tesla was a clever guy - but he was also, in many ways a typical clueless nut-job! This is annoying because many of the full-time nut-jobs out there latch on to every word he said and insist that it must all be true (I'm not putting User:Edison into that category of course!). In truth, maybe 10% of the things Tesla worked on were sheer first-class brilliance - the other 90% were WAY out in la-la-land. His theories on resonance mostly fall into the 90% category. He extrapolated his small scale experiments into gigantic thought-experiments without really thinking about what he was saying - and came up with some ridiculous claims as a result. The Mythbusters attacked some of those in one of their shows - and while the results they had were somewhat surprising, they all fell FAR short of Tesla's claims. Notably, his ideas about resonance...that if a system oscillates - and you add just a tiny bit of energy into each oscillation at just the right moment - then the energy and the size of the oscillation will build up without limit until the object disintigrates. Sure, there are some kinds of resonance that can destroy things in impressive ways...opera singers breaking wine glasses, the spectacular failure of the Tacoma Narrows Bridge (1940), etc. But unaccountably (for such a smart guy) he completely failed to realise that if the amount of mechanical energy the system dissipates due to internal friction, heating, etc during the course of one oscillation exceeds the amount of energy you are adding with each oscillation - then the amount of oscillation won't increase at all. And that's why your idea fails...sure, all of the people jump - and the earth oscillates a little bit. However, the planet is so big that all of the energy from that jumping has been absorbed LONG before it's time to jump again...so no resonance will build up. SteveBaker (talk) 18:01, 7 August 2009 (UTC)[reply]

Indeed I wonder if this idea is based in part on the myth that China is exactly on the oppposite side of the planet of the US (common in cartoons for example). This isn't true, that's the Indian Ocean [3] [4] Nil Einne (talk) 09:25, 7 August 2009 (UTC)[reply]

It's quite impressive how little of the inhabited land surface of the Earth has an antipodal point that's also on land and inhabited.

The closest noticeable land to my antipodes is some little dot in the French Southern and Antarctic Lands. But even that is several hundred miles from the right spot. --Trovatore (talk) 23:09, 7 August 2009 (UTC)[reply]

We actually have an article about this which I partially remembered but didn't find at the time but have now, it's Antipodes. These maps are perhaps better for finding them in general terms and for visualising them File:Antipodes LAEA.png & File:Antipodes rect2160.png. Parts of China do have antipodes but they're in South America of course not north. Nil Einne (talk) 09:51, 8 August 2009 (UTC)[reply]

At the very least, the Americans would all get a little bit of exercise, which is not a bad deal. Googlemeister (talk) 13:30, 7 August 2009 (UTC)[reply]