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August 13

BaBar Experiment

What is the consensus on the mysterious particle produced by the BaBar experiment a few months ago? Thanks *Max* 00:21, 13 August 2007 (UTC)[reply]

Y(4260) is believed to be a psi particle, possibly due to its unusual behaviour, it is hypothesised that it consists of 4 quarks in a diquark-antidiquark pair. I can't really offer you much more information than can be dug up by googling the right words. ΦΙΛ Κ 19:08, 17 August 2007 (UTC)[reply]

Y(4260) is not a J/psi (it decays to one); nor is it expected to be in the charmonium family, because certain expected decays are not present. Other experiments have obtained similar results. From what I can find on arXiv (arXiv:hep-ex/0702012), theorists are calling it a "hybrid meson with valence partons charm-anticharm-gluon". Needless to say, you have to be an expert in the field (which I am not) to fully appreciate this. - mako 22:49, 17 August 2007 (UTC)[reply]

Ok, well [1][2] led me to believe that it was charmonium or similar. And [3] gives a four quark interpretation of it as well. I don't know much about this, but it seems to be in conflict with what you said. Especially as to my knowledge, partons aren't literally compatible with the current standard model, and are only used as sort of accounting tokens in working out certain reactions.ΦΙΛ Κ 23:35, 17 August 2007 (UTC)[reply]

Since 2005, results from other experiments have come in, together with constraints on what the particle/resonance could be (search "4260" on arXiv for the latest). So opinions have changed. Regarding partons, they show up all over the place in the high energy physics literature. - mako 23:44, 21 August 2007 (UTC)[reply]

Japanese Vehicle Horsepower Restrictions

I know that Japanese auto manufactuere's adhere to a 280bhp horsepower limitation. Does that apply to foreign vehicles as well? Could I import a vehicle into Japan that exceeded that 280 bhp or does it just apply to domestic models?

Hmm, I read somewhere that they only advertised powers up to 280 hp, but that the cars in reality could develop more. I'm not sure, though. —Bromskloss 10:43, 13 August 2007 (UTC)[reply]

Yeah, I've heard the same thing, if I remember correctly I've read it in quite reputable car magazines. They have a gentleman's agreement if you like (or perhaps it's a government sanctioned agreement) that cars can only have a limited maximum horsepower. If the cars are over that horsepower then they simply understate it to the agreed limit, or usually just below the maximum. I believe they stick with these stated figures on exported models in other countries as well, I spose it would look a bit suss if the exported models had more horsepower than the identical domestic product. Kind of contrary to manufacturers in many other countries that tend to overstate their horsepower. It is an interesting question on what they do on cars imported to Japan that are known to have greater horsepower; do they pretend it's less, really limit it, or just not worry? As far as I know they just leave them at whatever they really are, but I could be wrong. --jjron 13:18, 13 August 2007 (UTC)[reply]

On the subject of Japanese manufacturers sticking with stated figures on exported models, the Mitsubishi GTO (sold as the 3000GT and Dodge Stealth in North America) was rated at 280 PS in Japan but 300-320 PS (depending on model year) in North America. I don't think there were hardware differences, though I could be wrong. I think Japanese manufacturers only stuck to the 280 PS "limit" in Japan and rated cars at their true output for export.

Here is a little news blurb from Car and Driver titled Japan Dumps 276-hp Pact]. It appears that, as Jjron says, it's a gentleman's agreement, not a law. I don't know anything about Japanese import laws, so I don't know how practical it would be to import a powerful car, but I don't think there's any law requiring that cars sold in Japan have no more than 280 PS. TomTheHand 14:38, 13 August 2007 (UTC)[reply]

I believe some of them have a restrictor plate installed somewhere in the intake that limits the amount of air and fuel the engine can receive. Once you purchase the car, you can take the restrictor off. I might be crazy, though. --Mdwyer 14:45, 13 August 2007 (UTC)[reply]

I don't have a source, but I'd be surprised if a modern car did something like that. Generally when they tune an engine to different outputs for different applications, it's done by tweaking the engine control software. Friday (talk) 16:53, 13 August 2007 (UTC)[reply]

I realize this is probably a dead topic and nobody will read this, but I found an article on Japanese Wikipedia that appears to discuss the 280 PS "limit". Here's the article: [4] and here's a Google translation: [5]. The translation is iffy, and the article seems to be unsourced, but maybe it's of some use to someone. It seems to imply that the 280 PS limit was actually set by the Ministry of Transport. TomTheHand 04:07, 16 August 2007 (UTC)[reply]

This is Wikipedia. There's always someone reading. A.Z. 04:29, 17 August 2007 (UTC)[reply]

Ohm's law

We know that Ohm's law is valid only when the temperature of the system is constant but we can observe that the temperature of the wire through which the current passes,changes with timei.e heat liberated =i²R.dt where dt is the time interval.How can we say that Ohm's law is valid in our real world conductors.218.248.2.51 08:04, 13 August 2007 (UTC)Ecclesiasticalparanoid[reply]

You are not taking into account that heat will be lost by the wire via conduction, convection and radiation. The conductor will reach as state of equilibrium very quickly. In most situations the conductors do not get sufficiently warm for there resistance to be markedly affected and so Ohm's law is seen to hold. In some situations of course the conductor does get very hot. A filament light bulb comes to mind. In that case Ohm's law certainly does not hold. Theresa Knott | The otter sank 08:27, 13 August 2007 (UTC)[reply]

You're saying Ohm's law doesn't hold for a light bulb filament, even in a steady state? Why not? —Keenan Pepper 09:04, 13 August 2007 (UTC)[reply]

Ohm's law relates resistance, current, and voltage, where voltage and current are variable. A "steady state" would involve all three being constant, so the law becomes meaningless if you confine yourself to looking at steady states. Someguy1221 09:29, 13 August 2007 (UTC)[reply]

I think you got it a bit wrong. Voltage, current and resistance are all variables, but neither have to vary with time. Steady state just means that things don't vary with time. —Bromskloss 10:41, 13 August 2007 (UTC)[reply]

It's not wrong. If you only look at a particular steady state, you can't see the variation, and so Ohm's law remains meaningless. If you're comparing steady states, that's something different. Someguy1221 20:27, 13 August 2007 (UTC)[reply]

Oh whoops, I see I wrote it slightly wrong the first time. Someguy1221 20:31, 13 August 2007 (UTC)[reply]

What Theresa Knott meant (I suppose) was that Ohm's law doesn't hold for the bulb unless you restrict yourself to steady state. (Alternatively, you could introduce a model for the resistance as a function of the current that has so far flown through the filament.) —Bromskloss 10:41, 13 August 2007 (UTC)[reply]

Ohms law doesn't hold through transient conditions. If you make a sudden voltage increase - and there is an inductor in the circuit - then Ohms law won't hold until the circuit reaches steady state. Ohms law does apply to light bulbs - providing the value for 'R' that you plug into the equation is the resistance of the bulb at the actual operating temperature. If you plug in the resistance when the thing was cold, you'll obviously get the wrong answers.

Ohm's Law always holds true: I always equals V / R. What doesn't hold true is any assumption that resistance is a constant.

Atlant 12:15, 13 August 2007 (UTC)[reply]

With real-world electrical wiring, the conductors may be loaded to say 80% of their rated current. Insulated wires will not reach an equilibrium temperature anything like instantaneously. The wire temperature will increase for an extended period, from several minutes to several hours, depending on how well ventilated the conduit, raceway, or cabinet containing them is. The resistance will change over the same period. At any point in time, Ohm's Law will apply, just as it applies when a variable resistor is included in a circuit. In some conductors, resistance increases with temperature, while in others it decreases. Edison 15:03, 13 August 2007 (UTC)[reply]

Ohm's law is an approximation, even in the best of circumstances. Of course, R can change with temperature. But even if I keep the temperature constant, and vary currents and measure voltage very carefully, I will find that the linear approximation is just that; an approximation. It might be a very good one, depending on the material, and the range of voltages and currents considered and so on. Why would anyone expect that this would not be true?--Filll 16:57, 13 August 2007 (UTC)[reply]

No, Ohm's Law is not "an approximation", at least once you're above the current levels where quantum effects (such as shot noise) start to matter. If you've correctly analyzed your circuit (and in doing so, accounted for every source of resistance and every source of EMF {voltage}), Ohm's law will apply essentially exactly.

Atlant 17:02, 13 August 2007 (UTC)[reply]

Ohms Law is a Law. It is always true. R, V and I can be a funtion of frequency, time and temperature but Ohm's law is always correct. Capacitors have a resistance that varies as a function of frequency/time. Same for inductors. The Law is always correct and the other variables will be modified so that it is true. Ohms law is provable from first principles. It is as fundamental and F=ma. --Tbeatty 17:14, 13 August 2007 (UTC)[reply]

Ohms law is true BY DEFINITION - because the definition of an ohm is: the electrical resistance offered by a current-carrying element that produces a voltage drop of one volt when a current of one ampere is flowing through it.. If some strange physical effect is happening at some voltage and current in some weird material or other then it doesn't matter because the definition of an ohm is "that thing that makes Ohms law work" - and therefore the resistance of the material (as measured in ohms) will always be what Ohms law predicts. The mistake that many respondants are making is in assuming that resistance is a fixed property of a material when it's clearly not. Resistance can change in all manner of weird ways as a function of any number of physical parameters - but Ohms law cannot possibly be untrue. SteveBaker 18:55, 13 August 2007 (UTC)[reply]

Hmm, interesting take but I am not sure I fully agree. Here is my understanding:

Ohm's Law states that the relationship between the applied voltage and induced current is linear and calls the proportionality constant (whatever it may be) "resistance". As has been sated above, this constant may depend upon many other factors, including temperature, frequency of applied voltage etc and none of this constitutes any violation of the law.
However the law does break down (or rather is considered inapplicable) for non-ohmic devices, such as a simple P-N diode in which the V-I curve is non-linear. Of course one can technically still keep saying that V=IR with R now also depends upon V, but IMO that semantic defense is not very valuable, because on the same basis I can define a new Abecedare law which says, $V=I^{2}R$ that by definition holds true as long as I allow R to be a function of V.

In summary my main points are, (1) Ohm's law is fine as long as the resistance depends on factors other than the voltage and the current, (2) It is not a universal fundamental law, but a empirically derived observation that AFAIK (but I am not sure about this) can be justified on the basis of Maxwell's equations for conductive materials and some voltage/current regimes. A final note: I know that there are several versions of (so-called) non-linear Ohm's laws and generalized Ohm's law that are used in several disciplines, but in my comment here I am referring solely to the

I\propto V

formulation. Abecedare 20:58, 13 August 2007 (UTC)[reply]

Certainly Ohm's law was formulated for conductors as that is the easiest to measure. However, the proof of Ohm's law is not limited to simple conductors and from that derive the more complex impedances that give rise to the I-V curves of diodes and other such devices. --Tbeatty 22:30, 13 August 2007 (UTC)[reply]

You can always make Ohms appear law true if you postulate varying resistance--SpectrumAnalyser 19:54, 14 August 2007 (UTC)[reply]

But by doing so you completely lose the value of it. Theresa Knott | The otter sank 19:57, 14 August 2007 (UTC)[reply]

seperation of sugar solution from pure water

(reformatted since I think this is supposed to be a new question from an anonymous user SteveBaker 11:57, 13 August 2007 (UTC))[reply]

One way to separate sugar from solution is to boil the solution - the water boils off - the sugar stays behind. You can condense the steam back into water and you'll be left with a bunch of sugar crystals. You need to turn the heat off as soon as all of the water is gone or else you'll burn the sugar. SteveBaker 11:57, 13 August 2007 (UTC)[reply]

Then it becomes the separation of caramel from a sugar and water solution. :P Lanfear's Bane

Yeah, I've done this, you just get a sticky toffee mess, not sugar crystals. You can boil off the water with ionic solutions like salt and get left with the crystals, but not with sugar. Perhaps if you just left the water to evaporate naturally it would be more likely to work, but not sure. --jjron 13:21, 13 August 2007 (UTC)[reply]

You can boil off some of the water and then chill the remaining solution and/or let more evaporate without heating to get rock candy. Just gotta avoid caramelization, which is certainly doable if you don't get it too hot—"boiling" represents an increasingly higher temperature as the water evaporates. DMacks 14:56, 13 August 2007 (UTC)[reply]

I have no idea if this would work, but what if you used a double boiler, such as those fondue pots, with the sugar water in the inner pot? This would probably waste a ton more energy but not burn the sugar, assuming the inner pot will still have evaporation? --Wirbelwindヴィルヴェルヴィント (talk) 17:59, 13 August 2007 (UTC)[reply]

If I understand correctly, heating the sucrose-water will cause some of the bonds to break, converting sucrose into simpler sugars (glucose? fructose?). Eg, you create a simple syrup, which is chemically different than sugar water. So, evaporating off the water isn't not going to make sucrose again. I'm just guessing, here, but I suppose it is possible that you could put the solution in a vacuum, and evaporate off the water without heat. That might work to get sucrose back out of sugar water. --Mdwyer 18:56, 13 August 2007 (UTC)[reply]

According to the caramelization page and the page it cites when talking about the temperature of caramelization of sucrose, sucrose does not appear to decompose appreciably below 160 °C. The double-boiler trick might work (ref says that 100 °C what happens to a sucrose solution is water-evaporation), and of course one can speed evaporation by reducing the pressure at a given temperature. DMacks 22:49, 13 August 2007 (UTC)[reply]

The process of refining sugar from sugarcane involves pulling it out of solution a number of times. As suggested above, the industrial separation process involve placing it in a vacuum, where the solution is boiled at a low temperature and begins to evaporate, forming a syrup. The syrup is then fed into a large, sterile pan and further slowly evaporated in a very controlled way, resulting in crystalisation. The crystals are then spun down and collected. Its true that the exact ratio and types of sugars you put into the solution is lightly to be altered though. Rockpocket 23:01, 13 August 2007 (UTC)[reply]

Gravitation

1.hoe does the force of gravitation between 2 objects change when the distance between them is reduced to half?

I fixed it. See Newton's law of universal gravitation. - Capuchin 13:29, 13 August 2007 (UTC)[reply]

G is inversely proportional to the square of the distance between the masses. Do your own homework? Plasticup ^T/C 14:05, 13 August 2007 (UTC)[reply]

Erm? G is constant? Did you mean gravitational force? :) - Capuchin 14:08, 13 August 2007 (UTC)[reply]

Oh, how embarrassing Plasticup ^T/C 18:31, 13 August 2007 (UTC)[reply]

Plasticup means 'g' - not 'G'. But even that is wrong because 'g' is the accelleration due to earth's gravity - not between two arbitary bodies. The force due to gravity is G x M₁ x M₂ / R² - where G is the universal gravitational constant (not the same thing as 'g'!), M₁ and M₂ are the masses of your two objects and R is the distance between their centers. So the force decreases as the square of the distance. So if you halve the distance, you quadruple the gravitational force. SteveBaker 18:45, 13 August 2007 (UTC)[reply]

No, Plasticup was just wrong. Plasticup hasn't taken a physics class in rather a long time. Plasticup ^T/C 18:49, 13 August 2007 (UTC)[reply]

Infinite light box

Consider a box with all six internal sides being mirrors or another highly reflective material with no imperfections. Suppose you shone a light into this box, then closed it up so that none of the light could escape (of course this can't be done because the speed of light is too fast, but imagine it could be), would the light remain in the closed box forever, reflecting from mirror to mirror, or would the light dissipate over time because it is expending energy by reflecting from place to place? Think outside the box 13:43, 13 August 2007 (UTC)[reply]

A perfect reflection would require no engery loss, so I suppose the light would bounce around forever. Assuming perfect mirrors, perfect construction, etc. Plasticup ^T/C 14:07, 13 August 2007 (UTC)[reply]

That perfect mirror is a bit of a practical problem. And you'd have to choose a wavelength that doesn't interact with whatever fills the empty space inside the box. OTOH, closing the box isn't the problem that Think seems to think. Light is fast but not instantaneous. With a big enough box, slamming it shut while a bright light is shining in would still trap a lot of photons. See also the last time we pondered a question like this. DMacks 14:15, 13 August 2007 (UTC)[reply]

Thanks for the link DMacks, interesting stuff Think outside the box 14:28, 13 August 2007 (UTC)[reply]

To me, this is a sort of paradox. Once the box is shut, how do you know if there is any light in it? You could have a sensor in it, but that would absorb light and mess up the whole thing. The only way to check is to open it and try to grab a few photons as they race out. If you fail to grab any, you would claim that there was no light - which may not be the case. If you do grab some, how do you know they came out of the box and not from the surrounding environment? So, you could just get any old box and put on a shelf and tell people that it holds some photons, but they can't open it to look because they'll let the photons out. -- Kainaw^(what?) 14:45, 13 August 2007 (UTC)[reply]

Schrödinger's light box? Lanfear's Bane

Well, you could theoretically use such a box for long-term storage of light. Strobe lights are too complicated, lets store bursts of light in tiny mirrored boxes. 69.95.50.15 16:04, 13 August 2007 (UTC)[reply]

Assuming we're quite familiar with the box beforehand, we can weigh it to determine if it has an above-thermal amount of photon energy in it, although we won't be able to count the photons unless we also know their spectrum. The interesting bit is that in theory we could do this without any advance knowledge, as I believe the effective inertial and gravitational masses of photons differ (because of their intrinsic momentum; see this relevant question from May). --Tardis 16:23, 13 August 2007 (UTC)[reply]

This opens up a whole new question... The photons would expectedly be bouncing all sides of the box equally. They don't sit on the bottom of the box. So, the collision with the sides of the box should not create extra downward force to measure as an increased weight in the box. The photons that are floating around inside the box aren't propelling themselves against the box or atmosphere inside the box, so they won't increase the weight of the box. There cannot any sort of heat/radiation discharge to measure - or it wouldn't be perfect mirrors. So, there's nothing to measure there. -- Kainaw^(what?) 17:07, 13 August 2007 (UTC)[reply]

The weight comes about because photons fall. They will strike the bottom of the box more directly and the top more obliquely (or even fail to strike it at all if they are travelling very nearly horizontally at the bottom); moreover, they will have more energy when striking the bottom because of gravitational redshift, although I'm not sure that the redshift's effect on the weight isn't a fiction caused by combining Newtonian mechanics for the box with GR for the photons. --Tardis 21:29, 13 August 2007 (UTC)[reply]

This question comes up over and over again here on the science desk (at least once a month!) - some people specify a spherical container - others just want a pair of precisely aligned parallel mirrors - other a cuboid box. But the answer is always the same - no surface is a 100% perfect mirror - they always absorb some small percentage of the light at each reflection. Hence the light bounces around inside the box for a while - gradually being absorbed and turned into heat. For reasonable sizes of box (ie one the order of a few meters), the light will be gone within a few microseconds - if you make a box that's a few light-years across, the light would be around for thousands of years before being absorbed. But with a hypothetical perfect mirrored interior, yes, the light bounces around forever. As the original light 'beam' disperses, gradually the individual photons will all be heading off in utterly random directions - and their wave-like behavior will result in interference and such - but since there is nowhere for the energy to go - the light would still be there centuries later. But there are no perfect mirrors - so it's not a particularly useful concept! SteveBaker 17:24, 13 August 2007 (UTC)[reply]

See Optical cavity. We have some of the most perfect mirrors in the world here at the 40-meter LIGO prototype lab, so that "optical ringdown" measurements become possible. The optical cavity is filled with laser light, and then the laser is turned off. The light immediately begins leaking out of the cavity, but the half-life of a photon in the cavity can be several milliseconds. If you keep making the mirrors better and better you can increase this ringdown time arbitrarily. There's no fundamental limit, just a practical one (no one knows how to make mirrors that good). —Keenan Pepper 19:52, 13 August 2007 (UTC)[reply]

Electrons in String Theory

In string theory, are electrons represented by open strings, or closed strings? MrRedact 15:00, 13 August 2007 (UTC)[reply]

As I recall, all strings of known particles are open. The theoretical graviton is alleged to be a closed string. 151.152.101.44 20:19, 13 August 2007 (UTC)[reply]

That's not correct, some theories posit that (at least) all bosons are closed strings (e.g. photons). Whether or not an electron is necessary open or closed, I don't know. Dragons flight 21:01, 14 August 2007 (UTC)[reply]

A string physicist I know is fond of explaining how he can concoct a theory complicit with all observations of quantum phenomena, but in which particles are actually elephant shaped (one of the reasons he gave up on string theory). Make any claim about the strings, there is probably a theory somewhere to match it. Now if only we could observe these little buggers...Someguy1221 22:12, 14 August 2007 (UTC)[reply]

box fan

I have a symmetrical box fan that takes in air from one side and blows it out the other. Why is it that when I put it in the middle of a room I can feel a strong airflow on the "blow" side of the room but feel no airflow on the "suck" side of the room? Shouldn't they be equal? --⁪froth^t 19:16, 13 August 2007 (UTC)[reply]

I've noticed this too - I believe it's because on the inlet side, the air is being sucked in over a wide angle - but on the output side, it's coming out in a fairly narrow-angled 'cone'. Hence the volume of air going in is the same as going out (well, more or less) - but the speed of the air on the outlet side is higher because that same volume of air is coming out of a smaller area. Using a small length of thread held between your fingertips to visualise the airflow will quickly convince you that this is the case. SteveBaker 19:40, 13 August 2007 (UTC)[reply]

SteveBaker says that the volume is "approximately" equal on intake and outflow. I disagree; the conserved quantity is of course mass flux. Applying the standard equations (PV=nRT) and noting that pressure is inversely proportional to velocity squared... an equal mass flow in and out, but at significantly different velocities, may have a huge difference in volume. I do agree with the conical flow-shaping hypothesis, that seems to explain the perceived difference in flow. Nimur 19:48, 13 August 2007 (UTC)[reply]

Well the change in volume is irrelevant once the fan has been running for a few seconds and the velocity is the same on both sides. I guess the cone theory makes sense though- but what would cause it to do that? The shape of the blades? --⁪froth^t 20:01, 13 August 2007 (UTC)[reply]

I think the basic issue here is that it's possible (and rather easy) to create a directed stream of a fluid, but there's no such thing as a "directed suck".

There's nothing magical about the shape of the fan blades, other than that they create a directed flow. Imagine a tall tank of water with a small hole on the side, at the bottom. Water shoots out of the hole at high velocity in a concentrated stream. But if you were swimming inside the tank, you probably wouldn't be aware of any significant flow towards the hole unless you were right next to it. (At which point you might get sucked against it and trapped and drown, so don't try this at home. :-) ) The reason you're not aware of much flow is that water is migrating towards the hole from all directions, to supply the flow exiting via the hole. And the same holds for the air in the room on the "suck" side of the box fan. --Steve Summit (talk) 23:16, 13 August 2007 (UTC)[reply]

Indeed. If you wanted to create a strong airflow on the intake side, the way to do it would be to force the air into a narrow channel by enclosing the fan in a pipe of the same shape. Then the flow would not arrive from all directions, but only through the pipe -- in fact, you would have a wind tunnel. But the effect outside the pipe would be no different than before. --Anonymous, August 13, 2007, 23:40 (UTC).

Hm? For the analogy to hold, the hole at the bottom would have to be the entire floor of the tank. Like I said it's a symmetrical box fan so the "hole at the bottom" is the same size as the "tank of water". --⁪froth^t 02:02, 14 August 2007 (UTC)[reply]

The analogy's not perfect, no, but I'm not sure what you're talking about with respect to "the entire floor of the tank". For the analogy to hold perfectly, you'd need a wall down the middle of your room, with a box-fan-sized hole in it, with the box fan in the hole.

But ignore the analogy, if you like: the point is that on the "suck" side, air can and does flow from all directions to supply the flow through the fan, while on the "blow" side, it's concentrated into a relatively narrow and well-defined stream. --Steve Summit (talk) 02:09, 14 August 2007 (UTC)[reply]

My gut feeling tells me that the air flow would still be different in the tube. Suppose there were a tube on both sides. Wouldn't the airflow be smoother on the suck side? Alas wind tunnel doesn't seem to cover this. DirkvdM 08:03, 14 August 2007 (UTC)[reply]

You mean a long tube with a fan in the middle? It's possible that passing through the blades of the fan would cause turbulence in the air current in the output end of the tube, but the rate of flow would be identical in both sections of the tube. Maelin (Talk | Contribs) 08:00, 19 August 2007 (UTC)[reply]

Btw, could this be seen as a proof of the irreversibility of time? If you'd turn time around, physics would no longer apply because the behaviour of air before and after the fan is reversed. DirkvdM 08:03, 14 August 2007 (UTC)[reply]

Not really: reversing time once the fan was running smoothly would involve blowing a strong wind at what was previously the output side of the fan and driving the fan with it, creating a turbulent, slow-moving mess on the other side (exactly as before). --Tardis 15:11, 14 August 2007 (UTC)[reply]

Realistic deaths in films?

How realistic are TV/movie death scenes? I know that the closed eyes are pretty implausible, but what else? Like, shouldn't the sphincter relax upon death, and wouldn't that be messy? And what about the speed of it all? People are often seen dying from a single gunshot, say to the abdomen, is this realistic? Wouldn't dying -- especially from a gunshot -- hurt a hell of a lot more than shown on TV? And last, but not least, are there any realistic death scenes at all?

Actually been meaning to ask this, but as a follow up question, what would the rate of blood loss be? It seems a lot of film tone down blood flow, or unrealistically increase ala Kill Bill. Is there a good example of blood loss in media? Zidel333 20:21, 13 August 2007 (UTC)[reply]

Well this resource (http://www.moviedeaths.com/deaths/) and its forums may help. I haven't looked much but I suspect it will go into detail about the variety of ways. Without science/medical reasoning I would say that no deaths in movies are not very real. I heard the sphincter-part was a myth so might be worth checking on snopes (quick search couldn't find anything on it). I think gun wounds vary in movies ranging from the somewhat odd seeming "no affect at all" style 'hero gettng shot' to the 1-shot-kill no matter where in the body 'hero does the shooting' variations. I expect the people that slip away do so in a similar style to movies. Blood loss wise i've no idea - though I understand exit wounds for bullets are much larger than entry wounds and this is rarely the case in movies (just like hiding behind a wood crate is enough to stop a bullet - to that end (http://www.intuitor.com/moviephysics/) is a good site). ny156uk 21:34, 13 August 2007 (UTC)[reply]

Deaths in film are rarely there to depict death. They are there for plot momentum. The hero has to kill a bunch of thugs. The bad guys have to kill a lot of innocent people. The head bad guy has to survive multiple deadly hits and, usually, blow up in the end. If you want to see real death, volunteer at your local emergency room. You'll see very little comparison between real trauma and what is on television or the movies. -- Kainaw^(what?) 22:33, 13 August 2007 (UTC)[reply]

The mechanisms of bullet wounds have been extensively discussed; I bet we've covered it on the RDs once or twice. Here's a quick synopsis by our mentor Cecil Adams.

Blood loss can be pretty alarming. Rob Cockerham investigates this in his characteristic style at "How much is inside?". I once saw the aftermath of some poor guy on a motorcycle who had done battle with an eighteen wheeler and lost, and there was at least that much blood on the pavement. --Steve Summit (talk) 23:09, 13 August 2007 (UTC)[reply]

Some war movies attempt to recreate realism, Saving Private Ryan is one example which has been both praised and criticised for it's often gruesome realistic depiction. Vespine 23:35, 13 August 2007 (UTC)[reply]

If you watch the video where politician Budd Dwyer shoots himself in the mouth, I think you'll find that it's very bloody indeed. --Sean 23:43, 13 August 2007 (UTC)[reply]

Well that answers my question quite well. Thank you ...I guess? 00:45, 14 August 2007 (UTC)

That is both disturbing and fascinating (not to mention tragic). I'm amazed that I had never even heard of Dwyer before now. In addition the the amount of blood, the speed of the whole thing is remarkable - I guess we are so used to slow-mo, and dramatic staggers before falling, after gun-shot wounds to the head in the fictional media. Rockpocket 07:22, 14 August 2007 (UTC)[reply]

No relation... --Mdwyer 14:11, 14 August 2007 (UTC)[reply]

Fire diamond

I had long been under the impression that numbers on fire ~~triangle~~ diamond warning labels only went up to 4, but to my utter shock last week, I saw one with warning numbers as high as 7 and 8 on the side of a large container. I even went and looked through the fire safety manual after seeing this, to find out what those numbers mean, but the manual only listed hazards up to number 4. What the heck was in there? 151.152.101.44 20:38, 13 August 2007 (UTC)[reply]

NFPA 704 is explicit that the numbers go up only to 4. The signs you saw were noncompliant. Perhaps someone was trying to underscore the danger involved? ←Ben^B4 21:52, 13 August 2007 (UTC)[reply]

Maybe the container contains a corpse that is dissolving and is on fire? That way it would exceed the 4s since they are all about potential for harm. DMacks 22:01, 13 August 2007 (UTC)[reply]

Do you remember what it was? Also, is this "Fire triangle" the NFPA 704 one? If not, it could easily be something else which does go that high. Alternately, someone may just be incompetent. 68.39.174.238 23:18, 13 August 2007 (UTC)[reply]

I posted the original question logged out. The fire diamond was in the format of the NFPA 704, aside from having such high numbers. Although maybe it was just to underscore the danger, as Ben suggested, "If you open this container, you will die." I was sort of hoping someone else saw something like this before...Someguy1221 23:32, 13 August 2007 (UTC)[reply]

Oopsie, I meant to say "diamond," and not "triangle." That's how well I can think when I only have 5 hours sleep. Someguy1221 23:35, 13 August 2007 (UTC)[reply]

The only time I've seen this I believe is in that Seinfeld episode The Pothole. The only reason I remember that episode at all is because the barrel of turpentine at the end has a flammability rating of "8", which is impossible. Yeah my brain is full of useless stuff like that. Comes in handy sometimes, like on this desk :) Of course, I have trouble remembering where I parked my car... --Bennybp 00:33, 14 August 2007 (UTC)[reply]

I've seen something like this a couple of times - in both cases with chemicals shipped directly from overseas. I think (this is a reasonable conjecture but YMMV) we eventually determined that the issue was most likely an incorrect merging/grafting of the DOT Classification of Hazardous Materials coding system into the NFPA diamond. For example, one of the chemicals was a batch of granular polymer beads; the white NFPA box had a numeric "9" in it (obviously not NFPA approved). -- MarcoTolo 00:33, 14 August 2007 (UTC)[reply]

You know, on second thought, I think it might have been DOT warnings arranged in a diamond. Thanks! Someguy1221 00:50, 14 August 2007 (UTC)[reply]

Health fairs at libraries

How many libraries have done health fairs in the last two years? May I have a name of a few?

I'm thinking you will need to say where, i.e., your locale; what city you're in, before anyone is likely to be able to help you with this. Having said that, see http://www.google.com/search?&q=library.health.fair ←Ben^B4 22:50, 13 August 2007 (UTC)[reply]

Skin pigmentation in higher latitudes.

I was wondering why such non-Caucasoid peoples that live in relatively high latitudes aren´t light-skinned, like the high Arctic people like the Chukchis, Evenks, and Inuit, and the Aborigines that live in Australia. While I know that the Asians that live in the high arctic are obviously lighter-skinned than the Aborigines, my question was why they don´t have blond hair and blue eyes. While I have read of some high arctic peoples having blue eyes, freckles, and relatively blond hair, those are only rare genetic mutations, or contributions from the Scandinavians who explored the Arctic. I have also read that the reason they aren~t lighter skinned is because the mongoloid race doesn~t have the capacity to change like the Caucasoid race has. Oh, and about the Aborigines, they lived below 40 S Latitude longer than anyone else did, so why aren~t they at least lighter than the rest of the Australian aborigines. —The preceding unsigned comment was added by 200.176.110.253 (talk • contribs) 22:32, 13 August 2007 (UTC)[reply]

There is selection against light skin in the tropics, but there is no selection against dark skin elsewhere, other than self-selection based on tribal instincts, xenophobia, etc. among the light skinned. Dark skinned migrants from the tropics to higher latitudes are likely to stay dark skinned if there are no indigenous light-skinned inhabitants. ←Ben^B4 22:55, 13 August 2007 (UTC)[reply]

We don't actually know for sure, as Ben suggests, that there is "no selection against dark skin elsewhere", although it is true in the skin pigmentation loci studied there was no evidence for positive selective pressure for pale skin alleles. However, there are other genes effecting skin pigmentation so far undiscovered, and until we can study those in the human population, we can't be definitive in that answer. We don't even know why Europeans do have blonde hair and blue eyes, so its beyond our current knowledge to know why other groups at Northern latitudes don't. It may simply come down to genetic drift and sexual selection. Rockpocket 23:16, 13 August 2007 (UTC)[reply]

Dark skin in low UV environments would be selected against because they would develop Vit D3 deficiency, surely? Aaadddaaammm 00:44, 14 August 2007 (UTC)[reply]

That's what I've always heard. I've heard that vitamin D deficiency is actually occasionally a problem for people with dark skin growing up in, for example, Britain. Skittle 00:52, 14 August 2007 (UTC)[reply]

That is "widely assumed". We know that variations at MC1R result in pale skin and we know that pale skin can generate more D3, but there is no evidence to support the fact the MC1R was under such evolutionary pressure. To quote the most extensive study :

There is a popular hypothesis that fair skin in Europeans has been positively selected to increase sensitivity to UV radiation and that, in northern latitudes, this adaptation is needed to increase UV radiation–induced vitamin D synthesis and to prevent rickets... However, we found no statistical evidence that MC1R diversity has been enhanced by selection, either in its apparently high levels or in its haplotype frequency–distribution patterns. [6]

The vitamin D deficiency story is nice, but the facts don't back it up. Rockpocket 01:04, 14 August 2007 (UTC)[reply]

Light skin is an adaptation to allow sufficient vitamin D3 to be produced during periods of lesser amounts of sunlight. However, most of the peoples of the extreme north have diets that are heavy in seal and fish oils. Fish oil is an excellent source of vitamin D. Hence, people who evolved in an environment where they have a fish-heavy diet did not need to evolve light coloured skin. SteveBaker 03:58, 14 August 2007 (UTC)[reply]

Wouldn't fish have been a (the?) major food source for the lightest-skinned people around, the Nordic peoples? (Of course, that doesn't disprove what you say, but it is somewhat salient.) DirkvdM 08:18, 14 August 2007 (UTC)[reply]

People living on the tundra get plenty exposure to UV as the snow refracts it everywhere. That's why you see polar explorers wearing sunglasses to prevent snow blindness. Inuits still have to cover their skin, so its not a great source of vitamin D, which fish oil takes care of. Bendž|Ť 09:53, 14 August 2007 (UTC)[reply]