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February 7

2008 studies show Earth will be swallow up because of the tidal interaction. Without considering loss of sun's mass/gravity to make it expand orbit, how big will sun be at first RGB in 5 billion years? is it roughly size of earth orbit little less than 1 AU or greatly than 1 AU. is it less than 200 solar radius or is is 250 solar radius? Does sun shrink between first and second RGB? My astronomy teacher display sun shrinks after first RGB then expands again at second RGB? How much will sun shrink between first and second RGB? about the size today or less than sun's size today. Becuase second RGB is 1.2 AU, or 260 solar radius. The article didn't mention the first RGB.--69.229.39.25 (talk) 00:37, 7 February 2012 (UTC)[reply]

I thought some scientist thought Europa was once filled with globes of liquid oceans and Titan might been habitatable at one time.[1] I wonder how they do it. Five billion years back in the history was sun bigger and brighter than now or was it smaller and fainter than now (I never paid attention in class, I just daydreamed). Europa and Titan is too small, I am surprised to hear Europa can handle off an atmosphere. How could Mars be blue and wet 4 billion years ago. Was sun bigger back then or was it smaller. Mars had alot of atmosphere back then was sun smaller, or was it bigger. Why did Mars lose its atmosphere?--69.229.39.25 (talk) 01:49, 7 February 2012 (UTC)[reply]

5 billion years ago the Sun did not even exist yet, let alone Jupiter and its moons.

The presence or absence of an atmosphere has little to do with the size or mass of the body. Uranus is 63 times more voluminous than Earth, yet it has almost no atmosphere. Titan is 1/15th the size of Earth, yet its atmosphere is actually denser than Earth's. The critical factor is whether or not there is a magnetosphere to protect the atmosphere from solar wind.

Atmosphere_of_Mars lists a few possible causes for Mar's current thin atmosphere.99.245.35.136 (talk) 02:47, 7 February 2012 (UTC)[reply]

Venus has no magnetosphere, it can still hold atmosphere. I don't know if Titan and Pluto even have magnetosphere. When Pluto is closer to sun it actually have more magnetosphere, that is weird. Yes, Uranus has atmosphere.--69.229.39.25 (talk) 05:10, 7 February 2012 (UTC)[reply]

Uranus has no atmosphere? It's a gas giant. All of it is atmosphere. --140.180.7.220 (talk) 05:16, 7 February 2012 (UTC)[reply]

I don't think magnetosphere may entirely protect the atmosphere. Venus has no magnetic field, its atmosphere is 90xs stronger than Earth. Titan's gravity is like one-seventh that of Earth compare to moon which is one-sixth of earth. I thought the discussion is gravity holds the atmosphere, I forgot we talked about it before. I thought all the sources I found on Europa is reliable, could be the author is just biased about something, sometimes I am just not careful, they may just privode speculations to trick dummies.--69.229.39.25 (talk) 06:49, 7 February 2012 (UTC)[reply]

The early sun was fainter, about 70% of modern. See faint young sun paradox. The presence or absence of an atmosphere is influenced by several factors: the mass of the planet/moon, the distance from the sun (affecting both surface temperature and solar wind), the composition of the atmosphere (heavier gases are more persistent than light gases), and the presence or absence of a magnetosphere. See atmospheric escape. Both Mars and Earth probably had a thicker atmosphere in the early solar system, but the Earth is much better at holding on to it due to its higher mass, so most of the Earth's atmosphere is still here while most of Mars' early atmosphere has been lost. Dragons flight (talk) 08:15, 7 February 2012 (UTC)[reply]

Does anyone know a good easy (but rigorous) introduction to special relativity? Money is tight (talk) 03:26, 7 February 2012 (UTC)[reply]

Tipler's Modern Physics pulls no punches, but only covers elementary relativity. On the other hand, it does so in an application-centric way, rather than a purely theoretical derivation from first principles; so it's immediately useful to the experimentalist or engineer (as much as any relativity knowledge can be useful to the applied sciences). It reads at a level suitable for an advanced freshman or sophomore (university-level) physics student. It also covers many other topics besides relativity. Jackson's Electrodynamics also has a chapter where he derives Lorentz transforms, but it's more suitable for the advanced physics student. Personally, I think the best way to "rigorously" learn special relativity is to "rigorously" learn classical electrodynamics; after this foundation, special relativity is the obvious consequence. Relativistic gravitation, on the other hand, is a whole different subject (...that is, not special relativity); there's not a lot you can do to prepare for it, but it does help to have an advanced background in the analytic mathematical tools of classical physics. Nimur (talk) 05:11, 7 February 2012 (UTC)[reply]

We do have an article Introduction to special relativity here. Graeme Bartlett (talk) 12:10, 7 February 2012 (UTC)[reply]

I learned it when I was in high school from Lillian Lieber's The Einstein Theory of Relativity, which I still think is wonderful if you can find a copy. Looie496 (talk) 18:49, 7 February 2012 (UTC)[reply]

I was pretty impressed by Ray d'Inverno's Introducing Einstein's Relativity, although I didn't go far with it, since I don't know anything about Maxwell's equations. It seemed fairly rigorous, although it started with easier introductions (the initial proof of the Lorentz transformations was somewhat simple, then it covered the orthodox method, afaicr). Also the Schaum's guide, but that one leaves you to do most of the work (proving E=mc^2 is one of the exercises). IBE (talk) 22:05, 7 February 2012 (UTC)[reply]

On a walk tonight, I saw a "star" in the western sky; given that it was brighter than anything in Orion, visible behind me, given that it was visible at all even though it was the Las Vegas sky and slightly overcast in that direction, I'm guessing it was a planet. (I don't think it was Sirus, as it was opposite Orion in the sky.) My question is that I could see a blue light from the top of it. It wavered in and out, disappearing for a few minutes, and was never clearly distinguishable from the star, but it was visible for most of an hour. It and the star were stable in position for that hour, so it wasn't a plane or anything of the like. It was maybe 30 degrees up in the sky, well above the mountains, so it wasn't a land based light. Am I correct in assuming it was some sort of chromatic aberration of the atmosphere or something? That's the conclusion I'm left with. (Yes, I thought UFO; besides all the other problems with that conclusion, it was stable, and you'd think a bright flying object a few miles from Nellis would get intercepted in the time I watched it. Unless the Air Force had someone hover in place for over an hour... but I don't regard that as a serious possibility.)--24.120.231.24 (talk) 03:38, 7 February 2012 (UTC)[reply]

Um, Venus? In tonight's Las Vegas sky, Venus would be to your west, pretty near the horizon, between Pisces and Aquarius. Someguy1221 (talk) 04:33, 7 February 2012 (UTC)[reply]

Venus, Jupiter, and Mars are all visible in the night sky in the Northern Hemisphere right now, and all are as bright or brighter than Sirius (the bright star that Orion's Belt points at, and the brightest star in the sky). Planets are pretty easy to spot if you can find the ecliptic, which is roughly the plane of the solar system, and corresponds to the path the sun seems to follow as it goes through the sky. At night, the moon and planets also lie roughly along the ecliptic, so they're easy to spot, especially if you are spotting one of the three planets brighter than Sirius; Venus is currently quite bright, and obviously the brightest thing after the moon. --Jayron32 05:23, 7 February 2012 (UTC)[reply]

Okay, thanks; but I wasn't really asking about the object itself. It seemed to be white, but the top of it sparkled with a bright blue light. That's why I thought it was an airplane at first glance, because that blue light isn't part of the natural hue of the night sky and it was so close to the other light (presumably Venus). I'm guessing the blue light was about a 2 magnitude by itself.--Prosfilaes (talk) 05:49, 7 February 2012 (UTC)[reply]

Hot air balloon \ or weather balloon? \ Chinese lantern? SkyMachine ⁽⁺⁺⁾ 07:11, 7 February 2012 (UTC)[reply]

No, it was almost certainly Venus. The real question is why the top edge looked blue. Did the bottom edge look red? If so, that's a sign of chromatic aberration of your glasses and/or eyes. --140.180.7.220 (talk) 07:25, 7 February 2012 (UTC)[reply]

I didn't notice a red bottom edge, though I wasn't looking. My glasses are pretty thick high-index plastic. I tried looking at Venus from different angles, and didn't notice the change in effects I usually see in chromatic aberration. It doesn't thrill me as an answer, but it does seem like the best fit.--Prosfilaes (talk) 13:11, 7 February 2012 (UTC)[reply]

You have heard/read of dealer cutoff switches - placed on cars purchased by customers with bad credit so that the dealer (or one in charge of the lien) can disable the car if the customer falls behind on payments (or start making annoying beeping sounds, disable the radio, etc. to entice the delinquent to pay up). Natch, they are triggered remotely, or even on a preset timer.

So, could "warranty cutoff switches" be placed in expensive equipment so that shortly after the warranty runs out, the timer running out causes the "switch" to foul up a critical component that is expensive to replace. (Perhaps the timer would choose a random time to toggle the switch in order to not create too many coincidences, but still take place shortly after the warranty ends.)

Why haven't I heard of such a switch? Wouldn't various manufacturers stand to make a killing on either expensive replacement parts, or just an upgrade to a newer product? I thought planned obsolescence would perhaps involve it, wouldn't you?

So on what products would there be "warranty cutoff switches" and yet, why haven't I heard of them on news and consumer reports? --70.179.174.101 (talk) 07:57, 7 February 2012 (UTC)[reply]

"Why haven't I heard of such a switch"? Two possibilities: (A) they don't exist. Or (B) they don't want you to hear about them. Why do you think that we'd be able to say which alternative was the right one? AndyTheGrump (talk) 08:01, 7 February 2012 (UTC)[reply]

You may be interested in Planned obsolescence 157.193.175.207 (talk) 08:23, 7 February 2012 (UTC)[reply]

If this wasn't disclosed, and you own the item outright, I believe they could be charged with a crime. After all, they are intentionally damaging your property. This would be "willful destruction of property", would it not ? This assumes that you own the item, versus the car which you don't own until you complete payments. And, of course, once it got out that they did such a thing, nobody would ever buy from them again. StuRat (talk) 08:42, 7 February 2012 (UTC)[reply]

It would be almost impossible to keep such a device secret long-term from the flapping lips of disgruntled workers, ex-employees, probing repairmen and techies, and the like. The manufacturer responsible would also gain a reputation for unreliability and poor quality if their "expensive equipment" repeatedly failed just after the warranty expired. Combine that progressive loss of face with their shattered reputation once they were inevitably outed, and why would any manufacturer with anything but the shortest possible timeframe in mind ever consider doing such a thing? As Stu says, no one would ever buy from them again, and their business would quickly fail through market forces, if not through government intervention. --jjron (talk) 10:01, 7 February 2012 (UTC)[reply]

Are you sure you don't own the car until you complete the payments? If you have a loan secured to something, you still own it unless you default on the payments. Are car loans different from normal secured loans? --Tango (talk) 13:20, 7 February 2012 (UTC)[reply]

The Quicken program stops working after a certain number of months, in that it will no longer download any data from banks or credit cards. You could, in principle, manually enter every transaction, but downloading is a major part of the functionality.This brute force way of getting you to buy the newer version is not present on most software, although vendors have historically cut off any tech support of old software, and have not provided free updates so the software works with newer operating systems. If a device has an internal clock, it is possible for it to shut down automatically after some "lifetime." This is common in carbon monoxide detectors,with a 5 year sensor life, as described in the instructions for this model: "Sensor Life Monitor: Internal clock starts once lithium battery is activated. Visual and audible signals notify you when sensor life has expired. " Edison (talk) 16:06, 7 February 2012 (UTC)[reply]

Right, and there they have a justification for the action and notify the consumer, making it all legal. Another example is a toothbrush with bristles designed to dissolve over time, supposedly to let you know when it's germy and needs to be replaced. This is a rather iffy justification, as, of course, any toothbrush can be sterilized by soaking it in bleach or by several other methods. StuRat (talk) 20:30, 7 February 2012 (UTC)[reply]

I don't know where you got that "germy" business from: a toothbrush's germ load will reach a steady-state level a few days after the first time you put it in your mouth. The reason to replace a toothbrush every few months is mechanical wear causing the bristles to lose their scrubbing power. --Carnildo (talk) 02:09, 8 February 2012 (UTC)[reply]

I've seen ads specifically claiming the reason to replace a toothbrush frequently is "germs". The bristles seem capable of lasting for years, provided they aren't designed to wear out quickly. StuRat (talk) 03:09, 8 February 2012 (UTC)[reply]

I'm no lawyer, but it seems to me there'd be some serious liability issues if the device were to be disabled at a particularly bad time, e.g. resulting in injury or death. Clarityfiend (talk) 21:57, 7 February 2012 (UTC)[reply]

Presumably the dealer cut-off switch wouldn't disable the engine, as that would inevitably leave stalled cars on the freeway, creating a dangerous situation and making it difficult to recover the car. They might disable the starter, though, meaning it couldn't be restarted once stopped. This could still kill people, though, like if somebody stops the engine out in the middle of nowhere on a subzero night (say for a "snuggle" with their significant other). It could also leave cars stranded in places like gas stations, with some rather irate business owners suing them for the loss in business. Overall, it's just a terrible idea. StuRat (talk) 07:19, 8 February 2012 (UTC)[reply]

Rather then random speculation which seems to be going on here, a simple search for 'dealer loan cut off switch' finds plenty of refs [2] [3] [4] most of which seem to link back to [5]. It seems clear they are real, but don't cut off the engine while the car is in motion and there is an option to extend the driving period in an emergency (although no description of how this works). Also the consumer signs and ticks a 5 page disclosure form the driver gets in car warnings for several days before the cut off is activated. It seems some people voluntarily sign up for these devices even when they aren't required to get a loan to ensure better loan terms. One of the companies making such devices has a website [6]. Of course stopping the car from working isn't unique to such scenarios, they're sometimes used for anti theft purposes. While immobilisers (which mean the car just never starts) may be more common, I believe there are some systems which allow the car to be driven for a while to be stopped later. We do have articles Kill switch and Anti-theft system which briefly describe these and in particularly mention their usage in Bait cars. Probably they are more common in Anti-hijack systems then anti theft. Nil Einne (talk) 08:33, 8 February 2012 (UTC)[reply]

But the OP's question was never whether or not such devices existed in automobiles with the owner's consent. It's whether or not such devices are intentionally (but unknowningly to the consumer) placed in other expensive equipment in order to result in expensive repairs or replacement shortly after the warranty expires. --jjron (talk) 11:44, 8 February 2012 (UTC)[reply]

I know that hence why I didn't reply until now (I regard it a silly question as with quite a number of questions of the OP). But it doesn't change the fact StuRat was in fact randomly speculating on the dealer cutoff switch in cars when a simple search finds info on such devices. Also my sourced answer does in fact provide a big hint to the OP's question. If these car loan devices require a 5 page disclosure form and have plenty of warnings and other such considerations, this speaks somewhat to the likelihood of someone putting in such a device without telling the consumer where it randomly disables without the consumer knowing why. (I didn't think this needed to be mentioned, since it should be obvious and is something others have already hinted at anyway.) Nil Einne (talk) 20:58, 8 February 2012 (UTC)[reply]

I don't see how anything you found disagrees with anything I've said. I just didn't see the point in listing sources which say that cut-off switches don't stop the engine mid-drive, when common sense tells us as much (and especially since that wasn't even the OP's question). StuRat (talk) 23:14, 8 February 2012 (UTC)[reply]

Philosophically, any car should have a "warranty cutoff switch", in the sense that the manufacturer should end the warranty at about the time when things usually start going wrong, so as to make the warranty sound attractive while not paying too much. But the "cutoff switch" might as well be the things that actually go wrong, of which there is no shortage. I've heard of a timing belt (as opposed to a timing chain) working quite effectively as such a "switch". I suppose a dishonest dealer might merely neglect to mention it requires a change, which indeed is consistent with such a story as I heard it, but can you prove that? And of course, it's technically your responsibility to RTFM, which (if true) is the beauty of the scheme. Wnt (talk) 20:50, 8 February 2012 (UTC)[reply]

There is a related problem that some disreputable dealers will try to "run out the clock" on the warranty by returning cars either completely unfixed (for problems which occur infrequently) or with a temporary fix, knowing that by the time the consumer returns, the warranty will have expired. StuRat (talk) 23:14, 8 February 2012 (UTC)[reply]

does ink and salt dissolve at the same rate and different temperature — Preceding unsigned comment added by 119.235.88.180 (talk) 11:20, 7 February 2012 (UTC)[reply]

no. Ink has much larger molecules and may even be a suspension. Graeme Bartlett (talk) 12:22, 7 February 2012 (UTC)[reply]

And ink molecules may be soluble in liquids in which salt is not soluble. But I think we've done enough of what sounds like your science homework for now. DMacks (talk) 15:25, 7 February 2012 (UTC)[reply]

Did the OP mean to ask the same rate at different temperatures? This seems plausible to me, unless there's some sort of cap or floor on rate of dissolution. 216.197.66.61 (talk) 23:17, 7 February 2012 (UTC)[reply]

I wanted to derive the equation (like, in Cartesian coordinate system) for electric field lines (for dipoles)and it turned out to be huge and not elegant at all. I just wanted to check if there's any online source to check it, or anybody here who knows the answer...-Irrational number (talk) 15:43, 7 February 2012 (UTC)[reply]

...please?--Irrational number (talk) 18:09, 7 February 2012 (UTC)[reply]

Does Dipole#Field from an electric dipole help? I'm pretty sure the field around a dipole is never going to look elegant in any conventional co-ordinate system since it has an annoying asymmetry in it - this is why for some applications it's best to use [dipole co-ordinates]. Smurrayinchester 18:27, 7 February 2012 (UTC)[reply]

My main reason to look for that was to be able to draw the lines of field for a dipole in "graph"(the program I mean), which was not successful... I had tried p orbital cross section before, so I guess I was underestimating the dipole?Is there a way for me to do that?--Irrational number (talk) 03:17, 8 February 2012 (UTC)[reply]

If you're satisfied with generating a vector field, you could just do the vector addition of of fields generated by the positive and negative ends of the dipole. This also assumes you're happy to model the dipole as two point particles some distance apart. But one of the few things I clearly recall from my college electromagnetism course is that there is no elegant formula for for the field about a dipole, although there are elegant approximations to the field once you are significantly further from the dipole than its own length. Someguy1221 (talk) 07:27, 8 February 2012 (UTC)[reply]

Having re-studied your exhaustive articles on "Black hole" and especially the segment on evaporation, cross checking the article on "Hawking radiation", having studied both the DVD on "Hawking's Universe" as well as the corresponding book, especially his "The Universe in a Nutshell", and having studied Kip Thorne's "Black Holes & Time Warps", I have noted a possible discrepancy between the sources and the published articles in Wikipedia: Stephen Hawking himself noted, especially in his book "The Universe in a Nutshell" that the discussed radiation just LOOKS as if the black hole would emit that radiation, but that this is just a picture where the mathematics happen to describe this specific behaviour of virtual particles near an event horizon, seemingly being radiated away by the black hole.

However, in physical reality, the black hole behaves "as usual" (or remaining "hairless"), especially not emitting any kind of radiation (or virtual particle). The consequence would be that a black hole would never evaporate, at least not via this kind of process.

My suggestion would be to expand your publications along those lines, provided you concur with the a.m findings. 87.184.34.215 (talk) 18:00, 7 February 2012 (UTC) edited for correction of main source 80.132.241.226 (talk) 20:14, 7 February 2012 (UTC)[reply]

I am not aware of any currently-available observational techniques that could detect Hawking radiation, so whether any theory predicts its existence is moot: it's below the noise-floor of what we can currently measure using 2012 equipment. On the other hand, we can measure all sorts of other effects of black holes (rather, relativistically supermassive objects): among these, we have pulsar radiation due to accretion; we have several observational signatures associated with active galactic nuclei (usually postulated to be related to relativistic effects of the massive nucleus), and of course, we have gravitational lensing. In fact, the lack of direct observationm of Hawking radiation is still consistent with theoretical predictions: the theory predicts it should be a magnitude much smaller than these other effects. What this means: every time astronomers look at a distant active galactic nucleus or other massive deep-sky object - and don't see Hawking radiation - they are confirming that our theoretical understanding of general relativity is pretty good. Nimur (talk) 20:12, 7 February 2012 (UTC)[reply]

Thanks! As for an observational technique that can detect Hawking radiation, and considering that this radiation could be called a derivative of the Casimir effect, this is well known and established. I do not doubt the existence of virtual particle pairs nor their interaction with a black hole, one of the pair (as a very rare event), not finding a partner to annihilate, falling into the black hole. It is the consequence I'm looking at (please see below, since the second answer deals with this aspect).

It does not really matter if the Hawking radiation comes from "inside" the black hole, or from "just outside". There are two alternative "simplified to my level" explanations for it - the "virtual particle" explanation, and the "uncertain speed of light" explanation. The fist one depends on virtual particle pairs, one escaping, and one dropping into the black hole. It's hard to understand how that would lead to evaporation, but it does. The reason is that the particle that drops into the hole has a negative total energy (i.e. the negative potential energy it has from materializing just inside the Schwarzschild radius is larger than its positive rest mass/energy). So, by falling into the black hole, it decreases the energy (and, amazingly, by just the energy of the escaping particle, so that the conservation of mass/energy is maintained). The second explanation is that particles in the black hole have to follow the uncertainty principle. Since their location is constrained by the Schwarzschild radius, their impulse has to be uncertain. But stuff falling into a black hole will travel (in the limit) at the speed of light. So, by applying the uncertainty principle, some of the particles will have a slightly higher speed, and can escape. Unsimplifying this to Hawing's or Thorne's level requires a good approximation of the TOE ;-) --Stephan Schulz (talk) 20:42, 7 February 2012 (UTC)[reply]

Thanks! However, the concept of negative energy in the realm of quantum electrodynamics is hardly supported, considering eg. the predicted vacuum energy in the order of some 10E113 J/m³. Besides this tremendous energy I have not found any scientific source (or explanation) for a negative energy (which should not be confused with postulates or even predictions). Not to mention classical physics, where either a frequency f or a mass m would need to be negative to maintain the formulae for E=m*c² and E=h*f. But even considering the virtual particle pair to come along with positive and negative energies, uncertainty does not predict more than a 50% probability for one of the particles to be positive or negative. Statistically, the odds should level out with time. So, if any (and looking at the magnitude of the effects), it might be that a black hole remains unaffected by this effect. One option to have a closer look at the process is the LHC. I remember people having been scared that (micro) black holes would be produced in the tests. The formation of such black holes was confirmed by CERN, but nevertheless not presenting any danger to earth. Are informations available on real formation(s) of such black hole(s) and their behaviour? (Still especially looking at some "evaporation".)

The energy of one of the virtual particles is not negative because of "negative mass", but because of its potential energy. The potential energy is always negative if you are inside any gravity well. But at the Schwarzschild radius, its absolute value is the same as (the energy equivalent of) its rest mass. So if a virtual particle pair gets created on the event horizon, but one particle just inside, the other just outside the black hole, the inside particle has negative total energy simply because of its location. --Stephan Schulz (talk) 21:11, 8 February 2012 (UTC)[reply]

Thanks again! However, would this not imply that all matter crossing the event horizon would have negative potential energy, this in turn leading to an even accelerated "evaporation" of a black hole? Without further detailed calculations, and being aware that also the kinetic energy of virtual particles may be negative (as opposed to real matter), I'm quite sure that this problem cannot be solved in quantummechanics.

It may be the other way around, i.e. that real black holes don't exist. Count Iblis (talk) 23:41, 7 February 2012 (UTC)[reply]

Thanks - but I would not bet against them, as Stephen Hawking already HAS lost his bet against Kip Thorne on this issue. 87.184.58.106 (talk) 17:57, 8 February 2012 (UTC)[reply]

You're reminding me of fuzzball (string theory), but I think that's something else again... Wnt (talk) 18:58, 8 February 2012 (UTC)[reply]

Thanks a lot! This reference (and some further information) might very well solve the problem (of mass being "radiated" away from a black hole), especially since supplementary strings apparently always add to the string ball, forming what is called a black hole without referring to some singularity. I recently started a brane description of black holes, so far being unaware that this has already been done. 80.132.232.112 (talk) 14:01, 9 February 2012 (UTC)[reply]

If someone {like Alberto Contador eat food contaminated with Clenbuterol, would this Clenbuterol show up in a blood test? 212.170.181.95 (talk) 20:46, 7 February 2012 (UTC)[reply]

Clenbuterol#Use as performance-enhancing drug mentions people testing positive in blood tests in that way, so I guess the answer is "yes". --Tango (talk) 22:36, 7 February 2012 (UTC)[reply]

Perhaps the OP meant to ask, could such excuses be true? After all, they only allege their food had been poisoned. 216.197.66.61 (talk) 23:16, 7 February 2012 (UTC)[reply]

The report on Contador did say that it could be true in theory, but they discounted this explanation because clenbuterol contaminated meat is rarely found in Europe. Count Iblis (talk) 23:34, 7 February 2012 (UTC)[reply]

What I thought about this case is that Contador indeed used Clenbuterol, however, he tried to cover it/eliminate it with another substance, which didn't work perfectly, therefore, thorough testing could still find some traces of it. 88.8.79.238 (talk) 23:50, 7 February 2012 (UTC)[reply]

Looking around I found a site "Dopeology", which despite the corny name sounds like it makes an effort to get hard data. [7] It says that the typical dose for an athlete taking the drug is 50-100 micrograms/day (somewhat above the standard treatment dose listed in our article). It says the amount allowed to be in meat in the U.S. and Europe is 100 ng/kg (isn't that just a cheery thought?) and in Mexico and other countries it could be more. And the required detection threshold is 2 ng/ml in urine - but the most sensitive equipment can detect 5-10 pg/ml, and there is no lower threshold. Under these circumstances it is plausible that a person can eat a half kg of meat (50 ng), pee a standard liter of urine, and have a detectable 50 pg/ml of the drug in it. Now the slop for this is only a factor of 10, in which all the processes of metabolism, dilution over time, cleaner meat, more fluids consumed, etc. have to work. But certainly I can picture an innocent running afoul of the test. Wnt (talk) 19:10, 8 February 2012 (UTC)[reply]

Is there a way to speed it up? Or does it really have to incubate overnight in the fridge to taste better? 216.197.66.61 (talk) 23:05, 7 February 2012 (UTC)[reply]

Gelatin forms when animal connective tissue (collagen and stuff like that) is heated to near boiling in a water solution. Gelatin takes time to form and disolve into the water, so it definately has to simmer for a while. The gelatin also does have to cool to form proper cross-linkages that give that unctuous mouthfeel associated with gelatin; that's why soups and stews and chilis always taste better after being refrigerated overnight and then reheated; the solution has to cool and rest for a while so the gelatin can "set up". Unfortunately, it's not a step that can be rushed or skipped, if gelatin formation is what you are going for. --Jayron32 06:00, 8 February 2012 (UTC)[reply]

A family member of mine recently underwent a fairly modest surgery. He was asked not to eat for about a day before the surgery and wasn't fed in the hospital for that period. I've never understood why this is. Surely you want people as strong as possible for surgery, not weakened by a short fast? Thanks, 130.88.172.34 (talk) 23:42, 7 February 2012 (UTC)[reply]

Under general anaesthesia a patient can vomit (or just leak) stomach contents, which they may then breathe (ref). -- Finlay McWalterჷTalk 23:47, 7 February 2012 (UTC)[reply]

That clears that up, thanks. 130.88.172.34 (talk) 23:48, 7 February 2012 (UTC)[reply]

Relevant Wikipedia articles: Preoperative fasting, nil per os,pulmonary aspiration, aspiration pneumonia. --Carnildo (talk) 02:15, 8 February 2012 (UTC)[reply]

It is a problem though, especially with diabetics, who really shouldn't fast. StuRat (talk) 02:49, 8 February 2012 (UTC)[reply]

I believe that diabetics have the option of taking glucose tablets to maintain blood sugar while keeping their stomach contents relatively empty. Doctors should know of diabetes and be able to advise on what diabetics can do before going under general anesthesia. --Jayron32 05:58, 8 February 2012 (UTC)[reply]

Yes, and that (along with insulin) should prevent a blood sugar crisis. However, eating nothing but sugar certainly isn't the best thing for a diabetic (or anyone else, for that matter). Diabetics should probably be monitored in a hospital setting during this fasting period, as the chances of a blood sugar crisis are increased by eating glucose tabs alone, since the patient won't be accustomed to the rapid BS spike and collapse they cause. StuRat (talk) 07:04, 8 February 2012 (UTC)[reply]

We seem to be veering in to original research related to medical and surgical procedures. Here is an article by the American Diabetes Association on management of diabetic patients having surgery. It notes measures recommended with the variables of Type 1/Type 2/oral medications/insulin/major surgery/minor surgery. Edison (talk) 21:04, 8 February 2012 (UTC)[reply]

You can get all the nutrients you need via a saline drip. There are people whose bowels have been partially removed who permanently get all their nutrients from a saline drip. Some of them haven't tasted food for decades. Count Iblis (talk) 23:54, 8 February 2012 (UTC)[reply]

That seems quite risky, and they are reluctant to do this if any other option exists. StuRat (talk) 00:31, 9 February 2012 (UTC)[reply]

Surely such a drip would have contain a lot more than just salt, which is what "saline" implies. HiLo48 (talk) 20:39, 10 February 2012 (UTC)[reply]

Not really. Most thing administered by IV are either given along with a saline drip or a dextrose drip, to dilute it. Administering undiluted IVs causes problems, and they can't dilute with water alone, as pure water would flow into cells and pop them. StuRat (talk) 00:06, 11 February 2012 (UTC)[reply]