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I learned in dental school that nothing reduces the internal production of cholesterol -- 1 mg of the 1.7-1.8mg daily cholesterol necessary for an adult human male is produced by the body as a given. DRosenbach(Talk | Contribs)01:51, 31 May 2010 (UTC)[reply]
Exogenous substances can certainly reduce the internal production of cholesterol. See statin. However, the body will compensate for the decreased cholesterol production and shunt the available cholesterol to the most important functions. According to our bile acid article, "the body produces about 800 mg of cholesterol per day and about half of that is used for bile acid synthesis" so you can bet that bile acid production will still get a significant proportion of the available cholesterol. In addition, about 90% of bile acids are recycled. This is the mechanism of the cholesterol-lowering effect of bile sequestrant drugs. According to bile acid, "since bile acids are made from endogenous cholesterol, the enterohepatic circulation of bile acids may be disrupted to lower cholesterol. Bile acid sequestrants bind bile acids in the gut, preventing reabsorption. In so doing, more endogenous cholesterol is shunted into the production of bile acids, thereby lowering cholesterol levels. The sequestered bile acids are then excreted in the feces." This indicates that bile acid production is pretty important and will likely be maintained in the face of reduced cholesterol synthesis. That being said, it would probably be possible in extreme cases to get such dramatic reduction of cholesterol that bile acid production might eventually be impaired. --- Medical geneticist (talk) 03:05, 31 May 2010 (UTC)[reply]
Is there any particular reason that bombarding the slick with napalm, or white phosphorus bombs near the point of orign, in order to burn the oil off before it reaches the coast hasn't been considered? Would this actually work? Or would it make an even bigger mess in the long term? --95.148.107.248 (talk) 01:18, 31 May 2010 (UTC)[reply]
They have been burning surface oil almost since day one [1] but the oil is only concentrated enough to burn in certain places...I believe the reason is that it's being released so far below the ocean surface that it diffuses too much to be effectively burned by the time it reaches the surface. Don't take my word for it though, just speculation. -RunningOnBrains(talk)01:27, 31 May 2010 (UTC)[reply]
Why not just stick something like an industrial-strength version of an umbrella or deflated balloon down the pipe to block it, is what I wonder. 92.28.254.179 (talk) 10:33, 31 May 2010 (UTC)[reply]
That won't work because the pressure in the pipe is so spectacularly high. The amount of friction it would require between your gadget and the walls of that smooth pipe would be spectacular (and oil is a pretty good lubricant!). Remember that the oil reserve is under 18,000 feet of solid rock - all of that rock is pressing down on the oil - so it squirts out with an amazing amount of force. Anything you put down that pipe just gets blown back out again. This is also not a small pipe - it is 21" in diameter. The recent (failed) "Top Kill" method was to pump heavier-than-oil drilling "mud" down the pipe to push down on the oil and reduce the pressure - but even that didn't work. SteveBaker (talk) 15:04, 31 May 2010 (UTC)[reply]
I'd love, actually, to hear from you, Steve, if you'd have any idea what might work. Outlandish as this idea might be. --Ouro (blah blah) 16:03, 31 May 2010 (UTC)[reply]
The only novel thing I've been able to come up with (which is probably completely impractical) would be to take four ships - each with a roll of some kind of fabric thats (let's say) 100' wide by at least 5,000' long and weighted at one end. The four ships would take position in a square formation above the outflow and slowly lower the weighted ends of the fabric into the sea. A heat-sealing device would seal the edges of the fabric together to make a tube some 400' in circumpherence that would then be lowered towards the sea-floor. This would contain the oil within a manageable space that could be pumped out continually until the relief wells finally relieve the pressure in three or four months time. The problem with this idea is that the ocean currents would push very strongly on the tube - so it probably wouldn't work for that reason.
Other oil wells have suffered similar fates in the past - it's instructive to read about the Ixtoc I oil spill - which was in much shallower water - and still took 9 months to finally cap. Due to shallower water and much less oil pressure, that should have been a much simpler task than the Deepwater Horizon spill. I'm not optimistic about the present situation!
The real problem here is that we should never have let such a well be drilled in the first place without adequate means to shut it off and multiple backups for that mechanism. Some problems can't be solved...this may turn out to be one of them. SteveBaker (talk) 13:10, 1 June 2010 (UTC)[reply]
Because the pipe (the riser) is incapable of holding the pressure if the gas were to stop flowing. What can hold that pressure is the BOP, and if I understand correctly what they're trying to do now is lower a second BOP on top of the first (broken one) and close that. TastyCakes (talk) 16:54, 31 May 2010 (UTC)[reply]
That's right, our folks are doing just that, but we're also drilling that relief well in case this doesn't work. (Had to give up the second relief well, something went wrong, I'm not quite sure what.) 67.170.215.166 (talk) 05:36, 1 June 2010 (UTC)[reply]
Probably because the pressure is too high. Consider that it is easy to attach a valve to a fire hydrant with no water in it, and then turn on the water; it would be a lot harder to attach it to the end when it was already gushing out, and this is considerably more difficult than a fire hydrant (both in terms of amount and nature of material going through, and because the thing is on the bottom of the ocean). Anyway, if there was an "easy" answer, don't you think BP would have done it by now? They don't exactly stand to benefit from dragging this out. --Mr.98 (talk) 16:54, 31 May 2010 (UTC)[reply]
Because the pipe would burst. Unless by "pipe" you mean blowout preventer, in which case that is essentially what they're trying to do now. But it's not that simple because huge amounts of fluid are coming out of it at the same time and they don't really know what's wrong with the existing BOP. TastyCakes (talk) 18:07, 31 May 2010 (UTC)[reply]
How much oil is actually down there on the other end of the pipe anyway? If they don't manage to successfully cap the leak, how long would it take before the pressures equalized and the stuff stopped spurting (or all the oil came out and ended up in the sea)? --Kurt Shaped Box (talk) 18:43, 31 May 2010 (UTC)[reply]
BP estimated there were 50 million barrels in the field (it's unclear if they meant recoverable or in place, which would change the number by a factor of about 2), but presumably primary recovery will be only a few percent of that (primary recovery is how long the well produces on its own pressure - when the pressure in the reservoir drops to water pressure at that depth, the well will stop flowing). Hopefully the relief well(s) will fix this before that (very high) limit is approached... As for how long it'll take for the pressure to equalize, its impossible to say because the actual size isn't really known and the rate of the spill isn't really known (somewhere between 10,000 and 100,000 bbl/day from what I've read). But if you could figure that out, you could ball park it by calculating say 4% of the reservoir and then dividing that by the rate per day. Which puts it at between 20 and 200 days by this very rough measure. TastyCakes (talk) 19:12, 31 May 2010 (UTC)[reply]
What about a T shaped pipe - attach one side of the T to the pipe with a valve on the other side fully open. Thus there is no pressure on the pipe. Once the pipe has been attached (jubilee clip ?!) the valve on the side can be closed and the flow diverted through the third pipe forming the T ... 87.102.77.88 (talk) 18:51, 31 May 2010 (UTC)[reply]
Incidentally I came across this yesterday on Nature, "Recently, an explosion on an offshore oil platform in the Gulf of Mexico caused both loss of life and a sizable and ongoing oil spill. We are asking Solvers worldwide to respond quickly with ideas and approaches to react to this very serious environmental threat." [2] As with others, I'm somewhat doubtful that there's anything you can propose that hasn't already been tried and I think the request is more for ways to reduce the negative effects rather then stop the spill but hey if you're sure you have an idea that no one else has thought of no reason you can't submit it Nil Einne (talk) 23:41, 31 May 2010 (UTC)[reply]
I live in an attic with dormers and it gets REAAAAAAAALLLY hot in the summer months -- so hot that the largest AC that fits in my living room window is a 12000BTU and it doesn't do nearly a good job of lowering the temp. I think it's because the dormers are the inside surface of the actual roof and the sun bakes my apartment. What can I do to achieve normal room temp in this case? DRosenbach(Talk | Contribs)01:54, 31 May 2010 (UTC)[reply]
Can you replace the blinds on your dormer windows with blackout ones? If you keep them closed during the day you will achieve a reduction in the temperature you achieve, compared to not using them. I confirmed this for myself during the UK hot summer a few years ago, when our south-facing living room became uninhabitable until I discovered this trick. --TammyMoet (talk) 13:21, 31 May 2010 (UTC) You might also like to see if you can get some of these: http://www.reflectiveblinds.com.au/faq.html --TammyMoet (talk) 13:26, 31 May 2010 (UTC)[reply]
I guess it depends on how drastic you want to get. Better insulation would be a good start - but getting it into the right places is tough. Shutting out light from windows might help some - but I'm thinking that the absorption of heat through the roofing material is the big problem. A lighter colored roof tile would help - but again, it's a pretty drastic solution. You could also cut incidental heat gain by looking for a more efficient refrigerator/freezer and making sure that your hot water tank is well insulated. If the temperature inside the attic is higher than outdoors - even for just a part of the day - then simply opening windows and using a decent fan to recirculate air would help - an experiment with a couple of thermometers over a 24 hour period would be pretty revealing. Make sure the filters on your air conditioner are clean - be sure to change/clean them regularly. 14:31, 31 May 2010 (UTC)
If there is space in the attic between your ceiling and the exterior roof, it can get quite hot under full sun.On a sunny 100 F/38C day, a sealed attic space might reach 130F/54C, causing the top floor living space to keep getting hotter all night even as the exterior temperature drops, as heat conducts through the ceiling, which might be of negligible R value unless insulated. If you are just a temporary renter, there is not much you can do. If your family owns the place, then some improvement is possible. If ventilators are placed in the eaves (screened and louvered metal) then cooler outside air can get into the attic. If exhaust vents are placed in the roof or eaves, then the air can get out. Bats of insulation can be placed between the ceiling joists, taking care not to block the ventilation path. A vapor barrier is suggested between the living space and the insulation, to prevent condensation. A power vent fan can be installed in the roof, which comes on when the attic temperature reaches some hot setpoint. Such a fan may have a high temperature shutoff to avoid providing forced draft if a fire occurs. Note that none of this ventilation refers to exhausting air from the living space. Edison (talk) 14:56, 31 May 2010 (UTC)[reply]
Sounds like good suggestions, though with 38 degree C day unvented ceiling would be 70 degrees C. Also, shading the roof with cheap cotton sheets or junk will work well, though messy. Also, put insulation on the inside and cover with cloth. (Polyester insulation is fine to live with if there are no flames). Also good because you can take it with you when you move. Putting a whirlybird vent on top of the roof to increase the ceiling ventilation wont make much difference inside the house, the ceiling temperature will go down maybe from 70 to 55, the room will still be very hot. Feel the ceiling and walls to see where the hottest parts are and then you will know where to concentrate on. Polypipe Wrangler (talk) 13:23, 2 June 2010 (UTC)[reply]
No, not right. It is possible to define comoving coordinates within special relativity. There is nothing fency about a comoving set of coordinates. it is simply a set of coordinates chosen in such a way that it moves along with the general flow of the matter of the universe called Hubble flow (universal expansion). As I said above, any given galaxy is likely to have some random movement (called peculiar velocity) added to the smooth movement given by the Hubble flow. Dauto (talk) 03:04, 2 June 2010 (UTC)[reply]
I know that the formula for gravitational acceleration is . Is this valid to all distances, or is there somehow a distance at which it's rendered invalid? In other words, do 433 Eros and Rigel have any absurdly-tiny effects on me? I've read Newton's law of universal gravitation, but I can't find anything to answer my question. Nyttend (talk) 02:54, 31 May 2010 (UTC)[reply]
I'm well aware, by the way, that we can't measure the effects of Eros' and Rigel's gravities if they do exist as far away as Earth. Nyttend (talk) 02:55, 31 May 2010 (UTC)[reply]
The answer depends somewhat on whom you are asking :) . Professor Mordehai Milgrom of Weizmann Institute has a theory that says it is not valid at all distances, but the "majority opinion" (if there is such a thing in physics) is that it is valid at all distances. --Dr Dima (talk) 03:06, 31 May 2010 (UTC)[reply]
Yes, as far as we can tell the gravity formula is valid for all distance and indeed rigel has a effect on you. But, as you pointed out, those distant forces are too small to be measured directly and we cannot be sure if the inverse square formula is really valid for all distances. Nevertheless we know that an effect exists (even if the formula chages at some distance) because we know that the sun (and other stars) orbit the center of the milkyway under its gravitational atraction. We also know that gravity acts at distances as far as galaxy superclusters or more, otherwise theose superclusters would not exist. Dauto (talk) 03:33, 31 May 2010 (UTC)[reply]
Hmm, interesting; I had no clue about that. Eros shouldn't take that long; it's going to be less than 0.2 astronomical units away in a couple of years. Nyttend (talk) 13:07, 31 May 2010 (UTC)[reply]
There is a quote that goes something like: "If you had an error in measurement of one atom radius in the position of an object many light years distant, it would make it impossible for you to correctly calculate the position of an atom of gas after some number of collisions." (I tried but I just can't find the original quote.) So yes, gravity from distant objects influences events on earth. (The person who said it actually calculated the various numbers, and I really wish I could find the quote again.) 11:22, 31 May 2010 (UTC)
There is no 'classical' cutoff - but it might be that there is some kind of quantum limit. However, because the force drops off as the square of the distance, the gravitational effects of almost everything out there is less than the gravitational field of (say) your pet cat when it's three feet away. At that point, the influences of all of these small (but nearby) things creates a shifting, totally chaotic mess that completely drowns out the influence of distant planets and stars. SteveBaker (talk) 14:20, 31 May 2010 (UTC)[reply]
According to General Relativity, the formula is never more than an approximation in any case -- GR replaces it with a formula that gives the curvature of space-time as a function of mass. Also, as another editor has pointed out, GR predicts that gravitational effects propagate at the speed of light. This implies that beyond a certain distance, the source of gravitational force will have moved to a different location by the time its effects arrive, so the formula will give the wrong result. Looie496 (talk) 15:42, 31 May 2010 (UTC)[reply]
But, since you're generally observing the source through another mechanism that propagates at the speed of light (i.e. light, or at least electromagnetic radiation), you can take that into account fairly easily. Confusing Manifestation(Say hi!) 23:54, 31 May 2010 (UTC)[reply]
Gravitational waves travel at the speed of light in GR, but there's no such lag in gravitational forces. Usually an object's gravitational field moves along with the object as though it was rigidly attached to it, and the gravitational force points toward the object's current location (with respect to any inertial frame, in the weak-field approximation). The gravitational attraction of the Sun doesn't point in the same direction as the Sun appears in the sky, because the latter is subject to aberration (which you can think of as a time-lag effect) while the former isn't. The difference is between waves and forces, not between electromagnetism and gravitation. If the Sun had a net electric charge then its electric field would point in the same/opposite direction as the gravitational field. -- BenRG (talk) 03:33, 1 June 2010 (UTC)[reply]
That's not correct BenRG, the force also travels at the speed of light. If it didn't you could rapidly move an object and measure the force to transmit information (not to mention energy) faster than light. Note that if the object is moving at a constant velocity, the force does too, such that when the force reaches the second body it is pointing at the current location of the object, and is not delayed. But if you suddenly move the object, the change propagates at the speed of light. Ariel. (talk) 10:38, 1 June 2010 (UTC)[reply]
May be I can help clarify what BenRG is saying. He is talking about a situation in which the source of the gravity field, say the sun, is moving at constant speed. In that situation, if ou look at the source you don't see it where it is now because there is, as you said, a delay caused by the finite speed of light (and gravity waves). But if you measure the gravity force (or the electromagnetic force) you will find out that the force points to where it is now. Now, as you said, information cannot travel faster than the speed of light so if the gravity source were to accelerate and change its path, there would be no way you could tell that had happened (assuming there hasn't been enough time for the information to reach you). In that case instead of the force pointing to where the object actually is, the force actualy points to where the object would have been had it not accelerated. It still doesn't point to where you see it. The article Liénard–Wiechert potential ought to have some information about that phenomenom. Dauto (talk) 03:56, 2 June 2010 (UTC)[reply]
My lab project involves research on the next generation of clean energy. I need to deposit a special organic film on a surface and the reaction should happen in a thin chemical solution.
I noticed that with the same chemical concentration, the reaction rate of a thin solution (like 5mm thick) is much slower, ~25%, than that of the bulk reaction. What may be the reason? How can I increase the reaction rate of a thin solution?
Hi, so you've got a reactive film applied to a surface and immerse the surface and film to a depth of 5mm in your solution? And that when you cover it much deeper in solution, it goes a bit faster? I can't think of any obvious causes for that, but some of the other users of the reference desks have much more chemistry-fu than I have. All I can suggest is that you help the products (assuming they're still in solution) move away from the film surface by circulating the solution with a pump or a stirrer. This should reduce the dependence of the reaction upon diffusion rate. Brammers (talk/c) 11:46, 31 May 2010 (UTC)[reply]
Surfaces of liquids (well water anyway) have more structure than the bulk, and a lower diffusion rate - if your liquid was 1mm thick and aqueous this might explain it.
What do you mean by reaction rate: the overall rate or the instantaneous rate? Did the solution contain an excess of reagent? if not the reduced concentration in the liquid might explain a reduced overall rate.94.72.235.30 (talk) 13:28, 31 May 2010 (UTC)[reply]
Simple possibility - if the reaction is exothermic, the surrounding solution may get heated fast enough to slow things down (order of 1000J must be released for this to occur). But this can honestly happen for a number of different reactions. For example, if large bubbles of gas are being released, maybe you're submerging it in too shallow a bath such that the bubbles are taking up too much space for the reagents to mix. In general, try changing depth and surface area but not volume. SamuelRiv (talk) 06:03, 1 June 2010 (UTC)[reply]
It is not entirely clear what you mean by that. A standing electromagnetic wave (say, in a cavity of a resonator) is trapped in space, but "propagates through time" as you put it. A traveling electromagnetic wave in vacuum, on the other hand, will always travel no matter what frame of reference you are in. In other words, if you try to chase a photon in vacuum, you will never succeed. --Dr Dima (talk) 06:30, 31 May 2010 (UTC)[reply]
I mean to be trapped in its point (say on a surface), not occupying any space as in the cavity where I believe it is still propagating but reflecting back and forth. The reason I ask this is to understand how energy is conserved as its power is varying with time and there are times electric and magnetic fields both tend to zero.--Email4mobile (talk) 07:01, 31 May 2010 (UTC)[reply]
I'm generally thinking about energy regardless if it were in a particle or wave form. I formulated a similar question several months ago (see here) and understood that energy is almost quantized in such case provided that transformation period is above Plank time. Here we are studying a wave and I don't know if energy has to be conserved within space-time or can also be conserved at a particular time (i.e.:Is power conservative?). These are some of the problems I still can't imagine beside relativity (I'll post some question about it later). Grateful for your assistance.--Email4mobile (talk) 08:09, 31 May 2010 (UTC)[reply]
I've often tried to think of a particle as a wave which has somehow lost its ability to propagate but then there are all theses silly subatomic particles that get in the way. 71.100.8.229 (talk) 08:55, 1 June 2010 (UTC)[reply]
There are no times that the E and B fields go to zero. An electromagnetic plane wave in vacuum just moves in some direction at the speed c; it doesn't grow or shrink or otherwise change as it propagates. You will detect changes in the E and B fields as the wave passes you, but that's because you're measuring different parts of it at different times. -- BenRG (talk) 09:50, 31 May 2010 (UTC)[reply]
Not sure if this answers your question, but a photon can not exist if it doesn't move. But, you can store the energy and properties of the photon (in an atom), and while stored it doesn't move. (And photons always travel at exactly the speed of light, in matter the photon is constantly being absorbed and re-emitted, this causes it to appear to slow down.) Ariel. (talk) 10:59, 31 May 2010 (UTC)[reply]
Ariel, in matter a photon does indeed travel slower than the speed of light because of the interactions of the electromagnetic field with the matter. The constantly absorbed and re-emitted picture doesn't really happen. Dauto (talk) 13:12, 31 May 2010 (UTC)[reply]
Email4Mobile, we have indeed means of "trapping" light using Negative index metamaterials, which are like if you took a prism (those triangles that you look through and everything gets rainbow-like) and pretty much turned the rainbows inside-out. This kind of technique is useful for simulating black holes and for computer information stuff (internet, your processor, etc). As I've heard it described, the method is to get your light beam into these "prisms" and then run it around in a circle enough times until it eventually slows down to a very manageable crawl. The photons themselves all move at the speed of light, but they are veering every whichway by the material, so they go in a very twisty path.
Incidentally, you may also be interested - one other thing metamaterials are nice for, aside from trapping light, is manipulating sound in such a way that we can create an actual sound-based black hole, or dumb hole. The article needs some work, but Cosmic Variance has a great summary. SamuelRiv (talk) 06:23, 1 June 2010 (UTC)[reply]
MedScape only checks for drug interactions and not for effect interactions. For instance mydriatic medications do not interact with Simvastatin purse but the effects of Phenylephrine, which dilates pupils, and the effects of Simvastatin which reduces Cholesterol are counter-indicated since Simvastatin indicates high cholesterol and restricted blood vessels and Phenylephrine which will dilate pupils and further restrict blood vessels should not be given for this consequential or secondary interaction. My question is do American and British hospital procedures require computer or manual checking for such drug effect secondary interactions I will call implication conflicts here? 71.100.8.229 (talk) 07:07, 31 May 2010 (UTC)[reply]
It was bedtime when I asked the question. I've corrected it to clarify I am speaking of dilating pupils using medication which has the side effect of restricting blood vessels. 71.100.8.229 (talk) 08:44, 1 June 2010 (UTC)[reply]
Let's say you're zapped millions of years back into the past to a time when the biosphere was much different from the way it is today. The biggest obstacle to survival (right after avoiding being eaten) would be to get something to eat. Obviously, meat would be a safe bet, since you can't really go too wrong with it (except in the rare cases where it's poisonous), but not only is meat difficult to acquire, it's also not something that you could easily live entirely off of.
So what kinds of plants would you be able to eat in prehistoric times? What do we know about the edibility of various extinct plant taxa? Is there a way you can test a plant for edibility?
Obviously, the answer would be different depending on what era of history you were dropped into. So I'll give five options here:
The essential thing the that link forgets is a saucepan (with lid for a smokey fire) so that you can cook vegetables and so on. A spoon would be useful too. 92.15.1.82 (talk) 16:53, 31 May 2010 (UTC)[reply]
Yeah, it's far from certain that you could find anything to eat in those times. Our biology has evolved for the kinds of plants and animals that are around today. We have no clue what missing trace elements there might be - or what poisons might exist in both plants and animals of that time. It's not certain that there would be nothing we could eat - but it's also not certain that there would be anything at all. As I pointed out in that previous discussion (thanks to Nil Einne for digging it out!) you don't have to look far back into the era of the dinosaurs to find that the dominant plant life was ferns. We don't (can't?) eat ferns. There were no grasses - hence no grains - and no flowering plants - hence no fruit. There might have been root plants that we could eat - but it's not like you're going to find recognisable stuff like carrots and potatoes - there would be roots...you'd have to experiment to find which ones were both edible and nutritious - and that's a dangerous thing when anything is potentially poisonous. But this is pure speculation - it's possible that there were things that we could eat and thrive on. But realise that our understanding of the biology of these ancient times derives primarily from fossils. You can't measure the chemical composition of a plant from a rough impression of it in a lump of rock!
Bottom line is: We don't know - but we certainly can't assume that "Obviously, meat would be a safe bet" - who knows what toxins were in their flesh or what crucial trace elements were missing? SteveBaker (talk) 12:46, 31 May 2010 (UTC)[reply]
"...to animals such as mice, rats, horses and cattle when ingested". In humans, the jury is still very much out. The article also says it can cause beriberi if eaten to excess, raw. 86.164.69.239 (talk) 21:29, 31 May 2010 (UTC)[reply]
As it would be unethical to do an experiment where you feed potentially carcinogenic food to people and see if it gives them cancer, then it has not been done. Since it is carcinogenic in four different mammals, its reasonable to generalise (in the abscence of more specific data) that it will be cancinogenic in humans as we are also mammals. 92.24.178.172 (talk) 10:49, 1 June 2010 (UTC)[reply]
I'd be really interested in a modern reliable source saying so, because the article could do with one. Even better if it quantified the risk, so you could compare it to the risk from (for example) eating burnt food. 86.164.69.239 (talk) 14:53, 1 June 2010 (UTC)[reply]
Lately all of my local markets have in fact being trying to sell fiddlehead ferns, which I found quite odd looking. The article picture is an accurate representation of what they look like—a cross between asparagus and dead snakes! :-) --Mr.98 (talk) 15:35, 31 May 2010 (UTC)[reply]
Actually gymnosperms date back even farther, to the Carboniferous period, which is part of the Paleozoic. So it seems likely that there would have been something like pine nuts then, which make excellent food. There have been human groups who used them as their main source of nutrition. Looie496 (talk) 15:35, 31 May 2010 (UTC)[reply]
True, but if humans have only come into their modern form within the last 40,000 years, then there's no telling what minor (or major) changes in composition could have occurred in pine nuts, ferns, and critters in the past 65 million years. Just as a thought though, if the species that diverged before that time are all edible today, wouldn't it be probable that they were edible back then too? Maybe not, as we have evolved with them... Falconusptc15:40, 31 May 2010 (UTC)[reply]
Just going back to the OP for a minute - do we actually know if 'meat' has changed, in terms of what it is, over the last however many million years? AFAIK (according to those survival manuals I read as a kid), nearly all meat from anything that lives on land is edible (sea may be different - there's some strange things down there), when cooked. Is it likely that meat from back when would be so fundamentally different as to be biologically incompatible with our systems? --Kurt Shaped Box (talk) 18:56, 31 May 2010 (UTC)[reply]
The very fact that all meat is edible now tells us that it must have been edible in the past. The only (likely) way that something can be shared between all animals is if it was a property of their common ancestor (who lived about a billion years ago). That means all animals have been edible for the past billion years. --Tango (talk) 20:23, 31 May 2010 (UTC)[reply]
And it really really helps that we're omnivores (they're very uncommon). We can definitely tell if past vertebrates were omnivores by their teeth, and thus we only find a select few of dinosaurs. So count yourself lucky, and remember your place in the food chain! SamuelRiv (talk) 06:30, 1 June 2010 (UTC)[reply]
Lately, I conducted an experiment in my house to verify a different, simple but similar evidence to that of Michelson–Morley experiment. A laser pointer was fixed to a wall, 3 meters high and vertically pointing to the ground such that incident light spotted on a graph paper (1x1 mm grids). Reducing temperature effects to minimum, I have been monitoring the light spot for few months (day and night) from time to time and the maximum offset achieved was less than 1 mm (perhaps due to temperature slight variation and different thermal coefficients). This makes me feel convinced about Michelson–Morley experiment.
If light were assumed moving with respect to ether (or absolute speed in space) it should of course have a maximum offset occurring within 12 hours, ; where x is the maximum shift (expected to occur within only 12 hours), h is the height or vertical distance light has propagated, v is Earth maximum known speed (let's say the same as galaxies'), and c is the light speed in air/vacuum. Given h=3m, v ≥ 220 km/s; thus x should be ≥ 4.4mm which is not verified in my experiment.
Eh? I don't understand how this reproduces Michelson-Morley? You don't describe how you're splitting the optical path...it sounds suspiciously like you're just shining a laser onto the floor and expecting to see the spot move...that's not how Michelson-Morley's experiment works. The changes in the speed of light that you'd see are of the order of a hundredth of a percent. Why would you expect that to produce a movement of the spot on the ground at all? Let alone a measurable one! To do this measurement (as our article points out) you need a beam splitter interferometer. The resulting interference fringes are so small that they can only be seen through a microscope. The error you're seeing is doubtless due to some kind of movement within the apparatus. To do this properly, you need a solid optical bench, a carefully constructed interferometer and a decent light microscope. SteveBaker (talk) 12:30, 31 May 2010 (UTC)[reply]
Sorry to tell it to you, but Steve is correct. In addition to what Steve said, This kind of experiment is so sensitive that the table top is usually placed floating on a tub of Hg to avoid vibrations. Dauto (talk) 13:03, 31 May 2010 (UTC)[reply]
I think he's assuming that as the light travels down toward the floor, the motion of the ether "sweeps" it along, and it moves sideways. Ariel. (talk) 14:36, 31 May 2010 (UTC)[reply]
I'm sorry for not explaining well. I've just uploaded a picture showing the experiment setup. The picture shows a house fixed to the Earth's surface (assume it on the Equator). The house is being observed 4 times a day referred to at states A, B, C, and D respectively. The laser pointer was fixed to the wall using some kind of epoxy and is powered by an external supply and confirmed no vibration all the time I was monitoring. I know I had the problems of beam angle which mad a spot of about 2mm on the ground and the house temperature wasn't that much affecting with time or I should have seen some significant deflect from high to low temperature times.
As you can see the experiment is totally different from Michelson–Morley experiment but the purpose and assumption are typically the same. We are trying to assume ether which will be detected when light passes through it at different angles.
Case A: The light is in the same direction of ether, increasing the relative speed of light but not deflecting from the centre.
Case B: Light is in normal direction to that of ether, so the resultant velocity is a vector of original light speed and relative ether-Earth velocity. This case is observed after about 6 hours of case A shifting incident laser spot from its centre. The expected offset/deflection from the center is .
Case C: After about 12 hours of case A and thus opposing the case by reducing the resultant speed but keeping laser spot at the centre.
Case D: After about 18 hours of case A. This is almost the same in case B but opposite deflection (see the astrisk, * in all cases).The expected offset/defection from the center is .
So the maximum possible offset should be .
As you can see, although it is not possible to predict center and offset from just one case but can be deduced after one complete rotation. I don't know what will go wrong with this experiment. Email4mobile (talk) 14:34, 31 May 2010 (UTC)[reply]
Yes but this is not a problem because we can replace the fixed house with the free rotating house in 3 dimensions (at least I don't have to wait for the 24 hours lol). On the other hand, I think we are also (but not sure) moving with the Milkyway partly in the third dimension but too slow angular velocity. Email4mobile (talk) 15:24, 31 May 2010 (UTC)[reply]
Perhaps you're still not following the explanations by Steve and Dauto. Previous experiments have been conducted so carefully that the margin of error is now smaller than a wavelength of light - that's why they used multipath interferometry. In order to "confirm" that there are no vibrations or other experimental errors, you're going to need to ensure that all of your errors are smaller than the wavelength of light. How are you doing this? For example, do you have confidence that all vibration amplitudes are smaller than a few nanometers? This is unlikely if you have epoxied the laser to the wall. You need a well-balanced optical bench. Before you can even hypothesize about a tiny effect, you need to build confidence that all other experimental errors with greater magnitude than your expected effect have been accounted for. Nimur (talk) 16:16, 31 May 2010 (UTC)[reply]
You're right Nimur. I may not be following because this experiment doesn't involve wavelength interferometry at all. Instead, I've to confirm that all sorts of errors are within 0.5mm at the spot for example; not nanometers. The major problem is the one mentioned by 94.72.235.30m the possibility of ether in the 3rd dimension and this can be done by a free rotating solid tube with the desired length and limited bend/vibrations to 0.0001L (L is the tube length). I might be wrong but feel this way is still easier to perform and if at least were true then it should be another method of verification rather than sticking in one method.Email4mobile (talk) 17:10, 31 May 2010 (UTC)[reply]
Decide what is the smallest deviation of the spot that you could reliably detect. Calculate what speed of ether wind could produce that deviation. Your simple experiment will disprove only that wind speed, which you will find is much faster than could be explained by Earth's movement through ether. Cuddlyable3 (talk) 21:35, 31 May 2010 (UTC)[reply]
I'm afraid that this experiment can never work even in principle because the ether theory doesn't predict the spot movement that you're looking for. Regardless of the motion of the apparatus through the ether, the light that emerges from the laser cavity will remain in line with the cavity. One way to see this is to note that the ether theory is equivalent to special relativity without length contraction or time dilation, and those don't have any effect on your setup. (Length contraction doesn't change the point of intersection of a line through the laser cavity with the floor.) -- BenRG (talk) 21:37, 31 May 2010 (UTC)[reply]
Hi
I want to measure the pressure and velocity of air flowing in a closed duct. What instruments should I use?
If it is a pneumatic conveying lift (the air carries fine solid particles with it), will it be ok to use a manometer and a pitot tube? or will the particles enter the apparatus and spoil the equipment? Are there any other instruments I can use?
Thank you —Preceding unsigned comment added by 122.169.188.160 (talk) 13:56, 31 May 2010 (UTC)[reply]
If you have lots of particles then an Ultrasonic Flowmeter would be perfect. Although I think the pitot tube will be OK (at least at first) because no air flows through it, it just measures pressure difference. But dust may eventually accumulate. If you filtered it, the filter will clog instead, so that doesn't help. You could blow a puff of air through the tube every once in a while to clean it. Ariel. (talk) 14:21, 31 May 2010 (UTC)[reply]
Thanks guys, but I need to measure in a closed duct (I can possibly drill a hole through the duct and insert a tube). In this case I dont think any of the conventional flowmeters as described immediately above will help, right? I dont have a very big budget so the ultrasonic flowmeter etc is out I'm afraid. Any other options? Thanks —Preceding unsigned comment added by 122.169.188.160 (talk) 04:04, 1 June 2010 (UTC)[reply]
as a black hole has greater gravity due to distance between particles decreasing, doesn't it seem common sense to assume the possibility that gravity is generated , like electricity from a generator, when the static fields of electrons & protons cross at the speed of light? if e=mc2 then m=e/c2
f=g(m1m2/d2) thus f=g(<EeEp>/<c4d2>)sine theta? —Preceding unsigned comment added by 98.22.137.1 (talk) 16:52, 31 May 2010 (UTC)[reply]
First you've got to deal with the problems of the statement "a black hole has greater gravity". A black hole doesn't have any more gravity than its constituent matter. Were you to take the sun and compress it into a black hole, the planets would continue to orbit just as they do now (because both can be adequately represented as point masses). There's only a difference if you examine the gravitational field from inside the radius of the initial object -- but that's no great physics revelation. As such, no, I don't think there's much common sense in the rest of your theory. — Lomn18:08, 31 May 2010 (UTC)[reply]
Also, you can't mix and match the formulas like that. E=mc² is a formulation from special relativity that relates to the interchangability of mass and energy (which shows that something with lots of energy will have a higher measured mass, and also states that if you could somehow transform mass into energy how much you'd get, but doesn't say how to do that), whereas F=GmM/r² is from Newtonian physics, and the very fact that this formula breaks down when you look at large scales (and these scales are measurable from Earth, including things like gravitational lensing and time dilation on an orbiting satellite) is why we need general relativity in the first place. Under GR, gravity isn't even a force any more - it's a deformation of space that causes straight lines to become curves. Confusing Manifestation(Say hi!) 23:50, 31 May 2010 (UTC)[reply]
I keep seeing an advert (with Sulu!) for some kind of 4-colour TV, that says it includes yellow as well as red, green and blue, allowing it to give truer colours. Presumably, this works by allowing the red, green and blue to be different, and thus allows you to accurately represent more of the normal human colour space? But wouldn't that require the image to have been filmed using the same system? Or are the colours in some film/TV encoded in some 'absolute' way that the TV then interprets as best it can? I don't quite see how that would work in this case.
No, that's 3 colours plus black - TV's get black by just turning off the pixel (they could have white as a separate colour, I suppose, but there wouldn't be much point - we print lots of black, we don't have lots of white on TVs). --Tango (talk) 17:11, 31 May 2010 (UTC)[reply]
In technical lingo, adding yellow may alter the gamut that the device can display; see that article for an explanation. Not having seen one yet, I'm not sure whether this is hype or something really useful. Looie496 (talk) 17:14, 31 May 2010 (UTC)[reply]
We do, of course, have an article on this technology: Quattron. It doesn't answer your question, though. It's a good question. I can't find an answer to it after a bit of googling - the official press releases don't touch on it, for example. I guess that at the moment they are just converting RGB colour information to RGBY information, which they may be able to do in a clever enough way to get some improvement, but they are probably hoping that the existence of such TVs will encourage the production of 4-colour video and by being first they will control the standard. --Tango (talk) 17:21, 31 May 2010 (UTC)[reply]
Thanks, that's quite helpful (all of it ^). So, the consensus in that last discussion is that it mostly allows a brighter display, with little benefit to the range of colours? Whereas the article suggests the makers claim it works by using something closer to the opponent process, thus supposedly more closely matching how humans see. If you encoded the same colours using red/green, blue/yellow, black/white as the basis instead of RGB, would that actually make a difference to how you saw it? Would that allow brighter displays? 86.164.69.239 (talk) 17:46, 31 May 2010 (UTC)[reply]
A typical RGB gamut Unfortunately, what Steve was saying in that thread is wrong. More phosphors can enlarge the gamut. Take a look at this image. That horseshoe shape is all the colours the human eye can see. The triangle is all the colours that you can get with a particular set of 3 phosphors. As you can see, it only covers a fairly small portion of the total colourspace. If you move the green phosphor down and to the right a bit (which Sharp's TV does) and introduce a new yellow phosphor then you get a quadrilateral that covers more of the colour space. Doing that won't actually improve the yellows much - we can get perfectly good yellows by mixing red and green, as you can see in the image. What's missing in the image that would be added is the blue-greens. That image shows only a very small range of colours around cyan, this 4-colour TV would have much more. The thing that makes this a gimmick isn't that the TV doesn't do what they say it does, it's that the TV signal doesn't contain the necessary information. --Tango (talk) 19:48, 31 May 2010 (UTC)[reply]
Tango: Why don't you actually look at the chromaticity diagram you just attached. Note how close the yellow on the diagram is to the edge of the envelope - even if what you said was true (it isn't) then the effect of extending the gamut in the yellow direction would be negligable. If they were really able to increase the gamut - why wouldn't they have added a cyan pixel? However, those chromaticity diagrams are very dubious ways to represent what's going on. But the killer fact here is that the transmission system in broadcast TV and the storage system in DVD's, etc are RGB based - without an additional 'Y' component created back in the TV/Film studio, this "fourth color" cannot possibly be doing anything meaningful - except (perhaps) to push the overall brightness up. The limitation in the color gamut of TV's is not the display - it's the cameras, mixers, recorders, transmission systems and decoders - and that's not something that's going to change anytime soon. This gizmo is pure advertising hype - and it should be thoroughly exposed as such. SteveBaker (talk) 12:45, 1 June 2010 (UTC)[reply]
In point of fact, MPEG, DV, H.264 and the like use a YCbCr space, and the YCbCr cube is much larger than the RGB cube that it contains (in part because it's oriented differently). So, in principle, you could expand the gamut by simply relaxing the restrictions on what part of the YCbCr space you're allowed to use in encoded video. That's what xvYCC (mentioned below) appears to do. There's no reason, that I can see, to use more than three components to describe an emissive color, even if the display ultimately uses four or more primaries. The perceptual color space is still three dimensional.
After much searching I finally found an image claiming to actually show the Quattron color primaries, here (also found here). I'm not sure I trust it. -- BenRG (talk) 03:13, 2 June 2010 (UTC)[reply]
They key thing isn't the addition of yellow, it's the moving of green. The addition of yellow is just necessary to fill in the gap left after you move green. We're all agreed that at the moment the signals don't contain any more information than can be seen on conventional TVs, but that could easily change in the future. --Tango (talk) 10:58, 2 June 2010 (UTC)[reply]
Ah, so that takes me back to my first thought: thanks. Is there any way they could extrapolate from the information in a normal or HD TV signal, to improve their colours? Or is it a complete bust? It just seems amazing to me that they could really be building an ad campaign around something they can't possibly deliver. 86.164.69.239 (talk) 21:24, 31 May 2010 (UTC)[reply]
I don't know enough about the TV encoding system to be sure. I expect they could do some clever stuff to extrapolate from the information they have, but I don't know how much improvement they would get out of it. --Tango (talk) 22:43, 31 May 2010 (UTC)[reply]
Without getting into whether or not there is actually sufficient information in any current video signal to extract to take advantage of this I would propose an alternative reason: They are doing it because they can:
It's a LCD - thus it's just an extra square in the mask - not actually much new technology
Getting 3 pixels in a square grid is tricky - but 4 is easy
The absolute brightness may be limited compared to a 3 pixel display - but this isn't an issue since LCDs are quite bright enough alreafy thank you
The actually color effect may be slightly 'kodachrome' - it distinguishes the display in the market, may produce extra 'impact' in a shop, and definately won't actually do sales any harm
It gives them a product to display at CES, and makes great advertising copy, and gives hardware sites something to write about.
Thus everyone wins, despite probably no real net improvement.
This article has some information about Quattron (near the bottom). The author is cynical but clearly knows his stuff. He asks the right question ("could it be that existing consumer HDTVs are unable to reproduce the standard sRGB/Rec.709 colour gamut, so Sharp’s fourth primary colour actually has something useful to do?") and answers it with experimental data. The answer is no: an older Sony LCD with three primaries correctly represents all of the colors in Rec.709. So if the Quattron's colors do look more "vivid", and it's not simply an increase of brightness, then it must be displaying them incorrectly. I wouldn't be surprised if they're deliberately oversaturated. That said, there's nothing wrong with the idea of using a larger set of primaries, as 87.* said above. I think the bit about the color opponent process is meaningless hype; I can't see how it could be anything else. -- BenRG (talk) 00:32, 1 June 2010 (UTC)[reply]
In addition to sRGB, HDTVs often support the xvYCC color space, which has a larger gamut. It's possible that the yellow subpixels would help in the xvYCC color space. Of course, xvYCC is not widely used at this time. -- Coneslayer (talk) 12:27, 1 June 2010 (UTC)[reply]
That is interesting. They might be making these TVs in anticipation of xvYCC becoming more widely supported. I expect their 4-colour TVs can display xvYCC colours better than a conventional TV. --Tango (talk) 15:24, 1 June 2010 (UTC)[reply]
I also doubt they cover the whole xvYCC gamut, but if they cover more of it than conventional TVs then they'll be able to display signals that support xvYCC better than conventional TVs (not that there are many such signals, but there may be more in the future). --Tango (talk) 19:20, 1 June 2010 (UTC)[reply]
Thanks for that, guess I was wrong. Was looking myself but got distracted by discussions of displays with xvYCC (and also was searching for xvYCC). I still wonder whether there's any real advantage though. It seems other xvYCC displays exist [4][5][6], if these already support the full gamut (and in theory there's no reason why they can't, xvYCC is still RGB based after all) then we hit the same issue. Nil Einne (talk) 19:41, 1 June 2010 (UTC)[reply]
Generally, "support" for xvYCC just means that the display accepts the xvYCC signal, and does not imply anything about the actual display gamut. My hunch would be that both 3-color and 4-color displays show less than the full xvYCC gamut, but that the 4-color display might show a larger fraction of the gamut. But unless somebody can dig up real measurements, that's just my hunch. (In a couple of months, I plan to build a new computer with a Radeon 5xxx GPU and connect it via HDMI to my Panasonic G10 plasma... the GPU apparently supports xvYCC, so maybe I can see if there's a visible or measurable difference in display gamut on my set.) -- Coneslayer (talk) 23:43, 1 June 2010 (UTC)[reply]
Why the standard value of "g = 9.8 m/s/s or 32 ft/s/s" is subjective and varies if expressed in unit of time other than second (e.g . m/half-sec/half-sec or m/hr/hr or m/min/min)?
Further falling velocity of object is increases at every fraction of second in given spacetime therefore acceleration is produced at every fraction of second rather than per second therefore I don’t understand why it is assumed by the experimenter that falling velocity of an object is constant vehemently/ specifically for one second duration in given spacetime. Is this is fixed by the nature for falling velocity to be increased constantly only after every second? 68.147.38.24 (talk) 17:50, 31 May 2010 (UTC)khattak#1-420[reply]
(edit conflict)Yes g is a constant, no it has nothing inherently to do with the units of meters or seconds. You can use other units (furlongs/fortnight/fortnight, if you prefer), the math will still work out but your velocity calculated at any given time will be in furlongs/fortnight, rather than the more familiar m/s or ft/s. Similarly, you don't have to calculate the velocity change for complete seconds, if you're accelerating at 20 m/s/s for .5 seconds, your final velocity is 10 m/s. TastyCakes (talk) 18:05, 31 May 2010 (UTC)[reply]
(ec) The standard value of g will be slightly different in different units due to rounding (9.8 m is 32.152231 ft, but it is false precision to keep that many decimal places after converting, so we quote it as 32 ft), but that's all. The acceleration is continuous. If an object starts from rest (stationary) and falls then after a tenth of a second, it will be going 0.98 m/s, after half a second, it will be going at 9.8/2 m/s=4.9 m/s, after a second, it will be going 9.8 m/s, and so on. It is quoted in seconds simply because that is our standard unit of time. --Tango (talk) 18:07, 31 May 2010 (UTC)[reply]
Gravitational acceleration is (close to) a constant and is not subjective. Time passes continuously and the OP needs to learn that the normal expression of acceleration g=9.8 m/s/s does not mean anyone thinks nature increases the speed in jumps of +9.8m/s every second. The OP can express the same acceleration as g=588m/min/min g=588m/min/s or g=35280m/min/min which is are both correct. However it is nonsense to attack the common use of the second as time unit as "contentious, not viable etc." as they are doing at Talk:Standard gravity. Cuddlyable3 (talk) 21:07, 31 May 2010 (UTC)[reply]
To answer the original question, scientists don't assume that the speed changes only once per second. If we only want to calculate the speed after one second, we can just multiply 9.8m/s/s * 1 s and get 9.8 m/s, but if we want to calculate the distance that the object travels, we can't just assume that it is still for one second and only starts moving at the end of the second. Instead, we have a more complicated formula (the equations of motion). These can be described as taking infinitely small fractions of a second into account. We can do this because of calculus, which was first invented by Isaac Newton for exactly this reason. 76.67.73.201 (talk) 05:28, 2 June 2010 (UTC)[reply]
Let's also recall that g is NOT constant: its numerical value on Earth varies with altitude, latitude and longitude, as well as with time, in whatever system of units of measurement it is expressed. The Gravitational constant (also known as 'Big G') is a true constant, but the question refers to the 'small g', which includes all applicable gravitational forces, including due to the Sun and the Moon. The field of science that studies these questions is called Gravimetry. The value 9.81 m/s/s is often quoted in schoolbooks and may be sufficient for most everyday applications but would be wholly inadequate in some fields of scientific research and technological development, such as geophysics (e.g., to predict tides) or space exploitation (e.g., to manage artificial satellites). Michel M Verstraete (talk) 21:42, 2 June 2010 (UTC)[reply]
While I love the antics of my beloved Earthicans, there are often jokes that I don't get. In 'Crimes of the Hot', Linda expresses the idea that turtles will be cooled off by windmills. Morbo responds that WINDMILLS DO NOT WORK THAT WAY!!
My question--while windmills clearly are designed for energy and not cooling things off, wouldn't they also do that? Aren't they sorta like over-sized fans? I'm sorry for my lack of scientific knowledge. Thanks gang!209.6.54.248 (talk) 20:05, 31 May 2010 (UTC)[reply]
They're like fans, except backwards. A fan consumes power to produce wind; a windmill consumes wind to produce power. --Trovatore (talk) 20:17, 31 May 2010 (UTC)[reply]
(ec) A windmill is a fan in reverse. A fan turns electricity into wind, and windmill turns wind into electricity (or some other useful form of energy). A windmill powered fan would be like a solar-powered torch - completely useless. The cooling power of the wind would be greater than the fan, due to inefficiencies. --Tango (talk) 20:20, 31 May 2010 (UTC)[reply]
Ya, but to boil it all down, I think the essence of the joke is that Linda was suggesting that the giant turbine could cool the turtles down, and Morbo was angrily pointing out that she has it backwards and that, although they look like big fans, wind mills don't create wind (they absorb it). TastyCakes (talk) 20:53, 31 May 2010 (UTC)[reply]
Part of the joke is also that Morbo, a huge alien who might eat you up, is pedantically irritated by Linda's weak grasp of physics. This mixture of lurking alien menace and pedestrian vulnerability is the running Morbo gag. 213.122.3.204 (talk) 19:29, 1 June 2010 (UTC)[reply]
hey, OP here--thanks for the clarification, but I'm wondering if you stood in front of a windmill, you're saying that it wouldn't be blowing air at you?209.6.54.248 (talk) 00:39, 1 June 2010 (UTC)[reply]
If you're infront of the windmill and the windwill is between you and where the wind is coming from - then you'll feel less wind (the wind gets slowed down, and as a consequence some goes round the windmill).87.102.77.88 (talk) 01:16, 1 June 2010 (UTC)[reply]
Neither fans nor windmills change temperatures (try sticking a thermometer in front of a fan and turn the fan on and off...the temperature reading won't change). The only reason fans make people feel cool is that they move the thin layer of air that's warmed by our body heat and trapped close to our skin by hair and clothing away - letting us feel the temperature that the air really is. When the air temperature is above body temp, fans actually make you feel hotter. Windmills remove energy from the air - thereby slowing it down - so they'll somewhat reduce the cooling feeling of the wind - but they won't change any temperatures (other than by friction and stuff - but that'll be negligable). I suppose that standing behind a windmill when the air temperature is above body heat would make you feel cooler. SteveBaker (talk) 03:06, 1 June 2010 (UTC)[reply]
Well, no, not quite. At 100% humidity, what you say would be true; the only way the wind cools you is convectively, and above body temperature (actually, above skin temperature, I think) it would actually heat you up.
But if you're not on the East Coast, the more important effect is likely to be evaporative cooling, where the wind strips away the humid air near your skin (made humid by your own sweat), and thereby allows more sweat to evaporate, carrying with it its latent heat of vaporization. This can cool you even when the air is significantly hotter than your core temperature. --Trovatore (talk) 05:06, 1 June 2010 (UTC)[reply]
Windchill is just the perceived drop in temperature when a wind is blowing due to the aforementioned removal of the thin warm air pocket above your skin. --Chemicalinterest (talk) 11:07, 1 June 2010 (UTC)[reply]
This is actually a good point, I'm not sure that ta cold-blooded, dry skinned animal would feel much "cooling" effect from a fan, certainly not on its shell. I'm sure that's not the joke intended by Futurama, though. The joke is simply that Windmills slow the wind down, not speed it up, even though they look like giant fans. Irony. APL (talk) 15:36, 1 June 2010 (UTC)[reply]
