|WikiProject Physics||(Rated Start-class, Low-importance)|
|WikiProject Temperature extremes||(Rated Start-class, Mid-importance)|
Why does the V overlap the 0 in the formula? Rmhermen 14:25, Aug 14, 2003 (UTC)
- 1 Main Image
- 2 For what range of values is the formula valid?
- 3 Skin, "feel like"
- 4 Frostbite
- 5 Inanimate Objects
- 6 Random Question
- 7 Watts per Square Meter wind chill factor
- 8 Also known as Wind Shield?
- 9 Graduses
- 10 Rewrite
- 11 Wind grill effect
- 12 Changed formula for original model
- 13 Removed NPOV tag
- 14 technical inaccuracy re: fluid mechanics
- 15 partial rewrite of explanation
- 16 Celsius/Fahrenheit
- 17 Subject is properly dressed and dry?
- 18 Windchill for warm temperatures?
- 19 Windchill for wet conditions? (watersports)
- 20 Paragraph on "Extraterrestrial Wind Chill"
- 21 Removing worlwide view tag
- 22 Reality check!
- 23 Contradiction
- 24 additions
I made a new image with celcius and km/h. For the regions of frostbite times I used the old image and wolframalpha. Would appreciate it if someone touched it up and uploaded it. I realize with 1 decimal place it looks odd but I didn't want to remove it since each block is in 2.5 degrees.
Here is the link: http://i.imgur.com/UjrUp5H.png
Many thanks. For those of us living outside of the USA, the use of scientific units is welcome. Mr Fahrenheit is old and stupid and his dumb-ass scale should be confined to the dustbin. It's like having a unit of distance that starts at minus seventeen. "How far is it to town?" "Minus twelve Blerghs." — Preceding unsigned comment added by 126.96.36.199 (talk) 19:14, 2 November 2014 (UTC)
For what range of values is the formula valid?
The wind chill formula seems on the face of it to only be valid for certain range of values. E.g., it seems strange that when V=0 it doesn't reduce to the ambient air temperature.
Also, doesn't wind chill apply in other circumstances such as when riding a motorcycle? For this situation rather higher temperatures and velocities might be interesting. Does this same formula apply. And if the effect is mostly related to evaporation, what effect does clothing have on the perceived temperature? -- Jake 21:45, 9 May 2006 (UTC)
- E.g., if you google ("wind chill" motorcycle) there are other formulas that seem to be used. How do they relate? Jake
Well one thing I am interested in is the wind "chill" at higher temperatures. I ride a motorcycle and as most of the riders out there can tell you over about 90 or so degrees the felt temp increases with speed. Does anyone have a table for that?188.8.131.52 01:33, 12 July 2007 (UTC)
- If you simply put the wind chill equation on the wind chill chart in the article into a spreadsheet, you will find that the equation explains the increase of temperature. When the temperature is over 60 F the wind causes the "real feel" temperature to increase; the increase is nearly imperceptible though, less than five degrees F until the temperature is over 90 degrees F.
- The wind chill formula is not applicable at 60 degrees F, nor at 90 degrees F.
- Note: Windchill Temperature is only defined for temperatures at or below 50 degrees F and wind speeds above 3 mph.
- I entered the new (2001) formula in a spreadsheet. The results near 90 degrees are clearly erroneous.-Ac44ck (talk) 19:20, 22 October 2008 (UTC)
- The wind chill formula is not applicable at 60 degrees F, nor at 90 degrees F.
- As for the invalidity of the equation at V=0, the equation is written assuming that there is a wind, because without a wind, the concept of wind chill is irrelevent. When the wind velocity is zero, several of the assumptions made in deriving the equation may no longer be valid.
- Clothing does make a big difference in wind chill. The type of cloth is important as is the number of layers. The cloth creates a barrier which decreases the wind speed that contacts your skin; the more it can decrease the wind speed the less wind chill you will feel. Yesterday, the actual temperature was -50 degrees F with a wind speed of 25 MPH, creating a "real feel" of -91 degrees F on Alaska's north slope. With my coat open to the wind, the real feel was unbearably cold; however, with the coat closed to the wind, it was bearable. (-91 degrees F and people are talking about global warming? send it my way). —Preceding unsigned comment added by DWL37 (talk • contribs) 19:01, 16 January 2008 (UTC)
Skin, "feel like"
In the first sentence of the article, it is claimed that skin is the only relevant thing that can feel wind chill, but later it is mentioned that it can apply even to inanimate objects, albeit with different formulae. Is the latter correct?
In a similar vein, the warning at the end seems to be sort of POV—the clause "but 32 °C can correctly be said to feel like 32 °C only at 0% humidity" is neither patently obvious nor supported by a citation. Indeed, it seems quite incorrect. Thoughts? /blahedo (t) 00:34, 1 August 2006 (UTC)
- i would disagree that it's not obvious. the idea they're trying to get across is that, if you are at 32 °C with x humidity, then what you are actually feeling is temperature+humidity. to feel only the temp, you would need to remove the other variable. in practice, charts/graphs would perhaps be more effective if they were based on some avg humidity, but that seems to me like it would muddy the concepts. for the same reason, i don't like the benchmark wind-speed idea in the article. if you look at a painting through sunglasses, you don't know what the colors of the painting actually look like, you only know what they look like filtered through the dark lenses. --dan 07:38, 30 November 2006 (UTC)
- Going into way too much mathematical detail, the temperature of your skin is what you feel, and the temperature of your skin is determined by a set of differential equations governing the rate at which your skin gains and loses heat from your blood, from the air, and from evaporating water (specifically: the partial derivative of thermal energy with respect to time is equal to the sum of the system's net generation of heat and the system's net transfer of heat. Expanding out these out in terms of temperatures, masses, models of heat transfer, etc gets messy rather quickly, but the general accounting equation holds with thermal energy). What the humidity does is reduce the rate at which your skin loses heat by reducing the rate at which water evaporates from your skin and/or sweat on top of your skin, causing less evaporating cooling for your skin. Thus, you get a higher steady state skin temperature with the ambient atmospheric temperature, since hotter skin sheds heat easier than cold skin. ThrustVectoring (talk) 06:15, 21 January 2009 (UTC)
Wind chill cannot apply to an inanimate object. There are no exceptions. Humans (and Americans in particular I think) always anthropomorphize. Eric Jensen
- Wind chill can be applied to an inanimate object, as long as the definition of skin is not limited to human skin. Wind causes the moisture on an object to evaporate, and the evaporation process removes heat from the surface on which the moisture was. A piece of fruit has moisture in it that keeps the surface moist as well. A piece of steel typically has no moisture in it as it is all solid; in a wind, the peice of fruit will lose more heat than will the piece of steel; however, the difference between the two will typically be to small to notice. It is interesting to note that the chill of the wind actually comes from evaporation of moisture, not from the wind. The wind is actually adding heat to a body through friction, but the evaporation of moisture removes more than the wind adds, at temperatures less than about 60 degrees F. —Preceding unsigned comment added by DWL37 (talk • contribs) 19:15, 16 January 2008 (UTC)
- Wind chill can apply to inanimate object, but only those that are either sufficiently thermally massive or contain an internal heat source. Wind chill refers to the difference in rate of cooling due to convective heat transfer from atmospheric wind. On the other hand, objects which vary in temperature moreso than humans do and vary in susceptibility to convective heat transfer are going to have wildly varying rates of cooling as compared to the zero-wind case, so it is pretty meaningless to assign a wind-adjusted 'feels like ____' temperature for inanimate objects. ThrustVectoring (talk) 06:15, 21 January 2009 (UTC)
- Also, for the 'wind heats things through friction' is limited by energy considerations (.5 * m * v^2); a gram of air going 25 miles per hour only contains enough kinetic energy to raise the temperature of a gram of water by .015 degrees Celsius (thank you google for calculating units, I google searched '.5 * 25^2 grams * (mph)^2 in calories').
It'd take quite a stiff breeze for air friction to really be significant at all, its pretty much thoroughly dominated by the better thermal coupling between air and skin caused by convective heat transfer. Alternatively, you can use the specs of a wind turbine to find the upper bound of energy content per unit area in wind. According to http://www.ge-energy.com/prod_serv/products/wind_turbines/en/15mw/specs.htm , you can get 1.5 MW out of a swept area of 3904 square meters. Random friction is probably much less efficient at slowing down wind than a turbine, so its a reasonable upper bound on the power is about 400 watts per square meter. Its about the same as a 60 watt lightbulb at a distance of 20 centimeters if you chug through the math. While wind should do the effects you are describing, it really does cause a lot more convective heat transfer as opposed to increased evaporation due to drying out the immediate atmosphere or heat generation through slowing down the wind, especially in sub 60 degree weather. Also, your theory of the relative magnitudes of the several ways of heat transfer/gain/loss due to moving air is relatively straightforward to empirically test. ThrustVectoring (talk) 06:53, 21 January 2009 (UTC)
The tables indicate that the adjusted temperature can be below 0 C even though the actual is over 0 C. Does this mean that an ice cube will not melt in +5 C if there is sufficient wind?
Windchill is a percieved measurement by the body on how cold it is. It is not something that is absolute and can be measured by instruments. Wind Chill is calcualted by pretending humans are a certin shapes etc and seeing how quickly the heat disapates from them in a series of lab experiments, at least from what i can understand.
So you can't use the charts for an Ice Cube as they were designed for humans, and the actual temperature does not change depending on the wind. So i guess no, the ice cube would still melt regardless of the wind. Philbentley 23:30, 15 November 2006 (UTC)
- also, as i understand it, the wind chill makes you feel colder by evaporating water off your skin. it heats up the water on your surface and when that water goes away you feel colder. presumably the wind would have a similar effect on an ice cube - i'm not sure how this would effect melting, per se, but in terms of the cube shrinking, it would probably be faster in higher wind. --dan 07:31, 30 November 2006 (UTC)
Frostbite is dependent on several factors including temperature. If the bollod circulation in an area is high the skin is warmed countering the effect of the cold and reducing the incidence of frostbite. Does the wind chill formula assume an inactive subject?Haans42 15:46, 5 February 2007 (UTC)
It would be useful to mention in the article that wind chill cannot reduce the temperature of an object below the ambient air temperature.(except if there is supercooling due to some evaproative effect) So if the wind chill is -60 and the coolant in your automobile engine coolant/antifreeze is only rated to -40 your engine block will not freeze as the common misconception. Haans42 15:46, 5 February 2007 (UTC) It would appear to me that the ice cube would disappear due to friction from the wind as apposed to melting however the friction could cause the melting? Contuka (talk) 10:08, 4 April 2009 (UTC) Contuka
- That's a clever question; however, as absolute zero is defined as the point where molecular motion ceases (temperature being directly proportional to how much the molecules are moving), there can be no wind-- as wind would require molecular movement, and therefore heat, and therefore a temperature above Zero. See the Absolute Zero article for more info. --BLP 16:16, 4 February 2007 (UTC)
- Even cleverer answer. Thank you for the quick response. 184.108.40.206 06:52, 5 February 2007 (UTC)
how can water freeze above 32degs?
Well, the molecular motion cited above refers to thermal agitation. It is correct that this motion stops at absolute zero, however the collective motion of atoms, e.g. translation of a solid (or a liquid if near zero) would be allowed.
However, wind can be defined as the relative motion of a gas with respect to a solid surface. the problem at absolute zero is that no gas would exist, it would condensate to a liquid. In this sense, the above answer may be correct.
But an important detail is that wind chill is related to human feelings. Since no human can survive at the absolute zero, I think that it is quite correct to say that no wind chill may exist at absolute zero :-) Cloren 20:50, 12 September 2007 (UTC)
Watts per Square Meter wind chill factor
I've added a sentence to the main article mentioning the use of Wind Chill Factor measured in Watts per square Meter that I remember being used in the 1980s. (I lived in Saskatoon Canada, so in January & February we would hear reports on the radio of "the wind chill is 2100 today, so bundle up!" If anyone can expand on it in the article, feel free. / Dzubint 14:06, 5 February 2007 (UTC)
I'm no expert on wind chill, but it seems to me that the article lacks precision. There is absolutely no mention of the fact that wind chill is a perceived temperature change, and not an actual temperature change (as I understand it). The section on 'definitions' does not give any definitions. After reading the article 4 or 5 times, I still don't have a clear picture of what is wind chill... To be fair, there are some very interesting and revealing pieces of information. Maybe lacks an 'encyclopedic' touch. —Preceding unsigned comment added by 220.127.116.11 (talk) 06:28, 5 December 2007 (UTC)
Also known as Wind Shield?
I've heard several people refer to wind chill as wind shield. Is this just a case of misheard words or is wind shield a proper term? —The preceding unsigned comment was added by 18.104.22.168 (talk) 04:48, 2 April 2007 (UTC).
- Misheard words. A windshield (follow the link) is something else. —Lowellian (reply) 01:36, 28 June 2007 (UTC)
ITS A CONSPIRACY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Why big chrart at the top in °F not in world standart °C? 22.214.171.124 18:23, 17 August 2007 (UTC)
- Agreed. Much of this article uses a very unusual writing style for an encyclopedia and reads more like a position paper or essay. Opinions about how things should be are sprinkled in with factual writing. The article also seems to have a self-referencing tone, as though it's patting itself on the back for how well it's written or how smart its explanations are. Very odd. As soon as I get some time during the hectic holidays, I'll start rewriting this myself, but if anyone gets time before I do, please feel free. --DavidGC (talk) 14:59, 23 December 2007 (UTC)
Wind grill effect
No mentioning of the wind grill effect? This isn't to be confused with the heat index related to humidity, but to the fact that the air feels hotter when there's wind if the air temperature is above body temperature (37°C). // Jens Persson (126.96.36.199 (talk) 09:06, 25 June 2008 (UTC))
- its the same effect, moving air has a better thermal coupling than still air, so the rate at which something exchanges heat with the atmosphere is the same with a lower temperature difference and still air as higher temperature difference with moving air. Its the same reason that forced-convection ovens cook at lower air temperatures. ThrustVectoring (talk) 05:18, 21 January 2009 (UTC)
Changed formula for original model
Page 113 of this document
suggests that the formula should be a product, not a fraction; that is:
- WCI = Wind chill index, kcal/m^2/hr
- V = Wind velocity, m/sec
- T = Air temperature, °C
The text talks about the units being revised to Watts per sq. metre. That is a rate of heat transfer, an increase in which would be perceived as colder. A colder (lower) air temperature would increase the rate of heat transfer — and the original WCI.
The quantity (33 - T) increases as air temperature decreases. The product in the formula above increases as air temperature decreases. That is consistent with what is suggested by the text.
The formula which was in the text was a fraction with (33 - T) in the denominator. The quotient in that formula decreases as air temperature decreases. That is not consistent with what is suggested by the text. -Ac44ck (talk) 16:30, 23 October 2008 (UTC)
Removed NPOV tag
I have several issues with the addition of this tag:
- The tagger created no section on this talk page to:
- Clarify the tagger's objections, or
- Provide a framework for addressing those objections.
- The edit summary for adding the tag mentions that the section is "biased toward USA". The section mentions North America and Canada; it says nothing about the USA.
- The edit summary makes an appeal for using metric units. Only metric units appear in the section. While kcal/m^2/hr may not be SI, it is metric; and the original formula was what it was — it need not be converted to the latest version of metric when:
- It isn't intended for current use anyway, and
- Bringing attention to "the strange sounding units" seems to be one of the reasons for including the formula.
- The edit was marked as "minor". Help:Minor edit says, "A minor edit is a version that the editor believes requires no review and could never be the subject of a dispute." It seems that the reasons for dropping the tag were to:
- Call for a review, and
- Raise disputes over issues that didn't seem to be of concern to anyone else.
technical inaccuracy re: fluid mechanics
The boundary layer of a fluid is ALWAYS infinitely thick, its merely a question of how far out you have to go in order to be close enough to the ambient condition to approximate it as equal. Equivalently, the layer of fluid that is still next to a solid object in a moving flow is basically infinitely thin, afterwards you have a continuum of velocities. So when the article talks about wind 'thinning' the boundary layer of air around the skin, the statement has zero meaning. Its true that the wind reduces the insulation of the air (equivalently, increases the thermal coupling between the air and the skin), its just that the explanation for it is talking about doing things to pseudosystematic terms which are undefinable. IMHO wind chill is much better explained by talking about convective heat transfer. Moving air = faster convective heat transfer out of your body = skin gets colder. ThrustVectoring (talk) 05:54, 21 January 2009 (UTC)
- This article shouldn't even been written in terms of "boundary layers". Everyone from Straight Dope to howstuffworks.com talks in terms of evaporative cooling, which is the more fundamental reason (i.e. wind chill would exist even if there was no "boundary layer"). If someone really wants to go this route, you should at least cite SOMETHING, SOMEWHERE.--188.8.131.52 (talk) 15:18, 7 March 2009 (UTC)
partial rewrite of explanation
I tried to rewrite the explanation paragraph. There may be something wrong, or it can be said better. But I think two things are clear: 1) It is more encyclopedic. How much more, I'm not sure. 2) And explanation of a cause should be the fundamental reason (a table isn't MERELY hard because wood is hard, there's a more fundamental reason). Whether or not this "insulating shell" exists is irrelevant to the fundamental explanation of wind chill. If you completely remove any insulation (like clothes) the phenomenon of wind chill still exists. You still get greater cooling, you still get maintenance of surface temperature, you still get more heat loss due to this maintenance, and you still get a greater risk to frostbite. Let's put it another way. Say the wind speed is so amazingly high in that formula that you have no "insulating shell" whatsoever. Do you still have wind chill? Of course. In fact you have so much that you're going to die. It's not the lack of an insulating shell that causes wind chill, but the extra loss of heat, and that's what we should concentrate on.--184.108.40.206 (talk) 16:32, 7 March 2009 (UTC)
This article refers only to fahrenheit, shouldn't it be mainly based on Celsius, seeing as that is the worldwide scientific temperature scale, and the majority of the world only uses Celsius —Preceding unsigned comment added by 220.127.116.11 (talk) 16:23, 14 December 2009 (UTC)
I would agree with this, and the main table should have units of Celcius and a fahrenheit table below the article. —Preceding unsigned comment added by 18.104.22.168 (talk) 23:23, 25 December 2009 (UTC)
Not a problem. Wind chill calculation has far more troubling issues than that. Where I live I use °C and MPH and always have to convert one or the other to fit. No big deal given that I have to write and check an entire formula too. In a small program, we only have to do this once. 22.214.171.124 (talk) 13:36, 7 May 2011 (UTC)
Subject is properly dressed and dry?
"The typical assumption is that the subject is properly dressed and dry." This statement is unsupported. Wind chill is typically described as the apparent temperature on exposed skin. So it would seem that it's used under the assumption that the subject is naked. If the air is very cold, and the subject is properly dressed and dry, then only a small area of skin is exposed, and the relevence of "wind chill" is blown all out of proportion, especially by the TV entertainers claiming to be meteorologists.
- There are huge differences (many degrees !) between skin types, and if skin is completely dry or not. So this thema of Wind chill is completely fake and formula is not objective, even if exposed by "scientists and experts". A temperature in degrees is a physical quantity, measured with an instrument, final point. One just has to distinguish between "wet temperature" and "dry temperature": wet temperature is obtained by physically measuring a wet object temperature under strong wind, and is often 5 to 10° lower than dry temperature (depending also of moisture, too). This, is scientific (but less fashion phenomena nor mass medias than "wind chill", limit Wikible). Regards, Jack ma (talk) 13:10, 11 March 2013 (UTC)
Windchill for warm temperatures?
I've tried to use Windchill formula for warm temperatures. For example it gives +22°C for all wind speeds at +22°C. It seems irrelevant, because it really gets cold in such conditions (I'm kitesurfing and experience wind a lot).
Windchill for wet conditions? (watersports)
Are there any formulas for counting experienced temperature in wet conditions? For example, all watersports users (surfing, kitesurfing, diving) understand, that being wet greatly reduces experienced temperature. But to what degree? Any clues? VladGenie (talk) 08:09, 27 January 2011 (UTC)
Paragraph on "Extraterrestrial Wind Chill"
This paragraph is ignorant and misleading - the reason the formula isn't applicable on other planets with 430 km/h winds is not just that it is on a different planet. Everything about the assumptions of the NWS windchill model (or other models) would be voided by this example. The NWS formula is only applicable to humans walking in the cold (in an Earth-like lower atmosphere) with their faces exposed, in a society where the wind speed is only measured at 10m off the ground, and where winds are of a class that would be experienced by human-shaped objects on inhabitable regions of Earth's surface. Even then it's a crude approximation of heat transfer (and, to a limited extent,physical response).
What would be more accurate and insightful would be to explain that a wind chill formula could be formulated for any object (with any surface area and makeup) and any class of wind (or, generically, fluid flow). Just point out that it would not necessarily be in any way similar to the NWS formula, and that outside of human experience, wind chill may not be that meaningful a measure. Thegeniusboy05 (talk) 03:20, 9 February 2011 (UTC)
- Has someone measured the wind chill temperature on another planet? If not, why is this paragraph here? - Ac44ck (talk) 04:16, 20 April 2011 (UTC)
You don't measure wind chill temperature, you measure temperature, and wind speed then you calculate. anyway another issue with this paragraph is that a wind chill factor below absolute zero is valid, wind chill refers to what it feels like. finally another real problem with this paragraph is that you might as well be calculating wind chill underwater based on flow speed, These formulas are for use in air the EARTH's air. if you want to calculate wind chill factor in a place with a very different air make up (say underwater) you need different formulas. so forget extrapolating to a different planet. you can't even extrapolate to super high altitudes. TL;DR Thegeniusboy05 is right and ac44ck needs to try THINKING. — Preceding unsigned comment added by 126.96.36.199 (talk) 22:25, 6 August 2011 (UTC)
Removing worlwide view tag
I removed the tag about this article not presenting a world-wide view.
If you have information about a Siberian wind chill scale, then by all means include it. A drive-by tagger doesn't motivate me to do that research.
Does the tagger know of other wind chill scales? If not, how do they know there is anything that can be used to "globalize" this article? If they do know of other scales, why didn't they mention them on this talk page?
It annoys me when editors use a tag that says, "discuss the issue on the talk page" but make no effort to start the discussion on said talk page.
This article talks about Antarctica, Canada, Australia, the US, and other planets. That's more than "worldwide," it's other-worldly.
I think the tag, which has been there for more than a year without generating any discussion, is pointless. I see no reason to keep the tag unless the tagger is prepared to mention a specific, glaring omission of some important wind chill scale that hasn't been mentioned. - Ac44ck (talk) 03:21, 21 April 2011 (UTC)
In the section "Extraplanetary application" someone said Citation needed. NO IT ISN'T! To suggest that it is in this case is gormless, the kind of 'thinking' of a person who needs a radio at a live sports event to tell them that it is actually happening! Some things are so self-evident that NO citation is needed, and to ignore this is an abuse of Wikipedia. Also, whoever wrote that section demonstrates that the NAWCI results in temperatures below absolute zero on Titan, and another someone even takes issue with that! Who cares if this is 'original research'? When the demonstration is one that any kid with half a brain can deduce by methods taught in high school, we DO NOT need to trouble the time and effort of the finest professors to say what we can see clearly on our own!
The NAWCI is also suspect at far more common earthly conditions too. I deeply distrust any index that tells us that the common conditions of negligible wind at temperatures close to freezing are supposed to feel like 13 degrees C! These conditions are felt far more widely around the world than ultra-cold, ultra-dry conditions for which the NAWCI was devised. It would be better to take the Australian method I saw mentioned, because it includes humidity, but there is no description of how it is done. Even if it is more complex, it would be a good addition, if fully expressed.
Any index used MUST NOT tell us nonsense about commonly experienced conditions. If that index also fails to take humidity into account it is especially weak because humidity is a relatively simple thing to measure, and any formula should handle that before it tries to handle variables like thermal resistance of some idealised human face. Omitting something so obviously relevant is equivalent to audio engineers agonising over psychoacoustics while ignoring more direct effects like phase and delay.
Even the old index has something important that the NAWCI lacks: a rise with temperature if the wind is strong enough. By my own 'original research' I can see the self-evidently obvious fact that if the actual temperature does not rise, an increase in wind speed can cause friction, and heating, beyond its ability to cool further. This DOES matter if a model is to take into account human skin behaviour.
My point is that some issues, like wind chill, are so hard to base in 'objectivity' that we can get lost if we don't reality check them. Reality checking is a thing we must do for ourselves. I know that such efforts appear to fly in the face of Wikipedia formalities, but that's the core point I'm making: get this wrong, and it makes a mockery of Wikipedia. If Wikipedia insists on citations for every last self-evident observation, it turns science into religion! Which is surely the reverse of the aims of Wikipedia. Wikipedia states that sources must be VERIFIABLE. But it doesn't dictate how. It does NOT stipulate that every last jot and tittle of reality needs some boffin to tell it to us before we can find courage to accept it. 188.8.131.52 (talk) 14:04, 7 May 2011 (UTC)
- First, the title of the section is wrong. Extraplanetary means "not on a planet." That would usually mean in space. There is no wind of the relevant type in space. Or in a space ship/station; again; no wind of the relevant type on a space ship/station of any size in existence.
- Second, it implies the wind chill is not applicable on any other planet. Are you certain that there are no other Earth-like planets with comparable atmospheres? Star Trek suggests there are many "Class M" planets. SETI doesn't seem opposed to the idea.
- Third, the whole section is silly. Who is going to expose their skin to the atmosphere on Titan? - Ac44ck (talk) 19:13, 7 May 2011 (UTC)
"It cannot, however, reduce the temperature of these objects below the ambient temperature, no matter how great the wind velocity (however a wet surface can become cooler than ambient temperature, due to loss of latent heat, for example, water sticking to a wet brick can freeze while the air temperature is above freezing point)."
Blatant contradiction between the sections inside and outside the brackets. Either the temperature can or can't fall below the ambient temperature due to wind chill. Would anybody care to clarify? — Preceding unsigned comment added by 184.108.40.206 (talk) 17:50, 27 March 2013 (UTC)
There may be no easy answer. Feelings of cold may be due to actual temperature, wind effect, humidity, and also the clarity of the sky, because radiation to space is strong when the sky appears to be much colder than it is due to low humidity. For example, it is easily observed in moderate climates that if the temperature falls enough for ice on a puddle, a bit of wind will reduce the chance that ice will form, despite an ambient temperature of zero. Any weather forecaster knows this, it is often mentioned that a slight wind keeps temperatures up. So despite evaporation (the main reason for wind chill, bearing in mind that WITHOUT evaporation, all the wind can do is increase the speed at which an object reaches air temperature). The only way that ice will form on a puddle within a couple of hours at freezing point is if the sky is clear, because the dew point will then be lower. This effect is usualyl stronger than wind chill! (Water, in a tray in the desert, will freeze at night, even though the temperature of the sand may still be warm, because the low humidity and clear sky combine to assist dramatic radiative cooling.
In short, wind chill cannot reduce temperature below ambient without evaporation, and even with evaporation in low humidity, it is unlikely to do so by enough to account for the wind chill figures usually given. ONLY radiative cooling to a clear sky, and the deep cold of space, can do that. Normal wind chill is really an expression of how it FEELS, not of how it is. Either way, the picture is a complex one, and most wind chill models focus on faces, ignoring extra bare skin, and have other gross flaws that ignore other, usually more significant; methods of natural cooling. Specifically, they take no account of humidity (except in the Australian method) and none at all of radiative cooling dependent on the sky conditions. 220.127.116.11 (talk) 03:25, 14 December 2014 (UTC)
I have not read accurately all the above and possibly I repeat something already stated, but perhaps in a more clear way.
AS DEFINED, the wind chill refers to the energy loss from an object at the given temperature, that therefore is never chilled nor heated because its temperature is constant by definition. It is a parameter to rate the weather conditions, not a quantity referring to a man being killed by the cold, whose body temperature decreases. It has the dimension of the temperature, but by no way may be considered a thermodynamic temperature.
Let H(v,T) the heat flux in the given conditions, the WCI (best called WC Equivalent Temperature, of course) is then defined by H(v0,WCI)=H(v,T). To have an insight, let us assume (as in the Siple-Passel model) that H(v,T)=f(v)*(T0-T), where T0 is the skin temperature. Then WCI=T0-[f(v)/f(v0)]*(T0-T). This solves the puzzle for the ice at 5C, because for ice T0=0 and therefore this WCI is positive. For the man dying by cold, this WCI is time-dependent because his T0 is.
By the way, the 2001 relation may be written as 30.95 - (0.6215+0.4865*v^0.16)*(28.69-T) that suggests a reference temperature of about 30C for the human skin.
I hope to have provided an help for someone, now I make some questions to see if someone may solve my doubts. The statement that the formula is DEFINED only "below 10 C and above 3mph" does really mean that the formula is accurate within 1C only in this range? The original Siple-Passel formula had troubles at low wind speed, that were fixed by assuming f(v)=f(4) for v<4 mph on the assumption that below this speed the wind has little cooling effect; this produces the desired WCI(0,T)=T. The hard with the original formula is the steep rise of the 2-root near 0 and the revised one has a far worse 6-root. Has someone suggested a similar fix?21:20, 4 December 2013 (UTC)2A00:1620:C0:64:21C:61FF:FE03:A4C (talk)
In Jerrard and Mc Neill; A Dictionary of Scientific Units there appears an early definition of the CHILL WIND (reverse order) used by the canadian army that is minus the Mc Millan Coefficient -- apparently an estimate of the energy flux rather than of the effective temperature
The 1974 edition of the Britannica gives an Hill cooling factor
H = [ 0.14 + 0.47*sqr(v) ] * (36.5-T) for dry skin
= [ 0.21 + 0.47*sqr(v) ] * (36.5-T) for perspiring skin
where H is the energy flux in milli-cal/(cm^2*s), T is the Celsius temperature and v is in m/s.
with this formula hot is 3, pleasant is 5 to 8 the cold starts at 15 and the extra cold at 20. pietro2A00:1620:C0:64:21C:61FF:FE03:A4C (talk) 22:47, 4 December 2013 (UTC)