Talk:Wind chill

Why does the V overlap the 0 in the formula? Rmhermen 14:25, Aug 14, 2003 (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?71.112.83.24 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.

http://www.weather.gov/os/windchill/index.shtml

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)

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)

Frostbite

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)

Inanimate Objects

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

Random Question

Can there be wind chill at absolute zero? —The preceding unsigned comment was added by 160.39.169.6 (talk) 17:16, 26 January 2007 (UTC).

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. 160.39.169.89 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 72.137.49.53 (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 71.48.123.175 (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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Graduses

Why big chrart at the top in °F not in world standart °C? 89.208.93.236 18:23, 17 August 2007 (UTC)

Rewrite

Someone needs to rewire this. Has the person who wrote this ever opened an Encyclopedia? —Preceding unsigned comment added by 128.146.33.195 (talk) 19:55, 14 December 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 (193.10.117.249 (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

http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19690003109_1969003109.pdf

suggests that the formula should be a product, not a fraction; that is:

$WCI=(10{\sqrt {V}}-V+10.5)\cdot (33-T_{a})$

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:

1. The tagger created no section on this talk page to:
   1. Clarify the tagger's objections, or
   2. Provide a framework for addressing those objections.
2. 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.
3. 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:
   1. It isn't intended for current use anyway, and
   2. Bringing attention to "the strange sounding units" seems to be one of the reasons for including the formula.
4. 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:
   1. Call for a review, and
   2. Raise disputes over issues that didn't seem to be of concern to anyone else.

The tagger has not made a substantive effort to "discuss and resolve the dispute". I am removing the tag, which seems to be quite erroneous. - Ac44ck (talk) 15:46, 24 December 2008 (UTC)

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.--75.187.32.194 (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.--75.187.32.194 (talk) 16:32, 7 March 2009 (UTC)

Celsius/Fahrenheit

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 77.249.163.15 (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 206.83.151.193 (talk) 23:23, 25 December 2009 (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.

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).

Are there any alternatives to this Windchill formula? VladGenie (talk) 08:07, 27 January 2011 (UTC)

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.