Re: Why does wind make the temperature colder?

Thanks for asking this question, because it happens to be one of my pet 
peeves. The Wind Chill Index is one of the most misleading and misunderstood 
pieces of information in all of public weather broadcasting. The primary 
misunderstanding is that wind chill has anything to do with ambient air 
temperature. What the index really measures is =rate= of cooling, 
specifically as compared to a container of warm water, intended to mimic 
roughly that of a human being. 

The reason it feels cooler on a windy day is that wind acts to remove energy 
from a surface. In the case of a human body, the wind is almost always 
cooler than the body's surface, unless you happen to be in a place where the 
wind temperature exceeds 37 degrees C. Try this simple experiment: roll up 
your sleeve, hold your arm 30 cm or so from your face, and blow on it. Your 
arm feels cooler at that spot, right? Has the air temperature suddenly 
dropped? No. What is happening is that the air movement is removing heat 
from that part of your body, making it feel cooler.

When the weatherman says the temperature is -10 degrees, with a wind chill 
of -30 degrees, the temperature is still -10 degrees. If you take a 
thermometer outside, that is what it will measure. What the wind chill 
indicates is that, at that temperature, with that wind speed, exposed human 
flesh will cool to -10 degrees as rapidly as if it were -30 and there were 
no wind. What applies to human flesh also applies to other things. On a 
windy winter day, the engine block on your vehicle will cool more quickly 
too, but it will never drop below the ambient air temperature.

A good test of how wind chill and ambient temperature are not directly 
related is to take a container of water outside on a windy day when the air 
temperature is slightly above freezing and the wind chill equivalent is 
given as below freezing. No matter how long you leave the water out, it will 
not freeze. All it will do is cool to air temperature faster than if there 
were no wind.

In Canada, meteorologists have tried to rectify this misunderstanding by 
using a more accurate description of wind chill. Heat loss and heat gain in 
the metric system is measured in watts per square metre. (In the British/
Imperial/American system, it would be something like calories per square 
inch.) When Environment Canada releases weather information, it provides the 
ambient air temperature, which is independent of wind speed, and the wind 
chill factor in watts/sq. metre. From the link listed below, here's an 
example.

With an ambient air temperature of -15 degrees C, and a wind speed of 20 km/
h, the wind chill factor is 1600 watts per square metre. That means that the 
wind is capable of removing heat from exposed parts of your body at that 
rate each second, making it feel as if the temperature were actually -26 
degrees C without wind. 

Unfortunately, Canadians, like most other people, are resistant to change, 
even when it's for the better. So most radio and tv weather reports provide 
the wind chill, not as a cooling rate, but rather as an equivalent 
temperature. Thus the misunderstanding is perpetuated, and people still 
think that a wind chill of -30 degrees means it's really that cold out.

For an excellent explanation of what wind chill is, what factors affect wind 
chill, and how to interpret wind chill, go to this very excellent site (but 
be aware that you will be dealing with the metric system - another example 
of what many people are unwilling to adapt to, despite its being for the 
better...)

 http://www.msc-smc.ec.gc.ca/cd/factsheets/windchill/index_e.cfm


And just as a by-the-way, here are the Fahrenheit/Celsius formulae for 
calculating the standard equivalent temperatures.

The formula the US National Weather Service uses to compute wind chill is:

    T(wc) = 0.0817(3.71V^0.5 + 5.81 -0.25V)(T - 91.4) + 91.4

Where:

    T(wc)= the wind chill
    V = in the wind speed in statute miles per hour and,
    T = the temperature in degrees Fahrenheit.

The formula to calculate a Celsius wind chill is: 

    T(wc) = 0.045(5.27V^0.5 + 10.45 - 0.28V) (T - 33) + 33

Where:

    T(wc)= the wind chill
    V = in the wind speed in kilometers per hour and,
    T = the temperature in degrees Celsius.

Hope this makes the picture clearer. Spread the word!