Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Wind Chill Factor Formula
Name: N/A
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: N/A


Question:
I would like to know what the actual formula is to calculate the wind chill factor. Or is there such a thing?



Replies:
I do not know if there is a formula, but I seem to recall (this was learned way back in high school so no guarantees) that it was evaluated by take the temperature of a wet thermometer at a particular wind speed and comparing that to a dry thermometer - this gives you a temperature decrease caused by the wind evaporating water that should closely approximate the effect of wind on the human body. I assume the wind chill factors quoted by weather people are derived from tables of this sort...

Check out:

http://observe.ivv.nasa.gov/nasa/earth/wind_chill_applet.html


Actually, different countries use different measures of wind chill and different units even... The relevant quantity is not the final temperature as I said earlier, but the rate of heat loss per unit area of exposed skin. There are formulas for heat loss (I think there is one ascribed to Newton) but basically in still air it goes linearly in the temperature difference between the inside (body temperature) and the air outside. When it is windy the heat loss is increased due to evaporation of water from the surface of your skin. The "wind-chill factor" then gives an equivalent temperature that it would have to be outside in still air for the same rate of heat loss that you get in still air with the current wind. I think it is possible that in a really strong wind the effective temperature could go below absolute zero, which would not make too much sense from a temperature perspective - just that the heat loss rate is greater than you could ever get by standing in still air no matter how cold. Anyway, that is the basic idea.

Arthur Smith



Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory