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I ride a motorcycle and it gets rather chilly when riding. I'm not interested in all the elaborate details, but please give me a simple rule of thumb to determine wind chill. For example 1/2 degree for every mile per hour, or something similar. I just want a decent approximation

2006-10-20 14:29:45 · 4 answers · asked by Billy 4 in Science & Mathematics Weather

4 answers

Okay, so I can only find really complicated equations for this, nothing simple. But I found, first, a canadian website that will calculate the wind chill for you (but only within certain fairly small temperature parameters)
http://www.msc.ec.gc.ca/education/windchill/windchill_calculator_e.cfm
And this US website that's just a table of temperatures/windspeeds and the corresponding wind chill
http://www.nws.noaa.gov/om/windchill/index.shtml

2006-10-20 14:37:56 · answer #1 · answered by kundalinicat 2 · 0 0

Not a simple answer, wind chill depends on air temperature, wind speed, and humidity. And it works on heating you up on the other side of things also. Best solution is to wear leather or something that does not allow the air to circulate to your skin. Wind chill only effects things that have moisture on them like skin. However, leather is not a good insulator so you have to put something on inside the leather to keep you warm. If you ride too long you will cool your core temperature and go into hypothermia. Reactions slow, don't make turns, etc. Can be bad.

Also as your velocity increases you lose heat out of every square inch exposed wether it is wet or not. Only way to stop it is to insulate.

2006-10-20 22:07:05 · answer #2 · answered by RobertB 5 · 0 0

There is no rule of thumb except that after 40 M.P.H. of wind there is very little change from the wind chill of 40 M.P.H. The faster you go the less the increase in wind chill temperature.

2006-10-20 22:31:00 · answer #3 · answered by Anonymous · 0 0

It's not simple, but see http://en.wikipedia.org/wiki/Wind_chill

You may be able to get a good approximation by simplifying the formula.

2006-10-20 21:36:26 · answer #4 · answered by gp4rts 7 · 0 0

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