# How Cold I am, a weather math question

**URL:** <https://boards.straightdope.com/t/how-cold-i-am-a-weather-math-question/41976>\
**Category:** Factual Questions\
**Created:** [November 20, 2000, 4:35pm UTC](https://boards.straightdope.com/t/how-cold-i-am-a-weather-math-question/41976 "2000-11-20T16:35:27Z")\
**Posts on this page:** 4\
**Page:** 1

<div class="post-metadata">

**Author:** ![handy](https://avatars.discourse-cdn.com/v4/letter/h/b5a626/32.png) [@handy](https://boards.straightdope.com/u/handy)\
**Post date:** [November 20, 2000, 4:35pm UTC](https://boards.straightdope.com/t/how-cold-i-am-a-weather-math-question/41976/1 "2000-11-20T16:35:27Z")

</div>

I’m trying to make a math formula that would indicate how cold it is for someone surfing. There are some factors; but I don’t know what order to put them in:

1. air temperature 2) wind chill factor 3) wet chill factor
2. water temperature

Is there a way to formulate this? Here are some practice numbers close to reality. air temp= 55, water temp= 52, wind speed=15mph  
thanks

---

<div class="post-metadata">

**Author:** ![Crafter\_Man](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/crafter_man/32/458_2.png) [@Crafter\_Man](https://boards.straightdope.com/u/Crafter_Man)\
**Post date:** [November 20, 2000, 5:23pm UTC](https://boards.straightdope.com/t/how-cold-i-am-a-weather-math-question/41976/2 "2000-11-20T17:23:00Z")

</div>

Here are some links which may prove helpful (courtesy of [dogpile.com](http://dogpile.com)):

[http://www.usatoday.com/weather/wchilfor.htm](http://www.usatoday.com/weather/wchilfor.htm)  
[http://www.nsac.ns.ca/mp/mp100/modules/cases/newton/data/csdata3.htm](http://www.nsac.ns.ca/mp/mp100/modules/cases/newton/data/csdata3.htm)  
[http://www.crh.noaa.gov/techpapers/arp19/19-07.html](http://www.crh.noaa.gov/techpapers/arp19/19-07.html)  
[http://www.woodrow.org/teachers/mi/1993/28stra.html](http://www.woodrow.org/teachers/mi/1993/28stra.html)

Have fun!

---

<div class="post-metadata">

**Author:** ![bizerta](https://avatars.discourse-cdn.com/v4/letter/b/3be4f8/32.png) [@bizerta](https://boards.straightdope.com/u/bizerta)\
**Post date:** [November 20, 2000, 10:01pm UTC](https://boards.straightdope.com/t/how-cold-i-am-a-weather-math-question/41976/3 "2000-11-20T22:01:50Z")

</div>

You might want to also consider releative humidity. On a dry day, the evaporation of water (if the surfer is not IN the water) is a substantial consumer of heat.

In the problem that you proposed, the fact that the water temperature is 52 degrees probably outweighs all other factors combined (and then some). I can easily stay outside today in 45 degree weather for about 15 minutes without any problem. I once was immersed in 58 degree water and I was in pain within 2 minutes.

---

<div class="post-metadata">

**Author:** ![handy](https://avatars.discourse-cdn.com/v4/letter/h/b5a626/32.png) [@handy](https://boards.straightdope.com/u/handy)\
**Post date:** [November 21, 2000, 1:28am UTC](https://boards.straightdope.com/t/how-cold-i-am-a-weather-math-question/41976/4 "2000-11-21T01:28:27Z")

</div>

bizerta, that’s because water takes heat from your body ten times as fast as air does. So, 2 minutes in the water =20 in air…brrrrr