When you combine a wet sub-tropical airmass with a weak upper-level low pressure center you get a pretty unsettled weather pattern through the weekend. Scattered rain showers with occasional thunderstorms will carry us through early next week. Snow levels will be quite high, likely staying above the major mountain passes, although with the unstable conditions brief columns of snow could drop to the upper roadways here and there. Temperatures drop to the 60s through Tuesday after which another colder low could drop snow levels into the foothills by midweek.

Winds up here aren’t as damaging as they are at sea level because the lesser density of air at elevation creates less force against an object for comparable wind speeds. But is there a way to correlate the force of winds at different altitudes? Yes, you can do that. We have about 15% less air in Reno than you do in San Francisco, which means that you have 15% fewer air molecules hitting your fence during a 100 mph wind. At first glance, you might think that a 100 mph here is equivalent to an 85 mph wind at sea level. But that’s an easy mistake to make (I’ve made it myself.)

But there’s a problem with that logic. Since force of wind increases exponentially with speed, the math gets a little more complicated. So how strong a wind at sea level would equal the force of a 100 mph wind at our elevation?

(WARNING: HIGH SCHOOL MATH AHEAD)

The formula for force of a wind is: Force (lbs/square foot) = (wind speed)^{2} X (.0027). The .0027 is a fudge factor that works at sea level. The .0027 factor has to be reduced to 15% less at our altitude (.002295). So a 100 mph wind at our altitude has a force of (100)^{2 }X (.002295) = 22.95 pounds/square foot. To find an equivalent sea level wind speed, plug this figure into the original formula and back out the wind speed, which comes to about 92 mph. This can work for any elevation, but you need to adjust the .0027 fudge factor by the percentage of atmosphere you have left. For instance, at 10,000’ elevation you have 30% less air, so the figure becomes (.0027) X (.7) = .00189, which when plugged back into the equation makes the force of a 100 mph wind equivalent to 83 mph at sea level.

* *

### Like this:

Like Loading...

*Related*