While the danger of flooding has subsided, there is still some punch left in this storm, especially if you are trying to get over the mountains. A Winer Storm Warning is still in effect until Saturday at 5 pm for the mountains around Lake Tahoe, with total storm snow accumulations which could total several feet in some locations on the crests.

In the valleys, colder air will move in dropping snow levels down to the valley floor by Saturday, although it is still unlikely we will see any significant accumulations in the lower valleys. High temperatures will drop to the mid-40s in Reno Saturday, before climbing back up into the 50s on Sunday with clearing skies.

Dan asked: *“I’ve often wondered why we don’t get more damage than we do when we get these strong winds. Perhaps the air density at this altitude makes comparable velocities less damaging than at lower altitudes. If that is the case, is there a method to calculate the actual force or perhaps differentiate damage from a 100 mph wind at sea level and at 5000 ft. elevation? Something like “our 100 mph wind is like a 75 mph wind at sea level.”*

He’s right… 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’ll use his example. 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.