Despite the fact that it’s mid-November, it’s been surprisingly warm the past few weeks with temperatures in the 60s and 70s. I’ve found that once temps get in the low 60s I have to start layering my clothing when I go out for a ride. (Recently I discussed my lack of motivation for riding in cold weather.) Today, I enjoyed a nice 15 mile ride with temperatures in the low 50s. I know the good-weather days are going to end soon as Winter moves in. While I was riding today, I wondered what exactly the wind chill was. My math skills leave a lot to be desired, so I turned to Google to search for a good explanation on wind chill and how it’s calculated. This chart from the National Weather Service was pretty helpful:
Apparently, this is the formula the U.S. National Weather Service uses to calculate wind chill:
T(wc) = 0.0817(3.71V**0.5 + 5.81 -0.25V)(T – 91.4) + 91.4
T(wc) is the wind chill, V is wind speed in miles per hour and T is the temperature in degrees Fahrenheit.
I have no idea what this means, nor how to calculate the numbers. (I was an English major – math frustrates me.) Maybe some of you can make more sense of it. Or perhaps I can just use a calculator like this one or this one. I found some good information at IceBike.org. I wondered how wind speed differed from cycling into the wind. According to IceBike, cycling into still air is the same as wind speed. Okay, I guess that makes sense. But does the wind chill factor change if a cyclist rides at 15-mph into a 10-mph headwind on a 40-degree day? Or is that the same as what’s stated on the chart? I’m still confused.
Today’s food journal:
— Organic LF yogurt with granola = 5 points
— Power Bar = 3 points
— 1 cup LF chocolate milk = 3 points
Dinner: (at restaurant, so I’m estimating points as best I can)
— Black bean lasagna = 13 points
— Iced tea = 0 points
— Skinny Cow ice cream = 2 points
Life’s a journey. Enjoy the ride!
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