The Physics of Skiing Into the Wind
It’s been a windy autumn out here on the edge of the prairie. And by "windy", I mean wind that comes out of Canada, gains speed across the flat and treeless Dakotas, and hits my little town of Ashby with its full force. On a recent rollerski down the Central Lakes Trail, a gust struck me so hard that it brought me to a complete stop, as though I was suddenly facing a steep incline.
I've had other experiences that have taught me to be careful with the wind as I ski on the Trail, but the most painful was a 50km roundtrip ski to Fergus Falls and back. It was a beautiful day with a light headwind when I started. I was looking forward to the same nice ski back home with the benefit of a little tailwind. When I turned around, I realized I had a problem. The wind had changed directions at some point during my 2hr trip out (or maybe I misjudged the conditions when I started), but I actually faced a pretty strong headwind for the entire trip back. I was never so happy to see our water tower on the horizon as I finally made my way back to Ashby, and I didn't have much left in my legs by the time I got home. (It was so bad I almost had to call my wife from the trail for an embarrassing rescue.)
Until recently, I'd just assumed the extra work due to a headwind was roughly cancelled out by the help I'd get from the tailwind during an out-and-back rollerski. However, I've also noticed that a roundtrip ski on my typical route takes longer on windy days than it does on calm days. This discrepency got me interested in the effects of headwind and tailwind on my skiing, and I thought of a couple questions I wanted to answer:
Q1: At what point does the power to overcome the wind drag exceed the power to overcome rolling friction?
Q2: How do the power and energy requirements of a roundtrip ski on a windy day compare to the same workout on a calm day?
Background
Like a lot of physics problems, this one starts with Newton's Laws. In this case his second -- the sum of forces applied to an object is equal to that object's mass times its acceleration. I'm assuming constant speed (no acceleration), which means that the sum of forces is equal to zero. It turns out this little equation -- the sum of forces is equal to zero -- is an incredibly versatile tool. In fact, I spend about eight weeks every year teaching my Statics class how to apply this single equation in a variety of mechanical design situations, such as designing a stone arch, determining the proper size for an axle, or figuring out when a piece of material will break. The same equation works for a skier, too. That's the nice thing about Newton -- his laws work no matter what. The trick is determining which forces are present. In the case of a stone arch, it's just the two forces of gravity and friction. For a skier, there are four forces involved: push (from the skier's legs and arms), friction (between the ski/wheels and ground), drag (wind resistance), and gravity (due to hills).
Equation
For this case, I'll assume flat ground to eliminate gravity. Since it's more common to work in power (Watts) instead of force (Newtons) for this type of problem, I'll also multiply everything by velocity. Finally, I'll move the applied power to the other side of the equation to end up with the formula (highlighted below) that I'll use to answer my questions.
The power required to maintain constant speed on flat ground is equal to the sum of the power lost due to friction with the ground and the power lost due to wind resistance.
Q1: At what point does the power to overcome the wind drag exceed the power to overcome rolling friction?
A1: Wind drag and rolling friction are approximately equivalent when rollerskiing into a headwind of 10km/hr at a pace of 3 min/km.
As the apparent wind speed on the skier increases, the friction due to air drag becomes the dominate force given the fact that the apparent wind speed is squared in the derived equation above (highlighted). The exact point where wind becomes more powerful than rolling friction is determined by the interaction between u and k. Using k = 0.3, u = 0.025, N = 800 (my weight in Newtons), and v = 5.5 m/s (a pace of 3 min/km), the crossover point is for an actual head wind speed of 2.75 m/s (10 km/hr). [See further discussion below for more on u and k.]
The equations show that on a calm day, about 70% of my power is applied to rolling friction and 30% goes toward wind resistance. Skiing into a headwind around 10 km/hr at a pace of 3 min/km results in power being divided approximately evenly between rolling friction and wind resistance, and at any wind speed above that, and wind quickly becomes the dominate force.
Of note, the weather service observes wind speed at 10m above the surface, so the actual wind speeds experienced on the ground are usually less than than values reported on your favorite weather app.
Q2: How do the power and energy requirements of a roundtrip ski on a windy day compare to the same workout on a calm day?
A2: Assuming the same average pace for the round trip on a flat course, a windy day requires about 50% greater maximum power and 25% more energy (energy is equal to power x time) than does a calm day.
The reason for the extra pain on a windy days goes back to the fact that wind friction increases as the square of wind speed. When the wind is at your back, it eliminates most of the wind friction, but it doesn’t give you a push unless the wind speed is very high, and even then, the effect would be minimal.
The first two rows of the spreadsheet below show the power and energy requirements for a 10km roundtrip ski in calm air. The bottom two rows are for the same 10k roundtrip on a windy day. I assume that the first leg of the journey is directly into a 20 km/hr headwind and the second leg is back along the same route with a 20 km/hr tailwind. While the total time remains the same in both examples, I assume a slower speed into the headwind and higher speed with the tailwind.
Comparison to Actual Data
The chart below shows the weather, along with my pace and heartrate data from a recent 10km ski (5km into the wind, 5km with the wind at my back). It took me 21 minutes to ski into the wind and 14 minutes to ski back, and it is easy to spot the point I turned around. My speed went from about 4:00/km to of 2:23/km while my heart rate dropped from the 170s to 156 at the turn. This is a route I can ski in just over 30 minutes on a calm day.
It makes sense that my time was about 4 minutes longer than on a calm day. Since I was skiing hard both days, applying 50% more power wasn't really an option for me on the windy day, so the total time was bound to increase.
More on k and u
The range of values for k found in literature for a roller skier is quite wide, but a middle-of-the-range value seems to be about k=0.3.* It’s a difficult quantity to measure even with a wind tunnel with sensitive testing equipment, and it is also dependent on the shape of the object on which the drag is occurring. Every time a skier moves her arms or changes the relative angle of her body to her legs, the value of k changes. Compare that to a cyclist who attempts to maintain the same tucked position mile after mile or a car that can hold the same shape forever, and it explains why there is so much variability in reported values for skiing. Beyond shape, there are several other components that determine k, including area (the amount of surface area facing the wind), air density (light air is easier to move through than heavy air), and turbulence (the path and swirl of air behind a skier).
It is slightly easier to find u, but there are still a lot of unknowns, such as the type of wheels and smoothness of pavement. Various papers suggest a range of u = 0.02 to 0.025 for rollerskiing. I’ll use the higher end of the estimate to better approximate the condition of the trails I’m usually on.
* I combined the drag coefficient, area, and air density to arrive at this single number, k.
