If you haven’t read part 1 of this series yet, you probably want to take a look at it to make sure you’re up to speed on the nuances of average speed. In that post, we revealed why going fast up hills on a bike is the major determinant of speed on any course, regardless of how fast you can go downhill. In short, even if you ride right back down the same way you climb on a mountain, your descent will never “make up” for time lost going up. In this post, we carry that basic principle over to a discussion of wind resistance.
We all like to think that we can buy gear or come up with a strategy to “cheat the wind.” It’s a favorite refrain of triathletes. I hate to be the bearer of bad news, but you can’t out-cheat the greatest cheater in all of racing. The math proves it.
From our discussions in the book, we’re all now pretty familiar with the relevant equations for air resistance. (Don’t worry, you don’t need to do the math! You just need to understand what the math proves. If formulas make your brain hurt, just stick with me here and it’ll all come together.)
Fair = 1/2CDAρv2
Pair = 1/2CDAρv3
Fair = Force of air resistance
CD = Drag coefficient
A = Surface area of the cyclist/bike
ρ = Air density (constant at sea level)
v = velocity
For most of the situations discussed in the book, we assumed zero wind speed. So velocity was always easy. Now we have to develop a way to calculate the athlete’s velocity relative to the wind. In other words, think of yourself riding your bike past someone standing on the side of the road. If you are traveling at 20mph, then that’s your speed relative to them (and all other stationary objects to include the road, which is how we measure speed).
But what if you pass someone who’s riding in the same direction as you at 10mph? You go by them slower, so to them it looks like you’re only riding at 10mph because your speed is relative to them. And finally, consider an athlete riding in the opposite direction as you. If you’re going 20mph and they’re going 10mph, you’re going to look like you go by them at 30mph. Simple! This is exactly how it works with wind, and now that we understand that I promise not to go any farther into the theory of relativity.
Back to wind speed, we need our force and power equations to account for this relative velocity based on whether the wind is blowing with us or against us. If it’s a tailwind, then the situation is like the cyclist who’s going the same way as you. That means the wind velocity is positive, since we always consider our direction of travel as the positive direction. A headwind situation is just like the cyclist going in the opposite direction as us, so that wind must be considered a negative (just like all resistive forces).
So, headwinds are negative numbers and tailwinds are positive numbers. We know from our above examples that a headwind means greater relative velocity and a tailwind means lower relative velocity, so that establishes the relationship for calculating the total relative velocity.
vcyclist – vwind
This would make our revised equation for the force of air resistance to be:
Fair = 1/2CDAρ(vcyclist – vwind)2
Now, there’s a little problem here due to the velocity squared term. In the case where the tailwind velocity is greater than our own velocity, we would have a negative relative velocity, which would mean the wind is actually pushing us forward. It’s no longer air resistance but air assistance. Because that push on our backs is the opposite of pushing on our front, that would mean a negative force of air resistance. Our equation can’t produce a negative number, though. Multiplying a negative number by itself turns it into a positive number. Without further belaboring the discussion with math, here’s how the physicists get around it.
Fair = 1/2CDAρ(vcyclist – vwind)|vcyclist – vwind|
Because “x2” simply means “multiply the number by itself,” we rewrite the equation to explicitly denote that multiplication, with a little bit of a twist. Those funny looking straight bars around the second relative velocity term denote “absolute value.” In other words, regardless of whether the result of the subtraction is a positive or a negative one, you always consider it as a positive. Now the equation is ready to compute negative forces for strong tailwinds while still finding the correct value for headwinds.
So how do we use this? We begin by applying an easy relationship. Power is equal to force multiplied by velocity.
P = Fvcyclist
In this case, our velocity is not that relative to the wind. It’s whatever your bike computer is telling you. We incorporate our previous force equation to get
P = Fairvcyclist
P = 1/2CDAρ(vcyclist – vwind)|vcyclist – vwind|vcyclist
At this point, we have to throw one more item into the calculation. Because we now have wind pushing us along, we are going to start hitting some pretty high speeds from a cycling perspective. That means rolling resistance becomes a significant factor. We include it in the equation simply by adding the term
Froll = μmg
Because P = Fv, we can include it into our power equation by
P = 1/2CDAρ(vcyclist – vwind)|vcyclist – vwind|vcyclist + (μmg)v
Let’s say we have winds strong enough to be a race factor. How about a steady 8mph (4.4m/s) right along the length of an out-and-back course half-iron distance course? Our athlete will pedal at a constant 230 watts. How does he do?
Because the equation is a cubic function, we use a computer to crunch the numbers, so don’t worry about trying to do the math at this point. Just sit back and observe the trends, which is where the real lesson is in all of this.
First, let’s establish a baseline. With no winds at all, he travels at a constant speed of 23.2mph. Now let’s get down to earth and see what wind does to him. Using our equation, here’s his out-and-back performances under different conditions.
At a constant wind speed of 8mph we get the following:
Speed into the headwind = 18.4mph Time to travel 28 miles = 1.52 hours
Speed with the tailwind = 28.6mph Time to travel 28 miles = 0.97 hours
Total time on the bike = 2 hours, 29 minutes Average velocity = 22.5mph
Not bad! But what happens if we turn the winds up just a bit to 10mph?
At a constant wind speed of 10mph we get the following:
Speed into the headwind = 17.4mph Time to travel 28 miles = 1.61 hours
Speed with the tailwind = 29.9mph Time to travel 28 miles = 0.93 hours
Total time on the bike = 2 hours, 32 minutes Average velocity = 22mph
At a constant wind speed of 11mph we get the following:
Speed into the headwind = 16.7mph Time to travel 28 miles = 1.67 hours
Speed with the tailwind = 30.8mph Time to travel 28 miles = 0.90 hours
Total time on the bike = 2 hours, 34 minutes Average velocity = 21.8mph
At a constant wind speed of 12mph we get the following:
Speed into the headwind = 16.3mph Time to travel 28 miles = 1.71 hours
Speed with the tailwind = 31.5mph Time to travel 28 miles = 0..89 hours
Total time on the bike = 2 hours, 36 minutes Average velocity = 21.5mph
The trend is clear. As the wind velocity increases, you go slower. Though not as pronounced as our uphill-downhill situation, nature still refuses to give you an equalizer.
All of this leads to a new question, though. Our investigation here kept power constant. But we’re athletes, not robots. We can change our effort to adjust to local conditions. That said, how should we change to get the best results? Is there a way to modulate our bike effort to get better results?
As it turns out, there is. That’s coming up in the third installment of this series.
If you’re interested in getting faster, you’ll be fascinated by FASTER: Demystifying the Science of Triathlon Speed. In Faster, astronautical engineer and triathlon journalist Jim Gourley explores the science of triathlon to see what truly makes you faster—and busts the myths and doublespeak that waste your money and slow down your racing. With this knowledge on your side, you can make simple changes that add up to free speed and faster racing.