Biking in the wind

Since i've been biking to work a lot, i've noticed that biking into the wind seems to be just as hard as biking up a hill. Often times, i'll wonder why it seems like my legs just don't have the power they should, and then i'll look around and see that the trees are swaying. Sometimes, i look up and there's no wind, and that is somewhat disheartening. Overall, though, i can definitely feel the effects of the wind.

So how strong of a wind equates to what grade of a hill? Seeing as how i've taken a couple of physics courses, i'm going to try to figure it out myself. Let's see (feel free to close your browser now)...

First, the hill - let's determine the extra energy i get when powering (or puttering) up a hill.

If my mass is m and i'm going up the hill at an angle θ and velocity v, then the height increase in time t is:

h = v*t*sin(θ)

Now, the potential energy equation is PE = m*g*h. g is the gravitational acceleration = 9.8 m/s^2. So the additional energy i've gained is:

E = m*g*v*t*sin(θ) = 9.8*m*v*t*sin(θ) (m/s^2)

So, let's leave that for now. Now we need to determine the energy lost from wind resistance. The drag equation is:

D = (1/2)*ρ*(V^2)*A*Cd

• ρ is the density of air ≈ 1.2 kg/m^3
• V is the wind velocity
• A is the cross-sectional area of me and my bike. i'm going to say that my bike and i present a 1m by 1m cross section (i'm short, but really wide), so A = 1 m^2
• Cd is the drag coefficient, which is apparently 0.9 for a bicyclist

So that's a force, but we need energy, so i'm going to use E = F*d, and d = v*t, so we get:

WE = 0.5*1.2*(V^2)*1*0.9*v*t = 0.54*(V^2)*v*t (kg/m)

Equating the two equations, we get:

0.54*V^2*v*t (kg/m) = 9.8*m*v*t*sin(θ) (m/s^2)
V^2 = 18.1*m*sin(θ) (m^2/kg*s^2)
V = √(18.1*m*sin(θ))

The final equation units comes out to m/s, which is good. At least the units are right. Let's try some actual numbers in there to see what kind of results we get.

Since i'm already a massive square, let's also assume i'm 100 kg, for ease of use.

V = √(1810*sin(θ))

For different values of θ:

• θ = 0º, V = 0 m/s
• θ = 5º, V = 12.6 m/s (28 mph)
• θ = 10º, V = 17.7 m/s (40 mph)
• θ = 20º, V = 24.9 m/s (56 mph)

That seems a bit extreme. Really, though, the V should include my velocity on my bike, and should be balanced out by my riding through no wind on the other side (there will still be drag). So maybe it would kind of work out closer to what i think should be the case. But i'm too lazy to work all that out right now. And i need to go do some other things. So maybe next time.

The other explanation is that i'm just a wimp (oh no, wind!). That's definitely possible.