I (think) I know the answer to this, but I wanted to get others' takes on it.

Riding a loop with no wind, vs. riding the same loop with wind (such that half the loop, you get a headwind vs. tailwind for other half).

Does this equal out, or does the wind version use more energy/power to maintain the same overall time?

I actually don't mean keeping the same speed over the whole loop, but the same time around the whole thing. Thus I'd expect speed to drop while heading into the wind, and increase the other way.

  • 5
    Obviously you haven't ridden a bike much. The only time you ever get a tailwind is when you're climbing a steep hill. (Seriously, it does not "equal out", since wind resistance increases with the "square" of apparent windspeed.) Commented Apr 2, 2014 at 17:58
  • Maybe head on over to physics.stackexchange.com Commented Apr 2, 2014 at 18:00
  • 4
    Tailwind? What is this tailwind you speak of? Commented Apr 3, 2014 at 1:23
  • 1
    tailwind = normal headwind at twice the groundspeed ;)
    – boomhauer
    Commented Apr 3, 2014 at 18:49
  • I ride a lot, just offroad in the trees, so not a lot of wind :)
    – boomhauer
    Commented Apr 3, 2014 at 20:20

5 Answers 5


If the effect of wind resistance was linear with respect to speed, then yes, the forces would balance out. That is to say: if pedaling into a 20 mph wind was only twice as hard as pedaling into a 10 mph wind, the forces would balance.

But this is not the case, as @Daniel mentions. The effort to overcome the 20mph wind is more than twice that needed for the 10mph wind. If we simplify your loop into an out-and-back (half directly into the wind, half with it at your back), say the wind is a constant 10 mph, and that you cycle to maintain 20mph, you can get an intuitive idea of how this works.

With those assumptions, for half your trip you would be fighting a 30mph headwind (10 wind + 20 from your own speed). The second half the "headwind" would be only 10mph (20 from your own speed - 10 from the wind). This is compared to a day with no wind, in which the whole trip would have a headwind of 20mph. The jump from 20mph to 30 requires much more energy (what is the last time you hit 30mph?) than the energy saved from dropping from 20 to 10mph.

Drag varies as the square of velocity. This means that velocity is the dominant factor in the equation (as opposed to drag coefficient).

From wikipedia:

For high velocities (or more precisely, at high Reynolds number) drag will vary as the square of velocity. Thus, the resultant power needed to overcome this drag will vary as the cube of velocity. The standard equation for drag is one half the coefficient of drag multiplied by the fluid mass density, the cross sectional area of the specified item, and the square of the velocity.

For an excellent treatment of the forces affecting a cyclist, see Rainer Pivit's article. The graphs are especially helpful.

  • 1
    This is a good and conventional answer but incomplete. The fuller answer, which may not be what the original questioner intended, depends on the the shape of the loop and how the rider's aero drag varies with yaw angle. Imagine a loop shaped like a triangle, with two outbound legs heading upwind but not straight upwind, and the inbound leg heading straight downwind. Now, add this twist: at non-zero yaw, some (but not all!) bikes/riders experience lower drag than at zero yaw. That is, they "sail." In this case, they can go faster on the loop with some wind than in a calm.
    – R. Chung
    Commented Apr 3, 2014 at 12:17
  • I had not considered "sailing", and even eliminated the possibility in my simplification. In this case, you are not talking about lower drag, but riding at an angle to the wind such that the rider's body acts as a sail and transfers the wind into forward momentum. Basically like a sailboat in close haul or beam reach. I wonder how well the wheels act as a keel.
    – superdesk
    Commented Apr 3, 2014 at 14:07
  • 1
    Effective drag can decrease with non-zero yaw (that is, the effective CdA can decrease). This has been shown not only for components like wheels and frames in wind tunnels but also for riders on complete bikes. That said, this is a small effect and it depends on the yaw angle (which depends on the loop, the rider's speed, and the the wind speed and direction) so from a practical perspective it's rarely important. I've upvoted your answer because it addresses the OP's direct question -- I was only mentioning this for completeness' sake.
    – R. Chung
    Commented Apr 3, 2014 at 14:55
  • 1
    By chance, Damon Rinard (who translated the Pivit article referenced above) happened to answer a question about decreasing drag with non-zero yaw earlier today.
    – R. Chung
    Commented Apr 4, 2014 at 1:39
  • 1
    I'm skeptical about the relevance of "sailing". For any significant power gained you'd have to lean over really hard to counter the sideways force. Land yachts have three wheels and a wide (6ft+) wheel base for good reason! Commented Apr 4, 2014 at 19:05

@Superdesk has the right answer, but I thought I'd add some math to give an impression of the size of the difference.

As others have stated, drag is a quadratic function of relative wind speed. This is why you need a lot more effort to go from 0 to 10 km/h than from 20 to 30 km/h on a bike.

Suppose that you like to bike at 20km/h. If you go for a ride on a day with no wind, you have a relative wind speed of 20km/h the whole way (you move 20 km/h faster than the air around you). Suppose that you lose 1000 Cal. to drag on this ride.

Now suppose you go riding on a day with a 10 km/h wind, and ride a loop that's half as long.

For the first half of the loop, with a tail wind, your relative wind speed is 20-10 km/h = 10 km/h. But because drag is quadratic, this means you actually lose just one fourth the energy you would at a 20km/h wind speed. So you lose 125 Cal to drag on the first half of the ride (1000/2 = 500 for only riding half way, 500/4 because of the reduced drag).

On the second half of the loop, with a head wind, your relative wind speed is 20+10 = 30 km/h. Again, because drag is quadratic, this means you don't expend 50% more energy, you expend 125% more energy. So you lose 1125 Cal. to drag on the second half of the ride (1000/2 = 500 for only riding half way, 500 *9/4 = 1125 for the quadratic drag)

So in total on the windy day, you expend 1250 Cal, or 25% more riding the same distance, at the same ground speed, just for a light breeze! Also interesting to note that you'll lose 90% of the energy on the way back!

This actually gets worse very fast as the wind speed climbs. Where I live we often have 40km/h sustained winds, which means 450% more energy loss for the same ground speeds. This is when it takes all your energy to ride downhill!

  • You've just described my commute on a bad day!
    – Chris H
    Commented Apr 3, 2014 at 9:10

The energy usage could only even out if the aerodynamics of your riding position from the front of the bike equaled the aerodynamics of your riding position from the rear of the bike. Think about your riding position on the bike.

  1. Going into the wind you are low on the bike, but the handle bars hit the wind first. The wind next hits you and your form cups the wind.
  2. Wind resistance goes up with the square of speed of the wind. If the wind is against you at 10 mph (ground speed) and you are traveling (ground speed) at 10 mph, The wind is effectively moving at 20 mph relative to you. That comes out to 4 times the wind resistance compared to you just pedaling at 10 mph with no wind.
  3. Going with the wind, even if you are sitting up, your back is slanted in the direction the wind is traveling and your shoulders are rounded, giving you a much better aerodynamic profile and when the wind comes from the opposite direction.

The end result is you don't break even, but it certainly is fun going with the wind.

  • Aerodynamics "from the rear of the bike" doesn't play a roll at all if the wind speed is less than the travel speed (since the wind relative to the rider would still be a headwind).
    – superdesk
    Commented Apr 2, 2014 at 19:14
  • 1
    It doesn't equal out because wind resistance vs speed is not linear. Commented Apr 2, 2014 at 19:35
  • Hicks, he said "you don't break even".
    – paparazzo
    Commented Jul 19, 2014 at 21:42

This is similar to a flat course versus a hill.
The square factor in the drag is the drag.

Let's assume a 20 miles out and 20 miles back
You can hold 20 mph in no wind
Your area A is constant
cd is constant
p is constant

On a dead still day your work is 2 hours 1/2 cd p A (20^2) dropping out all the constants 2 hours (20^2)

Lets assume a 10 mph head wind
You exactly even things wind resistance out with 10 mph out and 30 mph back
You have 20 mph net wind out and 20 mph net wind back
Problem there is time
time out is 20 mile / 10 mph = 2 hours time back is 20 mile / 30 mph = 2/3 hours (2 + 2/3) / 2 = 8/3 / 2 = 8/6 = 4/3
So exactly 1/3 more work because you have the same effort for 1/3 longer

Now say you want to hump it for the same time 1.5 hours out plus .5 back
Out you are doing 13.33 mph + 10 mph wind for 23.33 mph wind resistance
Back you are doing 40 mph - 10 mph wind for 30 mph wind resistance
Both ways you are fighting more net wind resistance
( 1.5 hours (((20/1.5)+10)^2) + .5 hours (((20/.5)-10)^2) ) /
2 hours (20^2)
same as 3/2 * ((40/3+10)^2) + 1/4 * ((40-10)^2) / 2 * (20^2)
equals 1.6
You want to hump it in and out at the same time you are going to spend about 1.6 as much work in the a 10 mph wind.


It doesn't equal out.

An illustrative hypothetical:

  • You have a top speed of 30mph.
  • The wind is blowing at >= 30mph.

Regardless of how fast you bike the tailwind half of the loop, it takes ∞ time to bike the headwind half, which is probably more time than in the no-wind case.

  • Headwinds are tough, but I've never had an infinitely long ride! Commented Apr 4, 2014 at 19:07
  • @JamesBradbury Yeah I try not to go out on days like that.
    – blahdiblah
    Commented Apr 4, 2014 at 20:37

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