Out-and-back time trial in the wind: why a headwind costs more than a tailwind gives back
Wind on an out-and-back
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Minutes lost by a 40 km out-and-back time trial in a 14 km/h wind, even though the return leg has a tailwind.
Model calculation: constant 300 W, CdA 0.24 m², 78 kg, pure head/tailwind.
At 300 W and no wind, the example 40 km time trial takes 55.4 minutes. With a 14 km/h wind, the outbound leg into the wind goes at 34.7 km/h and the return with the wind at 52.9 km/h, and the total rises to 57.3 minutes. The fast return does not make up for the slow outbound leg.
Why doesn’t a tailwind make up for a headwind?
For two reasons that add up. The first is time: into the wind you go slower and spend longer suffering that wind; with the wind you go faster and the benefit lasts less. The second is drag physics, which depends on the square of air speed1: a headwind raises air speed a lot, and a tailwind lowers it less than it seems because you are going faster.
The loss grows much faster than the wind: 28 seconds at 7 km/h, almost 2 minutes at 14 km/h and over 4 minutes at 22 km/h.
How should power be distributed to lose less?
The answer is to push harder into the wind and ease off with it. An extra watt yields more where you are slow, because you spend more time there. Power distribution models predicted it3,4 and a laboratory study with simulated wind confirmed it: varying power with the wind improved time compared with holding it constant2.
With a 14 km/h wind and the same time-weighted average power, riding 320 W into the wind and 271 W with it saves about 16 seconds. Riding 350 and 234 W saves about 30, but that split is far more demanding physiologically and is not always sustainable in practice.
- How much to vary: it is limited by the reserve above critical power. Pushing into the wind spends W′ and easing with it rebuilds it (see W′bal).
- Crosswind counts too: what matters is the component along the direction of travel and the angle at which it reaches the rider.
- Wind changes during the event: plan with the wind at the time you will pass each section (see wind by time of passage).
Rule of thumb: into the wind, about 5 to 10 % above average power; with the wind, below it, as far as the W′ reserve allows.
What is known and what is not
That wind does not even out on an out-and-back is straightforward physics1, and that varying power helps is supported by models and experiments2,3,4. What is uncertain is the real wind on the day: forecasts are wrong, and wind at rider height is lower than at weather stations.
Frequently asked questions
Does the wind even out on an out-and-back time trial?
No. With a pure head/tailwind you always lose time compared with a windless day, even though half the course has a tailwind.
Should you push harder into a headwind?
Yes, somewhat above average, and ease off with the wind. An extra watt yields more where you are slow.
References
- Martin JC, Milliken DL, Cobb JE, McFadden KL, Coggan AR. Validation of a mathematical model for road cycling power. J Appl Biomech. 1998;14(3):276-91. doi:10.1123/jab.14.3.276
- Atkinson G, Brunskill A. Pacing strategies during a cycling time trial with simulated headwinds and tailwinds. Ergonomics. 2000;43(10):1449-60. doi:10.1080/001401300750003899
- Swain DP. A model for optimizing cycling performance by varying power on hills and in wind. Med Sci Sports Exerc. 1997;29(8):1104-8. doi:10.1097/00005768-199708000-00017
- Boswell GP. Power variation strategies for cycling time trials: a differential equation model. J Sports Sci. 2012;30(7):651-9. doi:10.1080/02640414.2012.654397