{"id":460,"date":"2014-08-08T17:54:00","date_gmt":"2014-08-08T21:54:00","guid":{"rendered":"https:\/\/mikaelstaer.com\/clients\/cbeals\/dev\/?p=460"},"modified":"2015-06-08T20:07:51","modified_gmt":"2015-06-09T00:07:51","slug":"asktrinerd-headwinds-tailwinds","status":"publish","type":"post","link":"https:\/\/mikaelstaer.com\/clients\/cbeals\/dev\/2014\/08\/asktrinerd-headwinds-tailwinds\/","title":{"rendered":"#AskTriNerd: Headwinds & Tailwinds"},"content":{"rendered":"
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“I’m wondering about wind in an out-and-back situation on a bike. The ideal situation in a straight out-and-back would be zero wind, while having a headwind then tailwind or vice versa would not be desirable?”<\/i>\u00a0– Mike<\/i><\/p>\n<\/blockquote>\n

Great question. The science of\u00a0aerodynamics is often complex and counterintuitive. The answer to many aero\u00a0questions is “it depends” and this one is no exception. Mike is partially\u00a0correct, but surprisingly windless conditions are not necessarily the fastest\u00a0on an out-and-back course. In this post, we\u2019ll examine headwinds\/tailwinds and\u00a0I\u2019ll save a special case with crosswinds for next time.<\/p>\n

But first, a word on the #AskTriNerd\u00a0series. I find it frustrating that popular “science and tech” writing tends to gloss over the underlying science, instead relying on hand-waving and unreferenced claims that “studies have shown…”. At the same time, I recognize that many people are more interested in the bottom line than the dorktastic details. To keep these posts accessible to a broader audience, I’ll do my best to explain things in layman’s terms. But I’ll also include some literature references or calculations to appease nerds like me. Feel free to skip, skim or scrutinize this extra info as you see fit.<\/p>\n

Let’s consider a flat, straight\u00a0out-and-back course with the wind blowing parallel to the course (i.e.<\/i>\u00a0always in or opposite the direction of motion). Let’s assume\u00a0that the wind speed and direction remain constant over the duration of the ride. In this case, Mike is correct that any headwind\/tailwind would be slower than no wind at all.<\/p>\n

Intuitively, you might think that an equal\u00a0tailwind and headwind would cancel each other out. Shouldn’t the tailwind give\u00a0back as much speed as the headwind takes away? But if you’ve ridden on a windy day, you’ve probably noticed that this not the case. It\u2019s a sad fact that the wind takes more than it\u00a0gives.<\/p>\n

To show this, I did some calculations (details below) to find out how tailwinds and headwinds affect the speed of a cyclist. With the assumptions I made, a cyclist holding a constant power output of 250W would go about 40 km\/h with no wind. The graph below shows a cyclist’s speed with different wind speeds. Negative numbers represent headwinds and positive numbers represent tailwinds. The different curves show various constant power outputs.<\/p>\n

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To examine these results another way, I plotted the magnitude of the\u00a0change in speed from windless conditions over a range of headwind and tailwind\u00a0 speeds for the 250W cyclist. In the graph below, notice how the curve showing the speed increase\u00a0 due to a tailwind is always greater than the curve showing the speed decrease for an equivalent headwind. In other words, a tailwind adds more to your instantaneous <\/i>speed than an equivalent headwind takes away.<\/p>\n

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Hold on! Isn’t that the opposite of what I’m trying to prove?<\/p>\n

This result surprised me until I realized that you would spend more time riding into a headwind than riding with a tailwind on an out-and-back course. The net effect is still a slower average speed than if there were no wind. To\u00a0demonstrate this, I calculated the average speed of the 250W cyclist for an out-and-back course (of any\u00a0distance) over a range of wind speeds. It’s clear that any headwind\/tailwind results in a slower average speed than no wind.<\/p>\n

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Calculations<\/b><\/h2>\n

To simplify this problem, let\u2019s assume that the course is flat so that no work is done against gravity. Let\u2019s also neglect the accelerations at the start and at the turnaround and assume the cyclist travels at one constant speed into the headwind and another constant speed with the tailwind. Finally, we\u2019ll neglect frictional losses in the drivetrain and wheels, which are typically just a few percent.In this simplified case, a cyclist traveling at constant speed must work against two forces: the force of\u00a0aerodynamic drag, Fdrag<\/i>, and the much smaller force of rolling resistance, Frr<\/i> (arising from the interaction between of the tires and the road surface).<\/p>\n

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Fdrag<\/i> is proportional to the apparent wind speed squared and Frr<\/i> is independent of speed. These forces are described by the following expressions:<\/p>\n

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Where \u03c1<\/i> is the density of air, Cd<\/i> is the drag coefficient, A<\/i> is the frontal area of the cyclist, g<\/i> is the acceleration due to gravity, W<\/i> is the mass of the cyclist and bike, and Crr <\/i>is the coefficient of rolling resistance. The variable v<\/i> is the apparent wind speed, given by the vector subtraction of the velocity of the wind minus the velocity of the cyclist.<\/p>\n

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For future reference in the next post, the angle between the apparent wind and the direction of motion is called the yaw angle. Notice that the yaw angle depends on 1) the cyclist’s speed, 2) the wind speed, and 3) the wind angle relative to the cyclist. In this case, because we are dealing with headwinds and tailwinds, the yaw angle is always zero.<\/p>\n

A cyclist generating power P<\/i> working against the total force F<\/i> will maintain a constant speed v_cyclist<\/i>.<\/p>\n

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Combining these expressions, we get an equation of motion for the cyclist.<\/p>\n

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Ew, gross… an implicit cubic equation! Rather than trying to solve this beast analytically, I plugged it into my favourite math software (Wolfram Mathematica) to create the graphs above. Here are the values I used for the constants:<\/p>\n

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To determine the cyclist\u2019s average speed on the out-and-back course as a function of wind speed, let\u2019s say that our cyclist completes the headwind section in time th<\/i> with speed vh<\/i> and the tailwind section in time tt<\/i> with speed vt<\/i>. Since these two sections are equal in distance d<\/i>, we can relate the two speeds and times.<\/p>\n

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The average speed vavg<\/i> is given by the time-weighted average of vh<\/i> and vt<\/i>.<\/p>\n

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We can use these two expressions to solve for vavg<\/i> and eliminate time, giving the function shown in the last graph.<\/p>\n

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While working on this post, I stumbled upon a similar analysis on Wired.com<\/a>. I borrowed some ideas, but added some details (e.g.<\/i>, rolling resistance) and simplified the derivation.<\/p>\n

Questions or comments? Please join the conversation about this post on\u00a0Facebook<\/a>,\u00a0Twitter<\/a>\u00a0or in the comments below.<\/b><\/p>\n

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