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If I'm riding at 20mph on the flat with no wind, and then a 10mph wind kicks up, if I'm heading directly into it I'll slow down by X mph, and if I'm heading directly with it (i.e. pure tailwind) I'll speed up by Y mph. (As an aside, are X and Y likely to be the same? From experience I'm assuming not)

If this is on a perfectly circular course, is the effect of the side wind (I should have said 'the angle of the wind') smoothly proportional? I.e. assuming constant power output, when doing a lap and so experiencing each angle of wind for the same amount of time, would my speed show a smooth trend from 20-X mph, through 20mph, to 20+Y mph, and then back again, maybe like a sine wave? Or do side winds mess with the cyclist in a way that pure head and tailwinds don't?

For extra credit, do you know where the wind would be coming from for me to be travelling at my original speed of 20mph?

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  • 1
    As normal the tests have already been performed, results and a decent read can be found here researchgate.net/publication/…
    – Dan K
    Nov 24, 2019 at 14:14
  • Thanks, I think I understood some of that. :) It only seemed to consider a 90 degree side wind though, whereas I was curious about the effects of other angles, e.g. a north-westerly wind when travelling south.
    – Wilskt
    Nov 24, 2019 at 14:44
  • A related factor is how constant the wind is. If its regular and relatively unvaried, then that's better than a choppy wind that comes and goes. I remember one race where I was "leaning on the wind" while crossing an open ridge area, and thought "gosh I hope it doesn't stop blowing."
    – Criggie ♦
    Nov 25, 2019 at 9:35

4 Answers 4

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The answer to your questions is not straighforward, as shown below.

The easiest to answer is your first question: are X and Y the same? No, they are not. This follows from the formula for the air-drag power Pa = ½.rho.Cd.V.(V-Vw)² , where rho is air density, Cd is drag coefficient, A is frontal area, V is forward speed, Vw is wind speed.

Suppose that for a certain power input your speed in the absence of wind is V0. Now if you keep power constant, we have for a tailwind a forward speed Vt : Vt.(Vt - Vw)² = V0³. For a headwind we have a forward speed Vh : Vh.(Vh + Vw)² = V0³.

For any V0 and Vw, you can calculate Vt and Vf by solving a cubic equation. It is easy to verify by substitution that for a no-wind speed of 20 mph and a wind speed of 10 mph, the headwind speed becomes 13.95 mph and the tailwind speed 27.16 mph (neglecting rolling resitance). So Y>X.

However, I find it more elucidating to consider a return trip along a road heading straight in the direction of the wind first and returning in a tailwind on the same road. Suppose that the trip without wind takes a time T0. It can be shown that the total time Tw is always larger than T0 and that the ratio Tw/T0 does not depend on A, Cd and rho nor the distance : Tw/T0 = (V0/Vh +V0/Vt)/2.

A few years ago I made a graph where the ratio Tw/T0 is plotted against the wind speed, assuming a wind-free speed of 20 mph, shown below.

enter image description here

The graph shows that a wind always increases the return time. At 10 mph wind speed the return time increases by 8.5%. At 15 mph wind, the increase is about 20%. This calculation assumes that the rolling resistance is small compared to the aerodynamic resistance, which is roughly the case at 20 mph.

Your question about the effect of a side wind is much more difficult to answer, because the wind-exposed area plays a big role. This takes a much more elaborate study. There is a free-access paper on the internet by Osman Isvan that investigates this question in depth.

See research paper Wind speed, wind yaw and the aerodynamic drag acting on a bicycle and rider

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  • Thanks, this is why I asked it on Stackexchange. :)
    – Wilskt
    Nov 27, 2019 at 15:51
  • That article explained a lot, thank you. Do you know whether Barry's article 'Aerodynamic performance and riding posture in road cycling and triathlon' is similarly available? The only reference I can find is for Sage Journals etc and I don't have an academic access any more. Would be interested in reading the detail.
    – Wilskt
    Dec 1, 2019 at 13:14
  • Unfortunately, I also couldn't download that article free of charge. There is a new article by Nathan Barry in the ISEA 2018 proceedings that is free of charge (A New Method for Analysing the Effect of Environmental Wind on Real World Aerodynamic Performance in Cycling). Nathan Berry also contributed to aerogeeks.com/wp-content/uploads/2018/07/… , which although it is a cycle brand publication discusses a lot of aerodynamic topics.
    – mathieu van rijswick
    Dec 2, 2019 at 14:24
  • Thanks, I:m interested in the 'horizontal forearms on the hoods is most aerodynamic' statement, and whether, if you kept the exact same body/arm position but raised the bars so your hands were now on the drops, the effect would be the same. None of the summary articles on the web are really clear on it, maybe because they've only read the abstract too!
    – Wilskt
    Dec 3, 2019 at 13:00
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For constant input pedaling power, a sidewind will slow a cyclist down.

The sidewind exerts a force on the cyclist, so they have to steer into the direction of the wind to keep on their intended course. The action of the tires against the road producing a counter force also creates a force opposite to the direction of travel.

Unfortunately headwinds have more effect than tailwinds, because aero drag force is proportional to velocity squared.

[Aero drag force = ½.ρ.Cd.A.v.2][1]

For a rider with Cd 0.4, area 1m2, air density 1.225 kg/m3

Kph  Drag (N)
30   30
25   11.8
20   7.5

You can see the difference in force is greater for an increase of velocity than a decrease in velocity of the same amount.

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I think this was what I was looking for, just didn't quite know how to search for it: https://www.sheldonbrown.com/brandt/wind.html

The diagrams show that if you have a perpendicular crosswind, if you turn 10-15 degrees so that it is coming from slightly behind you then that is about the same as no wind at all, in terms of energy expended vs speed.

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Mathieu van rijswick has already given an excellent overview of the headwind/tailwind situation. I won't repeat that. However, what's missing is a good answer for side winds.

TL;DR: Even a perfectly perpendicular side wind slows you down.

Why? Let's start with a brutal simplification: We model the cyclist as a single vertical cylinder so that they have exactly the same drag coefficient in all directions. Yes, this is an oversimplification, but it serves our purposes.

Now that we have a perfectly rotationally symmetric cyclist, we can easily calculate the effective wind speed v_e that they experience by adding the vectors of the cyclists speed v_c and the wind speed v_w:

cyclist -> o
           |\
           | \
   -v_c -> |  \ <- v_e
           |   \
           |    \
           |     \
           v------X
              ^
              |
             v_w

This is the speed, not the force. The force is quadratic with the effective wind speed. So if we have a 20km/h rider with a side wind of 20km/h, the effective wind speed will be v_e = sqrt(v_c^2 + v_w^2)km/h = 28km/h. This effective wind speed will put a force exactly twice as big on the cyclist as the headwind without any side wind.

cyclist -> o
           |\
           | \
   -v_c -> |  \ <- v_e
           |   \
           |    \
           |     \
           v------X
              ^    \
              |     \
          v_w = v_c  \
                      \
                       _|
           effective force is increased
        from no-wind scenario by factor 2

This force splits into two components: One is perpendicular to the direction of travel and thus irrelevant to the work the cyclist must put out. The other one is parallel to the direction of travel and slows the cyclist down. Since the effective wind has a 45° angle with the direction of travel, each of the two components of the force gets sqrt(1/2) = 70% of the effective force. So, the total drag force experienced by our cylindrical cyclist in the side wind scenario is 2*sqrt(1/2) = sqrt(2) = 1.41 times the drag in the no-wind scenario.

At first, this result seems a bit counter-intuitive. However, it becomes more easier to understand when you consider that the side wind increases the amount of air that the cyclist interacts with. And the more air the cyclist needs to accelerate, the more drag they experience.

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  • There is an additional complicating factor and that's due to effective CdA not being constant across yaw angles, and with some set ups CdA drops as yaw increases, to a point. This results in a scenario where it can be favourable to have a little cross wind. Disk wheels can in some circumstances act like sails, providing net forward thrust.
    – alexsimmons
    Dec 1, 2019 at 23:14
  • @alexsimmons That's why I focused on the cylindrical cyclist. Yes, pretty much any form that deflects a slight cross wind towards the back provides a small forward force. It's how some wind turbine work. Yet, the biggest drag factor with a bike sits on the saddle, and that one's not formed in a way that allows cross wind to have any positive effect. So, I guess the situation is actually worse than for the cylindrical cyclist: Once you have a cross wind, the wind-exposed area increases quite drastically, easily making up for any slight gains from the wheels.
    – cmaster - reinstate monica
    Dec 2, 2019 at 7:59

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