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Recently we've had some pretty strong winds here -- today the weather forecast said 20mph gusting to 30mph, but a few weeks ago the gusts were supposedly over 40mph. The forecast direction was within a few degrees of into my face in both cases. The day it was gusting to 40mph (from a baseline of 30mph), just walking into the wind took considerable effort more often than not.

Even allowing for the wind funneling along the road etc. we can perhaps knock a bit off that wind speed, but even so, I was making headway into the wind. I know what riding at 30mph on the flat feels like, and 40mph downhill (in still air).

By these speeds, rolling resistance should be irrelevant compared to air resistance, shouldn't it (especially as I'm on flat bars)?

How is it possible to make enough headway into a headwind equal to your maximum riding speed that you can still stay upright? Lets assume 5mph ground speed into a 30mph wind (and flat ground).

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Chris H
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    By the way, I've seen Tactics for riding in strong winds -- interesting, but that's about "how to ride?" rather than the theory. – Chris H Jan 29 '16 at 16:54
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    The equation for bicycle drag is well understood though not always well known. In particular, the aerodynamic drag component of total drag includes airspeed as you're well aware -- but it also depends on air density. The reason why you can cycle into a strong headwind is because air isn't very dense. You would have a much harder time moving against a water current of 5 mph than an air current of 30 mph. – R. Chung Jan 29 '16 at 17:00
  • Theres also your cross-section in the direction of the wind. And that changes depending on your riding position. But your riding position also changes the power output you get. – Batman Jan 29 '16 at 17:21
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    What is the question? Why would you not be able to ride into a wind equal to your maximum speed in still air? – paparazzo Jan 29 '16 at 17:24
  • @R.Chung you certainly would (I'm familiar with kayaking) – Chris H Jan 29 '16 at 17:31
  • @Frisbee the net wind speed becomes the sum of the forward speed and the real wind speed. If the latter is the absolute fastest you can ride (or even slightly more) and the former is greater than zero, you're riding into more air drag than you can in still air. – Chris H Jan 29 '16 at 17:33
  • @Batman, yes I considered that. I don't believe I could achieve a more aerodynamic position battling into a strong wind than spinning in still air. Maybe I'm wrong by enough to make the difference. – Chris H Jan 29 '16 at 17:34
  • I am still not following what is the question here? Are you saying you have empirical evidence you can ride into wind that exceeds still ground speed beyond what can be accounted for with rolling resistance? A weather forecast of wind speed is not solid empirical data. – paparazzo Jan 29 '16 at 17:38
  • @Frisbee, you're right, it's not. So my evidence isn't conclusive. But we have: forecast >> I can ride in still air; the air on my face feels like riding downhill at 40 (and the difference is huge between that and 30); walking was difficult (Beaufort scale says "inconvenient" at 32-38mph). – Chris H Jan 29 '16 at 19:42
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    Ah, I think I'm beginning to understand. Is your question, "if I can ride a maximum speed of X in calm wind, why can I still move forward when faced with a headwind of X?" – R. Chung Jan 29 '16 at 19:47
  • In general, above about 10mph your rolling resistance is negligible compared to air resistance, so if you can't do 30mph on the flat in still air you will have difficulty moving at all in a 30mph headwind, if it's really head-on. And sometimes a crosswind is even worse. The main reason you can even stay upright in such a wind is that you're willing to exert yourself that hard, whereas you're not when it comes to doing 30mph, at least not for long. – Daniel R Hicks Jan 29 '16 at 20:47
  • A crosswind is certainly worse unless you've got the road to yourself. That was my take on the timing resistance as well. – Chris H Jan 29 '16 at 21:15
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    I tweaked the title to make it clearer, hope that's ok – Móż Jan 30 '16 at 00:35
  • @DanielRHicks you are exerting force on the ground not the air. Riding into a 40mph headwind is not equal to riding against a 40mph conveyer belt. – JamesRyan Jan 30 '16 at 03:05
  • @JamesRyan - You are exerting force on the pedals. (Riding a 40mph conveyor belt is easy, since you only have to deal with rolling friction.) – Daniel R Hicks Jan 30 '16 at 03:28
  • @DanielRHicks science says you are wrong. Try to understand why or don't but there is no point in arguing with reality. – JamesRyan Jan 30 '16 at 03:36
  • I'll admit that you've almost got me convinced. – Daniel R Hicks Jan 30 '16 at 03:55

5 Answers5

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You can do it, because your bike is connected to the ground.

The work done when moving an object is proportional to distance and resistance force (which consists of air resistance and rolling resistance for bikes). The air resistance depends on air speed (ground speed + wind speed), but distance depends only on ground speed. Going slower reduces the energy spent, even when air speed remains high.

As an extreme example, consider just standing in place in wind. You are not doing any work at all, but your air speed is still high. When going against wind, you will soon find yourself doing more work. And when going downwind, you are doing negative work!

For objects not connected to ground things get different. A balloon in 40 MPH wind would drifting 40 Mph downwind in ground coordinates and would have to use the same power to stay in place in ground coordinates as it would use to move 40 Mph in still air.

A numerical example may help. We know the power equation for bicycles. For a bicycle on flat ground, at steady speed so that there is no acceleration or deceleration, in calm wind conditions, for typical rolling and aerodynamic coefficients of drag for a flat bar bike (Crr ~ .005 and CdA ~ 0.4 m^2), and a total mass of 85 kg, the power required to go 30 mph (13.33 m/s) is:

.005 * 85 * 9.8 * 15.56 + 0.5 * 1.2 * 0.4 * 13.33 ^3 = 625 watts.

However, what would the required power be to travel at 5 mph (2.22 m/s) into a 30 mph headwind? In that case, airspeed will be 13.33+2.22 = 15.56 m/s but ground speed will be only 2.22 m/s.

.005 * 85 * 9.8 * 2.22 + 0.5 * 1.2 * 0.4 * (15.56^2) * 2.22 = 140 watts.

R. Chung
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ojs
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  • You are wrong with regard to a basic detail. The energy expended in this case is power multiplied by time, not force multiplied by distance. The amount of power required to barely move forward in a 30 mph wind is only slightly less than the power required to go 30 mph in still air. – Daniel R Hicks Jan 29 '16 at 23:24
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    Power equals work divided by time. Power multiplied by time equals work. – ojs Jan 29 '16 at 23:33
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    See also: http://www.physicsclassroom.com/class/energy/lesson-1/power, "Another formula for power" – ojs Jan 29 '16 at 23:35
  • Nice edit. I was thinking about doing a simple plot without units or rolling resistance, but real world units are always better. – ojs Jan 30 '16 at 00:08
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    @DanielRHicks The energy expended in this case is power multiplied by time, not force multiplied by distance. No, those are the same (see next comment though...) Energy is Joules - kg * m^2/s^2. So power in watts is J/s. Easy. But what is force? Newtons. A Newton is one kg * m/s^2 - kilogram time meters per second squared. So, what's a Newton-meter (a force times a distance)? A kg * m^2/s^2. AKA a Joule - energy. See https://en.wikipedia.org/wiki/Joule – Andrew Henle Jan 30 '16 at 01:05
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    To be precise, energy would be the integral of power over time, or the integral of the dot product of the force and velocity vectors over time, since power can vary, and the force and direction of movement can also vary. – Andrew Henle Jan 30 '16 at 01:08
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    The power a cyclist produces is the torque on the crank times the cadence. I don't know about you, but when I'm riding into a headwind I gear down, so that a turn of the crank is not getting me nearly as far as in still air. – Daniel R Hicks Jan 30 '16 at 01:54
  • Yes, the power input to system through cranks is the same as the power expended to wind resistance, rolling resistance and changes in kinetic and potential energy. I, Chung and Henle are attempting to explain how it is calculated. – ojs Jan 30 '16 at 07:57
  • OK that might be correct. Please modify so I may remove my down vote. – paparazzo Jan 30 '16 at 18:15
  • I certainly like this approach -- if we consider a rider who can track stand/balance with the brakes on (or even a tricycle) no power should be expended at zero ground speed. Do you have a citation for the formula with {V_air^2}{V_ground} in it? I looked last week and couldn't find anything like that. – Chris H Feb 01 '16 at 09:56
  • Not exact reference, but you arrive at it when you plug together W=Fs, F=kV_air^2 (k is constant that depends on cross section area and shape) and P=W/t. – ojs Feb 01 '16 at 16:53
  • @ChrisH The reference you're looking for is Equation (2) in this document. – R. Chung Feb 02 '16 at 18:46
  • Can we modify or remove paragraph 3, you do work standing still to not fall over in the wind and I wouldn't say you do negative work downwind, but obv you do get more out for your effort in thanks to wind in your direction. Wording doesn't fit with the other sound examples. Also if you can include the equation for calcs in answer that would be great. – Bent Spoke Cycle Repair Feb 11 '16 at 07:35
  • No, we can't do that modification because that would be wrong. In physics, the word "work" has a very specific meaning that does not fully correspond to your everyday understanding. – ojs Feb 11 '16 at 20:06
  • Ok thanks for making that clear without being patronising – Bent Spoke Cycle Repair Feb 14 '16 at 08:57
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    To be more clear, the muscles internally do some work in order to keep the person not falling over, but that work is not done against the wind. A rigid person-shaped mannequin standing in wind would not do work. – ojs Feb 14 '16 at 13:23
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You can do this because of the gearing of the bike. When you're riding at a slower ground speed, if you shift to a lower gear to keep your pedal RPM the same, then the same force on the pedals produces a higher thrust at the tire. Even if you do not shift, it is easier to produce higher force on the pedals at lower RPM.

The strength of a cyclist is generally measured by the power they can produce. Power = Force X Velocity. In this case, the velocity is measured with respect to the ground, because the bicycle drivetrain works by pushing on the ground (via the rear tire). So if riding 30 MPH in still air on a flat ground, neglecting rolling resistance, requires 600 W of power, then riding 5 MPH into a 25 MPH headwind (same drag force) will require (5/30) * 600 = 100 W.

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  • Gidday and welcome to SE Bicycles - that's a good first answer. – Criggie Jan 30 '16 at 00:13
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    @frisbee Its a good first answer because its not a useless one-liner. And it has encouraged some discussion.

    The answer breaks down a little because the wind is not connected to the ground, so there's a non-zero air "resistance" even when bike's ground speed is zero. So if you're going the other way are you doing work when not peddling- depends on whether your frame of reference is the ground, or is the wind a "second rider" on your bike?

     – Criggie Jan 30 '16 at 04:18
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    @frizbee The velocity in force*velocity is the component of velocity parallel to the force and relative to the thing you are pushing against. For a bike traveling in a straight line on a fixed surface this is equivilent to ground speed. Obviously for a bike on a trainer it is not because the bike is no longer pushing against the ground. – Peter Green Jan 30 '16 at 04:21
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    We have three interconnected systems here. The pedals convert up/down motion into rotary motion. The gears allow us to trade torque for speed. The drive wheel converts rotary motion into linear motion and exerts a force on the ground propelling you forwards. The important system here is the drive wheel. If the ground speed is lower then the outside of the drive wheel is moving slower which means it takes less power to apply a given force. – Peter Green Jan 30 '16 at 05:01
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    @Frisbee: when climbing there is another resistive force in play: gravity. This is why climbing is harder than riding on level ground. – ojs Jan 30 '16 at 08:02
  • @ojs Yes climbing is another resistive force. Is it less power because of lower ground speed as asserted in this answer? – paparazzo Jan 30 '16 at 09:55
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    Yes. To pick some random example numbers, 10 % climb for a person and bike of 80 kg together produces a resisting force of 10% * 80 kg * 9.81 m/s^2 = 78,48 kgm/s^2, which is the same unit as N. Power equals speed times force. Now, climbing this hill at 2 m/s uses 78,49 N * 2 m/s = 156.96 W. Climbing the same hill at 5 m/s uses 78,49 N * 5 m/s = 392 watts. As you see, lower ground speed uses less power. – ojs Jan 30 '16 at 10:07
  • No, the answer does not assert "same amount of leg power in the wind is somehow less power", the leg power is equal to power used against wind, rolling resistance, etc. Riding 5 mph into 25 mph head wind is less work than riding 30 mph in still air because of different ground speed. – ojs Jan 30 '16 at 10:26
  • Is you problem the difference in power lost to air resistance? Peter Green already tried to explain it, but I'm putting it to simpler terms. There is indeed power lost that does not come from the cyclist. It comes from the wind, which is powered by temperature differences caused mostly by the sun. In a wind tunnel the physics between ground, bike and air stay the same, but the power comes from fan. – ojs Jan 30 '16 at 10:44
  • Let us continue this discussion in chat. – paparazzo Jan 30 '16 at 10:45
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Partly it's due to the way that wind speed is measured. The standard for wind speed is to measure it 10 metres above ground level. Closer to the ground, an effect called the boundary effect kicks in and the wind speed is slower (in fact, the wind speed on the ground is effectively zero).

According to this site, wind speed on a flat grassy plain can be calculated as V=Vref((H/Href)^0.142). For a Vref of 30mph at an Href of 10m (excuse the mixed units, they drop out as the equation is dimensionless), the wind speed at 1m off the ground would only be 21mph.

However, the link also suggest an exponent value of 0.333 at the outskirts of a town, or of 0.5 in a city, corresponding to 1m-height wind speeds of 14mph or 9.5mph respectively.

So, the answer to why you can still pedal forwards into a 30mph headwind when you would normally be at equilibrium pedaling forwards at 30mph is that in the case of the headwind, the actual windspeed measured at your riding position will only be around 20mph or less.

Tethera
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    Now that's really interesting. I wonder how it relates to the observations in the Beaufort scale related to walking into the wind. I'm starting to think I want an airspeed sensor on the bike! – Chris H Jan 29 '16 at 21:14
  • Gidday and welcome to SE Bicycles - that's a good and interesting first answer. – Criggie Jan 30 '16 at 00:20
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    "the wind speed on the ground is effectively zero" Which is why leaves and litter on the ground never actually move when its windy. Oh, wait. – David Richerby Jan 30 '16 at 08:26
  • Lol, I knew someone would pick up on that. Actually, it's exactly why leaves and litter on the ground tend to get lifted almost directly upwards, being picked up by the bernouilli pressure differences, and then start moving with the wind. But it was more of an aside, and only really relevant at microscopic distances. – Tethera Feb 01 '16 at 18:49
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I have a degree in chemical engineering and we study this not just in pipe flow but in a fluidized catalytic bed and when you will lose catalyst out the chimney. In my chemical engineering studies we have never treated particle speed versus wind speed differently.

According to Galilean invariance you should get the same wind resistance in any frame of reference. It is only relative. Think about this we are spinning and rotating around the sun.

Wind resistance is the net. 30 mph in still air is exactly the same as 5 mph in 25 a mph head wind.

Gearing make this possible but that is not the stated question. The question is regarding wind resistance alone does:

30 mph (speed) + 0 mph (wind) = 5 mph (speed) + 25 mph (head wind)

The answer is yes they are the same. The proof is Galilean invariance.

Put 30,0 and 5,25 into this calculator. Both sets of numbers give same relative velocity (30) and the same WATTS.

FLO Cycling - How Velocity Affects Drag

When calculating drag, velocity is not simply the speed at which you are travelling on your bike. Velocity is the combination of the speed at which you are travelling on your bike and the velocity of the wind. This combination of velocities is know as relative velocity.
Head Wind In this example the cyclist is travelling at 15mph and the wind is travelling in the opposite direction at 5mph. The relative velocity is therefore equal to...
Rider Speed - Head Wind
(15mph) - (-5mph) = 20mph

Drag Forces in Formulas

The power required to overcome the total drag is:
P = Ftotal v where v : velocity in m/s
The formula for air resistance strictly applies only with no wind. With any wind the vector sum of wind due to motion of the bicycle plus true wind is to be taken instead of v;

Drag (physics)

refers to forces acting opposite to the relative motion of any object moving with respect to a surrounding fluid

v is the speed of the object relative to the fluid

Drag coefficient

u, is the flow speed of the object relative to the fluid

proportional to the square of the relative flow speed between the object and the fluid

Drag Force

v is the speed of the body relative to the fluid

Drag Force and Drag Coefficient

U is the relative velocity of the fluid with respect to the particle

If you drop a rock the terminal speed from gravity should be exactly the same as the wind speed from a fan it takes to hold it in air.

If the third V in power V³ is ground velocity and not relative velocity I am not finding any reference that states that. Let's assume that is true:
Vs1 is velocity still
Vs2 is velocity in wind
Vw is velocity of wind
Vs1^3 = (Vs2 + Vw) * (Vs2 + Vw) * Vs2
Vs1^3 = (Vs2^2 + 2*Vs2*Vw + Vw^2) * Vs2
Vs1^3 = Vs2^3 + 2*Vs2^2*Vw + Vw^2*Vs2
if Vs1 = 30 and Vw = 25 then Vs2 = 16
Able to ride 16 mph into a head 25 mph wind does not seem right to me but I am just not so sure anymore

The only possible difference is wind is slightly disturbed so it is going to have some turbulence. But at even a small speed you are into turbulent flow (Reynolds number).

Because of the gusting and turbulent nature of wind it will feel faster than the sensation of riding in still air.

Assume 30 net mph is 600 watts

  • At 30 mph in 52 x 11 gear
    cadence of 80
    I am delivering 600 watts

  • At 5 mph in 26 x 33 gear
    cadence of 80
    I am delivering 600 watts

paparazzo
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  • I think you make a particularly good point about the turbulence. The Galilean invariance is roughly what I had in mind when I asked the question, though I'd forgotten the term. Maybe it's a sum of poor estimation, rolling resistance being a bigger factor than I reckoned, and being able to apply more effort for brief periods than sustained. – Chris H Jan 29 '16 at 20:26
  • Comments are not for extended discussion; this conversation has been moved to chat. – jimchristie Jan 31 '16 at 15:28
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I assume your question is 'How is it possible that I can still riding upright, even though the headwind is greater than my maximum riding speed?'

The answer is:

  • The gust/wind is not always in constant direction, and certainly is not a constant flow (speed). So the average headwind speed is less than you would assume. This is especially true when you are in an urban area, when the wind direction changes constantly.
  • Even if there is a constant headwind of 30 mph (your max riding speed), your speed will not suddenly goes to zero. Your speed would gradually decreasing, from the speed of you and your bicycle just before the gust, until it reaches zero. The analogy to this situation is when you are riding at constant 30 mph and stop pedaling.

So to sum up, the average wind speed does not have enough drag (force) to null the speed of you and your bicycle, which is constantly 'replenished' by your pedaling.

Nhân Lê
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  • I can't disprove any of that, though I do know that there were even stronger bits that were as effective as my brakes (in the wet). Also pedestrians (observed plus me walking just before riding) and the Beaufort scale as in my comment under the Q. – Chris H Jan 29 '16 at 19:45
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    I have found gusts stopping me while pedalling down a 10% grade in the north of Scotland... (Berriedale for anyone who knows the hill) – user_1818839 Jan 29 '16 at 21:55
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    @BrianDrummond You should have turned around at the bottom and gone straight back up the hill with a tailwind. Imagine the strava segments! – Criggie Jan 30 '16 at 04:14
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    @Criggie Nice idea, but thanks to Berriedale's hairpin bends, you lose in both directions! – user_1818839 Jan 30 '16 at 11:53
  • Any hill in scotland is just like that, you either go 50 mph or 5 mph, because of the wind and gradient.. – Nhân Lê Jan 30 '16 at 11:57