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Headwind and Power

Old 12-31-21, 07:41 PM
  #1  
cyclezen
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Headwind and Power

so, in the spirit of the season - 'off-season' for some - a thread we can all present our As and Os...
...so, I've been wonderin, for so very long...
given the same rider, on the same bike
if you're riding at 10 mph (16 kph) into a 15 mph headwind (24 kph) , are you using the equal/same 'power' as if you were riding at 25 mph (40 kph) ???
is it an equal relationship, and is it linear ?
what other factors might influence this greatly ? - are rolling resistence and impedance nominal or significant players in this drama ?
So Bill Nyes _ The Science Guys in our midst - what is the reality?
not a scientist ? have an opinion? lets hear it !
My wishes for more power to all for the coming year
Ride On
Yuri

Last edited by cyclezen; 12-31-21 at 10:08 PM. Reason: error of omission...
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Old 12-31-21, 08:19 PM
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Have a play around with this website. You'll see the relationship is not as simple as you think. If you are a 75kg rider with 10kg bike, you would need to produce just 42.1W to go 16kph. Add in a 24kph headwind and you would need to produce 161.3W to sustain 16kph. If that wind disappeared and you were still producing 161.3W, you'd be traveling at 28.4kph, not 40kph.

https://www.omnicalculator.com/sports/cycling-wattage

This page will give you similar information but represented in a different way, and includes some helpful information to answer your aerodynamic drag questions.

https://www.gribble.org/cycling/power_v_speed.html

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Old 12-31-21, 09:03 PM
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Ground effect versus ďanemometerĒ wind speed presented by the local weather station is not equivalent.

So, best bet is, local weather station says Ď10mph windí you get a third that actual ground level true wind effect.

They donít hide the weather station anemometer at 3 feet off the ground behind a hedge.
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Old 12-31-21, 10:26 PM
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High school math quiz.

If two cyclists of equal power and the same weight with the same bikes were to ride 10 miles and one route was completely flat and the other had a 2 mile hill at 5% with a 2 mile descent at 5%, who would cover the 10 miles first?

Would the time lost climbing the hill be made up for on the descent?
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Old 01-01-22, 12:46 AM
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No. Force is the same. But power is the dot product of force and velocity. Velocity is how fast you are moving. So 25 mph in still air is more power than 15 mph into 10 mph headwind.

Sorry. Typed too fast first time. Power, not work.

Otto

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Old 01-01-22, 12:50 AM
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if have a 60 mph tailwind, how fast will it make me go?

i'm not drunk, just drinkin.
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Old 01-01-22, 01:32 AM
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Originally Posted by cjenrick View Post
if have a 60 mph tailwind, how fast will it make me go?

i'm not drunk, just drinkin.
If you have the gears for it, 60 mph would only require you to overcome drivetrain and tire losses assuming the surface is flat and pretty smooth. Even so, that might be in the vicinity of 150-200 watts. Pay attention and hold on tight if there are gusts or wind shifts. 😊

Otto
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Old 01-01-22, 08:59 AM
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According to bicyclecalculator the answer is NO (but maybe YES :-)

Riding 100mph in 0mph wind takes 21508w
Riding 50mph in 50mph headwind takes 10754w
Riding 1mph into a 100mph headwind takes 219w (this seems completely wrong)

and

Riding 100mph in 100mph tailwind takes only 182w (seems reasonable, because relative wind speed is zero)
Riding 50mph in 50mph tailwind takes only 91w

Meaning, as expected, at high speed power to overcome rolling resistance is negligible compared to power to overcome wind resistance.

However, I can offer no explanation as to why 100 + 0 mph takes so much more power than 50 + 50, when relative wind speed seem to be all that matters. Imo they should be MUCH closer.
Tbh, I have a suspicion its an anomaly/mistake in the calculator.

bikecalculator.com/

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Old 01-01-22, 09:04 AM
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Are we being tested on this? It wasn't listed in the syllabus.
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Old 01-01-22, 10:25 AM
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Originally Posted by Racing Dan View Post
However, I can offer no explanation as to why 100 + 0 mph takes so much more power than 50 + 50, when relative wind speed seem to be all that matters. Imo they should be MUCH closer.
Tbh, I have a suspicion its an anomaly/mistake in the calculator.

bikecalculator.com/
Simple. Itís just noting the difference between force and power.

Relative wind speed is what determines the force of air drag acting on bike and rider. But to require power (the rate at which you are doing work) that force must act through a distance and the faster it moves (higher velocity) against that force the more power is needed.

Examples:
1) there is a 30 mph wind. If I just stand there on my bike at the appropriate angle, there is some force acting on me, but I donít move myself and the bike, so I do no work.

2) 15 mph headwind but I ride into it at 15 mph. Same force is acting on me and the bike with the same relative speed but Iím moving at 15 mph against that force. That takes a considerable amount of work. For me itís probably in the neighborhood of 350 watts because donít ride low and am not very aero

3) still air and I manage to ride (briefly) at 30 mph. Same force is acting on me and bike but Iím moving 30 mph against that force requiring twice as much power as 2). Might be 700 watts, explaining why it would be only a brief moment.

Does that help?

Otto
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Old 01-01-22, 10:47 AM
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Originally Posted by rsbob View Post
High school math quiz.

If two cyclists of equal power and the same weight with the same bikes were to ride 10 miles and one route was completely flat and the other had a 2 mile hill at 5% with a 2 mile descent at 5%, who would cover the 10 miles first?

Would the time lost climbing the hill be made up for on the descent?
You never recover the time lost climbing vs descending simply because of the extra time you have to spend climbing vs descending. It always brings your overall average speed down. The flat route will always be quickest overall. Tale the extreme example of a mountain col. It would take you an hour to climb 10 km at 10 kph and then 10 mins to descend at 60 kph. Your overall average speed would be 20 km/ 1.167 hours = 17 kph. Considerably slower than riding on the flat with an equivalent power.

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Old 01-01-22, 10:58 AM
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@ofajen

Possibly. Ill have to think about it.


To my mind its the same as submarine either going 10mph (om the map) in a body of water with zero current or 0mph (on the map) with 10mph current dead against it. To do that would require the exact same power in either scenario.


I dont think 1/ is relevant. Its like calculating the power of a fence post. Nonsensical. I would however be relevant if it was a ball. What power would it take to maintain position in a 30mph "head"wind? 0w because its not moving (on the map) or the power to overcome the force of the 30 mph wind? If its 0w Im sure you could get a high paying job in an off shore operation :-)


2/ 3/ I believe you are conflating map speed with relative wind speed.
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Old 01-01-22, 11:06 AM
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Originally Posted by ofajen View Post
Simple. Itís just noting the difference between force and power.

Relative wind speed is what determines the force of air drag acting on bike and rider. But to require power (the rate at which you are doing work) that force must act through a distance and the faster it moves (higher velocity) against that force the more power is needed.

Examples:
1) there is a 30 mph wind. If I just stand there on my bike at the appropriate angle, there is some force acting on me, but I donít move myself and the bike, so I do no work.

2) 15 mph headwind but I ride into it at 15 mph. Same force is acting on me and the bike with the same relative speed but Iím moving at 15 mph against that force. That takes a considerable amount of work. For me itís probably in the neighborhood of 350 watts because donít ride low and am not very aero

3) still air and I manage to ride (briefly) at 30 mph. Same force is acting on me and bike but Iím moving 30 mph against that force requiring twice as much power as 2). Might be 700 watts, explaining why it would be only a brief moment.

Does that help?

Otto
Spot on explanation. Power = Force x Velocity
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Old 01-01-22, 11:18 AM
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Originally Posted by Racing Dan View Post
@ofajen

Possibly. Ill have to think about it.
Sorry, but you need to think about this more. Iíve explained it accurately and the best I can but here is Jobst Brandtís much better if lengthier discussion along with many interesting graphs. This should help.

https://www.sheldonbrown.com/brandt/wind.html

Also, Brandt apparently borrowed from this earlier article by Osman Isvan:

https://docs.google.com/file/d/0B-pZ...uERuLOFBZ1sPAg

Otto

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Old 01-01-22, 11:19 AM
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Originally Posted by Racing Dan View Post
@ofajen

Possibly. Ill have to think about it.


To my mind its the same as submarine either going 10mph (om the map) in a body of water with zero current or 0mph (on the map) with 10mph current dead against it. To do that would require the exact same power in either scenario.


I dont think 1/ is relevant. Its like calculating the power of a fence post. Nonsensical. I would however be relevant if it was a ball. What power would it take to maintain position in a 30mph "head"wind? 0w because its not moving (on the map) or the power to overcome the force of the 30 mph wind? If its 0w Im sure you could get a high paying job in an off shore operation :-)


2/ 3/ I believe you are conflating map speed with relative wind speed.
No, you are confusing the power of the head wind with the power of the cyclist. The cyclist only sees the force component of the head wind power. The power the cyclist requires to move forward against that headwind increases with the headwind force, but is still Force x Velocity of the cyclist. It's just that the Force increases with headwind.

So a stationary cyclist has to resist the headwind force to prevent being blown backward, but he doesn't produce any power at the cranks until he starts pedalling forward. He is just like a fence post. There is certainly power in the airflow, but the fence post only experiences a force derived from that power. It doesn't produce any power of its own.
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Old 01-01-22, 11:25 AM
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Originally Posted by Racing Dan View Post
According to bicyclecalculator the answer is NO (but maybe YES :-)

Riding 100mph in 0mph wind takes 21508w
Riding 50mph in 50mph headwind takes 10754w


However, I can offer no explanation as to why 100 + 0 mph takes so much more power than 50 + 50, when relative wind speed seem to be all that matters. Imo they should be MUCH closer.
Tbh, I have a suspicion its an anomaly/mistake in the calculator.
Coming back to this. Riding at 100 mph against the same relative headwind force takes exactly twice as much power, because Power = Force x Velocity and in this case 100 mph is twice as fast as 50 mph. There is no mistake in the calculator.
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Old 01-01-22, 11:35 AM
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Interesting discussion and thanks for the links. Our daily wind speed average is ~15 mph and gusts to 20-25 is common. Direct head winds suck but the tail winds are wonderful and make the struggle worth the price. My normal route is straight into the wind along the ocean and tails back, this translates to 10 out and 30 back, great stuff. The difference between wind and hills is that you know the distance to the hill top.
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Old 01-01-22, 11:59 AM
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Originally Posted by PeteHski View Post
Coming back to this. Riding at 100 mph against the same relative headwind force takes exactly twice as much power, because Power = Force x Velocity and in this case 100 mph is twice as fast as 50 mph. There is no mistake in the calculator.

That may very well be how bikecalculator calculated it, but Im still convinced its wrong. Applying the same argument to the U-boat example, power to maintain position (0mph on the map) would always come to 0W no matter what speed or direction of the current, because the way You figure velocity in "Power = Force x Velocity" it would be zero, as long as the boat is not moving in relation to the map. That simply isnt true. V is the relative velocity of the fluid, be it water or air.
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Old 01-01-22, 12:05 PM
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Originally Posted by Racing Dan View Post
That may very well be how bikecalculator calculated it, but Im still convinced its wrong. Applying the same argument to the U-boat example, power to maintain position (0mph on the map) would always come to 0W no matter what speed or direction of the current, because the way You figure velocity in "Power = Force x Velocity" it would be zero, as long as the boat is not moving in relation to the map. That simply isnt true. V is the relative velocity of the fluid, be it water or air.
A submarine in water is a different system. Our bike wheels apply force to the ground, not the air. You need to solve the physics of the relevant system. If we pedaled to power a propellor and flew through the air, that would be yet a different system but more like the submarine in water.

Otto
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Old 01-01-22, 12:26 PM
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It may help if you think of the difference between using a pole to propel a punt versus paddling a boat. Different systems. The punt is more like a bike, since the force is applied to the bottom surface of the waterway, not the water.

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Old 01-01-22, 12:38 PM
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Originally Posted by Racing Dan View Post
That may very well be how bikecalculator calculated it, but Im still convinced its wrong. Applying the same argument to the U-boat example, power to maintain position (0mph on the map) would always come to 0W no matter what speed or direction of the current, because the way You figure velocity in "Power = Force x Velocity" it would be zero, as long as the booat is not moving in relation to the map. That simply isnt true. V is the relative velocity of the fluid, be it water or air.

I'm not a phycicist but this is the umpteenth time this issue has been raised on bf. One thing that always goes wrong with people's intuitive sense of what the right answer is is that they analogize to airplanes and uboats, where the vehicle is actually suspended in the moving fluid (air is a fluid for these purposes). When you're bicycling, the ground you're resting on isn't moving relative to the bike, so while you will have to do work to keep your uboat "hovering" as "standing" still is actually moving forward, the same is not true on the bike.

Maybe this clears up the intuitive confusion--analogize it to running. I can run several mph into a 25 mile headwind, but I can't get anywhere near 25 mph running when it's dead calm.

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Old 01-01-22, 01:31 PM
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Im not the confused one - you are. You continue to conflate map speed with relative speed.

when Im cycling the ground IS moving in relation to me or Im standing still, not cycling. However No one claims standing still is like hovering a boat. Standing is like anchoring a boat or digging in a fence post. Not relevant as I too said, many posts ago.

Hovering a boat is like riding the bike no matter the mechanism to generate propulsion is either a propeller in water or a wheel on the road. You are just confused because you cant ride a bike at 0mph but you can "hover" a uboat at 0mph, even in a strong current. - Thus you erroneously equate hovering the boat to standing still, feet on ground. To calculate power to hover or sail forward in a Uboat or cycle forward (not account for rolling resistance), in all cases you relative speed to air or water.
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Old 01-01-22, 01:37 PM
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Originally Posted by livedarklions View Post
I'm not a phycicist but this is the umpteenth time this issue has been raised on bf. One thing that always goes wrong with people's intuitive sense of what the right answer is is that they analogize to airplanes and uboats, where the vehicle is actually suspended in the moving fluid (air is a fluid for these purposes). When you're bicycling, the ground you're resting on isn't moving relative to the bike, so while you will have to do work to keep your uboat "hovering" as "standing" still is actually moving forward, the same is not true on the bike.

Maybe this clears up the intuitive confusion--analogize it to running. I can run several mph into a 25 mile headwind, but I can't get anywhere near 25 mph running when it's dead calm.
So scroll past. You don't have to reply to every post if you see it pop up again.
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Old 01-01-22, 03:03 PM
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Originally Posted by Racing Dan View Post
To calculate power to hover or sail forward in a Uboat or cycle forward (not account for rolling resistance), in all cases you relative speed to air or water.
Actually that is how you calculate the force of air drag. It is proportional to s**2, where s = the relative wind speed = u + v, where u = wind speed and v = rider speed, in the simple case of a direct headwind.

The power requirement to overcome that force is proportional to s**2 x v, where the rider speed, v, is simply the riderís speed over the riding surface.

If you refer back to Brandtís article, the variable names are different but the calculation is the same. Drag force = D and is proportional to W ** 2, where W is the relative wind speed and the power requirement, P is proportional to U x W **2, where U is the rider speed (relative to riding surface.).

Otto

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Old 01-01-22, 03:17 PM
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Originally Posted by Siu Blue Wind View Post
So scroll past. You don't have to reply to every post if you see it pop up again.

You misunderstood. That wasn't a complaint, I was just trying to explain where I'd picked up my info. I apologize if that wasn't clear.
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