Headwind and Power
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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
...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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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
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
Last edited by pennpaul; 12-31-21 at 08:22 PM.
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Ground effect versus “anemometer” wind speed presented by the local weather station is not equivalent.
So, best bet is, local weather station says ‘10mph wind’ you get a third that actual ground level true wind effect.
They don’t hide the weather station anemometer at 3 feet off the ground behind a hedge.
So, best bet is, local weather station says ‘10mph wind’ you get a third that actual ground level true wind effect.
They don’t hide the weather station anemometer at 3 feet off the ground behind a hedge.
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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?
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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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
Sorry. Typed too fast first time. Power, not work.
Otto
Last edited by ofajen; 01-01-22 at 01:24 AM.
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Otto
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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/
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/
Last edited by Racing Dan; 01-01-22 at 09:39 AM.
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Are we being tested on this? It wasn't listed in the syllabus.
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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/
Tbh, I have a suspicion its an anomaly/mistake in the calculator.
bikecalculator.com/
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 don’t 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 I’m moving at 15 mph against that force. That takes a considerable amount of work. For me it’s probably in the neighborhood of 350 watts because don’t 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 I’m 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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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?
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?
Last edited by PeteHski; 01-01-22 at 10:57 AM.
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#12
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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.
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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Simple. It’s 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 don’t 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 I’m moving at 15 mph against that force. That takes a considerable amount of work. For me it’s probably in the neighborhood of 350 watts because don’t 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 I’m 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
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 don’t 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 I’m moving at 15 mph against that force. That takes a considerable amount of work. For me it’s probably in the neighborhood of 350 watts because don’t 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 I’m 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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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
Last edited by ofajen; 01-01-22 at 11:24 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.
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.
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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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.
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.
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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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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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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.
Otto
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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.
Otto
Otto
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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.
Last edited by livedarklions; 01-01-22 at 12:46 PM.
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#22
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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.
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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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.
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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The power requirement to overcome that force is proportional to s**2 x v, where the rider speed, v, is simply the rider’s speed over the riding surface.
If you refer back to Brandt’s 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
Last edited by ofajen; 01-01-22 at 03:27 PM.
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