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Headwind and Power

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Headwind and Power

Old 01-01-22, 03:29 PM
  #26  
Racing Dan
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Stop acting as if you are lecturing me. You are not. I know the difference between force and power.


You can believe what you want, nothing I say can make you reconsider and Im not playing any further. Try playing with some numbers and you soon realize the way you mix rider speed (map speed) and relative speed will give you some obvious erroneous results as pointed out many posts ago.
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Old 01-01-22, 03:38 PM
  #27  
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Originally Posted by Racing Dan View Post
Stop acting as if you are lecturing me.
That works.

Otto
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Old 01-01-22, 03:59 PM
  #28  
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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?
Is there bacon involved?
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Old 01-01-22, 04:11 PM
  #29  
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do those calculators take into account spoke drag? riding 50 in a 50 tailwind would take a lot of juice to overcome the fan effect of the spokes me thinks,

to illustrate turn your bike upside down and hand crank the pedals to see how fast you can spin the wheels.
and notice how unbalanced bike wheel are.
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Old 01-01-22, 04:16 PM
  #30  
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Originally Posted by cjenrick View Post
do those calculators take into account spoke drag? riding 50 in a 50 tailwind would take a lot of juice to overcome the fan effect of the spokes me thinks,

to illustrate turn your bike upside down and hand crank the pedals to see how fast you can spin the wheels.
and notice how unbalanced bike wheel are.
That drag is just because the bike is upside down. The same thing happens with an upside down boat.
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Old 01-01-22, 04:31 PM
  #31  
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Originally Posted by cjenrick View Post
do those calculators take into account spoke drag? riding 50 in a 50 tailwind would take a lot of juice to overcome the fan effect of the spokes me thinks,

to illustrate turn your bike upside down and hand crank the pedals to see how fast you can spin the wheels.
and notice how unbalanced bike wheel are.
Good question. Id expect the air drag term to just have a coefficient for frontal surface area times relative wind speed squared times velocity of the rider. Probably no allowance in that term for spoke drag at high speed. I also doubt it gets put into the terms for rolling losses or drivetrain losses.

Otto
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Old 01-01-22, 05:13 PM
  #32  
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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.
In your submarine analogy, if no power was applied by the propellor, then the submarine would simply flow along with the current as there are no other forces opposing the flow. So the sub would have a relative speed against the current of zero and zero power output. When you power up the propellor you are applying a force directly against the flow of the moving water (not the fixed sea bed/map reference point), so the relative speed of the water against the propellor is what matters in that case. So a sub moving at 10 mph through still water is the same as a sub moving at 0 mph against a 10 mph current. Power required is the same in both cases.

A bicycle is different because there is friction between the wheels and ground and you are applying power against the ground, not directly against the air flow. So if you just sit stationary on your bike in a headwind, there is no power required on your part to stay in position unless the wind force is so strong as to overcome the friction against the ground. When you apply power to the cranks you move forward relative to the ground where you are applying force. The relative air speed only affects the force opposing your movement. So riding at 50 mph in a 50 mph headwind produces the same opposing wind force as riding at 100 mph with no headwind. But your velocity relative to the ground (where you are applying your force) is double at 100 mph and therefore requires double the power. The opposing wind force remains the same in both cases, but you are pedalling twice as fast.

Now if you were riding along in mid air with some sort of propellor, then your power equation would be the same as for the sub i.e. Power = force x air speed. Ground speed would be irrelevant.

Think about it. If power required was simply a matter of relative airspeed then with a 30 mph tailwind you would be bowling along at 50+ mph with relatively little effort. Similarly with a 30 mph headwind you wouldn't be able to move forward at all.

@ofagen's equations in post #25 show the relationship between the Drag Force (dependent on relative airspeed) and Power (dependent on both drag force and rider ground speed).

Last edited by PeteHski; 01-01-22 at 05:35 PM.
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Old 01-01-22, 07:06 PM
  #33  
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Originally Posted by Racing Dan View Post
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.

So if I'm on a stationary bike pedaling like crazy going nowhere, I'm working harder if I have a fan blowing in my face? This makes no sense.

​​​​The question is how much effort am I putting into the propulsion of the bicycle under the various conditions. We can have an infinite number of combinations of bicycle and wind speed to produce the same 30 mph difference on level ground between them, a continuum from headwind 30 mph/bike 0 mph to bike 30 mph/headwind 0 mph. We'll ignore the 0 mph bike scenario because that seems to bother you so much, let's assume we're in the bike 2 mph/headwind 28 mph. What you're failing to notice is that the bicyclist's efforts are producing very near zero of that differential, that's as opposed to the 30 mph cyle/0 mph headwind scenario. In that scenario the cyclist's effort is producing 100% of the differential. In other words, the cyclist is not only resisting the differential, she's actually generating it. So, basically, in scenario one, the wind is doing almost all of the work relative to the rider, while in scenario two, the rider is doing all of the work relative to the air. Same 30 mph differential, completely different energy inputs.

BTW, if you tether the submarine to an anchor point yes, you are making the submarine analogous to us non-floating cyclists. Congratulations.

Also, if you factor out friction with the ground, your bike isn't going anywhere.
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Old 01-01-22, 09:40 PM
  #34  
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Originally Posted by livedarklions View Post
So if I'm on a stationary bike pedaling like crazy going nowhere, I'm working harder if I have a fan blowing in my face? This makes no sense.
Since the bike is stationary, there is no work, hence power, involved in moving it. Adding the headwind might double it, but two times zero is still zero.

While correct, the above's not really on point. The force to pedal the stationary bike doesn't depend on the environment, neglecting possible changes in the resistance unit with temperature if not using a smart trainer, so a fan makes no difference.

Last edited by asgelle; 01-01-22 at 09:51 PM.
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Old 01-02-22, 06:25 AM
  #35  
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An App for the psychics challenged.


https://www.gribble.org/cycling/power_v_speed.html
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Old 01-02-22, 06:29 AM
  #36  
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Originally Posted by asgelle View Post
Since the bike is stationary, there is no work, hence power, involved in moving it. Adding the headwind might double it, but two times zero is still zero.

While correct, the above's not really on point. The force to pedal the stationary bike doesn't depend on the environment, neglecting possible changes in the resistance unit with temperature if not using a smart trainer, so a fan makes no difference.

See the part you quoted where I said "this makes no sense"? I was responding to an argument we were confused because we couldn't cycle at 0 mph. Point is that a " hovering" submarine resisting a current is actually "moving" relative to the water it's floating on. We're not "resting" on the air the way the sub is floating in the water, so unless we're actually going forward relative to the ground, the air flow makes no difference.

BTW, the notion we're not doing work on a trainer is obviously wrong. The "distance" is really measured in turns of the rollers rather than miles-- our work is moving the rear wheel, as it always is on a bicycle. If you think about it, distance in miles when we're riding outside is really a proxy measure for how many times we've rotated the rear wheel.

Last edited by livedarklions; 01-02-22 at 07:29 AM.
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Old 01-02-22, 07:43 AM
  #37  
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Originally Posted by asgelle View Post
Since the bike is stationary, there is no work, hence power, involved in moving it. Adding the headwind might double it, but two times zero is still zero.


How do people manage to say things that are obviously wrong?

Di you really think power meters on stationary bikes are showing zero?

How do generators work?

What is torque?

Last edited by njkayaker; 01-02-22 at 07:46 AM.
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Old 01-02-22, 08:36 AM
  #38  
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Originally Posted by njkayaker View Post
How do people manage to say things that are obviously wrong?
You seem to manage quite easily. Do you understand that while the pedals on a stationary bike are moving, the bike as a whole is not?
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Old 01-02-22, 09:09 AM
  #39  
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Originally Posted by asgelle View Post
You seem to manage quite easily. Do you understand that while the pedals on a stationary bike are moving, the bike as a whole is not?


You ignored the rest of my post and just repeated the same stupid nonsense.

Generators are stationary too.

Work is being done with moving the pedals. Why do you keep ignoring that?

https://www.engineeringtoolbox.com/w...ue-d_1377.html

Last edited by njkayaker; 01-02-22 at 09:20 AM.
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Old 01-02-22, 09:10 AM
  #40  
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Originally Posted by njkayaker View Post
Work is being done by the pedals moving. Why do you keep ignoring that?
Because the issue was the effect of a fan blowing on a stationary bike.
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Old 01-02-22, 10:27 AM
  #41  
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A fan blowing on a stationary bike has no effect on the power required to pedal it, which is solely dependent on the resistance of the trainer and how fast you pedal it. Trainer power is proportional to pedal force x cadence
You can sit on your trainer with a fan blowing and you need to produce zero power to resist it. I can't believe people are even questioning this!??
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Old 01-02-22, 10:33 AM
  #42  
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Originally Posted by njkayaker View Post


How do people manage to say things that are obviously wrong?

Di you really think power meters on stationary bikes are showing zero?

How do generators work?

What is torque?
In this case Torque = pedal force x crank arm length
Power = Torque x cadence

The fan is a complete red herring. You are not pedalling against the force of the fan, you are pedalling against the resiistance force of the trainer. The force of the fan is resisted statically and requires no extra work/power to overcome.
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Old 01-02-22, 10:59 AM
  #43  
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Tangential, but interesting:

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Old 01-02-22, 12:11 PM
  #44  
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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.
It's not a mistake in the calculator, it's just a consequence of how kinematics work. A simpler way to visualize this is not to compare 100 + 0 with 50 + 50, but with 0 + 100.

In 0 + 100, the power requirement is zero. The bicycle needs to have some force against the ground to oppose the force from the wind, but this will be achieved with everything nonmoving, so no work is being done.

However, what if we put the bicycle onto a 100mph treadmill?

In order to counteract the force from the wind, the bicycle still needs to produce a force with the ground (now a treadmill). But since the treadmill is moving, putting force on it in the direction of that motion implies work being done: the treadmill can be thought of as acting as a reaction drive. Actually: the force is the same as in the 100 + 0 case since the relative wind speed is the same, and the velocity of the reaction material is equal to the forward velocity of the bicycle from the 100 + 0 case, so the resulting power (F*v) is the same.

Also observe: 0 + 100 on a treadmill is the same exact scenario as the original 100 + 0 case, if you choose the bicycle as your frame of reference.

Last edited by HTupolev; 01-02-22 at 12:15 PM.
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Old 01-02-22, 12:23 PM
  #45  
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Originally Posted by PeteHski View Post
A fan blowing on a stationary bike has no effect on the power required to pedal it, which is solely dependent on the resistance of the trainer and how fast you pedal it. Trainer power is proportional to pedal force x cadence
You can sit on your trainer with a fan blowing and you need to produce zero power to resist it. I can't believe people are even questioning this!??
Exactly. I brought up the fan just to demonstrate how absurd the other poster's assertions about air speed vs. forward speed differential always being critical to defining work was. He said we weren't able to understand the relationship because it's impossible to ride a bike without going forward, and it's my understanding that some people, including yourself, actually perform that supposedly impossible feat on a regular basis. I didn't just take it for granted that no one would think I was claing air in the face of a stationary bike rider would increase the work, I said explicitly it "made no sense". For some reason, asgelle seemed to think it necessary to explain to me why the thing I said made no sense made no sense. Unfortunately, he then proceeded to make an absurd statement about a stationary bike rider not doing any work.
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Old 01-02-22, 12:24 PM
  #46  
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Originally Posted by HTupolev View Post
It's not a mistake in the calculator, it's just a consequence of how kinematics work. A simpler way to visualize this is not to compare 100 + 0 with 50 + 50, but with 0 + 100.

In 0 + 100, the power requirement is zero. The bicycle needs to have some force against the ground to oppose the force from the wind, but this will be achieved with everything nonmoving, so no work is being done.

However, what if we put the bicycle onto a 100mph treadmill?

In order to counteract the force from the wind, the bicycle still needs to produce a force with the ground (now a treadmill). But since the treadmill is moving, putting force on it in the direction of that motion implies work being done: the treadmill can be thought of as acting as a reaction drive. Actually: the force is the same as in the 100 + 0 case since the relative wind speed is the same, and the velocity of the reaction material is equal to the forward velocity of the bicycle from the 100 + 0 case, so the resulting power (F*v) is the same.

Also observe: 0 + 100 on a treadmill is the same exact scenario as the original 100 + 0 case, if you choose the bicycle as your frame of reference.
Yeah the treadmill analogy is useful in understanding. You can also use it to show the fundamental difference between 100 mph road speed with no headwind vs 50 mph road speed with a 50 mph headwind. In the first case you run the treadmill at 100 mph with a 100 mph headwind and in the second case you run the treadmill at 50 mph with the same 100 mph headwind, therefore halving the power requirement. The air speed and therefore drag force is the same in both cases, but in the second case you only have to push against it at half the ground speed (treadmill speed in this case)
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Old 01-02-22, 12:33 PM
  #47  
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Originally Posted by asgelle View Post
You seem to manage quite easily. Do you understand that while the pedals on a stationary bike are moving, the bike as a whole is not?
Originally Posted by asgelle View Post
Because the issue was the effect of a fan blowing on a stationary bike.

No the fan blowing on the stationary rider wasn't the issue, you completely missed the point of my post.
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Old 01-02-22, 12:34 PM
  #48  
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Originally Posted by HTupolev View Post
It's not a mistake in the calculator, it's just a consequence of how kinematics work. A simpler way to visualize this is not to compare 100 + 0 with 50 + 50, but with 0 + 100.

In 0 + 100, the power requirement is zero. The bicycle needs to have some force against the ground to oppose the force from the wind, but this will be achieved with everything nonmoving, so no work is being done.

However, what if we put the bicycle onto a 100mph treadmill?

In order to counteract the force from the wind, the bicycle still needs to produce a force with the ground (now a treadmill). But since the treadmill is moving, putting force on it in the direction of that motion implies work being done: the treadmill can be thought of as acting as a reaction drive. Actually: the force is the same as in the 100 + 0 case since the relative wind speed is the same, and the velocity of the reaction material is equal to the forward velocity of the bicycle from the 100 + 0 case, so the resulting power (F*v) is the same.

Also observe: 0 + 100 on a treadmill is the same exact scenario as the original 100 + 0 case, if you choose the bicycle as your frame of reference.
Omg! Just stop :-) This was addressed several times before. Including by me in #13 and Im sure in other posts too since you all seem to throw this at me, even if I made no claim to the contrary.

This thread has devolved in to nonsense and Im out of here as I said in #27
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Old 01-02-22, 12:45 PM
  #49  
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Originally Posted by HTupolev View Post
It's not a mistake in the calculator, it's just a consequence of how kinematics work. A simpler way to visualize this is not to compare 100 + 0 with 50 + 50, but with 0 + 100.

In 0 + 100, the power requirement is zero. The bicycle needs to have some force against the ground to oppose the force from the wind, but this will be achieved with everything nonmoving, so no work is being done.

However, what if we put the bicycle onto a 100mph treadmill?

In order to counteract the force from the wind, the bicycle still needs to produce a force with the ground (now a treadmill). But since the treadmill is moving, putting force on it in the direction of that motion implies work being done: the treadmill can be thought of as acting as a reaction drive. Actually: the force is the same as in the 100 + 0 case since the relative wind speed is the same, and the velocity of the reaction material is equal to the forward velocity of the bicycle from the 100 + 0 case, so the resulting power (F*v) is the same.

Also observe: 0 + 100 on a treadmill is the same exact scenario as the original 100 + 0 case, if you choose the bicycle as your frame of reference.

Maybe the best way to look at this is that we're really discussing countervailing work. If there's a headwind, that indicates atmospheric conditions are doing the work of pushing air into our face. When there's no wind, any air resistance is caused by our work of pushing our face through the air.

When there's a tail wind, some portion of the work propelling the bike is being down by the wind. There isn't less work to keep the bike at 30 mph in a tailwind, we're just doing less of it ourselves.
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Old 01-02-22, 12:47 PM
  #50  
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Originally Posted by Racing Dan View Post
Omg! Just stop :-) This was addressed several times before. Including by me in #13 and Im sure in other posts too since you all seem to throw this at me, even if I made no claim to the contrary.
I wasn't intending to "throw it at you" or imply that you believe something contrary to it. I brought it up because I know that you understand it, to use it as a tool to explain the part that you seemed to be having a hard time understanding.
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