Wind Force Equal: Explaining the Answer to Why C?

[rcgldr]

imagine that you're riding a bike on the surface of a very long treadmill ...

Situation (windless day) analogy - Wind speed wrt ground is zero. Treadmill speed wrt ground is zero. Rider is pedaling the bike at 25 kph wrt treadmill.

Situation (headwind day) analogy - Wind speed wrt ground is zero. Treadmill speed wrt ground is 20kph. Rider is pedaling the bike at 5 kph wrt treadmill.

I don't get it - you're saying you're driving off the treadmill.

In my earlier post I mentioned a very long treadmill, in this case it would have to be at least 25km long. It was just my attempt to make an analogy using a theoretical situation, not a real one.

 

[A.T.]

frame of the vehicle... how the rider interacts with the road surface is irrelevant.

No, it is not irrelevant. The rider exerts the same force on the ground in both cases. But the distance the ground moves is different.

work = force * distance

 

[D H]

Besides, the instructor didn't mention this interaction with the ground. He instead computed the work done as perceived by an observer fixed with respect to the ground. The exact same analysis would indicate that the airplane is doing more work on the calm day or that canoeist is doing more work while paddling on the lake.

These would indeed be the correct answers if the question had asked which scenario involves more effort from the perspective of the ground observer. That is not what the question asked, however. The question was clearly asking about effort expended from the perspective of the bicyclist (or the airplane pilot or the canoeist).

 

[D H]

work = force * distance

The distance is the same in the wind frame.

 

[rcgldr]

The distance is the same in the wind frame.

So that means the change in energy of the air using the two different wind frames is the same. Since there is also a force applied to the ground, that will change the energy of the earth. This process can be considered to be a conversion of potential energy from the rider into kinetic energy of a closed system: earth, air, rider, and bicycle (assuming no losses such as heat). The total kinetic energy added to the system on a windless day will be more than on a windy day regardless of the frame of reference (as long as the frame of reference is inertial (no acceleration)). It's easiest to see this using the rider as the frame of reference, decrease of energy of the air is the same (drag slows the air relative to rider) on both days, but increase of energy of the Earth is greater on the windless day (due to force times distance of the surface of Earth that moved under the rider).

 

[A.T.]

Besides, the instructor didn't mention this interaction with the ground.

I guess he assumed that everyone knows how a bicycle works.

The question was clearly asking about effort expended from the perspective of the bicyclist

Exactly. And in the frame of the bicyclist the work done on the ground by the bicyclist is different for the two cases.

 

[D H]

I guess he assumed that everyone knows how a bicycle works.

Do you? Do any of you? Do you really think that riding on the windless day takes five times the energy?

Here's yet another way to look at this nonsensical answer. Suppose that instead of riding for the same amount of time on the two days, the bicyclist instead rode the same distance on the two days. By the logic espoused in this thread (and by a teacher), the force is the same on the two days, the distance on the ground is the same on the two days, so work ("effort") is the same on the two days.

In other words, there is no such thing as wind resistance. This is of course nonsense.

 

[rcgldr]

Suppose that instead of riding for the same amount of time on the two days, the bicyclist instead rode the same distance on the two days. By the logic espoused in this thread (and by a teacher), the force is the same on the two days, the distance on the ground is the same on the two days, so work ("effort") is the same on the two days.

The work done would be the same in this case. On the first day the speed and power are 5x that of the second day, but on the second day the time spent is 5x the time spent on the first day.

Assume the force is 1,000 N.

energy = power x time = (force x speed) x time

First day = 1000 N x 25000 m / hour x 1 hour = 25000000 joules

Second day = 1000 N x 5000 m / hour x 5 hours = 25000000 joules

 

[D H]

Once again, work and energy are frame dependent quantities. Pick a nonsense frame and you get a nonsense answer. You have picked a nonsense frame. There are two sensical frames here, a frame in which the wind speed is zero (the wind frame), and a frame in which the vehicle's speed is zero (the vehicle frame). Both of these will give the same answer. If you use a motor to power the vehicle instead of a person, this answer will agree with the amount of fuel consumed.

 

[jim hardy]

These two situations are not the same "effort," whether you look at effort as work or power. Here we're dealing with a conservative force, so work is easily calculable (work is path independent for conservative forces). The work done in hauling the bag of cement up five flights of stairs is five times that of the work done in hauling it up just one flight of stairs. To perform those different amounts of work in the same amount of time requires different amounts of power as well.

Effort is ambiguous term - it can be force applied by legs or it can be degree of exhaustion at end of the hour which will be in proportion to work done.

Distance on calm day was 5X distance on windy day, drag force on bicycle was same both days..

Originally Posted by D H

If you ride a bicycle at all, you know you need to shift to a lower gear when riding into a stiff headwind due of physiological effects.

The rear tire makes 5X more revolutions on the calm day, but as DH mentions, the rider could use a 5X lower gear on the windy day. In this case rate of pedal rotations and the number of pedal rotations would be the same for both days, but on the windy day with the 5X lower gear, only 1/5th of the torque would be used.

1/5th the torque for same number of pedal-revolutions in that hour would be one fifth the work.
But I ride a one speed coaster brake "Cruiser".

I stand by my stairway analogy
though it may not have been a satisfactorily clear answer.

old jim

 

[rcgldr]

a frame in which the vehicle's speed is zero (the vehicle frame).

Using the vehicles frame of reference, the rider's power is generating a forward force against the wind, slowing the wind down, decreasing it's energy. The rider's power is also generating a backwards force against the surface of the earth, which is moving at -25 kph on the first day and at -5 kph on the second day. Assuming the pedals are geared to move at the same angular speed on both days, then the torque on the pedals on the first day is five times the torque on the pedals on the second day. The riders energy input equals torque times the angular displacement of the pedals, or torque x angular speed x time.

If time spent riding is the same, in this case 1 hour, for both days, then same amount of negative work is done on the wind (slowing it down) for both days, and more positive work is done on the Earth on the first day.

If the time spent riding on the first day is 1 hour and the time spent riding on the second day is 5 hours, then more negative work is done on the wind (slowing it down) on the second day, and the same amount of work is done on the Earth if you ignore the fact that the Earth would be accelerating. If the Earth's acceleration was taken into account, then I assume that the total energy output would be the same for both days, since the total energy input by the rider is the same for both days. This could probably be done by using the rider's impulse (force x time) to determine the change in momentum of the earth, and then using the momentum change to determine the gain in energy.

 

[willem2]

Once again, work and energy are frame dependent quantities. Pick a nonsense frame and you get a nonsense answer. You have picked a nonsense frame. There are two sensical frames here, a frame in which the wind speed is zero (the wind frame), and a frame in which the vehicle's speed is zero (the vehicle frame). Both of these will give the same answer. If you use a motor to power the vehicle instead of a person, this answer will agree with the amount of fuel consumed.

You don't do much cycling, do you?

Doing 50 kph with no wind is top-professional racing

Doing 15 khp with a 35 kph headwind is easy for for a healthy untrained person, because it only takes 30% of the power.

The power to push an object with force F, is F * (relative speed).

It would be instructive for you to find the error in all your posts.

 

[Michael C]

Once again, work and energy are frame dependent quantities. Pick a nonsense frame and you get a nonsense answer.

Well, since all frames are valid, there are really no nonsense frames: there are just frames which make the calculations easier or harder.

There are two sensical frames here, a frame in which the wind speed is zero (the wind frame), and a frame in which the vehicle's speed is zero (the vehicle frame). Both of these will give the same answer. If you use a motor to power the vehicle instead of a person, this answer will agree with the amount of fuel consumed.

In both these frames you need to take the speed of the ground into consideration. Pushing with a certain force against something moving at a particular speed needs more power than pushing with the same force against the same thing moving at a lower speed. Therefore b is the correct answer.

 

[A.T.]

Suppose that instead of riding for the same amount of time on the two days, the bicyclist instead rode the same distance on the two days. By the logic espoused in this thread (and by a teacher), the force is the same on the two days, the distance on the ground is the same on the two days, so work ("effort") is the same on the two days.

This is correct. In that case he would do the same work in both cases.

In other words, there is no such thing as wind resistance.

Of course there is resistance. That is why he has to do work in the first place. But the force of resistance is the same and the distance is the same, so work is also the same here.

Not so in the original question.

 

[jim hardy]

Let's change the analog.

Suppose the cyclist and his cycle were on a small moon someplace where there's no wind drag
but they weighed(by some fortuitous coincidence of local gravity and their mass) exactly the same as the wind drag in OP's postulate.

If he climbed straight up for one hour at 5 kph
and next day climbed straight up at 25 kph,
against same force both days,

only way he could interpret them as equivalent would be from the psychological exhilaration of the changing scenery, in my book.

 

[A.T.]

Let's change the analog.

Suppose the cyclist and his cycle were on a small moon someplace where there's no wind drag
but they weighed(by some fortuitous coincidence of local gravity and their mass) exactly the same as the wind drag in OP's postulate.

If he climbed straight up for one hour at 5 kph
and next day climbed straight up at 25 kph,
against same force both days,

only way he could interpret them as equivalent

Equivalent? Climbing 1h at 25 kph will result in a potential energy gain 5 times greater than climbing 1h at 5 kph. Where will that extra energy come from, if not from him doing more work?

The physical exhaustion is of course not simply linear here, because muscles "waste" energy and get tired, even if they do no work, just apply a static force. But I don't think the simple qualitative question was about that.

 

[jim hardy]

We agree.

But I don't think the simple qualitative question was about that.

Indeed - i don't know what was teacher's intent, unless to make people contemplate interrelation of force, work, velocity , distance and time.

At that he succeeded . We've been at it for days !

 

[russ_watters]

Do you? Do any of you? Do you really think that riding on the windless day takes five times the energy?

I must ask again: are you playing devil's advocate here? Because while you are correct in your analysis of what the wind sees, you are wrong about that being the power the rider is applying to the pedals: the pedals, gears and wheels "make" distance over the ground happen, they don't directly make distance against the wind happen. The wind only provides force for the rider to push against.

Again, your analysis suggests that the rider will expend energy standing still on a windy day or I can get a workout by sitting on my couch with a fan pointed at me.

 

[russ_watters]

I actually liked the treadmill example, but only if we use a treadmill in the way it is meant to be used: to keep whatever is on it stationary. So a few scenarios:

Scenario 1: Treadmill moving at 25km/h, fan in your face blowing at 25 km/h. This is equivalent to the first scenario in the OP. Drag force is Fd. Power delivered by the biker to the bike pedals is 25*Fd.

Scenario 2: Treadmill is moving at 5km/h, fan in your face is blowing at 25 km/h. This is equivalent to the second scenario. Drag force is Fd again, since the wind in your face is the same. Power delivered by the biker to the bike pedals is 5*Fd. The treadmill is moving 1/5 as fast, so the biker is spinning the pedals 1/5 as fast.

Scenario 3: Treadmill is not moving, fan in your face is blowing at 25 km/h. You'll have to put your feet down unless you have good balance. Whether through your feet on the ground or through your feet pushing on the pedals, you have to apply the drag force, which is still Fd. But your feet aren't moving, so you are applying no power to the bike.

Note that in all three scenarios, the relative wind speed seen by the biker is the same and the force of drag is the same, so as far as the wind knows, the power being applied to it (the wind) is the same. But as I have shown, that is not the same as the power applied by the biker to the pedals of the bike. Where does that wind power go? Well that's the trick: the rider is not applying a force directly to the wind, he's simply transferring it from the ground to the wind by applying a force with his legs onto the ground (either through the drivetrain of the bike or directly by standing). But that doesn't really answer the question: where does it go? In all three cases, all of the wind's energy is dissipated as aerodynamic drag and converted to heat, on the cyclist, by the ground.

The power being applied by the person to the drivetrain of the bike is between the ground and the person. The wind provides a force, but the power applied by your legs to the drivetrain is the same if that force is provided by a bungee cord attached to the wall, a person in front of you pushing backwards, or any other static means.

 

[russ_watters]

We also shouldn't overlook another obvious set of examples involving a standard stationary [exercise] bike.

A standard stationary bike sits on feet. It has normal pedals, a fixed gear ratio and a flywheel instead of regular wheels. Pedal power is dissipated at the flywheel, with a brake. Older bikes use a mechanical brake, which directly converts the pedal power to heat. Newer bikes use an electromagnetic brake, which converts the pedal power to electricity, then heat. Either way, the result is the same: all of the pedal power is converted to heat and not transferred to anything else (ground, wind, whatever).

Now consider a set of scenarios involving or not involving a 25kph wind, provided by a fan. Since in DH's reasoning it is the power dissipated between the person and the wind that shows up in the person's sweaty t-shirt, a 25 kph wind should produce the same sweaty t-shirt regardless of how fast the person is pedaling or even if they are pedaling at all. But of course, that's not what happens. As the power equation demands, pedal power is a function of the rpm of the pedaling and the force (torque) applied by the brake. If you hold the torque and rpm constant, the power applied by the rider to the bike (and thus manifest by his sweaty t-shirt) stays constant regardless of if the fan is on or off or even if you replace it with a person pushing on your chest.

 

[Per Oni]

Relative speed gives the force.
The same applies for flow in water: F=constant x ρ x V(relative)^2 x area. But force doesn’t equate to energy. If the cyclist decides not to cycle but just to stand still (windy or not) no energy is required. If he decides to ride then: Power = F x V (cyclist) = constant x V(relative)^2 x V(cyclist).
Since V(relative) is the same in both rides but in answer b, V(cyclist) is 5 times as high as in answer a, the power in b is 5x as high. Energy is power times 1 hour in both cases, therefore case b supplies 5 times more energy than case a.

 

[chingel]

Where does the energy go? He moves with the same speed through the wind, so he displaces the same amount of air, there is no friction in the bicycle so the only force is of the wind and since the drag factor is the same, he is pushing the Earth with the same force for the same amount of time in both cases, F=ma and the Earth gets the same acceleration. Should the Earth's energy increase just because the cyclist is moving when he applies the force? But what about a=F/m?

 

[rcgldr]

Where does the energy go? he is pushing the Earth with the same force for the same amount of time in both cases

Same impulse equals same change in momentum, so from the riders frame of reference, on the windless day the Earth surface speed increased from 25 kph to 25+ΔV kph, and on the windy day from 5 kph to 5 + ΔV kph. Assuming the Earth only rotates faster due to surface speed increase, the energy change is relative to 1/5 m (V1² - V0²) = 1/5 m (50 x ΔV) on windless day and = 1/5 m (10 x ΔV) on windy day. So there is 5x energy added to the Earth on the windless day versus windy day (if the force x time are the same).

 

[sophiecentaur]

Let's get it right. Energy / work is Force times Distance not time. Ft is momentum change!

 

[Per Oni]

This is not a simple question at all.
Flow and turbulence can be very complicated, in addition we’re dealing with a human body hence our picture becomes very clouded.

In cases such as this one I’ll try to reduce the variables as much as possible. So I imagine comparing the energy needed for my car to be parked up in a 25 kph wind and then the same car to drive 25 kph on a windless day. I’d be p*** off if I had to refill fuel just for standing still. So what’s the difference if the forces are the same?
I’d imagine that if we have to ignore tire friction, engine heat etc then when driving, the fuel energy gets transferred to higher air vortices in front and behind the car. This higher turbulence will eventually end up in a higher air KE, ie heat up the surrounding air.

I cannot see how mass comes into this, unless you mean the mass of air involved.

Edit: I just realized I’ve left out another subtle but important factor. When I’m driving the car it’s the car which is causing the turbulence, when parked the wind provides the energy.

 

[chingel]

Same impulse equals same change in momentum, so from the riders frame of reference, on the windless day the Earth surface speed increased from 25 kph to 25+ΔV kph, and on the windy day from 5 kph to 5 + ΔV kph. Assuming the Earth only rotates faster due to surface speed increase, the energy change is relative to 1/5 m (V1² - V0²) = 1/5 m (50 x ΔV) on windless day and = 1/5 m (10 x ΔV) on windy day. So there is 5x energy added to the Earth on the windless day versus windy day (if the force x time are the same).

Didn't quite catch that. How did you get that the energy change is relative to that?

 

[rcgldr]

Same impulse equals same change in momentum, so from the riders frame of reference, on the windless day the Earth surface speed increased from 25 kph to 25+ΔV kph, and on the windy day from 5 kph to 5 + ΔV kph. Assuming the Earth only rotates faster due to surface speed increase, the energy change is relative to 1/5 m (V1² - V0²) = 1/5 m (50 x ΔV) on windless day and = 1/5 m (10 x ΔV) on windy day. So there is 5x energy added to the Earth on the windless day versus windy day (if the force x time are the same).

Didn't quite catch that. How did you get that the energy change is relative to that?

Assume that all the energy change in this situation is angular, and v is surface velocity of Earth relative to rider, and that Earth is a solid uniform sphere.

E = 1/2 I ω²
for solid uniform sphere: I = 2/5 m r²
ω = v/r
E = 1/2 (2/5 m r²) (v/r)² = 1/5 m v²

The increase in energy = (1/5 m) ((v + Δv)² - v²) = (1/5 m) (v² + 2 v Δv + Δv² - v²) ~= 1/5 m 2 v Δv

 

[chingel]

OK, I follow that now, the Earth seems to have more energy because the rider is going faster, but according to a different frame from the rider's frame, let's say in a frame originally at rest to the Earth before the bicycle started riding, where does the energy go?

 

[A.T.]

let's say in a frame originally at rest to the Earth before the bicycle started riding, where does the energy go?

Into the air. On a windless day he definitively adds KE to the air. By going slowly upwind he eventually even removes KE from the air. But he cannot use that energy. It is converted into turbulence.

 

[rcgldr]

In a frame originally at rest to the Earth before the bicycle started riding, where does the energy go?

Keep in mind that the rider is generating 5 times the power into the system on the windless day versus the windy day, the force at the rear tire is the same, but the speed on the windless day is 5x that of the windy day. Also the frame of reference initially at rest with the earth, is inertial and doesn't accelerate with the surface of the earth.

So the impulse to the Earth is the same, resulting in the same increase in velocity and energy. On the windless day, energy is added to the air by the rider. On the windy day, energy is extracted from the air (and transferred to the earth). Total energy of this system is conserved (assuming no losses), so after 1 hour, the total kinetic energy is increased by the decrease in rider potential energy, which is 5x more on the windless day. In this frame, the total energy change of the Earth is the same on both days. On the windless day, the rider provides all of the energy added to the Earth and wind. On the windy day, the rider supplies only a fraction of the energy added to the Earth and the slowing the wind provides most of the energy added to the earth.