Wind Force Equal: Explaining the Answer to Why C?

[909kidd]

Against the Wind

One windless day you ride your bicycle at 25 km/h for one hour on a flat road. The next
day, you ride 5 km/h into a 20-km/h headwind for one hour on a flat road. Assuming that
the force of the wind (the primary "friction" in this case) scales with the square of the
relative velocity of the bike and wind, the wind's force is the same on both days. Which
ride seemed to take more effort on your part?
(a) The ride on the windy day was harder.
(b) The ride on the windless day was harder.
(c) Both rides were about the same.
Explain your answer.

This not Homework or anything. My class was discussing this Question, and must of use came up with the C), but not a lot of us came with the same Why?

I assumed that Force in both cases are Zero, but the work done on day one would be greater, because I am going at a 5x the speed on the first day. I know my answer is wrong. But can anyone explain why the answer is C?

 

[rcgldr]

russ watters has posted the correct answer below.

 

[909kidd]

But if the wind is blowing directly at you at 20KM/hr, then wouldn't the 5km/hr day be harder, since the first day is windless?

 

[russ_watters]

The answer is b, even if we ignore friction in the bike. Power is force times speed, so while the force is the same, the speed in b is much higher, so the power is much higher.

 

[rcgldr]

It might be easier to understand this if you imagine that you're riding a bike on the surface of a very long treadmill, and moving at 25 kph with zero wind. Situation (b) (windless day) day would correspond to the treadmill not moving, so the rider is supplying all the power in order to go 25 kph. Situation (a) (windy day) would correspond to the treadmill moving at 20 kph, supplying most of the power, with the rider only having to supply the power to move an additional 5 kph relative to the treadmill.

 

[russ_watters]

Er, no. If there is no wind and the bike is moving at 25kph, then that's like the treadmill moving at 25kph, with a fan blowing in your face at 25kph. In the other situation, you're riding 5kph with a fan blowing at 25kph.

 

[sophiecentaur]

It's a pity the word "effort" was used in the question because it is a bit ambiguous in this context. Effort is a term used for the Force applied to a Machine and is an 'instantaneous' quantity,
If the question is referring to the total Energy involved then pedalling against the wind with the same force as in still air will involve going slower - hence taking longer over the journey. This will involve more energy because you are doing the same amount of work against the wind per second - but for longer. Muscles are impossible to analyse accurately but you could imaging that you might drop a gear to go against the wind so your 'effort' force was the same in both cases. Hence, the Energy input would be proportional to the time taken. I'd go for 'a' - with my reading of the question.

 

[rcgldr]

It might be easier to understand this if you imagine that you're riding a bike on the surface of a very long treadmill, and moving at 25 kph with zero wind. Situation (b) (windless day) day would correspond to the treadmill not moving, so the rider is supplying all the power in order to go 25 kph. Situation (a) (windy day) would correspond to the treadmill moving at 20 kph, supplying most of the power, with the rider only having to supply the power to move an additional 5 kph relative to the treadmill.

If there is no wind and the bike is moving at 25kph, then that's like the treadmill moving at 25kph, with a fan blowing in your face at 25kph. In the other situation, you're riding 5kph with a fan blowing at 25kph.

Apparently I worded that badly, so a redo:

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.

 

[D H]

The answer is b, even if we ignore friction in the bike. Power is force times speed, so while the force is the same, the speed in b is much higher, so the power is much higher.

The windless day (answer b) is harder? That doesn't make sense. Ride a bike the same distance against a considerable headwind versus a calm day. You'll feel a much more tired on reaching the destination on the windy day compared to the calm day. However, this is not a problem of riding equal distances with respect to the ground. It is a problem of riding for equal amounts of time.

The correct answer is c.

The problem here is one of mixing reference frames. Energy (and thus work and power) are frame-dependent quantities. Ground speed is irrelevant since we're ignoring rolling friction, so using ground speed as the basis for reasoning is a red herring. Much better is to use a frame moving with the flowing air. This is the same as the ground frame in the case of a windless day, but not on the windy day.

In both cases (windy versus windless), the bicyclist is moving at 25 km/hr relative to the wind frame, making the aerodynamic force against the bicyclist is the same in both cases. The force exerted by the bicyclist is also the same in both cases by Newton's third law. In both cases, the distance traveled in this wind frame is the same, 25 kilometers. Same force, same distance, same amount of time means the work against the non-conservative aerodynamic drag is the same in both cases.

Another way to get an apples-to-apples comparison is to look at things from the perspective of the bicyclist. The bicyclist is exerting the same amount of power in both cases. Work is average power times time, so once again, the answer is c.

 

[rcgldr]

In both cases, the distance traveled in this wind frame is the same, 25 kilometers.

Except the point of application of the force is at the ground. If the rider was driving a propeller to move the bicycle, then the power would be the same, but instead the rider power is output to the ground, in which case I think my treadmill analog may help explain the situation.

 

[D H]

Except the point of application of the force is at the ground.

That's an irrelevant fact because we are ignoring ground friction here.

If the rider was driving a propeller to move the bicycle, then the power would be the same, but instead the rider power is output to the ground, in which case I think my treadmill analog may help explain the situation.

Your treadmill adds yet an unnecessary complication to the already unnecessary complication added by considering ground speed. Your propeller-driven bicycle is in fact a much better analogy.

Here's another analogous situation, using a canoeist rather than a bicyclist. On the first day, the canoeist paddles five kilometers across a lake in one hour. On the next day, the canoeist again paddles for one hour, but this time paddles a kilometer upstream in a river flowing at four km/hr. To an observer standing still on the lake shore / river shore, the paddler will have performed more work on the first day compared to the second. To the canoeist, the two days are of course identical.The question asked "Which ride seemed to take more effort on your part?" It did not ask "Which ride seems to take more effort from the perspective an observer fixed with respect to the ground?" Energy is a frame dependent quantity. One has to be very careful that one is comparing apples to applies when looking at energy.

 

[sophiecentaur]

Except the point of application of the force is at the ground. If the rider was driving a propeller to move the bicycle, then the power would be the same, but instead the rider power is output to the ground, in which case I think my treadmill analog may help explain the situation.

I think it adds to the confusion, actually (but I see where you're coming from).
If you assume that the cyclist can't tell what is causing the 'moving air' effects (energy loss) then he is constantly using the same Power in overcoming them. The journey into the wind will take him longer - hence more energy is needed for the whole journey.

 

[anuragkanase]

W=F x s
F= m x a
here a=(a1 due to motion - a2 of wind)
So the answer goes b.
S=SPEED/T
So s is inversely proportional to T.
Answer goes a.
Taking over the overall situation, the whole Work done in Unit TIME, we are equating both thee equations. So the answer and the CORRECT answer is C.
The answer will totally change when you consider aerodynamics.
P. S. Correct me if I am wrong.

 

[D H]

I think it adds to the confusion, actually (but I see where you're coming from).
If you assume that the cyclist can't tell what is causing the 'moving air' effects (energy loss) then he is constantly using the same Power in overcoming them. The journey into the wind will take him longer - hence more energy is needed for the whole journey.

The two journeys do not cover the same ground distance in this problem. They are instead of the same time duration. The same amount of power expended during the same amount of time means the same amount of energy will be expended in the two cases -- from the bicyclist's perspective. Other observers will have different thoughts because energy is a frame dependent quantity.

 

[chingel]

Since the wind is at the same speed relative to him in both cases, the drag force is the same. But since he is pushing the ground, in one case he has to push the ground a longer distance, since the ground is moving faster relative to him. Work is force over a distance so he has to do more work in case b.

 

[sophiecentaur]

A touch of the RTFQ, here. Woops!
If he's doing the same amount of work for the same time against 'air' then the other friction effects will be greater if he travels further (force times distance). That must mean that the longer journey over the ground (b) must use up more total energy.

a, b or c, depending on how carefully or pickily you read the question. Excellent conversations piece.

 

[D H]

A touch of the RTFQ, here. Woops!
If he's doing the same amount of work for the same time against 'air' then the other friction effects will be greater if he travels further (force times distance). That must mean that the longer journey over the ground (b) must use up more total energy.

Read the question even more carefully. This answer, b[/b], is an answer to the question "Which ride seems to take more effort from the perspective an observer fixed with respect to the ground?"

That is not the question that was asked. The question that was asked was "Which ride seemed to take more effort on [the bicyclist's] part?"

 

[sophiecentaur]

Read the question even more carefully. This answer, b[/b], is an answer to the question "Which ride seems to take more effort from the perspective an observer fixed with respect to the ground?"

That is not the question that was asked. The question that was asked was "Which ride seemed to take more effort on [the bicyclist's] part?"

The sports commentator? It would all depend on what sums the ground-based observer was doing. I wouldn't have to be on the bike to get the answer right. (Assuming I'd bothered to RTFQ properly) I would be 'empathising' with that poor cyclist.

 

[jim hardy]

I think i agree with Russ in #4. and Sophie.
In both cases he senses a 25mph headwind... so the force he applies to the pedals is the same. As in walking up a flight of stairs.

But the pedals made 5X more revolutions on the calm day.

So it's semantics - what is meant by "seems to take more effort" ?
I might be able to carry an eighty pound sack of cement up one flight of stairs in perhaps a minute.
But if I take same cement bag and try to run up five flights in that same minute, my pacemaker will go into overdrive,
Even though my legs are applying same "effort".

Isometrics may be good for the muscle but not so much for cardiovascular.

Nice brain teaser indeed.

 

[D H]

But the pedals made 5X more revolutions on the calm day.

Says who? 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.

So it's semantics - what is meant by "seems to take more effort" ?

I would take it too mean "how much energy did the bicyclist expend?"

It's probably best to separate physiological work from physical work here. To see the difference, extend your arm straight out, palm up, with a book resting on your palm. Hold your arm steady. Even though the physical work is identically zero, your arm will quickly tell you otherwise.

So, instead of a bicyclist, think of the question in terms of a motorized vehicle that is subject to zero rolling friction and is outfitted with an ideal continuously variable transmission, and with "effort" interpreted as "amount of gas consumed."

I might be able to carry an eighty pound sack of cement up one flight of stairs in perhaps a minute. But if I take same cement bag and try to run up five flights in that same minute, my pacemaker will go into overdrive, Even though my legs are applying same "effort".

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.

 

[rcgldr]

But the pedals made 5X more revolutions on the calm day.

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.

Since the original post mentions effort versus time, it doesn't matter than the rider only covered 1/5th the distance (relative to the ground), on the windy day.

 

[909kidd]

Okay, so today we discussed the question again and the answer was approved by the Professor.
W=F*s
No Wind: 1hr= 25km
Wind: 1hr=5km
25/5= 5km

Day 1 with 25km/hr worked 5x harder then the 5km/hr therefore the Answer was, (b) The ride on the windless day was harder.

 

[russ_watters]

It's a pity the word "effort" was used in the question because it is a bit ambiguous in this context. Effort is a term used for the Force applied to a Machine and is an 'instantaneous' quantity...

The word "effort" implies to me, 'that which makes you sweat', which is power and the dictionary agrees.

Muscles are impossible to analyse accurately but you could imaging that you might drop a gear to go against the wind so your 'effort' force was the same in both cases.

Yes, the force at the ground is the same, but dropping a gear means dropping the force provided by your legs because of the improved gear ratio. So again, moving slower means lower power.

 

[russ_watters]

Apparently I worded that badly, so a redo:

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.

I don't get it - you're saying you're driving off the treadmill at 25kph, which would likely make you crash. Why would you do that?

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.

Again, you're driving the bike off the front of the treadmill.

I don't see how invoking the treadmill in these contexts is useful.

 

[sophiecentaur]

My point was that the only valid objective measurement using muscles would be when the gears are enabling you to use identical power input for both cases.

 

[russ_watters]

The windless day (answer b) is harder? That doesn't make sense. Ride a bike the same distance against a considerable headwind versus a calm day. You'll feel a much more tired on reaching the destination on the windy day compared to the calm day.

The OP says nothing about distance, only speeds and time.

However, this is not a problem of riding equal distances with respect to the ground. It is a problem of riding for equal amounts of time.

Yes, so why did you bring it up...?

The correct answer is c.

The problem here is one of mixing reference frames. Energy (and thus work and power) are frame-dependent quantities.

I see no such problem: we're talking about the bike rider in both cases. By your reasoning, sitting on my couch with a fan in my face requires effort!

I don't know if you are saying this to play devil's advocate, but since other posters have made a mess of this, I think it would be best to make the correct answer clear before doing that.

Says who? 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.

Changing gear ratios will only confuse the situation so it is best not to do it: Power supplied in anyone scenario is the same regardless of the gear.

 

[russ_watters]

My point was that the only valid objective measurement using muscles would be when the gears are enabling you to use identical power input for both cases.

? You're begging the question. "What is the power input?" is the question we are trying to answer.

And besides which, your formulation doesn't provide equal power input. Your example doesn't support your intended formulation.

[edit] Didn't read all your posts though - it seems you basically got it figured out.

 

[D H]

The OP says nothing about distance, only speeds and time.

Exactly. Because rolling friction is deemed a non-factor, my contention is the only speed that counts is the speed relative to the wind. The speed with respect to the wind is the same in the two scenarios, the time is the same in the two scenarios, so "effort" as perceived by the rider is the same in the two scenarios. The speed relative to the ground is a red herring, just as ground speed is irrelevant to the problem of how much fuel an airplane consumes. All that matters is air speed and time.

Yes, so why did you bring it up...?

Because some other user who later read the fine question brought it up earlier.

I see no such problem: we're talking about the bike rider in both cases. By your reasoning, sitting on my couch with a fan in my face requires effort!

Changing gear ratios will only confuse the situation so it is best not to do it: Power supplied in anyone scenario is the same regardless of the gear.

Agreed.How does this problem differ from an airplane flying for one hour in calm air at 50 km/hr versus an airplane flying for one hour at a ground speed of 10 km/hr against a wind blowing at 40 km/hr (relative to the ground), or from my previous example of a canoeist (see post #11)?

 

[A.T.]

How does this problem differ from an airplane

Look at it from the frame of the vehicle:
- An airplane is not doing any work on the ground, therefore ground speed is not relevant.
- A bicyclist is doing positive work on the ground, therefore ground speed is relevant.

or from my previous example of a canoeist

Same as the airplane. He is doing work on a medium that moves at the same speed relative to him in both cases. But the bicyclist is doing work on the ground which moves at different speeds relative to him in the two cases.

The answer is b.

 

[D H]

Look at it from the frame of the vehicle:
- An airplane is not doing any work on the ground, therefore ground speed is not relevant.
- A bicyclist is doing positive work on the ground, therefore ground speed is relevant.

No, it isn't.

Look at it once more from the frame of the vehicle. In a real bicycle, the rider is slowing the wind down a bit and is changing the Earth's rotation rate by an imperceptibly small (but non-zero) amount. Here, rolling friction is non-existent, so that all that the rider is doing slowing the wind down a bit. The work the rider is doing against the wind is the same in the two scenarios. Since rolling friction is non-existent, how the rider interacts with the road surface is irrelevant.

 

[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.