# Wind Force Equal: Explaining the Answer to Why C?

• 909kidd
In summary, the question discusses two different scenarios where a person is riding a bicycle on a flat road. In the first scenario, the person rides at a speed of 25 km/h for one hour on a windless day. In the second scenario, the person rides at a speed of 5 km/h for one hour against a headwind of 20 km/h. The question asks which ride seemed to take more effort. Many people initially answered that both rides were about the same, but the correct answer is b) the ride on the windless day was harder. This is because power, which is equal to force times speed, is a more accurate measure of effort in this case. The power required to move at a speed of 25
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.

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?

russ watters has posted the correct answer below.

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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?

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.

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.

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.

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 said:
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 said:
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.

russ_watters said:
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 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.

D H said:
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.

rcgldr said:
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.

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.

rcgldr said:
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.

W=F x s
F= m x a
here a=(a1 due to motion - a2 of wind)
S=SPEED/T
So s is inversely proportional to T.
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.

sophiecentaur said:
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.

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.

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.

sophiecentaur said:
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?"

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D H said:
chingel said:
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.

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.

jim hardy said:
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.

jim hardy said:
But the pedals made 5X more revolutions on the calm day.

D H said:
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.

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.

sophiecentaur said:
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.

rcgldr said:
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.

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.

D H said:
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 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.
D H said:
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.

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sophiecentaur said:
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.

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

russ_watters said:
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)?

D H said:
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.
D H said:
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.

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A.T. said:
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 said:
imagine that you're riding a bike on the surface of a very long treadmill ...

rcgldr said:
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.

russ_watters said:
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.

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D H said:
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

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).

A.T. said:
work = force * distance
The distance is the same in the wind frame.

D H said:
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).

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