Where is the 0 point?

[Paul2]

Forgive me if this is a simple question.

Where is the direction from which wind blowing on a bicyclist will not hinder nor help the bicyclist on a straight path?

Assume that the wind blowing from directly behind a bicyclist will push the bicyclist forward +X amount, and the wind blowing from directly in front of the bicyclist will help -X amount (that is, hinder as much as the wind blowing from behind helps).

Because the effect of the wind goes from +X to -X, and because the direction of the wind can be altered continuously through a semi-circle going from directly behind to directly in front (the semi-circle's endpoints are on the bike's path), there must be a direction at which the effect of the wind is 0.

Common sense says that that direction is when the wind is perpendicular to the path of the bicycle. But I then considered what the effect of *no wind* would be. That effect is obviously 0. But how could no wind blowing be the same effect as a perpendicular wind? Surely the perpendicular wind would have some small effect. That is, the perpendicular wind will push the bicyclist to the side to some small extent, so with the bicyclist putting the same effort and energy into going forward, the bicyclist's path will be slightly angled compared to what it would be if there was no wind. If you consider that these two paths are lines from the center of a circle going to the circle, you'll see that the line tilted by the perpendicular wind does not go as far forward in the direction of the bike with no wind as the bike with no wind does (call this "the Difference"). Which means a perpendicular wind does not have a 0 effect, but hinders the bike slightly. This means that, in order to get a 0 effect, the wind would have to help the bike forward slightly, just enough to offset the Difference above. So the 0 point is not at the perpendicular, which is contrary to common sense.

Do I have this right?

[Roboramma]

Sounds about right to me.

One thing that this makes me wonder is whether the direction at which 0 would be reached would be different for different wind speeds.

If I am reading you right, zero is where the energy input by the rider over a certain distance in a certain time is equal to what it would be if there were no wind at all, correct?

[Badly Shaved Monkey]

This leads to something I have wondered about as I cycle.

If I cycle at 10mph against a 0-20mph gusty headwind I feel 10-30mph wind speed against me. If I cycle at 30mph with a 0-20mph tail wind, I feel 10-30mph against me.

Obviously I know which is which because I am doing the cycling, but if I was blindfold and pulled on a smooth trolley is there an intrinsic difference in the way that wind rises, falls and gusts that would enable you to tell whether you were going slowly against a headwind or faster with a tail wind?

[fuelair]

Forgive me if this is a simple question.

Where is the direction from which wind blowing on a bicyclist will not hinder nor help the bicyclist on a straight path?

Assume that the wind blowing from directly behind a bicyclist will push the bicyclist forward +X amount, and the wind blowing from directly in front of the bicyclist will help -X amount (that is, hinder as much as the wind blowing from behind helps).

Because the effect of the wind goes from +X to -X, and because the direction of the wind can be altered continuously through a semi-circle going from directly behind to directly in front (the semi-circle's endpoints are on the bike's path), there must be a direction at which the effect of the wind is 0.

Common sense says that that direction is when the wind is perpendicular to the path of the bicycle. But I then considered what the effect of *no wind* would be. That effect is obviously 0. But how could no wind blowing be the same effect as a perpendicular wind? Surely the perpendicular wind would have some small effect. That is, the perpendicular wind will push the bicyclist to the side to some small extent, so with the bicyclist putting the same effort and energy into going forward, the bicyclist's path will be slightly angled compared to what it would be if there was no wind. If you consider that these two paths are lines from the center of a circle going to the circle, you'll see that the line tilted by the perpendicular wind does not go as far forward in the direction of the bike with no wind as the bike with no wind does (call this "the Difference"). Which means a perpendicular wind does not have a 0 effect, but hinders the bike slightly. This means that, in order to get a 0 effect, the wind would have to help the bike forward slightly, just enough to offset the Difference above. So the 0 point is not at the perpendicular, which is contrary to common sense.

Do I have this right?

There is no O effect wind - all wind movement is applying force to the bicycle/bicyclist and force is non-zero. You could make an argument, though, if you changed the question to how much wind force would be needed to just overcome the resistance of air to the motion of the bicycle (more trouble to calculate but calculateable) so that air overall had a zero net force on the bicycle - if the wind was coming straight from behind the cyclist (any side wind will have a negative effect in that situation).


Edit e for a, punctuation.

[epepke]

Forgive me if this is a simple question.

Where is the direction from which wind blowing on a bicyclist will not hinder nor help the bicyclist on a straight path?

Assume that the wind blowing from directly behind a bicyclist will push the bicyclist forward +X amount, and the wind blowing from directly in front of the bicyclist will help -X amount (that is, hinder as much as the wind blowing from behind helps).

Because the effect of the wind goes from +X to -X, and because the direction of the wind can be altered continuously through a semi-circle going from directly behind to directly in front (the semi-circle's endpoints are on the bike's path), there must be a direction at which the effect of the wind is 0.

Common sense says that that direction is when the wind is perpendicular to the path of the bicycle. But I then considered what the effect of *no wind* would be. That effect is obviously 0. But how could no wind blowing be the same effect as a perpendicular wind? Surely the perpendicular wind would have some small effect. That is, the perpendicular wind will push the bicyclist to the side to some small extent, so with the bicyclist putting the same effort and energy into going forward, the bicyclist's path will be slightly angled compared to what it would be if there was no wind. If you consider that these two paths are lines from the center of a circle going to the circle, you'll see that the line tilted by the perpendicular wind does not go as far forward in the direction of the bike with no wind as the bike with no wind does (call this "the Difference"). Which means a perpendicular wind does not have a 0 effect, but hinders the bike slightly. This means that, in order to get a 0 effect, the wind would have to help the bike forward slightly, just enough to offset the Difference above. So the 0 point is not at the perpendicular, which is contrary to common sense.

Do I have this right?


A side wind, say from the right, would tend to try to blow the bicycle over to the left. If the bicycle is going forward, then that would tend to force the handlebars to turn to the left, which the cyclist would attempt to compensate for by turning to the right. Very rapidly, an equilibrium would be achieved, where the cyclist were leaning slightly into the wind, holding the handlebars straight and going straight. After that, no compensatory force need to be applied to the handlebars, as gravity from being at an angle would just compensate.

[Ziggurat]

After that, no compensatory force need to be applied to the handlebars, as gravity from being at an angle would just compensate.

I don't think that's quite what he's asking, because that's not the end of the story. The biker will lean into a perpendicular wind, and will feel no direct force forward or backward from this wind. However, the effective weight on the tires at this point is slightly larger than the weight of the rider and bike alone, and this means that the rolling resistance of the tires (an effective friction) is going to increase because of the wind. So the rider will indeed be slowed down even by a purely perpendicular wind.

I suspect a full answer is going to depend on a lot of details, like the wind speed, the rider speed, the rider/bike weight, and the dependence of the rolling friction of the tires on the force applied to the tires. The exact answer should vary depending on these quantities, and possibly others as well.

[Paul2]

Sounds about right to me.
If I am reading you right, zero is where the energy input by the rider over a certain distance in a certain time is equal to what it would be if there were no wind at all, correct?Yes.

[Paul2]

There is no O effect wind - all wind movement is applying force to the bicycle/bicyclist and force is non-zero. You could make an argument, though, if you changed the question to how much wind force would be needed to just overcome the resistance of air to the motion of the bicycle (more trouble to calculate but calculateable) so that air overall had a zero net force on the bicycle - if the wind was coming straight from behind the cyclist (any side wind will have a negative effect in that situation).


Edit e for a, punctuation.By the 0 point I mean the direction in which the wind would come from at which it would neither help nor hinder the bike, so that the bike would travel as far in the direction in which the bike is pointing as if there was no wind, given an arbitrary amount of energy and effort from the rider.

I think we should limit the discussion to an idealized vehicle and not get into distinctions between forces on the tires, the handlebars, etc.

My main point is that the true answer is not at all obvious. Common sense, by trying to find the half-way point between helping the bike along in its path (a direct tailwind) and hindering it (a direct head wind), leads one to the perpendicular as the answer, but that answer is incorrect, for the reasons in my first post. It's the truth that is not the same as common sense that is my interest.

[epepke]

I don't think that's quite what he's asking, because that's not the end of the story. The biker will lean into a perpendicular wind, and will feel no direct force forward or backward from this wind. However, the effective weight on the tires at this point is slightly larger than the weight of the rider and bike alone, and this means that the rolling resistance of the tires (an effective friction) is going to increase because of the wind. So the rider will indeed be slowed down even by a purely perpendicular wind.

I suspect a full answer is going to depend on a lot of details, like the wind speed, the rider speed, the rider/bike weight, and the dependence of the rolling friction of the tires on the force applied to the tires. The exact answer should vary depending on these quantities, and possibly others as well.

In that case, it would almost entirely be due to pushing the tire slightly to the side, as the side force would not increase the downward force. This would waste some energy in distorting the rubber as it went around. However, with an idealized tire, after the wheel had been turned to compensate, there would be no extra force that the rider would have to exert. Even if there were, the only actual work with the force would happen as the wind changed.

[rwguinn]

By the 0 point I mean the direction in which the wind would come from at which it would neither help nor hinder the bike, so that the bike would travel as far in the direction in which the bike is pointing as if there was no wind, given an arbitrary amount of energy and effort from the rider.

I think we should limit the discussion to an idealized vehicle and not get into distinctions between forces on the tires, the handlebars, etc.

My main point is that the true answer is not at all obvious. Common sense, by trying to find the half-way point between helping the bike along in its path (a direct tailwind) and hindering it (a direct head wind), leads one to the perpendicular as the answer, but that answer is incorrect, for the reasons in my first post. It's the truth that is not the same as common sense that is my interest.

It all depends on why you ask the question.
If it is for a first-year physics, Trigonometry, Analytical Geometry,or similar class where you are doing introductions to vector analysis, the answer is "That direction where Vwind*cos(alpha)=0",where alpha is the angle measured from the direction of the bicycle. What the problem is looking for is the Dot product, not the Cross product. so, perpendicular to the velocity of the bicycle would be correct.

There is such a thing as making the problem harder than it needs to be.

In reality, there is no "0 point" for all the reasons givenin responses above

[Paul2]

In reality, there is no "0 point" for all the reasons givenin responses aboveSurely there has to be a 0 point if the effect of the wind continuously varies from +100% to -100%, eh?

[Dilb]

Surely there has to be a 0 point if the effect of the wind continuously varies from +100% to -100%, eh?

True, as that's actually the Intermediate Value Theorem (http://mcraefamily.com/MathHelp/CalculusTheorem1IntermediateValue.htm). However, it's possible that there isn't a 0 point, because you asked a real world question, not a calculus question.

For example, you might say that a 0 point should really be entirely equivalent to having no wind, both in terms of physical effort (i.e. forces acting over distances) and physiological effort (the strain that the muscles apply). It could be that a cross wind causes no physical effort, but because the bike is leaning sideways slightly the rider feels he is applying more force.

Now it would also be true that at some different wind angle the wind helps the rider enough that he's back to zero physiological effort, but because these no-effect points occur at different places, there's no overall point where it's exactly like having no wind. Depending on the exact factors, we could even imagine the physiological 0-point occurs while riding in to the wind, because it cools the rider down.

[Molinaro]

Zero is when tailwind speed = bike's forward speed.

[Roboramma]

Zero is when tailwind speed = bike's forward speed.
No, I don't think that's right. First, because it deals with a different question - he isn't talking about varying the speed, he's talking about varying the direction.
Secondly, because if the tailwind speed = the bike's forward speed, the bike will not be experiencing a force backward due to wind resistance. If there were no wind, it would, so the bike will go faster with the same energy input than it would without that tailwind.

I'm curious if epepke's response is correct - that a perpendicular wind is actually the same as no wind from a perspective of energy expenditure.

I think the problem is easier to consider if we talk about a car rather than a bike:
Two identical cars set out at a constant speed from points A to B that are linked by a long straight highway. Both cars take 1 hour to make this trip. Car 1 sets out on Monday when there is no wind. Car 2 sets out on Tuesday and there is a strong wind. What direction, relative to the direction of travel of the car, would the wind have to be blowing at for both cars to use the same amount of gas? Would this direction vary with the size of the car, or the speed of the wind?
Would anything else affect this?

Is this the same problem?

[Paul2]

I think the problem is easier to consider if we talk about a car rather than a bike:
Two identical cars set out at a constant speed from points A to B that are linked by a long straight highway. Both cars take 1 hour to make this trip. Car 1 sets out on Monday when there is no wind. Car 2 sets out on Tuesday and there is a strong wind. What direction, relative to the direction of travel of the car, would the wind have to be blowing at for both cars to use the same amount of gas? Would this direction vary with the size of the car, or the speed of the wind?
Would anything else affect this?

Is this the same problem?Yes, I think this is the same problem.

[rjh01]

Is a cross wind the same as no wind? Imagine a ship. With a cross wind from the left side the ship will want to go right. The only way to stop this is to head slightly to the left. Hence energy is being used. I think the answer will be the same for a car or a bike.

What angle would the wind be blowing to be the same as no wind? That depends on how much wind resistance there is when there is a side wind compare to a tail wind.

[Zep]

There is no zero point in the problem as posed. There is at least one minimum-value point.

My simplified thinking:

If the vehicle was in a vaccuum and weightless (theoretically), it would need no force to maintain a constant velocity (once it had obtained that velocity).

Therefore, ANY force acting on it in any direction will alter its velocity in some way.

(Have we all thought of Newton yet? Good.)

In real life, on the road in a straight line, the vehicle's velocity is subject to the force vectors of air and rolling resistance.

Any sideways force on the vehicle, such as wind, introduces a force vector acting at angles to the vehicle's direction, and thus it will affect its velocity.

The change in velocity is the sum of all those force vectors, which will result in a certain force in a certain direction. Thus the vehicle's velocity WILL change.

The intention of the vehicle is to maintain the original velocity and direction, so to overcome the summed force vector exactly, there needs to be an equal and opposite force (i.e. from steering - more accurately, resistance to sliding by the tyres - and engine power, reduced air-resistance, etc).

So the question becomes: At what point is the summed force vector including an additional wind force the same as if there were no wind force. The answer is obviously only if the wind force is zero, i.e. no wind (one trivial solution). Not even if the vehicle's velocity is zero, i.e. it's not moving, is it the same - the tyres have to exert a reactive force to prevent the vehicle sliding away in the breeze...or exert friction if it does slide. ;)

Of course, you could exhaustively search the envelope of performance of many theoretical cars, roads, tyres, winds, etc. combinations. And the results won't be linear, because many of these forces interact and are interdependent AND have non-linear functions (friction, for example). However it would be fair to say that there are no "zero" points in that envelope except the one I indicated.

[Roboramma]

So the question becomes: At what point is the summed force vector including an additional wind force the same as if there were no wind force. The answer is obviously only if the wind force is zero, i.e. no wind (one trivial solution). Not even if the vehicle's velocity is zero, i.e. it's not moving, is it the same - the tyres have to exert a reactive force to prevent the vehicle sliding away in the breeze...or exert friction if it does slide. ;)

Wait, if the car has no velocity, and a wind blows on it, then the force (from the tires or whatever) will be exactly equal and opposite to the force due to the wind (at least for winds that aren't incredibly strong). So in that scenario the summed total force is zero - and it will have zero net effect on the car's velocity, meaning that the car will have to use zero total energy to maintain it's velocity.

If this is true in this situation, I don't see how it can't be true in situations where the velocity of the vehicle is greater than zero. In fact, looking at the argument that Paul2 is making, it seems necessary that it is.

[Crispy Duck]

Zep - agreed. I fly for a hobby, so accounting for the wind is very important in navigation and fuel planning. In Paul2's original question, there will be a zero-point at which the crosswind force has no fore-and-aft component - the position of this point will vary depending on the bike's forward speed. However, even at the zero-point, the rider will have to lean into the wind. I'm not convinced that this will require any extra energy expenditure in the bicycle case, but it does in the aircraft case.

A common brainteaser in flying is this - assume that you fly from point A and go 100 miles directly north to point B, and then back again. If the wind is northerly at 10 miles per hour, will the trip take less time, the same time, or more time, than if there was no wind?

Many people intuitively guess that the wind will 'cancel out' in this situation, but it doesn't. Any wind at all will always lead to a longer trip duration than nil wind. By analogy with the OP, there is no true 'zero point'.

I think.

[Roboramma]

However, even at the zero-point, the rider will have to lean into the wind. I'm not convinced that this will require any extra energy expenditure in the bicycle case, but it does in the aircraft case.
In the aircraft case, is there no point at which there will be no extra energy expenditure for a straight line journey? (not including return?)
Say the wind is blowing directly from behind. In that case, the energy expenditure should be less (than no wind). Now rotate the wind 1 degree. I think it likely that it will still be less, that is, while you will have to slightly change the angle of the aircraft to accommodate for the wind, the boost that the wind gives you will more than make up for any energy loss to that angling.
What if it was 2 degrees, or three? At some point the energy gain from the wind should drop to zero.

Many people intuitively guess that the wind will 'cancel out' in this situation, but it doesn't. Any wind at all will always lead to a longer trip duration than nil wind. By analogy with the OP, there is no true 'zero point'.Neat, but anyway, it's a different problem. Basically you're saying that you will lose more speed or energy from a 10 mile per hour (or any value) wind straight ahead than you will gain from one directly behind. Similarly that no matter the angle of the wind, whatever gains it gives you, if you rotate it 180 degrees, you will lose more than you gained.
The problem with a constant wind from a constant direction is very different, so I don't think that this analogy really informs it much.

[Timothy]

Many people intuitively guess that the wind will 'cancel out' in this situation, but it doesn't. Any wind at all will always lead to a longer trip duration than nil wind. By analogy with the OP, there is no true 'zero point'.

But the OP isn't about a round trip (which is essentially two trips with opposite wind direction), it's about the conditions of wind direction needed to expend equal energy. Perhaps people are getting hung up on the phrase "zero point."

Call it "wind angle such that equal energy is expended."

Clearly the answer is highly dependent on the shape of the vehicle/rider, and can't be universally answered without knowing those details. As an extreme case for demonstration purposes, consider a rider who has a large 45 degree flat surface somewhere on himself or the bike. A direct 90 degree crosswind will either propel him forward (and to the side) or hinder him back (and to the side), depending on which orientation the surface has to the wind.

So, no simple answer without knowing the exact conditions.

Now factor in changing body shape while pedalling, flapping clothing, etc., and it's no longer a static problem, further moving it from the realm of being able to answer it with a single, easily explained number.

- Timothy

[Crispy Duck]

I played around with the aircraft version of this problem. As the OP suggested, it is clearly the case that, as the wind moves from a pure tailwind to a pure headwind, there must be a point where the aircraft's ground speed is equal to its airspeed (as a constant airspeed equates to a constant energy expenditure, this is a reasonable proxy for the bicycle case). For example, for an aircraft travelling at 100 knots with the wind blowing at 30 knots from the west, a heading of around 350 degrees gives a track (course) over the ground of about 007 degrees, with a groundspeed of 100 knots.

This puzzled me a bit, as it implies that the aircraft is unaffected by the wind and makes the same distance in the same time. However, looking at the figures in the example above, the reason for this becomes clear. The course of the aircraft over the ground is slightly to the east of north, which means that the strong westerly wind is blowing the aircraft slightly east, even as it fights personfully into the wind. The 'blowing east' factor is balancing the 'fighting the headwind' factor.

If the pilot were maintaining a track of 360 (ie due north), the groundspeed would indeed reduce to around 97 knots, correctly reflecting the fact that some energy is being used to fight the crosswind.

Does this have any relevance to the bicycle case? I'm not sure. :)

[Art Vandelay]

Let's say the rider wants to experience a force f, the wind is exerting a force w, and the rider applies a force f' which is constrained to have the same magnitude, but not the same direction, as f. The fact that f and f' have the same magnitude means that they can be represented as two different radii on the same circle. Let O be the center of that circle, A be endpoint for f, and B be the endpoint for f'. Since f=f'+w, w is the chord going from f' to f. If the angle between f and w is x, then x is the measure of angle OAB, which is equal to the measure of angle OBA. So measure of angle AOB is pi-2x.
sin((pi-2x)/2)=AB/2OA=|w|/2|f|
sin((pi-2x)/2)=sin(pi/2)cos(-x)+cos(pi/2)sin(-x)=cos(-x)+0=cos(x)
cos(x)=|w|/2|f|
x=arccos(|w|/2|f|)

One thing that this makes me wonder is whether the direction at which 0 would be reached would be different for different wind speeds.Seems pretty obvious to me that it should be. Take the extreme case where the wind’s force is twice the rider’s. Then the rider should ride directly into the wind. But for smaller winds, that’s obviously not the zero point.

Obviously I know which is which because I am doing the cycling, but if I was blindfold and pulled on a smooth trolley is there an intrinsic difference in the way that wind rises, falls and gusts that would enable you to tell whether you were going slowly against a headwind or faster with a tail wind?I think so. For one thing, wind speeds near the ground should be slightly lower than those far away from the ground. The larger the relative velocity between wind and ground, the more pronounced this will be.

In that case, it would almost entirely be due to pushing the tire slightly to the side, as the side force would not increase the downward force. This would waste some energy in distorting the rubber as it went around. However, with an idealized tire, after the wheel had been turned to compensate, there would be no extra force that the rider would have to exert. Even if there were, the only actual work with the force would happen as the wind changed.With idealized tires etc., a rider with no wind would not have to exert any force, since there would be no friction. In that case, you get a divide-by-zero error in my formula.

It all depends on why you ask the question.
If it is for a first-year physics, Trigonometry, Analytical Geometry,or similar class where you are doing introductions to vector analysis, the answer is "That direction where Vwind*cos(alpha)=0",where alpha is the angle measured from the direction of the bicycle. What the problem is looking for is the Dot product, not the Cross product. so, perpendicular to the velocity of the bicycle would be correct.No, that’s not the right answer. The question is not where the wind will exert no force in the direction of travel, but where it will not affect the total force at all.

So the question becomes: At what point is the summed force vector including an additional wind force the same as if there were no wind force. The answer is obviously only if the wind force is zero, i.e. no wind (one trivial solution). No, because the direction of the force that the rider supplies can be changed.

Many people intuitively guess that the wind will 'cancel out' in this situation, but it doesn't. Any wind at all will always lead to a longer trip duration than nil wind. By analogy with the OP, there is no true 'zero point'.No, the analogy doesn’t hold. But problems can often be illuminated by looking at extreme cases. In your example, if the wind is 100mph, then obviously a round trip would take an infinite amount of time, so obviously the wind doesn’t cancel out.

[Zep]

Wait, if the car has no velocity, and a wind blows on it, then the force (from the tires or whatever) will be exactly equal and opposite to the force due to the wind (at least for winds that aren't incredibly strong). So in that scenario the summed total force is zero - and it will have zero net effect on the car's velocity, meaning that the car will have to use zero total energy to maintain it's velocity.

If this is true in this situation, I don't see how it can't be true in situations where the velocity of the vehicle is greater than zero. In fact, looking at the argument that Paul2 is making, it seems necessary that it is.Correct, up to the highlighted bit. For the summed total forces to be zero, the other force vectors, when summed, have to cancel out the single force vector of the wind. If not, the car's velocity will be affected by the wind (the net sum is non-zero, so the velocity will change). Given that the wind force is non-zero, they too must be non-zero.