# Does a 90 degree cross wind slow you down compared with no wind?



## eniveld (Dec 6, 2012)

A head wind slows you down. A tail wind speeds you up. How would a 90 degree cross wind affect your speed as compared with absolutely no wind?

Is this a FAQ?


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## AndreyT (Dec 1, 2011)

Deleted. Have to think about it.


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## QuiQuaeQuod (Jan 24, 2003)

eniveld said:


> A head wind slows you down. A tail wind speeds you up. How would a 90 degree cross wind affect your speed as compared with absolutely no wind?


Cross winds slow people down, that's for sure. They are not neutral.

It takes energy to maintain a straight line in a cross wind, energy that does not go into forward motion. And that tends to tire you out faster as well.

Also, cross winds tend to push you around, so less of a straight line, harder to maintain the "smooth" line.

With more than one person, cross winds make it harder to draft, and really hard to draft on public roads (since you can't form an echelon).


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## Pirx (Aug 9, 2009)

QuiQuaeQuod said:


> Cross winds slow people down, that's for sure. They are not neutral.
> 
> It takes energy to maintain a straight line in a cross wind, energy that does not go into forward motion. And that tends to tire you out faster as well.


Not a good answer but the final conclusion is somewhat correct, with a twist: 
Yes, _in general_ cross wind will slow you down. The way to think about this is to ask about the drag force as a function of the angle of attack in a coordinate system that is fixed to the rider, and parallel to the total velocity vector (cross wind minus bicycle velocity). What we find in a case like that is that, for an oblong symmetric obstacle, most of the time the drag coefficient will be minimal at zero degrees angle of attack, meaning no cross-wind. Since both the drag coefficient and the total velocity increases when you increase the cross-wind velocity at constant forward speed of the bicycle, this means that the total drag increases as well.
[Note for the expert: One has to be careful to make a correction for the fact that, in aerodynamics, the drag is defined as the force antiparallel to the relative wind velocity vector, whereas here we are only interested in the force component antiparallel to the bicycle; this latter force will include an aerodynamic lift component as well, see below. The end result, at least in the absence of significant lift, still works out as increased drag.]

However, at a fluid dynamics conference I attended last year I have seen a presentation by Mavic engineers who showed results for a new aero wheel prototype they have that has _negative_ drag at certain (small) angles of attack. 
[Note: The reason this is possible is exactly the aerodynamic lift component I mentioned in my remark above.]
This means it feels like the wheel is actually providing a (very small) propulsion force in this situation. The effect was extremely small, and I have some doubts as to the accuracy of their results but, nevertheless, this shows that decreased drag in cross winds is not impossible. It is a rare exception, however.

P.S.: Oh, and of course QQQ's remark on drafting is spot on.



AndreyT said:


> Deleted. Have to think about it.


Thinking about it is not good enough in this case. This is a non-trivial problem.


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## Roland44 (Mar 21, 2013)

eniveld said:


> A head wind slows you down. A tail wind speeds you up. How would a 90 degree cross wind affect your speed as compared with absolutely no wind?
> 
> Is this a FAQ?


I am not a physicist but from my personal experience I am pretty sure that any wind who isn't a tail wind slows you down.


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## Pirx (Aug 9, 2009)

Roland44 said:


> I am not a physicist but from my personal experience I am pretty sure that any wind who isn't a tail wind slows you down.


Physics agrees with your experience most of the time. But, there are exceptions. These exceptions are important, for example for sailboats having to make progress against the wind. They do so by "crossing" against the wind, exploiting the effect I described above. Alas, cyclists are not sailboats...


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## eniveld (Dec 6, 2012)

Thank you Pirx - Could you please elaborate on where the force against the direction of the bicycle comes from, when the wind is 90-degrees?

Edit: I understand how you might get some lift, but why would that slow you down?


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## QuiQuaeQuod (Jan 24, 2003)

Pirx said:


> Not a good answer but the final conclusion is somewhat correct, with a twist:
> Yes, _in general_ cross wind will slow you down. The way to think about this is to ask about the drag force as a function of the angle of attack in a coordinate system that is fixed to the rider, and parallel to the total velocity vector (cross wind minus bicycle velocity). What we find in a case like that is that, for an oblong symmetric obstacle, most of the time the drag coefficient will be minimal at zero degrees angle of attack, meaning no cross-wind. Since both the drag coefficient and the total velocity increases when you increase the cross-wind velocity at constant forward speed of the bicycle, this means that the total drag increases as well.
> [Note for the expert: One has to be careful to make a correction for the fact that, in aerodynamics, the drag is defined as the force antiparallel to the relative wind velocity vector, whereas here we are only interested in the force component antiparallel to the bicycle; this latter force will include an aerodynamic lift component as well, see below. The end result, at least in the absence of significant lift, still works out as increased drag.]
> 
> ...



Well said.

However, I was not addressing the aero component so much as the physiological and psychological ones in the pure (90 deg) cross wind situation. Basically, leaning sideways against the wind, which is often gusting and causing constant adjustments... and the fact that cross winds suck. 

I hope you could follow my technical language there.


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## Pirx (Aug 9, 2009)

That's not too easy to do in just words, but I'll try to get you to make a drawing that should clarify things. Here we go:

Draw an arrow indicating the relative wind direction to the bicycle (with no cross wind that arrow will be parallel to the bicycle, and pointing towards it if the arrow is in front of the bicycle; if there's a cross wind component, then the arrow will be at an angle relative to the direction of the bicycle). Now draw an arrow that starts at the bicycle, and points in the direction of the relative wind (i.e., it is parallel to the one you have drawn first). This arrow indicates the aerodynamic drag force. Now draw an arrow that is at a right angle to this one, also starting at the bicycle position. This indicates what is defined as the aerodynamic lift. Now "add" these last two vectors together as follows: You move the second one (the one for the lift) such that it starts at the tip of the one for the drag force, and then extend it from there. Now draw an arrow that starts at the bicycle and ends at the tip of your "lift vector". This is your total force.
Now all you have to do is check if this total force vector "leans" forward or backward relative to the direction the bicycle is moving in. Play around a little with the relative strength of the lift and drag forces and you'll see that when the lift force is quite large and the drag force small, you can have a situation where you get a propulsive force (total force arrow leaning forward relative to the direction of the bike).

P.S.: Answering the question you added while I was writing this: You misunderstand: This lift force can in fact _drive you forward_.


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## Pirx (Aug 9, 2009)

QuiQuaeQuod said:


> However, I was not addressing the aero component so much as the physiological and psychological ones in the pure (90 deg) cross wind situation. Basically, leaning sideways against the wind, which is often gusting and causing constant adjustments... and the fact that cross winds suck.
> 
> I hope you could follow my technical language there.


Yep, and nothing you said was wrong. Sorry if I came across differently.


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## xxl (Mar 19, 2002)

OK, so a cross wind slows one down; what about a happy wind?


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## skepticman (Dec 25, 2005)

HED used to have an apparent wind calculator on their site. A copy of it is still available at the link below. There is quite a bit of apparent head wind even even with a partial tail wind on whatever wheel was used for their calculator, but it's probably similar for most wheels.

HED Cycling - Yaw Calculator


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## Kerry Irons (Feb 25, 2002)

eniveld said:


> A head wind slows you down. A tail wind speeds you up. How would a 90 degree cross wind affect your speed as compared with absolutely no wind?
> 
> Is this a FAQ?


If you model the rider as a simple cylinder (lots of simplification there) then you start to get some assist from a partial tail wind when the direction is about 20 degrees to the rear (from direct side wind). Because the body is not a cylinder and the bike is very complicated, you can get into arguments about what exactly that number is, but 20-30 degrees from a pure crosswind is a pretty good estimate.


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## QuiQuaeQuod (Jan 24, 2003)

Pirx said:


> Yep, and nothing you said was wrong. Sorry if I came across differently.


Didn't come across that way to me. I was just clarifying what I said given you saying things about what I did not say, and the fact that "partially correct" can mean part of the answer or part right, part wrong.


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## rm -rf (Feb 27, 2006)

A 90 degree wind will appear to be from a front angle if your bike is moving.









(found here.)

But even an apparent 90 degree wind (which would be coming from a tail wind angle) would use up energy. The wind would push your bike tires sideways, and it takes force to stay riding in a straight line.


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## AndreyT (Dec 1, 2011)

rm -rf said:


> A 90 degree wind will appear to be from a front angle if your bike is moving.
> 
> 
> 
> ...


Well, that picture is not enough to explain anything.

It is indeed true that when you move in perpendicalar true wind (let's call it "crosswind"), you will experience the _apparent wind_, which will include a _headwind_ component.

But the question here is: does this _headwind component_ of apparent wind depend on the speed of the crosswind?

The answer is: no, it doesn't.

This is actually something that is immediately seen from a simple diagram on the piece of paper and basic trigonometry. (In fact, this should actually be obvious without any pictures.) If the bicycle is moving forward at 25 mph, the headwind component of the apparent wind will always be _exactly 25 mph_. 

One more time: the headwind component of the apparent wind depends only on the ground speed of the bicycle. It does not depend on the strength of the crosswind. If the bicycle moves at 25 mph (ground speed), then 0 mph crosswind, 25 mph crosswind, 100 mph crosswind or 1000 mph crosswind will all result in apparent wind with exactly 25 mph of headwind component. 

So, from the basic force diagram point of view limited to _apparent wind_ considerations, perpendicular true wind *does not* slow the cyclist down, regardless of how strong that crosswind is. That's actually what the above picture is intended to illustrate, once one understands it correctly. 

And that's what brings us to the essence of the original question. Again, from the force diagram perspective, perpendicular true wind does not affect a bicycle rider. So, the question is: what about other factors?

Obviously, crosswind will change the airflow picture, turbulence etc. and might result in an increase in drag. But that is something that is dependent on too many minute details and cannot be answered in general.

One of the factors that is indeed worth the consideration is what you mentioned below:



rm -rf said:


> But even an apparent 90 degree wind [...] would push your bike tires sideways, and it takes force to stay riding in a straight line.


When a cyclist rides in a crosswind, all they have to do (at the first sight) is to lean the bike into the wind in order to compensate for the lateral force induced by the crosswind. If that's indeed all that's needed, then no, it does not use energy. The force that counteracts the wind in this case is lateral friction and the ground reaction force, i.e. nothing you have to spend your energy on.

The question that remain though is whether any crosswind, regardless of strength, makes the traction path of the bicycle to slip to the side. I.e. whether any crosswind will force the cyclist to make trajectory correction "into the wind" (a bit like an airplane would be forced to do). Does the crosswind strength have to be above certain critical value to overcome the lateral traction capacity of the tire in order force the cyclist to steer into the wind? Or does even the smallest breeze already require countersteering? The answer is not obvious to me.


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## Winn (Feb 15, 2013)

eniveld said:


> A head wind slows you down. A tail wind speeds you up. How would a 90 degree cross wind affect your speed as compared with absolutely no wind?
> 
> Is this a FAQ?


Physics or whatever the answer is an absolute yes I ride in cross winds a lot and they will slow you down. Go ride in one and you will know.


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## azpeterb (Jun 1, 2006)

Winn said:


> Physics or whatever the answer is an absolute yes I ride in cross winds a lot and they will slow you down. Go ride in one and you will know.


That's the bottom line and I agree...cross winds are the devil. I appreciate all of the fancy physics talk in some of the other posts here but my feeble mind doesn't operate on that level.


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## obed (Jan 12, 2014)

A crosswind slows me down. If it has strong random gusts, it can spook me too.


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## DrSmile (Jul 22, 2006)

I am still waiting for a frame manufacturer to design a down, head and seat tube that have flexible trailing edges that can form a wing shape similar to a sail. In theory this should provide actual forward force from any crosswind. Not sure about headwind since the fluttering around of the trailing edge may cause turbulence. Also a strong crosswind may push you off the road even more with this design...


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## junior1210 (May 2, 2013)

DrSmile said:


> I am still waiting for a frame manufacturer to design a down, head and seat tube that have flexible trailing edges that can form a wing shape similar to a sail. In theory this should provide actual forward force from any crosswind. Not sure about headwind since the fluttering around of the trailing edge may cause turbulence. Also a strong crosswind may push you off the road even more with this design...


10 minutes after such a frame is released, UCI will outlaw it, then nobody will build another.


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## azpeterb (Jun 1, 2006)

obed said:


> A crosswind slows me down. If it has strong random gusts, it can spook me too.


One of the strangest sites I've seen involving bicyclists was a few years ago on an extremely windy day. I don't remember the exact wind speed but sustained winds were probably in the 30-40 mph range with stronger gusts. There were 2 cyclists riding south along the interstate frontage road as I was driving on the highway and the wind was absolutely howling from the west....and they were leaning into the wind at about a 60-degree angle in order to keep from being blown over, and it worked. I've never seen guys lean so far and not fall over, meanwhile going in the direction they wanted to go. That was hardcore.


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## eniveld (Dec 6, 2012)

I just discovered this most excellent article on this topic from Jobst Brandt: Headwinds, Crosswinds and Tailwinds by Jobst Brandt


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## AndreyT (Dec 1, 2011)

eniveld said:


> I just discovered this most excellent article on this topic from Jobst Brandt: Headwinds, Crosswinds and Tailwinds by Jobst Brandt


Yep. The graph in Fig. 3 shows that the effect of perpendicular crosswind is minuscule in comparison to what the naive "apparent wind" analysis would "predict" (e.e the naive and incorrect "when you're riding in crosswind you're riding into the apparent wind and that slows you down"). The headwind component of crosswind, as I said before, is exactly zero.

The losses come from such factors as disturbed airflow profile and, since this is empirical data, from distorted ergonomics (when riding a bicycle slanted against the crosswind).

P.S. Another interesting and simple (yet, non-obvious to many) fact from the article is that riding from point A to point B against headwind will not be compensated by riding later from point B to point A with same tailwind. The losses in the A-B leg are greater than the gains in the B-A leg.


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## Pirx (Aug 9, 2009)

eniveld said:


> I just discovered this most excellent article on this topic from Jobst Brandt: Headwinds, Crosswinds and Tailwinds by Jobst Brandt


That article ignores the dependence of the drag coefficient on angle of attack, and the existence of lift. However, if we model the rider as a simple cylinder (as in Kerry Iron's remark), then such an analysis is correct within the confines of the given approximation. Note how the given curves compare to Kerry's recollection: This shows that you need a cross-wind that is only more than 10 degrees off the exact vertical direction towards a tailwind to get some benefit. And, of course, you can clearly see the increased power required in an exact crosswind.



AndreyT said:


> The headwind component of crosswind, as I said before, is exactly zero.


That is a statement about kinematics that, while trivially correct, is irrelevant to a question about forces.


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## AndreyT (Dec 1, 2011)

Pirx said:


> That is a statement about kinematics that, while trivially correct, is irrelevant to a question about forces.


This is an incorrect assertion. That statement has obvious relevance to the question of about forces. One can only claim "quantitative irrelevance" (if there's such a thing) since, again, the headwind component of crosswind is exactly zero.


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## Alex_Simmons/RST (Jan 12, 2008)

The answer is not always straightforward as the coefficient of drag can drop with some crosswinds.


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## eniveld (Dec 6, 2012)

Just got a new intuitive understanding of this I thought I'd share for whoever stumbles on this. This is not a physics explanation, but somehow this really helps me.

If you have a 90-degree cross wind coming from your right side, you're going to need to lean to the right to compensate, right? You'll need to expend some amount of effort to counter the force pushing you left, in order to keep going straight. That extra effort will detract from your forward speed.

And that, to my limited mind, is the most intuitive explanation of this issue I've read anywhere.


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## Soaring Vulture (Jun 25, 2013)

I'm not sure about the effect on speed but I know crosswinds have a substantial effect on enjoyment; they kill it. Of course, headwinds kill both speed *and* enjoyment.


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## Pirx (Aug 9, 2009)

AndreyT said:


> This is an incorrect assertion. That statement has obvious relevance to the question of about forces. One can only claim "quantitative irrelevance" (if there's such a thing) since, again, the headwind component of crosswind is exactly zero.


The OP said he is considering pure crosswind, meaning wind velocity at a 90-degree angle to the direction the bicycle travels in. You come back with the glorious insight that "The headwind component of crosswind, as I said before, is exactly zero". Since that is nothing but re-stating the OPs original premise, it is indeed trivial, and since it adds exactly zero additional information or insight, it is irrelevant to the question relating to drag.



eniveld said:


> Just got a new intuitive understanding of this I thought I'd share for whoever stumbles on this. This is not a physics explanation, but somehow this really helps me.
> 
> If you have a 90-degree cross wind coming from your right side, you're going to need to lean to the right to compensate, right? You'll need to expend some amount of effort to counter the force pushing you left, in order to keep going straight. That extra effort will detract from your forward speed.
> 
> And that, to my limited mind, is the most intuitive explanation of this issue I've read anywhere.


Unfortunately, this insight of yours is entirely wrong. An "effort" that is at a right angle to the direction of travel does not entail any power loss at all. Consider a force against a stationary object: The required work for this is zero, even though it takes a physiological effort.


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## eniveld (Dec 6, 2012)

Would not that physiological effort mean that less physiological effort is available to push forward? What am I missing?


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## redondoaveb (Jan 16, 2011)

Physics, schmisics, ask these guys!


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## Pirx (Aug 9, 2009)

eniveld said:


> Would not that physiological effort mean that less physiological effort is available to push forward? What am I missing?


That's a fair question. However, in this case the leaning does not take any physiological effort, since you will simply use gravity to provide the balancing force, not your muscles.


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## eniveld (Dec 6, 2012)

Pirx;An "effort" that is at a right angle to the direction of travel does not entail any power loss at all. Consider a force against a stationary object: The required work for this is zero said:


> Although there may be no work loss, does this not take effort from the bike rider? And does not this effort take away from what could otherwise be put into forward speed?
> 
> I'm still confused, btw.


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## jbinbi (Jan 7, 2013)

Pirx said:


> P.S.: Answering the question you added while I was writing this: You misunderstand: This lift force can in fact _drive you forward_.


This is exactly how a sailboat works. When the wind is coming 'abeam' 90 deg perpendicular to the boat, you would expect the boat to be driven sideways.

If the boat didn't have a keel, this is what would happen. However the keel in layman's terms moves the boat forward. There is some slip sideways as well, so not all the energy from the wind is transferred forward.

If you could think of the wheels as the keel, and you as the sail, if you had the proper aerodynamics to provide 'lift', a crosswind could move you forward.


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## tomato coupe (Nov 8, 2009)

AndreyT said:


> Well, that picture is not enough to explain anything.
> 
> It is indeed true that when you move in perpendicalar true wind (let's call it "crosswind"), you will experience the _apparent wind_, which will include a _headwind_ component.
> 
> ...


You have be be careful with your vector diagram because this is a non-linear problem. The vectors that are important in this problem are the force vectors that result from the apparent wind, NOT the wind vectors themselves. The correct way to analyze the forces are:

1. Add the "true wind" and "man-made wind" vectorially to get an "apparent wind."
2. Calculate the force that results from the apparent wind. It will be in the same direction as the apparent wind, but its magnitude will be proportional to the square of the apparent wind.
3. Determine the vector component of the force that is opposite to the direction of travel. 

If you go through the math, you'll find that a pure cross wind does, in fact, add to the force that opposes your motion.

Cheers


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## Pirx (Aug 9, 2009)

tomato coupe said:


> It will be in the same direction as the apparent wind,


No, in general it won't. For the restrictive assumption that it does, a very nice, complete analysis has already ben quoted and linked to. You might want to read this thread.


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## rm -rf (Feb 27, 2006)

Pirx said:


> No, in general it won't. For the restrictive assumption that it does, a very nice, complete analysis has already *been quoted* and *linked to*. You might want to read this thread.


The Jobst Brandt link?

Or the HED Cycling Yaw Calculator?

Can you explain in a little more detail?

EDIT -- I saw the comments about the Jobst Brandt charts:


AndreyT said:


> Yep. The graph in Fig. 3 shows that the effect of perpendicular crosswind is minuscule in comparison to what the naive "apparent wind" analysis would "predict" (e.e the naive and incorrect "when you're riding in crosswind you're riding into the apparent wind and that slows you down"). The headwind component of crosswind, as I said before, is exactly zero.
> 
> The losses come from such factors as disturbed airflow profile and, since this is empirical data, from distorted ergonomics (when riding a bicycle slanted against the crosswind).
> 
> ...snip...


But Chart 1 shows that a 90 degree cross wind that is 80% of the speed of the rider will require about 1.2 times more power than no wind at all. 

That's less than I would have guessed, but it's still substantial, and I don't consider it "miniscule".

EDIT2--
The HED calculator shows an apparent wind speed of 28 mph for a 20 mph rider with a 90 degree 20 mph wind. So that sounds pretty extreme. But it is at a wide effective angle, so the effect of that wind is much less.


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## PBL450 (Apr 12, 2014)

The OP asks if a 90 degree crosswind slows you down, the question isn't asking about force? The answer is, it depends... There are thousands of factors including both physical and psychological factors that can slow you down regardless of the physics at play. The fact that a strong 90 cross wind is psychologically unnerving can slow you down because you use greater caution. The crosswind could be carrying smoke, water particles or sand (I often experience one of these three) and reduce visibility as it gets behind your glasses. If the wind velocity is variable, as it likely is in the real world, then leaning into the wind to compensate is not a constant posture but one that requires continuous repositioning. That could significantly slow you down. The physics are important, and they explain the greater proportion of the answer, but they don't fully answer the original question.


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## tomato coupe (Nov 8, 2009)

Pirx said:


> No, in general it won't. For the restrictive assumption that it does, a very nice, complete analysis has already ben quoted and linked to. You might want to read this thread.


AudreyT's instincts tell her that a "pure" crosswind does not slow a cyclist, and she has a reasonable argument to support her stance. I was trying to convey to her why, in this case, her instincts are leading her astray. I will repeat my main point:

"You have be be careful with your vector diagram because this is a non-linear problem. The vectors that are important in this problem are the force vectors that result from the apparent wind, NOT the wind vectors themselves."

I have read the rest of the thread, and I have looked at Jobst Brandt's article. The calculation I made agrees with the one in his article. His model is a simple one, but it correctly shows that a pure crosswind does, in fact, slow you down. 


You have elected to use a more complex model. I have no qualms with that, but I do have a few comments:

1. Cyclists are not sailboats. They are big doughy blobs. (Brandt refers to this as "bluff.") Drag forces on big doughy blobs are generally parallel to the direction of the apparent wind. That is the assumption I made in my calculation, as did Brandt in his.

2. Brandt's simple model correctly predicts that a pure crosswind will slow you down, without including aerodynamic lift.

3. Simple models often lend to a better understanding of a subject than do more complex models because, by definition, they are simpler.

4. You made some very nice comments about the effects of aerodynamic lift in your first post:

"Note for the expert: One has to be careful to make a correction for the fact that, in aerodynamics, the drag is defined as the force antiparallel to the relative wind velocity vector, whereas here we are only interested in the force component antiparallel to the bicycle; this latter force will include an aerodynamic lift component as well, see below. *The end result, at least in the absence of significant lift, still works out as increased drag.*

I completely agree with this comment, and I think it illustrates that the inclusion of aerodynamic lift is not necessary in order to understand the effect of the crosswind.

5. Perhaps you will find my comments to AudreyT more palatable with a little editing:

-- Add the "true wind" and "man-made wind" vectorially to get an "apparent wind."
-- Calculate the force that results from the apparent wind. *In a simple model* it will be in the same direction as the apparent wind, but its magnitude will be proportional to the square of the apparent wind.
-- Determine the vector component of the force that is opposite to the direction of travel. 

Cheers


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## GearDaddy (Apr 1, 2004)

rm -rf said:


> The Jobst Brandt link?
> 
> Or the HED Cycling Yaw Calculator?
> 
> ...


Interesting discussion!

It seems like the Jobst Brandt explanation is pretty hard to argue against, since it backs up its mathematical explanation with real world test results. The interesting part is the calculation of the "Relative Wind Speed":

Relative_Wind_Speed	=	SQRT((U+V×COS(α))2+(V×SIN(α))2)

where 'α' is the angle of the real wind to the rider, 'V' is the wind velocity, and 'U' is the rider's speed.

So, the formula clearly shows that the speed of the crosswind does in fact result in a larger relative wind speed, i.e. it results in more drag on the rider. Even more interesting is that a direct (i.e. 90 degree) crosswind will in fact result in an increase in the relative wind speed, thereby slowing down the rider.

So, take the example of a rider going 20mph and a wind speed of 10 mph. According to this formula the relative wind speed at 90 degree crosswind is 22.36mph. At an 80 degree crosswind (+10 towards headwind) the relative wind speed is 23.86mph. It's not until you get to a 114 degree crosswind (-14 towards tailwind) that you would "break even", so to speak, to get to a relative wind speed of 20mph.

I must say that this makes sense from my experience, i.e. direct crosswinds do slow you down, and it's not just a psychological effect.

Then there's this whole thing about some wheel technologies allowing for a "sail effect" where even a slight headwind/crosswind can actually "lift" you forward. I don't doubt that this effect is possible, but it can only serve to decrease the overall drag on the rider slightly, and doesn't come near counteracting the additional drag as calculated for the relative wind speed. As was aptly pointed out before, bicycle riders are far from behaving entirely like a sail, and are still misshaped blobs or "bluff" objects for the purposes of calculating their drag.


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## Winn (Feb 15, 2013)

Ok this all getting too technical so I thought I would add my own diagram to help










Please note this is on my official "work" paper so it makes it true. Obviously the guy with no wind is faster based on the evidence in the diagram above


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## GearDaddy (Apr 1, 2004)

Winn said:


> Ok this all getting too technical so I thought I would add my own diagram to help
> 
> 
> 
> ...


Yes, very well demonstrated!

I would think that those riders could significantly reduce the effects of a crosswind if they could reduce the size of their heads!


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## tomato coupe (Nov 8, 2009)

Winn said:


> Ok this all getting too technical so I thought I would add my own diagram to help
> 
> 
> 
> ...


Are they riding on 23mm or 25mm tires?


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## tomato coupe (Nov 8, 2009)

Winn said:


> They're booth running 23's because that's fastest  BTW it's not to scale sorry I should have said something. The surprising part is no one noticed the guy on the right is an amputee.


Okay, they looked like 25's in the drawing ...


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## Winn (Feb 15, 2013)

They're booth running 23's because that's fastest  BTW it's not to scale sorry I should have said something. The surprising part is no one noticed the guy on the right is an amputee.


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## jbinbi (Jan 7, 2013)

wait wait wait. while i don't doubt the veracity of the drawing, you are introducing a whole new variable here. This guy is clearly bald, and that changes the coefficient of drag tremendously (just ask Michael Phelps how many razors he goes through).

Clearly a crosswind will affect a bald guy much differently than a guy who has hair and a helmet, especially one with vents. In fact, as I write this, I think if I had a foil type helmet, and I can turn my head appropriately to the wind, I might actually get lift and INCREASE my speed.


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## Winn (Feb 15, 2013)

jbinbi said:


> Clearly a crosswind will affect a bald guy much differently than a guy who has hair and a helmet, especially one with vents. In fact, as I write this, I think if I had a foil type helmet, and I can turn my head appropriately to the wind, I might actually get lift and INCREASE my speed.


Sounds legit to me. Let us know how it works out and be sure to use an equation similar to this Relative_Wind_Speed	=	SQRT((U+V×COS(α))2+(V×SIN(α))2) so everything is clear


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## PBL450 (Apr 12, 2014)

For the first time in my life I truly understand the principles of physics. Thank you Winn! Everyone understands things differently... For me, it was the sweat and miserable face that made it all come together. The perfect description of pertinent factors.


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## Robert1 (Mar 27, 2012)

Completely different. A sail uses lift to cause force against the concave part of the sail. Same thing as a plane wing. 

Don't over think the bike thing. Anyone that has ridden long enough knows that tail wind is best, head wind is worse and cross wind is somewhere in between.




Pirx said:


> Physics agrees with your experience most of the time. But, there are exceptions. These exceptions are important, for example for sailboats having to make progress against the wind. They do so by "crossing" against the wind, exploiting the effect I described above. Alas, cyclists are not sailboats...


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## Pirx (Aug 9, 2009)

GearDaddy said:


> It seems like the Jobst Brandt explanation is pretty hard to argue against, since it backs up its mathematical explanation with real world test results. The interesting part is the calculation of the "Relative Wind Speed":
> 
> Relative_Wind_Speed	=	SQRT((U+V×COS(α))2+(V×SIN(α))2)
> 
> where 'α' is the angle of the real wind to the rider, 'V' is the wind velocity, and 'U' is the rider's speed.


Well, you missed about half of the story, which is the fact that the drag force corresponding to the above velocity is not parallel to the bike, so if you're interested in the force there's a factor of COS(α) that comes into play. Qualitatively your musings are correct, however.



Robert1 said:


> Completely different. A sail uses lift to cause force against the concave part of the sail. Same thing as a plane wing.


Do yourself a favor and don't try to explain to me what you feel is "completely different". You may have seen that I had specifically discussed the lift-related picture. Lift doesn't "cause a force", it _is_ a force. That force is not "against the concave part of the sail". And most airplane wings do not have "concave parts".


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## AndreyT (Dec 1, 2011)

GearDaddy said:


> The interesting part is the calculation of the "Relative Wind Speed":
> 
> Relative_Wind_Speed	=	SQRT((U+V×COS(α))2+(V×SIN(α))2)
> 
> ...


You are confused.

Nobody is arguing with the fact that relative wind (aka apparent wind) does depend on the speed of the crosswind, even if the crosswind is directly perpendicular to the direction of travel. By definition, relative wind vector is the sum of real wind vector and man-made wind vector. So the dependency of the crosswind is there, no argument about it. This is actually the *starting* point of this discussion, not the end of it.

Your "Even more interesting is that a direct (i.e. 90 degree) crosswind will in fact result in an increase in the relative wind speed..." is not interesting at all - that basic fact is what's given from the very beginning and understood by everyone.

Meanwhile, your "...thereby slowing down the rider" is just naked jumping to conclusions. We are not interested in baseless assertions. We want to know the mechanics of _how_ the slowdown happens. Your "thereby" is certainly not enough. (I'd be able to prove anything if it were.)

Now, again, yes, crosswind does affect relative wind. But the question is whether the relative wind component introduced by perpendicular crosswind can slow down the rider. And if it can, then how (!). The more-or-less exact mechanics of this effect is what we want to know.

The formula you quoted, 

Relative_Wind_Speed = SQRT((U+V×COS(α))2+(V×SIN(α))2)

by itself does not not answer that question at all. Moreover, this formula, if taken by itself, actually says exactly the opposite: that perpendicular crosswind *does not* slow the rider (contrary to what you managed to naively conclude from it). (Note also that in our case 'COS(α)' is 0 and 'SIN(α)' is 1, meaning that the formula can be greatly simplified for our case.)

The formula above is simply a purely kinematic (geometric) expression that gives us the magnitude (length) of the relative wind vector. Now, from the _purely kinematic_ point of view, in order to calculate how much of that relative wind vector actually works against the rider (what Brandt calls _inline_ part of the force), you have to multiply that formula by 'COS(β)' (referring to the picture in the article).

But once you multiply it by 'COS(β)', the formula will immediately collapse on itself and fold down to

Inline_Relative_Wind_Speed = U+V×COS(α)

Now if you take into account that for perpendicular crosswind 'COS(α)' is zero, all you left with is just 'U', which is man-made wind only. I.e. the formula says that there's no kinematic crosswind dependence at all.

So, no, while Brandt article does claim that the crosswind dependency is there, that dependency is not explained by a naive "'V' is in the above formula, thereby slowing you down". The factors that actually introduce the real dependency are in the next formulas Brandt uses (his squaring of 'W' is the main point here). It is absolutely necessary to involve the square dependence of drag on speed in order to arrive at Brandt's result.


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## Pirx (Aug 9, 2009)

Correct. If drag depended linearly on speed, the effect would not exist. Of course, basic dimensional analysis _requires_ a quadratic dependence on velocity. So does momentum conservation under the assumption of flow similarity.


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## tomato coupe (Nov 8, 2009)

AndreyT said:


> You are confused.
> 
> Nobody is arguing with the fact that relative wind (aka apparent wind) does depend on the speed of the crosswind, even if the crosswind is directly perpendicular to the direction of travel. By definition, relative wind vector is the sum of real wind vector and man-made wind vector. So the dependency of the crosswind is there, no argument about it. This is actually the *starting* point of this discussion, not the end of it.
> 
> ...


So far, so good -- but you're about to contradict yourself.




> Moreover, this formula, if taken by itself, actually says exactly the opposite: that perpendicular crosswind *does not* slow the rider (contrary to what you managed to naively conclude from it).


If the former statement is correct (which it is) then you cannot make the latter statement. (What's good for the goose is good for the gander.)




> (Note also that in our case 'COS(α)' is 0 and 'SIN(α)' is 1, meaning that the formula can be greatly simplified for our case.)
> 
> The formula above is simply a purely kinematic (geometric) expression that gives us the magnitude (length) of the relative wind vector. Now, from the _purely kinematic_ point of view, *in order to calculate how much of that relative wind vector actually works against the rider *(what Brandt calls _inline_ part of the force), you have to multiply that formula by 'COS(β)' (referring to the picture in the article).


You can't do this. Multiplying the relative wind vector by cos(B) does not tell you how much of the relative wind vector is working against the rider. You much first calculate the FORCE that results from the relative wind vector, and then multiply the force by cos(B). (You state this yourself in the final sentence of your post.)




> But once you multiply it by 'COS(β)', the formula will immediately collapse on itself and fold down to
> 
> Inline_Relative_Wind_Speed = U+V×COS(α)
> 
> Now if you take into account that for perpendicular crosswind 'COS(α)' is zero, all you left with is just 'U', which is man-made wind only. I.e. the formula says that there's no kinematic crosswind dependence at all.


You're going in circles here. You've combined a headwind and a crosswind to get an relative wind vector. Now you're just decomposing the relative wind vector back into a headwind and crosswind. 




> So, no, while Brandt article does claim that the crosswind dependency is there, *that dependency is not explained by a naive "'V' is in the above formula*, thereby slowing you down". The factors that actually introduce the real dependency are in the next formulas Brandt uses (his squaring of 'W' is the main point here). *It is absolutely necessary to involve the square dependence of drag on speed in order to arrive at Brandt's result.*


I don't know what a "naive" velocity is, so I don't understand your first sentence. Your final sentence makes for a good summary, however.


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## tomato coupe (Nov 8, 2009)

Pirx said:


> Correct. *If drag depended linearly on speed, the effect would not exist.* Of course, basic dimensional analysis _requires_ a quadratic dependence on velocity. So does momentum conservation under the assumption of flow similarity.


Yahoo!


Now for the nitpicking part that everyone else can/should ignore:

Why would dimensional analysis require a quadratic dependance?

Cheers


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## Pirx (Aug 9, 2009)

tomato coupe said:


> I don't know what a "naive" velocity is, so I don't understand your first sentence.


Watch where he put the double quotes, then you should be able to understand his sentence. Hint: It's not the "velocity" that's naïve.


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## tomato coupe (Nov 8, 2009)

Pirx said:


> Watch where he put the double quotes, then you should be able to understand his sentence. Hint: It's not the "velocity" that's naïve.


Okay, I got it now ...


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## Robert1 (Mar 27, 2012)

Actually yes it is, the force is perpendicular to the sale. The use of the word concave was for explanation. Yes wings are not concave, but the top part is convex. I'm sure you understand that lift is caused by the difference in air pressure on each side of a sale or wing because of the distance the air has to travel across each side is different thus causing difference in speed/pressure resulting in lift. Now please explain to me how the hell that applies to a cyclist? You're not really suggesting that a cyclist can experience lift are you?



Pirx said:


> Well, you missed about half of the story, which is the fact that the drag force corresponding to the above velocity is not parallel to the bike, so if you're interested in the force there's a factor of COS(α) that comes into play. Qualitatively your musings are correct, however.
> 
> 
> 
> Do yourself a favor and don't try to explain to me what you feel is "completely different". You may have seen that I had specifically discussed the lift-related picture. Lift doesn't "cause a force", it _is_ a force. That force is not "against the concave part of the sail". And most airplane wings do not have "concave parts".


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## GearDaddy (Apr 1, 2004)

Pirx said:


> Well, you missed about half of the story, which is the fact that the drag force corresponding to the above velocity is not parallel to the bike, so if you're interested in the force there's a factor of COS(α) that comes into play. Qualitatively your musings are correct, however.


Criminy, u peoplz r so frick'in nitpicky over semantics! :mad2:

Yes, I didn't elaborate on the rest of the Jobst Brandt calculations for drag and power, because it wasn't necessary. The interesting part, i.e. the part that actually answers the OP's question, is demonstrated in the apparent wind speed calculation:
W (Apparent_Wind_Speed)	=	SQRT((U+V×COS(α))2+(V×SIN(α))2)

It shows that an increasing crosswind velocity (even at 90 degrees) results in an increasing apparent wind speed to the rider. The other accompanying formulas for drag and power clearly show that increasing apparent wind speed results in increasing drag and power to the rider. Duh!


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## Tschai (Jun 19, 2003)

Does any of this affect the airspeed velocity of an unladen swallow?


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## BikeLayne (Apr 4, 2014)

I was reading an article in one of the bike mags a few months ago. It said that a wind from anyplace withing 200 degrees in front/side slows you down and a tailwind anywhere withing 160 degrees behind speeds you up. It was an article about testing bikes and they said that usually the bikes are testing in winds that do not vary more then 15 degrees from dead front. It also said that some testing is being done to reflect real life riding. Basically on the road with positions that people actually ride in. We have mostly side winds around here. The proportion to winds that slow you down is out of proportion to the winds that speed you up. It does not seem fair. But it keeps our air clean and nice. It's worth it.


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## GearDaddy (Apr 1, 2004)

OK people, it seems that some are struggling with an intuitive explanation for what is accurately described in these formulas. Try this analogy on for size: AIR DENSITY.

Air density is in fact a real factor in computing drag on the rider, and it is not included as a significant factor in the formulas (You better believe you can pedal pretty damn fast in a vacuum!). The overwhelming force the rider has to overcome is wind resistance. Wind resistance increases in more dense air. Think about headwinds/crosswinds as if you are having to push through varying degrees of air that is more dense than if there was no wind at all. At a certain angle where you are encountering a tailwind/crosswind the air that you have to push through is less dense, allowing you to go faster than when no wind is present. Verstehe?


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## GearDaddy (Apr 1, 2004)

Tschai said:


> Does any of this affect the airspeed velocity of an unladen swallow?


You betcha. African and European!


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## Winn (Feb 15, 2013)

Tschai said:


> Does any of this affect the airspeed velocity of an unladen swallow?





GearDaddy said:


> You betcha. African and European!


But only if he's drafting you


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## Winn (Feb 15, 2013)

I've been thinking about this and based on some of the excellent answers I have realized I was missing a part that seems to be contested. Since there is clearly still confusion I have prepared diagram 2 in hopes of completely dispelling all mystery here and satisfying all queries (unqueried most notably by the OP). Hope this helps


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## scottma (May 18, 2012)




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## Pirx (Aug 9, 2009)

Robert1 said:


> I'm sure you understand that lift is caused by the difference in air pressure on each side of a sale or wing because of the distance the air has to travel across each side is different thus causing difference in speed/pressure resulting in lift.


I happen to work in the field, and no, I do not understand this. The first part of your sentence (about lift being caused by pressure differences) is a tautology and thus trivial, the second part is flat-out wrong and blatant nonsense. Those pressure differences have absolutely nothing whatsoever to do with "distance of travel". Hint: Zero-thickness airfoils create lift, too, despite the fact that there can be no "difference in travel".



Robert1 said:


> You're not really suggesting that a cyclist can experience lift are you?


Of course I am, and of course s/he is. It's typically not much, but it is, in general, non-zero. Look, this is not the place for me to teach aerodynamics to someone without the background to understand the theory, so I'll just leave it at that.


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## Pirx (Aug 9, 2009)

GearDaddy said:


> Try this analogy on for size: AIR DENSITY.


Ahrrgghh, that really hurt. No further comment.


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## AndreyT (Dec 1, 2011)

Robert1 said:


> I'm sure you understand that lift is caused by the difference in air pressure on each side of a sale or wing because of the distance the air has to travel across each side is different thus causing difference in speed/pressure resulting in lift.


That explanation of the nature of aerodynamic lift, despite being quite popular, is patently bogus. The nature of lift is significantly more complicated.










Here's a good online resource with a rather comprehensive explanation

 3  Airfoils and Airflow


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## GearDaddy (Apr 1, 2004)

Pirx said:


> Ahrrgghh, that really hurt. No further comment.


Sorry about that. I was just trying to come up with a simple analogy to help understand that this is a aerodynamics (i.e. fluid dynamics) problem where a solid object is moving through a fluid field (i.e. a bike moving through the air). The "naive" argument that was presented by AndreyT suggested that the crosswind force was not a fluid dynamic problem, i.e. it is as if the bike rider is in a vacuum and simply has a force pushing on the rider from the side.


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