It seems that a weather vane will rotate in order to minimize energy and thus orient itself parallel to the wind.

What I do not understand is why it is implied that the weather vane arrow should point in the direction of the wind.

I do not understand why the arrow pointing in the opposite direction of the wind is also not a minimum energy solution.

enter image description here

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    $\begingroup$ The design is what it is because of the names of the winds. Boreas gave boreal weather. $\endgroup$
    – user137289
    Jan 14, 2020 at 9:36
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    $\begingroup$ @user391339 Clarification please: It seems you are asking why the arrow would point in the direction of the wind while you feel it should be wrong. If this is what you are asking, you are correct that this should be wrong, but misunderstanding a weather vane. It points in the direction the wind is from, not in the direction it is blowing. $\endgroup$
    – dlb
    Jan 14, 2020 at 19:47
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    $\begingroup$ It looks like the pictured weathervane would cause the arrowhead to point upwind. It should be noted that airflow on weather maps show arrows with the arrowhead pointing downwind $\endgroup$ Jan 14, 2020 at 22:16
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    $\begingroup$ This doesn't seem to be your actual question, but your second sentence makes it seem like you're asking why the vane arrow points in the opposite direction of the wind flow. The answer to that is that winds are named based on their direction of origin, not their direction of travel. That's why (in the northern hemisphere) a southernly wind tends to be warm, and why weather reporters say things like "winds coming out of the west". $\endgroup$
    – MooseBoys
    Jan 15, 2020 at 5:30
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    $\begingroup$ Yes, the arrow pointing the opposite way is a solution to an ideal analysis. But since wind is a chaotic phenomenon it is never in a purely single direction for any non-trivial period of time. So even if it was pointed the opposite way for an instant it will always end up pointing towards the wind in the real world (due to the larger surface area of the tail.) $\endgroup$
    – CramerTV
    Jan 15, 2020 at 18:06

5 Answers 5


The vane has to be designed so that it has a preference to point in the right direction. In the example that you included, this is implemented by the flag at the back providing a broader cross section than the arrow head and also by the rooster standing slightly to the back half of the arrow.

You are correct that if the vane became perfectly anti-aligned to the wind, it might stay there for a bit. That solution, however, is an unstable equilibrium solution. If the wind shifts even a little, assuming the vane is well-designed, it should snap around to the proper direction for the reasons above.

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    $\begingroup$ The arrow tail was obvious to me, but I hadn't thought about the rooster. Now I'm wondering if you can buy trick weather vanes that have the rooster aligned so that the vane points away from the wind... $\endgroup$
    – Michael
    Jan 16, 2020 at 1:44
  • $\begingroup$ The "problem" of wind switching 180° is rather rare if not impossible in nature. Usually earths rotation alone will prevent a complete 180° switch .. so the weather vane always has time to align $\endgroup$
    – eagle275
    Jan 17, 2020 at 14:43
  • $\begingroup$ Re, "trick weather vanes," I don't think you'll get rich by selling those. $\endgroup$ Jan 17, 2020 at 15:55
  • $\begingroup$ I am very frustrated that the community receives this answer so well...Come on guys you can do better than that! If this was a test question for a physics exam a student answering with this would get a flat zero. HOW is it that this rigid object has a stable solution with the wind aligned to one direction, and has an unstable direction when its anti-aligned to the other?? Obviously it has to do with the surface area difference between sides, but you're not explaining how that feature causes this effect. $\endgroup$ Jan 18, 2020 at 1:31
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    $\begingroup$ The question was a "Hot Network Question" on the right for a while, so it attracted more attention. I'd guess the answer got votes b/c it's understood that this is not a physics student asking or answering an exam question. There's some value - apparently the voting agrees - to answering the question at the level that it's asked. Also, you seem to have misread the question, which, as written, essentially asks why the anti-alignment is not a equilibrium. The answer to that is that it is an equilibrium, just unstable. Apparently what you felt was "obvious" was not so to some. @StevenSagona $\endgroup$
    – Brick
    Jan 19, 2020 at 19:21

When the wind blows perfectly parallel to the wind vane's long axis, there is no rotational force on the wind vane. When the wind direction is not parallel to the long axis of the wind vane it will exert a turning force on the vane until the wind vane is parallel to the wind direction.

The force of the wind on any part of the wind vane depends on its (surface area) X (its distance from the fulcrum/rotational axis). Any area of the wind vane that is exactly on the fulcrum will result in no rotation.

In designing your wind vane therefore, the part that you desire to be pointing away from the direction from which the wind is blowing should have the greatest product of (surface area) X (distance from the fulcrum).

In the image posted with your question, the rooster and the arrow head at their positions and sizes are merely decorative. A wind vane with only a tail of similar shape to that in your image, attached to the fulcrum would work just fine. Users would need to understand that it is pointing in the direction in which the wind is going, not where it is coming from. It also follows that if the arrowhead were large enough in area and far enough from the fulcrum, then it would point away from the direction from which the wind is coming.

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    $\begingroup$ The rooster and arrowhead are not purely decorative: they also indicate the direction. The arrowhead alone would suffice, but the rooster makes it more obvious. $\endgroup$
    – OrangeDog
    Jan 14, 2020 at 17:56
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    $\begingroup$ @OrangeDog You could say it's more correct to say the rooster and arrowhead are merely for visual effect. I meant that they are not necessary to determine the direction in which the wind vane will point when the wind blows, in the context of my final paragraph. $\endgroup$
    – Dlamini
    Jan 15, 2020 at 0:10
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    $\begingroup$ The important point that seems to have been missed is that the arrow's tail is larger than its head. Any slight deviation from perfectly parallel to the wind will exert opposite forces on both head and tail, but the tail, being larger, has more force exerted on it. If the tail fell off, the arrow would point the other way. (I'm ignoring the rooster, since it looks roughly symmetrical. $\endgroup$
    – jamesqf
    Jan 15, 2020 at 4:48
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    $\begingroup$ @jamesqf Sorry, but the rooster itself is a long way from symmetrical - its tail is huge, and the whole thing is slightly behind the axis of rotation. $\endgroup$
    – MikeB
    Jan 17, 2020 at 12:34
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    $\begingroup$ @MikeBrockington that's true, but it's also very close the the axis, therefore contributing minimal torque. I'd be curious which wins. $\endgroup$
    – spectras
    Jan 17, 2020 at 14:40

While it's possible to analyze this in terms of least energy, it's not very straightforward. That's because it's not the weather vane's energy that is being minimized, but the wind's. The wind does work on the weather vane, which then causes friction, converting the energy to heat.

Suppose the weather vane starts out perpendicular to the wind and rotating. One side will be coming towards the wind, and therefore doing work on the wind. The other side will be going the same direction as the wind, and wind will be doing work on it. Work is force times distance, and the distance a rotating object travels is proportional to the radius, so the wind will be doing net work on the vane if the side moving in the same direction as the wind has a larger mass*radius (i.e. moment of inertia) than the other side. Thus, minimizing the energy of the wind means the side with larger moment of inertia moves in the direction of the wind. This means it moves towards pointing to the side the wind is blowing. Once it reaches the side the wind is blowing, any perturbation away from that side results in a torque back.

BTW, systems don't always move to the configuration with lowest energy. Rather, they tend to move in the direction in which energy is decreasing, which means that once they reach a point where the energy isn't decreasing in any direction, they stop. For instance, if you put a ball on the top of a hill perfectly balanced, it will stay there. This is an equilibrium, because the derivative of energy with respect to position is zero. But it's an unstable equilibrium, because the second derivative is negative.

The weather vane pointing the opposite direction is also an unstable equilibrium. The forces are equal, but any perturbation will result in amplifying the perturbation until the weather vane is pointing the "right" way.

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    $\begingroup$ For a complex object like the pictured vane, given turbulence effects, is it actually guaranteed that there exists an unstable equilibrium point in the anti-sense direction? Or is it the case that the point exists, but the asymmetrical turbulence forces will, with high probability, push it off "immediately"? $\endgroup$ Jan 14, 2020 at 22:12
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    $\begingroup$ @LawnmowerMan Well, a real-world vane also experiences friction. This acts against any rotation, increasing the stability of both equilibria. So even given turbulence, you could probably observe a vane pointing 180 degrees off in sufficiently light winds, if it does not cause enough imbalance to overcome the friction. See also Cort Ammon's answer. You will therefore want a vane that moves lightly. $\endgroup$
    – sigma
    Jan 15, 2020 at 22:08
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    $\begingroup$ @LawnmowerMan We can use Intermediate Value Theorem type logic to argue that if there is one position where the vane goes one way, and a position where it goes the other way, then there is a position where it goes neither way. Stuff like changing wind directions does complicate it, though. $\endgroup$ Jan 15, 2020 at 22:19
  • $\begingroup$ @Acccumulation at that point you might want to throw in the "local minimum" concept in your third paragraph as well, it does not apply to the situation but that would make the equilibrium explanation complete. $\endgroup$
    – spectras
    Jan 17, 2020 at 14:45

When we model aerodynamic forces like this, one of the very useful tools is a "center of pressure." The center of pressure is a point where you can pretend all of the aerodynamic forces are exerted on that point and get the correct behavior. It can move as the angle of attack (AoA) changes, meaning the angle between the vane and the wind, but at any position you can analyze it as though all of the forces are on a single point.

Note that the tail of the vane has a much larger square, compared to the triangle. It has a larger surface. Generally speaking, this means the center of pressure will be closer to the tail.

Think about what happens when the vane moves. If the vane points its head to the left of the wind, the tail is kicked out to the right. This means the tail is now a square that isn't perfectly in line with the wind. Thus the airflow will push the tail a bit to the left, "trying" to straighten it out. We can treat it as though this force was from the center of pressure, pushing the center of pressure to the left. Since the center of pressure is behind the vane's center of rotation, this will rotate the whole vane to be more straight.

When the center of pressure is behind the center of rotation, we call this a stable configuration. In airplanes the center of rotation is the center of mass, so you'll typically see stability phrased as "center of pressure behind center of mass." In our case, we're anchored to the roof, so we rotate around that instead.

So what happens if the vane is pointed backwards? If the vane is backwards and kicked to the left, the pressure on the tail (which is upwind) is to the left. Since the center of pressure is now in front of the center of rotation, this will cause the vane to swing around further, until the vane points in the right direction. The weather vane is unstable when "flying" backwards.

You can get the situation you describe, where the vane points the wrong way. Say the wind is coming from the north, and then dies down. The weathervane is thus pointing north. If the wind shifts and picks up from the south, precisely, the side to side forces are zero. However, if it shifts just a little bit off of south (and winds are never perfect), for just a brief moment, that will turn the weathervane a bit. Now, even if the wind returns to being perfectly from the south, the weathervane is no longer pointed directly north. The instability will cause the vane to swing around so that there's no longer that metastable point from the south.

In practice, the bearings on the vane are not perfect, so you have to get a wind that's reasonably far from the south before you'll actually move the vane. However, winds are also decidedly not smooth. They have small gusts which will drive the vane out of that metastable point, and then it will point in the correct direction.


Wind vane will always point to the direction from which the wind is blowing.

I.E., if wind is blowing from East to West, then the arrow of the wind vane points to 'East'.

The pointed end of the arrow offers least resistance to wind. Therefore, the arrow achieves the state of equilibrium by pointing itself against the wind (Direction from which the wind is blowing).


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