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Why does a symmetrical weather wane work?

  1. Oct 18, 2009 #1
    Let's say we have a symmetrical weather wane that's free to rotate about its vertical axis, and a uniform fluid flows in some horizontal direction d. Then, I assume, the wane will align itself parallel to d, but why? Is there a simple argument or does one have to carry out a numerical simulation of Navier-Stokes?

    I noticed that if one models the fluid as a stream of pellets that interact with the wane in an elastic manner but don't interact with each other, then the wane won't rotate at all: the torque is zero.
     
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  3. Oct 19, 2009 #2
    If the vane (I think that's the right word) was perfectly symmetrical and the fluid flow was perfectly uniform and laminar and perpendicular to the vane, then I think the vane wouldn't rotate. But this is hardly the case in reality. The weather vane will rotate until it comes to a stable equilibrium and for a plate, this will be approximately when the plate is parallel to the flow stream.

    I would assume that a weather vane has an aerodynamic centre that is not in line with its pivot access, so that the vane rotates and finds a stable equilibrium in a flow instead of rotating continuously. The aerodynamic centre is the point on the vane around which the lift force works.

    A plate at an angle to a free stream will generate lift which will rotate the plate around its aerodynamic centre, if it is not fixed, until such a time as it is parallel to the flow and no more lift is generated (perfect scenario)
     
  4. Oct 19, 2009 #3
    You're right of course, "vane" is the word, thanks.

    I guess what I'm asking is: why is the situation with the vane being orthogonal to the flow not a stable equilibrium? If we deviate a bit from it, so that the vane-flow angle is a bit smaller than 90 degrees, then as you say there's a lift force created. Does that force pull stronger on one side of the vane than on the other, so as to decrease the angle further? If so, I just don't see the reason right now. Maybe some sort of conservation-of-angular-momentum trick?

    It's also fun to think about the problem the other way round: the fluid is stagnant, you have a vane that can freely rotate around its shaft, and then you pull the shaft through the fluid, with the shaft orthogonal to the pulling direction. Will the vane line up parallel to the direction in which you're pulling? In this scenario I'm not at all sure what would happen.
     
  5. Oct 19, 2009 #4

    rcgldr

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    I had the impression that a vane with a centered axis would rotate continously in a flow, at least in a gas flow, (not sure about a fluid flow) probably because lift effects are higher at the leading edge than the trailing edge in the real world. For example, drop a paint stick, ruller, or yard stick out the window of a building, and the object will rotate rapidly.
     
  6. Oct 19, 2009 #5
    It seems to be that the very concept of a perfectly symmetrical vane is an abstraction that can't be realized in the real world. If either side is marginally larger than the other, and the flow is not perfectly laminar, being orthogonal to the flow is not a stable equilibrium: the vane will try to rotate so that the larger side is aligned with the flow.
     
  7. Oct 19, 2009 #6

    DaveC426913

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    I think before you go further with surmising why it happens, you should be sure you know, in fact, what does actually happen.

    What does a symmetical weather vane actually do?

    i.e.:
    Do you know this for a fact?
     
  8. Oct 19, 2009 #7
    That's true, in the absence of perfect symmetry, the 90-degree situation is certainly not stable.

    But suppose the vane is already almost parallel to the flow, let's say angle=15 degrees or so. Then my question would be: is the torque acting to increase or decrease the angle? I guess that answer shouldn't depend on perfect symmetry (unless, of course, there's no torque in the perfectly symmetrical case).
     
  9. Oct 19, 2009 #8

    rcgldr

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    In the real world air, because of real world effects (vortices, detachment of flow), the lift at the leading edge is higher than the lift at the trailing edge, resulting in a "upwards" pitch torque. If this torque and momentum effects are sufficient, the vane will end up rotating. If there isn't enough momentum, the vane osciallates back and forth. Unless the pivot point is far enough forwards, and there's some inherent dampening, the vane will not be stable. Imagine a solid flag, although it's mounted at the leading edge, turbulence will still result in some oscillation unless the wind speed is relative slow. For a windmill, the vane is well aft of the pivot point, and the angular momentum of the main rotor dampens any yaw reactions.

    Symmetrical air foils produce a pitching moment whenever they produce lift, but the effective center of lift doesn't move versus angle of attack as it would for a cambered airfoil. Helicopters generally use symmetrical airfoils because the ideal pivot point remains at the same location regardless of angle of attack, reducing the stress on the rotor head and in the rotor blades.

    I've been trying to find a video showing a ruler or paint stick spinning as it falls, but no luck so far. These can be thrown upwards to see the effect.
     
    Last edited: Oct 19, 2009
  10. Oct 20, 2009 #9
    if it is a real world situation; there are three possibilities:

    1. The vane will rotate about it's pivot point continuously finding a dynamic equilibrium.
    2. The vane will remain perpendicular to the flow, if all the factors in the flow field combine perfectly to allow this, but I would assume this the least likely in a real world situation, unless the pivot has a very high friction factor, maybe.
    3. The vane will not be perfectly symetrical, and if this factor is large enough, the vane will rotate and find an equilibrium that is slightly oscillating (depending on the dampening) parallel to the main flow field.

    If it is a perfect situation with a wind vane that is perfectly symmetrical, perfectly flat, with no imperfections on its surface and with zero friction in the pivot and a perfectly laminar flow stream that approaches the plate exactly normal to the plate, and we assume the created vortices are identical and opposite and we assume that the turbulence generated behind the plate is perfectly symmetrical then I guess it could remain perpendicular to the flow.
     
  11. Oct 20, 2009 #10

    DaveC426913

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    Why are we discussing the reason for an event when we don't even know if the event occurs?

    We need facts.

    This reminds me of the pre-Galilean understanding of gravity. Everyone simply assumed that larger objects fell faster than smaller objects. It never occurred to anyone to see what was actually happening and then discuss that.
     
  12. Oct 20, 2009 #11
    Ok, do a test, let us know what you find out.

    I'm basing my assumptions on what I know about fluid dynamics.
     
  13. Oct 20, 2009 #12

    DaveC426913

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    The OP is asking a question as to why something happens. Is that what happens?
     
  14. Oct 20, 2009 #13
    If it is true that the lift is always larger at the leading edge, then the situation with the vane being parallel to the flow (angle of attack=0) is an unstable equilibrium: deviate a bit from it, and the resulting torque will pull you farther away. On the other hand, the situation of the vane being perpendicular to the flow would be a stable equilibrium: deviate a bit from it, and the torque pulls you back.

    NASA has a http://www.grc.nasa.gov/WWW/K-12/airplane/ac.html" [Broken], where they say that the AC is located about 25% of the airfoil's length from the leading edge for slow air speeds, and closer to 50% for supersonic speeds. Further, for symmetrical airfoils like our vane, the torque ("aerodynamic moment") about this AC point is zero, independent of the angle of attack. That certainly requires that the lift at the leading edge is a lot higher than the lift at the trailing edge, for all angles of attack, and that the torque about the airfoil's midpoint always acts to restore the perpendicular equilibrium.

    All this then tells me that for real-world weather vanes it's crucial to have a fair amount of asymmetry, or they won't work.

    Is there some intuitive way to understand why the lift is larger at the leading edge, without resorting to experiment?
     
    Last edited by a moderator: May 4, 2017
  15. Oct 20, 2009 #14

    rcgldr

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    At sub-sonic speeds, the air has more time to respond to effects of the trailing edge of a lift producing solid than then leading edge, so accelerations are less. The flow over the upper leading edge involves a significant component of centripetal acceleration to flow first upwards then downwards over the upper surface, but at the trailing edge, the flows are near parallel to the surfaces. The flow is initially upwards because the lower pressure zone above and higher presssure zone below the lifting surface diverts the air in front of the lifting surface upwards before it reaches the leading edge of the lifting surface.

    The torque is always zero at the AC, but for cambered air foils the AC moves with respect to angle of attack.
     
  16. Oct 20, 2009 #15
    Well, on the NASA page they define AC as the point where the moment doesn't vary with angle; then they claim that the location of AC doesn't vary with angle either, and that the moment is zero at AC for symmetrical airfoils.

    In any event, something still seems fishy to me. In the symmetrical case where the AC is 25% away from the leading edge and there's no moment around it, what happens if the angle of attack approaches 90 degrees? Clearly, at 90 degrees we have a moment about AC, since one side of the airfoil is three times as long as the other side. So this statement "moment is zero at AC" only works for small angles of attack?
     
  17. Oct 20, 2009 #16

    rcgldr

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    In the real world, the AC also moves with changes in angle of attack on a symmetrical airfoil, but the movement of the AC is relatively small for the normal range of angle of attacks (say from -13 to +13 degrees for a model helicopter rotor, not sure if this applies to the 20 degrees AOA possible with some delta wing designs).
     
  18. Oct 21, 2009 #17
    It depends on the weather vane construction. Most weather vanes are constructed using this common design guideline: "The design of a weather vane is such that the weight is evenly distributed each side of the axis of rotation, but the surface area is unequally divided..."

    I assume you mean a standard weather vane with a flat plate or symmetical airfoil (as in symmetrical cross section) as the operating surface that is pivoted at a point where more surface area occurs on one side of the pivot point. If you mean symmetric weather vane when the pivot point is exactly in the middle of a symmetrical plate, then the weather vane will not align itself with the flow.

    A flat plate or symmetrical airfoil does not produce lift at zero angle of attack in a perfect flow stream, this would be the wind vanes equilibrium point. If the plate (or symmetrical airfoil) is not parallel to the flow, it will have an angle of attack corresponding to the angle that it is incident to the flow. This will allow the plate to generate lift. This lift will act at the centre of pressure of the plate, which is positioned 25% of the length of the plate, from the leading edge, as proved by thin-airfoil theory (devised by German mathematician Max Munk and further refined by British aerodynamicist Hermann Glauert and others). If the centre of pressure of the weather vane is behind it's pivot point, then the resulting moment from the lift will rotate the vane around it's pivot and align the vane with the flow.

    In the special case where there are large angles of attack, the airfoil or plate is effectively stalled. But now, at an angle of attack of 90° for example, the drag force of the plate causes it to rotate about the pivot until the angle of attack reduces sufficiently for the lift force to take over again until the plate finds it's equilibrium parallel to the flow.

    Am I getting there yet or am I still misunderstanding the question?
     
  19. Oct 21, 2009 #18
    Very good.

    Now, this raises a problem. According to your explanation, the angle of attack of 0° is an unstable equilibrium, because the aerodynamic center is 25% of the length from the leading edge, and therefore it's ahead of the pivot point. I'm not an expert on hydrodynamics but I'm inclined to agree.

    Is there a stable equilibrium point?

    You're saying that 90° is not stable, though you don't offer any reasoning why it's not. Is there a certain angle in between 0 and 90 that is stable?
     
  20. Oct 21, 2009 #19
    I think my explanation was not clear then. The pivot point has to be ahead of the centre of pressure. That is why the design guidelines recommend that one area is much larger than the other on either side of the pivot. When this is the case, there will be a stable equilibrium at 0° angle of attack when the plate produces zero lift. If it produces lift, the lift force will act through the centre of pressure (AC for a symmetrical airfoil) and a moment will be generated around the pivot point.

    at 90°, the drag force causes the system to be unstable (The areas of the plate are unequal on each side of the pivot, so the drag force on one side of the pivot is larger than the other, causing a moment around the pivot and rotation of the plate) and it will try to find it's stable equilibrium, parallel to the flow at 0° angle of attack.

    For a symmetrical cross section, with a pivot that is ahead of the centre of pressure, the only stable position is at 0° angle of attack ie. parallel to the flow.
     
  21. Oct 21, 2009 #20
    I will soon resort to sketches, if this is still unclear.
     
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