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Bernoulli's Law and Aerodynamic Lift

  1. May 6, 2003 #1
    Please click here first for some illustrations (otherwise it might be difficult to get my point across).

    The aerodynamic lift on the wing of an airplane (airfoil) is generally explained by the argument that the faster speed of the air along the top of the wing leads to reduced air pressure there and hence produces a lift (Bernoulli's Law).
    Using this argument, one should also expect a lift for a symmetric wing profile as shown in Fig.1.
    However, if one considers the problem from a microscopic point of view, one comes to a different conclusion: upward and downward forces should exactly cancel for a symmetric wing profile.
    This is easy to see if one simplifies the situation and replaces the curved wing surface by two plane sections (Fig.2):
    If the wing is stationary, the pressure on all parts of the wing is identical, i.e. there is no lift. If the wing is moving in the indicated direction, the front half of the upper wing surface experiences an increased pressure because of the increased speed and number of air molecules hitting it (due to the orientation of the surface, this creates a downward force). On the other hand, the rear half experiences a reduced pressure because the of the reduced speed and number of air molecules hitting it (creating a lift). Overall, there is consequently no lift, but only an anti-clockwise torque.
    It is obvious that an overall lift is only achieved if the rear section of the wing has a larger area than the front section, i.e. one would get the maximum lift for the profile in Fig.3 (and this is (schematically) actually how airplane wings seem to be designed (see for instance http://www.zenithair.com/kit-data/ht-87-6.html).
    On the other hand, the reverse situation (Fig.4) should lead to a downward force, although Bernoulli's Law would again predict a lift.

    Note: the above arguments assume that the lower surface of the wing is always parallel to the velocity vector, i.e. the pressure acting on it is unchanged; by varying the 'attack angle' of the wing the amount of lift can of course be arbitrarily be changed and one could even generate a lift for the bottom image.
    Last edited by a moderator: May 7, 2003
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  3. May 6, 2003 #2


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    Ah, but you are ignoring the reason why Bernoulli's equation works: an assumption of incompressible flow. This assumption is valid up to ~Mach 0.3.

    If the flow is incompressible, then when the streamlines reach a disturbance (i.e. the airfoil), the streamlines must become smaller. In other words, since you can't shove more mass into a smaller space, the flow must speed up to get out of the way.

    In your first image (symmetric about the c/2 point), the flow will slow down again once you pass the c/2 point, increasing the pressure and obtaining zero lift.

    This is not a simplification by any means. A triangle in a flow is not an accurate simplification unless you are going at supersonic speeds. Nature abhors discontinuities. The sudden change in angle on the top of the wing will cause flow seperation and turbulence, resulting in a huge drag (and because of the geometry, a downward force).

    Using that airfoil is anything but obvious. Trying to model a 'wing' with a flat front plate cannot be done easily. The air will come practically to a full stop and then get pushed about to either side, again causing major turbulence.

    How would Bernoulli predict a lift again?

    What is Bernoulli's equation please?

    Show for me where in that equation it says how much lift you have obtained.
    Last edited: May 6, 2003
  4. May 7, 2003 #3
    Re: Re: Bernoulli's Law and Aerodynamic Lift

    I can not see how air can be incompressible (especially as it is not confined into a limited space) but this is not a crucial point for my argument anyway (Click here for my illustrations):

    if you neglect complications like turbulence etc. (i.e. assume very small velocities), then it should be clear that for a symmetric wing profile there can not be an overall lift, because the wing area exposed to the airstream (higher pressure) is equal to the wing area shielded from the airstream (lower pressure). The velocity of the airstream parallel to the wing surface can not possibly have any effect on the lift as it can not transfer any momentum to the wing (i.e. it does not change the pressure). This is contrary to what the usual argument regards the connection between the speed of the airstream and the pressure on the wing is (which is supposed to explain the aerodynamic lift). If anything, the causal connection is in fact reverse: the airstream speeds up because the motion causes a high pressure region (front) and a low pressure region (back) to develop over the wings, and the air will be accelerated from the high to the low pressure.

    Yet, standard theories do not assume the asymmetry of an airfoil to be crucial to generate lift (see http://hyperphysics.phy-astr.gsu.edu/hbase/fluids/airfoil.html)(under 'Flying Upside-Down') which, on the basis of the above arguments, appears to be a rather unfounded assumption.

    However, one not only has to question the Bernoulli- interpretation of the aerodynamic lift here, but also the one based on Newton's law of action and reaction: this is not a problem of action at a distance, but the only way forces are excerted on the wing during flight is by the changed speed and number of air molecules hitting it. Both the 'downwash' and the higher airspeed are separate consequences of this rather than the cause.

    Again, I can not see that a symmetrical (front-back) wing profile could generate any overall lift. Neither should one of course expect any lift if the profile is top-bottom symmetric (assuming zero attack angle).
    Last edited by a moderator: May 11, 2003
  5. May 7, 2003 #4
    Wing does not have to be asymmetric to provide lift. Take a piece of paper and blow over it - you get lift regardless if it is symmetricly bent or not. Key is to have more velocity over it than under it.
  6. May 7, 2003 #5
    It is difficult to blow uniformly across a bent a piece of paper, however if you put a sheet of paper flat on a table, fix it to the edge of the table and blow over it from the edge: it will not move one millimeter.
  7. May 7, 2003 #6


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    Re: Re: Re: Bernoulli's Law and Aerodynamic Lift

    It isn't incompressible... but for up to Mach 0.3 you can treat it as such with negligible (<5%) error. (this value is derived from thermodynamic principles)

    For speeds higher than M 0.3, Bernoulli cannot be used to determine pressures accurately, and you need to resort to thermodynamics to obtain values.

    But it does change the pressure.

    Bernoulli is derived directly from conservation of momentum (IIRC) applied to fluids, assuming incompressible flow.

    The pressure changes as a direct result of conservation of mass and momentum (again IIRC)

    Correct! We've got a simple "cause/effect" nitpicking issue here. The real answer is that they are all tied together, but everything does line up as expected

    If you look at the 'hyperphysics' website you cited and look at the plot of p[oo] - p vs. position on the wing it will show that (p[oo] is the freestream static pressure). The pressure is highest at the leading edge of the wing. It is smallest along the back edge of the top. The air is then accelerated as it passes over the front tip and slows down again as it proceeds to the rear end of the wing. This meshes exactly with what happens in actual tests. The vast majority of the lift is produced along the front 1/3 of the wing.

    I wish that site had an overlay of flow velocity vs. position so you could compare with the pressure graphs. If you're interrested in hitting a library, I can find an example in one of my textbooks for you.

    No, asymmetry isn't crucial... but for a symmetrical airfoil you do need an angle of attack to generate lift.

    After thinking about the pictures you provided I realized that I had made an error. Assuming the flat bottom airfoil at 0 degree angle of attack is not a simple case.

    The stagnation point will not occur like it does for a traditional wing. Since you have a razor thin leading edge and no angle of attack, the stagnation point may occur on the top surface of the wing, and thus cause a downward lift component. If you have any positive angle of attack, it will form on the bottom, but predicting where on the bottom is not possible without tests or computer simulations.

    If you round out the front tip a bit, you can generate lift. As I mentioned above, almost all of the lift is generated along the front 1/3 to 1/2 of the airfoil (depending on actual shape). When calculating statistics for wing box structural dimensions, you don't even need to take into account the area near the rear of the wing (except for structural components holding up the ailerons).

    Unless the stagnation point is directly at the front of the wing (which will only occur for a symetric about the chord line airfoil at zero AoA), the streamlines near the leading edge on the bottom will become fatter and the streamlines (near LE) on the top will compress. The parts to the rear of the wing just don't contribute much lift. Air is a fluid, and fluids will stabilize with the surrounding envirronment, which is why they don't add much; the flow has stabilized.

    I have already had a discussion very similar to this one earlier in the year, and there is quite a bit of misunderstanding on what causes lift. There is even some junk science out there. The previous thread was started based on the incorrect assumption that the Bernoulli effect was based on the difference in curvature from top to bottom of the wing. This begged the question: "How do paper airplanes fly?"

    As soon as the semester is over in about three weeks, I will be writing an article on www.physicspost.com about the cause of lift from a technical perspective. I will be covering low speed (incompressible) flow, conservation of mass, momentum, and energy, and Bernoulli's equations and how they lead to lift.
    Last edited: May 7, 2003
  8. May 8, 2003 #7
    Re: Re: Re: Re: Bernoulli's Law and Aerodynamic Lift

    A symmetrical airfoil with a non-zero angle of attack is effectively asymmetrical (with regard to the airstream). This is exactly the point I am trying to make.
    From many publications one gets the impression that the motion of air parallel to a surface would create an under-pressure (and therefore a corresponding force). This is simply not true (see my response to Alexander's 'paper sheet'- example above). A change of pressure means a change of momentum transfer to the object, and for this to happen some parts of the surface must to some degree be orientated normally to the airstream. It is obvious that for a symmetric wing profile (for zero attack angle) the resultant upward- and downward forces will be identical (i.e. no overall lift).

    If the angle of attack is non-zero, even a plane wing will produce lift (the center of mass of the paper airplane would have to be adjusted accordingly, so that the airstream supports the wing from below).
    In a sense, gliders work actually just the other way around compared to propelled airplanes: they convert height (i.e.potential gravitational energy) into forward motion, rather than forward motion into height (lift).
    Last edited by a moderator: May 11, 2003
  9. May 8, 2003 #8
    You can't lift paper lying close to table surface (because you have practically vacuum under it so to speak and atmospheric pressure won't let you lift it unless your lift force is more than atmospheric pressure.

    Elevate slightly your paper - say, put a pack of sigarets under it's center. And blow hard - with a few m/sec at least (to get noticeable lift).

    You may just hold paper in hands and blow over one surface, again there is plenty of lift regardless paper shape.

    You'll see that symmetry does not matter - all it matters is difference in velocities above and below.
  10. May 8, 2003 #9


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    Maybe I'm missing the point here, but a symetric airfoil at zero aoa does in fact produce zero lift. So what's the problem?
  11. May 8, 2003 #10


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    Re: Re: Re: Re: Re: Bernoulli's Law and Aerodynamic Lift

    Not so.

    If a glider had no lift, it would drop like a rock.

    I'm with russ. I've lost what point you're trying to make in this thread.

    You should really look up a derivation of bernoulli's equation and a technical description about how lift is produced.

    Introducing a disturbance into a flow causes the streamlines to compress, which causes the flow to increase velocity, which causes the pressure to decrease. Pressure does not depend on what direction the flow is going. It is a point property and equal in all directions.

    For a symmetric airfoil at AoA=0, the flows compress equally on both top and bottom, producing no lift.

    Change in pressure does correspond to a change in momentum. No current methods debate that.

    And of course some portion of the wing needs to be oriented normal to the flow. If there weren't, it wouldn't be a wing.
  12. May 9, 2003 #11
    I am sorry, but if I put a sheet of paper flat on my fingers and blow over it as hard as I can, it stays firmly on my fingers. You need the paper surface to be (asymmetrically) curved to generate any lift or use a non-uniform airflow.
  13. May 9, 2003 #12
    I am not sure if we are talking about the same kind of symmetry here: this is not about top-bottom symmetry (for which Bernoulli's law would also predict a zero lift), but front-back symmetry, for which Bernoulli's law would predict a lift. However, from an elementary consideration of the change in momentum transfer to such an airfoil there should be no lift.
  14. May 9, 2003 #13


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    In that case, a lift forms for 2 reasons.

    1) The flow seperates towards the trailing end of the airfoil, especially with a large drop-off and non-parallel (to the flow) trailing edge.

    2) The flow balances itself a little as it progresses towards the trailing edge.

    Like I stated above, the vast majority of the lift is caused in the leading half of the airfoil. The aerodynamic center can easily be approximated (as a first cut) at the quarter-chord point for practically any airfoil.

    For those reasons, the symmetry on the trailing edge doesn't really matter in the calculations.
    Last edited: May 9, 2003
  15. May 9, 2003 #14
    Because you paper now bends backward.

    Hold it horizontally and make it SYMMETRICALLY bent down. Blow strong over top - it lifts.
  16. May 9, 2003 #15


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    Try it. The piece of paper IS curved due to gravity and only has air moving over the top. It is not a symetrical airfoil.

    Btw, there seems to be some confusion as to what "symetrical" and "asymetrical" are referring to. They are referring to the cross section of the airfoil. So a "symetric" airfoil is symetric along the entire width (chord) of the airfoil. There is nothing to make the airflow any different on top than it is on bottom at 0 aoa.
    Last edited: May 9, 2003
  17. May 10, 2003 #16


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    Indeed there is. In aviation terminology, symetric airfoil allways means top/bottom symetry. In this thread, it is not being used that way at all (see illustrations).

    However, I will repeat my post in the previous thread on this subject. The thing to remember about Bernoulli's law and aerodynamic lift is that it's all about air flow. If airflow is disrupted, lift is generated very inefficiently or not at all.

    The object in Fig.1 will indeed generate lift. Anyone who's thrown a Frisbe can tell you that.

    Fig.2 will also generate lift, but the virtex formed at the downwind side of the peek will decrease the efficiency of the lift. BTW; These vortices can be used to create usefull lift also, that's how the F-117 "Stealth Fighter" flies.

    The object in Fig.3 doesn't create lift because the airflow is hopelessly disrupted, (NOT because of a failing of Bernoulli's law).

    Fig.4 shows an object that will, beleve it or not, create lift. It would also create horrendous amounts of drag, and the lift-to-drag ratio would make it a terribly inefficient wing. But try it, it will lift (some).

    Which brings me back to a rather significant point; now that you have your proposed designs, and have formed hypotheses predicting their behavior, you should go to experimentation. If you cut the shapes out of styrofoam, or make them out of peper or something, you can turn a fan on them and see which ones get pushed strait back, and which ones get lifted up. Might be fun. (Have you ever done the experiment with the spool of thread and the sheet of paper? That one is just cool!)
    Last edited: May 10, 2003
  18. May 10, 2003 #17
    You can not have a lift without some disruption of the airflow. A wing with zero drag would also have zero lift because no momentum would be transferred to it in the first place (the total momentum transferred to a surface is proportional to the sine of the angle the surface forms with regard to the airstream, and of this a factor sine goes into the drag and a factor cosine into the lift, i.e. the drag force is proportional to sin&sup2;(a) and the lift force is proportional to sin(a)*cos(a) (which by the way has a maximum at 45 deg, with the drag having the same value for this angle). Obviously, you can only have a zero drag for a=0, but this would also make the lift zero.

    I am not exactly familiar with frisbees, but I doubt that you could create a lift with a zero angle of attack. A lift only developes if the airstream can support the underside of the frisbee, i.e. you have to tilt it somewhat when you throw it (I found a website confirming this conclusion)

    I cannot see how the profile of Fig.4 could generate any lift. The additional momentum created by the airstream would, apart from the drag, have a vertical component that points down but not up. Racing cars use similar aerodynamic profiles in order to have better contact with the ground, not in order to take off.

    The problem with your proposed experiments is that if you are using materials so light that they easily lift, they are also easily blown away. One would need to fix the object to a sensitive scale and in this way verify the lift force.
    Anyway, observing a lift in a given case does not help to understand it, in particular as there may be complicating factors like turbulence etc.
    Last edited by a moderator: May 11, 2003
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