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Airfoil Lift

  1. Sep 10, 2010 #1
    I’ve been reading about airfoil lift and have found it rather difficult to find an answer to this question.

    Is the lift generated by an airfoil primarily dependent on accelerating static air gas particles down?
  2. jcsd
  3. Sep 10, 2010 #2


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    That is an awfully complicated question that can be answered with a simple "yes" if you aren't looking for an explanation of how that actually happens...
  4. Sep 10, 2010 #3
    Another way of seeing it is that the air pressure underneath the wing is much greater than above it.
  5. Sep 11, 2010 #4


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    And I'll throw in an explanation too.

    First an example in a setup where gravity is not a factor. You're in a single seater hovercraft, therefore no friction. Next to you a supply of 5 kg weights. To propel yourself to the left you throw out weights to the right. Whatever direction you're throwing the weights, since you don't have any grip on the ground you will propel yourself somewhat in the opposite direction; the action-reaction principle.

    This action-reaction principle is how rocket engines and jet engines achieve propulsion. The exhaust gases leave the nozzle at tremendous velocity. (The higher the speed of the exhaust the bigger the yield in propulsion per kilogram of spewed out exhaust.)

    The jet aircraft Harrier has the ability to use its jet propulsion for hovering, the exhaust is redirected downwards.

    Finally, to the lift of aircraft wings in normal flight. An aircraft sustains a constant altitude by accelelaring air mass down. It's the action-reaction principle.

    A source of confusion is that some people may mistake a side effect of the process for the essential physics. In the course of accelerating air mass down it is inevitable that there is larger air pressure beneath the wing than above. There's always that pressure difference, but it's not the principle of wing lift.
    Last edited: Sep 11, 2010
  6. Sep 11, 2010 #5
    I considered that if an aeroplane is flying at a steady altitude for a given amount of time, that the impulse of the aircraft (mass of craft x gravitational acceleration x time) would equal the impulse or change in momentum of all the air forced downwards over the duration (e.g. one ton of air accelerated to 1m/s downwards).

    Is this a correct assumption?
  7. Sep 11, 2010 #6


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    It looks correct to me.

    The nice thing about it is that it embodies the fact that an aircraft does in fact transfer impulse. The aircraft makes large quantities of air mass drop down, instead of dropping down itself.

    Then again, you don't know how much air mass is accelerated, and neither do you know to what velocity it is accelerated. Estimate one and you can calculate the other, but they remain estimations.
  8. Sep 11, 2010 #7


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    Yes. The impulse imparted by the plane onto the air is eventually transmitted through the air onto the surface of the earth. Think of it this way, the plane and atmosphere are a closed system, so the downforce on the earth will equal the total weight of the plane and the air (assuming the model is not accelerating vertically and after a stable state is achieved).
  9. Sep 11, 2010 #8
    Yea, thats where I was about to go. Im trying to get an understanding of flight efficiency. As KE= MV^2 the more the impulse is based on the quantity of air moved rather than the velocity of the air moved, the more efficient flight is.
  10. Sep 17, 2010 #9
    I believe we need to take care in how we ascribe downward acceleration to wing lift.
    If we look at the flow over the top of the wing we see that there is indeed a downward mass acceleration that is equal to the pressure relief contribution.
    This downward acceleration is a component of the acceleration of the flow curvature, which is locally normal to the surface. It is the changing of the velocity vector. The mass inertia of the air resists this normal acceleration resulting in a reduction in its ability to apply static pressure. Static pressure has NO directionality. The only directionality of force upon the wing is normal to the surface elements.
    While the up and down components of the reaction are equal, they are produced by the accelerations normal to the surface.
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