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Aerospace Lift Force Calculations

  1. Jul 28, 2011 #1
    Hi, I'm right now coursing IB (equivalent to A-levels) and one of the compulsory essays we must do is called Extended Essay. It's an essay you do on a topic you choose. I choose aerodynamics. Created a "wind tunnel" out of plastic and made it rigid with hard plastic pieces which were partially flexible, and I also created a wing out of balsa wood. Now, I calculated the lift assuming the acceleration was uniform to then use the uniformly accelerated equations, which were then displayed in force diagrams. However my results when plotting lift generated vs angle of attack it gave me the optimum angle 45ยบ, and the shape of the graph was an inverse x^2, clearly it is not the correct graph.

    I know the results must be innacurate due to my assumptions, but does anyone know how to calculate the lift with another methodology?

    I also use the modern lift equation but I want experimental datas :)

  2. jcsd
  3. Jul 28, 2011 #2
    You say you calculated lift assuming the acceleration was uniform. What acceleration are you talking about?

    In a typical wind tunnel test the wing should be fixed as the air rushes past it at a constant velocity. The forces acting on the wing are then usually measured with a force balance. The air accelerates locally over the wing but I do not believe that is the acceleration you are referring to.

    The calculation of lift for an arbitrary body usually requires the use of panel methods or more sophisticated techniques.
  4. Jul 29, 2011 #3
    I haven't said much about the actual experiment... my bad, I'll explain. I made two holes on each side of the wind tunnel, where I put two wooden cylinders. Then there is a squared platform which attaches both cylinders at both ends of the platform where the wing will lay. Then in that platform there are two vertical cylinders. At the side of the wing there are attached to it two small circles which go through the vertical cylinders. So to calculate the force I time how long it takes the wing to go from the platform (resting point) to the top of the vertical cylinders. Therefore I have displacement, time, assuming acceleration is uniform, I can calculate the acceleration using one of the uniform accelerated formulas. Then newton second law nad I find the resultant force upwards, which is the lift. Of course there is friction but I can't get into that much detail due to the fact I do not have available any cool apparatus, no sensors, nothing. I built an anemometer to calculate the velocity of the fan, to then compare my experimental results with theoreticla results (Lift equation).

    The whole set up is hard to explain, tell me if you understood.

    THank you very much.
  5. Aug 15, 2011 #4
    I'm doing my extended essay also on lift and i thought also about doing your experiment but isn't there to much friction in order to get accurate results ?
  6. Aug 15, 2011 #5
    YKobe23, yes there is a lot of friction I must say, however I think it's the same in all AoA's, so therefore the only problem is I get less lift than there really is. However I might be wrong, because my results right now are wrong, so...


    P.S What are you doing it on now then your EE?
  7. Aug 15, 2011 #6


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    One problem I foresee is that your wing is so close to this platform that the platform will affect your results. If it starts out resting on the platform, you won't even get the thing to lift off.

    You would be better off just attaching the wing to the horizontal cylinders that leave the tunnel and Gavin those cylinders on a vertical track instead of just a hole. Attach some springs or rubber bands of known (or measurable) spring constant to cylinders to hold it at the bottom of this track and measure how far it moves against that spring force.
  8. Aug 15, 2011 #7
    Boneh3ad, You are right in that the wing is too close to the platform, however I did introduce some wood pieces so some air could get underneath the wing, making it lift, however I tried already the idea of a spring, using a newtonmeter, the smallest I found, and the force upwards is too small to distinguish the force produced at different angles :/
  9. Aug 15, 2011 #8
    is the friction the same whatever the AoA ? I think if you increase the angle of attack the amount of air passing under and over the wing increases, and therefore more molecules hit the wing and doesn't that create more friction ? or am i wrong ?
    Well i started with Daniel Bernouilli's principle in order to explain lift and now i wanted to find mathematically and experimentally the lift formula. And then to finish explain drag, and all that stuff...

  10. Aug 15, 2011 #9
    Normally yes, however in my experiment due to the fact the resting point of my wing is on a wooden platform, this restricts the air flowing under the wing, therefore no matter what AoA the air molecules under will be nearly consistent, which is what I think causes my data to be wrong. However the friction your talking about is tiny, air molecules cause friction, but the worst is the friciton caused by the poles when the wing is lifted.
  11. Aug 15, 2011 #10


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    Even with a little space between the wing and the platform, you will have significant wall effects. The majority of the airflow is likely going to just pass under the platform as opposed to going between it and the wing except for the case where the distance between the wing and the platform is significant. That is why in a real wind tunnel, you don't see the test articles being installed too close to the ceiling or floor. You end up with a large amount of blockage and it will significantly affect your results.
  12. Aug 21, 2011 #11
    Can someone help me on how to calculate lift distribution function of linearly tapered wing. I have to design a composite UAV wing.my email is cloudiuschuene@gmail.com
  13. Aug 29, 2011 #12
    If you don't allow air to freely flow beneath the wing, then you don't have a wing. You have an air trap. If you don't have a chord length of space between the wing and the "bottom" of the wind tunnel (or really, a lower surface) then you are ruining your experiment. I would go much further than a chord length, in fact.

    You are creating a mini wind tunnel beneath the wing which is not only causing a TON of drag (screwing up your results) but is also speeding up the air flowing under the wing, which causes a less significant pressure drop, which again, screws up your data.
  14. Aug 29, 2011 #13
    Okay I get the idea my data is screwed, so if I fix the airflow below I should be okay? I'll try that then and see how it goes.

    Another question: Anyone used javafoil?

    Thanks to everyonee.
  15. Aug 30, 2011 #14
    Can someone tell me if Bernouilli's principle ( 1/2p(v1)^2+ P1= 1/2p(v2)^2+ P2 ) correctly explain the theory of lift ?
  16. Aug 30, 2011 #15


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    No. It is impossible to fully explain lift without viscosity. Bernoulli can explain why the pressures are different as a result of the differing velocities, but it can't explain why the velocities are what they are.
  17. Aug 30, 2011 #16
    Bonehead, I agree that Bernoulli can't FULLY explain lift. But it can explain the pressure differential that defines it. It can provide a correct understanding of lift even if it doesn't accurately or fully describe it, no?
  18. Aug 30, 2011 #17


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    No. It can get you a pressure from a velocity field. That is about it. If you already know the velocity field, then yes, Bernoulli can get you the correct (or nearly correct) value for lift (or lift coefficient). However, it still does not explain lift. It also will not provide a complete understanding.

    Try and explain lift using only Bernoulli. The first thing you will say is there is a pressure differential. My question then would be "Why is there a pressure differential?" The answer is that from Bernoulli's equation, the air speed over the top, which is smaller than that over the bottom surface, leads to a lower pressure on top. "Alright, so why is the velocity on top faster?" At this point, you can't answer the question using Bernoulli. I invite you to try, but I am telling you it can't be done. Lift cannot be properly described without viscosity. That is why in potential flow around an airfoil, to get the correct lift you have to add a vortex at the trailing edge.
  19. Aug 30, 2011 #18
    Do you think for IB (similar to A levels) Bernoulli is enough ? What does completely explain the lift theory then to explain the velocities? Thank you
  20. Aug 30, 2011 #19
    I agree that Bernoulli is insufficient to fully explain the value of lift. And I also agree that in order to properly explain velocity fields you need your viscosity. But to explain the theory of lift, ie why a wing creates lift, it will serve as a pretty good description.

    What I mean is that if you make the assumption or observation that air is travelling faster over the wing than beneath, you can adequately explain lift with Bernoulli.

    Think of the spoon and faucet water experiment. Run water from a faucet and hold a plastic spoon neer the water. You will feel the spoon tend toward the water. You don't need to know ANY values to demonstrate theory.

    That being said. I would still say no. It can't correctly explain lift in a complete fashion.
  21. Aug 30, 2011 #20


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    I don't know anything about IB, so I couldn't tell you what is and isn't sufficient.

    I can tell you where lift actually comes from though. In an inviscid flow over a lifting body with a sharp trailing edge (as is the case with nearly all airfoils), the flow would start from the forward stagnation point and travel around the wing and leave from a rear stagnation point that depends on angle of attack. If the stagnation point just happens to not be at the trailing edge, the flow will just wrap around and find the rear stagnation point. This results in no lift and is predicted by inviscid theory.

    However, in real life with a viscous fluid, for the fluid to wrap around the sharp edge like that would require the velocity to go to infinity. Clearly this can't happen. The incredibly fast speeds around that trailing edge lead to the formation of the starting vortex, which is soon shed from the airfoil, which then has an equivalent circulation around it since vorticity must be conserved. This bound vortex, as it is called, is what enforces the sharp trailing edge as the rear stagnation point and what leads to lift. The bound vortex circulates in such a way that the upper surface of the airfoil sees faster moving air than the lower surface, leading to lift.

    This is often called the Kutta condition, and the value of this bound circulation can be used with the Kutta-Joukowski Theorem to directly calculate lift.
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