There is no "lift"!

Discussion in 'Hydrodynamics and Aerodynamics' started by jehardiman, Jul 25, 2013.

  1. jehardiman
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    jehardiman Senior Member

    Ok, I understand it is an "in your face" title...but it is one of my soapboxes/touchstones that I came to an understanding with many years ago...

    So lets start on a discussion of a "a unified propulsion theory", i.e. there is no real difference between an oar, a sail, and a propeller....

    There is no "lift"...there is only "drag" in the direction you want to go.

    1) Premise: When taking the scalar ("dot") product of the fluid inflow vector to the body force vector, the scalar ("dot") product is always non-negative for steady state.

    2) Primise: The only expection to (1) is due to an energy gradient (gravitionational acceleration or system work input) where there is not a steady state condtion and only when movement is is the direction of the energy gradient.

    Comments...
     
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  2. DCockey
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    DCockey Senior Member

    It depends on the definition of lift and the reference frame used. Perhaps you could state which your comment refers to?
     
  3. jehardiman
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    jehardiman Senior Member

    Exactly what it says..."the scalar ("dot") product of the fluid inflow vector to the body force vector. the scalar ("dot") product is always non-negative for steady state."
     

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  4. jehardiman
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    jehardiman Senior Member

    Yes, but "lift" is a fictious force...it does not exist in and of itself, it is only a function of the Body Force. This is the point I wish to make, and where to start from....There is no real "lift".
     
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  5. DCockey
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    DCockey Senior Member

    The usual definition of drag is the component of hydrodynamic force in the direction of the inflow.

    The usual definition of lift is the component of hydrodynamic force normal to the direction of the inflow.

    I added vectors showing the lift and drag components of the body force vector in your diagram.
     

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  6. DCockey
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    DCockey Senior Member

    Looks like you responded to my post while I was making a small correction.

    If you are stating that lift is not an independent force then I agree. I've always considered lift to be a component of the body force.
     
  7. jehardiman
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    jehardiman Senior Member

    So if "lift" never goes "forward" into the flow (assuming you concur with Primise (1)), and the Body force is always scalar positive to inflow, then any lift is "-work" done on the system. (i.e. force * distance on the output side is -work...note to all, agrees with the 2nd Law...entropy increases)
     
  8. DCockey
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    DCockey Senior Member

    Yes, for steady flow. Unsteady flow be more complicated.
     
  9. Mr Efficiency
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    Mr Efficiency Senior Member

    " There is no lift ! " Wonderbra advertising contradicts this assertion, imo.
     
  10. DCockey
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    DCockey Senior Member

    Also, the "inflow" is the inflow relative to the particular "body" associated with the body force. For a sail the inflow is the apparent wind, for a propeller blade or an oar blade the relevant inflow is the direction of the water relative to the blade. So for the sail, propeller blade, oar blade, etc the "inflow" will generally not be the negative of the vessel's velocity through the water.
     
  11. jehardiman
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    jehardiman Senior Member

    Yes, and go on....

    Prove me wrong, reply, mathematicly, in vector algebra...show your reference planes...don't leave a hand wave where a mathematical proof will serve....lol. I really want somebody (other than me) to explore this...I have so much data I want to understand where this data goes.. leads to....lol...This is why I tell people to go back to Prandlts original papers on propeller theory...lots of asumptions were made...lol. IS THERE A REAL DIFFERENCE!
     
  12. Petros
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    Petros Senior Member

    Sir Issac Newton called it a force; F=ma. it does not matter that you call it lift, thrust, suction, or a force, it is the same thing, caused by accelerating a mass. It is only convention that is is called lift.

    All of the useful work, no matter what you call it, is a force.
     
  13. daiquiri
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    daiquiri Engineering and Design

    I guess that Jehardiman's point here is, the power is a scalar product of two vectors: "hydrodynamic force" and "water speed".

    Since the "water speed" vector does not have components perpendicular to itself, it means that only the component of the "force" vector parallel to the water speed (i.e. drag) is involved in the computation of the power, and lift has no place in it.

    So we get back again to my statement in the original thread about oars, that oars are devices which work by creating drag...
     
  14. Petros
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    Petros Senior Member

    Daiquiri, you are incorrect, an oar does not work by creating drag.

    consider a old round ww2 style parachute, they have a L/D ratio of about 3 or 4 to one, the object of a parachute is to slow the decent of the parachutist. without the lift generated by the canopy the parachute would have to be about 4 times larger to safely land a human. but also consider that both the lift and drag vector is in a useful direction for slowing the decent. but that is not to say that both of these components is drag. one is caused by the potential flow and perhaps vortex flow over the canopy, the other is caused by simple drag through the air. Both serve to slow the decent, but both are generated by very different means.

    The same can be said for a paddle blade, there is a large component of lift caused by the potential flow over the blade being accelerated over the surface, and a component of drag coming off the blade. Both contribute to the propelling the row boat forward, but both are generated by very different means. to confuse the two means you do not understand fluid mechanics. when you heave against a oar, you are pulling mostly against the lift or force generated by the blades with the vector pointing forward.
     

  15. philSweet
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    philSweet Senior Member

    jehardiman,

    You're fine as long as the entire propulsion system is one piece and no attention is paid to its stresses or design. But suppose you have a paddle, and you want to minimize its weight and momentum. In order to do that, you need to look at the forces that the paddle encounters and the orientation of the paddle to those forces. Lift is one of those forces. Once the component is designed, feel free to chuck it. I think the idea of disposing of it when considering the entire assembly can save a lot of mental anguish.

    With respect to efficiency, I suspect the only devices suitable for assigning an efficiency based on the Eta_p = 2/(1+Ve/Vs) are ones that can be described without referencing lift.
     
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