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Old Sep 14, 2012, 09:09 PM
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Originally Posted by ShoeDLG View Post
When you push a car, the force you are applying is in the direction of the car's motion... it does work. A wing's lift force is perpendicular to the motion of the wing... it cannot do any work. When you lift water, you move it in the direction of the applied force. A wing does not move in the direction of the lift force.
Yup--
What does the work which keeps a glider aloft at a constant altitude?
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Old Sep 14, 2012, 09:40 PM
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If we're talking about an idealized glider with no drag, then there is no power (and no work) required to keep it airborne.

If we're talking about a real glider with drag, then the only way it can stay aloft at constant altitude is if it is in air that is moving upward. The glider is extracting energy from the upgoing air at rate equal to the drag force times its airspeed.
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Old Sep 14, 2012, 10:10 PM
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An efficient glider will be in level-flight equilibrium when the air is rising at a rate approximately equal to its airspeed times the drag divided by the lift:

v = Airspeed*D/L

where v is the speed at which the air is going up. Because the lift is approximately equal to the weight in this case:

v = Airspeed*D/W

The drag force acting on the glider is therefore approximately equal to the glider's weight times the air's vertical speed divided by the glider's airspeed:

D = v*W/Airspeed

The power being supplied to the glider is its airspeed times the drag:

P = D*Airspeed, which is from the above expression for D is: P = v*W

In other words the glider is extracting energy from the air precisely at the rate it would be losing potential energy if the air were not going up. This just shows that the upgoing air is supplying energy (doing work on the glider) at the required rate.
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Old Sep 14, 2012, 10:31 PM
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If you're more comfortable with the idea of the glider supplying energy to the air (ultimately in the form of heat), then you could view a glider flying at constant altitude from the perspective of a neutrally buoyant balloon rising with the air. From the balloon's perspective the air is still and the glider is falling. The same equations as above show that, in the balloon's reference frame, the glider's rate of potential energy loss is its airspeed times the drag:

v*W = D*Airspeed

The work done by gravity in this case has to supply energy at the same rate at which the drag is (ultimately) heating up the air. The lift force is not doing any work on the air (adding any energy to it).
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Old Sep 14, 2012, 11:07 PM
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Thanks for detailing it nicely
I was of course interested in the work provided by upgoing air
and the work done by gravity.
My point being - something has to do work to keep it aloft and the common comment would be, "lift" is doing the work.
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Old Sep 15, 2012, 02:07 AM
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I'm not sure if you're agreeing, pulling my leg, or missing the important idea completely.

People do sometimes confuse the term "lift" - referring to the buoyant motion of the air used by a glider to stay aloft with the "lift" force acting on the glider's wings... to the point where someone was suggesting it's a good idea to fly faster in a thermal because the wings create more lift when you fly faster... believing that will somehow make the glider go up faster.

The vertical "lift" associated with thermals is not the same, nor even analogous to, the aerodynamic "lift" acting on the wing.

When you "lift" an object from the floor to a shelf, that action is neither the same, nor analogous to the aerodynamic "lift" acting on the wing.

The "lift" is not the same as "to lift" which itself is entirely different from the "lift". This discussion was amusing, but is no longer providing much of a "lift", and is becoming work. I hope the previous explanation worked because I need to find a lift to work (where I need to use the stairs because the lift no longer works). Your obfuscation of this simple idea (I give you credit for deliberateness here) is ironic given your propensity to chastise others for complicating what appears straightforward to you. Fly more, cogitate less. ;-)
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Old Sep 15, 2012, 05:15 AM
greg
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Quote:
Originally Posted by richard hanson View Post
Thanks for detailing it nicely
I was of course interested in the work provided by upgoing air
and the work done by gravity.
My point being - something has to do work to keep it aloft and the common comment would be, "lift" is doing the work.
as my high school chemistry teacher described it, you can push and pull all day on the lab bench bolted to the floor and do no work because it hasn't moved, even though you are exhausted.

a magnetic field, drawing current from the power grid is doing no work even though it is holding something to the ceiling.
how is the force of a magnetic field generated from electricity -- how is lift generated from the movement of a wing through air ?

greg
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Old Sep 15, 2012, 07:24 AM
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Originally Posted by ShoeDLG View Post
I'm not sure if you're agreeing, pulling my leg, or missing the important idea completely.

People do sometimes confuse the term "lift" - referring to the buoyant motion of the air used by a glider to stay aloft with the "lift" force acting on the glider's wings... to the point where someone was suggesting it's a good idea to fly faster in a thermal because the wings create more lift when you fly faster... believing that will somehow make the glider go up faster.

The vertical "lift" associated with thermals is not the same, nor even analogous to, the aerodynamic "lift" acting on the wing.

When you "lift" an object from the floor to a shelf, that action is neither the same, nor analogous to the aerodynamic "lift" acting on the wing.

The "lift" is not the same as "to lift" which itself is entirely different from the "lift". This discussion was amusing, but is no longer providing much of a "lift", and is becoming work. I hope the previous explanation worked because I need to find a lift to work (where I need to use the stairs because the lift no longer works). Your obfuscation of this simple idea (I give you credit for deliberateness here) is ironic given your propensity to chastise others for complicating what appears straightforward to you. Fly more, cogitate less. ;-)
Long ago -I found, that one does not really understand a subject till they can have fun with it .
Lift is interesting to me because the concept , defininition and the uses of the word can mean different things---to me
Perhaps to others .
Lift AND drag -are another interesting/amusing study.
Are they opposites or really the same thing .
I enjoyed the exchange -
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Old Sep 15, 2012, 07:26 AM
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Originally Posted by ciurpita View Post
as my high school chemistry teacher described it, you can push and pull all day on the lab bench bolted to the floor and do no work because it hasn't moved, even though you are exhausted.

a magnetic field, drawing current from the power grid is doing no work even though it is holding something to the ceiling.
how is the force of a magnetic field generated from electricity -- how is lift generated from the movement of a wing through air ?

greg
Pressure differences .
My std answer and I 'm sticking to it
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Old Sep 15, 2012, 09:15 AM
Texas Buzzard
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Apples and Oranges. Lift has to be a result of a force.

Quote:
Originally Posted by DPATE View Post
I seem to have offended you; that was not my intention.

http://en.wikipedia.org/wiki/Work_%2...al_calculation
You have to take into account the direction of the motion and the direction of the force by using the dot product of these vectors. Since lift is defined to be perpendicular to the freestream flow it does no work on the object (because the dot product of perpendicular vectors is zero). Any work done on the object by the resultant aerodynamic force is via drag but not lift.
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DPATE, You Didn't even Try To Answer my Questions. Why?
You didn't offend anyone. I was posting to you just like I wpold talk to my Physics Students in class. I was trying to wake you up.

For instance: To illustrate your "definition(s)". Please draw an inclined plane of aprox. 20 dregrees, then draw a block of wood (which has mass) resting on the inclined plane.
Draw a horizontal vector ( your airstream!) which represents a Force that is pushing on the block of wood.
Allow that Force to be grater than any frictional forces. Apply enough Horizontal Force to shove that wooden block UP the plane through a distance.

QESTIONS:
1. Did the wooden block gain Potential Energy?
2. Was the block of wood acted on by a vertical resultant of the horizontal force?
3. WAS WORK DONE ON THE WOODEN BLOCK?
4. IF THE WOODEN BLOCK IS ALLOWED TO SLIDE DOWN THE INCLINED PLANE AFTER REMOVING THE HORIZONTAL FORCE, WAS THE FORCE OF GRAVITY ACTING IN A VERTICAL DIRECTION?
5. FINALLY. As the block was made to slide UP that plane by the Horizontal Force - 1.WAS WORK BEING DONE?; 2. Was Work Done?

PLEASE ANSWER even though you may now want to. It is a simple but fundamental exercise.

Addendum: You seem to say that DRAG causes the upward vector we represent as "lift". Does YOUR Drag have a Horizontal component? If you say YES then you have Contridicted yourself haven't you?

Sorry to ask you to do simple H.S. vector diagrams, but we have to be on the "same frequency".

Your original question asked what happens to the air when one moves a piece of paper that is not horizontal through the air AND WHY?
How would you answer that question as a "graduate student"?
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Last edited by Texas Buzzard; Sep 15, 2012 at 09:23 AM.
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Old Sep 15, 2012, 09:47 AM
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Mind if I watch?
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Old Sep 15, 2012, 10:09 AM
Grad student in aeronautics
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Quote:
Originally Posted by Texas Buzzard View Post
.................................................. .................................................. ..................
DPATE, You Didn't even Try To Answer my Questions. Why?
You didn't offend anyone. I was posting to you just like I wpold talk to my Physics Students in class. I was trying to wake you up.
The first thing we learn in grad school is humility. I put that title under my name because that's what I am. If you can't respect that, then this "conversation" is over.

As for your elementary questions, I think that exercise is best left to you.
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Old Sep 15, 2012, 02:25 PM
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Going back to sleep eh? ...lazy graduate students...
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Old Sep 15, 2012, 02:46 PM
Grad student in aeronautics
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Quote:
Originally Posted by ShoeDLG View Post
Going back to sleep eh? ...lazy graduate students...
Okay, I'll respond. Get ready for some sarcasm.

Here is what I learned in 2006:
Work = int( F*dr) from position one to position two. Where * is the dot product, F is the force vector, and dr is the differential position vector. The dot product of two perpendicular vectors is zero.

Here is what I learned in 2007:
Lift is defined as the component of the resultant aerodynamic force perpendicular to the freestream and drag is defined as the component parallel to the freestream. See below.


Combining these two pieces of knowledge gives the following result:
By definition, lift does no work on the object because lift is perpendicular to the direction of motion.
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Old Sep 15, 2012, 03:19 PM
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How could you leave out that in 2012 you learned that wings make lift by knocking air down? With that gem there isn't much need for any of the complicated stuff you wasted your youth on.
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