Sep 15, 2012, 03:58 PM
Ascended Master
Palmdale, CA
Joined Oct 2000
13,499 Posts
There is one (1) force on a fixed plate in a moving airstream, or a moving plate in a fixed airstream.
We can resolve that force into the directions of interest. The resolution of the force up (or down) perpendicular to the direction of the airstream we call Lift.
The resolved force in the direction of the airstream we call Drag. With a good combination of air flowing over the plate and area of the plate, the object in question will move. Restrained, as in a wind tunnel, the forces on the supports for the model can be resolved into the desired directions.. Up(or down) is called Lift, and (restrained by the support) in the direction of the airflow we call Drag.
Flying machines can make Lift with motion forward into an air mass, using a propulsion system to "thrust" it forward. And by cleverly making the flying machine light and stiff, with lots of area that is also cleverly designed to permit the machine to rise off the ground with sufficient forward motion, provided by the propulsion system.

Images

 Sep 15, 2012, 04:11 PM Registered User United States, UT, Salt Lake City Joined Oct 2007 7,183 Posts Bear with me - If I see the directions of all the arrows correctly----- the highest pressure is under the plate -along the FORCE line and the lowest is above that plate -along that line. The forces on the plate will change the relative direction ofthe FORCE line as speed and AOA change. The other two may remain the same . I am assuming the lift and drag arrows are relative to the drawing (left right /up down) not the wing or stream It makes sense to me if that is the case.
Sep 15, 2012, 04:47 PM
Ascended Master
Palmdale, CA
Joined Oct 2000
13,499 Posts
THE force on the plate is resolvable into whatever frame of reference one desires.
Typically the lift force component is taken as perpendicular to the airstream, and the drag force component parallel to the airstream.
Wind tunnel balances do this force resolution so that the designer doesn't need to know the magnitude of the force on the object nor its direction relative to the object, and its sin or cosine when dealing with the design.
Makes things easier to compute.
It's possible to determine the angle of the object relative to the airstream and find that angle where there is only the drag component, with no lift.
A flat plate or a symmetrical airfoil that's usually when it's aligned with the airstream.
A cambered surface, the zero lift line is generally with a negative angle of attack... F'rinstance the zero lift angle for a Clark-Y or similar semi-symmetrical shape is about 4 degrees below level.
On an airfoil polar, it's where the Cl line passes through the angle of attack zero line.
There will always be drag with motion.

Images

Sep 15, 2012, 05:57 PM
United States, GA, Atlanta
Joined Oct 2010
464 Posts
Quote:
 Originally Posted by ShoeDLG 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.
That would have been a sarcasm overload!
Sep 16, 2012, 02:13 AM
Registered User
Germany, BW, Stuttgart
Joined Mar 2012
857 Posts
Quote:
 Originally Posted by Sparky Paul THE force on the plate is resolvable into whatever frame of reference one desires.
This is certainly true, but as you point out, some choices are better than others. Why not define the lift force as the component of aerodynamic force acting perpendicular to the chord, and the drag as the component acting along the chord?
Seems like a pretty small and inconsequential shift from the traditional definitions.

The plot below shows CL and CD as a function of AOA for a rectangular wing with a Clark Y airfoil. The solid lines show the traditional resolution axes (perpendicular and parallel to the air flow). The dashed lines show the same force resolved perpendicular and parallel to the chord. The CL behavior doesn't change dramatically, but the CD behavior changes completely. In this coordinate system the "drag" has to become negative in order for the "lift" and "drag" components to sum up to the correct net aerodynamic force. I think we'd all agree that the traditional definition of drag far better captures the concept we are referring to when we say "drag".

Images

Sep 18, 2012, 12:22 PM
Texas Buzzard
McAllen,Texas
Joined Mar 2004
1,034 Posts
Where are the definitions that you use?

Quote:
 Originally Posted by DPATE By definition lift does no work. Drag does, lift doesn't.
Mr. DEPATE,
If you were writing a paper for a Philosophy class I suppose that your "definitions" would be called "apriori statements".

Using your way of explaining things is analagous to one saying that a
God caused the thunder storms and eclipses. One then has to ask what is your "first cause" that produces these effects/

I really am sorry that I am not communicating well to you. I will try for the third time.

If you simply say that an effect is truley factual BECAUSE A DEFINITION
SAID IT WAS leaves something to be desired. Give your authority for that definition.

For the second time you fail to answer simple questions that a grad student
in aeronautical engineering should easily answer. Are you a Social Science Major? I have a Masters in Chemistry with a Physics Minor attached. But hey, I have known practising Electrical engineers who can't solve ordinary problems in Static Electricity.
Sep 18, 2012, 12:32 PM
Registered User
United States, UT, Salt Lake City
Joined Oct 2007
7,183 Posts
Quote:
 Originally Posted by Texas Buzzard Mr. DEPATE, If you were writing a paper for a Philosophy class I suppose that your "definitions" would be called "apriori statements". Using your way of explaining things is analagous to one saying that a God caused the thunder storms and eclipses. One then has to ask what is your "first cause" that produces these effects/ I really am sorry that I am not communicating well to you. I will try for the third time. If you simply say that an effect is truley factual BECAUSE A DEFINITION SAID IT WAS leaves something to be desired. Give your authority for that definition. For the second time you fail to answer simple questions that a grad student in aeronautical engineering should easily answer. Are you a Social Science Major? I have a Masters in Chemistry with a Physics Minor attached. But hey, I have known practising Electrical engineers who can't solve ordinary problems in Static Electricity.
Asking the same question over and over and getting the same results -- leads one to think that :
A- you are asking the wrong questions
B- you are asking the wrong person
C- You are speaking different languages.
I would love to see some of these guys in front of a jury and being asked to explain in simple understandable terms , "what is lift?"
 Sep 18, 2012, 12:43 PM Grad student in aeronautics United States, GA, Atlanta Joined Oct 2010 464 Posts It's D for David. Feel free to call me David. Because we define lift to be the force perpendicular to the free stream velocity, and because a force acting perpendicular to the motion of a body does no work on that body, lift does no work.
Sep 18, 2012, 12:46 PM
Registered User
Germany, BW, Stuttgart
Joined Mar 2012
857 Posts
Quote:
 Originally Posted by Texas Buzzard I have a Masters in Chemistry with a Physics Minor attached.
If you can't understand that the component of force acting perpendicular to an object's motion doesn't do any work on the object, then perhaps it's time to detach that Physics Minor.
 Sep 18, 2012, 01:03 PM Registered User United States, UT, Salt Lake City Joined Oct 2007 7,183 Posts OK---- try that on a jury----- Not disagreeing that it has validity. It simply is too much of a leap from basic question. Been there
Sep 18, 2012, 01:25 PM
Registered User
Germany, BW, Stuttgart
Joined Mar 2012
857 Posts
Quote:
 Originally Posted by richard hanson OK---- try that on a jury----- Not disagreeing that it has validity.
I thought this forum was for discussing science. I shudder to think of where we would be if the litmus test for scientific merit was the appeal of an idea to a jury.
Sep 18, 2012, 02:06 PM
Registered User
United States, UT, Salt Lake City
Joined Oct 2007
7,183 Posts
Quote:
 Originally Posted by ShoeDLG I thought this forum was for discussing science. I shudder to think of where we would be if the litmus test for scientific merit was the appeal of an idea to a jury.
Science at what level?
it is a model form-
as for appeal to a jury-
You missed the point
It is what can you present that can be readily grasped by non technically oriented folk Not stupid - just not technically sophisticated .
I have likely noted it before but an old friend who was head of Enginering at a larg university told me that of a given graduating class , only about 10% of the degreed engineers were of much value as engineers .
I had brought up the subject because of the engineering crew I had foisted on me to do detail work .
Nice people -good at math -no good at innovative stuff
 Sep 18, 2012, 02:18 PM Registered User Germany, BW, Stuttgart Joined Mar 2012 857 Posts The term "work" has a very specific meaning in Physics. The work done on a body is the force applied to a body times the displacement of the body in the direction of the applied force. It follows directly from this definition, that motion of the body perpendicular to an applied force will result in no work done by that force. You don't need to be technically sophisticated to grasp that. Last edited by ShoeDLG; Sep 18, 2012 at 02:38 PM.
 Sep 18, 2012, 03:22 PM Grumpy old git.. Who me? Aberdeen Joined Mar 2006 12,009 Posts David / Shoe You are truly wasting your proverbial breath. You know that you are right, I know that you are right. But no matter how you explain these most simple principles of basic physics (such as 'work') some people cant, or wont grasp it. Best just accept it and move on, it saves a lot of frustration all round. Steve
Sep 18, 2012, 03:22 PM
Sink stinks
United States, GA, Atlanta
Joined Apr 2005
4,549 Posts
Quote:
 Originally Posted by ShoeDLG The term "work" has a very specific meaning in Physics. The work done on a body is the force applied to a body times the displacement of the body in the direction of the applied force. It follows directly from this definition, that motion of the body perpendicular to an applied force will result in no work done by that force. You don't need to be technically sophisticated to grasp that.
And there's a very good reason for this definition. Forces perpendicular to the motion of the body do not alter the energy state of the body. Thus, they do no work. If this were not true, you could say goodbye to the study of dynamics and many other fields.