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proppastie

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Looking at Gottingen 535 airfoil data: There is data point for C.P. (Center of pressure?) at 92% of chord from leading edge at -6 deg, and the C.P. curve is extended to -150% of chord from leading edge. So how is the center of pressure not on the wing, off in space so to speak, or what does that mean. If my flapperon is for example at 90% from leading edge seems like that is a lot of lift/force on the flapperon and fabric portion of the wing when at negative angles of attack, without even talking about gust loading.
 

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Jay Kempf

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Could be a function of the way the balance beam was set up in the test rig. In other words they picked a datum/pivot off the wing.
 

StarJar

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Like Jay said, this is an abstract average. And C.P. is different than moment, because it has no 'quantity'. The actual force could be very small, and probably is, at -6 deg. on this particular airfoil. Additionally, most of that force could be from a downward force on the nose of the airfoil, which in effect would move the C.P. rearward.
The actual twisting force, or Cm could be very small, and nothing to worry about.
Now, if your Cm came out to 100%+, you would be in big trouble, when you multiply it by speed, it would get real big.
At -6 AOA, if your lift was +1oz. or -1oz., Then who cares where 1oz. of twisting force is applied, because it is so small.
Good question..it got me thinking.
 
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Autodidact

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Here's a good illustration of the why (attachment); at low angles of attack, a highly cambered airfoil does have downward lift at the front and upward lift farther back, which is a couple and creates a moment. A symmetrical airfoil only has downward lift at negative angles of attack, so that you can divide lift on a cambered airfoil into two parts: that due to camber (basic lift) and that due to AoA (additional lift).

It's important to see that the % of chord values in the airfoil graph do not go negative as they move down the page so that at very low lift coefficients (which also happens at high speed if the wing loading is kept constant) the center of lift is out behind the wing because of the moment (couple) created:

pressure_distribution_airfo.gif
 

Birdman100

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Like Jay said, this is an abstract average. And C.P. is different than moment, because it has no 'quantity'. The actual force could be very small, and probably is at -6 deg. on this particular airfoil. Additionally, most of that force could mostly be a downward force on the nose of the airfoil, which in effect would move the C.P. rearward.
The actual force, or Cm could be very small, and nothing to worry about.
Good question..it got me thinking.
CP, like name suggests is point, and like that the only relevant info is its position. What really happens around the airfoil is pressure field, and that pressure field can act only on airfoil surface (pressure is scalar and every part of air has its value, but we can feel and measure that pressure only at some surface).

But in calculations, for the reason of simplicity and practicality we substitute all that pressure with corresponding forces (lift and drag) and one moment.

If we for whatever reason want to substitute lift, drag and moment with only one force that will still have the same mechanical effect on the airfoil we can do that by putting that force in center of pressure.

So, CP is CALCULATED (or imagined if you like) point in which we can put one force (lift) so we have still the same mechanical picture. Of course that point doesnt exist in reality it is purely mathematical.

The fact CP goes backward (with decreasing AoA) and in one moment it goes far beyond TE just show us we have much moment acting on the wing (airfoil). Try to see where CP is at zero lift AoA;)
 

proppastie

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At -6 AOA, if your lift was +1oz. or -1oz., Then who cares where 1oz. of twisting force is applied, because it is so small.
looking at the lift equation, then when CL=0 Lift=0 and very little lift force
 

Birdman100

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looking at the lift equation, then when CL=0 Lift=0 and very little lift force
CL=0 and Lift=0 is trivial case so it is not of an interest (that is when your craft is in a garage)
In (horizontal) flight CL is never 0, but it can be low due to high speed! That case should worry you as you still have a lot of lift, actually if flying horizontal your LIFT=WEIGHT and that is not +/- 1oz but several hundred pounds!

If you fly fast, that means your AoA is low (can be negative as well) and the CP is far behind meaning that moment is huge and so wing twist.

At -6 AOA, if your lift was +1oz. or -1oz., Then who cares where 1oz. of twisting force is applied, because it is so small.
Good question..it got me thinking.
First, its not 1oz and second it is very important where is applied. Far enough and you have destructive moment. Thats why we have VNE

"Give me a lever long enough and a fulcrum on which to place it and I shall move the world" - Archimedes
 

proppastie

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CL=0 and Lift=0 is trivial case so it is not of an interest (that is when your craft is in a garage)
In (horizontal) flight CL is never 0, but it can be low due to high speed! That case should worry you as you still have a lot of lift, actually if flying horizontal your LIFT=WEIGHT and that is not +/- 1oz but several hundred pounds!

If you fly fast, that means your AoA is low (can be negative as well) and the CP is far behind meaning that moment is huge and so wing twist.



First, its not 1oz and second it is very important where is applied. Far enough and you have destructive moment. Thats why we have VNE

"Give me a lever long enough and a fulcrum on which to place it and I shall move the world" - Archimedes
Now that you have answered the CP out in space away from the wing question.

Trying to understand:.....CL=0 at -9 degrees. Cd=0 at -6 to -10 degrees. Going down hill for sure, but is the wing seeing forces,or does the lift and drag equations not apply. I am thinking of the "vomit Comet" and "zero Gs", is that shown on these curves?
 

StarJar

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CL=0 and Lift=0 is trivial case so it is not of an interest (that is when your craft is in a garage)
In (horizontal) flight CL is never 0, but it can be low due to high speed! That case should worry you as you still have a lot of lift, actually if flying horizontal your LIFT=WEIGHT and that is not +/- 1oz but several hundred pounds!

If you fly fast, that means your AoA is low (can be negative as well) and the CP is far behind meaning that moment is huge and so wing twist.



First, its not 1oz and second it is very important where is applied. Far enough and you have destructive moment. Thats why we have VNE

"Give me a lever long enough and a fulcrum on which to place it and I shall move the world" - Archimedes
Yes, I see your point and agree 95%.
All I'm saying is the Center of Pressure has no meaning for the structural design, and one must refer to the Cm.
The aircraft can move through the air with zero lift, and this is the zero lift on the lift polar, which may be at -6 degrees.
But of course you are right that the airplane can not sustain itself at zero lift.

If you fly fast, that means your AoA is low (can be negative as well) and the CP is far behind meaning that moment is huge and so wing twist
.

If the CP is far behind, the Cm could be very small, or very large. You don't know untill you look at the Cm. It could be a pebble, sitting far behind, or a boulder. The CP tells you no magnitude.

But I stand corrected, that it is not neccessarilly small, like I said. But also, probably not out of the ordinary for airfoils of that shape.
 
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StarJar

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Now that you have answered the CP out in space away from the wing question.

Trying to understand:.....CL=0 at -9 degrees. Cd=0 at -6 to -10 degrees. Going down hill for sure, but is the wing seeing forces,or does the lift and drag equations not apply. I am thinking of the "vomit Comet" and "zero Gs", is that shown on these curves?
OK, I've been a little off-track, and Birdman is right, this might be be very high twisting force, but what I was tyring to say, not abnormally high for a cambered airfoil. I guess I was egsagerating to show the difference btween Cm and CP.

If you do not have a Cm polar, I think you have to use a formula that use CP and Lift instead.

When you say drag=0, that can't be right because there is always drag.

But drag will not be needed in finding the twisting force.
 

proppastie

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OK, I've been a little off-track, and Birdman is right, this might be be very high twisting force, but what I was tyring to say, not abnormally high for a cambered airfoil. I guess I was egsagerating to show the difference btween Cm and CP.

If you do not have a Cm polar, I think you have to use a formula that use CP and Lift instead.

When you say drag=0, that can't be right because there is always drag.

But drag will not be needed in finding the twisting force.
No Cm, but looking at others about -.09, I do not know how to use that yet. Cd is coefficient drag of airfoil, flat plate drag, or induced drag, I do not know how to calculate yet. But in my ignorance I still see a zero lift AoA.
 
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Autodidact

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Trying to understand:.....CL=0 at -9 degrees.
There is both negative and positive lift @ -9 deg, and the are equal to each other. But they are located at different places on the chord and still create the moment.

Cd=0 at -6 to -10 degrees. Going down hill for sure, but is the wing seeing forces,or does the lift and drag equations not apply.
The lift and drag equations always apply, see above for the lift; the drag does not go to zero for the Gottingen airfoil you posted - it's lowest value is about 0.018...
 

Topaz

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...But in my ignorance I still see a zero lift AoA.
Zero-lift angle of attack (for the airfoil) is important because the Cm at that AoA is the one you'll use as a baseline in trim calculations. For a wing, as opposed to a 2D airfoil, there are additional drag factors beyond the Cd of the airfoil. There is induced drag, which is present any time the wing is developing lift. There is additional parasite drag from real-world protrusions like rivet heads, control horns, mass balances, etc.; deviations from the "perfect" airfoil (low and high spots, or generally out of contour). There is interference drag where struts, fuselage, and nacelles join the wing. Control surface and flap deflections create their own drags and moment contributions when they're deflected. In some cases (ailerons in particular), these are not symmetrical across the span in a given flight condition, which creates a torque in the wing structure all by itself. There are also inertial loads from accelerated flight such as in turns, pull-ups, gusts, landing impact, etc. I recall hearing of at least one aircraft where the negative-g load case on the wing was imposed by the specified worst-case landing impact for the gear design when the aircraft was full of fuel, rather than by aerodynamic loads. Gust loads can be quite significant, and should always be looked at. FAR 23 gives a nice, relatively simple method for that.

Developing a real-world loading case for a wing (or any airframe component) is a matter of looking at all of these loads and combinations of loads, and determining the worst-case inside the flight envelope that applies the highest load to that particular component, allowing you to size it to that worst-case load. You have to have a very clear understanding of the flight envelope, aerodynamics, and requirements imposed on your aircraft before you can even begin to hope to develop an accurate loads analysis for it. Typically loads analysis happens after the completion of the conceptual/configuration design phase that is the subject of Dan Raymer's Aircraft Design: A Conceptual Approach. His Simplified Aircraft Design for Homebuilders takes a much-simplified, more "cookbook" approach that's really only useful as a primer for the larger book, IMHO. But it might be a good place to start in order to get a broad-brush overview on the task. Another good resource, especially for learning how to calculate pitch-axis moments and to develop trim plots, would be John Roncz's series of articles in Sport Aviation from 1990, entitled, "Designing Your Homebuilt". I'm sure there are others in the same vein available.

Once loads and structural analysis are complete, the design of the aircraft can move into the detailed design phase, where you start drawing the individual detailed components and systems, and figure out all the ways they'll need to be individually manufactured and assembled.
 
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Norman

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No Cm, but looking at others about -.09, I do not know how to use that yet.
post #4 pretty much covered it but since the discussion is still ongoing I guess a different picture is needed. The moment coefficient is a constant about the aerodynamic center which is near, but usually not at, 25% of the chord. It results from the fact that the sum of all vertical pressures on the airfoil do not act through the AC. Specifically if you total the vertical component of all the vectors on both surfaces at zero lift you'll get a picture like the attached drawing. Even though there's no net vertical force there's still a moment and since a moment is a force times a leaver you have a multiply by zero problem ie the moment is non-zero but there's no vertical force on the lever so the length of the lever must be infinite. Obviously you can't do math with infinities so aerodynamic labs switched to moment coefficient. Often airfoil data is shown relative to the 25%c point and the Cm curve will be all over the place. The Cm at the AoA for zero lift should be pretty close to the constant measured about the true aerodynamic center. This little spreadsheet will find the correct X position of the AC but the AC also has a Y offset and I don't know how to find that.
Cd is coefficient drag of airfoil, flat plate drag, or induced drag, I do not know how to calculate yet. But in my ignorance I still see a zero lift AoA.
Cd is what's called "form drag" it's sometimes also called "zero lift drag" It's the sum of pressure drag and friction and does not include lift induced drag.
 

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proppastie

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post #4 pretty much covered it but since the discussion is still ongoing I guess a different picture is needed. The moment coefficient is a constant about the aerodynamic center which is near, but usually not at, 25% of the chord. It results from the fact that the sum of all vertical pressures on the airfoil do not act through the AC. Specifically if you total the vertical component of all the vectors on both surfaces at zero lift you'll get a picture like the attached drawing. Even though there's no net vertical force there's still a moment and since a moment is a force times a leaver you have a multiply by zero problem ie the moment is non-zero but there's no vertical force on the lever so the length of the lever must be infinite. Obviously you can't do math with infinities so aerodynamic labs switched to moment coefficient. Often airfoil data is shown relative to the 25%c point and the Cm curve will be all over the place. The Cm at the AoA for zero lift should be pretty close to the constant measured about the true aerodynamic center. This little spreadsheet will find the correct X position of the AC but the AC also has a Y offset and I don't know how to find that. Cd is what's called "form drag" it's sometimes also called "zero lift drag" It's the sum of pressure drag and friction and does not include lift induced drag.
One of my problems is I do not have the airfoil data for the Carbon Dragon, I found the Super Floater on this site (thanks guys) and the performance and profile looked close so I decided to use it for calculations. The data is as posted with no Cm. However the Cp is shown. So if your AC is the same as Cp as shown in the data I can reverse calculate using your spread sheet, to find the Cm of the Gottingen.
 

proppastie

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Often airfoil data is shown relative to the 25%c point and the Cm curve will be all over the place. The Cm at the AoA for zero lift should be pretty close to the constant measured about the true aerodynamic center. This little spreadsheet will find the correct X position of the AC but the AC also has a Y offset and I don't know how to find that.
I got a chance to look at the "airfoil data" page, right now mostly over my head, however when comparison to say NACA 4421 in Abbott and Von Doenhoff the Cm data is "all over the place" and whereas the latter is a straight line at -.085 or so. Using the data I mounted and your spread sheet at CL=0 your AC =.25 where as my data shows at CL=0 CP at 140% from the LE. So I take it AC is not CP, or different reference points is causing a problem for me.
 

StarJar

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I got a chance to look at the "airfoil data" page, right now mostly over my head, however when comparison to say NACA 4421 in Abbott and Von Doenhoff the Cm data is "all over the place" and whereas the latter is a straight line at -.085 or so. Using the data I mounted and your spread sheet at CL=0 your AC =.25 where as my data shows at CL=0 CP at 140% from the LE. So I take it AC is not CP, or different reference points is causing a problem for me.
Wouldn't you rather work with Cm? It's the measurement used in all modern design books, with fairly simple formulas to find your pitching moment.
Here's the Goe 535 on Airfoiltools.com. You can look at the Cm at Reynolds number 1 million, or you can copy the the coordinates of the Goe 535, from Airfoiltools, and paste them directly into the free JavaFoil program to create custom polars for your RN, or even get a 3d estimate at chosen Aspect Ratios. You can even input cord lengths, and mach numbers for more accuracy.
 

proppastie

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Wouldn't you rather work with Cm? It's the measurement used in all modern design books, with fairly simple formulas to find your pitching moment.
Here's the Goe 535 on Airfoiltools.com. You can look at the Cm at Reynolds number 1 million, or you can copy the the coordinates of the Goe 535, from Airfoiltools, and paste them directly into the free JavaFoil program to create custom polars for your RN, or even get a 3d estimate at chosen Aspect Ratios. You can even input cord lengths, and mach numbers for more accuracy.
I will give it a try I did not have that link until today. I am still confused as why one source shows the Cm constant about .08-.1 and this site shows it not.


I still plan on using the Irv Culver airfoil, however finding one close will help. The Super Floater also has the USA 35-A with R:3,190,000.
 
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