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View Full Version : Aerodynamic Center / Neutral Point position : demonstration in a wind tunnel


manucordier
17th Apr 2012, 12:54
Hello,

I am having a hard time to find a footage on youtube (or elsewhere) which demonstrates the position of aerodynamic center (AC, 25%MAC) and neutral point (NP).

We can read in many books on aerodynamics that while the angle of attack (AoA) is changing on the wing taken alone, the pitching moment of the wing about the AC remains constant (thus AC is our pivot in this experience).

The reason for this, according to what we can read in the books, is because the lift (L) is effectively applied at the center of pressure (CP, which position changes for different AoA) while the change in lift (Delta L) takes place at the AC (which position is a constant: 25%MAC). Moreover, we know that any change in a force applied on the pivot (AC) of a body does not modify its total moment.

I suppose this could be easily shown in a wind tunnel by setting a pin through the wing at 25% of the MAC. Then you increase the angle of attack of the relative airflow. During the change in AoA, the pitching moment should remain constant. However once the variation of AoA comes to an end and the AoA remains constant, all the lift takes effectively place at the CP. So the pitching moment must change and generate a pitch up or down (depending on the position of the CG compared to that of the CP).

I want to see that in a wind tunnel.

The same goes for the Neutral Point, because the NP is, in some way, the AC of the whole airplane.

Could anyone provide me with a document (text, photos, movie) showing that experience in a wind tunnel.

Thank you very much.

Mad (Flt) Scientist
17th Apr 2012, 17:11
I dont believe it would be possible to do what you describe DIRECTLY.

If the aerofoil were free to rotate about the (estimated) AC/NP, the first problem would be that it WOULD rotate - under its own weight, if nothing else. So it'd be quite an effort to provide a balanced aerofoil.

The second problem is that you vary angle of attack in a wind tunnel by moving the model, not the wind. So you actually need to have control over the position of the aerofoil, rather than leaving it free to rotate.

The closest you could do would be to constrain the position though a pivot at the AC and measure the resulting moment on the pivot (perhaps by some simple force gauge) and compare wind on and wind off (which, very simplistically, is what a "real" wind tunnel does when accounting for what's called tares).

It might not be quite as compelling but it'd convince anyone it was proerly explained to i expect.

manucordier
17th Apr 2012, 19:46
Thanks for yout answer Mad (Flt) Scientist. It makes sense.

If anyone else has something to add, it is more than welcome.

cwatters
21st Apr 2012, 15:37
Not sure if this helps. The pivot point for a stabilator is typically close to the AC so that control forces don't change dramatically with AOA.

Stabilator - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Stabilator)

Obviously you may want to adjust the control force but rather than move the pivot point for the whole tail you can do that by providing an antibalance tab.

antibalance tab: Definition from Answers.com (http://www.answers.com/topic/antibalance-tab)

cwatters
21st Apr 2012, 16:34
Regarding..

However once the variation of AoA comes to an end and the AoA remains constant, all the lift takes effectively place at the CP. So the pitching moment must change and generate a pitch up or down (depending on the position of the CG compared to that of the CP).

Perhaps I missunderstand what you are saying but that's not correct. The pitching moment doesn't change just because the AOA stops changing.

I think this bit..

the lift (L) is effectively applied at the center of pressure (CP, which position changes for different AoA) while the change in lift (Delta L) takes place at the AC

might be confusing you.

A wing generates forces and torques. Neither ever act at or about a single point. For example the lift force is allways distributed over the surface of the wing. To some extent the AC and CP are just mathamatical constructs that make the maths easier. They don't change anything in the real world any more than changing from Cartesian (xyz) to Polar (r, θ) cordinates does.

For example lets say you want to write one equation for the vertical forces acting on a wing and another for the torques. If you choose the leading Edge as the origin then both equations feature the AOA. If you choose the AC as the origin only the one for the vertical forces features the AOA. In effect you have simplified the equations without having to do extra mathematical steps. However nothing has changed in the real world so you would get the same answers eventually.

So for the purposes of this thread.. If the AOA stops changing then the pitching moment also stops changing.

abdul aero
21st May 2012, 08:59
why cog is expressed interms of chord / mean aerodynamic chord.?

how about, expressing it by fuselage reference line?

can u please help me in this regard.

John Farley
21st May 2012, 11:48
abdul aero

To quote a CG as being at (say) 88.32 meteres from some datum tells me nothing wihtout extra information.

To say a CG is at 18% MAC or 31.6% MAC immediately makes me think "Cor that is a long way forward or a long way aft"

Mad (Flt) Scientist
22nd May 2012, 16:45
agree 100% with JF but...

often the CG calculation, done by the weights engineer, will use fuselage stations or some other convenient geometric reference (since he may be having to consider the weight and position of things a long way from the wing, and doing calcs with "-150% mac" is a bit silly really.

So the weight and cg calc will consist of defining a whole bunch of masses at a whole bunch of reference positions (often aft of some notional reference, just in front of the nose, so that everything is the same sign). Then all the individual moments are calculated, and then summed, and then divided by the total mass to get a cg position, in the same reference.

THEN, to make it a useful number for pilots, aerodynamicists and other strange people, it'll be converted into a %mac (or something similar) so that it becomes a "sensible" number.