2
$\begingroup$

Center of pressure(CoP) is point where total force vector acting at. If you hold wing at this point, wing will not rotate(moment=0). Example hanging point at hanglider.

I see at some diagrams for very low and negative AoA-s that center of pressure is out of wing physical dimensions. At diagram below for AoA 0.5°, -3°, 4° What does it mean in reality, no place where can I hold wing without rotation? Or if I extend wing with some stick and hold at this position, wing will not rotate?

At first I will tell that for every AoA, must be place on wing where moment is zero.But obviusly I am wrong. Hard to imagine to me how pressure distribution around wing must be to shift CoP so much out.

That also mean if you dont know what you are doing, you can design so bed hanglider wing that will shift CoP out wing dimension for some AoAs. This is dangerous situation, theoretically speaking. enter image description here

$\endgroup$
8
  • $\begingroup$ That's why the modern way to analyze the situation is treat the lift force as acting at the wing's aerodynamic center, and also include a pitching moment term. Still, the question appears to be answerable. $\endgroup$ Dec 15, 2021 at 14:28
  • $\begingroup$ "My logic tell me that for every AoA, must be place on wing where moment is zero, it can not be out of wing dimensions" is wrong. Surely, it's not out of the realm of imagination that, at some incidence angle, no matter where you place your finger on the airfoil, the airfoil would always rotate? In that scenario, the CoP is out of the dimension. $\endgroup$
    – JZYL
    Dec 15, 2021 at 14:34
  • $\begingroup$ @JZYL So math again showing absolute truth. That mean you can design bed hanglider wing that will not have correct hangpoint position for some AoAs? Hmm this is scary. $\endgroup$
    – Jurgen M
    Dec 15, 2021 at 15:04
  • $\begingroup$ @JurgenM No, that's why you have tail or you use reflex airfoil with elevon, which guarantee feasible trim point at least until buffet/stall. $\endgroup$
    – JZYL
    Dec 15, 2021 at 15:07
  • $\begingroup$ @JZYL Yes but in theory you can make so bed wing that will shift CoP out of wing.I talk theoretically $\endgroup$
    – Jurgen M
    Dec 15, 2021 at 15:09

1 Answer 1

1
$\begingroup$

If you want to be able to trim a hang glider to have zero pitch trim force (pilot doesn't need to push or pull on control bar) with the airfoil shown above, at 0.5 degrees angle-of-attack, assuming drag on the pilot's body is negligible, the "hang point" would have to be vertically in line with the circle on the diagram that is just aft of the trailing edge of the wing, labelled "0.5 degrees". Connecting the hang strap right at the circle would accomplish this, but that is not only possible configuration-- it could also connect above, or below, the circle.

If drag on the pilot's body is not negligible-- so that when the pilot lets go of the control bar, the hang strap angles backwards rather than running straight up and down-- then the extended line of the hang strap will have to pass through the circle on the diagram that is just aft of the trailing edge of the wing, labelled "0.5 degrees". Connecting the hang strap right at the circle would accomplish this, but that is not only possible configuration. It could be attached below the circle and slightly aft of it, or above the circle and slightly forward of it. The key point is that the extended line of the hang strap must pass through that circle.

(Note that we're only talking about trim, not stability. Assume for simplicity that we have a straight, non-swept, constant-chord wing. Such a wing would not be stable-- the angle-of-attack would not tend to stay constant-- at least with a conventional airfoil cambered in the usual direction-- but we can still discuss the trim characteristics at any given angle-of-attack, where the forces and torques are at least momentarily in balance. Hang glider stability dynamics have been discussed elsewhere on ASE, and are fertile ground for additional future questions.)

Now imagine that we flip the wing upside down, so that we have an airfoil with "negative camber", and wish to fly at a 3-degree angle-of-attack. Now to trim the glider for hands-off flight, the extended line of the hang strap will have to pass through the circle labelled "-3" on the diagram. Again, this can be accomplished by attaching the hang strap right at the "-3" circle, but this is not the only possible configuration.

It seems somewhat counter-intuitive that the center-of-pressure could lay outside the physical dimensions of the airfoil, but it is possible. The reason that this happens is that any non-symmetrical airfoil generates a pitching moment. If the airfoil is cambered in the normal direction, the pitching moment acts in the nose-down direction. Even at the zero-lift angle-of-attack, the airfoil generates a nose-down pitch torque. The modern way to analyze the forces and torques generated by an airfoil is to treat the Lift and Drag forces as acting at a Center of Pressure which is defined to be fixed in location, at the quarter-chord point of the wing. Any nose-up or nose-down aerodynamic pitch torque that would exist if the wing were fixed on a pivot at this Center of Pressure, is expressed as a Pitching Moment Coefficient.

$\endgroup$
2
  • $\begingroup$ You can extend with stick to circle 0.5 and put hang strap here and when wing increase AoA it will stall and kill yourself. $\endgroup$
    – Jurgen M
    Dec 15, 2021 at 16:58
  • $\begingroup$ @JurgenM-- see my comments in answer about hang glider stability-- despite your comments, hang gliders do indeed have cambered airfoils, yet are indeed stable in pitch. I suspect that both you and I know the reasons that this is possible, but that's a whole 'nother topic entirely. $\endgroup$ Dec 15, 2021 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.