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The picture below shows the location and movement of centre of pressure when angle of attack is increased. Consider the aircraft is trimmed at the second airfoil in the picture. Thus the horizontal stabilizer has to provide a downforce to trim the aircraft. Now if the angle of attack increases, the centre of pressure moves towards the .25c point thus reducing the moment. If the stabilizer is still providing downforce, then the aircraft is unstable. How does the stabilizer give positive static stability?

figure 1

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  • $\begingroup$ What you show is a 2D view of the center of pressure. But the center is not the same on the whole wingspan and the wing can be swept. What will count is the overall torque for the lifting surfaces, the whole wings and the fuselage too (aerodynamic center). See here. That said, the inversion you describe exists, just think of aerobatics. $\endgroup$
    – mins
    Aug 13, 2016 at 9:45
  • $\begingroup$ See also this answer for an explanation. When the AoA on the wing changes, it does equally so at the tail, so the downforce is diminished as AoA goes up. $\endgroup$
    – Peter Kämpf
    Aug 13, 2016 at 10:02
  • $\begingroup$ @PeterKämpf, Even if the downforce is diminished as the AoA goes up, it still doesn't provide the pitch down moment necessary for positive stability. The above figure is true for a rectangular wing with no washout, sweep etc. right? In the pages you have linked, there is a comment that says that the tailplane does not need to produce a downforce in a conventional tail. How will the aircraft be stable then ? $\endgroup$
    – Sherlock_Dumbledore
    Aug 13, 2016 at 14:13
  • $\begingroup$ @GokulJ Yes it does. Did you read the first paragraph of this answer? What is it you don't get? The relative lift change at the tail is bigger, and the direction of lift at the tail is secondary. $\endgroup$
    – Peter Kämpf
    Aug 13, 2016 at 22:41
  • $\begingroup$ I'm sorry sir but I don't get how having less lift per area gives a higher lift change in the tail than in the wing $\endgroup$
    – Sherlock_Dumbledore
    Aug 15, 2016 at 6:27

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