In response to Peter Kämpf's answer to this question, why did moving the CG back on the Wright Flyer improve its stability?


2 Answers 2


It did not improve stability, but flyability.

The Wright Flyer had the CG located aft of the neutral point and was unstable in pitch. A longitudinally unstable aircraft will deviate from its trim point, and when flown by a human pilot, will have a wavy flight path. The Wrights called this undulation. It is caused by the delay in the pilot's response and his tendency to overcompensate, so the aircraft is oscillating around the trim point (= the angle of attack where all moments around the pitch axis are in balance). This delay is more problematic when the eigenfrequency of the motion is higher. All Zeppelins were unstable in pitch above a certain airspeed and in yaw over the full speed range, but nobody bothered. These things were so big that any motion needed a long time to develop and deviations from the trim condition were easy to correct.

With aircraft, things are different, because they are more maneuverable and much smaller. The characteristic parameter here is the short period mode which characterizes the pitch motion. Its frequency is around 1 Hz for small aircraft and can be approximated like this: $$\omega_{\alpha} = \frac{v_{\mathrm{trim}}}{i_y} \cdot \sqrt{\left(\frac{c_{m\alpha}}{\mu} + \frac{c_{L\alpha}\cdot c_{mq}}{\mu^2}\right)}$$

$\omega_{\alpha} \:\:\:$ Eigenfrequency of the short period mode
$v_{\mathrm{trim}}\:$ trimmed airspeed
$i_y \:\:\:\:\:$ Radius of inertia around the pitch axis
$c_{m\alpha} \:\:$ pitch moment gradient over angle of attack
$\mu \:\:\:\:\:\:$ reduced mass. $\mu = \frac{2\cdot m}{\rho \cdot S \cdot l_{\mu}}$
$c_{L\alpha} \:\:$ lift coefficient gradient over angle of attack
$c_{mq} \:\:$ pitch damping coefficient
$m \:\:\:\:\:$ mass
$\rho \:\:\:\:\:\:$ density of air
$S \:\:\:\:\:\:$ reference area (normally wing area)
$l_{\mu} \:\:\:\:\:$ mean aerodynamic chord

To answer the question, all it needs is the first factor of the equation. The frequency increases with airspeed and goes down with the pitch inertia of the plane. By adding a weight at the back of their airplane, the Wrights increased this pitch inertia, thus lowering the eigenfrequency and making it easier to control. Make no mistake: The Flyer became even more unstable, but the instability became easier to correct because the deviations from the trim point built up more slowly.

Modern replicas of the Flyer have their CG ahead of the neutral point and are naturally stable.

  • 1
    $\begingroup$ +1 excellent explanation. I did not realize that sometimes making a plane slightly more unstable improves its handling (as you mentioned, due to increased inertia). I've noticed this myself on one of my RC planes with CG so far aft that neutral trim required down elevator. I couldn't explain why it flew for a very long time. $\endgroup$
    – slebetman
    Aug 11, 2014 at 3:08
  • $\begingroup$ What about mentioning PIO? $\endgroup$
    – yankeekilo
    Feb 16, 2015 at 20:15
  • $\begingroup$ @yankeekilo: That is what the Wrights called undulation. I used their terminology because I find it very descriptive. But, yes, this was an early case of PIOs, and they got less intense when the aircraft's eigefrequency was shifted away from the eigenfrequency of the control system (this being Wilbur or Orville plus the Flyer's pitch control). $\endgroup$ Feb 16, 2015 at 20:33

For a plane to be stable, small control input should lead to small corrections. Also, small air currents should lead to small deviations.

Now consider what happens when the Flyer pitches a bit down. It picks up speed, and the wing generates more lift. If the CG is to far forwards, the wing lift generates torque which pitches the nose even further down. In other words, that small change is amplified.

Moving the CG aft reduces the torque change on pitch changes, which means that small changes remain small - and thus the plane is stable.

  • 1
    $\begingroup$ Moving CG forward makes the aircraft stable. For both planes with regular tail or canard (Flyer was a canard). The reason is that with forward CG the forward surface flies with higher angle of attack. The Flyer had the CG too far aft and was not stable. (-1, this is incorrect explanation). $\endgroup$
    – Jan Hudec
    Aug 11, 2014 at 6:56
  • 1
    $\begingroup$ @JanHudec: I can't find a good source which states that the Flyer had its canard at a higher AoA. With a canard generating upward lift, you must have a CG between the canard and the wing (obviously, else the two torques cannot cancel at all). Moving it a bit forward or afterwards between those limits alters the balance of torques, but you cannot generalize and say that the CG forwards is stable. Just put the CG at the location of the canard and stability becomes impossible. $\endgroup$
    – MSalters
    Aug 11, 2014 at 14:20
  • 2
    $\begingroup$ Relative AOA of the control surface and the main wing is a function of the centre of gravity, not the other way around. Because the pilot adjusts the canard to balance the craft. Flyer had the canard at lower AoA than main wing, because it was not stable. But moving it's CG forward would make pilot increase AoA of the canard and therefore improve stability. But Wrights never moved it forward far enough to make it stable. And as Peter Kämpf correctly explains, moving it forward a bit made it less controllable and moving it aft a bit made it more controllable. Though less stable. $\endgroup$
    – Jan Hudec
    Aug 11, 2014 at 15:09
  • $\begingroup$ Aircraft with CG at the location of the "canard" could easily be stable. But it's "canard" would be the main wing producing all lift and the "wing" would be a stabilizer flying at 0 AoA in steady flight. Such configuration would be stable just fine. $\endgroup$
    – Jan Hudec
    Aug 11, 2014 at 15:12
  • $\begingroup$ @JanHudec: That's a bit of a naming issue then: I use canard to mean a control surface forward of the wing, which can generate torque in order to alter the pitch as needed. To generate torque, you need a lever, and the lever is by definition 0 if the CG is located at the canard. You can of course have the wing at the CG if the control surface (canard or tail) generates 0 torque under normal conditions. $\endgroup$
    – MSalters
    Aug 11, 2014 at 15:21

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