For a dutch roll mode, can having too small of a roll or yaw stiffness destabilize it?

I would think not because you want a lower value to maintain it?


To estimate whether the dutch-roll mode goes unstable, we can assess the real-part of the dutch-roll eigenvalue: negative is stable and positive is unstable. From Etkins, Dynamics of Flight, one illustrative estimate (although not necessarily numerically accurate) is as follows:

$$n_{DR}\approx\frac{1}{2} \left[ \frac{Y_v}{m} + \frac{N_r}{I'_z} + \frac{I'_zL_v}{I'_xN_v}\left(\frac{N_p}{I'_z} - \frac{g}{u_0} \right) \right] $$

where $n_{DR}$ is the real-part of the eigenvalue, $Y_v$ is the dimensional side force sideslip derivative, $N_r$ is the dimensional yaw moment damping, $L_v$ is the dimensional negative roll stiffness (dimensionless form: $C_{l_{\beta}}$), $N_v$ is the dimensional yaw stiffness (dimensionless form: $C_{n_{\beta}}$), $N_p$ is the yaw-roll cross-coupling moment, $u_0$ is the trim airspeed, $g$ is gravitational acceleration, $m$ is aircraft mass, all the $I'$ are inertia terms about the respective axes. Note that all the inertia terms are about the principle axes for simplicity.

The first term ($Y_v$) is always negative, and so is the second term ($N_r$). Ignoring $N_p$ for now, positive roll stiffness requires $L_v$ to be negative, and positive yaw stiffness requires $Y_v$ to be positive; this means that the third term is net positive. Therefore, we can draw the following conclusions:

  1. Dutch-roll damping worsens at lower airspeed
  2. Roll stiffness worsens the dutch-roll damping
  3. Yaw stiffness improves the dutch-roll damping

To answer your question: yes, it's entirely possible for dutch-roll to be unstable due to sufficiently low yaw stiffness.

  • $\begingroup$ Please explain why $Lv$ is expressed as "negative roll stiffness" and $Nv$ as only "yaw stiffness". Is the negative sign added to the roll stiffness term for the equation, or does it mean a direction of roll (away from level/returning to level)? $\endgroup$ – Robert DiGiovanni Mar 18 '20 at 13:45
  • $\begingroup$ @RobertDiGiovanni Pleas read third paragraph. $\endgroup$ – JZYL Mar 18 '20 at 13:47
  • $\begingroup$ I would expect $Lv$ to be numerically positive, because $-g/u0$ creates the negative in the eigenvalue calculation. $Np$ certainly does make it less negative! $\endgroup$ – Robert DiGiovanni Mar 18 '20 at 13:57
  • $\begingroup$ @RobertDiGiovanni Huh? These are stability derivatives $\endgroup$ – JZYL Mar 18 '20 at 13:59
  • $\begingroup$ Yes, and safe planes are design detivatives. Notice how roll/yaw coupling (from swept wings) makes the third term less negative. $\endgroup$ – Robert DiGiovanni Mar 18 '20 at 16:41

Your Answer

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