I'm wanting to build a laterals-only kinematic model in MATLAB considering only the ailerons for control inputs and was wondering if anyone has a rough guideline for the relationship between their deflection and the resulting roll moment.

EDIT: The answer from @PeterKämpf on the following question helped a lot (Thankyou Peter) but I couldn't understand the difference between p, dimensionless roll rate and wx, rolling speed in rad/sec. How can a roll rate be dimensionless?

His answer is here: How I determine numerically the roll rate of an aircraft?

Sorry, I couldn't comment on the actual post, I don't have enough reputation.

  • $\begingroup$ The lateral and directional modes are usually coupled, so much so that the usual distinction is between longitudinal and lateral-directional stability. I trust you are aware? $\endgroup$ – AEhere supports Monica Feb 22 '19 at 8:27
  • $\begingroup$ @AEhere I wasn't, I have only learnt about longitudinal and lateral dynamics, I didn't know there was a third to consider. Thank you I'll take a look. $\endgroup$ – Rim3rd Feb 22 '19 at 11:52
  • $\begingroup$ @Rim3rd this might be a translation issue, but I mean the yaw-roll coupling. In most cases your aileron-only input will cause adverse yaw, which will feed back into your roll response. $\endgroup$ – AEhere supports Monica Feb 22 '19 at 12:09
  • $\begingroup$ @AEhere Ah I see, yes thank you, I was aware. I'm only looking for an equation relating aileron angle with roll rate at this stage. $\endgroup$ – Rim3rd Feb 22 '19 at 12:16

Using a linear theory, the roll moment coefficient can be expressed as:

$$ C_l =C_{l0}+C_{l\beta}\beta+C_{l\delta_{\alpha}}\delta_{\alpha}+C_{l\delta_{r}}\delta_{r} $$

, where:
$\beta$ : slip angle
$\delta_{\alpha}$ : aileron deflection
$\delta_{r}$ : rudder deflection

This doesn´t really say much, as we are simply defining this coefficient in a way that is convenient. This equation just says that $C_l$ is equal to a linear composition of the effects of assymetry, slip angle, and aileron and ruder deflections, each weighed by a stability derivative. Some of these stability derivatives have proper names or are trivially discarded:

  • $C_{l0}=(C_l)_{\beta=\delta_{\alpha}=\delta_r=0}=0$ , for symmetric aircraft without engine torques

  • $C_{l\beta}$ is the dihedral effect, or the effect of slip on roll. It is the sum of contributions from the wing-body assembly, the HTP and the VTP. The wing-body effect can be further split into: the geometric dihedral of the wing, wing sweep and vertical position (high or low).

    $C_{l\beta} =(C_{l\beta})_{WB}+(C_{l\beta})_{HTP}+(C_{l\beta})_{VTP}$

    $C_{l\beta}=-\frac{a_w \Gamma}{4}-\frac{C_L}{4}sin(2\Lambda)+(C_{l\beta})_{\text{wing height}}+(C_{l\beta})_{HTP}-a_v\eta_v\frac{S_v}{S}(1+\frac{\partial\sigma}{\partial\beta})\frac{h_v}{b}$

    $(C_{l\beta})_{HTP}$ is obtained the same way as the WB contribution, although with different adimensionalizers. For general reference, wingtips above the root are stabilizing, positive sweep is stabilizing and the wing height effect depends on the fuselage fraction ahead and behind the wing, but usually a high wing should be stabilizing.

  • $C_{l\delta_{\alpha}}$ is the roll control power or the effect of ailerons on roll. Is is calculated directly from the Prandtl wing theory by considering the impact of aileron deflection on the lift distribution

  • $C_{l\delta_{r}}$ is the effect of rudder on roll


Your original question asked for the roll rate, and Peter Kämpf does indeed provide an answer, but it is probably no more useful to your particular need than my very first equation.

If you are interested in the roll rate, you will need to solve the lateral-directional system. You can simplify it somewhat by allowing fixed controls and level fight, thus obtaining a stability cuartic. This has, for most configurations, two real and two complex conjugate solutions, corresponding to the three lateral-directional eigenmodes: Roll Subsidience (large negative real root), Spiral Divergence (small real root, sometimes even >0) and Dutch Roll (complex conjugate root). Since the large negative real eigenvalue has a much greater modulus than the rest, its mode (Roll Subsidience) will dominate for short periods, and you can obtain a good approximation of the roll transfer function by doing $\Delta\beta=\Delta\hat{r}=\Delta\delta_r=0$ and $C_{l\hat{\dot{\delta}}_\alpha}\cong0$:

$$ G_{\hat{p}\delta_\alpha}=\frac{\Delta\hat{p}(s)}{\Delta\delta_\alpha(s)}=\frac{C_{l\delta_\alpha}}{\hat{I_x}}\frac{1}{s-(C_{l\hat{p}}/\hat{I_x})} $$

...in the Laplace plane. Keep in mind the parameters are non-dimensional.

  • $\begingroup$ Thank you for taking the time to help me with such a detailed response! You've given me a lot to read up on and it's much appreciated. I have voted up your answer but apparently I don't have the reputation for it to display. Thanks again! $\endgroup$ – Rim3rd Feb 26 '19 at 10:32
  • $\begingroup$ @Rim3rd No need, it was a welcome reminder for myself as well. I used one of my professor's books on the subject, which afaik has not been translated into English, but you can get a good, if simpler, overview from this resource: courses.cit.cornell.edu/mae5070/DynamicStability.pdf $\endgroup$ – AEhere supports Monica Feb 28 '19 at 13:06
  • $\begingroup$ This is a great resource! Thanks very much! $\endgroup$ – Rim3rd Mar 5 '19 at 16:35

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.