For the sake of this question let's assume that the considered aircraft is an A320. The question is what is the sign of the aerodynamic $C_{n_q}$ coefficient? In another word, when the aircraft starts rolling, will the aircraft experience sideslip and if yes in which direction (and most of all why) ?

Instinctively, I would say that if the right wing (seen from behind) goes down (positive roll rate), it will be delayed compared to the left one, however I already read somewhere that the coefficient tends to be zero (no sideslip induced). I even heard some aeronautical engineers say that it is the opposite...

If needed, we can set that the x-axis is pointing backward and z down.

  • $\begingroup$ How do you end up with x-axis pointing backward, but positive roll meaning right wing down. That is pretty non-standard in my experience as it conflicts with the right hand rule. Also you don't define the direction of your y axis, which is pretty essential if you want to discuss the sign of the $C_{n_q}$ coefficient. And since you define the direction of p non-standard, do I need to assume q is also non-standard? $\endgroup$
    – DeltaLima
    Sep 26, 2014 at 9:55
  • $\begingroup$ Please add a list of symbols and definition of you axis frame of reference, include signs of rotations. For me, the letters you use don't correspond with the description you give. $\endgroup$
    – DeltaLima
    Sep 26, 2014 at 10:07
  • $\begingroup$ You did not say what causes the roll. If the aircraft is thrown around by a gust you will not see adverse yaw. In case of hitting an updraft unsymmetrically, the uplifted wing is pulled forward by the updraft. If the pilot commands the roll, adverse yaw (means opposite to what is needed in a coordinated turn) will result. And, please, no capital cs for the coefficients! $\endgroup$ Sep 27, 2014 at 5:49

3 Answers 3


To answer what you wrote: $c_{n_q}$ is the yawing moment induced by a pitching motion. In a conventional, symmetric configuration this is zero for the airframe and has a small value for propeller aircraft, depending on the rotation direction of the propeller. If the tractor propeller is spinning clockwise (when seen by the pilot), a positive pitch motion $q$ (nose up) will cause a small negative sideslip angle $\beta$. A positive pitch acceleration will add a precession moment which similarly pulls the aircraft's nose to the right. But I think you want to know something else.

To answer what you meant: $c_{n_p}$ is the yawing moment induced by a rolling motion. The axis system you propose (with z pointing up) is used to measure relative positions within the airframe, but for flight mechanics we prefer to put the x-axis in flight direction, y to the right and z down. A positive rolling motion $p = \omega_x\cdot\frac{b}{2\cdot v_{\infty}}$ (or $p = \omega_x\cdot\frac{b}{v_{\infty}}$, both definitions are used) is right-wing-down, and a positive yawing moment means the aircraft's nose moves to the right, causing negative side slip angles.

When a positive rolling motion happens in still air, the resulting yawing moment is negative, pulling the aircraft's nose left, as Federico said. Thus, $c_{n_p}$ is negative.

But the reason is different: When the pilot deflects the ailerons and causes a difference in lift on the left and right wings, and consequently a difference in induced drag, the yawing moment caused by that difference in induced drag is called $c_{n_\xi}$, the yawing moment due to aileron deflection. $c_{n_\xi}$ is positive.

Once the aircraft is in a steady roll, the roll damping $c_{l_p}$ will fully compensate the differential lift caused by aileron deflection. Now the rolling motion will increase the local angle of attack on the downward moving wing the farther you are away from the center, and decrease it on the upward moving wing. Since lift is perpendicular to local airflow, the air force vector on the downward moving wing is pointing slightly forward, and that on the upward moving wing slightly backwards. Please note that the total lift produced by each wing is roughly the same! The rolling moment of both is exactly zero once the roll rate is steady. However, the aileron deflection will decrease lift on the downward moving wing such that it just compensates for the increased angle of attack due to the rolling motion, and the same holds for the upward moving wing with its trailing-edge-down aileron deflection.

What is different is not the magnitude, but the direction of lift on both wings. This pulls the downward-moving wing forward and the upward moving wing back, causing a negative yawing moment in a positive rolling motion. That is why your conventional $c_{n_p}$ is negative.


First, a clarification. The standard reference frame for and aircraft, w.r.t. a person sitting in the pilot seat, is:

  • x axis positive ahead of the aircraft
  • y axis positive in the right wing direction
  • z axis positive towards down

In addition, to be clear about the conventions used in the rest of this answer:

  • positive roll is when the right wing is down
  • positive sideslip $\beta$ is when the pilot sees the air coming from the right side.

And, for completeness, $C_{N_p}$ is the aerodynamic derivative of yaw rate $r$ (not $\beta$) as function of roll rate $p$.

While $C_{N_q}$ would be the one due to pitch rate $q$, but that would not be within your specified range of conventional aircraft.

Now to your question.

when the aircraft starts rolling, will the aircraft experience sideslip and if yes in which direction (and most of all why) ?

TL;DR version:

Yes, it experiences sidelip and for positive roll rate $p$ the sideslip $\beta$ will be negative.

Extended version:

  • Positive roll rate means that the right wing is rotating down and the left wing up.
  • This translates in a momentarily higher Angle of Attack $\alpha$ seen by the right wing w.r.t. the left wing
  • Because of this difference in $\alpha$ seen by the wings, the right wing will produce slightly more lift (and this is the direct cause of the Roll Subsidance)
  • Nevertheless, aileron deflection will make the left wing produce significantly more lift
  • More lift means more induced drag
  • More induced drag means that the left wing will tend to "lag" and the left wing will move ahead.
  • The plane has thus rotated towards the left side, achieving a negative $\beta$

The amount of $\beta$ will obviously depend on the specific aircraft.

You might find more by looking for information about the Dutch Roll.

  • $\begingroup$ I think you’re right all the way up to induced drag part. Here I think you got it reversed. Even though there is slight lift difference between two wings due to different local AOAs, the bigger chunk of lift difference comes from aileron deflection. That surpasses the former. Therefore left wing will have larger induced drag than right wing. And this will cause left yawing motion in a right roll in a conventional aircraft, which we call adverse yaw. The way you described it sounds like proverse yaw. $\endgroup$
    – Kolom
    Oct 7, 2019 at 4:08
  • $\begingroup$ @Kolom I think you might be right, I'll revise $\endgroup$
    – Federico
    Oct 8, 2019 at 10:57

Instinctively, I would say that if the right wing (seen from behind) goes down (...), it will be delayed compared to the left one

You are correct.

This answer is aimed at the case where the ailerons are centered, and the local airmass is uniform, and the aircraft is rolling for some unknown reason (such as retained rotational inertia in the roll axis from a previous disturbance.)

The right wing will experience an increased angle-of-attack due to the rolling motion, and thus will experience more drag.

If we are talking about a steady-state rolling motion (constant roll rate) due to deflected ailerons, then the exact opposite is true -- see https://www.av8n.com/how/htm/yaw.html#sec-adverse-yaw for more.

Likewise if an updraft is striking the rising wing and causing the rolling motion.


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