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An answer on ASE states

  • The presence of steady-state non-zero body yaw and roll rate induces side force ($C_{y_p}$ and $C_{y_r}$).

What is a non-mathematical explanation of why this is so?

Is it correct to say that the side force components described above would exist even if the body were somehow following the same trajectory, and experiencing the same yaw and roll rotation rates, in a vacuum? I.e., that they do not directly stem from aerodynamic effects?

Also, do the sideforce components created by a non-zero body yaw and/ or roll rate depend upon at what physical location on the aircraft we are measuring the sideforce, i.e. upon exactly where the slip-skid ball (inclinometer) is located? If so, might there be a location where the measured sideforce resulting from the yaw and/or roll rate is zero even though the aircraft is yawing and/or rolling?

quiet flyer
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1 Answers1

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Let's say we have a thin, straight and rigid stick traveling in a 2D circular path. We will attach our reference of frame to the CG of the stick, aligned with its natural forward. If it has a forward speed of $u$ and the circular path has a radius of curvature of $R$, then it has a yaw rate, $\dot{\psi}=u/R$, all of which are expressed in the said reference frame.

stick

At locations forward or aft of the CG, different parts of the stick will experience different lateral speeds due to the yaw rate, when measured using our reference frame: $u(x) = \dot{\psi}x$. The intuition is that the stick is spinning, so top and bottom will experience the largest speed while the centre will have none.

If you had attached a fin near the of the stick, then it will experience a side airspeed, effectively giving it an aerodynamic incidence angle. As a result, there will be aerodynamic forces and moments, simply from the motion of rotation.

From an analysis perspective, we usually don't re-calculate the effects on every surface due to local speed changes. Instead, the stability derivatives are measured in the body axis (well, usually in the stability axis but easily transformed to the body axis). So we can easily lump the sum of the aerodynamic effects of the individual portions into a single derivative with respect to the rate of rotation about that axis.

JZYL
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  • The original answer quoted in my question also referenced yaw damping, which would seem to me to related to the aerodynamic torque associated with the aerodynamic sideforce generated by specific parts of the aircraft, like the fin, having a sideways velocity with respect to the direction of the flight path. But I'm still not seeing how a steady (as opposed to increasing or decreasing) yaw rate automatically generates a sideforce. I'll make an edit to the question that may possibly help express the source of my confustion a bit better. – quiet flyer Jul 10 '20 at 17:43
  • @quietflyer Of course there would be no aerodynamic forces/moments in a vacuum. – JZYL Jul 10 '20 at 18:07
  • I just didn't understand whether the components were an inherent aspect of the rotation rates and trajectory, or something specifically generated by aerodynamic consequences following from the rotation rates/ trajectory. And still don't understand which is the case, I guess. – quiet flyer Jul 10 '20 at 21:13
  • @quietflyer It's all aerodynamic due to change in local airflow from the rotation. A constant rate of rotation, as mentioned, will induce different velocities at different parts of the body, relative to the air. – JZYL Jul 10 '20 at 22:57
  • @JZYL.Is above explantion with stick ,reason why car in turn feel side wind coming form outside or reason is that car has some yaw angle which cause side wind?Is this two reasons("stick" and yaw angle) same or separated phenomens? –  Aug 02 '20 at 07:16
  • @Noah Car is different in that the slip angle is determined by the relative cornering of the front and rear wheels. Depending on the CG and arrangement, the total vehicle slip can be adverse or proverse. – JZYL Aug 02 '20 at 11:38
  • @JZYL,If put wind indicators at your stick on picutre,at front,back and at center,will all 3 indicators point prependicular to their radius of curvature? –  Aug 02 '20 at 11:46
  • @Noah You mean to their respective local radius of curvature? Yes, but your vane will have to be infinitesimal, otherwise it will experience local airflow that's different from the point of rotation you're trying to measure. – JZYL Aug 02 '20 at 11:55
  • @Noah This is going way beyond the comment section. This should be posted as a new question, maybe on physics SE. The article is basically agreeing with this answer. – JZYL Aug 02 '20 at 12:21