In the book Dynamics of flight, the authors present how the pitch rate q and the angle of attack $\alpha _x$ may vary:

enter image description here
Source: Dynamics of flight: Stability and control,
Bernard Etkin, Lloyd Duff Reid, 1996.

Why is the pitch rate q null while the angle of attack changes, and vice versa?

  • $\begingroup$ No sir not exactly , i'm asking about the concept of relation between pitch rate and angle of attack . $\endgroup$ – SAMER BABIKIR Feb 9 '17 at 15:32

Why the pitch rate q is zero while angle of attack is changing and vice versa?

It generally does not (*). You misunderstood the statement in the Etkin's book.

The image you posted come from the chapter about "the $q$ derivatives" $C_{L_q}$, $C_{M_q}$ and $C_{h_q}$.

The book is quite explicit:

These derivatives represent the aerodynamic effects that accompany rotation of the airplane about a spanwise axis through the C.G. while $\alpha$ remains zero

It then presents

Figure (b) shows the general case in which the flight path is arbitrary

(this is the figure you posted)

This should be contrasted with the situation illustrated in figure (a), where $q=0$ while $\alpha$ is changing.

The book in no way implies that either one case or the other is true, it is only presenting contrasting academic examples to illustrate the concept of what the $q$ derivatives are.

(*): I said generally because the book itself present one case where you CAN have constant $q$ and $\alpha = 0$, the steady pull-up. Again, this in no way implies that you always have

q is zero while angle of attack is changing and vice versa

| improve this answer | |

'q' is the pitch rate....rate of change of pitch angle, i.e d(theta)/dt. Pitch angle is the angle between aircraft reference line and horizon..... In the first image, the pitch angle of the aircraft is zero (as nose of aircraft is aligned with the horizon)....So, pitch rate is zero....

Angle of attack is an angle between aircraft reference line and velocity vector...In the second image, angle of attack is zero as the aircraft nose is aligned with the velocity vector .....

| improve this answer | |
  • 1
    $\begingroup$ this explains the terms employed, but does not actually answer the question asked (or explain the misunderstandings behind the question) $\endgroup$ – Federico Jun 7 '17 at 10:16

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.