6
$\begingroup$

I am reading "Introduction to Aircraft Aeroelasticity and Loads" by Jan R. Wright and Jonathan E. Cooper.

The book said: In steady bank turn (as shown in picture below), The rate of turn is given by: $\omega_{turn} = V/r $ (I understand this) enter image description here

And so the aircraft experiences steady yaw and pitch rates given by':

$n_{yaw}= \omega_{turn}cos \phi$

$q = \omega_{turn}sin \phi$

I don't understand 2 equations above. I think the yaw rate would be equal to the turn rate and the pitch rate is zero.

$\endgroup$
  • $\begingroup$ The yaw rate and pitch rate are in the frame of the airplane, not in the frame of the earth. If the pitch rate is zero, then the plane will fly a inclined circular path (one side of the circle is higher) when viewed on earth. $\endgroup$ – kevin Sep 10 '17 at 10:35
6
$\begingroup$

Turn rate is referenced to earth axes, yaw rate to aircraft axes, hence the $cos\Phi$:

  • If you turn at $\Phi$ = 0, your rate of turn is the yaw rate and the pitch rate is zero.
  • If you turn at $\Phi$ = 90°, your yaw rate is zero and your pitch rate = rate of turn.

In order to maintain a steady turn and not lose altitude the pilot does need to pull on the stick when turning..

$\endgroup$
  • 1
    $\begingroup$ The pilot needs to pull for two reasons: 1. Trim for higher lift, 2. Compensate pitch damping. $\endgroup$ – Peter Kämpf Sep 10 '17 at 20:25

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.