Why are yaw rate and pitch rate obtained by these equations?

Question

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)

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.

Answer 1

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..