I'm doing a problem on Flight Mechanics in which a banked turn is performed. The center of gravity of the aircraft is always within an horizontal plane.
In order to write the dynamic equations, I have to assume some simplifications and conditions. One of them is the following:
The thrust’s sideslip angle is equal and with opposite sign than the aeroplane’s sideslip angle, that is: ($\nu = −\beta$).
Side slip angle ($\beta$) is defined as the angle made by the velocity vector to longitudinal axis of the vehicle at the center of gravity in an instantaneous frame.
Thrust side slip angle ($\nu$) is defined as the angle made by the thrust vector to longitudinal axis of the vehicle at the center of gravity in an instantaneous frame. https://en.wikipedia.org/wiki/Slip_(aerodynamics)#Sideslip_angle
Then, I have to write the dynamic equations for the symmetric flight and for the non-symmetric flight.
My questions are:
- In my opinion, the condition written above should not be accounted for the symmetric flight, so $(\beta = \nu = 0)$. Is this affirmation correct?
- For the non-symmetric case, I have drawn the following scheme of velocity and thrust. But I'm not sure if it's correct because it doesn't make sense, despite that is what the condition says: $(\nu=-\beta)$. Is the scheme correct according to the condition? (The scheme is the top view of the aircraft.)