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.)

Thrust-speed scheme.

  • 1
    $\begingroup$ In your graphic, what do subscripts b and I mean? Usually b is body and I is inertial. But then why shouldn't thrust be parallel to the body x-axis? $\endgroup$
    – DeltaLima
    Sep 26, 2015 at 14:48
  • 1
    $\begingroup$ Yes! It means that. The reason why thrust is not parallel to the axis is because the statement says that the thrust has a sideslip angle $\nu$ and I think this angle is measured between thrust and X-axis. Please correct me if I'm wrong. @DeltaLima $\endgroup$
    – Airman01
    Sep 26, 2015 at 15:14
  • $\begingroup$ If I understand the situation correctly, the aircraft is flying with the body axis aligned with the inertial axis and slipping to the right. E.g. the aircraft is on close final, aligned with the runway and correcting for a crosswind from the right. $\endgroup$
    – DeltaLima
    Sep 26, 2015 at 15:39
  • $\begingroup$ Can you state the definition of thrust's sideslip angle and airplane's sideslip angle? Between which two vectors/axis each is measured? $\endgroup$
    – Jan Hudec
    Sep 26, 2015 at 19:52
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    $\begingroup$ @Airman01, the diagram appears to be correct according to that definition, but with engines fixed to the airframe I have to wonder what could cause the change of thus defined $\nu$. $\endgroup$
    – Jan Hudec
    Sep 26, 2015 at 20:45

1 Answer 1


Relying on physics, causing a mass to execute a circular path requires a suitable centripetal force. In an airplane in a banked balanced turn, this comes from the inward component of lift due to the bank angle.

However, if the turn is not properly co-ordinated, there can be additional contributions to the centripetal force from the power plant thrust no longer being directed tangentially and sidethrust on the airframe due to slip.

The extreme case is the flat turn where the wings remain level and the power plant thrust has an inward component due to the sideslip.

That said, apart from an airplane having vectoring thrust, the only reason for a thrust sideslip angle I can think of is that many single airscrew airplanes have built-in sidethrust to counteract engine torque. However there is no requirement for the engine sidethrust angle to equal the slip angle, so in that case the assumption that the two angles are the same cannot be valid.

The thrust line of an airscrew doesn't change with slip. The slip causes asymmetric thrust on the blades which precesses gyroscopically so the airscrew creates a couple which tends to increase the slip angle.

  • 1
    $\begingroup$ Would a twin with an engine out & some bank into the operating engine (common assumption & technique to lower Vmca) produce the angles being discussed here? $\endgroup$
    – Ralph J
    Jan 1, 2016 at 15:44

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