enter image description here

During straight and level flight, coordinated flight is assumed when there is no net lateral force (no slip or no skid). But this concept totally breaks down when it comes to turning, in a co-ordinated or unco-ordinated turn; there will always be a net lateral force due to centripetal force provided by the horizontal component of the lift force. The centrifugal force (inertial force) found in the books, in an attempt to explain it doesn't make sense to me as the centrifugal force is an imaginary force. So how is co-ordinated flight achieved in term of the forces? Thanks in advance.

  • 2
    $\begingroup$ Centrifugal force exists in the reference frame of the aircraft. If you do the analysis there, you need it, if you are doing it in the reference frame of the air, you don't. Since General Relativity, inertial forces are now considered just as real as any other forces, and include gravitational force, which by principle of equivalence (the core postulate of GR) is locally indistinguishable from acceleration of the reference frame. Note I didn't say gravity, because gravity actually means sum of all inertial forces (includes centrifugal force due to Earth rotation). $\endgroup$
    – Jan Hudec
    Commented Jul 5, 2020 at 20:55
  • 2
    $\begingroup$ … doing analysis in the aircraft frame of reference is usually a bit nicer since there aircraft is stationary so all the aerodynamic and inertial forces cancel out. Plus it tells you what you'd feel if you were sitting in the aircraft, because you feel exactly the sum of inertial forces (a.k.a local gravity; includes gravitational forces). Of course the inertial forces just correspond to the acceleration as viewed from the reference frame of air, but since that frame is not in free fall, there are still some inertial forces and you have a more cases than in the aircraft one. $\endgroup$
    – Jan Hudec
    Commented Jul 5, 2020 at 21:08
  • 3
    $\begingroup$ Oh boy, here we go again... $\endgroup$ Commented Jul 6, 2020 at 1:27
  • 1
  • $\begingroup$ Maybe this will inspire me to draw those diagrams... $\endgroup$ Commented Jul 7, 2020 at 15:34

2 Answers 2


To get a deeper understanding beyond the superficial, it helps to break down the forces and moments present on an aircraft that may affect its rigid-body motion:

  • Aerodynamic forces: These are the forces and moments exerted by the airflow on the aircraft

  • Ground forces: These are the forces and moments exerted by the ground on the aircraft, transmitted through tyres and landing gear. Not applicable when it's flying.

  • Propulsion: Forces and moments due to direct thrust. For simplification, let's assume that thrust acts inline with the forward axis.

  • Gravitational forces: Gravity pulls the aircraft towards the ground. It's rather special because everything, be the aircraft structures, you, me, or the accelerometers, get pulled at the same rate ($g$)1. This is distinctly different than the other two types of forces, which are only present when airflow affects exposed areas, or when contact with the ground has been made.

  • Inertial forces: These are fictitious forces and moments that are required to maintain non-uniform motion. This includes your centripetal force. Inertial forces must always be equal to the sum of all the aforementioned external forces.

First of all, let's consider any turn a steady-state maneuver (thereby ignoring the transients like rolling in and rolling out), which means that the vector sum of all the external forces, including gravity, must sum to the inertial forces. As you've correctly pointed out, the sum of aerodynamic forces + gravitational forces must be equal to the centripetal force, which in the turning plane is provided by lift, side force and gravity. This must hold true for any steady-state turn, whether coordinated or not.


There are two ways to define a coordinated turn. With all engines operating, they are approximately equivalent:

  1. A zero-sideslip turn
  2. A ball-centered turn: we are going to use this definition

What ball-centered means is that there are no aerodynamic forces acting laterally on the airplane: lift must provide all the centripetal force required. Ball centered provides the best average feel for the occupants, since the forces are directly inline with the floor, and there's no side force causing sway. Since everything feels gravity at the same rate, occupants or the ball cannot detect gravity.

For illustration:


For the more math oriented, in inertial frame, Newton's second law is stated as:

$$\vec{F_{i}}+m\vec{g_i}=m\frac{d\vec{V_i}}{dt} \tag{1}$$

The entire right hand side is considered to be inertial forces. However, if the measurements occur in a rotating frame (on an airplane, for example), then we need to express everything into the body frame. The left-hand side is easy:

$$\vec{F_{b}} = C_{bi}\vec{F_{i}} \tag{2}$$

$$\vec{g_b} = C_{bi}\vec{g_i} \tag{3}$$

where $C_{bi}$ is the rotation matrix transforming a vector from inertial frame to body frame.

The right-hand side requires some adjustments, because the Euler angles of the body themselves are changing:


Apply chain rule, and we have:

$$\frac{d\vec{V_b}}{dt}=\frac{dC_{bi}}{dt}\vec{V_i}+C_{bi}\frac{d\vec{V_i}}{dt} \tag{4}$$

It can be mathematically shown the following identity:

$$\frac{dC_{bi}}{dt}\vec{V_i} = -\vec{\omega_b} \times \vec{V_b} \tag{5}$$

Substitute (5) into (4), and we get:

$$C_{bi}\frac{d\vec{V_i}}{dt} = \frac{d\vec{V_b}}{dt} + \vec{\omega_b} \times \vec{V_b}$$

Finally, if we multiply both sides of (1) by $C_{bi}$ and simplify with (2), (3) and (5), we get:

$$\vec{F_{b}}+m\vec{g_b}=m \left( \frac{d\vec{V_b}}{dt}+\vec{\omega_b} \times \vec{V_b} \right) \tag{6}$$

(6) is the Newton's second law in a rotating frame. The entire right-hand side are still inertial forces (same as (1)), except now we also have the cross-product, which produces the centripetal inertial forces:

$$\vec{\omega_b} \times \vec{V_b} = \begin{bmatrix}p \\ q \\ r\end{bmatrix} \times \begin{bmatrix}u \\ v \\ w\end{bmatrix} = \begin{bmatrix} qw - rv \\ ru - pw \\ pv - qu \end{bmatrix} \tag{7}$$

You will readily identify terms like $ru$, which is very close to the familiar $a_c=\omega V$ for a restricted 2D centripetal motion (whence $p=0$ and $q=0$).

One final note, the right-hand side is fictitious in that they are not real forces! They are a result of the kinematic motion itself, and must require the left-hand side (which are the real forces), to sustain.

1: Technically, this is only true locally, because Earth is a sphere and does not exert a uniform field at different altitudes. But at the range of altitudes that airplanes will be flying, this is a rather good approximation.

  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Jamiec
    Commented Jul 20, 2021 at 8:29

It is helpful to understand a basic concept in Physics to understand this. The answer to this depends on what frame of reference you are measuring things in. In what is called an "Inertial" frame of reference, only "real" forces need be considered, as only "real" forces cause acceleration (F=ma). In an inertial frame of reference the only real forces acting on an aircraft are the aerodynamic force (the pressure of the atmosphere on every square inch of the surface of the airframe), and the thrust produced by the engines. PERIOD.

But we all eat, sleep, walk, and fly on the surface of the earth, which is an accelerated frame of reference, due to the acceleration of Gravity, the rotation of the earth, the motion of the earth around the sun, etc. etc., and in that accelerated frame of reference, in order to explain the apparent motion of any body, other "fictitious" forces, such as *centrifugal force, the force of gravity itself, and other inertial forces, must all be considered to make the math work out. (In other words, as mentioned in another answer, to make the forces cancel out when the aircraft is in a "steady state").

*Note: Centrifugal force is often added on diagrams of aircraft in turns to help explain the apparent "balance" of forces for the aircraft that is stable in the diagram. But the force, and the "stability" are both fictitious, because the diagram is depicted in a turning and rotating frame of reference.

Basically, these fictitious forces must be added in to "subtract out" the acceleration of the frame of reference itself, because without considering them, the answer you get would only be calculating the acceleration of the aircraft within an inertial (zero-G) frame of reference, and we generally want to know our acceleration in the frame we are otherwise examining (generally the earth-frame).

  • $\begingroup$ Re "Basically, these fictitious forces must be added in to "subtract out" the acceleration of the frame of reference itself... and we generally want to know our acceleration in the earth-frame."-- this answer might be a bit more clear if it explained that to calculate the acceleration in the earth-frame (or the airmass frame), the only "fictitious" force that must be added is gravity, not the "centrifugal force" that we typically see included in diagrams in flight training manuals. These diagrams are typically based on the aircraft reference frame, not the ground or airmass reference frame. $\endgroup$ Commented Jul 7, 2020 at 17:44
  • $\begingroup$ (Which is why they have no explanatory power. The only thing they really "explain" is that the apparent net force as seen from the aircraft reference frame is always zero. They don't actually explain why the ball sits where it does in a skidding or slipping turn.) $\endgroup$ Commented Jul 7, 2020 at 17:46
  • $\begingroup$ @quiet flyer, I only mention centrifugal force as a "fictitious" force because I have seen documents and texts with diagrams that attempt to explain the balance of forces in turns, coordinated and uncoordinated, where the diagram is done in the frame of reference of the aircraft itself, which, obviously, IS an accelerated frame of reference. $\endgroup$ Commented Jul 8, 2020 at 12:48
  • $\begingroup$ I have edited my answer to make this clear. $\endgroup$ Commented Jul 8, 2020 at 12:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .