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See this image from page 5-24 of the FAA's Pilot's Handbook of Aeronautical Knowledge (2016 edition).

For the purpose of this question, let us take the Newtonian perspective that Weight is an actual force.

For simplicity, not that it necessarily matters, assume zero airmass movement, i.e. no wind.

For simplicity, assume that the Thrust vector is exactly equal in magnitude and opposite in direction to the Drag vector, though that may not always be exactly true in reality.

Assume that the intent of the illustrator is that any forces that are not included in the illustration are cancelled out by equal and opposite forces that are also not included in the diagram. For example, assume that Thrust is cancelled by Drag.

Overlook the oddity of using blue to represent the vector that is the vector sum of the orange vectors as far as the "Lift" vectors are concerned, but not following this convention for the other remaining vectors. (Perhaps the intent of the illustrator is to reserve blue for the forces that are actually real forces, and are neither the result of the decomposition of other forces, nor are fictitious forces or the vector sums of fictitious forces and real forces?)

Assume that the intent of the illustrator is to represent "Load" as the vector sum of "Weight" and "Centrifugal force", just as "Lift" is the vector sum of its vertical and horizontal components.

Assume that the vector labelled "Lift" is intended to represent the component of the aerodynamic force generated by the wing which acts orthogonal to the direction of the flight path in the direction that is "upwards" in the aircraft's reference frame, i.e. parallel to the vertical fin.

Take as a "given" that the orientation of the vector labeled "Load" is correct, i.e. that this vector indicates the direction that a pendulum suspended in the aircraft would orient itself in each case. It represents the "apparent inertial loading" experienced by the aircraft and contents.

Assume that the direction of the instantaneous trajectory is aimed straight out of the page toward the reader. Note that the nose of the aircraft may be yawed to point slightly to the left or right of the direction of the instantaneous trajectory, although the illustrator has not really shown that, except via the shaded arrows by the wingtip that suggest the aircraft is moving sideways in relation to the direction that the nose is pointing. The intent of the question is to ask whether the actual force vectors are depicted correctly, not to ask about whether the aircraft's orientation in space has been depicted correctly.

The actual question(s):

A) The inclusion of the "Centrifugal force" vector indicates that the illustrator is adopting a reference frame fixed on the aircraft, which is obviously not a valid inertial reference frame. With that as a "given", are the diagrams correct? If not, what is wrong with them? Shouldn't the net force be zero in this reference frame?

B) If we deleted the vector labeled "Centrifugal force" and the vector labeled "Load", would the diagrams correctly illustrate each situation as seen from the earth reference frame? (For simplicity, assume that the earth is a valid inertial reference frame from the Newtonian viewpoint, i.e. disregard the fact that the earth is actually orbiting the sun, etc.) If not, why not, i.e. what would be wrong with the diagrams? Note that if we deleted the vector labeled "Centrifugal force" and the vector labeled "Load", then all the force vectors illustrated in each of the three diagrams would be identical.

C) It seems possible that some additional real aerodynamic force should be included in some of the diagrams. Is this the case? What is that force? What should the "corrected" diagrams look like?

Related question on PSE: Turning an airplane - coordinated turn and inclinometer ("the ball")?

Link to a similar question as the present one, on Aviation Stack Exchange (no well-received answers have been provided as of yet): What is missing from these diagrams of the forces in slips and skids?

I decided to ask the closely related question on Physics SE because I wanted to utilize the different expertise of a different community who might approach the problem from a different perspective.

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In the reference frame of the airplane the sum of physical and fictitious forces has to be always equal to zero, thus we can readily tell that the second and third diagrams labelled "slipping turn" and "skidding turn" are incorrect.

I think the missing component is the lateral drag in the horizontal direction. The value of this component of drag would depend on the relative lateral velocity of the airplane with respect to the surrounding air. In the "slipping turn" diagram, lateral drag should act in the same direction as the centrifugal force, so as to balance the horizontal component of lift. In the "skidding turn" diagram lateral drag should act in the opposite direction in order to balance the centrifugal force.

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  • $\begingroup$ This good answer might possibly be further improved by replacing the phrase "lateral drag" with the phrase "lateral force due to the 'sideways lift' created by the fuselage and other surfaces moving sideways through the air" $\endgroup$ – quiet flyer Apr 30 at 19:56
  • $\begingroup$ or the phrase "Aerodynamic forces perpendicular to Lift-drag plane" $\endgroup$ – Charles Bretana May 1 at 0:02
  • $\begingroup$ And, I would posit, the better solution would be to just cant (tilt) the Lift force so that it is no longer perpendicular to the wings, and call it Total Aerodynamic force. If necessary for clarity, it could also be shown in components, one perpendicular to the wings, and one perpendicular to the Lift-Drag Plane (parallel with the wings) $\endgroup$ – Charles Bretana May 1 at 0:05
  • $\begingroup$ Lift is defined as the component of aerodynamic force perpendicular to the flow velocity, while drag is defined as the component parallel to flow velocity. Thus, I think it is safe to call the horizontal force due to the lateral velocity of the airplane a drag force. On a separate note, the existence of a lateral velocity means that the lift is no longer necessarily in the plane of the paper (or the computer screen), so that's one more issue with the diagrams. $\endgroup$ – Tofi May 1 at 0:51
  • $\begingroup$ @Tofi -- wouldn't a force parallel to the flow velocity be aimed straight away from the viewer, and therefore not visibile in the diagram? Given the assumed direction of the trajectory as stated in the question. Perhaps you feel that the direction of the trajectory should be described differently? $\endgroup$ – quiet flyer May 1 at 0:59

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