In a banked turn, why does the center of gravity, rather than the center of lift, follow the path of a circular arc?

Question

I'm a post-solo student pilot who sometimes gets hung up on fundamental points of aerodynamics.

"As long as the aircraft is banked, the side force is a constant, unopposed force on the aircraft. The resulting motion of the center of gravity of the aircraft is a circular arc." (https://www.grc.nasa.gov/www/k-12/airplane/turns.html)

I'm having a difficult time understanding why the center of gravity is translated along this circular arc, when the turning force that pulls the airplane through the turn is acting upon the airplane's center of lift, aft of the CG. It makes intuitive sense to me why the airplane rotates about the CG on all of its axes, but all translation / movement being located at the CG is more difficult for me to visualize.

My intuition is that the turning force acting aft of the CG would cause the plane to rotate (pitch) about the CG, rather than turn. I know that in straight and level flight, this tendency is counteracted by the downward force exerted on the horizontal stabilizer. I'm assuming this point is also relevant in answering my question, but it still strikes me as counterintuitive.

The PHAK describes the relationship between CG, CL, and tail as a "lever," balanced on the CL, with the weight of the CG and the downforce of the tail on either side. If the force that turns the airplane stems from the balance point of that lever, I would expect that to be the point that follows the direction of the turn, rather than the weight on one side of it!

Thanks for any insight, and I hope that I was clear in my phrasing of this question.

Cheers.

Answer 1

I'm having a difficult time understanding why the center of gravity is translated along this circular arc, when the turning force that pulls the airplane through the turn is acting upon the airplane's center of lift, aft of the CG. It makes intuitive sense to me why the airplane rotates about the CG on all of its axes, but all translation / movement being located at the CG is more difficult for me to visualize.

The truth is that in turning flight, every point on the aircraft is following a circular arc. There's no reason to single out the CG. Note however that the circular arc followed by each point on the aircraft is not the same. For example, the inboard wingtip follows an arc of smaller radius than the outboard wingtip does. This is important to understanding the "overbanking" tendency that some aircraft experience-- since the outboard wingtip covers more distance (circumference) in a given unit time than the inboard wingtip, the outboard wingtip experiences a higher airspeed than the inboard wingtip, and so tends to generate more lift than the inboard wingtip.

We can think of the instantaneous linear velocity (speed and direction) of any point on the aircraft as the result of the aircraft's instantaneous linear velocity at the CG, acting tangent to the circumference of the turn, plus an additional linear velocity component due to the aircraft's rotation about the CG. But guess what -- we're just using the CG as a handy reference point here. It's equally valid to pick any other point on the aircraft as our reference point. For example, we could pick a reference point on the inboard wingtip, measure its instantaneous linear velocity, and then calculate the instantaneous linear velocity of every other point on the aircraft by treating the problem as if the aircraft were rotating around the inboard wingtip, rather than rotating about the CG. Either way of treating the problem would be equally valid, and either way would give the same result for the instantaneous linear velocity of any point on the aircraft. For example, we'll still find that the outboard wingtip is moving through the air faster than the inboard wingtip, and we'll still get the same number for the difference in instantaneous linear velocity between the inboard wingtip and the outboard wingtip. There's no rule that says we must pick the CG of the aircraft as our reference point or "pivot point" for these sorts of calculations. It's not really valid to say that a moving, rotating body rotates about its CG and not about some other pivot point. The choice of a "pivot point" is arbitrary, and need not be at the CG.

If the "center of lift" (which you seem to be defining as where the upward lift from the wing acts, without regard to where the upward or downward lift from the tail acts) were somewhere other than at the CG, you would be perfectly valid in choosing that center of lift as the reference point for the sort of calculation we've been describing here. But you could equally well choose the CG for your reference point. It's arbitrary.

However, your question does show a misconception that the "center of lift" of the entire aircraft is somewhere other than at the CG, at least in the fore-and-aft sense. Remember, the horizontal tail can generate either downward lift or upward lift. As far as the static balance of pitch torques is concerned, downward lift generated by the tail has the same effect as upward lift generated by a canard wing, and the situation can be analyzed as if the aircraft had an uplifting canard wing rather than a downlifting tail. Either way, the "effective center of lift" of the wing-tail system, i.e. of the entire aircraft, must be co-located with the CG, at least in the fore-and-aft sense, or a net pitch torque would be created, and the aircraft would be experiencing an increase or decrease in the rate of pitch rotation.

(Disclaimer-- we're assuming here that thrust and drag are equal and opposite and act in line with each other. An engine mounted high above the aircraft on a tall pylon, for example, would tend to generate nose-down pitch torque that could be offset by moving the center of lift of the entire aircraft ahead of the CG.)

There's one other significant twist though. If we are interested in calculating the balance of pitch torques on a rotating body, rather than simply calculating instantaneous linear velocities at various points on the body as described above, then if we choose any point other than the CG as our reference point, then we have to introduce a fictitious pseudoforce called "centrifugal force" which acts at the CG and is generated by the CG's curving trajectory around the center of the turn. Choosing the CG as our reference point avoids the need for introducing this pseudoforce, so "centrifugal force" need not enter our calculations.

Answer 2

Well, are you forgetting to use the tail?

The "center of gravity" model can take some time to fully fathom, but it works well to describe rotation on all three axes.

Let's try pitch. Since CG is ahead of wing CP (center of pressure), CP will try to pitch the nose down. So how do we "rotate" to take off? More up elevator!.

Flying a gentle full circle the plane actually does one rotation around the yaw axis with some help of the rudder. Steeper turns use more elevator.

We can see that the elevator and rudder provide the torque (+ or minus) that controls rotation in pitch and yaw. Rotation on the roll axis is with ... the ailerons.

In a turn the tail coordinates the rate of rotation with the change in flight path (the "turn") caused by the horizontal component of the lift vector (the constant, unopposed side force). The result is the nose of the plane follows the curvature of the circle as much as possible, reducing drag.

It helps to keep one a bit more "on the ball".

Answer 3

The centre of gravity is the reference point of the rigid body. The six degrees of freedom are modeled as three linear translations of the CoG, and three rotations around the CoG.

You’re right in that the centre of lift could be taken as the reference point as well, or any other point for that matter. It’s just that rotations of a body suspended in air take place around the CoG, and it is the movement of the rigid body that defines the reference point.

Answer 4

The airplane follows an arc in the turn because of the vertical fin. When you bank, it introduces a lateral vector to the wing's lift force. The lateral vector makes it displace sideways as it's moving forward. Without the fin to provide a weathervaning effect, it would simply continue more or less pointed on its original heading, but on a slewed track, so that the airflow is striking the body from ahead and to one side.

Think of a helicopter hover taxiing, moving forward very slowly. You apply left cyclic to tilt the rotor disc a bit as you move forward. The machine continues pointing where it was pointing, but starts to move laterally because the rotor's lift vector was tilted. If it was pointing North, it would continue pointing North, but actually proceed in a straight line Northwest.

It won't actually start to arc around with a heading change until the lateral flow hitting the tiny little fin at the back starts to point it if you're going fast enough, or you move the tail yourself with some pedal input. When you bank the airplane, the same forces are at work.

On the plane, the offset flow causes the fin (the fixed fin, not the rudder) to generate a local lateral lift force, a weathervane effect, realigning the body into the flow. The slewing or side slipping action followed almost immediately by a realignment by the fin results in a turning arc.

It's desirable to have a slight lag in this activity, because this initial side slipping action is required for dihedral effect to work, to provide lateral stability. Dihedral effect requires that a little bit of lateral sideslip be allowed to develop when a bank is induced, to create the differential lift that tends to restore the plane back to level flight, before the fin starts to do its realignment function. Fin sizing has to take this into account; too big and the plane starts to align its yaw axis to the flow instantly when a bank occurs and you end up with a spiral tendency, and too small, the plane slithers and slides around.

Note that it's the fixed fin, not the rudder doing this. If you have your feet on the floor, the rudder just trails in the airflow, not doing anything. If the plane has no adverse yaw, you can leave your feet on the floor while maneuvering and the plane will maintain coordinated flight because the fin itself is able to continuously weathervane the body into the flow.

The C of G is simply the point at which any rotational movements are centered. Pitch forces come into it mostly to compensate for the loss of the purely vertical component of lift that was reduced when the bank was induced, to keep from descending. Elevator inputs change the trim force balance to increase the AOA that will be sought by the plane's static pitch stability forces.

Since increasing AOA increases the lift vector perpendicular to the body, the increase raises both the vertical and lateral lift forces, increasing both the lift required to keep from descending, and the turn rate resulting from the lateral lift vector. So turn rate is influenced by both bank angle and elevator input, because both factors influence the lateral component that wants to displace the airplane sideways, with the pitch factor only a minor part of it until the bank angle gets steep. As the bank steepens, more and more of the total turn rate is coming from pitch rate from the elevator (take it to the extreme and imagine a turn at near 90 degrees bank - nearly all the change in heading is from pitch since you are basically flying a loop laid flat).

While all that is going on, the fin just sits there keeping the body aligned into the flow, with the rudder chipping in to help out via your feet to compensate for adverse yaw or bumps, as required.