# Why does a banked aircraft create sideslip?

Below is an excerpt from Raymer (Aircraft Design) where he explains the dihedral effect. He states that a banked aircraft (due to say a roll disturbance) will have a sideslip which leads to a restoring moment back to equilibrium. What is the physical explanation for why there is a sideslip?

The side slip happens because the lift vector is tilted. Think of a hovering helicopter rotor tilting; the machine drifts to the low side of the rotor disc because the thrust vector of the rotor is tilted. The wings of the plane are just making thrust, but moving forward at the same time. Tilt the thrust line by banking, and a lateral thrust component emerges and the body moves laterally in response (the helicopter in forward flight is doing the same thing - you move the cyclic to tilt the rotor, and the machine moves laterally while going forward because the rotor's thrust line was tilted).

The introduction of the lateral thrust component means a reduction of the vertical thrust component, and unless total lift is increased, the body descends as it's moving sideways, so you have to raise the nose in the turn (or pull more collective pitch, or tilt the rotor back slightly, in the helicopter). And since the lateral movement created by tilting the lift vector is occurring at the same time as the body is moving forward, the movement describes an arc. A turn.

The nose follows the turn because the big weathervane at the back, the vertical fin, sees a lateral angle of attack resulting from the sideslip induced by the tilting of the lift vector, and it seeks to keep the fin's lateral AOA close to zero and so the tail follows more or less in line with the arcing path of the turn. And this is where we get to dihedral effect and sideslip.

You don't want the fin to do this without a little bit of lag. This is the rub; for dihedral effect to work, some sideslip must be present to create the differential lift that creates the restoring or righting force. So the fin needs to be sized to be effective at passively keeping the tail aligned in the airflow while the airplane moves laterally as result of being banked, but not so big as to be instantaneous at doing it; it needs to tolerate a bit of sideslip before it starts to work, so that dihedral effect can do its thing when small disturbances occur.

If you make the vertical fin too big, a bank induced by a disturbance is followed instantly by a yaw created by the incipient sideslip, to keep the tail aligned in the airstream, and nearly no sideslip occurs. Plus the yawing movement means the outside wing is moving faster, tending to increase the bank more. This airplane will tend to stay at its bank angle, or increase its bank angle and want to spiral.

Fin sizing has to find a sweet spot where just enough sideslip is tolerated to make dihedral effect work, but not too much. Too big, and the airplane wanders off into a turn with the slightest disturbance because any amount of induced bank from a bump is following by immediate weathervaning toward the low wing. Too small, and lateral stability is good, but the nose hunts and wanders in bumps.

My own airplane is a homebuilt Pazmany PL-2, and it's a little under-finned, which combined with the inertial mass of fuel in the wingtips, makes the nose wander more in bumps than a lot of airplanes. It has very little spiral tendency however and you can bank it 15 degrees in smooth air and leave it (too much bank for dihedral effect to work), and it will just stay like that, and a spiral dive takes long time to develop.

Incorrect explanations abound for why an aircraft tends to sideslip when banked, in the absence of any "coordinating" rudder input. Often the fact that the weight or gravity vector has a component pointing towards one wingtip is invoked. See for example the diagram in Martin Simons' "Model Airplane Design".

The truth is that this situation exists in a coordinated turn as well as in a sideslip, and so this cannot be the reason for the sideslip. According to Newton's principles, a steady sideways force tends to drive a curvature in a trajectory, not a linear sideways motion. It is not correct to say that the weight or gravity vector tends to cause the aircraft to slide toward the low wingtip whenever the aircraft is banked, just because the weight or gravity vector contains a component that points toward the low wingtip.

Here is one reason that an aircraft tends to sideslip when banked-- banking tends to cause a turn (a curvature in the flight path), and the resulting curvature in the flight path "bends" the airflow or relative wind. In other words, as the aircraft starts to yaw in response to the changing direction of travel, different parts of the aircraft are moving through the airmass in different directions at different speeds at any given instant in time, and so different parts of the aircraft experience different velocities (speed and direction) in the local relative wind, which can be described as a "curve" or "bend" in the direction of the relative wind. In the absence of any rudder input, the vertical tail experiences a relative wind component that tends to oppose the turn and slow the yaw rotation rate. This tends to cause a sideslip.

If this is unclear, consider a pinwheel, or an aircraft in a flat spin. As the body rotates, clearly, different parts of the body are moving through the airmass in different directions at any instant in time.

This phenomenon can also be described as "yaw damping".

This is also the phenomenon that is described as the "long-tail slip" effect in Section 8.10 of John S. Denker's "See How It Flies" website.

If the aircraft's fuselage could bend like a banana to conform to the curving path of the turn, then every part of the aircraft from the nose to the tail could be streamlined to the local direction of the curving relative wind, and this phenomenon would not occur.

Rotational inertia in the yaw axis also plays some small role in driving sideslip as the bank angle and turn rate increase.

Explanations that attempt to account for the relationship between banking and and sideslip, without discussing the curvature in the flight path that generally results from banking, are invariably incorrect.

• Awesome answer! exactly correct! Proof is that if rudder was in front of the aircraft center of pressure (assuming the lateral instability could be controlled some other way), the induced sideslip would be in the opposite direction. In the F-4 on Check rides, we had to perform a maneuver where we pulled the nose up to 45-60 deg nose high, then unloaded to zero AOA, and rolled 360 deg as the flight path followed a ballistic arc, ending up pointed 45-60 deg down. As the aircraft rolled, to keep the ball centered, you had to input rudder towards the ground, with 60-90 degs bank angle lead. – Charles Bretana Aug 30 '20 at 17:16
• This was necessary because as the flight path changed, continually bending towards the ground, the aircraft nose position was always lagging behind (upwards) from the the flight path, (which was bending downwards). So If you rolled left, then in the first 90 deg of roll, when the aircraft was in 90 deg left bank, the flight path was to the left of the nose, and you needed extra left rudder to "kick" it down, to the left, to align it with the moving flight path vector. Then in the last part, when in a right bank, you needed right rudder, to kick the nose down to the right, for the same reason. – Charles Bretana Aug 30 '20 at 17:20

The act of rolling the aircraft creates adverse yaw, which generates sideslip. This is a transient effect which is explained in this answer.

In a steady coordinated turn (ball centered), however, sideslip will also be present. This is a steady-state effect. There are a few contributing factors:

• There is steady-state non-zero rudder, which is necessary to zero out the yaw damping moment ($$C_{n_r}$$), yaw-roll damping moment ($$C_{n_p}$$) and adverse aileron moment ($$C_{n_{\delta_a}}$$). This rudder deflection creates an aerodynamic side force ($$C_{y_{\delta_r}}$$).
• The presence of steady-state non-zero body yaw and roll rate induces side force ($$C_{y_p}$$ and $$C_{y_r}$$).

In a ball centered turn, the sum of aerodynamic side force is zero, which physically demands a non-zero sideslip to generate the necessary aerodynamic side force to cancel.

Together:

$$C_{y_\beta}\beta+C_{y_{\delta_r}}\delta_r=C_{y_p}\theta\frac{\dot{\psi}b}{2V}-C_{y_r}\cos{\phi}\frac{\dot{\psi}b}{2V}$$

This is with all engines operating, so no thrust asymmetry is present. And this steady-state sideslip is small for ball-centered turn.