7
$\begingroup$

I understand that when the aircraft is put into some bank angle, the lift vector now acts at that angle instead of directly against gravity, so we can pitch the aircraft up to increase AoA to increase lift so that the vertical component of lift cancels out weight. Then we are just left with a horizontal component of lift acting perpendicular to the forwards direction of the aircraft. I don't understand why this doesn't just cause the aircraft to continue moving forwards but with some sideslip velocity.

If we complete a full 360 degree turn with some bank angle the aircraft must experience 360 degrees of rotation about its centre of gravity, so where does this rotation come from?

$\endgroup$
4
  • 1
    $\begingroup$ Welcome to aviation.stackexchange.com! $\endgroup$
    – DeltaLima
    Jan 4 at 18:57
  • 1
    $\begingroup$ "I don't understand why this doesn't just cause the aircraft to continue moving forwards but with some sideslip velocity." It will if you counter the direction of bank with opposite rudder. $\endgroup$ Jan 4 at 19:03
  • $\begingroup$ The airplane will also rotate around the pitch axis, the more the steeper its bank angle is. By coordinating pitch and yaw rotation, a circular path can be achieved. $\endgroup$ Jan 5 at 13:37
  • $\begingroup$ Related -- aviation.stackexchange.com/q/34271/34686 $\endgroup$ Jan 5 at 15:44

6 Answers 6

9
$\begingroup$

You are absolutely right that the horizontal force perpendicular to the direction of flight cause a side slip.

The rotation comes from the vertical stabilizer which as soon as a side slip occurs generates a aerodynamic moment which rotates the aircraft into the direction of flight. This ensures that the horizontal component of the lift vector remains perpendicular to the (aerodynamic) direction of flight.

$\endgroup$
3
  • 1
    $\begingroup$ Technically, the side-slip is a secondary effect - the perpendicular force has no vertical component, by definition. The slip is actually caused by the other aero forces being out of balance. $\endgroup$
    – MikeB
    Jan 5 at 13:45
  • 2
    $\begingroup$ I am not sure I understand your comment @MikeB. The slide slip is caused by the sum of forces (weight and aerodynamic) resulting in a lateral force which changes the direction of the flight path. When the flight path changes direction (turn) but the aircraft heading remains the same, the result is a sideslip. $\endgroup$
    – DeltaLima
    Jan 5 at 15:37
  • $\begingroup$ It might be easier to understand if you try drawing out the vectors, before and after the turn has been initiated. Use arbitrary values, and you will see that after the turn starts, the VERTICAL component reduces - this is where the problem starts. $\endgroup$
    – MikeB
    Jan 8 at 9:25
5
$\begingroup$

Key point: in a steady-state turn, net torque on the airplane must be zero. Net force on the airplane however is not zero-- there must be a horizontal force toward the center of the turn. This is called a "centripetal" force. This comes from the banked wing.

This centripetal force plays the same role as does the steady pull of the sun's gravity upon the earth or any other planet-- it causes the moving body's trajectory to curve.

Our intuition from driving cars, steering shopping carts, etc leads us astray because with these vehicles it seems that the net torque acting on the vehicle might not be zero even in a steady-state turn, when we are going around and around in a circle of constant radius, at a constant speed and turn rate. But that's an illusion that comes from factors like the castor of the wheels, etc. Net torque is in fact zero in a steady-state turn even in these vehicles, even if we find ourselves having to constantly exert a torque on the steering wheel, push bar of the shopping cart, etc due to these additional complicating factors.

It's true that as we are initiating a turn, or more specifically as the vehicle's rate of rotation about the yaw axis is increasing rather than staying constant, there must be some net torque exerted on the vehicle to overcome the vehicle's yaw rotational inertia. In an airplane, we can create that torque with the rudder, or we can allow the airplane to sideslip briefly, in which case the force of the sideways airflow against the fin will generate the torque needed to overcome yaw rotational inertia and increase the vehicle's yaw rotation rate as needed to end the sideslip.1

To maintain a steady-state sideslip, which is characterized by a sideways airflow over the aircraft, in most airplanes we need to hold a constant rudder input toward the high wingtip.2 In this case "downwind" yaw torque from the rudder, and the "weathervane" (or "upwind") yaw torque caused by the sideways airflow over the rear fuselage, fixed vertical fin, etc, are essentially "fighting" each other so that the net yaw torque is zero.3 In some sense we can think of the banked wing as "dragging the airplane sideways through the air", but it's important to recognize that the net force on the airplane is actually zero in any linear, constant-speed sideslip-- the horizontal force from the banked wing is opposed by an equal and opposite horizontal force from the sideways airflow colliding with the side of the fuselage and other surfaces of the aircraft.4

Footnotes:

  1. In actual practice, in most airplanes most of the sideslip we see when entering a turn without using the rudder comes from aerodynamic effects (e.g. "adverse yaw"), not yaw rotational inertia. Yaw rotational inertia may even act to limit the maximum amount of sideslip caused by these aerodynamic effects (e.g. "adverse yaw"), when entering a turn without using a rudder. But ultimately, if we are not using the rudder to generate a proverse yaw torque-- say for example if we are only using exactly as much rudder is needed to neutralize adverse aerodynamic torques (e.g. "adverse yaw") and no more-- then something must generate some net yaw torque to overcome yaw rotational inertia and start the aircraft rotating about its yaw axis, and this something will most often be the force of the sideways airflow against the vertical fin as the aircraft sideslips to some small degree. Of course, as an aircraft enters a turn without using the rudder, it will generally not be obvious to the pilot what percent of any sideslip that temporarily develops is due to adverse aerodynamic torques, and what percent is simply due to yaw rotational inertia.

  2. We're focussing on the "big picture" here, and ignoring the slight sideslip due to aerodynamic damping in the yaw axis that we'll typically see even in a steady-state turn if the rudder is exactly centered. This can be further compounded, or can be countered, by aerodynamic yaw torques generated by the deflected ailerons, if some aileron deflection is needed to hold the bank angle constant. So it's only to a first approximation that keeping the rudder exactly centered will result in zero sideslip in a steady-state turn, where yaw rotational inertia is no longer a factor and aerodynamic yaw torques associated with a changing bank angle are also no longer a factor.

  3. Note that only the side area of the fuselage that is aft of the CG contributes to the stabilizing, "weathervane", "upwind" yaw torque during a sideslip. Side area of the fuselage that is ahead of the CG contributes a yaw torque in the opposite direction, essentially "helping" the rudder keeping the nose yawed to point in a different direction than the airflow is actually coming from, i.e. a different direction that the aircraft is actually travelling through the air. An interesting manifestation of this: in the Challenger ultralight, which has large removable cockpit doors forward of the CG of the aircraft, a full rudder input will command a significantly larger sideslip angle when the doors are installed than when the doors are absent and the airflow can pass through the cockpit. Similarly, aircraft on floats, which have a substantial surface area forward of the CG of the aircraft, often have additional tail area added as well in the form of a ventral fin to restore the desired yaw stability characteristics.

  4. It really makes a lot more sense to think of an intentional, linear (non-turning) sideslip as being "caused" by the deflected rudder, not by the banked wing. The banked wing is just what cancels the sideforce from the airflow hitting the side of the fuselage, so that the net force can be zero and the flight path can be linear rather than curving. As a demonstration, we could hold full right rudder and slowly vary the bank angle from a right bank to wings-level to a shallow left bank to a steeper left bank, and to a first approximation we'd have the same angle of sideways airflow over the aircraft the whole time. But the flight path would vary from a tight skidding right turn (caution!), to a slow wings-level right turn (which still can be called a "skidding" turn), to a situation where the left wing is lower but the flight path is slowly curving to the right (still a form of "skid"?), to a linear slip with the left wing low, to a slipping left turn. From the point of view of the "yaw string", these are all manifestations of sideslip, and the deflected rudder is the fundamental cause of the sustained sideways airflow over the aircraft, not the banked wing.

$\endgroup$
7
  • 1
    $\begingroup$ snall point but Newton's first law applies to rotations and moments in exactly the same way as it applies to forces and velocities. Just as it is true that a body in motion will remain in motion at a constant velocity unless a force is applied, a body in rotation at a constant angular velocity will remain in rotation at the same angular velocity unless a moment is applied to it. $\endgroup$ Jan 5 at 0:10
  • 1
    $\begingroup$ I didn't get into it here in this answer, but the idea that some sustained net force (such as the sideways from the banked wing) could result in some sort of steady-state, linear flight path is basically an Aristotelian concept, rather than a Newtonian one. Steady-state linear sideslips can only be properly understood when we realize that the sideways force from the banked wing is opposed by another force (as explained in answer), and the net force is actually zero -- . And once we realize that the sideways airflow over the aircraft is playing a key role in the forces at play, (ctd) $\endgroup$ Jan 5 at 15:13
  • $\begingroup$ (ctd) then we have to think about how that sideways airflow affects the torques as well. We'll soon realize that since aircraft have "weathercock" or "directional" stability, a linear (non-turning) sideslip generally won't persist for long unless the rudder is deflected. Exceptions exist, where strong dihedral stability requires strong aileron deflection to maintain a constant bank angle while slipping, and the adverse yaw characteristics of the ailerons are such that the aileron deflection itself is sufficient to create a strong, "adverse", "downwind", yaw torque, sustaining the slip. (ctd) $\endgroup$ Jan 5 at 15:18
  • $\begingroup$ (Ctd) In such an a/c you could actually find yourself in a sustained, constant-banked, steady-state sideslip with the stick or yoke deflected to create a roll torque toward the low wingtip and prevent the bank angle from decreasing, and the rudder centered. I've seen some evidence of such behavior in some some sailplanes. $\endgroup$ Jan 5 at 15:27
  • $\begingroup$ Although you did not explicitly reference my comment in your comments, if they were in response to mine, let me clarify... The reason I posted it is because I feel that the fundamental aspect of the concept of a "sideslip" is not the change in the direction of aircraft velocity, it is the [lateral] mis-alignment of the aircraft's orientation with it's velocity. Understanding that, is it not clear that sideslip is not caused, (explained), by forces, but by the distribution of forces relative to the CG, i.e., by moments, as you expertly explain in your additional comments. $\endgroup$ Jan 6 at 14:15
3
$\begingroup$

You are NOT right. An aircraft in flight has six degrees of freedom. You must analyze them all to understand the motion. The aicraft is in motion in three dimensions. It can move, in each of these three directions, and it can also rotate in each of those three directions.

To understand the motion, you need to analyze the total forces acting on the aircraft, and resolve them ( add them up vectorially), into a single force representing the sum of all of them. That vector sum of all forces then determines the direction and magnitude of the acceleration the aircraft will experience. if the orientation or direction of that total force is to one side, the aircraft will turn. I don`t mean it will rotate, we'll get to that later. I mean it will turn. It's velocity vector will change direction.

The other three degrees of freedom control what the aircraft attitude is, i.e., how it rotates or what direction it is pointed in. This is controlled by the torques or moments created by the distribution of the forces on the airframe relative to the center of mass (often called center of gravity or CG) - this is called a torque, or moment. (Sit on one end of a seesaw with your 50 lb granddaughter on the other end)

In the scenario you describe, the horizontal component of the lift vector creates a horizontal component of acceleration, and that causes the velocity vector to change direction. If there was no rudder, the aircraft would still change direction of motion, but it's orientation (the direction the nose is pointed in), might not change. That is controlled by the distribution of the forces on the airframe relative to the CG. If more force is pushing to the right 5 feet ahead of the CG than is pushing right 5 feet behind the CG, then the aircraft will Yaw to the right.

Basically, the moment is the product of the force vector times the distance between the CG and where the force is pushing on the airframe. You add up all the moments and determine the total moment to see what the impact of all of them on the aircrafts tendency to rotate, (pitch, yaw or roll).

Because stability about the vertical axis (directional stability) is so important, almost all aircraft are designed with aerodynamic surfaces that provide this. We call them vertical stabilizers. And most of them have adjustable sections (rudders) to allow the pilot to control them. This is what affects and controls sideslip, not the horizontal component of lift.

$\endgroup$
1
$\begingroup$

why this doesn't just cause the aircraft to continue to move forward but with some sideslip velocity

we are left with a horizontal component of lift acting perpendicular to the forward direction of the aircraft

Let's start there: 3 force vectors that the aircraft is producing acting against gravity and drag.

We roll into the bank, increasing Angle of Attack (and Power) to produce a larger lift vector to maintain adequate vertical lift. Mathematically, Lift = 1/cosine bank angle × vertical lift, and the horizontal lift component is sine bank angle × Lift.

The horizontal component and the thrust component act on the plane, the result is a change in course of the aircraft.

Now think of what a weathervane does when the wind changes. In a relativistic sense (as in wind tunnels) the wind on the stationary weathervane and the plane flying through the air will have the same effect: the weathervane or the plane will try to align itself to the new "relative wind" direction.

Either the wing, outside aileron, or inadequate rudder (slip) or excessive rudder (skid) can cause the relative wind to be either inside or outside the aircraft direction of flight, respectively.

The rear fuselage of the aircraft and empennage are designed the "weathervane" into the relative wind, and the rudder provides "fine tuning" for "coordinated" flight.

So, the "circular path" is a result of the horizontal lift component --> changes relative wind --> creates sideforce on aircraft --> torques the tail around the Center of Gravity to realign with the relative wind, over and over again.

Notice, as the pilot controlling the plane, you can move the position of the "ball" with either ailerons or rudder. Usually, we bank a turn to change course, and "clean it up" by positioning the rudder properly to give just the right amount of yaw to coordinate the turn.

keeping the relative wind as close to the nose as possible minimizes drag

$\endgroup$
2
  • $\begingroup$ So, it seems the trick is to rotate the aircraft around the turn with the rudder instead of using sideforce (from uncoordinated flight) to keep the nose pointed into the relative wind. An object in orbit can turn without rotating because there is no drag. $\endgroup$ Jan 6 at 3:39
  • $\begingroup$ Could one imagine trying to "coordinate" the rotation (degrees/second) of the satellite with the orbit time to always have a view of the earth? $\endgroup$ Jan 6 at 3:50
1
$\begingroup$

The answers here are correct but let me try to illustrate this in an intuitive way that will “sink in” for most readers even without a physics understanding.

You “get” the part about the bank causing a horizontal component to the wing’s lift, in the direction of the turn. From that starting point, take the wings and the bank out of it. Imagine a simple arrow in flight, which is similarly starting to fly “sideways” just like the banked airplane is starting to fly sideways. What would happen to that arrow? The tail feathers or “fletching” would cause the arrow to tend to turn into that wind, just like a weather vane.

If you bank the plane left, the aircraft does start to move left, and the tail, just like on the arrow, weather vanes the airplane into the turn.

The technical reason for this is that the CP (center of pressure) is well behind the CG (center of gravity) but hopefully, even if you’ve never heard of either of these concepts, the arrow example helps you form the correct intuition in your mind.

$\endgroup$
0
$\begingroup$

Banked flight indeed produces a side force. A side force produces sideslip. Unless it passes through the center of gravity, it will also produce a rotation. Ideally, the total effect is such that a passenger not looking out of the window does not notice anything. Effective gravity should still be directly head to feet. Rudder may be required to sustain the illusion. This may help you to understand some of the responses that you received.indeed

$\endgroup$
2
  • $\begingroup$ "A side force produces sideslip." -- not always! "Unless it passes through the center of gravity, it will also produce a rotation." -- substitute "torque" for "rotation". Not the same thing! $\endgroup$ Jan 6 at 1:56
  • $\begingroup$ Forces do not themselves produce sideslip, if a force is applied to a body so that it passes directly through the body's center of mass, the body will experience an acceleration, and will change its velocity, but will not change its orientation, because no torque or moment will have been generated. It is the torque or moment created by the vertical stabilizer or rudder displacement or p-factor or propellor slipstream that creates sideslip. $\endgroup$ Jan 6 at 14:08

You must log in to answer this question.

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