# How Airplanes correct for the Coriolis Effect

Wednesday, September 2, 2020 - 22:39 | Author: wabis | Topics: Knowlegde, Physics, Aviatic I show with an interactive model how the Coriolis Effect emerges in the rotating reference frame of the earth. I show how airplanes take the Coriolis Effect into account and how they manage to keep up with different tangential speeds at different latitudes. I provide all equations and calculators to calculate the corrections airplanes have to apply. I show that the Coriolis Effect is small enough so it can be ignored in flight planning.

## Model of Earth's Coriolis Effect

Read How Coriolis Forces emerge below for the explanation of the Coriolis effect.
Use the mouse wheel to zoom. Click and drag or touch to rotate the view. Press Shift to rotate both views in sync.

## Left Model

The left Model shows the earth as seen from a non-rotating, inertial reference frame in space with the origin at the center of the earth. In this reference frame the earth is rotating.

Space latitude and longitude: are the coordinates of the bullet or airplane with respect to the non-rotating, inertial reference frame of the space, depicted by the blue Space Grid.

Distance: is the distance the object has traveled with respect to space.

Speed, Heading: is the speed and heading of the object in the space coordinate system.

Acceleration (Accel): is the resulting acceleration acting on the object in the inertial reference frame of space. This acceleration causes the object trajectory as seen in the space reference frame. The acceleration is decomposed into a component along the direction of flight of the object along the space trajectory, a vertical component and a component to the right with respect to the direction component.

This acceleration is the sum of all forces divided by the mass of the object. Because the space coordinate system is an inertial reference frame there are no inertial forces like Coriolis and centrifugal forces. But in the case of the controled flight there are the additional forces the airplane produces to counter the Coriolis and Centrifugal effect. The resulting acceleration vector defines the 3D motion of the object in the inertial reference frame of space.

## Right Model

The right Model shows the same scene as seen from a co-rotating, non-inertial reference frame. The co-rotating reference frame is attached to the earth, so the earth does not rotate in this reference frame.

Earth latitude and longitude: are the coordinates of the bullet or airplane with respect to the rotating reference frame. This are the coordinates used on earth.

Distance: is the distance the object has traveled on earth.

Speed, Heading: is the speed and heading of the object as measured on earth.

Acceleration (Accel): is the resulting acceleration acting on the object in the rotating reference frame of the earth. This acceleration causes the object trajectory as seen in earths reference frame. The acceleration is decomposed into a component along the direction of flight of the object along the trajectory on earth, a vertical component and a component to the right with respect to the direction component.

This acceleration is the sum of all forces, i.e. all physical forces plus the inertial Coriolis and Centrifugal forces, divided by the mass of the object. In the case of the controled flight there are the additional forces the airplane produces to counter the Coriolis and Centrifugal effect. The resulting acceleration vector defines the 3D motion of the object in the rotating reference frame of the earth.

It is assumed that the bullet and the airplane produce a lift force of the necessary magnitude, that it counters gravity but leaves a net centripetal force to hold it at a constant distance above the surface. This net force acts in both reference frames with the same magnitude. An airplane trimmed for a certain altitude automatically creates the correct lift force.

Coriolis Acceleration: The Coriolis acceleration exist only in the rotating reference frame of the earth. The Coriolis acceleration has 2 components, one acting horizontal perpendicular to the direction of motion in earths reference frame and one acting vertical. Coriolis Accel only displays the horizontal component of the Coriolis acceleration. The vertical component of the Coriolis effect is called the Eötvös effect , which has no influence on the flight path.

The Centrifugal Acceleration is contained in the effective gravity vector. The effective gravity vector is the vector sum of the gravitiy vector pointing to the center of the earth and the centrifugal acceleration vector. The effective gravity vector does not point to the center of the earth, except at the poles and at the equator, but is always acting perpendicular to the ellipsoidal shape of the earth. Because the lift force acts exactly opposite to the effective gravity vector, the centrifugal acceleration is automatically canceled if the airplane flies level with respect to the ellipsoidal surface. In the calculations above the correction for the the centrifugal acceleration is contained in the lift force and not explicitely displayed.

Deviation: shows how much the bullet (dark blue dot) deviates due to the Coriolis effect from a track without Coriolis forces (light blue dot).

Bank Angle: shows how much an airplane has to bank to create a side force that cancels the Coriolis force. This produced side force provides the acceleration to keep up with the latitude dependent tangential speed of the surface of the earth with respect to space.

## How Coriolis Forces emerge

Fig1: Paths of a Bullet (blue dot) in the different Reference Frames of Space (red line) and the rotating Earth (black line). The bullet is fired at the green dot due south.
Fig2: ISS Path in Earths rotating Reference Frame

Lets assume no air friction and only a small centripetal force, so that a bullet in motion stays exactly at the same height above the surface of the earth. A bullet fired from a gun attached to the rotating earth will enter a perfect orbit around the earth (red line in Fig1), as soon as the bullet leaves the gun. It's speed and the orientation of the orbit is determined by the speed at which the bullet leaves the gun plus the tangential speed of the gun on the rotating earth. So the bullet inherits the additional tangential momentum from the gun. After the bullet leaves the gun and enters the orbit, it moves in orbit indefinitely.

In the inertial (non-rotating, non-accelerating) reference frame of space the orbit of the bullet is a perfect circle around the earth, which does not change its orientation in space, like a gyroscope. The earth rotates underneath it independently. The bullet and the orbit are decoupled from the rotating earth, if we assume that there is no atmosphere co-rotating with the earth.

In the rotating reference frame of the earth however, i.e. from the perspective of an observer on earth, the same bullet traces a weird path over the surface of the earth (black line in Fig1). The shape of this path depends on earth rotation rate, the speed of the bullet and the initial position and direction the bullet is fired from. Play with the Distance slider in the animation above to see the path emerge.

If the bullet has the orbital speed of the ISS then the path on earth is a sinusoidal figure like it is published and shown in Fig2. If the bullet is much slower, the path looks very unfamiliar, because we never see objects orbiting the earth with such slow speeds. Real bullets will slow down due to friction and fall to the earth in relatively short distances. So we can only observe the initial part of the deviation from a straight line, a very short path of the black line in Fig1, starting at the green dot where the gun is placed.

Now what has this to do with the Coriolis Effect?

According to Newton's first law of motion, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force, any object that changes direction and/or speed is reacting to forces by definition. Forces, no matter what causes them, are the only reason why objects change their movement. In the rotating reference frame of the earth, the orbiting bullet does change its direction and speed all the time. So in this reference frame there must be additional forces acting on it. This forces are called Inertial forces. The Inertial forces are differentiated into the Centrifugal force and the Coriolis force.

Inertial forces do not appear in the inertial reference frame of space. They only emerge in rotating reference frames. Centrifugal and Coriolis forces are a consequence of inertia in rotating frames of reference. All physical forces, like (classical) gravity and lift, are the same in every reference frame.

On the rotating earth the Centrifugal and the Coriolis forces are vectors pointing in different directions. The Coriolis force acts always perpendicular to the direction of motion. So it has a side and a vertical component. The Centrifugal force acts always away from the axis of rotation, so it can have a component in any direction as viewed from the moving object, depending on its position and orientation.

Because Inertial forces appear only in non-inertial reference frames, they are sometimes called fictitious or pseudo forces. But in the non-inertial reference frame they act and change trajectories of objects like physical forces, such as thrust or wind, so they have to be treated like any other forces in calculations of motions.

Note: A bullet does not leave earth's reference frame when it leaves the gun and starts to follow an orbital path. All objects are in all reference frames all the time. But the motion, the apparent path that an object traces, depends also on the motion of the reference frame in which the object is observed. So an observer in the inertial reference frame of space sees another track than an observer in the rotating reference frame of the earth. As a consequence the 2 observers see different forces acting on the same object at the same time. Motions and inertial forces are always relative to a certain reference frame.

## Properties of Coriolis Forces

• Coriolis Forces only emerge in rotating reference frames.
• The Coriolis force is one of the terms that emerges from the transformation of Newton's law of motion into a rotating reference frame. Other terms are the Centrifugal force and the Euler force, see Equation of Motion in a rotating Reference Frame.
• Coriolis Forces are called fictitious or pseudo forces to distinguish them from physical forces like thrust, friction, classical gravity...
• Inertial forces like the Coriolis forces are real. If airplanes do not correct for Inertial forces, their motion deviates from a straight path over the rotating surface of the earth.
• Every single moving object is affected by the Coriolis Effect, even single atoms. That's the reason why we have rotating weather systems.
• A typical mean magnitude of the Coriolis force acting on an airplane is about the same as of a weak 12 kt crosswind, calculated for an A320 at cruising altitude at 45° latitude, see Calculation of Crosswind Forces.
• The corrections for the Inertial forces produce exactly the acceleration needed for an airplane to accelerate to different tangential speeds at different latitudes.
• The horizontal component of the Coriolis acceleration on a sphere depends on the speed of the object in the rotating reference frame, the rotation rate of the reference frame and the latitude of the object, see Calculation of the Coriolis Force.
• The horizontal component of the Coriolis Force always acts to the left or right with respect to the direction of motion, see (1). Another component acts vertical. The whole Coriolis force is a 3D vector.
• At the equator the horizontal component of the Coriolis Force is zero and the vertical component is maximal.
• The horizontal component of the Coriolis Force acts to the right in the northern hemisphere and to the left in the southern hemisphere. This causes different storm rotation patterns north or south of the equator.

## Airplanes do Correct for Coriolis Forces

Fig3: Calculated deviation from track, if no Coriolis corrections are applied

Contrary to idealized bullets with no friction, airplanes are connected to the co-rotating atmosphere by means of control surfaces like the wings and have engines to control lift and speed. Airplanes make constantly small adjustments to counter any forces, such as wind and Coriolis forces, that try to push the airplane away from the planned track, see Evidence for the Coriolis Correction of Airplanes. If no corrections would be made, the Coriolis forces would push the airplane away from the track, to the right in the northern hemisphere, shown in Fig3.

What about the Centrifugal acceleration? They have components in any direction. So don't they affect the flight path of an airplane?

The earth has an ellipsoidal shape due to the Centrifugal forces. The effective gravity vector is a combination of the gravity vector pointing to the center of the earth and the vector of the Centrifugal acceleration. The effective gravity vector points always perpendicular to the ellipsoidal surface everywhere, not to the center of the earth.

If an airplane would fly level with respect to a spherical earth shape, the centrifugal forces would have an influence on the flight path. But if the airplane flies level with respect to the ellipsoidal surface of the earth, the lift vector is exactly opposite to the effective gravity vector and implicitly contains a component that cancels the Centrifugal force. So the airplane does automatically correct for the Centrifugal force when flying wings level to the surface. It only has to take the Coriolis force into account explicitly.

How big is the correction for Coriolis? How do airplanes correct for Coriolis?

To counter the Coriolis acceleration, the airplane simply has to bank slightly to the opposite side, without changing heading. How much?

At the equator the needed Coriolis correction is zero. At a mean latitude of 45° north and a speed of 250 m/s = 486 kt (kt = knots = nautical miles per hour), the Coriolis acceleration is only 0.0257 m/s2 to the right. The airplane has to create an equal acceleration to the left to stay on track.

The needed bank angle for such a sideways acceleration is for any airplane the same: only a minute 0.15°.

This bank angle is not noticable. The pilot or autopilot applies this bank angle automatically as he is constantly correcting for any deviation from the planned track. There is no need to plan for such corrections, because they are too small to have any influence on the flight calculations and the necessary corrections for crosswind are bigger and unpredictable.

Does the Coriolis correction have any influence on flight calculations?

The counterforce to Coriolis, produced by banking the airplane slightly, is borrowed from the lift force. The lift force of an airplane always points perpendicular to the surface of the wings upwards. When an airplane banks, this lift force vector banks to one side too. The lift force can then be decomposed into a vertical component and a horizontal component. The horizontal component counters the Coriolis effect or causes the airplane to curve. The vertical component has to stay constant to counter the waight of the airplane. This means that as an airplane banks to one side, the lift force has to increase. This is achieved by increasing the angle of attack. An increase of the angle of attack increases the drag. So the thrust has to be increased to overcome the increase in drag. More thrust means more fuel consumption.

So how much more fuel is needed due to Coriolis corrections? I did the calculation for an Airbus A320.

The thrust has to be increased only 0.00021 %. This has not to be taken into account in any calculations.

See below for the detailed calculations.

## Evidence for the Coriolis Correction of Airplanes

The bank angle of an airplane is displayed on the Attitude Indicator (AI). But the bank angle needed to correct for Coriolis is too small to be noticed on an AI. Additionally a mechanical AI aligns itself to the combined force of effective gravity and the vertical component of the Coriolis force. So if the airplane banks to correct for the Coriolis force, it banks exactly as much as the AI deviates from true level due to the Coriolis force. So on a mechanical AI we can never see the bank angle of the Coriolis correction.

An electronic AI however, that is fed by an Inertial Navigation System (INS), always is aligned perfectly with true level and not affected by the Coriolis force, because the INS corrects for Coriolis internally. So if the bank angle were big enough, it would show up in the AI.

Some airplanes, like the Bombardier Global Express, have a maintenace mode, in which the bank angle of the airplane can be displayed to 0.1° accuracy in digital format (see Roll indication in Imgage 2). This is just enough to measure the bank angle above a certain latitude in cruise speed.

The pilot Wolfie6020 made a video where we can see the bank angle displayed on a display. The angle varies due to corrections the airplane constantly makes to correct for fluctuations in the air stream, but the mean value shows a bias that is expected for the Coriolis correction. Wolfies Video showing the Roll Indicator ## Calculation of the Coriolis Force

The horizontal component of the Coriolis force depends on the mass and speed of the airplane, earth's rotation rate and the latitude. The Coriolis force acts always to the right in the northern hemisphere and to the left in the southern hemisphere. We can calculate the Coriolis acceleration instead of the Coriolis force. The connection is a = F / m. This way we get rid of the mass term m:

(1)
where'
 $a_\mathrm{coriolis}$ ' =' 'Coriolis acceleration. Positive values act to the right, negatives to the left. $v$ ' =' 'ground speed of the airplane = true air speed when assuming no wind $\omega$ ' =' '2π/(24·3600 s) = earth rotation rate $\varphi$ ' =' 'latitude of the airplane, 0 = equator

Note: as the airplane changes speed and latitude, the needed corrections constantly change and get smaller as the airplane approaches the equator, where the Coriolis effect is zero, due to the sine of the latitude in the equation.

This simplified equation is derived from the Equation of Motion in a rotating Reference Frame for an object flying over the surface of a rotating sphere and only taking the left/right components into account. Use my Coriolis Calculator for your own Coriolis calculations.

## Calculation of the Bank Angle

The airplane can maintain its heading and apply a small bank angle to create a counterforce to the Coriolis force. The necessary bank angle depends on the latitude φ, the ground speed v (= true air speed, assuming no wind), the earth rotation rate ω and the gravitational acceleration g:

(2)
where'
 $\beta$ ' =' 'bank angle $F_\mathrm{coriolis}$ ' =' 'Coriolis force $F_\mathrm{vertical}$ ' =' 'Flift · cos(β) = vertical component of the lift force $F_\mathrm{lift}$ ' =' 'm · g / cos(β) = lift force $m$ ' =' 'mass of airplane $g$ ' =' '9.806 m/s2 = gravitational acceleration $v$ ' =' 'ground speed (= true air speed, assuming no wind) $\omega$ ' =' '2π/(24·3600 s) = earth rotation rate $\varphi$ ' =' 'latitude in radian, 0 = equator

Positive bank angles mean banking to the right, negative to the left.

As the airplane flies along the longitude track due south, it's latitude changes and so does the required bank angle change over time.

Space: As viewed from space, the force produced by banking the airplane to the left is the only side force acting on the airplane. Any sideforce acting on a moving object in an inertial frame of reference like space creates a curved path. The calculated bank angle yields exactly the necessary left curve to follow the longitude track of the rotating earth.

Earth: In the non-inertial, rotating reference frame of the earth the bank angle produces no curve, because the produced side force to the left is exactly equal in magnitude but opposite in direction to the Coriolis force, so they cancel each other. The airplane flies a straight great circle route in the reference frame of the rotating earth.

The latitude dependent banking provides the side acceleration needed for the airplane, so that it maintains the same tangential speed as the ground and the atmosphere, at any latitude it reaches.

## Calculating the Thrust Increase

The Coriolis counterforce is borrowed from the lift force. To maintain altitude with a bank angle β, the lift force has to be increased by a factor of 1/cos(β). This is achieved by increasing the angle of attack (increasing pitch), which increases the induced drag force slightly. So we have to increase the thrust accordingly to maintain speed.

How much additional thrust is needed to compensate the Coriolis force?

The following calculations are for an Airbus A320 with the parameters as listed below.

The needed thrust increase to compensate for the Coriolis force is only ΔT = 0.0002 %.
So we don't have to account for Coriolis in fuel calculations!

Interesting: For a curve with a bank angle of 15° you have to increase the thrust only about 2.1 % to maintain lift.

Note: The horizontal lift component LiftHoriz. is used to counter the horizontal Coriolis force. It accelerates the airplane sideways exactly the amount necessary to keep up with the different tangential speeds of the surface of the earth with respect to space at different latitudes.

#### Equations

Here are the equations and parameters used:

(3)
where'
 $T(\beta)$ ' =' 'thrust as a function of the bank angle β $\rho(h)$ ' =' '0.409736 kg/m3 = air density at h = 33,000 ft $A_\mathrm{W}$ ' =' '122.6 m2 = wing area A320 $v$ ' =' '230 m/s = true air speed = ground speed (no wind) $c_\mathrm{d0}$ ' =' '0.02, $c_\mathrm{d1}$ = 0.04 = drag coefficients A320, measured in simulator MS FSX $m$ ' =' '62,000 kg = mass of the aircraft $g$ ' =' '9.81 m/s2 = gravitational acceleration

Here is the required thrust increase to maintain a bank angle of 0.15°:

(4)
(5)
(6)
where'
 $T_0$ ' =' 'Thrust at bank angle β = 0° $T_1$ ' =' 'Thrust at bank angle β = 0.15° (needed at latitude φ = 45°) $\Delta T$ ' =' 'required thrust increase in % to maintain lift at the bank angle of 0.15°

Equation (3) is derived from the following basic equations used in aviation.

Lift equation:

(7)
(8)
where'
 $F_\mathrm{L}$ ' =' 'lift force $\rho(h)$ ' =' 'air density at altitude h from barometric equations $c_\mathrm{L}(\alpha)$ ' =' 'lift coefficient depending on angle of attack α $A_\mathrm{W}$ ' =' 'wing area $v$ ' =' 'true air speed of airplane $m$ ' =' 'mass of airplane $g$ ' =' '9.81 m/s2 = gravitational acceleration $\beta$ ' =' 'bank angle

Calculating drag coefficient from lift coefficient:

(9)
where'
 $c_\mathrm{d}(\alpha)$ ' =' 'drag coefficient depending on angle of attack α $c_\mathrm{d0}$ ' =' 'static drag coefficient (e.g. 0.02 for A320) $c_\mathrm{d1}$ ' =' 'dynamic drag coefficient (e.g. 0.04 for A320) $c_\mathrm{L}(\alpha)$ ' =' 'lift coefficient depending on angle of attack α, see (8)

Thrust equation:

(10)
where'
 $T$ ' =' 'thrust force $F_\mathrm{D}$ ' =' 'drag force $\rho(h)$ ' =' 'air density at altitude h from barometric equations $c_\mathrm{d}(\alpha)$ ' =' 'drag coefficient depending on angle of attack α, see (9) $A_\mathrm{W}$ ' =' 'wing area $v$ ' =' 'true air speed of airplane

## Equation of Motion in a rotating Reference Frame

The simple equation for the vertical component of the Coriolis effect acting on an object moving on the surface of the rotating earth (13) is derived from the Coriolis term of the equation of motion (11) for a reference frame attached to the surface of a rotating sphere.

Equation of Motion in a rotating frame of reference: 

(11)
where'
 $\mathbf{a}$ ' =' 'total acceleration of mass m. a is a 3D vector $\sum \mathbf{F}$ ' =' 'vector sum of the physical forces acting on the object relative to the rotating reference frame $m$ ' =' 'mass of the object $\mathbf{\Omega}$ ' =' 'rotation vector (magnitude ω) of the rotating reference frame relative to an inertial reference frame $\mathbf{v}$ ' =' '3D velocity of mass m relative to the rotating reference frame $\mathbf{r}$ ' =' 'position vector of mass m relative to the rotating reference frame

The Coriolis Term in this equation is the following:

(12)
where'
 $\mathbf{a}_\mathrm{coriolis}$ ' =' 'Coriolis acceleration term. a is a 3D vector $\mathbf{\Omega}$ ' =' 'rotation vector (magnitude ω) of the rotating reference frame relative to an inertial reference frame $\mathbf{v}$ ' =' '3D velocity of mass m relative to the rotating reference frame

From this we can derive the horizontal component of the Coriolis acceleration $a_\mathrm{coriolis}$ for an object moving along the surface of a sphere:

(13)
where'
 $a_\mathrm{coriolis}$ ' =' 'horizontal Coriolis acceleration component, acting to the left (negative value) or right (positive value) with respect to the direction of motion $\omega$ ' =' '2π / (24·3600 s) = rotation rate of the earth $v$ ' =' 'speed of object relative to the surface of the rotating earth $\varphi$ ' =' 'latitude of the object

There is also a vertical component of the Coriolis acceleration, which is ignored for our calculations, because it does not push the airplane to the side.

## Calculation of Crosswind Forces

Now I want to calculate the amount of crosswind that creates the same force as the Coriolis effect. We can calculate the force of crosswind acting perpendicular from the side on an aircraft as follows:

(14)
where'
 $F_\mathrm{cw}$ ' =' 'crosswind force $\rho(h)$ ' =' 'air density at altitude h (sea level ρ0 = 1.225 kg/m3) $c_\mathrm{d}$ ' =' 'drag coefficient of airplane side profile $A_\mathrm{d}$ ' =' 'side cross section area of airplane $m$ ' =' 'mass of airplane $w$ ' =' 'crosswind speed from one side perpendicular to the flight direction of the airplane

The Coriolis force acting from one side is:

(15)

To calculate how much crosswind produces the same side force as the Coriolis effect, we can set the two forces Fcw and Fcoriolis equal and solve the resulting equation for the crosswind speed w. So for an A320 at cruising altitude and 45° latitude the Coriolis force is equivalent to a crosswind speed of:

(16)
where'
 $w$ ' =' 'crosswind speed corresponding to the Coriolis force at the following parameters for an A320: $v_\mathrm{gs}$ ' =' '230 m/s = 447 kt = Ma 0.77 = ground speed of airplane $\omega$ ' =' '2π/(24·3600 s) = earth rotation rate $\varphi$ ' =' '45° = latitude $m$ ' =' '62,000 kg = mass of the airplane $\rho(h)$ ' =' '0.41 kg/m3 = air density at cruising altitude h = FL 330 = 10,058 m $c_\mathrm{d}$ ' =' '1.0 = drag coefficient of airplane side profile (estimated 0.5...1.5) $A_\mathrm{d}$ ' =' '170 m2 = side cross section area of an A320

Note: Airplanes correct for crosswind not by banking the airplane, but by changing the heading slightly into the wind.

## Implementation

The accelerations listed in the animation are not calculated using the Equation of Motion in a rotating Reference Frame. But rather they are derived from the calculated track itself. The tracks emerged simply from transforming an orbit onto a rotating sphere. I checked the derived accelerations from geometry with the calculation using the Equation of Motion in a rotating Reference Frame and came to the same results.

Calculating an acceleration from geometry involves calculating 3 nearby points P0, P1 and P2 with a distance of Δd = v · Δt for a very small time difference Δt. Then a good approximation for the acceleration is:

 (17)

## References

Coriolis force ; Wikipedia
https://en.wikipedia.org/wiki/Coriolis%5Fforce
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