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Guidance, Navigation and Control · Theory · M2

M2-T1 — Frames for Flight: ECI, ECEF, Geodetic, NED, Body and Wind

A frame is a choice of axes and an origin. Choosing the simplest one that does not lie to you is half of modelling a vehicle.

  • Lesson 1 of 5
  • Intermediate
  • Module 2: Modelling a Vehicle for GNC
  • Theory
  • About 18 minutes to read

Hopper hops 100 m, a transport aircraft flies 10 000 km, and a satellite goes round the Earth every 90 minutes. They cannot all use the same frame: a flat map is exact for the first, hopeless for the second, and a rotating Earth is a nuisance for the third. This lesson sets out the frames a GNC engineer needs, shows how to convert between them, and puts numbers on the question that decides which one to use: how big is the error if the Earth is treated as flat and still?

Learning objectives

After this lesson, you should be able to:

  • Name the frames used in flight (ECI, ECEF, geodetic, NED, body and wind axes) and say what each is used for.
  • Convert a position between geodetic, ECEF and local NED coordinates.
  • State how the velocity and acceleration of a point differ between an inertial and a rotating frame.
  • Estimate the error of a flat, non-rotating Earth for a given distance and speed, and decide whether it can be neglected.

The intuition

If you have read UAV M1-T4: UAV coordinate frames, you already know the body frame, the local NED and ENU frames and why a mismatch between them ruins a flight. That page treats the Earth as flat and still, which is right for a multirotor over a field. This lesson starts where that assumption stops: it adds the frames that an aircraft, a spacecraft or a long-range vehicle must use, and the wind axes that aerodynamics needs.

ECEF axes and the local north-east-down frame at a point of latitude phiA circle representing a meridian cross-section of the Earth. The ECEF z axis points up through the north pole and the x axis points right through the equator at the prime meridian. A point P on the surface at latitude phi has a local frame: north is tangent to the surface towards the pole, down points to the centre of the Earth, and east points out of the page. The angle between the equator and the line to P is the latitude. z (north pole) x (equator, prime meridian) Earth centre φ P north down east (out of the page) ECI: does not rotatewhere Newton's laws hold ECEF: turns with the Earthat 7.29e-5 rad/s; GNSS positions NED: origin at a chosen point Pwhere Hopper lives Body: attached to the vehiclesensors and actuators live here
A meridian cross-section of the Earth. A point P at latitude φ has its own north-east-down frame, whose axes are fixed by the geometry: north along the surface towards the pole, down towards the centre, east out of the page.
The frames of flight
FrameOriginAxesTurns?Used for
ECI (Earth-centred inertial)Centre of the EarthFixed relative to the distant stars; z through the north poleNoOrbits, star trackers, any use of Newton's laws without extra terms
ECEF (Earth-centred, Earth-fixed)Centre of the Earthz through the north pole, x through the equator at the prime meridianYes, with the EarthGNSS positions, long-range navigation
GeodeticNone (angles and a height)Latitude, longitude, height above the reference ellipsoidYes, with the EarthMaps, charts, people
Local NED (north, east, down)A chosen point on or near the groundTangent to the ellipsoid at the originYes, with the EarthShort-range navigation, guidance, the whole of Hopper's range
BodyThe vehicle's centre of massFixed to the vehicle, e.g. forward, right, downYes, with the vehicleSensors, actuators, equations of rotation
Wind (air-path)The vehicle's centre of massFirst axis along the air-relative velocityYes, with the airflowLift, drag and sideforce

Which frame for which job

No frame is the right one in general. Each is the simplest one for some job, and a GNC system moves quantities between them many times per second.

  • Sensors measure in the body frame. An accelerometer or a gyroscope reports components along its own axes, and a thruster pushes along the vehicle's own axis. Every sensor value and every actuator command starts or ends in the body frame.
  • Navigation states live in a local or an Earth frame. Over a few kilometres a local NED frame is exact enough. Over continents the state is kept in ECEF or in geodetic coordinates, and a local frame is attached wherever it is needed.
  • Guidance works in the frame of its target. A landing pad is easiest in the local NED frame of the pad; a rendezvous is easiest in a frame attached to the target spacecraft (module 11).
  • Aerodynamics works in the wind frame. Lift is perpendicular to the airflow and drag is along it. The angles between the wind and body frames, the angle of attack α\alpha and the sideslip β\beta, are what an aircraft's aerodynamic coefficients depend on.
  • Orbits and star sensors use an inertial frame. A satellite's motion is simple in ECI and complicated in a frame that turns once a day.

When the Earth may be flat and still

Two approximations are involved, and they are independent. Treating the Earth as flat replaces the curved surface with its tangent plane; the error grows with the square of distance. Treating it as not rotating drops the extra terms that appear in a rotating frame; the error grows with speed and with the square of time. Hopper, 100 m and 20 s long, can neglect the first and has to look at the second. The worked example below puts numbers on both, and the same arithmetic tells you for any other vehicle which frame it needs.

Choose the simplest frame whose error is below the budget of the vehicle. The error of a flat, still Earth is two short formulas, and a navigation engineer evaluates them before choosing.

Definitions

Geodetic coordinates (ϕ,λ,h)(\phi, \lambda, h)
Latitude ϕ\phi and longitude λ\lambda in radians, and height hh in metres above the reference ellipsoid. Latitude is the angle between the equatorial plane and the local vertical (the normal to the ellipsoid), not the line to the centre.
Reference ellipsoid (WGS 84)
Equatorial radius a=6 378 137 ma = 6\,378\,137\ \text{m}, flattening f=1/298.257223563f = 1/298.257223563, and first eccentricity squared e2=f(2−f)e^2 = f(2 - f). This is the ellipsoid GNSS uses.
Earth rate Ω\Omega
7.2921150 × 10⁻⁵ rad/s: one turn per sidereal day of 86 164 s, relative to the stars.
Attitude angles (φ,θ,ψ)(\varphi, \theta, \psi)
Roll, pitch and yaw of the body relative to the local NED frame, in the aerospace order: yaw about down first, then pitch, then roll.
Angle of attack and sideslip (α,β)(\alpha, \beta)
The direction of the air-relative velocity in the body frame, with body-frame components (u,v,w)(u, v, w): α=atan2⁡(w,u)\alpha = \operatorname{atan2}(w, u) and β=arcsin⁡(v/V)\beta = \arcsin(v/V) for airspeed VV.

Geodetic coordinates become an ECEF position through the radius of curvature in the prime vertical NN:

Geodetic to ECEF
pECEF=((N+h)cos⁡ϕcos⁡λ(N+h)cos⁡ϕsin⁡λ(N(1−e2)+h)sin⁡ϕ),N=a1−e2sin⁡2ϕ\mathbf{p}_{\text{ECEF}} = \begin{pmatrix} (N + h)\cos\phi\cos\lambda \\ (N + h)\cos\phi\sin\lambda \\ \bigl(N(1 - e^2) + h\bigr)\sin\phi \end{pmatrix}, \qquad N = \frac{a}{\sqrt{1 - e^2 \sin^2\phi}}
The reverse conversion has no closed form that is simple, and is done by a short iteration (a few steps reach millimetres). The library function ecef_to_geodetic does it.

A vector given in ECEF components is turned into north, east and down components at latitude ϕ\phi and longitude λ\lambda by the matrix whose rows are the three unit vectors:

ECEF to local NED
RNED←ECEF=(−sin⁡ϕcos⁡λ−sin⁡ϕsin⁡λcos⁡ϕ−sin⁡λcos⁡λ0−cos⁡ϕcos⁡λ−cos⁡ϕsin⁡λ−sin⁡ϕ)\mathbf{R}_{\text{NED}\leftarrow\text{ECEF}} = \begin{pmatrix} -\sin\phi\cos\lambda & -\sin\phi\sin\lambda & \cos\phi \\ -\sin\lambda & \cos\lambda & 0 \\ -\cos\phi\cos\lambda & -\cos\phi\sin\lambda & -\sin\phi \end{pmatrix}
Row one is north, row two east, row three down, each written in ECEF components. A position offset p−p0\mathbf{p} - \mathbf{p}_0 from the origin is turned into NED the same way.

Finally, a frame that turns changes what an observer measures. For a point with position r\mathbf{r} and velocity vE\mathbf{v}_E measured in the turning ECEF frame, and the Earth's rotation vector Ω\boldsymbol{\Omega}:

Velocity and acceleration in an inertial and in a rotating frame
vI=vE+Ω×raI=aE+2 Ω×vE+Ω×(Ω×r)\mathbf{v}_I = \mathbf{v}_E + \boldsymbol{\Omega}\times\mathbf{r} \qquad \mathbf{a}_I = \mathbf{a}_E + 2\,\boldsymbol{\Omega}\times\mathbf{v}_E + \boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r})
Subscript II is the inertial (ECI) value and EE the value measured in ECEF. The term 2Ω×vE2\boldsymbol{\Omega}\times\mathbf{v}_E is the Coriolis term, the last is the centripetal term. The next lesson derives this rule for any rotating frame.

Derivation: the NED matrix, the curvature of the ground and the Coriolis drift

Three results of this lesson can be derived in a few lines each. Together they give the numbers used in the example.

The rotation from ECEF to NED

Start with the ECEF axes and turn them twice.

  1. Rz(λ)=(cos⁡λsin⁡λ0−sin⁡λcos⁡λ0001)\mathbf{R}_z(\lambda) = \begin{pmatrix} \cos\lambda & \sin\lambda & 0 \\ -\sin\lambda & \cos\lambda & 0 \\ 0 & 0 & 1 \end{pmatrix}

    Turn the axes about the polar axis by the longitude. The first axis now lies in the meridian plane of the point, pointing outwards along the equator, and the third axis still points at the north pole.

  2. Ry(−π2−ϕ)=(−sin⁡ϕ0cos⁡ϕ010−cos⁡ϕ0−sin⁡ϕ)\mathbf{R}_y(-\tfrac{\pi}{2} - \phi) = \begin{pmatrix} -\sin\phi & 0 & \cos\phi \\ 0 & 1 & 0 \\ -\cos\phi & 0 & -\sin\phi \end{pmatrix}

    Turn the axes about the second axis (east) by −(90∘+ϕ)-(90^\circ + \phi). In the meridian plane, north at latitude ϕ\phi is the direction (−sin⁡ϕ,0,cos⁡ϕ)(-\sin\phi, 0, \cos\phi) and down is (−cos⁡ϕ,0,−sin⁡ϕ)(-\cos\phi, 0, -\sin\phi). These are the first and third rows.

  3. RNED←ECEF=Ry ⁣(−π2−ϕ)Rz(λ)\mathbf{R}_{\text{NED}\leftarrow\text{ECEF}} = \mathbf{R}_y\!\left(-\tfrac{\pi}{2} - \phi\right)\mathbf{R}_z(\lambda)

    Apply the longitude turn first and the latitude turn second. Multiplying the two matrices gives the matrix of the definitions above. The program in the example checks this product against the library matrix to machine precision.

How far the tangent plane leaves the ground

  1. (RE+h)2=RE2+d2(R_E + h)^2 = R_E^2 + d^2

    A straight line leaves the surface of a sphere of radius RER_E along a tangent. At a distance dd along the tangent from the point of contact, the line is at distance RE+hR_E + h from the centre, by Pythagoras.

  2. h=RE2+d2−RE≈d22RE(d≪RE)h = \sqrt{R_E^2 + d^2} - R_E \approx \frac{d^2}{2 R_E} \qquad (d \ll R_E)

    Solve for the height. For distances much smaller than the radius, the first term of the series of the square root is enough: the height grows as the square of the distance.

With RE=6.37×106R_E = 6.37 \times 10^6 m this gives 0.8 mm at 100 m, 7.9 m at 10 km and 785 m at 100 km. A flat local frame is exact for Hopper, adequate for a multirotor, and wrong for an aircraft: after a hundred kilometres a navigation that ignores it puts the vehicle most of a kilometre off in height.

The sideways drift from the Earth's rotation

  1. aE=aI−2 Ω×vE−Ω×(Ω×r)\mathbf{a}_E = \mathbf{a}_I - 2\,\boldsymbol{\Omega}\times\mathbf{v}_E - \boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r})

    Solve the acceleration rule of the definitions for the acceleration seen on the turning Earth. A body with no force on it has aI=0\mathbf{a}_I = 0 (apart from gravity, which the centripetal term is usually folded into).

  2. ∣2 Ω×vE∣=2 Ω v sin⁡ϕ(horizontal motion)\lvert 2\,\boldsymbol{\Omega}\times\mathbf{v}_E \rvert = 2\,\Omega\,v\,\sin\phi \quad (\text{horizontal motion})

    For horizontal motion at latitude ϕ\phi only the vertical component of the Earth rate, Ωsin⁡ϕ\Omega\sin\phi, turns the velocity sideways.

  3. Δ=12 (2 Ω vsin⁡ϕ) t2=Ω v t2sin⁡ϕ\Delta = \tfrac12 \,(2\,\Omega\,v\sin\phi)\,t^2 = \Omega\,v\,t^2 \sin\phi

    A constant sideways acceleration for a time tt displaces the body by half the acceleration times the time squared. This is the size of the error from ignoring the Earth's rotation.

Both errors grow with the square of something. Curvature grows with the square of distance, rotation with the square of time. A short, slow vehicle can ignore both; a long, fast one can ignore neither.

A worked example

Hopper's test range is at latitude 35° north, longitude 117° west, 700 m above the ellipsoid. Three range beacons are surveyed as offsets from the range origin. Convert them to ECEF as a GNSS receiver would report them, check the round trip, and then decide whether the flat, non-rotating approximation is good enough for Hopper and for two other vehicles.

  1. Convert the beacons. Each beacon is a north, east, down offset from the origin. The library turns it into an ECEF position (the offset rotated into ECEF and added to the origin's ECEF position) and back; the round trip returns the offset to within 10⁻⁹ m.
  2. Check the rotation. Compose Ry(−90∘−ϕ) Rz(λ)\mathbf{R}_y(-90^\circ - \phi)\,\mathbf{R}_z(\lambda) from the two elementary rotations and compare with the matrix the library uses. They agree to 10⁻¹⁶.
  3. Curvature. A point 100 m from the origin along the tangent plane is 0.8 mm above the ellipsoid. At 10 km the figure is 7.9 m and at 100 km it is 786 m, matching the d2/2REd^2/2R_E formula.
  4. Rotation. For Hopper at 10 m/s for 20 s the drift Ωvt2sin⁡ϕ\Omega v t^2 \sin\phi is 0.17 m. That is not nothing: it is more than a quarter of the 0.60 m navigation budget of M1-T4: From Mission Requirement to GNC Budget. A light aircraft at 60 m/s for an hour, and an airliner at 250 m/s for an hour, drift by 33 km and 136 km if the rotation is ignored.
  5. Decide. For Hopper, the local NED frame is exact in curvature (0.8 mm) and the Earth rotation term needs to be included in the navigation filter, or else budgeted at 0.17 m. For the aircraft, both effects are far too large: it navigates in ECEF or geodetic coordinates and includes the rotation. At the equator the ground itself moves at Ωa\Omega a = 465 m/s relative to the stars, which is why launches go east.
import math
from gnckit import (geodetic_to_ecef, ecef_to_geodetic, ned_to_ecef, ecef_to_ned, ned_from_ecef,
                    rot_y, rot_z, mat_mul, OMEGA_EARTH, WGS84_A)

lat0, lon0, h0 = math.radians(35.0), math.radians(-117.0), 700.0    # origin of the test range
beacons = {'B1': (60.0, -40.0, 4.0), 'B2': (-55.0, 30.0, 2.5), 'B3': (10.0, 85.0, 6.0)}   # north, east, down (m)
for name, ned in beacons.items():
    ecef = ned_to_ecef(ned, lat0, lon0, h0)
    back = ecef_to_ned(ecef, lat0, lon0, h0)
    err = max(abs(a - b) for a, b in zip(ned, back))
    print(f'{name}: ECEF ({ecef[0]:.2f}, {ecef[1]:.2f}, {ecef[2]:.2f}) m, round trip error {err:.0e} m')

R = mat_mul(rot_y(-math.pi / 2 - lat0), rot_z(lon0))                 # the two rotations of the derivation
diff = max(abs(R[i][j] - ned_from_ecef(lat0, lon0)[i][j]) for i in range(3) for j in range(3))
print(f'R_y(-90 deg - lat) R_z(lon) differs from the NED matrix by {diff:.0e}')

print('a straight line along the local tangent plane leaves the ground by:')
for d in (100.0, 10e3, 100e3):
    p = ned_to_ecef((d, 0.0, 0.0), lat0, lon0, h0)
    h = ecef_to_geodetic(*p)[2] - h0
    print(f'  {d / 1000:7.1f} km out: {h:10.4f} m above the ellipsoid')

print('sideways drift of a free body from Earth rotation, Omega v t^2 sin(lat):')
for name, v, t in (('Hopper', 10.0, 20.0), ('light aircraft', 60.0, 3600.0), ('airliner', 250.0, 3600.0)):
    print(f'  {name:15s} v = {v:5.0f} m/s, t = {t:6.0f} s: {OMEGA_EARTH * v * t * t * math.sin(lat0):10.2f} m')
print(f'inertial speed of the ground at the equator: {OMEGA_EARTH * WGS84_A:.1f} m/s')
B1: ECEF (-2374842.94, -4660803.61, 3638315.27) m, round trip error 3e-10 m
B2: ECEF (-2374811.08, -4660895.25, 3638221.93) m, round trip error 4e-10 m
B3: ECEF (-2374743.85, -4660884.45, 3638273.16) m, round trip error 3e-10 m
R_y(-90 deg - lat) R_z(lon) differs from the NED matrix by 1e-16
a straight line along the local tangent plane leaves the ground by:
      0.1 km out:     0.0008 m above the ellipsoid
     10.0 km out:     7.8651 m above the ellipsoid
    100.0 km out:   786.4311 m above the ellipsoid
sideways drift of a free body from Earth rotation, Omega v t^2 sin(lat):
  Hopper          v =    10 m/s, t =     20 s:       0.17 m
  light aircraft  v =    60 m/s, t =   3600 s:   32523.78 m
  airliner        v =   250 m/s, t =   3600 s:  135515.76 m
inertial speed of the ground at the equator: 465.1 m/s
The drift row for the aircraft is the size of the effect if the rotation were ignored, not a flight path: a real vehicle is steered and corrects for it.

In robotics

The choices in this lesson show up as concrete decisions in every vehicle.

  • Hopper. The beacons are surveyed once in geodetic or ECEF coordinates (a GNSS receiver on a fixed mast gives them in ECEF) and converted to the local NED frame of the range origin. The IMU's body-frame measurements are rotated into NED using the attitude estimate. Guidance and control then see only a flat, local world, and the Earth's rotation is a small term in the navigation filter.
  • Aircraft. The navigation state is held in geodetic coordinates or in ECEF, and the equations include the transport rate (the rate at which the local NED frame turns as the vehicle flies over the curved Earth) and the Earth rate. Module 6 derives them. Aerodynamics uses wind axes: α\alpha and β\beta come from air data and the attitude (see UAV M5-T1: UAV equations of motion for the wind triangle).
  • Spacecraft. Orbits are propagated in ECI, where gravity is the only force. A star tracker reports attitude relative to ECI, and a ground-station pass needs the ECI to ECEF rotation Rz(Ωt)\mathbf{R}_z(\Omega t) to know where the ground is.
  • Ground robots. A mapping robot has no use for any of this: a map frame and an odometry frame are enough (see SLAM M2-T3: Rotation Matrices, Quaternions, and Pose Representation). The Earth frames only matter when a GNSS receiver is the source of the position.

The typical bugs are the same everywhere and are worth knowing by name: mixing degrees and radians; using geocentric latitude (the angle at the centre) where geodetic latitude is meant, which is out by up to 0.19° (about 21 km) near 45°; swapping NED and ENU, which negates the vertical; and applying a rotation in the wrong direction. The Linear Algebra track's Three-Dimensional Rotation Matrices and Transforming Points Between Frames cover the mechanics of the transformations, and the habit of naming every rotation by what it turns into what (as in RNED←ECEF\mathbf{R}_{\text{NED}\leftarrow\text{ECEF}}) prevents most of the bugs.

Common misconceptions

“Latitude is the angle between the equator and the line from the centre of the Earth.”

That is geocentric latitude. Geodetic latitude, the one on maps and in GNSS, is measured from the local vertical, which on an ellipsoid does not point at the centre. The two differ by up to 0.19° near 45° latitude, which is about 21 km on the ground. Always say which one a variable holds.

“A flat local map is fine for any drone.”

It is exact for 100 m and good for a field, but the tangent plane leaves the ground by 8 m at 10 km and 786 m at 100 km. A long-range drone that plans in a flat frame will be metres off in height at the far end, and kilometres off in a continental flight. Use the formula before deciding.

“The Earth's rotation only matters in space.”

It matters on the ground too. The sideways drift of an unsteered body is Ωvt2sin⁡ϕ\Omega v t^2 \sin\phi: 0.17 m for Hopper in a 20 s hop at 35° latitude, a quarter of its navigation budget. It is also what a gyroscope at rest measures: 15 degrees per hour, which sets the limit of gyrocompassing in module 6.

“ECI and ECEF are the same frame up to a rotation, so either will do.”

They differ by a rotation that changes with time, and so the velocities differ by Ω×r\boldsymbol{\Omega}\times\mathbf{r}, up to 465 m/s at the equator. Newton's law has its simple form only in ECI, and GNSS gives positions in ECEF, so a system that uses both must convert with the right time.

“Newton's laws hold in the body frame because the vehicle is where the forces act.”

They hold only in an inertial frame. In the body frame, which turns, extra terms appear (the next lesson derives them), and in ECEF the Coriolis and centripetal terms appear. The frame that makes the equations simplest is not always the frame in which the sensors measure.

Key takeaways

  1. The frames of flight are ECI, ECEF, geodetic, local NED, body and wind axes. Each is the simplest one for some job.
  2. Sensors and actuators work in the body frame; navigation states in a local or Earth frame; aerodynamics in the wind frame; orbits in ECI.
  3. The NED rotation is Ry(−90∘−ϕ) Rz(λ)\mathbf{R}_y(-90^\circ - \phi)\,\mathbf{R}_z(\lambda): a rotation by the longitude followed by one by the latitude.
  4. A frame that turns adds the Coriolis and centripetal terms. For a body moving at speed vv, the Coriolis drift over tt is Ωvt2sin⁡ϕ\Omega v t^2 \sin\phi.
  5. A tangent plane leaves the ground by d2/2REd^2/2R_E: 0.8 mm at 100 m, 7.9 m at 10 km, 786 m at 100 km.
  6. Name every rotation by what it turns into what, and say whether an angle is geodetic or geocentric, in degrees or radians. Most frame bugs are one of these.

References and further reading

Each source is marked free or paid. Free to read does not mean free to reuse: the course resources page lists every source with its licence.

  • P. D. Groves, Principles of GNSS, Inertial, and Multisensor Integrated Navigation Systems, 2nd edition, Artech House, 2013, the chapter on coordinate frames, kinematics and the Earth. The reference for this lesson. Paid book.
  • ESA Navipedia, the pages on reference frames and coordinate systems . A short, careful treatment from the European Space Agency. Free to read.
  • MIT 16.07 Dynamics (OpenCourseWare) , the lectures on rotating frames. Free; licence CC BY-NC-SA 4.0.
  • B. L. Stevens, F. L. Lewis and E. N. Johnson, Aircraft Control and Simulation, 3rd edition, Wiley, 2016, the chapter on kinematics and dynamics, for body and wind axes. Paid book.

Knowledge check

Three questions. Write an answer, then reveal the explanation. Your answers stay in your browser.

Question 1

A vehicle is steered along a straight line in a local NED frame that was set up at its starting point. How high above the ground, in a flat-earth approximation, is the end of a 50 km flight, and what does that mean for a long-range navigation system that works in such a frame?

Explanation

The tangent plane leaves the ellipsoid by d²/(2R_E) = (50 000)² / (2 × 6.37 × 10⁶) = 196 m. A navigation system that keeps position in a single flat frame would be nearly 200 m off in height at the end of the flight (and more in the horizontal if the path is curved), so a long-range vehicle must either work in ECEF or geodetic coordinates, or re-attach the local frame as it moves. For 100 m, the same formula gives 0.8 mm and the flat frame is exact.

Question 2

A stationary GNSS antenna sits at latitude 35°. About how fast does it move relative to the distant stars, and in which frame would this speed be zero?

Explanation

About 380 m/s. The ground at the equator moves at Ωa = 465 m/s in ECI, and at latitude 35° the radius of the circle the point travels on is smaller by cos 35° = 0.82, so the speed is about 465 × 0.82 = 381 m/s. It is zero in ECEF, the frame that turns with the Earth. This is the difference between the two velocities, vI = vE + Ω × r, which a navigation system that mixes the two frames gets wrong by hundreds of metres per second.

Question 3

A GNSS receiver reports the velocity of a vehicle in NED as (3.0, 0.0, −2.0) m/s. Is it going up or down? A colleague who works in ENU says the same vehicle has velocity (0.0, 3.0, +2.0). Are the two consistent?

Explanation

It is climbing at 2 m/s: the third component of NED is down, so −2.0 means 2 m/s upwards, and it is moving north at 3 m/s. The colleague is consistent: in ENU the components are east, north, up, giving (0.0, 3.0, +2.0), the same velocity with the first two components swapped and the third negated. The two conventions differ by exactly that swap and sign, and a signal that is read in the wrong convention shows a vehicle moving east and descending: a very common bug.

What’s next

Next lesson

M2-T2 — Rigid-Body Equations of Motion for Any Vehicle

Derive the translational and rotational equations in body axes with the transport theorem, and reduce them to the planar model of Hopper.

Nothing below is required to finish this lesson.

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