Strapdown Inertial Navigation Examples: Earth Rotation and Navigation Frames

This companion to INS06 turns the frame geometry into calculations that can be checked by hand and then reused as software unit tests. The emphasis is on transformations among ECI, ECEF, and local NED coordinates, Earth rate, transport rate, geodetic position kinematics, and the frame-rate subtraction required by strapdown attitude propagation.

The notation follows the main INS series. A passive direction cosine matrix C_a^b transforms components from frame a into frame b,

b b a v = C av ,

(1)

and ω_ab^c denotes the angular velocity of frame b relative to frame a, resolved in frame c.

The WGS-84 constants used throughout are

1 2 a = 6378137 m, f = --------------, e = f(2 − f), 298.257223563

(2)

−5 ΩE = 7.292115 × 10 rad∕s.

(3)

The ellipsoidal radii of curvature are

a RN = ∘-----------2--, 1 − e2sin ϕ

(4)

a(1 − e2) RM = -----2---2---3∕2. (1 − e sin ϕ)

(5)

For the local NED frame,

|------⌊----------------------------------⌋--| | − sinϕ cos λ − sinϕ sin λ cos ϕ | |Cne = ⌈ − sin λ cos λ 0 ⌉ .| | − cos ϕcos λ − cosϕ sinλ − sin ϕ | ----------------------------------------------

(6)

Figure. ECI and ECEF share the same origin but not the same orientation. A passive coordinate transformation changes the components used to describe a physical vector while leaving the physical vector unchanged.

1 Exercises

Exercise 1: An inertially fixed position observed in ECEF

At t = 0, let ECI and ECEF axes coincide. An inertially fixed point has

[ ] i 7000 0 r = 0 km.

(7)

After t = 1800 s, use

⌊ cos 𝜃 sin 𝜃 0⌋ e ⌈ ⌉ C i = − sin 𝜃 cos𝜃 0 , 𝜃 = ΩEt, 0 0 1

(8)

to:

compute the Earth rotation angle 𝜃;
compute r^e = C_i^erⁱ;
show that an inertially fixed vector has the ECEF coordinate derivative
$dre- e e dt = − ωie × r .$

Exercise 2: Transforming an ECEF velocity into NED

∘ ∘ ϕ = 40 , λ = − 105 ,

(9)

a velocity is represented in ECEF coordinates as

⌊ ⌋ 120 ve = ⌈− 80⌉ m ∕s. 30

(10)

Construct C_eⁿ.
Compute vⁿ = C_eⁿv^e.
Verify that the transformation preserves vector magnitude.

Figure. The local NED axes are a basis attached to the geodetic position. ECEF and NED coordinates describe the same physical vector using different basis vectors.

Exercise 3: Earth rate in NED and in a yawed body frame

A level vehicle is stationary at geodetic latitude

ϕ = 35∘.

(11)

Its heading is ψ = 60^∘, measured clockwise from North. For a level vehicle use

⌊ ⌋ cosψ − sin ψ 0 Cn = ⌈ sin ψ cos ψ 0⌉ . b 0 0 1

(12)

Compute ω_ieⁿ.
Express the components in degrees per hour.
Compute the same Earth-rate vector in body coordinates,
$ωb = Cb ωn . ie n ie$
Explain why the physical rate has not changed even though all three components changed.

Exercise 4: Transport rate for a moving vehicle

ϕ = 45 ∘, h = 2000 m,

(13)

a vehicle has

vN = 150 m ∕s, vE = 250 m ∕s.

(14)

Compute R_M, R_N, and

⌊ ⌋ --vE---- | RN + h | ωn = ||− --vN---|| . en |⌈ RM + h|⌉ vE-tan-ϕ − RN + h

(15)

Express the result in both rad/s and degrees per hour, and compare its magnitude with the Earth rotation rate.

Exercise 5: Geodetic position rates

Continue Exercise 4 and let

vD = − 3 m ∕s.

(16)

Compute

dϕ, dλ-, dh-, dt dt dt

(17)

and express latitude and longitude rates in degrees per hour.

Figure. North, east, and down velocity components drive latitude, longitude, and height rates through the local ellipsoid geometry.

Exercise 6: From inertial gyro rate to body rate relative to NED

Continue Exercises 4 and 5. The vehicle remains level with heading

ψ = 30∘.

(18)

Its body executes an additional yaw rotation relative to NED of

2∘∕s

(19)

about body down. Assume no roll or pitch rate relative to NED.

Compute
$n n n ω in = ωie + ωen.$
Transform the reference-frame rate into body coordinates.
Construct the ideal gyro measurement ω_ib^b.
Recover ω_nb^b using the strapdown reference-rate subtraction.

Figure. Earth rate and transport rate combine to form the inertial angular rate of the navigation frame. That reference rate must be transformed to body axes before it is subtracted from the inertial gyro measurement.

Exercise 7: DCM chain and round-trip consistency

ϕ = 30 ∘, λ = 20 ∘,

(20)

a level vehicle has heading ψ = 45^∘.

Construct C_eⁿ and C_bⁿ.
Form
$Ceb = CenCnb .$
Transform the body-frame forward velocity
$b [ ]T v = 100 0 0 m ∕s$

into ECEF.
Transform the ECEF result back to NED and verify that it matches the direct body-to-NED transformation.
Check det C_b^e and (C_b^e)^TC_b^e.

Exercise 8: Derivative consistency between ECEF and NED

ϕ = 45∘, λ = 30∘, h = 2000 m,

(21)

use

⌊ 150⌋ n ⌊ 0.5 ⌋ n ⌈ ⌉ dv-- ⌈ ⌉ 2 v = 250 m ∕s, dt = − 0.2 m ∕s . − 3 0.1

(22)

The derivative relation is

|----------------------------| |dvn- n n ndve- | | dt = − ωen × v + C e dt .| ------------------------------

(23)

Compute ω_enⁿ × vⁿ.
Solve for dv^e∕dt.
Substitute the result back into the NED derivative equation and verify the original dvⁿ∕dt.
Explain why simply applying C_eⁿ to dv^e∕dt would be incorrect.

Exercise 9: Stationary Earth-fixed attitude consistency

A level vehicle is fixed to Earth at

∘ ϕ = 60

(24)

with heading ψ = 90^∘. Since the vehicle is stationary relative to ECEF,

ωnen = 0.

(25)

Compute ω_ieⁿ.
Compute the ideal gyro measurement
$ωb = Cb ωn . ib n ie$
Show that
$b b b n ω nb = ωib − C nω ie = 0.$
Verify that the two terms in
$n dC-b-= Cn [ωb ]× − [ωn ]×Cn dt b ib ie b$

are equal, giving dC_bⁿ∕dt = 0.

Figure. A stationary Earth-fixed gyro measures Earth rotation. Correct subtraction of the navigation-frame inertial rate produces zero body rate relative to local NED and therefore constant local attitude.

Exercise 10: Why local NED becomes ill-conditioned near the poles

∘ ϕ = 89.9 , h = 0,

(26)

a vehicle travels due east at

vE = 100 m ∕s, vN = 0.

(27)

Compute R_N.
Compute dλ∕dt and express it in degrees per hour.
Compute ω_enⁿ.
Identify the component that becomes large as ϕ → 90^∘.
Explain why this is a coordinate-frame problem rather than a physical singularity in the vehicle motion.

2 Worked solutions

Solution 1: An inertially fixed position observed in ECEF

The Earth rotation angle after 1800 s is

𝜃 = ΩEt = (7.292115 × 10− 5)(1800 ) = 0.13125807 rad.

(28)

Therefore

|--------------| |𝜃 ≈ 7.52053 ∘. | ---------------

(29)

The ECEF components are

⌊ ⌋ ⌊ ⌋ cos𝜃 sin 𝜃 0 7000000 re = ⌈− sin𝜃 cos 𝜃 0⌉ ⌈ 0 ⌉ , 0 0 1 0

(30)

|----⌊----------------6-⌋-----| | 6.9397861 × 10 | re ≈ ⌈ − 9.1617046 × 105⌉ m. | | 0 | -------------------------------

(31)

The point is fixed in ECI, but ECEF rotates underneath it. Hence its ECEF coordinate derivative is

⌊ ⌋ dre e e e 0 ----= − ωie × r , ω ie = ⌈ 0 ⌉ . dt ΩE

(32)

The cross product gives

|--------------------------| | ⌊ ⌋ | |dre − 66.8082 | |-dt-≈ ⌈ − 506.0572⌉ m ∕s.| | 0 | ---------------------------

(33)

This result is a useful reminder that a coordinate derivative can be nonzero even when the physical vector is fixed in inertial space.

Solution 2: Transforming an ECEF velocity into NED

For ϕ = 40^∘ and λ = −105^∘,

⌊ 0.1663657 0.6208852 0.7660444 ⌋ n ⌈ ⌉ Ce ≈ 0.9659258 − 0.2588190 0 . 0.1982669 0.7399421 − 0.6427876

(34)

Then

vn = Cneve,

(35)

which gives

|-----⌊-----------⌋------| | − 6.72560 | vn ≈ ⌈ 136.61662 ⌉ m ∕s.| | | -------−-54.68697---------

(36)

The negative down component means the velocity has an upward component of about 54.7 m/s.

Because C_eⁿ is orthogonal,

n e ∥v ∥ = ∥v ∥.

(37)

Numerically,

|------------------------------| |∥ve∥ = ∥vn∥ ≈ 147.3092 m ∕s.| -------------------------------

(38)

This is one of the simplest and most useful DCM implementation checks.

Solution 3: Earth rate in NED and in a yawed body frame

Earth rate in NED is

⌊ Ω cosϕ ⌋ n ⌈ E ⌉ ωie = 0 . − ΩE sin ϕ

(39)

At 35^∘,

|------⌊----------------⌋--------| | 5.97335 × 10− 5 | ωn ≈ ⌈ 0 ⌉ rad∕s.| | ie −5 | --------−-4.18259-×-10------------

(40)

In degrees per hour,

|----------------------| | ⌊ 12.3209 ⌋∘ | | n ⌈ ⌉ | |ω ie ≈ 0 ∕h. | ---------− 8.6272------|

(41)

For ψ = 60^∘,

⌊ ⌋ cosψ sin ψ 0 Cbn = (Cnb )T = ⌈− sinψ cosψ 0⌉ . 0 0 1

(42)

Thus

|------⌊-------------−-5⌋--------| | 2.98668 × 10 | ωbie ≈ ⌈− 5.17307 × 10−5⌉ rad∕s,| | − 4.18259 × 10−5 | ----------------------------------

(43)

|-----⌊----------⌋-----| | 6.1605 ∘ | ωb ≈ ⌈ − 10.6702⌉ ∕h.| | ie | ---------− 8.6272-------

(44)

The physical angular-velocity vector has not changed. Only its coordinate components changed because the basis changed from NED to body axes.

Solution 4: Transport rate for a moving vehicle

At 45^∘ latitude,

RM ≈ 6367381.82 m, RN ≈ 6388838.29 m.

(45)

Substituting h = 2000 m, v_N = 150 m/s, and v_E = 250 m/s gives

|----------------------------------| | ⌊ 3.91185 × 10 −5 ⌋ | | n ⌈ − 5⌉ | |ω en ≈ − 2.35502 × 10− 5 rad∕s. | ---------−-3.91185-×-10------------|

(46)

In degrees per hour,

|------------------------| | ⌊ 8.06877 ⌋ ∘ | | n ⌈ ⌉ | |ω en ≈ − 4.85757 ∕h. | ---------−-8.06877-------|

(47)

Its magnitude is

∥ωn ∥ ≈ 6.01259 × 10−5 rad∕s ≈ 12.4019 ∘∕h. en

(48)

Earth rotation has magnitude about 15.0411^∘∕h. Therefore the transport rate in this fast-moving example is of the same order as Earth rate. It cannot be neglected in a precise local-level navigator.

Solution 5: Geodetic position rates

The latitude rate is

dϕ-= -------150--------≈ 2.35502 × 10−5 rad∕s. dt 6367381.82 + 2000

(49)

Converting to degrees per hour,

|------------------| |dϕ | |--- ≈ 4.85757∘∕h. | -dt----------------

(50)

The longitude rate is

dλ- = ------------250------------ ≈ 5.53245 × 10− 5 rad ∕s, dt (6388838.29 + 2000)cos 45∘

(51)

|------------------| |dλ-≈ 11.41096 ∘∕h.| -dt-----------------

(52)

Finally,

|--------------------| |dh | |---= − vD = 3 m ∕s.| -dt------------------

(53)

The sign follows directly from the NED convention. Negative down velocity means climbing.

Solution 6: From inertial gyro rate to body rate relative to NED

At 45^∘ latitude,

⌊ 5.15630 × 10− 5⌋ n ⌈ ⌉ ω ie ≈ 0 −5 rad∕s. − 5.15630 × 10

(54)

Adding the transport rate from Exercise 4,

|------⌊----------------⌋---------| | 9.06815 × 10− 5 | ωn ≈ ⌈− 2.35502 × 10− 5⌉ rad∕s. | | in − 5 | --------−-9.06815-×-10-------------

(55)

For a level vehicle at heading 30^∘,

⌊ ⌋ 6.67574 × 10−5 Cbnωnin ≈ ⌈ − 6.57358 × 10 −5⌉ rad ∕s. − 9.06815 × 10 −5

(56)

The commanded body-relative yaw rate is

2 ∘∕s = 0.03490659 rad∕s.

(57)

Therefore the ideal inertial gyro measurement is

⌊ ⌋ 0 ωbib = Cbn ωnin + ⌈ 0 ⌉ , 0.03490659

(58)

|-----⌊-----------------⌋--------| | 6.67574 × 10− 5 | ωb ≈ ⌈ − 6.57358 × 10−5⌉ rad∕s.| | ib − 2 | ---------3.48159-×-10-------------

(59)

Now subtract the navigation-frame inertial rate in body coordinates:

|---------------------⌊-----------⌋--------| | 0 | |ωb = ωb − Cb ωn = ⌈ 0 ⌉ rad∕s.| | nb ib n in | -----------------------0.03490659-----------

(60)

The original 2^∘∕s yaw rate relative to NED is recovered exactly.

Solution 7: DCM chain and round-trip consistency

At ϕ = 30^∘, λ = 20^∘,

⌊ ⌋ − 0.4698463 − 0.1710101 0.8660254 Cne ≈ ⌈ − 0.3420201 0.9396926 0 ⌉ . − 0.8137977 − 0.2961981 − 0.5

(61)

For a level heading of 45^∘,

⌊0.7071068 − 0.7071068 0⌋ n ⌈ ⌉ Cb ≈ 0.7071068 0.7071068 0 . 0 0 1

(62)

The body-to-ECEF transformation is

e e n Cb = C nC b ,

(63)

which evaluates to

|-----⌊---------------------------------------⌋--| | − 0.5740763 0.0903868 − 0.8137977 | |Ceb ≈ ⌈ 0.5435406 0.7853854 − 0.2961981 ⌉ . | 0.6123724 − 0.6123724 − 0.5 | -------------------------------------------------|

(64)

For v^b = [100, 0, 0]^T m/s,

|-------------⌊----------⌋-------| | − 57.40763 | |ve = Cevb ≈ ⌈ 54.35406 ⌉ m ∕s.| | b | ----------------61.23724----------

(65)

Directly in NED,

⌊ ⌋ 70.71068 vn = Cnb vb = ⌈70.71068 ⌉ m ∕s. 0

(66)

Transforming the ECEF result back gives

⌊ ⌋ 70.71068 Cneve ≈ ⌈ 70.71068⌉ m ∕s, 0

(67)

up to numerical roundoff.

Finally,

|----e-------------eT--e------| det-Cb-≈-1,-----(Cb)--Cb-≈-I.--

(68)

This single example checks matrix order, handedness, transpose convention, and norm preservation.

Solution 8: Derivative consistency between ECEF and NED

The transport rate is the same as in Exercise 4,

⌊ −5 ⌋ n 3.91185 × 10 ω en ≈ ⌈− 2.35502 × 10− 5⌉ rad∕s. − 3.91185 × 10− 5

(69)

Its cross product with velocity is

|----------------------------------| | ⌊ 0.00985028 ⌋ | | n n ⌈ ⌉ 2 | |ωen × v ≈ − 0.00575042 m ∕s .| ---------------0.01331215-----------

(70)

Rearrange the derivative relation:

e n Cn dv--= dv--+ ωn × vn. e dt dt en

(71)

Therefore

( ) dve e dvn n n ----= C n ---- + ω en × v . dt dt

(72)

Using ϕ = 45^∘ and λ = 30^∘ gives

|-------⌊-----------⌋--------| |dve − 0.2787323 2 | |----≈ ⌈− 0.3985063⌉ m ∕s .| | dt 0.2803948 | ------------------------------

(73)

Substitute this result back:

⌊ 0.5000000 ⌋ n n ndve- ⌈ ⌉ 2 − ω en × v + C e dt ≈ − 0.2000000 m ∕s . 0.1000000

(74)

The original NED derivative is recovered.

The extra cross-product term is essential because C_eⁿ is changing with time. Transforming the ECEF coordinate derivative alone would compare derivatives taken in different rotating bases.

Solution 9: Stationary Earth-fixed attitude consistency

At 60^∘ latitude,

|--------------------------------| | ⌊ 3.64606 × 10− 5⌋ | | n ⌈ ⌉ | ω ie ≈ 0 −5 rad∕s.| --------−-6.31516-×-10------------

(75)

In degrees per hour this is

⌊ ⌋ 7.52053 ∘ ⌈ 0 ⌉ ∕h. − 13.02595

(76)

At heading 90^∘,

⌊ ⌋ 0 1 0 Cbn = ⌈ − 1 0 0⌉ 0 0 1

(77)

apart from tiny floating-point representations of cos 90^∘. Thus the ideal gyro output is

|-----⌊-----------------⌋--------| | 0 | ωbib ≈ ⌈ − 3.64606 × 10−5⌉ rad∕s.| | − 6.31516 × 10−5 | ----------------------------------

(78)

Since the body is fixed relative to NED,

ωbnb = ωbib − Cbnωnie = 0.

(79)

The DCM equation gives

n dC-b-= Cn [ωb ] − [ωn ] Cn . dt b ib× ie × b

(80)

For this case both matrix terms evaluate numerically to

⌊ ⌋ 6.31516 × 10− 5 0 0 ⌈ 0 6.31516 × 10−5 − 3.64606 × 10−5⌉ 3.64606 × 10− 5 0 0

(81)

up to roundoff, so they cancel exactly:

|---n------| |dC-b-= 0. | --dt-------|

(82)

This is an excellent unit test for a local-level attitude mechanization. A stationary Earth-fixed IMU must not slowly rotate in the computed NED frame when Earth rate is modeled correctly.

Solution 10: Why local NED becomes ill-conditioned near the poles

At ϕ = 89.9^∘,

R ≈ 6399593.56 m. N

(83)

The longitude rate is

d-λ -----100---- dt = R cos89.9∘. N

(84)

Numerically,

|------------------| |dλ- ≈ 1846.70∘∕h. | -dt----------------|

(85)

The transport rate is

|----------------------------------| | ⌊ −5 ⌋ | | n 1.56260 × 10 | |ω en ≈ ⌈ 0 ⌉ rad∕s. | | − 8.95303 × 10− 3 | -----------------------------------

(86)

In degrees per hour,

⌊ ⌋ ∘ 3.2231 ωnen ≈ ⌈ 0 ⌉ ∕h. − 1846.6940

(87)

The large down-axis component comes from

vE tan ϕ − --------, RN + h

(88)

which grows without bound as cos ϕ → 0 in the local longitude parameterization.

The vehicle is not physically rotating thousands of degrees per hour in inertial space. The singular behavior comes from the local definition of North and longitude near the pole. An arbitrarily small displacement near the pole can correspond to a very large change in longitude and a rapid rotation of the local NED axes. This is why high-latitude inertial navigation often uses alternative local frames, wander-azimuth frames, or ECEF mechanization instead of conventional NED coordinates [1, 2].

3 What these examples establish

These calculations turn the frame hierarchy of INS06 into a set of concrete checks.

First, ECI, ECEF, NED, and body coordinates are representations of physical vectors, not different physical vectors. Proper DCM transformations preserve vector magnitude and compose in a strict order.

Second, a time derivative is more subtle than a coordinate transformation. When the destination basis itself rotates, the derivative contains an angular-rate cross product. Exercise 8 is particularly useful for testing this point in software.

Third, Earth rate and transport rate are reference-frame rates. They are not vehicle turn rates, but they enter the attitude mechanization because the navigation frame itself rotates relative to inertial space.

Fourth, position and velocity are coupled geometrically. North and east velocities rotate the local frame and therefore change the reference rate that must be removed from gyro measurements.

Finally, local NED coordinates have a geometric singularity near the poles. That singularity is a property of the coordinate system, not a physical divergence of the vehicle dynamics.

4 Implementation checks suggested by this problem set

A strapdown implementation should reproduce at least the following numerical behaviors:

C_eⁿ(C_eⁿ)^T = I and det C_eⁿ = 1;
vector norms are unchanged by ECEF/NED transformations;
the round trip vⁿ → v^e → vⁿ returns the original vector to numerical precision;
ω_enⁿ = 0 for zero horizontal velocity;
a stationary Earth-fixed IMU has ω_nb^b = 0, not ω_ib^b = 0;
the DCM derivative of a stationary Earth-fixed platform is zero after Earth-rate subtraction;
ECEF and NED vector derivatives agree only after the transport-rate derivative term is included;
the local NED longitude and transport-rate formulas should be guarded or replaced as the poles are approached.

These are small calculations, but together they catch many of the frame-order, transpose, sign, and reference-rate errors that otherwise appear later as unexplained attitude or position drift.

References

[1] David H. Titterton and John L. Weston, Strapdown Inertial Navigation Technology, 2nd ed., Institution of Electrical Engineers, 2004.

[2] Paul D. Groves, Principles of GNSS, Inertial, and Multisensor Integrated Navigation Systems, 2nd ed., Artech House, 2013.

[3] Christopher Jekeli, Inertial Navigation Systems with Geodetic Applications, Walter de Gruyter, 2001.

[4] Paul G. Savage, “Strapdown Inertial Navigation Integration Algorithm Design Part 1: Attitude Algorithms,” Journal of Guidance, Control, and Dynamics, vol. 21, no. 1, pp. 19–28, 1998.

[5] National Geospatial-Intelligence Agency, Department of Defense World Geodetic System 1984: Its Definition and Relationships with Local Geodetic Systems, NGA.STND.0036, Version 1.0.0, 2014.

"Strapdown Inertial Navigation Examples: Earth Rotation and Navigation Frames" is owned by bloftin.
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