Strapdown Inertial Navigation Examples: Effective Gravity and Navigation Equations

This companion to INS05 turns the gravity models and rotating-Earth equations into explicit calculations. The central bookkeeping distinction is

|----------------------------| -g-=-ggrav −-ωie ×-(ωie-×-r)-,|

(1)

where g_grav is gravitation from Earth’s mass distribution and g is effective gravity for a rotating Earth-fixed mechanization.

The examples use the WGS-84 constants

a = 6378137 m, f = ------1-------, Ω = 7.292115 × 10−5 rad∕s, 298.257223563 E

(2)

14 3 2 2 μ = 3.986004418 × 10 m ∕s , e = f(2 − f),

(3)

and the Somigliana constants

2 γe = 9.7803253359 m ∕s, k = 0.00193185265241.

(4)

The notation and sign conventions follow INS05 and standard inertial-navigation references [1, 2, 3, 4].

Figure. In the spherical model, gravitation points toward Earth’s center while the centrifugal acceleration points away from the rotation axis. Their vector sum is the effective gravity used by the rotating-Earth navigation equation.

1 Exercises

Exercise 1: Point-mass gravitation and effective gravity at the equator

At the WGS-84 semi-major radius R = a, approximate Earth as a spherical point mass.

Compute the point-mass gravitational acceleration magnitude
$μ-- ggrav = R2.$
Compute the equatorial centrifugal acceleration
$2 acf = ΩER.$
Compute the simple spherical effective gravity at the equator.
Compare it with WGS-84 Normal gravity at the equator and explain why the two are not expected to be identical.

Exercise 2: Centrifugal acceleration resolved in NED

Use a spherical Earth of radius R = a at latitude ϕ = 45^∘. For a point fixed to Earth, show that the centrifugal acceleration resolved in local NED is

⌊ 2 ⌋ n − ΩER sin ϕcos ϕ acf = ⌈ 0 ⌉ . − Ω2ER cos2ϕ

(5)

Compute the numerical vector. Then combine it with spherical gravitation

⌊ ⌋ 0 gn = ⌈ 0 ⌉ grav ggrav

(6)

to find the effective gravity vector, its magnitude, and its angular deflection from the local radial down direction.

Exercise 3: WGS-84 normal gravity from Somigliana’s formula

Evaluate

1 + ksin2ϕ γ(ϕ) = γe∘-------------- 1 − e2sin2 ϕ

(7)

at ϕ = 30^∘, 45^∘, and 60^∘.

Compute all three values.
Compute the difference between 60^∘ and 30^∘.
Explain why this latitude variation matters to a strapdown INS.

Figure. WGS-84 normal gravity increases with geodetic latitude. The variation is large compared with inertial-sensor error levels, so using one constant value of g is not a high-quality navigation model.

Exercise 4: First-order normal-gravity correction with altitude

At latitude 45^∘, use the simple first-order height approximation

( ) 2h- γ(ϕ,h ) ≈ γ (ϕ,0) 1 − a .

(8)

Find normal gravity at h = 2.00 km and calculate the reduction from the ellipsoid-surface value.

Exercise 5: Stationary accelerometer sanity check in NED

A level IMU is stationary relative to Earth at latitude 45^∘ and height h = 0. Neglect local gravity anomalies and use WGS-84 normal gravity.

Write gⁿ in NED.
Find the ideal accelerometer specific force fⁿ.
Substitute the values into the local NED velocity equation with v_ebⁿ = 0 and verify that the velocity derivative vanishes.
State the most common sign error this example exposes in navigation code.

Exercise 6: ECEF velocity equation for eastward motion at the equator

Use the simple spherical Earth model at the equator and longitude zero. Let

⌊ ⌋ ⌊ ⌋ R 0 re = ⌈ 0⌉ , ωeie = ⌈ 0 ⌉ , 0 ΩE

(9)

and suppose the vehicle has constant ECEF east velocity

⌊ ⌋ 0 veeb = ⌈200⌉ m ∕s. 0

(10)

Compute g_grav^e.
Compute the explicit centrifugal acceleration.
Compute effective gravity g^e.
Compute the Coriolis contribution −2ω_ie^e × v_eb^e.
Find the specific-force vector required for dv_eb^e∕dt = 0 in this idealized example.

Exercise 7: Proving the two ECEF gravity forms are equivalent

Start with

e dveb- e b e e e e e e dt = C bfib + ggrav − 2ω ie × veb − ωie × (ω ie × reb).

(11)

Use the definition of effective gravity to derive

dveeb e b e e e ----= Cbfib + g − 2ω ie × v eb. dt

(12)

Then explain precisely why adding another centrifugal term to the second equation would be incorrect.

Figure. The explicit-gravitation and effective-gravity ECEF forms are equivalent descriptions. The centrifugal contribution belongs in exactly one place.

Exercise 8: Full numerical NED velocity update

ϕ = 45 ∘, h = 1000 m,

(13)

a vehicle has

⌊ ⌋ 100 vneb = ⌈200 ⌉ m ∕s − 5

(14)

and transformed specific force

⌊ 0.600 ⌋ n b ⌈ ⌉ 2 C b fib = − 0.400 m ∕s . − 9.700

(15)

Use

-------a------ RN = ∘ -----2---2--, 1 − e sin ϕ

(16)

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

(17)

⌊ ⌋ n ΩE cosϕ ωie = ⌈ 0 ⌉ , − ΩE sin ϕ

(18)

and

⌊ --vE----⌋ | RN + h | n | --vN---| ω en = ||− RM + h|| . ⌈ vE tan ϕ⌉ − -R--+--h N

(19)

Compute

dvneb n b n n n n -----= Cb fib + g − (2 ωie + ω en) × veb. dt

(20)

Use the first-order altitude correction for γ(45^∘, 1000 m).

Figure. A local-level velocity update combines transformed specific force, gravity, Earth rate, transport rate, and the current velocity. Each term has a distinct physical origin.

Exercise 9: What double-counting centrifugal acceleration does

At the equator, suppose a programmer uses WGS-84 effective gravity but also subtracts a separate centrifugal term of magnitude approximately

acf = Ω2Ea.

(21)

Treat the resulting acceleration error as constant for 60 s.

Find the acceleration error magnitude.
Find the resulting velocity error after 60 s.
Find the resulting position error from rest using δr = δat².

Exercise 10: Gravity-model error accumulation

A gravity model has a constant vertical error of 50 mGal. Recall that

1 Gal = 0.01 m ∕s2, 1 mGal = 10− 5 m ∕s2.

(22)

Assume no aiding and neglect all other errors.

Convert the gravity error to SI acceleration units.
Find the vertical velocity error after 300 s.
Find the corresponding position error from rest.
Explain why this simple t² result is only a short-time approximation for a full navigation system.

Figure. A constant gravity-model acceleration error integrates once into velocity error and twice into position error. This is one reason gravity-model fidelity matters even when the accelerometers themselves are ideal.

2 Worked solutions

Solution 1: Point-mass gravitation and effective gravity at the equator

Using R = a = 6378137 m,

|-----------------| 3.986004418--×-1014 | 2 | ggrav = (6378137 )2 ≈ -9.79828548-m-∕s--.

(23)

The centrifugal acceleration at the equator is

2 −5 2 |---------------2| acf = Ω ER = (7.292115 × 10 ) (6378137 ) ≈ 0.03391571--m-∕s- .

(24)

In this simple spherical model, both vectors are radial and opposite. Therefore

|----------------| g = g − a ≈ |9.76436977 m ∕s2. eff grav cf ------------------

(25)

WGS-84 normal gravity at the equator is

2 γe = 9.7803253359 m ∕s .

(26)

The spherical point-mass estimate differs by about

2 9.78032534 − 9.76436977 ≈ 0.01596 m∕s .

(27)

The discrepancy is not a numerical mistake. The point-mass model ignores Earth’s oblateness and the way the reference ellipsoid and normal gravity field are constructed together. This comparison is a useful warning: subtracting a centrifugal term from a crude spherical gravitation model is not the same as evaluating the WGS-84 normal gravity field.

Solution 2: Centrifugal acceleration resolved in NED

At ϕ = 45^∘,

1 sin ϕ = cos ϕ = √---. 2

(28)

First compute

Ω2ER ≈ 0.033915706 m ∕s2.

(29)

Therefore

2 1- 2 2 acf,N = − Ω ER sinϕ cosϕ = − 2Ω ER ≈ − 0.016957853 m ∕s ,

(30)

and

a = − Ω2 R cos2ϕ ≈ − 0.016957853 m∕s2. cf,D E

(31)

Hence

|---------------------------| | ⌊ ⌋ | | n − 0.0169579 2 | a cf ≈ ⌈ 0 ⌉ m∕s . | | − 0.0169579 | ----------------------------

(32)

Using the point-mass value from Exercise 1,

⌊ ⌋ 0 gngrav ≈ ⌈ 0 ⌉ m∕s2, 9.7982855

(33)

|-----⌊------------⌋-------| | n − 0.0169579 2 | g ≈ ⌈ 0 ⌉ m ∕s .| | 9.7813276 | ----------------------------

(34)

The magnitude is

∘ ----------------------------- |---------------| |gn | = (− 0.0169579 )2 + (9.7813276 )2 ≈ 9.7813423 m∕s2 . -----------------

(35)

The angular deflection from radial down is approximately

( ) − 1 0.0169579- |------∘-| α = tan 9.7813276 ≈ -0.0993--.

(36)

The horizontal component points toward the equator. In a realistic ellipsoidal model, local geodetic down is defined so that the normal-gravity field is aligned with the ellipsoid normal, which is why this spherical result should be treated as a physics illustration rather than a production local-gravity model.

Solution 3: WGS-84 normal gravity from Somigliana’s formula

Using

e2 = f (2 − f) ≈ 0.00669437999014,

(37)

Somigliana’s formula gives

|-----------------| γ(30∘) ≈ |9.79324727 m ∕s2 , ------------------

(38)

|-----------------| γ(45∘) ≈ |9.80619777 m ∕s2 , ------------------

(39)

and

|-----------------| γ(60∘) ≈ |9.81917695 m ∕s2 . ------------------

(40)

The difference between 60^∘ and 30^∘ is

|-----------------| Δ γ ≈ 9.81917695 − 9.79324727 ≈ |0.02592968 m ∕s2 . ------------------

(41)

This is roughly 2.6 × 10⁻³g. A strapdown navigator that simply inserts a single constant value of gravity everywhere creates a deterministic acceleration error that can be much larger than the errors of a good inertial sensor.

Solution 4: First-order normal-gravity correction with altitude

At 45^∘,

γ(45∘,0) ≈ 9.80619777 m ∕s2.

(42)

At h = 2000 m,

( ) ∘ --4000-- γ (45 ,2000) ≈ 9.80619777 1 − 6378137 ,

(43)

which gives

|----∘-------------------------2-| -γ(45-,2000-) ≈-9.80004789--m-∕s-.|

(44)

The reduction is

|------------------------| -Δ-γ-≈-0.00614988--m-∕s2.|

(45)

The approximation captures the dominant inverse-square trend. More accurate normal-gravity formulas include ellipsoidal height dependence explicitly.

Solution 5: Stationary accelerometer sanity check in NED

At the surface and 45^∘ latitude,

⌊ ⌋ 0 gn ≈ ⌈ 0 ⌉ m ∕s2. 9.80619777

(46)

For a stationary supported IMU, INS03 showed that ideal specific force is the negative of effective gravity:

|-----⌊-------------⌋--------| | n 0 2 | |f ≈ ⌈ 0 ⌉ m ∕s .| | − 9.80619777 | -----------------------------|

(47)

Because v_ebⁿ = 0, the Earth-rate and transport-rate cross-product terms vanish. Therefore

dvneb n n -----= f + g = 0. dt

(48)

This is one of the strongest unit tests for a local-level mechanization. A common mistake is to use

⌊ ⌋ 0 gn = ⌈ 0 ⌉ − g

(49)

while simultaneously declaring the navigation frame to be NED. In NED, positive down means the gravity vector has a positive third component.

Solution 6: ECEF velocity equation for eastward motion at the equator

At longitude zero on the equator,

⌊ ⌋ ⌊ ⌋ − ggrav − 9.79828548 gegrav = ⌈ 0 ⌉ ≈ ⌈ 0 ⌉ m ∕s2. 0 0

(50)

The explicit centrifugal acceleration is

⌊ ⌋ Ω2ER − ωeie × (ωeie × re) = ⌈ 0 ⌉, 0

(51)

|------⌊------------⌋--------| | 0.033915706 | |aecf ≈ ⌈ 0 ⌉ m ∕s2.| | 0 | -----------------------------|

(52)

Thus the simple spherical effective gravity is

|-----⌊-------------⌋--------| | − 9.76436977 | |ge ≈ ⌈ 0 ⌉ m ∕s2.| | | -------------0---------------|

(53)

Now

⌊ ⌋ − ΩE (200) ωeie × veeb = ⌈ 0 ⌉ , 0

(54)

⌊ ⌋ ⌊------------⌋-------| 2ΩE (200) | 0.02916846 | − 2ωe × ve = ⌈ 0 ⌉ ≈ ⌈ 0 ⌉ m ∕s2.| ie eb | | 0 -------0--------------

(55)

For zero ECEF velocity derivative,

0 = fe + ge − 2ωe × ve , ie eb

(56)

|-----⌊-----------⌋--------| | e 9.73520131 2 | |f ≈ ⌈ 0 ⌉ m ∕s .| | 0 | ---------------------------|

(57)

The required outward specific force is slightly less than the magnitude of effective gravity because the Coriolis contribution in this geometry is outward.

Solution 7: Proving the two ECEF gravity forms are equivalent

Define effective gravity by

|------------------------------| |ge = gegrav − ωeie × (ωeie × reeb). -------------------------------

(58)

Substituting this definition into the explicit equation immediately gives

|--e---------------------------| |dveb= Cef b+ ge − 2ωe × ve .| --dt-----b-ib----------ie----eb--

(59)

No physics disappeared. The centrifugal contribution was grouped into the definition of g^e.

If a navigation program uses this effective-gravity form and then subtracts another

e e e ω ie × (ω ie × reb) ,

(60)

it counts the rotational contribution twice. This is a bookkeeping error, not a higher-fidelity model.

Solution 8: Full numerical NED velocity update

At 45^∘ latitude,

|--------------| RN ≈ -6388838.29-m--,

(61)

and

|--------------| RM ≈ -6367381.82-m--.

(62)

Earth rate resolved in NED is

⌊ ⌋ 5.1563 × 10−5 ωnie ≈ ⌈ 0 ⌉ rad ∕s. − 5.1563 × 10 −5

(63)

Using v_N = 100 m/s and v_E = 200 m/s,

⌊ 3.1300 × 10−5 ⌋ n ⌈ −5⌉ ωen ≈ − 1.5703 × 10 rad∕s. − 3.1300 × 10−5

(64)

Therefore

⌊ −4 ⌋ n n 1.34426 × 10 2ωie + ωen ≈ ⌈ − 1.5703 × 10− 5⌉ rad∕s. − 1.34426 × 10−4

(65)

The cross product with velocity is

⌊ ⌋ 0.0269637 (2ωn + ωn ) × vn ≈ ⌈− 0.0127704 ⌉ m ∕s2. ie en eb 0.0284554

(66)

The first-order altitude-corrected gravity is

|----------------| γ(45 ∘, 1000) ≈ |9.80312283 m ∕s2, ------------------

(67)

⌊ ⌋ 0 gn ≈ ⌈ 0 ⌉ m ∕s2. 9.80312283

(68)

Finally,

⌊ 0.600 ⌋ ⌊ 0 ⌋ ⌊ 0.0269637 ⌋ dvneb- ⌈ ⌉ ⌈ ⌉ ⌈ ⌉ dt = − 0.400 + 0 − − 0.0127704 , − 9.700 9.80312283 0.0284554

(69)

which gives

|-------⌊-----------⌋--------| | n 0.573036 | |dv-eb-≈ ⌈ − 0.387230⌉ m ∕s2.| | dt | -----------0.074667----------|

(70)

This example shows why the mechanization cannot be interpreted as “rotate accelerometer data and add g.” The rotating-frame correction is small compared with g, but it is not small compared with the accelerations a navigation system may be trying to resolve.

Solution 9: What double-counting centrifugal acceleration does

At the equator,

|----------------| δa ≈ Ω2 a ≈ |0.0339157 m ∕s2 . E -----------------

(71)

After t = 60 s, the velocity error is approximately

|----------| δv = δa t ≈ 0.0339157 (60) ≈|2.035 m ∕s . ------------

(72)

The corresponding position error is

δr = 1δa t2 ≈ 1(0.0339157)(60)2 ≈ |61.05-m-. 2 2 ----------

(73)

This large error develops in only one minute. The example demonstrates why the definition of the gravity vector in software must be paired unambiguously with the form of the navigation equation.

Solution 10: Gravity-model error accumulation

A 50 mGal error is

|----------------| δg = 50 × 10 −5 = 5.0 × 10− 4 m ∕s2 . -----------------

(74)

After 300 s,

|----------| δv = δg t = (5.0 × 10− 4)(300 ) = 0.150 m ∕s . ------------

(75)

The corresponding position error is

1- 2 1- −4 2 |------| δr = 2δgt = 2 (5.0 × 10 )(300 ) = 22.5-m-.

(76)

The t² result assumes a constant error direction in a simple Cartesian model. A real local-level INS couples gravity error, attitude error, Earth curvature, transport rate, and Schuler dynamics. Those effects will be developed later in the series. The short-time estimate is still valuable because it reveals the direct double integration mechanism.

3 What these exercises establish

The examples reinforce several implementation rules.

First, gravitation and effective gravity are not interchangeable symbols. The correct choice depends on the exact navigation equation being implemented.

Second, centrifugal acceleration is not a small conceptual detail. At the equator it is about

−2 2 3.39 × 10 m ∕s ,

(77)

which is large compared with precision inertial-sensor errors.

Third, WGS-84 normal gravity already represents the rotating reference ellipsoid. A local-level mechanization using normal gravity must not add the same centrifugal contribution a second time.

Fourth, the complete local NED velocity equation

|--------------------------------------| |dvneb- n b n n n n | | dt = C b fib + g − (2ω ie + ωen) × veb| ----------------------------------------

(78)

contains terms from sensor physics, gravity modeling, Earth rotation, local-frame motion, and geometry. Every term has now been derived or calculated explicitly.

Finally, even modest gravity-model errors integrate into navigation errors. Later articles will show how the full inertial error dynamics modify this simple growth and lead to Schuler behavior.

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] 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.

[5] Helmut Moritz, “Geodetic Reference System 1980,” Bulletin Geodesique, vol. 54, pp. 395–405, 1980.

"Strapdown Inertial Navigation Examples: Effective Gravity and Navigation Equations" is owned by bloftin.
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