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[parent] Strapdown Inertial Navigation Examples: Gyroscope Angular-Rate Sensing and Attitude Propagation (Example)

Strapdown Inertial Navigation Examples: Gyroscope Angular-Rate Sensing and Attitude Propagation

This companion to INS04 develops the gyroscope measurement equation into a set of concrete calculations. The central ideal measurement is

|----|
|ωbib,|
-----
(1)

which is the angular velocity of the body frame b relative to inertial frame i, resolved in body coordinates. The quantity required to propagate body attitude relative to a navigation frame n is instead

|--------------------|
|ωbnb = ωbib − Cbnωnin.|
---------------------
(2)

The exercises deliberately move through three levels of interpretation:

  1. the physical sensing mechanism of a gyro;
  2. the frame-rate bookkeeping required on a rotating Earth;
  3. numerical propagation of attitude from gyro samples.

Unless otherwise stated, use the WGS-84 Earth rotation rate

|----------------------------|
ΩE  = 7.292115 ×  10−5 rad∕s.|
------------------------------
(3)

Angles used inside trigonometric functions must be converted to radians. The notation follows INS04 and standard inertial-navigation references [1, 2, 3, 4, 5].

1 Exercises

Exercise 1: A gyro measures rate, not attitude

A vehicle begins with yaw

ψ0 =  − 20 ∘.
(4)

It then rotates at a constant positive yaw rate

ωz =  12∘∕s
(5)

for 7.5 s. Assume the navigation frame is inertially fixed for this short example and the motion is a pure rotation about the common vertical axis.

  1. Find the accumulated yaw change.
  2. Find the final yaw.
  3. If the same rate measurement were obtained but the initial yaw were unknown, could the final yaw be determined uniquely?
  4. Explain why this scalar example is simpler than general three-dimensional attitude propagation.

PIC

Figure. A constant rate produces a linearly changing yaw only in this special one-axis case. The gyro supplies the slope, while the initial attitude supplies the integration constant.

Exercise 2: Classical rotor gyro torque and precession

A symmetric rotor has spin-axis moment of inertia

I  = 2.50 × 10− 4 kg m2
 s
(6)

and spins at 18,000 rpm. Its spin-axis angular momentum is perpendicular to a forced precession rate of 3.00∘∕s.

  1. Convert the rotor spin rate to radians per second.
  2. Find the angular-momentum magnitude H = Isωs.
  3. Convert the precession rate to radians per second.
  4. Find the torque magnitude required to sustain the precession using
    τ =  Ω  × H.
       p

  5. Explain what part of this calculation gives the rotor gyro its usefulness as an inertial reference.

Exercise 3: MEMS Coriolis angular-rate sensing

A simplified MEMS gyro drives a proof mass with instantaneous velocity

vr = 0.080ex m ∕s.
(7)

The sensor rotates with

Ω = 120 ∘∕sez.
(8)

Use the apparent Coriolis acceleration

aCor,app = − 2Ω  × vr.
(9)

The proof mass is 0.50 mg and the sense-axis stiffness is 20.0 N/m.

  1. Convert the rotation rate to radians per second.
  2. Find the Coriolis acceleration vector.
  3. Find the corresponding sense-axis force magnitude.
  4. If the sense motion is treated quasistatically, estimate the displacement x = F∕k.
  5. Explain why reversing the drive velocity reverses the Coriolis signal even though the platform angular rate has not changed.

PIC

Figure. A MEMS rate gyro converts platform rotation into a transverse Coriolis response of a driven proof mass. The sign follows the cross product between the platform angular rate and the drive velocity.

Exercise 4: Earth rate resolved in NED

An IMU is stationary relative to the Earth at geodetic latitude

      ∘
ϕ = 45 .
(10)

Neglect transport rate. In the local North-East-Down frame, use

      ⌊ Ω   cosϕ ⌋
 n    ⌈   E      ⌉
ωie =       0      .
       − ΩE  sin ϕ
(11)

  1. Compute the North, East, and Down components in rad/s.
  2. Convert those components to degrees per hour.
  3. Verify that the magnitude remains ΩE.
  4. How much inertial rotation angle accumulates in one minute if Earth rotation is not removed?
  5. Explain why a stationary Earth-fixed gyro does not ideally read zero.

PIC

Figure. Earth rate is a physical angular-velocity vector. In local NED coordinates its North and Down components vary with latitude, while its magnitude remains ΩE.

Exercise 5: What a level body-mounted gyro reads

A vehicle is stationary relative to the Earth at latitude 40∘. It is level, but its forward axis is pointed 30∘ east of north. Thus its yaw is

       ∘
ψ  = 30 ,
(12)

with zero roll and pitch. Use

      ⌊                 ⌋
        cosψ   − sin ψ  0
Cnb = ⌈ sin ψ   cos ψ   0⌉ .
          0      0     1
(13)

  1. Compute ωien.
  2. Form Cnb = (C bn)T .
  3. Compute the ideal gyro triad output
    ωbib = Cbnωnie.

  4. Convert each body-axis component to degrees per hour.
  5. Explain why changing heading changes the individual gyro channels but not the angular-rate magnitude.

Exercise 6: From inertial body rate to body rate relative to navigation

At one instant a strapdown INS has the measured gyro vector

       ⌊       ⌋
  b      0.005
ω ib = ⌈− 0.002⌉ rad∕s.
         0.025
(14)

The navigation-frame rate is

      ⌊             ⌋
         6.0 × 10 −5
ωnin = ⌈  2.0 × 10 −5 ⌉rad ∕s.
        − 5.0 × 10− 5
(15)

The vehicle is level at heading 30∘, with C bn as defined in Exercise 5.

  1. Resolve ωinn in body coordinates.
  2. Compute
    ωbnb = ωbib − Cbnωnin.

  3. Compare the size of the reference-frame correction to the measured gyro rate.
  4. Explain why a small correction can still matter when attitude is integrated for a long time.

PIC

Figure. A gyro measures body rate relative to inertial space. Attitude relative to the navigation frame requires subtraction of the navigation frame’s own inertial angular rate after both vectors are expressed in the same coordinates.

Exercise 7: Transport rate of a moving local-level frame

A vehicle moves over the Earth at latitude 45∘ with

vN = 100 m ∕s,     vE = 200 m ∕s,     h = 0.
(16)

Use

RM  = 6,367,381.8 m,     RN  = 6,388,838.3 m,
(17)

and the NED transport-rate model

       ⌊          ⌋
          --vE----
       |  RN  + h |
  n    ||−  --vN---||
ω en = |   RM  + h| .
       ⌈  vE-tan-ϕ⌉
        −  RN +  h
(18)

  1. Compute ωenn in rad/s.
  2. Compute Earth rate ωien at 45∘.
  3. Form
    ωn  =  ωn +  ωn .
  in    ie    en

  4. Convert the components of all three rate vectors to degrees per hour.
  5. If the vehicle body is held exactly fixed relative to NED, what ideal angular rate must its gyro sense relative to inertial space?

Exercise 8: Exact constant-yaw attitude propagation

At time t = 0 the body and navigation frames coincide,

Cnb (0) = I.
(19)

The body rotates relative to the navigation frame at a constant rate

      ⌊      ⌋
          0
ωbnb = ⌈   0  ⌉ .
        10∘∕s
(20)

Assume that the relative rate remains exactly along the common Down axis.

  1. Find the yaw after 3.0 s.
  2. Write the exact Cbn(3 s).
  3. For a single time step Δt = 0.1 s, form the forward-Euler approximation
               (      b      )
Ck+1  ≈ Ck  I + [ωnb]×Δt  .

  4. Compare this one-step approximation with the exact 1∘ finite rotation.
  5. Evaluate Ck+1T C k+1 for the Euler step and identify the orthogonality error.

Exercise 9: Verify the full strapdown attitude equation numerically

At one instant,

      ⌊      ∘         ∘   ⌋
       cos30    − sin 30   0
Cnb =  ⌈sin30 ∘  cos 30∘   0⌉ ,
          0        0      1
(21)

and the navigation-frame rate is

      ⌊  5.0 × 10 −5 ⌋
  n   ⌈          −5 ⌉
ω in =    1.0 × 10 − 5 rad ∕s.
        − 4.0 × 10
(22)

Suppose the desired body rate relative to navigation is

       ⌊       ⌋
         0.002
ωbnb = ⌈− 0.001⌉ rad∕s.
         0.020
(23)

  1. Compute the ideal inertial gyro measurement
    ωbib = ωbnb + Cbnωnin.

  2. Evaluate
    dCn
---b-= Cnb [ωbib]× − [ωnin]×Cnb .
 dt

  3. Independently evaluate
      n  b
Cb [ω nb]×.

  4. Verify numerically that the two matrices agree.
  5. Explain what cancellation this demonstrates physically.

Exercise 10: A finite three-dimensional angular increment

During one IMU sample interval the integrated gyro increment is

       ⌊       ⌋
         0.010
Δ 𝜃 =  ⌈− 0.020⌉ rad.
         0.015
(24)

  1. Find the rotation magnitude 𝜃 = |Δ𝜃| in radians and degrees.
  2. Form S = [Δ𝜃]×.
  3. Use Rodrigues’ formula
    ΔC  =  I + sin𝜃S +  1-−-cos𝜃-S2
             𝜃         𝜃2

    to compute the exact finite-rotation update.

  4. Form the first-order approximation I + S.
  5. Compare the matrices and calculate (I + S)T (I + S) − I.
  6. Explain why finite-rotation propagation is preferable to repeatedly applying I + S without restoring orthogonality.

PIC

Figure. A practical strapdown attitude step begins with the inertial gyro measurement, removes reference-frame rotation, integrates a finite angular increment, and updates the attitude with a proper rotation matrix.

Exercise 11: Constant gyro bias and attitude drift

A stationary platform has a constant unmodeled gyro bias about one axis of

bg = 0.020 ∘∕s.
(25)

Ignore all other errors and suppose the navigation computer interprets this bias as real angular motion.

  1. Convert the bias to rad/s.
  2. Find the attitude error after 60 s.
  3. Find the attitude error after 300 s.
  4. Explain why even a small constant rate error is dangerous in an unaided inertial system.
  5. Connect the result qualitatively to the INS03 relation between tilt error and false horizontal acceleration.

2 Worked solutions

Solution 1: A gyro measures rate, not attitude

For this one-axis case,

dψ- = ωz.
 dt
(26)

Since the rate is constant,

Δψ = ωzΔt (27)
= (12∘∕s)(7.5 s) (28)
= 90∘ . (29)

Therefore

ψf = ψ0 + Δψ (30)
= −20∘ + 90∘ (31)
= 70∘ . (32)

If the initial yaw were unknown, the final yaw would also be unknown. The gyro supplies a change in orientation, not an absolute orientation reference.

The scalar calculation is unusually simple because the rotation axis is fixed. In general three-dimensional motion, successive rotations do not commute. For example, a 10∘ roll followed by a 10∘ pitch is not identical to performing the pitch first and the roll second. General strapdown propagation therefore uses DCMs, quaternions, or another finite-rotation representation rather than integrating three Euler-angle channels independently.

Solution 2: Classical rotor gyro torque and precession

First convert the rotor speed:

ωs = 18,000 rev
----
min( 2π rad)
  -------
  1 rev( 1 min )
  ------
   60 s (33)
= 1884.96 rad∕s . (34)

The angular momentum is

H = Isωs (35)
= (2.50 × 10−4)(1884.96) (36)
= 0.471239 N m s . (37)

The precession rate is

Ωp = 3.00 π
----
180 (38)
= 0.0523599 rad∕s . (39)

Since Ωp is perpendicular to H,

τ = ΩpH (40)
= (0.0523599)(0.471239) (41)
= 2.4674 × 10−2 N m . (42)

The key inertial property is conservation of angular momentum. In the absence of torque, the angular-momentum direction tends to remain fixed in inertial space. A vehicle rotation relative to that inertial direction therefore creates a measurable mechanical response. Modern strapdown gyros may use very different hardware, but the quantity being estimated remains an angular rate relative to inertial space.

Solution 3: MEMS Coriolis angular-rate sensing

Convert the rotation rate:

               |--------------|
Ω =  120-π--=  2.094395  rad∕s|.
        180    ----------------
(43)

The drive velocity is along +x and the angular rate is along +z, so

Ω × vr = (2.094395 )(0.080 )ey.
(44)

Therefore the apparent Coriolis acceleration is

aCor,app = −2Ω × vr (45)
= −2(2.094395)(0.080) ey (46)
= −0.335103 ey m∕s2 . (47)

The mass is

                       − 7
m =  0.50 mg =  5.0 × 10   kg.
(48)

Hence the force magnitude is

F = m|aCor,app| (49)
= (5.0 × 10−7)(0.335103) (50)
= 1.676 × 10−7 N . (51)

With stiffness k = 20.0 N/m, a quasistatic displacement estimate is

x = F-
 k (52)
= 1.676 × 10− 7
-------------
    20.0 (53)
= 8.38 × 10−9 m . (54)

This is only several nanometers, which helps explain why MEMS gyros require sensitive mechanical structures and electronics.

If the drive velocity reverses, vr →−vr. The cross product changes sign:

− 2Ω  × (− vr ) = +2Ω × vr.
(55)

The rotation has not changed, but the Coriolis response follows the oscillating drive velocity. Practical MEMS gyros demodulate that sense-axis response against the known drive motion to recover angular rate.

Solution 4: Earth rate resolved in NED

At ϕ = 45∘,

sin45 ∘ = cos 45∘ = √1-.
                     2
(56)

Therefore

ωien = ⌊      √ --⌋
   ΩE ∕  2
⌈     0 √ -⌉
  − ΩE ∕  2 (57)
= ⌊              −5 ⌋
   5.15630  × 10
⌈        0        ⌉
  − 5.15630 ×  10−5 rad∕s . (58)

To convert rad/s to degrees per hour, multiply by

180-× 3600.
 π
(59)

Thus

|----------------------|
|     ⌊  10.6356 ⌋ ∘   |
| n   ⌈          ⌉     |
ω ie ≈      0       ∕h .
--------−-10.6356-------
(60)

The magnitude is

|ωien| = ∘ ------------------------
  (ΩE cosϕ )2 + (ΩE sin ϕ)2 (61)
= ΩE∘ --------------
  cos2ϕ + sin2ϕ (62)
= ΩE . (63)

In one minute, the inertial rotation angle of the Earth is

Δ𝜃E = ΩE(60) (64)
= 0.00437527 rad (65)
= 0.250684∘ . (66)

A gyro attached to the ground is stationary relative to Earth, but Earth itself rotates relative to inertial space. Therefore the ideal inertial rate measurement is not zero.

Solution 5: What a level body-mounted gyro reads

At latitude 40∘,

      ⌊            ⌋    ⌊                ⌋
        ΩE  cos40∘        5.58608 × 10− 5
ωnie = ⌈      0     ⌉ =  ⌈        0       ⌉ rad ∕s.
        − ΩE sin 40∘      − 4.68728 × 10− 5
(67)

For ψ = 30∘,

               ⌊ cos 30∘   sin 30∘  0⌋
  b      n T   ⌈        ∘       ∘   ⌉
C n = (C b ) =  −  sin 30   cos30   0  .
                    0        0     1
(68)

Hence

ωibb = C nbω ien (69)
= ⌊                 ⌋
  4.83769 ×  10−5
⌈ − 2.79304 × 10 −5⌉
                −5
  − 4.68728 × 10 rad∕s . (70)

In degrees per hour,

|-----⌊----------⌋-∘---|
|        9.97845       |
ωbib ≈ ⌈ − 5.76106 ⌉ ∕h .
|       − 9.66821       |
------------------------
(71)

The body components changed because the sensor axes rotated relative to North and East. However,

   b      n
|ω ib| = |ω ie| = ΩE,
(72)

because an orthogonal coordinate transformation changes components but not vector magnitude.

This is a useful static-IMU test. If the IMU orientation is known and the gyro is sufficiently sensitive, its measured Earth-rate components should agree with the transformed model to within sensor errors.

Solution 6: From inertial body rate to body rate relative to navigation

Using the same 30∘ level attitude,

     ⌊                         ⌋
 b     0.8660254      0.5      0
Cn = ⌈   − 0.5    0.8660254   0⌉ .
           0           0      1
(73)

Transform the navigation-frame rate into body coordinates:

Cnbω inn = ⌊              −5 ⌋
   6.19615  × 10
⌈ − 1.26795 ×  10−5⌉
  − 5.00000 ×  10−5 rad∕s . (74)

Subtract this from the inertial gyro rate:

ωnbb = ⌊       ⌋
  0.005
⌈− 0.002⌉
  0.025 −⌊              −5 ⌋
   6.19615  × 10 −5
⌈ − 1.26795 ×  10  ⌉
  − 5.00000 ×  10−5 (75)
= ⌊             ⌋
  0.00493804
⌈− 0.00198732 ⌉
   0.0250500 rad∕s . (76)

The correction is much smaller than the dominant 0.025 rad/s body rate in this example. However, attitude is obtained by integration. A reference-frame rate error of only 10−5 to 10−4 rad/s can accumulate into a visible angle error over minutes or hours. For precision inertial navigation, small systematic rate terms cannot simply be discarded because they are smaller than the instantaneous vehicle maneuver rate.

Solution 7: Transport rate of a moving local-level frame

At 45∘, tan ϕ = 1. The transport-rate components are

ωen,Nn = ----200----
6,388,838.3 = 3.13046 × 10−5 rad∕s, (77)
ωen,En = −----100----
6,367,381.8 = −1.57050 × 10−5 rad∕s, (78)
ωen,Dn = −----200----
6,388,838.3 = −3.13046 × 10−5 rad∕s. (79)

Thus

|------⌊----------⌋--------------|
|        3.13046                 |
ωnen = ⌈− 1.57050 ⌉ × 10−5 rad∕s .
|       − 3.13046                |
----------------------------------
(80)

From Exercise 4,

      ⌊         ⌋
        5.15630
ωnie = ⌈    0    ⌉ ×  10−5 rad∕s.
       − 5.15630
(81)

Therefore

ωinn = ω ien + ω enn (82)
= ⌊          ⌋
  8.28676
⌈− 1.57050 ⌉
 − 8.28676 × 10−5 rad∕s . (83)

In degrees per hour,

|-------⌊---------⌋-----|
|         6.45704   ∘   |
|ωn  ≈  ⌈− 3.23940⌉  ∕h ,
|  en                   |
---------−-6.45704------|
(84)

|-----⌊----------⌋-----|
|        10.6356   ∘   |
ωn  ≈ ⌈     0    ⌉  ∕h ,
| ie                   |
--------−-10.6356-------
(85)

and

|----------------------|
|      ⌊ 17.0927 ⌋ ∘   |
| n    ⌈         ⌉     |
ω in ≈   − 3.23940   ∕h .
--------−-17.0927-------
(86)

If the body is held exactly fixed relative to NED, then

ωnb  = 0.
(87)

The body must still rotate relative to inertial space at the same angular rate as the local navigation frame. If body axes are aligned with NED,

|----------|
| b     n  |
-ωib =-ωin.-
(88)

The magnitude is approximately

  n                −4              ∘
|ωin| ≈ 1.1824 ×  10   rad∕s ≈ 24.39 ∕h.
(89)

This example shows why the local navigation frame is not merely Earth-fixed. It also rotates because the vehicle moves over the curved Earth.

Solution 8: Exact constant-yaw attitude propagation

The rate magnitude is

ω =  10∘∕s = 0.1745329 rad∕s.
(90)

After 3.0 s,

                    |----|
ψ = (10∘∕s)(3.0 s) =-30∘-.
(91)

The exact passive body-to-navigation DCM is

|--------------------------------------|
|          ⌊0.8660254     − 0.5    0 ⌋ |
| n        ⌈                         ⌉ |
|Cb (3 s) =     0.5     0.8660254  0   .
-----------------0----------0------1---|
(92)

For one 0.1 s step, the angular increment is

        ∘
Δ ψ =  1 =  0.01745329  rad.
(93)

Starting from Ck = I, forward Euler gives

         ⌊     1       − 0.01745329  0⌋
  Euler   ⌈                            ⌉
C k+1 =   0.01745329         1       0  .
               0             0       1
(94)

The exact 1∘ rotation is

         ⌊                            ⌋
           0.9998477   − 0.01745241   0
Cexka+c1t=  ⌈0.01745241    0.9998477    0⌉ .
               0             0       1
(95)

The one-step numerical difference is small, but the Euler matrix is not exactly orthogonal:

|-----------------⌊---------------------------⌋-|
|   EulerT  Euler    1.00030462       0       0  |
|(Ck+1 )  Ck+1  = ⌈      0      1.00030462   0⌉ .
|                        0           0       1  |
------------------------------------------------|
(96)

The diagonal error is second order in the angular increment. Repeated first-order propagation causes a DCM to drift away from the rotation group unless a finite-rotation update or periodic orthogonality restoration is used.

Solution 9: Verify the full strapdown attitude equation numerically

For the given 30∘ yaw,

     ⌊                         ⌋
 b     0.8660254      0.5      0
Cn = ⌈   − 0.5    0.8660254   0⌉ .
           0           0      1
(97)

Transform the navigation-frame rate to body coordinates:

         ⌊                 ⌋
            4.83013  × 10−5
Cb ωn  = ⌈ − 1.63397 ×  10−5⌉ rad∕s.
  n  in                   −5
           − 4.00000 ×  10
(98)

Therefore the ideal inertial gyro measurement is

ωibb = ω nbb + C nbω inn (99)
= ⌊             ⌋
  0.00204830
⌈− 0.00101634 ⌉
   0.0199600 rad∕s . (100)

Substitution into the full attitude equation gives

|-------⌊----------------------------------------⌋----|
|   n     − 0.0100000   − 0.0173205   0.000133975       |
|dC-b-≈ ⌈  0.0173205    − 0.0100000  − 0.00223205 ⌉ s−1.
| dt                                                  |
-----------0.0010000-----0.0020000---------0------------
(101)

Now evaluate the relative-rate form directly:

Cnb [ωbnb]×.
(102)

It produces the same matrix to numerical roundoff. The identity follows algebraically from

  b     b      b n
ω ib = ωnb + Cnω in
(103)

and the cross-product transformation identity

  b  n       b  n     n
[C nω in]× = C n[ωin]×C b .
(104)

When multiplied by Cbn, the contribution of the navigation-frame rate in the gyro measurement cancels the explicit

− [ωnin]×Cnb
(105)

term. What remains is exactly the rotation of body relative to navigation. This is the physical meaning of the full strapdown attitude equation.

Solution 10: A finite three-dimensional angular increment

The increment magnitude is

𝜃 = ∘ -------------------------------
  (0.010)2 + (− 0.020)2 + (0.015)2 (106)
= 0.0269258 rad (107)
= 1.54274∘ . (108)

The skew-symmetric matrix is

             ⌊                       ⌋
                0    − 0.015  − 0.020
S = [Δ 𝜃]× = ⌈0.015     0     − 0.010⌉ .
              0.020   0.010      0
(109)

Rodrigues’ formula gives

|----------⌊-------------------------------------------⌋-|
|           0.999687519   − 0.015098182   − 0.019922588   |
ΔC      ≈  ⌈0.014898194    0.999837510    − 0.010148783 ⌉ .
|   exact                                                 |
------------0.020072579----0.009848801-----0.999750015-----
(110)

The first-order approximation is

|---------------⌊------------------------⌋-|
|                   1    − 0.015   − 0.020  |
|ΔC   = I + S = ⌈ 0.015     1     − 0.010 ⌉ .
|   1                                      |
------------------0.020---0.010------1-----|
(111)

Its orthogonality defect is

|----------------⌊---------------------------------⌋-|
|                  0.000625   0.000200   − 0.000150  |
ΔCT  ΔC   − I =  ⌈ 0.000200   0.000325   0.000300  ⌉ .
|   1    1                                           |
------------------−-0.000150--0.000300---0.000500-----
(112)

The Frobenius norm of this defect is approximately

                    |-----------|
∥ΔCT  ΔC1  − I∥F  ≈ |1.03 × 10 −3.
     1              -------------
(113)

The exact Rodrigues update is a member of SO(3), so it preserves orthogonality and determinant +1 apart from numerical roundoff. The linear approximation is useful in derivations and error-state models, but repeated attitude propagation should use a proper finite-rotation update.

This example also shows why a three-component angular increment is not simply three independent Euler-angle increments. The vector Δ𝜃 defines one finite rotation through the exponential map. The coupling between axes appears in the S2 and higher-order terms.

Solution 11: Constant gyro bias and attitude drift

Convert the bias to radians per second:

bg = 0.020-π--
180 (114)
= 3.49066 × 10−4 rad∕s . (115)

For a constant unmodeled bias, the small-angle attitude error initially grows approximately as

δ𝜃(t) ≈ bgt.
(116)

After 60 s,

δ𝜃(60) = (0.020∘∕s)(60 s) (117)
= 1.20∘ . (118)

After 300 s,

δ𝜃(300) = (0.020∘∕s)(300 s) (119)
= 6.00∘ . (120)

This is dangerous because the attitude solution is used to rotate accelerometer specific force into the navigation frame. A tilt error causes part of the large gravity-related specific-force vector to leak into a horizontal channel. For a small horizontal tilt error,

δaH ≈  gδ𝜃.
(121)

Therefore a gyro rate bias can become an attitude error, then a false horizontal acceleration, then velocity and position errors after further integration. INS19 will derive this error chain in detail.

3 What these exercises establish

The worked examples above establish several habits that are central to strapdown navigation:

  1. A gyro measures angular rate, not attitude. An initial attitude and a rotation-propagation law are still required.
  2. Mechanical rotor gyros, MEMS vibrating-mass gyros, and optical gyros use different physics but target an angular rate relative to inertial space.
  3. A stationary Earth-fixed gyro can have a nonzero ideal output because the Earth rotates relative to inertial space.
  4. Earth rate must be resolved in the sensor coordinates before it can be compared with body-axis gyro measurements.
  5. A local navigation frame rotates because of both Earth rotation and vehicle motion over the curved Earth.
  6. The attitude-propagation rate is ωnbb, not directly the measured ω ibb.
  7. The full attitude equation and the relative-rate form are mathematically equivalent when frame-rate transformations are handled consistently.
  8. First-order DCM updates are useful approximations but do not preserve orthogonality exactly.
  9. Finite angular increments should be propagated with a proper rotation update such as Rodrigues’ formula or an equivalent quaternion update.
  10. Small systematic gyro errors accumulate in attitude and can couple into large translational navigation errors.

These results prepare the reader for INS05, where the gravity and effective-gravity models required by the translational mechanization will be developed in detail. They also prepare the mathematical ground for INS08, where quaternion attitude propagation will be derived from the same angular-rate measurements.

References

[1]   D. H. Titterton and J. L. Weston, Strapdown Inertial Navigation Technology, 2nd ed., IET, 2004.

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

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

[4]   A. Lawrence, Modern Inertial Technology: Navigation, Guidance, and Control, 2nd ed., Springer, 1998.

[5]   P. 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.

[6]   J. E. Bortz, “A New Mathematical Formulation for Strapdown Inertial Navigation,” IEEE Transactions on Aerospace and Electronic Systems, vol. AES-7, no. 1, pp. 61–66, 1971.


"Strapdown Inertial Navigation Examples: Gyroscope Angular-Rate Sensing and Attitude Propagation" is owned by bloftin.
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Keywords:  strapdown inertial navigation, gyroscope, angular rate, Earth rate, relative frame rate, transport rate, attitude propagation, direction cosine matrix, MEMS gyroscope, Coriolis sensing, rotor gyroscope, worked examples

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Physics Classification: 06.30.Gv (Velocity, acceleration, and rotation)
 07.07.Df (Sensors ; remote)
 02.20.Qs (General properties, structure, and representation of Lie groups)
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