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[parent] Strapdown Inertial Navigation Examples: Rotating-Frame Derivatives, Coriolis, and Centrifugal Acceleration (Example)

Strapdown Inertial Navigation Examples: Rotating-Frame Derivatives, Coriolis, and Centrifugal Acceleration

This companion to INS02 develops rotating-frame kinematics through explicit calculations. The goal is to make the transport theorem operational while preserving its physical meaning. The reader should be able to look at each term in

|----------------------------------------|
a  = a  + 2ω ×  v +  α × r + ω ×  (ω  × r)|
--i---r----------r------------------------
(1)

and explain where it came from, what direction it points, and whether it belongs to an inertial acceleration decomposition or to an apparent-force description in the rotating frame.

Throughout this companion, the angular-acceleration vector is denoted by α = dω∕dt to keep the rotating-frame notation compact.

The exercises are stated first so the article can be used as a self-study problem set. Complete worked solutions follow. The notation follows INS02 and standard inertial-navigation texts: i denotes an inertial frame, r a generic rotating frame, and ωir the angular velocity of r relative to i [1, 2, 3].

1 Exercises

Exercise 1: An inertially fixed vector seen from rotating axes

A frame r rotates counterclockwise about the common z axis relative to inertial frame i at the constant rate

ω  = 20∘∕s.
(2)

At t = 0 the frames are aligned. The physical vector

a = 5eix
(3)

is fixed in inertial space.

At t = 3.0 s:

  1. find the rotation angle of frame r;
  2. compute the components ar;
  3. differentiate those rotating-frame components directly with respect to time;
  4. verify the transport-theorem prediction
    (   )
  da-
  dt   =  − ωir × a;
      r

  5. explain why a vector can have a nonzero derivative of its coordinates in r even though its inertial derivative is zero.

PIC

Figure. A physical vector fixed in inertial space acquires changing numerical components in a rotating basis.

Exercise 2: A point fixed on a rotating turntable

A point is rigidly attached to a horizontal turntable at

     ⌊  ⌋
       2
rr = ⌈ 0⌉ m.

       0
(4)

The turntable rotates at constant angular velocity

      ⌊  ⌋
        0
ωir = ⌈ 0⌉ rad/s.
        3
(5)

The origins coincide and the point is stationary relative to the turntable.

  1. Compute the inertial velocity of the point.
  2. Compute its inertial acceleration.
  3. Show that the acceleration is directed toward the rotation axis.
  4. If Newton’s law is instead written in rotating-frame form, what apparent centrifugal acceleration is introduced?

Exercise 3: Radial motion and Coriolis acceleration

On a turntable rotating with

ω  = 2ez rad/s,
(6)

a small slider is instantaneously located at

r = 2ex m
(7)

and moves radially outward with rotating-frame velocity

v  = 1.5e  m/s.
 r       x
(8)

At that instant its rotating-frame relative acceleration is zero, and the turntable rate is constant.

  1. Compute the Coriolis term 2ω × vr.
  2. Compute the rotational centripetal term ω × (ω × r).
  3. Determine the total inertial acceleration.
  4. Compute its magnitude and direction in the xy plane.
  5. State the sign of the corresponding apparent Coriolis acceleration when Newton’s law is rearranged into rotating-frame form.

PIC

Figure. Radial motion on a rotating turntable. The Coriolis contribution is transverse to the radial velocity while the rotational centripetal contribution points inward.

Exercise 4: Angular acceleration and the Euler term

A rigid arm rotates about the z axis. At one instant,

                                     2
ω  = 1.5ez rad/s,     α = 0.8ez rad/s .
(9)

A point is fixed to the arm at

r = 3ex m.
(10)

  1. Compute the Euler acceleration α× r.
  2. Compute the rotational centripetal acceleration.
  3. Find the total inertial acceleration of the point.
  4. Explain physically why the Euler and centripetal contributions are perpendicular in this geometry.

Exercise 5: Translating and rotating origin

A moving coordinate frame r has an origin whose inertial acceleration is

       ⌊     ⌋
         1.2
A   =  ⌈− 0.4⌉ m/s2.
  O
          0
(11)

At a particular instant the frame has

     ⌊   ⌋                 ⌊    ⌋
       0                      0
ω =  ⌈ 0 ⌉  rad/s,     α = ⌈  0 ⌉ rad/s2.
      0.5                    0.1
(12)

A particle has rotating-frame state

    ⌊  ⌋              ⌊     ⌋                ⌊     ⌋
      4                  0.3                    0.05
r = ⌈ 1⌉  m,     v  = ⌈ − 0.2⌉ m/s,     a  = ⌈ 0.10⌉  m/s2.
                  r                       r
      0                  0                      0
(13)

Use

ai = AO  + ar + 2ω ×  vr + α × r + ω ×  (ω  × r)
(14)

to compute every contribution separately and then the total inertial acceleration.

PIC

Figure. For a translating and rotating frame, inertial acceleration is assembled from origin acceleration, relative acceleration, Coriolis, Euler, and rotational centripetal contributions.

Exercise 6: A point fixed to the rotating Earth is not inertially stationary

Approximate the Earth as a sphere of equatorial radius

                    6
RE  = 6.378137 × 10  m
(15)

rotating at

                   −5
ωie = 7.292115 × 10   rad/s.
(16)

Consider a point fixed to the Earth at the equator.

  1. Compute its inertial speed due solely to Earth rotation.
  2. Compute the magnitude of its rotational centripetal acceleration.
  3. Compare this acceleration with g = 9.81 m/s2 as a percentage.
  4. Explain why an Earth-fixed navigation frame is non-inertial even for an object that is stationary in ECEF coordinates.

Exercise 7: DCM derivative from angular velocity

At t = 0, frames r and i are aligned. Frame r rotates relative to i with

      ⌊  ⌋
        0
ωi  = ⌈ 0⌉ rad/s.
  ir
        2
(17)

Use the INS02 identity

C˙ir = [ωiir]×Cir
(18)

to answer the following.

  1. Compute Ċri(0).
  2. Use a single forward-Euler step with Δt = 0.01 s to approximate Cri(Δt).
  3. Compute (Cri)T C ri for that Euler approximation.
  4. Explain why ordinary Euler integration slowly destroys exact orthogonality.
  5. Show analytically that if Ċ = ΩC with ΩT = −Ω, then the exact solution preserves CT C = I.

PIC

Figure. Angular velocity generates the time derivative of the DCM. Exact rotational propagation remains on SO(3); a naive Euler step does not.

Exercise 8: Adding angular rates through nested frames

At one instant, frames i, e, n, and b are momentarily aligned so that all angular-rate components can be added directly without another coordinate transformation. Suppose

      ⌊    ⌋             ⌊     ⌋             ⌊    ⌋
         0                  0                 0.20
ωie = ⌈  0 ⌉ ,     ωen = ⌈ 0.05 ⌉ ,    ωnb =  ⌈  0 ⌉
       0.10                 0                   0
(19)

in rad/s.

  1. Compute ωin.
  2. Compute ωib.
  3. Explain why this simple component-wise addition would generally be invalid if the rates were supplied in different resolving frames.
  4. Relate the result to the later strapdown identity
    ωin =  ωie + ωen.

Exercise 9: Inertial terms versus apparent accelerations

At some instant a particle is described in a uniformly rotating frame by

r = 2ex  m,     vr = 1.5ex m/s,     ω  = 2ez rad/s.
(20)

Assume the particle is force-free in the inertial frame, so ai = 0.

  1. Starting from the acceleration transport equation, solve for ar.
  2. Identify the apparent centrifugal acceleration.
  3. Identify the apparent Coriolis acceleration.
  4. Explain why their signs are opposite those of the corresponding terms in the inertial acceleration decomposition.

Exercise 10: Implementation sanity check – degrees are not radians

A software routine evaluates the rotational centripetal magnitude using

ac = ω2r.
(21)

The physical angular speed is 15∘∕s and the radius is r = 1.2 m. A programmer accidentally passes the number 15 directly into a routine that expects rad/s.

  1. Compute the correct angular speed in rad/s.
  2. Compute the correct centripetal acceleration.
  3. Compute the erroneous acceleration produced by the software.
  4. By what factor is the result wrong?
  5. Propose at least three unit tests or interface safeguards that could catch this class of error in an inertial-navigation code base.

2 Worked solutions

Solution 1: An inertially fixed vector seen from rotating axes

The angular speed in SI units is

        π
ω =  20----=  0.34906585  rad/s.
       180
(22)

After 3 s,

𝜃 = ωt = 60 ∘.
(23)

For axes rotated counterclockwise through 𝜃, the passive transformation from inertial coordinates to rotating coordinates is

      ⌊ cos𝜃   sin 𝜃  0⌋
  r   ⌈                ⌉
Ci =   − sin𝜃  cos 𝜃  0  .
          0      0    1
(24)

Thus

ar = C irai (25)
= ⌊       ∘        ∘   ⌋
  cos60     sin 60   0
⌈− sin60 ∘ cos 60∘  0⌉
     0        0     1⌊  ⌋
  5
⌈ 0⌉
  0 (26)
= ⌊        ⌋
  2.5000
⌈− 4.3301⌉
     0. (27)

Because

        ⌊         ⌋
          5 cosωt
ar (t) = ⌈ − 5 sin ωt⌉ ,
             0
(28)

direct differentiation gives

          ⌊           ⌋
( da )      − 5ω sin ωt
  ---   = ⌈ − 5ω cosωt⌉ .
  dt   r         0
(29)

At t = 3 s,

|------------------------------|
|(   )     ⌊        ⌋          |
|  da-      − 1.5115           |
|  dt   ≈  ⌈− 0.8727⌉  units/s.|
|      r        0              |
--------------------------------
(30)

Now use the transport theorem. Since a is fixed in inertial space,

(   )
  da-  =  0.
  dt  i
(31)

Therefore

     (   )
      da-
0 =    dt   + ω  × a,
           r
(32)

so

(----)-------------|
| da               |
| ---   =  − ω × a.|
---dt--r------------
(33)

Resolved in r,

      ⌊           ⌋
            0
ωr  = ⌈     0     ⌉ ,
        0.34906585
(34)

and

−ω × a = −⌊     0     ⌋
⌈           ⌉
      0
  0.34906585 ×⌊    2.5   ⌋
⌈         ⌉
  − 4.3301
     0 (35)
≈⌊ − 1.5115 ⌋
⌈         ⌉
  − 0.8727
     0, (36)

which agrees with the direct differentiation.

The physical vector has not changed. Only the basis vectors used to report its coordinates have changed. A derivative of coordinate numbers is therefore not automatically the derivative of the physical vector itself.

Solution 2: A point fixed on a rotating turntable

Because the point is fixed in r,

vr = 0,     ar = 0.
(37)

The inertial velocity is

vi = ω × r (38)
= ⌊  ⌋
  0
⌈ 0⌉
  3 ×⌊ ⌋
 2
⌈0⌉
 0 (39)
= ⌊  ⌋
  0
⌈ 6⌉
  0 m/s. (40)

For constant ω,

ai = ω × (ω ×  r).
(41)

First,

        ⌊ 0⌋
        ⌈  ⌉
ω × r =   6  ,
          0
(42)

then

ai = ⌊0⌋
⌈ ⌉
 0
 3 ×⌊ 0⌋
⌈  ⌉
  6
  0 (43)
= ⌊− 18⌋
⌈    ⌉
  0
  0 m/s2. (44)

The acceleration is inward, toward the axis. Its magnitude is also

  2     2            2
ω  r = 3 (2) = 18 m/s .
(45)

When Newton’s law is rearranged to describe dynamics in the rotating frame, the corresponding apparent centrifugal acceleration is the negative of the rotational centripetal term:

|------------------------------------|
|                                  2 |
-acf,app-=-−-ω-×--(ω-×-r)-=-18ex-m/s--.-
(46)

It points outward.

Solution 3: Radial motion and Coriolis acceleration

The Coriolis contribution to inertial acceleration is

2ω × vr = 2(2ez) × (1.5ex) (47)
= 6ey m/s2. (48)

The rotational centripetal term is

ω × (ω × r) = −ω2r (49)
= −(2)2(2e x) (50)
= −8ex m/s2. (51)

Since ar = 0 and α = 0,

|----------------------|
|ai = − 8ex + 6ey m/s2.|
------------------------
(52)

Its magnitude is

        ∘ -----------  |--------|
∥ai∥ =    (− 8 )2 + 62 =|10 m/s2 .
                       ----------
(53)

Measured counterclockwise from the positive x axis, the vector lies in quadrant II:

                 (  )
       ∘      −1  6-     |-----∘-|
𝜃 = 180  − tan    8   ≈  143.13--.
(54)

In rotating-frame Newton equations the apparent Coriolis term has the opposite sign:

|----------------------------------|
|a      = − 2ω  × v  = − 6e  m/s2. |
--cor,app------------r-------y-------
(55)

This sign change is caused by algebraic rearrangement, not by a different cross-product convention.

Solution 4: Angular acceleration and the Euler term

The Euler term is

α× r = (0.8ez) × (3ex) (56)
= 2.4ey m/s2. (57)

The rotational centripetal term is

ω × (ω ×  r) = − ω2r = − (1.5)2(3)ex,
(58)

so

|--------------------|
|ac = − 6.75ex m/s2. |
---------------------
(59)

The point is fixed to the rotating arm, so vr = ar = 0. Therefore

|----------------------------|
|ai = − 6.75ex + 2.40ey m/s2.|
------------------------------
(60)

Its magnitude is

       √ -------------   |-----------|
∥ai∥ =   6.752 + 2.402 ≈  7.164 m/s2  .
                         ------------
(61)

The Euler term is tangential because it results from changing angular speed. The centripetal term is radial because it results from changing the direction of the tangential velocity. Hence they are perpendicular for planar rotation about a fixed axis.

Solution 5: Translating and rotating origin

Evaluate the terms one at a time.

The origin contribution is already given:

      ⌊      ⌋
         1.2
AO  = ⌈ − 0.4 ⌉.

          0
(62)

The relative acceleration is

     ⌊     ⌋
       0.05
ar = ⌈ 0.10 ⌉ .
        0
(63)

For the Coriolis term,

2ω × vr = 2⌊    ⌋
   0
⌈  0 ⌉
  0.5 ×⌊     ⌋
  0.3
⌈− 0.2⌉
   0 (64)
= ⌊     ⌋
  0.20
⌈ 0.30⌉
   0 m/s2. (65)

For the Euler term,

α× r = ⌊ 0 ⌋
⌈   ⌉
  0
 0.1 ×⌊ 4⌋
⌈  ⌉
  1
  0 (66)
= ⌊− 0.10⌋
⌈      ⌉
  0.40
    0 m/s2. (67)

For the rotational centripetal contribution,

                           |---------------|
                   ⌊  ⌋    ⌊       ⌋       |
                  2  4     | − 1.00       2 |
ω ×  (ω  × r) = − ω ⌈ 1⌉ =  ⌈ − 0.25 ⌉ m/s  .|
                     0     |   0           |
                           -----------------
(68)

Adding all five terms,

ai = ⌊ 1.2 ⌋
⌈     ⌉
 − 0.4
   0 + ⌊0.05⌋
⌈    ⌉
 0.10
   0 + ⌊ 0.20⌋
⌈     ⌉
  0.30
   0 + ⌊ − 0.10⌋
⌈      ⌉
  0.40
    0 + ⌊ − 1.00 ⌋
⌈       ⌉
  − 0.25
    0 (69)
= ⌊    ⌋
 0.35
⌈0.15⌉
   0 m/s2. (70)

This example is useful because no single term dominates by assumption. The inertial result emerges only after all kinematic contributions are assembled consistently.

Solution 6: A point fixed to the rotating Earth is not inertially stationary

At the equator, the perpendicular distance to the Earth’s spin axis is approximately RE. The inertial tangential speed is therefore

v = ωieRE (71)
= (7.292115 × 10−5)(6.378137 × 106) (72)
≈ 465.10 m/s. (73)

The rotational centripetal acceleration is

ac = ωie2R E (74)
≈ 0.03392 m/s2. (75)

Relative to g = 9.81 m/s2,

                       |-------|
100ac ≈ 100 0.03392-≈  0.346%  .
    g         9.81      ---------
(76)

Thus a point whose ECEF coordinates never change still travels around the Earth’s spin axis at hundreds of meters per second in inertial space. ECEF is therefore a rotating, non-inertial frame. This is precisely why Earth-rate and rotational terms enter terrestrial inertial navigation.

Solution 7: DCM derivative from angular velocity

At t = 0,

 i
Cr(0) = I.
(77)

The skew matrix is

        ⌊0  − 2  0⌋
        ⌈         ⌉
[ω ]× =   2   0   0  .
         0   0   0
(78)

Therefore

|--------⌊----------⌋--|
|          0  − 2  0   |
|C˙i(0) = ⌈ 2   0   0⌉ .|
|  r       0   0   0   |
-----------------------|
(79)

A forward-Euler step gives

Cri(Δt) ≈ I + Ċ ri(0)Δt (80)
= ⌊               ⌋
   1    − 0.02 0
⌈ 0.02     1    0⌉
   0      0    1. (81)

Now

           ⌊------------------⌋--|
           | 1.0004     0    0   |
(Cir)TCir = ⌈   0     1.0004  0⌉ .|
           |   0        0    1   |
           -----------------------
(82)

The matrix is already slightly non-orthogonal after one Euler step. Repeated steps accumulate this defect unless the integration method respects rotational geometry or the matrix is periodically re-orthonormalized.

For the exact differential equation

               T
C˙=  ΩC,     Ω   = − Ω,
(83)

consider

d
--
dt(CT C) = ĊT C + CT Ċ (84)
= (ΩC)T C + CT (ΩC) (85)
= CT ΩT C + CT ΩC (86)
= CT (ΩT + Ω)C (87)
= 0. (88)

Thus if CT C = I initially,

|----------|
|CT C  = I |
-----------
(89)

for all time in the exact solution.

Solution 8: Adding angular rates through nested frames

Angular velocity obeys the relative-rate chain rule. At the instant when the frames are aligned,

ω   =  ω  +  ω  .
  in    ie    en
(90)

Hence

|--------------------|
|      ⌊    ⌋        |
|         0          |
ωin =  ⌈0.05⌉  rad/s.|
|       0.10         |
----------------------
(91)

Similarly,

ωib =  ωin + ωnb,
(92)

so

|------⌊----⌋--------|
|       0.20         |
|ωib = ⌈0.05⌉  rad/s.|
|       0.10         |
----------------------
(93)

The geometric angular-velocity vectors may always be added in this way, but their numerical coordinate columns may only be added after being resolved in a common frame. If, for example, ωie were supplied in ECEF coordinates and ωnb in body coordinates, one or both would have to be transformed first.

This is the conceptual origin of the later navigation-frame rate

|----------------|
ωin-=--ωie +-ωen.-
(94)

Solution 9: Inertial terms versus apparent accelerations

For constant ω and a common origin,

a  = a  + 2ω ×  v +  ω × (ω × r ).
 i    r          r
(95)

Because the particle is force-free in inertial space,

ai = 0.
(96)

Therefore

ar = − 2ω × vr − ω ×  (ω × r).
(97)

From Exercise 3,

2ω  × vr = 6ey,
(98)

and

ω  × (ω × r) = − 8ex.
(99)

Thus

|--------------------|
ar = 8ex −  6ey m/s2.|
----------------------
(100)

The apparent centrifugal acceleration is

|------------------------------|
|acf,app = − ω × (ω × r) = 8ex. |
-------------------------------
(101)

The apparent Coriolis acceleration is

------------------------------
|a      = − 2ω ×  v  = − 6e .|
--cor,app------------r-------y--
(102)

Their signs are opposite because the transport equation first expresses inertial acceleration as a sum of kinematic contributions. To write Newton’s law as an equation for ar, those contributions are moved to the other side of the equation. The minus signs are therefore produced by rearrangement.

Solution 10: Implementation sanity check – degrees are not radians

Convert the physical rate to radians per second:

              |---------------|
ω =  15-π--=  0.2617994  rad/s .
       180    -----------------
(103)

The correct centripetal acceleration is

ac = ω2r (104)
= (0.2617994)2(1.2) (105)
≈ 0.08225 m/s2. (106)

The erroneous implementation instead computes

|--------------------------------|
|ac,wrong = (15 )2(1.2) = 270 m/s2. |
----------------------------------
(107)

The multiplicative error is

ac,wrong
  ac = (          )
  ---15----
  15π ∕1802 (108)
= (     )
  180-
   π2 (109)
≈ 3282.81. (110)

Useful safeguards include:

  1. require angular rates at software interfaces to use explicit SI-unit names such as omega_rad_s;
  2. add a unit test using a known case such as ω = 1 rad/s and r = 1 m, for which ac = 1 m/s2;
  3. add a degrees-to-radians conversion test with 180∘∕s = π rad/s;
  4. assert physically plausible rate bounds at interfaces where such bounds are known;
  5. compare the cross-product implementation ω × (ω × r) against the scalar −ω2r ⊥ for simple perpendicular test vectors;
  6. keep configuration files and telemetry metadata explicit about units rather than relying on comments or convention.

In inertial navigation, unit errors can enter inside repeated integrations and therefore become enormous state errors. Unit discipline is part of the physics implementation, not merely software style.

3 What these exercises establish

The examples above reinforce several results that will be reused throughout the strapdown series:

  1. A changing coordinate column does not necessarily imply a changing physical vector.
  2. A point fixed in a rotating frame generally has nonzero inertial velocity and acceleration.
  3. Coriolis acceleration appears whenever there is relative motion in a rotating frame.
  4. Euler acceleration appears when the frame angular velocity changes.
  5. The term ω×(ω×r) is inward in the inertial acceleration decomposition; the outward centrifugal term appears after rearranging Newton’s law in the rotating frame.
  6. A translating origin contributes its own inertial acceleration in addition to all rotational terms.
  7. DCM kinematics are generated by a skew-symmetric angular-rate matrix, and exact propagation preserves orthogonality.
  8. Angular-rate vectors from nested frames must be resolved in a common coordinate frame before their component columns are added.
  9. Earth-fixed coordinates are non-inertial even when a vehicle is motionless relative to the ground.

These points provide the mechanics foundation for INS03, where the measured accelerometer quantity will be distinguished from ordinary kinematic acceleration.

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]   H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.


"Strapdown Inertial Navigation Examples: Rotating-Frame Derivatives, Coriolis, and Centrifugal Acceleration" is owned by bloftin.
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