Strapdown Inertial Navigation: Time Derivatives in Rotating Frames

Strapdown Navigation: Time Derivatives in Rotating Frames

A vector can have the same geometric meaning while its numerical components change simply because the coordinate basis is rotating. This fact is the source of some of the most recognizable terms in inertial navigation: Earth-rate corrections, transport rate, Coriolis acceleration, and the distinction between inertial and Earth-fixed dynamics.

INS01 developed coordinate transformations at one instant of time. INS02 asks the next question:

|--------------------------------------------------------------------------| What---happens-when--the-basis-itself changes-while-we-differentiate-a-vector?

(1)

The answer is the transport theorem. For a physical vector a observed from an inertial frame i and a rotating frame r,

|----------------------------| |( da) ( da) | | --- = --- + ωir × a.| ---dt--i-----dt--r-----------

(2)

Here ω_ir is the angular velocity of frame r relative to frame i. This equation will be derived from the motion of the basis vectors rather than quoted as a rule.

Applying the theorem twice to position leads to

|------------------------------------------| -ai =-ar +-2ω-×-vr-+-ω˙×--r +-ω-×-(ω-×-r-),|

(3)

for frames sharing an origin. Coriolis, Euler, and rotational centripetal terms therefore arise from ordinary differentiation in a moving basis; they are not arbitrary corrections added to Newton’s laws.

This result is central to strapdown navigation. Earth-fixed coordinates rotate relative to inertial space, local-level coordinates rotate as a vehicle moves over the Earth, and the body frame rotates with the vehicle. The navigation equations are built by applying this same geometric theorem repeatedly [1, 2, 3].

Figure. An inertial basis i and a rotating basis r. Even when a physical vector is unchanged, its components in the rotating basis can vary because the basis vectors themselves rotate with angular velocity ω_ir.

1 Learning objectives

After completing this entry, the reader should be able to:

distinguish a time derivative of a physical vector from the time derivative of its coordinate components;
derive the first-order change of a vector under an infinitesimal rotation;
prove that a rotating unit basis vector satisfies
$( der) --k- = ωir × erk; dt i$
derive the transport theorem from a rotating orthonormal basis;
write the transport theorem in skew-matrix and coordinate form;
derive the relationship between relative and inertial velocity;
derive the complete rotating-frame acceleration relationship;
identify Coriolis, Euler, and rotational centripetal terms and distinguish them from the apparent centrifugal term;
extend the result to a rotating frame whose origin also translates;
explain why a point fixed in a rotating frame can have nonzero inertial velocity and acceleration;
explain why a vector fixed in inertial space can have changing coordinates in a rotating frame;
connect the transport theorem to DCM time derivatives and later strapdown attitude kinematics;
identify where Earth rotation and navigation-frame transport will enter the inertial-navigation equations.

2 Why an ordinary derivative is not enough

Suppose a physical vector a is expressed in the rotating basis

ℛ = {er1,er2,er3}.

(4)

Then

a = ar1er1 + ar2er2 + ar3er3.

(5)

Differentiate this expression as seen from inertial frame i:

( ) ( ) da d ∑3 r r --- = -- akek . dt i dt k=1 i

(6)

Applying the product rule gives

( ) ( ) da ∑3 r r ∑ 3 r derk --- = a˙kek + ak ---- . dt i k=1 k=1 dt i

(7)

There are therefore two distinct causes of change:

the scalar components a_k^r can change;
the basis vectors e_k^r can rotate.

The first term is what an observer attached to frame r calls the derivative of a. The second term is the contribution caused purely by the rotating basis.

This distinction is the central idea of the article.

3 Infinitesimal rotation of a vector

Consider a vector e attached to a rigid rotating frame. Over a small time interval Δt, suppose the frame undergoes the small rotation vector

Δ 𝜃.

(8)

The direction of Δ𝜃 is the instantaneous rotation axis and its magnitude is the small rotation angle in radians.

To first order, the change of any vector rigidly attached to the rotating frame is

---------------------------- | 2 | Δe--=-Δ-𝜃-×-e-+-O-(∥Δ-𝜃∥-).-

(9)

The direction follows from the right-hand rule. The cross product is perpendicular to both the rotation axis and the original vector, which is exactly the tangent direction in which the vector tip moves on the unit sphere.

Divide by Δt:

Δe--= Δ-𝜃-× e + O(Δt ). Δt Δt

(10)

In the limit Δt → 0,

Δ-𝜃- Δlitm→0 Δt = ωir,

(11)

|(---)-------------| | de- = ω × e.| | dt i ir | -------------------

(12)

Applied to every rotating basis vector,

--------------------- |( r) | | dek- = ωir × erk.| ---dt---i------------|

(13)

This basis-vector derivative is the geometric engine behind the transport theorem.

4 Derivation of the transport theorem

Return to

∑ 3 a = arkerk. k=1

(14)

The inertial derivative is

( ) ∑3 ∑ 3 ( r) da- = a˙r er+ ar dek- . dt i k=1 k k k=1 k dt i

(15)

Using the rotating-basis derivative,

( ) ∑3 ∑ 3 da- = a˙rkerk + ark (ωir × erk). dt i k=1 k=1

(16)

Because the cross product is linear,

∑3 r r ∑3 r r a k (ωir × ek) = ωir × a kek. k=1 k=1

(17)

Since the sum reconstructs the physical vector,

∑3 r r ak (ωir × ek) = ωir × a. k=1

(18)

Define

( ) 3 da- ∑ r r dt ≡ ˙akek. r k=1

(19)

Therefore

|(---)-----(---)-------------| | da- = da- + ωir × a.| | dt i dt r | -----------------------------

(20)

This is the transport theorem, also called the rotating-frame derivative theorem.

Figure. The inertial change of a vector contains two pieces: change of its components relative to the rotating frame and change caused by rotation of the basis itself.

5 Coordinate form of the transport theorem

INS01 introduced the skew-symmetric cross-product matrix

[ω] a = ω × a. ×

(21)

If every quantity is resolved in frame r, then the transport theorem becomes

|---------------------------| [( ) ]r | | da- = a˙r + [ωrir]×ar. | ----dt--i--------------------

(22)

The dot on a^r means the derivative of the coordinate column while the r basis is regarded as fixed. The second term restores the motion of that basis relative to inertial space.

This equation is often what actually appears in navigation software, but its meaning is clearest only after the geometric derivation above.

6 A diagnostic example: a vector fixed in inertial space

Suppose a is physically constant in inertial space. Then

( ) da --- = 0. dt i

(23)

The transport theorem gives

( ) da- 0 = dt + ωir × a. r

(24)

Hence

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

(25)

So an inertially fixed vector appears to rotate backward when observed from the rotating frame.

This is exactly what should happen. If a turntable rotates counterclockwise beneath an inertially fixed arrow, an observer standing on the turntable sees that arrow rotate clockwise.

The minus sign is therefore physical, not merely algebraic.

7 Velocity from the transport theorem

Let r be the position vector of a point measured from the common origin of frames i and r. Apply the transport theorem to r:

( dr) ( dr) --- = --- + ωir × r. dt i dt r

(26)

Define

( ) vi ≡ dr- , dt i

(27)

and

( ) v ≡ dr- . r dt r

(28)

Then

|v-=--v--+-ω--×--r.| --i----r-----ir-----|

(29)

This has a direct physical interpretation:

v_r is motion relative to the rotating frame;
ω_ir × r is the velocity caused solely by carrying the point around with the rotating coordinates.

7.1 Point fixed to a turntable

If the point is bolted to the rotating frame,

vr = 0.

(30)

Nevertheless,

v--=-ω---×-r.| --i----ir-----

(31)

Thus “at rest in the rotating frame” does not mean “at rest inertially.”

8 Acceleration: applying the theorem a second time

Differentiate the inertial velocity

vi = vr + ω × r,

(32)

where, for compactness in this section,

ω ≡ ωir.

(33)

The inertial acceleration is

( ) dvi- ai = dt . i

(34)

Therefore

( ) ( ) dvr- d- ai = dt + dt (ω × r). i i

(35)

Treat the two terms separately.

8.1 Derivative of the relative velocity

Apply the transport theorem to v_r:

( ) ( ) dvr- dvr- dt = dt + ω × vr. i r

(36)

Define

( ) dvr- ar ≡ dt . r

(37)

Then

( dvr ) ---- = ar + ω × vr. dt i

(38)

8.2 Derivative of the rotational velocity term

Using the product rule for a cross product,

( ) ( ) ( ) d- (ω × r) = dω- × r + ω × dr- . dt i dt i dt i

(39)

For the angular velocity vector itself,

( ) ( ) dω- dω- dt = dt + ω × ω. i r

(40)

But

ω × ω = 0,

(41)

so the inertial and rotating derivatives of this angular velocity vector are equal:

( ) ( ) dω- dω- dt = dt ≡ ˙ω. i r

(42)

Also,

( ) dr- dt = vr + ω × r. i

(43)

Therefore

( d ) -- (ω × r) = ω˙× r + ω × (vr + ω × r) . dt i

(44)

Expanding the last cross product,

( ) d- dt (ω × r) = ˙ω × r + ω × vr + ω × (ω × r). i

(45)

8.3 Collecting the terms

Substitution gives

ai = ar + ω × vr + ˙ω × r + ω × vr + ω × (ω × r).

(46)

Thus

|------------------------------------------| |ai = ar + 2ω × vr + ω˙× r + ω × (ω × r ).| -------------------------------------------

(47)

This is the fundamental acceleration relationship for two frames that share an origin.

Figure. The acceleration decomposition for a point observed from a rotating frame. Relative acceleration is supplemented by Coriolis, Euler, and rotational terms because the observer’s basis is itself moving.

9 Interpreting each acceleration term

The equation

ai = ar + 2ω × vr + ˙ω × r + ω × (ω × r)

(48)

contains four physically different contributions.

9.1 Relative acceleration

|ar| ----

(49)

is the acceleration measured relative to the rotating coordinates themselves.

9.2 Coriolis kinematic term

|--------| -2ω-×-vr--

(50)

appears only when the object moves relative to the rotating frame. The factor of two arose because one ω × v_r term came from differentiating the relative velocity and another came from differentiating the rotational transport velocity.

That factor of two is therefore not mysterious; it has a precise product-rule origin.

9.3 Euler term

|------| -˙ω-×-r-|

(51)

appears when the rotating frame’s angular velocity itself changes. In mechanics this is commonly called the Euler acceleration. It is unrelated to Euler Angles despite the shared name.

9.4 Rotational centripetal term

|------------| |ω × (ω × r)| --------------

(52)

is present even for a point fixed in the rotating frame.

For rotation about the z axis and a position perpendicular to that axis,

ω ⋅ r = 0.

(53)

Use the vector triple-product identity:

ω × (ω × r) = ω(ω ⋅ r) − ω2r.

(54)

Hence

|---------------------| ω × (ω × r) = − ω2r. | ----------------------

(55)

It points inward. In the inertial acceleration decomposition it is therefore a centripetal contribution.

10 Why centrifugal acceleration seems to have the opposite sign

The sign becomes different when Newton’s second law is written as an equation for the acceleration observed in the rotating frame.

Suppose the frames share an origin and the inertial frame obeys

F = mai.

(56)

Substitute the rotating-frame decomposition:

F = m [ar + 2ω × vr + ω˙× r + ω × (ω × r )].

(57)

Solve for ma_r:

|-------------------------------------------------| mar--=-F-−--2m-ω-×-vr-−-m-ω˙×--r −-m-ω-×-(ω-×-r-).--

(58)

The terms moved to the right-hand side are often described as inertial or apparent forces.

Thus the rotating-frame apparent accelerations are

|------------------| |aCor = − 2ω × vr, | -------------------

(59)

|----------------| |aEuler = −ω˙ × r,| ------------------

(60)

and

|--------------------| |acf = − ω × (ω × r).| ----------------------

(61)

Since the double-cross term points inward, its negative points outward. This outward term is the familiar centrifugal acceleration.

Therefore:

|------------------------------------------------------------------| |in inertial acceleration decomposition: + ω × (ω × r) is inward, | | in rotating-frame Newton equation: − ω × (ω × r) is outward. | --------------------------------------------------------------------

(62)

Keeping track of which equation is being written eliminates much of the usual sign confusion.

11 Generalization to a translating and rotating origin

So far the two frames have shared an origin. A more general moving frame has an origin O that translates relative to inertial frame i.

Let

RO

(63)

be the inertial position of the moving origin, and let

ρ

(64)

be the position of point P measured from O in the rotating frame. Then

R = R + ρ. P O

(65)

Differentiate once:

|--------------------------| | i i r | V-P-=--V-O-+-vP∕O-+-ω-×--ρ.-

(66)

Differentiate again:

-------------------------------------------------------- | i i r r | A-P-=--A-O-+-aP∕O-+-2ω--×-vP∕O-+-ω˙-×-ρ-+-ω-×-(ω--×-ρ).-

(67)

The new term

|--i-| -A-O-|

(68)

accounts for translational acceleration of the rotating frame’s origin.

This general form is important conceptually for local navigation frames. A local NED frame not only rotates; its origin moves with the vehicle over the curved Earth. Later lessons will organize those effects into Earth rate, transport rate, gravity, and position kinematics.

12 Turntable example: where Coriolis curvature comes from

Consider a horizontal turntable rotating at constant angular velocity

ω = ωez.

(69)

A puck moves radially outward relative to the turntable with

vr = uer.

(70)

The Coriolis contribution in the inertial acceleration decomposition is

2ω × v = 2(ωe ) × (ue ). r z r

(71)

Therefore

2ω × vr = 2ωue 𝜃.

(72)

In the rotating-frame Newton equation the corresponding apparent acceleration is

|----------------| -aCor-=-−-2ωue-𝜃.|

(73)

Thus an observer on the turntable sees the freely moving puck deflect sideways even when no real sideways force acts on it.

Figure. A puck moving across a rotating turntable. In inertial space its motion follows ordinary Newtonian dynamics; in rotating coordinates the same trajectory appears curved, and the rotating-frame equation contains the Coriolis term.

12.1 Numerical scale

Let

ω = 2.0 rad∕s, u = 1.0 m ∕s.

(74)

Then the Coriolis acceleration magnitude is

2 |aCor| = 2ωu = 4.0 m ∕s .

(75)

At radius

r = 0.50 m,

(76)

the centrifugal acceleration magnitude is

|acf| = ω2r = 2.0 m ∕s2.

(77)

These are large because the turntable rate is deliberately high. Earth’s rotation is much slower, but inertial navigation integrates small accelerations for long periods, so even modest rotating-frame terms matter.

13 A point fixed to a uniformly rotating frame

A second diagnostic case is even simpler. Suppose

vr = 0, ar = 0, ω˙ = 0.

(78)

The inertial velocity is

vi = ω × r,

(79)

and the inertial acceleration is

|------------------| |a = ω × (ω × r).| --i----------------

(80)

For r ⊥ ω,

|-----------| | 2 | ai-=-−-ω-r.--

(81)

The point therefore undergoes ordinary centripetal acceleration even though every coordinate of the point is constant in the rotating frame.

This example is a useful reminder:

|------------------------------------------------------------| -constant-coordinates-do-not-imply--zero-physical acceleration.

(82)

14 Connection with direction cosine matrices

INS01 defined the passive DCM C_rⁱ that maps rotating-frame coordinates into inertial coordinates:

ai = Cirar.

(83)

The columns of C_rⁱ are the rotating basis vectors expressed in inertial coordinates. Since each basis vector obeys

( ) derk r dt-- = ωir × ek, i

(84)

the entire DCM satisfies

---------------- |˙i i i | C-r-=-[ωir]×C-r.-

(85)

Differentiate

i i r a = C ra :

(86)

˙ai = C˙irar + Cir˙ar.

(87)

Using the DCM derivative then gives

˙ai = [ωiir]×Cirar + Cira˙r.

(88)

Since

Cirar = ai,

(89)

this is the coordinate-matrix form of the same transport theorem.

For the inverse DCM,

Cr = (Ci )T, i r

(90)

|----------------| |˙r r r | C-i-=-−-[ω-ir]×C-i.-

(91)

These identities are the direct bridge from rotating-frame mechanics to strapdown attitude propagation. INS07 will extend them to the case in which both the body frame and navigation frame rotate relative to inertial space.

15 Where this enters inertial navigation

The transport theorem appears repeatedly in the navigation problem.

15.1 Body frame

The IMU is physically attached to frame b. Its axes rotate with the vehicle. Gyroscope measurements describe that rotation, and attitude propagation determines how body-resolved measurements are transformed into a navigation frame.

15.2 Earth-fixed frame

Frame e rotates relative to inertial frame i with Earth angular rate

ωie.

(92)

Consequently, derivatives in ECEF coordinates are not inertial derivatives.

15.3 Local navigation frame

The local-level frame n rotates both because Earth rotates and because the vehicle moves across the curved Earth. Its total inertial angular rate is eventually decomposed as

|-n-----n-----n--| ω-in =-ωie +-ωen,-

(93)

where

ω_ieⁿ is Earth rotation resolved in n;
ω_enⁿ is navigation-frame transport rate relative to Earth.

INS06 and INS13 will derive these quantities in detail.

15.4 Velocity equation

Repeated use of the transport theorem ultimately produces the rotating-frame terms in the local-level strapdown velocity equation introduced in INS00:

n n b n n n n ˙veb = Cb fib + g − (2ωie + ωen) × veb.

(94)

INS02 does not yet derive this complete navigation equation. Its purpose is more fundamental: every rotating-frame term in that equation should now have a mechanical origin rather than appearing as unexplained notation.

Figure. The transport theorem is the common geometric source of the rotating-frame corrections that later appear in attitude and velocity mechanization.

16 Common sign and notation traps

16.1 Confusing ω_ir with ω_ri

By definition,

ωri = − ωir.

(95)

The transport theorem written here uses the angular velocity of the rotating frame r relative to inertial frame i:

( ) ( ) da da --- = --- + ωir × a. dt i dt r

(96)

Reversing the angular-rate subscripts reverses the sign.

16.2 Calling the double-cross term centrifugal in every equation

ai = ar + ω × (ω × r) + ⋅⋅⋅ ,

(97)

the double-cross term points inward for a perpendicular radius and is the physical centripetal contribution.

The outward centrifugal term appears only after this contribution is moved to the rotating-frame side of Newton’s law.

16.3 Forgetting the factor of two in Coriolis acceleration

The factor of two comes from two different derivatives. One term arises when v_r is differentiated in a rotating basis and the second when ω × r is differentiated.

16.4 Forgetting translational acceleration of the moving origin

The simple four-term relation assumes a common origin. If the rotating frame origin translates, add A_Oⁱ explicitly.

16.5 Mixing coordinate derivatives with physical-vector derivatives

The expression

a˙r

(98)

means the derivative of the r-resolved coordinate column. It is not generally the inertial derivative of the physical vector.

17 Implementation checks suggested by the physics

Even before writing a complete inertial navigator, several tests should be used to validate rotating-frame code.

17.1 Inertially fixed vector test

Choose a constant inertial vector and rotate the coordinate frame at known ω. The numerical rotating-frame coordinates should satisfy

˙ar = − [ωr ] ar. ir×

(99)

17.2 Rigidly rotating radius test

For a point fixed at radius r in a frame rotating at constant rate ω, verify

|vi| = ωr

(100)

and

2 |ai| = ω r.

(101)

The acceleration must point toward the rotation axis.

17.3 Radial-motion Coriolis test

Give a point known radial velocity u in a frame rotating at ω and verify that the Coriolis magnitude is

2ωu.

(102)

17.4 DCM derivative test

For a frame rotating at constant known angular velocity, verify numerically that

˙i i i Cr − [ωir]×Cr

(103)

approaches zero at the expected integration order.

These tests will later become useful unit tests for strapdown mechanization software.

18 Where INS03 begins

INS02 has explained how derivatives change in rotating coordinates, but it has not yet answered a different and equally important question:

|--------------------------------------------------------------------| -What--physical-acceleration-does-an-accelerometer--actually-measure?--|

(104)

A stationary accelerometer on a table reports approximately one g, while an accelerometer in ballistic free fall reports approximately zero. This seems paradoxical if one assumes an accelerometer directly measures kinematic acceleration.

INS03 will resolve that issue using proof-mass mechanics and Newton’s second law. The result will be the specific-force relationship

|----------| |f = a − g,| ------------

(105)

which provides the translational input used by a strapdown inertial navigator.

19 Summary

For any physical vector a observed from inertial frame i and rotating frame r,

|----------------------------| |( da) ( da) | | --- = --- + ωir × a.| ---dt--i-----dt--r-----------

(106)

The result follows because every rotating basis vector satisfies

|(----)--------------| | derk r | | ---- = ωir × ek.| ---dt---i------------

(107)

Applying the theorem to position gives

|----------------| -vi =-vr +-ω-×-r.-

(108)

Applying it a second time gives

|------------------------------------------| |ai = ar + 2ω × vr + ω˙ × r + ω × (ω × r).| -------------------------------------------

(109)

For a translating rotating origin O,

|-i------i----r------------r---------------------------| A-P-=--A-O-+-aP∕O-+-2ω--×-vP∕O-+-ω˙-×-ρ-+-ω-×-(ω--×-ρ).-

(110)

The same basis kinematics also gives the DCM derivative

|-i-----i----i-| C˙r-=-[ωir]×C-r.-

(111)

These relationships are the Newtonian rotating-frame foundation on which Earth-fixed and local-level strapdown navigation is built.

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] Herbert Goldstein, Charles Poole, and John Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.

[5] Donald T. Greenwood, Principles of Dynamics, 2nd ed., Prentice-Hall, 1988.

"Strapdown Inertial Navigation: Time Derivatives in Rotating Frames" is owned by bloftin.
(view preamble)

Strapdown Navigation: Time Derivatives in Rotating Frames

1 Learning objectives

2 Why an ordinary derivative is not enough

3 Infinitesimal rotation of a vector

4 Derivation of the transport theorem

5 Coordinate form of the transport theorem

6 A diagnostic example: a vector fixed in inertial space

7 Velocity from the transport theorem

7.1 Point fixed to a turntable

8 Acceleration: applying the theorem a second time

8.1 Derivative of the relative velocity

8.2 Derivative of the rotational velocity term

8.3 Collecting the terms

9 Interpreting each acceleration term

9.1 Relative acceleration

9.2 Coriolis kinematic term

9.3 Euler term

9.4 Rotational centripetal term

10 Why centrifugal acceleration seems to have the opposite sign

11 Generalization to a translating and rotating origin

12 Turntable example: where Coriolis curvature comes from

12.1 Numerical scale

13 A point fixed to a uniformly rotating frame

14 Connection with direction cosine matrices

15 Where this enters inertial navigation

15.1 Body frame

15.2 Earth-fixed frame

15.3 Local navigation frame

15.4 Velocity equation

16 Common sign and notation traps

16.1 Confusing ωir with ωri

16.2 Calling the double-cross term centrifugal in every equation

16.3 Forgetting the factor of two in Coriolis acceleration

16.4 Forgetting translational acceleration of the moving origin

16.5 Mixing coordinate derivatives with physical-vector derivatives

17 Implementation checks suggested by the physics

17.1 Inertially fixed vector test

17.2 Rigidly rotating radius test

17.3 Radial-motion Coriolis test

17.4 DCM derivative test

18 Where INS03 begins

19 Summary

References

16.1 Confusing ω_ir with ω_ri