DEM Modeling: Lecture 13 3D Rotations - Pharmahub...Quaternions •Describe an orientation using a...

C. Wassgren, Purdue University 1

DEM Modeling: Lecture 13

3D Rotations


3D Rotations• Newton’s 2nd Law:

• Easiest to put the rotational eqns of motion

in a body-fixed frame of reference (FOR)

aligned with the particle’s principle axes

– in a body-fixed FOR, the moments of inertia

don’t change due to changes in orientation

– aka Euler’s rotational eqns of motion

( )

( )

( )

( )

1

1

1

1

B B B B

B B B B

x x y z yy zz

xx

B B B B

y y z x zz xx

yy

B B B B

z z x y xx yy

zz

T I II

T I II

T I II

ω ω ω

ω ω ω

ω ω ω

− = ⋅ − × ⋅

= + −

= + −

= + −

ω I T ω I ωɺ

ɺ

ɺ

ɺ

where

• Ixx, Iyy, and Izz are the

principle moments of inertia

for the particle

• ωωωωB is the rotational speed of the particle in a body-fixed

FOR

• TB is the net torque acting

on the particle in the body-

fixed FOR

xG

yG

zG

zB

xB

yB

ωBx, T

Bnet,x

ωBy, T

Bnet,y

ωBz, T

Bnet,z

( )G Gd

dt⋅ =I ω T


3D Rotations…

• How should we describe the orientation of an object in

3D space?

• Three common methods

– direction cosines

– Euler angles

– quaternions

xG

yG

zG

zB

xB

yB

superscript “B” = body-fixed FOR

superscript “G” = global FOR


Direction Cosines

• Use the cosines of the angles that the body-fixed axes

make with respect to the global axes to describe the

orientation of the object

xG

yG

zG

xBB Gx yθ

B Gx xθB Gx z

θ

ˆ ˆ ˆ ˆcos cos cosB G B G B GB G G Gx x y zx x x y x z

θ θ θ= + +e e e e

cos cos cosˆ ˆ

ˆ ˆcos cos cos

ˆ ˆcos cos cos

B G B G B G

B G B G B G

B G B G B G

BG

B Gx x x y x zx x

B G

y yy x y y y zB G

z zz x z y z z

R

θ θ θ

θ θ θ

θ θ θ

=

=

e e

e e

e e

��

RBG ≡ rotation matrix from the Global to the Body FOR


Direction Cosines…

• The rotation matrix RBG is orthonormal

– the basis vectors forming the rows and columns of the matrix are

mutually perpendicular

– the magnitude of the basis vectors formed by these rows and

columns have a magnitude of one

– ⇒ RGB = (RBG)-1 = (RBG)T

– although the matrix RGB has nine terms, only three are

independent

ˆ ˆ

ˆ ˆ

ˆ ˆ

G B

x x

G GB B

y y

G B

z z

R

=

e e

e e

e e



• Example

– The unit vectors for the body-fixed FOR written in terms of the

global FOR are:

– Sketch both the body-fixed and global FOR axes. Let the global

and body-fixed FORs share the same origin.

– Write the rotation matrix for the transformation from the body-

fixed FOR to the global FOR.

– Express the vector vB = (0, 1, 0) which is given in the body-fixed

FOR in terms of the global FOR using the rotation matrix

determined in the previous part.

ˆ ˆ

ˆ ˆ

ˆ ˆ

B Gx y

B Gy z

B Gz x

=

=

=

e e

e e

e e



• Solution

( ) ( )1

cos cos cos 0 1 0

cos cos cos 0 0 1

1 0 0cos cos cos

0 0 1

1 0 0

0 1 0

B G B G B G

B G B G B G

B G B G B G

x x x y x z

BG

y x y y y z

z x z y z z

TGB BG BG

R

R R R

θ θ θ

θ θ θ

θ θ θ

−

= =

= = =

zG

xG

yB

yG

xB

zB

0 0 1 0

1 0 0 1

0 1 0 0

G GB BR=

=

v v

zG

xG

yB

yG

xB

zBv 0

0

1

G

∴ =

v


Euler Angles

• Decompose the orientation into three successive

rotations about various coordinate axes (e.g. the 1-2-3

sequence shown below)

x0, x1

y0

z0

η

z1

y1

η

η

λ

y2

x2

z2, z3

x3

y3

λλ

x0

y0

z0

x0

y0

z0

z3

x3

y3

x1

ξ

z1z2

x2

ξ

ξ

y1, y2


Euler Angles…

• The full sequence of rotations may be expressed as a

single rotation matrix

1 10 0 0

1 0 0

0 cos sin

0 sin cos

R η ηη η

= = −

a a a

2 21 1 1

cos 0 sin

0 1 0

sin 0 cos

R

ξ ξ

ξ ξ

− = =

a a a

3 32 2 2

cos sin 0

sin cos 0

0 0 1

R

λ λλ λ

= = −

a a a

3 30 0 32 21 10 0R R R R= =a a a

30

c c c s s s c s s c s c

c s c c s s s s c c s s

s s c c c

R

ξ λ η λ η ξ λ η λ η ξ λξ λ η λ η ξ λ η λ η ξ λξ η ξ η ξ

+ − = − − + −


Euler Angles…

• There are twelve possible Euler angle sequences that

could be used:. 1-2-1 (e.g. x0y1x2), 1-3-1, 2-1-2, 2-3-2, 3-

1-3, 3-2-3, 1-2-3, 2-3-1, 3-1-2, 3-2-1, 2-1-3, and 1-3-2.

• The rotation matrix is orthonormal.

• The Euler angles used to achieve a final orientation are

not unique. For example, the rotation matrix is identical

for (η, ξ, λ) = (135°, 135°, 135°) and (η, ξ, λ) = (-45°, 45°, -45°).


Euler Angles…

• Example

– A vector vB = (0, 1, 0) is expressed in FOR B. The

Euler angles for converting from FOR A to FOR B are

(η, ξ, λ)BA = (90°, 90°, 0°). Sketch FORs A and B and the vector v. Find vector v’s components in FOR A.


Euler Angles…

• Solution

xA

yA

zAz1

y1

x1

ηBA = 90°z2

x2

y1z1

y2

x1ξBA = 90°

λBA = 0°z2

x2

yBy2

xB

zB

v

zA

xA

yB

yA

xB

zB


Euler Angles…

• Solution...

( ) ( )1 TB BA A A BA B BA BR R R

−= ⇒ = =v v v v v

c c c s s s c s s c s c

c s c c s s s s c c s s

s s c c c

0 1 0

0 0 1

1 0 0

BA BA BA BA BA BA BA BA BA BA BA BA

BA BA BA BA BA BA BA BA BA BA BA BA BA

BA BA BA BA BA

R

ξ λ η λ η ξ λ η λ η ξ λξ λ η λ η ξ λ η λ η ξ λξ η ξ η ξ

+ −

= − − + −

=

( )�

0 0 1 0

1 0 0 1

0 1 0 0

BTBA

A

x

y

z

R

v

v

v

==

=

v

��

0

0

1

A

xA

y

z

v

v

v

∴ = =

v


Euler Angles…

• Euler angle rate of change

– consider a small rotation about each

of the Euler angle sequence axes

– express the rate at which the rotation,

∆θθθθ, occurs in terms of the body-fixed FOR (frame 3)

1 20 ˆˆˆzyx ξη λ∆ = + +∆ ∆∆ e eeθ

3

2 3 3

1 2 2

2

3 3

0

2

1

ˆ

ˆ ˆ ˆsin cos

ˆ ˆ ˆcos sin

ˆ ˆ ˆcos sin

ˆ

ˆ

ˆ

x

z

y x y

x x z

x x y

y

z

λ λ

ξ ξ

λ λ

=

= = +

= = +

= −

e

e e e

e e e

e e

e

e

e

e

x0, x1

y0

z0

η

z1

y1

η

η

λ

y2

x2

z2, z3

x3

y3

λλ

x1

ξ

z1z2

x2

ξ

ξ

y1, y2


Euler Angles…

• Euler angle rate of change…

( ) ( ) ( )3 3 3 3ˆ ˆ ˆcos cos sin cos sin cos sinx y zη ξ λ ξ λ η ξ λ ξ λ η ξ λ∆ = ∆ + ∆ + −∆ + ∆ + ∆ + ∆θ e e e

� � � � � � �3

33 3 3ˆ ˆ ˆcos cos sin cos sin cos sinx y z

d d d d d d d

dt dt dt dt dt dt dtη η ηξ ξ λ

η ξ η ξ η λξ λ λ ξ λ λ ξ

= = == = ==

= + + − + + +

ω

θe e e

ɺ ɺ ɺɺ ɺ ɺ

3

3

3

cos sin 0cos cos

sin cos 0

tan cos tan sin 1

x

y

z

λ λη ωξ ξξ λ λ ωλ ξ λ ξ λ ω

− ∴ = −

ɺ

ɺ

ɺ

3

3

3

cos cos sin 0

cos sin cos 0

sin 0 1

x

y

z

ω ξ λ λ ηω ξ λ λ ξω ξ λ

= −

ɺ

ɺ

ɺ

• if ξ → π/2 or 3π/2, then cosξ → 0

• The Euler angle formulation is not

well posed for all Euler angles.

• Other Euler angle sequences (e.g.

1-2-1) will also have singularities at

particular angles.

• This behavior is sometimes referred

to as “gimbal lock.”


Quaternions

• Describe an orientation using a single

rotation about a unit vector

θ

yG

xG

zG

ˆ Gn

xB

yB

zB

• Unlike direction cosines and

Euler angles, a quaternion

uses four quantities to

describe the orientation: (θ, nxG, ny

G, nzG). One of the

quantities is not independent.

• An Euler angle rotation

sequence can be thought of

as a sequence of three

quaternion rotations.


Quaternions…

• Consider a complex number

( ) ( )1 1 1 1 1 1 1, cos , sinz x y r rθ θ= = =v

y2 2

1tan

r x y

y

xθ −

= +

= x

r1θ1

( )2 2 2cos ,sinz θ θ=

( )( )( ) ( )( )

1 2 1 1 1 1 2 2

1 1 2 1 1 2

cos , sin cos ,sin

cos , sin

z z r r

r r

θ θ θ θ

θ θ θ θ

′ = =

= + +

v

We can rotate the vector v about the origin by an angle θ2 by multiplying z1 by the complex number z2 as given below.

Let the 2D vector v be represented by the

complex number z1

v

y

xθ1

vv’θ2

∴∴∴∴2D rotations may be performed using complex numbers


Quaternions…

• (Unit) quaternions are “hypercomplex” numbers

that can be used to perform 3D rotations

– originally proposed by Hamilton (1843)

( ) ( )( ) ( ) ( ) ( )

0 1 2 3

2 2

1 2 32 2 2 2

ˆ

ˆcos , sin

cos sin sin sin

q q i q j q k

n i n j n k

θ θ

θ θ θ θ

= + + +

=

= + + +

q

n

2 2 2 1i j k

ij ji k

jk kj i

ki ik j

= = = −

= − =

= − =

= − =

i

jk+

i

jk–

θ yG

xG

zG

ˆGn

xB

yB

zB


Quaternions…

• Some handy quaternion properties

2 2 2 2

0 1 2 3

ˆ

q q q q= + + +

=

q

qq

q

[ ][ ] [ ]( )( )( ) ( )( ) ( )

0 1 2 3 0 1 2 3

0 0 1 1 2 2 3 3 0 1 1 0 2 3 3 2

0 2 1 3 2 0 3 1 0 3 1 2 2 1 3 0

, , ,s t st s t

a ia ja ka b ib jb kb

a b a b a b a b i a b a b a b a b

j a b a b a b a b k a b a b a b a b

= = − ⋅ + + ×

= + + + + + +

= − − − + + + − +

− + + + + − +

ab u v u v v u u v

( ) ( )( ) ( )

, in general

+ = +

+ + = + +

=

≠

a b b a

a b c a b c

a bc ab c

ab ba


Quaternions…

• Some handy quaternion properties…

[ ][ ] ( ) [ ] ( )

( )

1

10 1 2 3 0 1 2 3

11 1

1,

if , , , , , then , , , ,s q q q q s q q q q

−

−

−− −

=

= = = − = − − −

=

qq 0

q v q v

p q qp

[ ] [ ] ( ) ( ) ( )1

ˆA vector rotated about unit vector by an angle is given by:

ˆ ˆ ˆ ˆ ˆ0, 0, 0, cos sin 1 cos

θ

θ θ θ−′ = = + × + ⋅ −

v n

v q v q v n v v n n

[ ] ( )[ ]( )1 2

1 1

2 1 1 2

ˆ ˆTwo successive rotations (first rotate using then rotate using ):

ˆ ˆ ˆ ˆ0, 0, − −′′ =

q q

v q q v q q [ ] [ ]( )12 1

ˆ ˆ ˆ ˆ ˆequivalent to 0, 0, where −′′ = =v r v r r q q


Quaternions…

• Some handy quaternion properties…

[ ] [ ]

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

1

2 2

2 3 0 3 1 2 0 2 1 3

2 2

0 3 1 2 1 3 0 1 2 3

2 2

0 2 1 3 0 1 2 3 1 2

ˆ ˆ0, 0,

1 2 2 2

where 2 1 2 2

2 2 1 2

R

q q q q q q q q q q

R q q q q q q q q q q

q q q q q q q q q q

−′ ′= ⇒ =

− + − + + = + − + − + − + + − +

v q v q v v


Quaternions…

• Example

– A vector v = (1, 0, 0) is rotated about the y-

axis by an angle of 90º. Determine the (unit)

quaternion encoding this rotation and show

that the rotated vector is v’ = (0, 0, -1). Also

find the rotation matrix corresponding to the

quaternion.


Quaternions…

• Solutiony

x

z

θ = 90°

v’

v

( ) ( ) ( )

( )2 2

1 1

2 2

ˆ ˆ ˆcos ,sin where =90 , 0,1,0

ˆ ,0, ,0

θ θ θ = =

⇒ =

q n n

q

�

[ ] [ ] ( ) ( ) ( )

( ) ( ) ( ) ( ){ }( )

( )

1

0

ˆ ˆ ˆ ˆ ˆ0, 0, 0, cos sin 1 cos

0, 0,1,0 1,0,0 1,0,0 0,1,0 0,1,0

0, 0,0, 1

θ θ θ−

=

′ = = + × + ⋅ −

= + × + ⋅

= −

v q v q v n v v n n

0��

( )0,0, 1′∴ = −v


Quaternions…

• Solution

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

2 2

2 3 1 2 0 3 0 2 1 3

2 2

0 3 1 2 1 3 0 1 2 3

2 2

0 2 1 3 0 1 2 3 1 2

1 2 2 2

2 1 2 2

2 2 1 2

q q q q q q q q q q


q q q q q q q q q q

− + − + = + − + − + − + + − +

0 0 1

0 1 0

1 0 0

R

∴ = −

�

1

2

3

0 0 1 1 0

0 1 0 0 0

1 0 0 0 1

R

v

v

v

==

′ ′ = = ′ − −

v

��

Note: (same result as before)


Quaternions…

• Using quaternions to change FORs

– Let qGB be the unit quaternion that rotates FOR G to FOR B.

– Since the vector v does not rotate with the FOR, it’s as if v

rotates in the opposite sense of qGB (imagine the movement of

v if standing in FOR B).

– Thus, to express v in FOR B, rotate vG using the inverse

quaternion qBG = (qGB)-1.

( ) ( )( ) ( )

1

1

ˆ ˆ0, 0,

ˆ ˆ0,

B BG G BG

GB G GB

−

−

=

=

v q v q

q v q

xG

yG

zG

zB

yB xB

v

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 2

2 3 0 3 1 2 0 2 1 3

2 2

1 2 0 3 1 3 0 1 2 3

2 2

0 2 1 3 0 1 2 3 1 2

1 2 2 2

where 2 1 2 2

2 2 1 2

B GB G

GB GB GB GB GB GB GB GB GB GB

GB GB GB GB GB GB GB GB GB GB GB

GB GB GB GB GB GB GB GB GB GB

R

q q q q q q q q q q


q q q q q q q q q q

=

− + + − +

= − − + + + − + − +

v v


Quaternions…

• Example

– A vector vG = (0, 1, 0) is expressed in FOR G.

FOR B is found by rotating FOR G about the

vector (1, 0, 0)G (expressed in FOR G) by an

angle of 180°. Determine the (unit) quaternion encoding the rotation from FOR G

to FOR B. Express the vector v in FOR B.


Quaternions…

• Solution

yG

xG

zGθ = 180°

vxB

yB

zB

( ) ( ) ( )2 2

ˆ ˆcos ,sin 0,1,0,0GB Gθ θ = = q n

( ) ( ){ }( ) ( )( ){ }( )( )( )

1

ˆ ˆ0, 0,

0, 1,0,0 0,0,1,0 0,1,0,0

0, 1,0,0 0,0,0, 1

0,0, 1,0

B GB G GB−

=

= −

= − −

= −

v q v q

( )0, 1,0B∴ = −v


Quaternions…• Quaternion rate of change

– consider a rotation about the body-fixed axes over a short time ∆t

– ωωωωB(t) is the rate of rotation about the body-fixed axes at time t– qGB(t) is the unit quaternion to go from FOR B to FOR B at time t

– the new quaternion may be found by multiplying the old quaternion by the

quaternion describing the incremental rotation

( )( ) ( )

( )( )

( )ˆ ˆcos , sin2 2

G GG

GB GB

G

t t t ttt t t

t

∆ ∆ + ∆ =

ω ωωq q

ω

( ) ( ) ( )1

2ˆ 0,GB GB Bt t t ∴ = q q ωɺ

• there is no singularity in the quaternion rate of change (recall that there was

using Euler angles)

• only unit quaternions encode rotations so the quaternion should be

normalized after it has been updated to prevent numerical “drift” resulting

from numerical precision errors

Note: ( ) ( ) 1

ˆ ˆ0, 0,G GB B GB−

= ω q ω q


Quaternions…

• Example

– A cylinder’s body-fixed FOR (see below for the orientation of the

body-fixed FOR) is found in the global FOR using the quaternion

qGB = (0, 1/sqrt(2), 0, 1/sqrt(2)). The cylinder rotates about its

body-fixed axes with speed ωωωωB = (0, 0, 1). Sketch the cylinder’s orientation in the global FOR and determine the rate of change

of the cylinder’s quaternion at this instant in time.

yG

xG

zG

xB

yB

zB


Quaternions…

• Solution

yG

xG

zG

zB

yB

xB

Determine the orientation of the body-fixed z-axis

[ ] ( ) ( ) ( )( ) ( )( ){ }( )( )( )

1

1 1 1 1

2 2 2 2

1 1 1 1

2 2 2 2

1 1 1 12 2 2 2

ˆ ˆ0, 0, 0,0,1

0, ,0, 0,0,0,1 0, ,0,

0, ,0, ,0, ,0

0, ,0,

GB GB−

′ =

= − −

= −

= + − +

z q q

( )1,0,0′∴ =z the body-fixed z axis points along the

global x axis

Perform a similar analysis for the body-fixed x-axis

[ ] ( ) ( ) ( )( ) ( )( ){ }( ) ( )( )

1

1 1 1 1

2 2 2 2

1 1 1 1

2 2 2 2

1 1 1 12 2 2 2

ˆ ˆ0, 0, 1,0,0

0, ,0, 0,1,0,0 0, ,0,

0, ,0, ,0, ,0

0, ,0,

GB GB−

′ =

= − −

=

= − +

x q q

( )0,0,1′∴ =x

( )

( )1 1

2 2

ˆNote: 180 and 1,0,1

ˆ 0, ,0,

Gθ = =

⇒ =

n

q

�

ˆGnθ = 180º


Quaternions…

• Solution…

yG

xG

zGzB

yB

xB

( )( )12

1 1 12 2 2

ˆ0,

0,0,0, 0, ,0,

GB B GB =

=

q ω qɺ

( )1 1

2 2 2 2,0, ,0GB∴ = −qɺ

ωωωωB


Summary

• 3D orientations and rotations are more complex

than their 2D counterparts

• Euler angles suffer from “gimbal lock” at

particular angles – shouldn’t use for modeling

rotational motion

• Quaternions have many advantages and are

commonly used in DEM applications

– don’t suffer from gimbal lock

– easy to correct for numerical drift

– multiple rotations can be represented by quaternion

multiplication

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