Quaternion differentiation - · PDF fileQuaternion differentiation Quaternion...
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Quaternion differentiation Quaternion differentiation’s formula connects time derivative of component of quaternion q(t) with component of vector of angular velocity W(t). Quaternion q(t)=(q 0 (t), q 1 (t) , q 2 (t) , q 3 (t)) determines attitude of rigid body moving with one fixed point, vector of angular velocity W(t)=( W x (t) , W y (t) , W z (t)) determines angular velocity of this body at point of time t. point of time. Norm of quaternion q(t) is unit, i.e. 1 ) ( ) ( ) ( ) ( 2 3 2 2 2 1 2 0 = + + + t q t q t q t q (1) Vector W(t) can be represented as quaternion with zero scalar part, i.e )) ( ), ( ), ( , 0 ( ) ( t W t W t W t W z y x = (2) Quaternion differentiation’s formula can be represented as ) ( ) ( 2 1 ) ( t q t W dt t dq = (3) Using quaternion multiplication rule )) ( ) ( ) ( ) ( ) ( ) ( ( 2 1 ) ( )) ( ) ( ) ( ) ( ) ( ) ( ( 2 1 ) ( )) ( ) ( ) ( ) ( ) ( ) ( ( 2 1 ) ( )) ( ) ( ) ( ) ( ) ( ) ( ( 2 1 ) ( 1 2 0 3 3 1 0 2 2 3 0 1 3 2 1 0 t q t W t q t W t q t W dt t dq t q t W t q t W t q t W dt t dq t q t W t q t W t q t W dt t dq t q t W t q t W t q t W dt t dq y x z x z y z y x z y x − + = − + = − + = + + − = (4) Below formula (3) brief and vigorous derivation follows. Let R 0 is any given vector (quaternion with zero scalar part) fixed in rigid body at initial moment of time t 0 , R t is the same vector (quaternion) at moment of time t. Then, obviously ) ( ) ( 1 0 t q R t q R t − = (5) If we differentiate (5) then dt t dq R t q t q R dt t dq dt dR t ) ( ) ( ) ( ) ( 1 0 1 0 − − + =
Transcript of Quaternion differentiation - · PDF fileQuaternion differentiation Quaternion...
Quaternion differentiation Quaternion differentiation’s formula connects time derivative of component of quaternion q(t) with component of vector of angular velocity W(t). Quaternion q(t)=(q0(t), q1(t), q2(t), q3(t)) determines attitude of rigid body moving with one fixed point, vector of angular velocity W(t)=( Wx(t), Wy(t), Wz(t)) determines angular velocity of this body at point of time t. point of time. Norm of quaternion q(t) is unit, i.e.
1)()()()( 23
22
21
20 =+++ tqtqtqtq
(1) Vector W(t) can be represented as quaternion with zero scalar part, i.e
))(),(),(,0()( tWtWtWtW zyx= (2) Quaternion differentiation’s formula can be represented as
)()(21)( tqtW
dttdq=
(3) Using quaternion multiplication rule
))()()()()()((21)(
))()()()()()((21)(
))()()()()()((21)(
))()()()()()((21)(
1203
3102
2301
3210
tqtWtqtWtqtWdt
tdq
tqtWtqtWtqtWdt
tdq
tqtWtqtWtqtWdt
tdq
tqtWtqtWtqtWdt
tdq
yxz
xzy
zyx
zyx
−+=
−+=
−+=
++−=
(4) Below formula (3) brief and vigorous derivation follows. Let R0 is any given vector (quaternion with zero scalar part) fixed in rigid body at initial moment of time t0, Rt is the same vector (quaternion) at moment of time t. Then, obviously
)()( 10 tqRtqRt
−= (5) If we differentiate (5) then
dttdqRtqtqR
dttdq
dtdRt )()()()( 1
01
0
−− +=
(6) From (5) and (6) we have
dttdqtqRRtq
dttdq
dtdR
ttt )()()()( 1
1−
− +=
(7) Because norm of quaternion q(t) is unit, i.e.
1)()( 1 =− tqtq (8) we have
0)()()()( 11 =+
−−
dttdqtqtq
dttdq
(9) It follows from (7) and (9) that
)()()()( 11 tqdt
tdqRRtqdt
tdqdt
dRtt
t −− −=
(10) Let
)()()( 1 tqdt
tdqtp −=
(11) Obviously
))(),(),(()())(())(()( 2210111 tqtqtqtqtqVtqStq −=+= −−−
(12) where S() =scalar part, V()=vector part of quaternion Scalar part S(p(t)) of quaternion p(t) equals
0)()()()(
))(( 33
22
11
00 =+++= q
dttdq
qdt
tdqqdt
tdqqdt
tdqtpS
because norm of quaternion q(t) is unit. It follows that p(t) is vector. Because Rt is also vector
])([2)()( ttt RtptpRRtp =− (13) where [ab]= cross-product of vector a and vector b.
On the other hand, variable dt
dRt is velocity of point of rigid body with one fixed point. Hence
])([ tt RtW
dtdR
=
(14). Because Rt is arbitrary vector, it follows from (10), (11), (13) and (14)
)()(2)(2)( 1 tqdt
tdqtptW −==
(15) It follows from (15) finally that
)()(21)( tqtW
dttdq=
(3) It is necessary to draw attention: in formula (3) angular velocity vector W(t)=(WX(t), WY(t), WZ(t)) is represented by projections on axes of unmoving system of coordinates. If we apply the projections on axes of moving system of coordinates for the same angular velocity vector then obviously:
)())(),(),(,0)(())(),(),(,0( 1 tqtWtWtWtqtWtWtW zyxZYX−=
(16) Where Wx(t), Wy(t), Wz(t) are projections of angular velocity vector on axes of moving system of coordinates. It follows from (3) and (16) that
))(),(),(()(
)()(21)(
tWtWtWtW
tWtqdt
tdq
zyx=
=
(17) Finally, consider example of formulas (3) and (17) using for so-called conic moving. In this case quaternion q(t) equals:
))0),cos(),sin((2
sin,0()(
))0),sin(),(cos(2
sin,2
(cos)(
))0),sin(),(cos(2
sin,2
(cos)(
1
ttdt
tdq
tttq
tttq
ωωβω
ωωββ
ωωββ
−⎟⎠⎞
⎜⎝⎛=
⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛=
⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛=
−
(18)
Accordingly (3) and (17) projections of angular velocity vector on axes of unmoving and moving systems of coordinates equals:
))1)(cos(),cos()sin(),sin()sin(()()(2))(),(),((
)))cos(1(),cos()sin(),sin()sin(()()(2))(),(),((
1
1
−−==
−−==
−
−
βωωβωωβω
βωωβωωβω
ttdt
tdqtqtWtWtW
tttqdt
tdqtWtWtW
zyx
ZYX
