Since dot products are commutative, one can see from equations above
RT = Q i.e. R −1 = RT = Q
The transformation given in equation Pxyz = RPuvw and P = QPXYZ is refer to as orthogonal transformation. UVW
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–问题1（运动学正问题）：已知关节角求机械手 末端执行器相对于参考坐标系的位置和姿态。 –问题2（运动学逆问题）：已知机械手末端执行 器相对于参考坐标系的位置和姿态求各个关节角。
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z
2.2 (Direct Kinematics Problem) 2.2.1 Rotation Matrices
o V
y
x
v v 设 p 在OUVW坐标系中静止并固定， p 可用OXYZ坐标系表示，也
可用OUVW坐标系表示。
v T pxyz = ( px p y pz )
v T puvw = ( pu pv p w )
v v p uvw → p xyz
There exists a transformation matrix We would like to find R.
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− sin θ cos θ 0
0 0 1
2.2.2 Composite Rotation Matrix α Example: A rotation of angle about the OX axis followed by a rotation of angle about the OZ axis θ ϕ followed by a rotation of angle about the OY axis.
1 0 Rxα = 0 cα 0 sα
U′ 1 Rxα = 0 0
Z
W
0 − sα = [ r1 r2 cα
pu p r3 ] v pw
Let us choose a point p fixed in the OUVW coordinate system to be (1, 0, 0)T
Z
W
p=
( pu
pv
p w)
T
V
Y O
U
p = ( pu
pv
pw )
T
X
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o
W
V y
– Reference coordinate system (OXYZ) – Rotated coordinate system (OUVW)
U x
z
W
P
U
A point P in the space can be represented by its coordinates with respect to both coordinate systems. 2011-3-7
iu k z p x jv k z p y kwk z pz
pu iu ix p =j i v v x p w k wi x
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iu j y jv j y kw j y
Q = R −1
the 3x3 rotation Rzθ for rotation about the OZ axis with θ
cos θ sin θ Rzθ = 0
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− sin θ cos θ 0
0 0 1
0 1 Rxα = 0 cos α 0 simα
v p= v p=
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p x ix iu p = j i y y u pz k z iu
Similarly
i x jv j y jv k z jv
R
ix k w pu p j y kw v k z k w pw
px p = r y 1 pz
The first column of the rotation matrix represents the principal axis iu with respect to the OXYZ coordinate system
旋转矩阵的几何解释： R = [r1 r2 r3 ] r1 —OUVW 坐标系中的U 轴单位矢量在OXYZ 坐标系中的各分量 r2 —OUVW 坐标系中的V 轴的单位矢量在OXYZ 坐标系中的各分量 r3 —OUVW 坐标系中的W 轴的单位矢量在OXYZ 坐标= 1 ix jv = ix k w = 0 j y iu = 0 j y jv = cos α k z iu = 0 k z jv = sin α 0 − sin α cos α k z k w = cos α
1 0 Rxα = 0 cα 0 sα 0 − sα cα
第二章 机器人运动学
(Robot Kinematics)
(Manipulator Kinematics)
哈尔滨工业大学控制科学与工程系 刘志远、 刘志远、刘海峰
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Degree of Freedom (DOF)
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end-effector
机器人各连杆视作刚体
End-effector
α
α
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p x ix iu p = j i y y u p z k z iu
i x jv j y jv k z jv
iu k w pu j y k w pv k z k w pw
v v p xyz = RPuvw
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v puvw = pu iu + pv jv + pw k w v pxyz = px ix + p y j y + pz k z v v puvw = pxyz v px = ix ⋅ puvw = ix ⋅ iu ⋅ pu + ix ⋅ jv ⋅ pv + ix ⋅ k w ⋅ pw v p y = j y puvw = j y ⋅ iu ⋅ pu + j y ⋅ jv ⋅ pv + j y ⋅ kw ⋅ pw r pz = k z puvw = k z ⋅ iu ⋅ pu + k z ⋅ jv ⋅ pv + k z ⋅ k w ⋅ pw
px p y pz
set
px 1 p = 0 y pz 0
X r11 r T R = 12 r13 1 ˆ ˆ r3 ] 0 = r1 0
Y r21 r22 r23
U V r11 R = r21 r31
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W r13 X r23 Y r33 Z
p = ( pu pv pw )
T
r12 r22 r32
column vectors
geometric interpretation of rotation matrices.
r1 = r2 × r3
R中的约束条件： r3 = r1 × r2 r2 = r3 × r1
r1 ⋅ r1 = r2 ⋅ r2 = r3 ⋅ r3 = 1
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det R = 1, det R = r1T (r2 × r3 )
例 (Example): Determine a transformation matrix Rxα for rotation about the ox axis with α angle.
g2 (t)
Joint angle
Link
g1 ( t )
Actuator
关节角
g (t ) = [ g1 (t ) g 2 (t ) K g n (t )]T
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g(t) = [g1(t)
g2 (t)]
T
。若为n自由度的机械手则
2.1 引言 引言(Introduction) Two problems: – Problem l (direct ( or forward ) kinematics problem ) : For a given manipulator, given the joint angle vector and the geometric link parameters where n is the number of degree of freedom , what is the position and orientation of the endeffector of the manipulator with respect to a reference coordinate system. – Problem 2 (inverse kinematics (or arm solution ) problem): Given a desired position and orientation of the end-effector of the manipulator and the geometric link parameters with respect to a reference coordinate system, can the manipulator reach the desired prescribed manipulator hand position and orientation? And if can, how many different manipulator configurations will satisfy the same condition?