Transcript Denavit-Hartenberg Convention.ppt

Denavit-Hartenberg Convention
Denavit-Hartenberg Convention
• Number the joints from 1 to n starting with the base and ending with
the end-effector.
• Establish the base coordinate system. Establish a right-handed
orthonormal coordinate system ( X 0 , Y0 , Z 0 ) at the supporting base
with Z 0 axis lying along the axis of motion of joint 1.
• Establish joint axis. Align the Zi with the axis of motion (rotary or
sliding) of joint i+1.
• Establish the origin of the ith coordinate system. Locate the origin of
the ith coordinate at the intersection of the Zi & Zi-1 or at the
intersection of common normal between the Zi & Zi-1 axes and the Zi
axis.
• Establish Xi axis. Establish X i  (Zi 1  Zi ) / Zi 1  Zi or along the
common normal between the Zi-1 & Zi axes when they are parallel.
• Establish Yi axis. Assign Yi  (Zi  X i ) / Zi  X i to complete the righthanded coordinate system.
• Find the link and joint parameters
Example I
• 3 Revolute Joints
Z3
Z1
Z0
Y0
O3
Y1
Link 1
Joint 1
Joint 3
O0 X0
Joint 2
Link 2
O1 X1 O2 X2
Y2
a0
a1
X3
d2
Link Coordinate Frames
• Assign Link Coordinate Frames:
– To describe the geometry of robot motion, we assign a Cartesian
coordinate frame (Oi, Xi,Yi,Zi) to each link, as follows:
• establish a right-handed orthonormal coordinate frame O0 at
the supporting base with Z0 lying along joint 1 motion axis.
• the Zi axis is directed along the axis of motion of joint (i + 1),
that is, link (i + 1) rotates about or translates along Zi;
Z3
Z1
Z0
Y0
O3
Y1
Link 1
Joint 1
Joint 3
O0 X0
Joint 2
Link 2
O1 X1 O2 X2
Y2
a0
a1
X3
d2
Link Coordinate Frames
– Locate the origin of the ith coordinate at the intersection
of the Zi & Zi-1 or at the intersection of common normal
between the Zi & Zi-1 axes and the Zi axis.
– the Xi axis lies along the common normal from the Zi-1
axis to the Zi axis X i  (Zi 1  Zi ) / Zi 1  Zi , (if Zi-1 is
parallel to Zi, then Xi is specified arbitrarily, subject only
Z3
to Xi being perpendicular to Zi);
Z1
Z0
Y0
Joint 3
O3
Y1
X3
d2
Joint 1
O0 X0
Joint 2
O1 X1 O2 X2
Y2
a0
a1
Link Coordinate Frames
– Assign Yi  (Zi  X i ) / Zi  X i to complete the righthanded coordinate system.
• The hand coordinate frame is specified by the geometry
On
of the end-effector. Normally, establish Zn along the
direction of Zn-1 axis and pointing away from the robot;
establish Xn such that it is normal to both Zn-1 and Zn
axes. Assign Yn to complete the right-handed coordinate
system.
Z3
Z1
Z0
Y0
Joint 3
O3
Y1
X3
d2
Joint 1
O0 X0
Joint 2
O1 X1 O2 X2
Y2
a0
a1
Link and Joint Parameters
• Joint angle  i : the angle of rotation from the Xi-1 axis to
the Xi axis about the Zi-1 axis. It is the joint variable if joint i
is rotary.
• Joint distance d i : the distance from the origin of the (i-1)
coordinate system to the intersection of the Zi-1 axis and
the Xi axis along the Zi-1 axis. It is the joint variable if joint i
is prismatic.
• Link length ai : the distance from the intersection of the Zi-1
axis and the Xi axis to the origin of the ith coordinate
system along the Xi axis.
• Link twist angle  i : the angle of rotation from the Zi-1 axis
to the Zi axis about the Xi axis.
Example I
Z3
Z1
Z0
Y0
Joint 3
O3
Y1
X3
d2
Joint 1
O0 X0
Joint 2
O1 X1 O2 X2
Y2
a0
a1
 i : rotation angle from Zi-1 to Zi about Xi
ai : distance from intersection of Zi-1 & Xi
D-H Link Parameter Table
Joint i
i
ai
di
i
1
0
a0
0
0
2
-90
a1
0
1
3
0
0
d2
2
to origin of i coordinate along Xi
di
: distance from origin of (i-1) coordinate to intersection of Zi-1 & Xi along Zi-1
i
: rotation angle from Xi-1 to Xi about Zi-1
Example II: PUMA 260
1
2
Z1
3
O1
1.
Number the joints
2.
Establish base frame
3.
Establish joint axis Zi
4.
Locate origin, (intersect.
of Zi & Zi-1) OR (intersect
of common normal & Zi )
X1
Z2 Z6
5.
Y1
O
Y3 2
Z5
Z4
Y6
O3

X
6
Y2 2
5
O
6
Z0
Y5
Y
X3 4
t
O5
X5 X6
O4 Z 3
X4
PUMA 260
4
Establish Xi,Yi
X i  (Zi 1  Zi ) / Zi 1  Zi
Yi  (Zi  X i ) / Zi  X i
Link Parameters
1
J
2
1
Z1
2
3
O1
3
i
1
2
3
4
5
6
i
ai d i
-90 0
13
0
8
0
90
0
-l
8
4
-90 0
X1
Z2 Z6
Y1
5
90 0 0
Y3O2
Z5
Z4 6
Y6
0
0 t
O3

X
6
2
Y2
5  : angle from X to X
O
i-1
i
6
Z0
i
Y5
about Zi-1
X 3 Y4
O5
X5 X6
 i : angle from Zi-1 to Zi
O4 Z 3
about X
i
X4
Joint distance
4 ai : distance from intersection
of Zi-1 & Xi to Oi along Xi
d i : distance from Oi-1 to intersection of Zi-1 & Xi along Zi-1
Transformation between i-1 and i
• Four successive elementary transformations
are required to relate the i-th coordinate frame
to the (i-1)-th coordinate frame:
– Rotate about the Z i-1 axis an angle of i to align the
X i-1 axis with the X i axis.
– Translate along the Z i-1 axis a distance of di, to bring
Xi-1 and Xi axes into coincidence.
– Translate along the Xi axis a distance of ai to bring
the two origins Oi-1 and Oi as well as the X axis into
coincidence.
– Rotate about the Xi axis an angle of αi ( in the righthanded sense), to bring the two coordinates into
coincidence.
Transformation between i-1 and i
• D-H transformation matrix for adjacent coordinate
frames, i and i-1.
– The position and orientation of the i-th frame coordinate
can be expressed in the (i-1)th frame by the following
homogeneous transformation matrix:
Source coordinate
Ti i 1  T ( zi 1 , d i ) R ( zi 1 , i )T ( xi , ai ) R ( xi ,  i )
Reference
Coordinate
C i
 S
 i
 0

 0
 C i S i
C i C i
S i
0
S i S i
 S i C i
C i
0
ai C i 
ai S i 
di 

1 
Kinematic Equations
q  (q1 , q2 ,  qn )
• Forward Kinematics
– Given joint variables
– End-effector position & orientation Y  ( x, y, z,  , , )
n
• Homogeneous matrix T0
– specifies the location of the ith coordinate frame w.r.t.
the base coordinate system
– chain product of successive coordinate transformation
matrices of Ti i 1
T  T T T
n
0
Orientation
matrix
 R0n

0
1 2
0 1
n
n 1
Position
vector
P0n  n s a P0n 


1  0 0 0 1 
Kinematics Equations
• Other representations
– reference from, tool frame
0
Treftool  Bref
T0n H ntool
– Yaw-Pitch-Roll representation for orientation
T  Rz , R y , Rx ,
C
 S

0

0
 S 0 0  C
C 0 0  0
0
1 0    S

0
0 1  0
0 S
1 0
0 C
0
0
0
0
0

1
1 0
 0 C

0 S

0 0
0
 S
C
0
0
0
0

1
Solving forward kinematics
• Forward kinematics
1 
 
 2
 3 
 
 4 
 5 
 
 6 
 px 
p 
 y
 pz 
 
 
 
 
 
• Transformation Matrix
 nx
n
T  y
 nz

0
sx
sy
sz
ax
ay
az
0
0
px 
p y 
pz 

1
Solving forward kinematics
• Yaw-Pitch-Roll representation for orientation
CC CSS  SC
 SC SSS  CC
T0n  
  S
CS

0
 0
 nx
n
T0n   y
 nz

0
Problem?
sx
sy
sz
ax
ay
az
0
0
px 
p y 
pz 

1
CSC  SS
SSC  CS
CC
0
  sin 1 (nz )
az
  cos (
)
cos 
nx
1
  cos (
)
cos 
1
Solution is inconsistent and ill-conditioned!!
px 
p y 
pz 

1
atan2(y,x)
y
x
 0    90
for  x and  y



for  x and  y
 90    180
  a tan 2( y, x)  



180




90
for  x and  y

  90    0
for  x and  y