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Lecture 2 (III-LT1)
Robot Kinematics (Ch. 5)
by
S.K. Saha
Aug. 07, 2015 (F)@JRL301 (Robotics Tech.)
PROPRIETARY MATERIAL. © 2014, 2008 The McGraw-Hill Companies, Inc. All rights reserved. No part of this PowerPoint slide may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this PowerPoint slide, you are using it without
permission.
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Pose  Position + Rotation
P
Z
W
p΄
p
Translation: 3
Rotation: 3
Total: 6
OM
V
M
o
O
Y
F
U
X
A moving body  Pose or Configuration
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Position Description
[p] F
p 
 x
 py 
p 
 z
. . . (5.8)
p = px x + py y + pz z
. . . (5.9)
0
0
1
 
 
 
[x]F  0 , [y ]F   1  , and [z ]F   0 
 
 
 0
0
 
 1 
 
. . . (5.10)
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Orientation Description
1. Direction cosine representation
2. Fixed-axes rotations
3. Euler angles representation
4. Single- and double-axes rotations
5. Euler parameters representation
I will illustrate the first TWO only
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Direction Cosine Representation
Refer to Fig. 5.12
p = puu + pvv + pww
. . . (5.12)
u = ux x + uy y + uz z
. . . (5.11a)
v = vx x + vy y + vz z
. . . (5.11b)
w = wx x + wy y + wz z
. . . (5.11c)
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Substitute eqs. (5.11a-c) into eq. (5.12)
p = (puux + pvvx + pwwx)x + (puuy + pvvy + pwwy)y
+ (puuz + pvvz + pwwz)z
. . . (5.13)
px = uxpu + vxpv + wxpw
. . . (5.14a)
py = uypu + vypv + wypw
. . . (5.14b)
pz = uzpu + vzpv + wzpw
. . . (5.14c)
[p]F = Q [p]M
. . . (5.15)
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[p]F = Q [p]M
. . . (5.15)




p 
p

 uT x vT x w T x 


u
v
w
 x
 x x x 

 u
T
T
T
[p]   p  , [p]M   p  , Q  u v w   u y v y w y 
F  y
 y y y  T
Tz wTz 
 v
u
z
v


 u v w  

p 
z
z
z


p


 w 
 z
.. . (5.16)
Orientation description 1
uTu = vTv = wTw = 1, and
uTv(vTu) = uTw(wTu) = vTw(wTv) = 0
Q is called Orthogonal
… (5.17)
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Due to orthogonality
u  v = w,
v  w = u, and
w  u = v . . . (5.18)
QTQ = QQT = 1 ; det (Q) = 1; Q1 = QT . . . (5.19)
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Interpretation 1
Example 5.6 Rotations [Elementary] (Fig. 5.13a)
Cα 


[u]F   Sα  ,
 0 


  Sα 


[ v ]F   Cα  ,
 0 


0
 
[ w ]F   0 
1 
 
. . . (5.20)
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C
Q Z   S
 0
 C
QY   0
 S
 S
C
0
0
1
0
0
0
1
S 
1
0 ; Q X  0
0
C 
. . . (5.21)
0
C
S
0 
 S 
C 
. . . (5.22)
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Non-commutative Property: An Illustration
Fig. 5.20 Successive rotation of a box about Z and Y-axes
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Non-commutative Property (contd.)
Fig. 5.21 Successive rotation of a box about Y and Z-axes
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Lecture 3
Robot Kinematics (Ch. 5)
by
S.K. Saha
Aug. 10, 2015 (M)@JRL301 (Rob. Tech.)
PROPRIETARY MATERIAL. © 2014, 2008 The McGraw-Hill Companies, Inc. All rights reserved. No part of this PowerPoint slide may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this PowerPoint slide, you are using it without
permission.
14
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• Orientation representations
– Non-commutative
• Direction cosines: Has disadv. of 9 param.
• Fixed-axes (RPY) rotations (12 sets)
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Homogeneous Transformation
P
Z
W
p΄
p
OM
V
M
o
O
Y
F
U
X
Task: Point P is known in moving frame M. Find P in fixed frame F.
Fig. 5.23 Two coordinate frames
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p = o + p
. . . (5.45)
[p]F = [o]F + Q[p’]M
[p] F   Q
 1   0 T

 
. . . (5.46)
[o]F  [p] M 
. . . (5.47)



1   1 
[p] F  T [p] M
. . . (5.48)
Homogenous Transformation
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T: Homogenous transformation matrix (4  4)
TTT  1
T
1
or
T1  TT
Q T
 T
 0
Q
. . . (5.49)
[o] F 

1

T
. . . (5.50)
Example 5.10 Pure Translation
1
0
T
0

0
Fig. 5.24 (a)
0
2
0 1 1

0 0 1
. . . (5.51)
0
1
0
0
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Example 5.11 Pure Rotation
Fig. 5.24 (b)
C 30o  S 30o

o
o
S
30
C
30
T
 0
0

0
 0
 3

1

0 0

2
 2

 1

3
 
0 0
2
 2

 0
0
1 0


0
0 1 
 0
0
0
1
0
0

0
0

1 
. . . (5.52)
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Non-commutative Property
Like rotation matrices homogeneous transformation
matrices are non-commutative, i. e.,
TATB  TBTA
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Denavit and Hartenberg (DH) Parameters
• Serial chain
- Two links connected
by revolute or
prismatic joint
• Four parameters
– Joint offset (b)
– Joint angle ()
– Link length (a)
– Twist angle ()
Fig. 5.27 Serial manipulator
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• Joint axis i: Link i-1 + link i
• Link i: Fixed to frame i+1 (Saha) / frame i (Craig)
DH
Saha
Variables
bi and i
[Screw@Z]
XiXi+1@Zi
Constants
ai and i
[Screw@X]
ZiZi+1@Xi+1
Craig
Xi-1Xi@Zi
ZiZi+1@Xi
Zi+1
Z’’’i
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• bi (Joint offset): Length of the intersections of the
common normals on the joint axis Zi, i.e., Oi and
Oi. It is the relative position of links i  1 and i.
This is measured as the distance between Xi
and Xi + 1 along Zi.
Zi+1
Z’’’i
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• i (Joint angle): Angle between the orthogonal projections of
the common normals, Xi and Xi + 1, to a plane normal to the
joint axes Zi. Rotation is positive when it is made counter
clockwise. It is the relative angle between links i  1 and i.
This is measured as the angle between Xi and Xi + 1 about Zi.
Zi+1
Z’’’i
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• ai (Link length): Length between the O’i and Oi
+1. This is measured as the distance between
the common normals to axes Zi and Zi + 1 along
Xi + 1.
Zi+1
Z’’’i
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• i (Twist angle): Angle between the orthogonal
projections of joint axes, Zi and Zi+1 onto a plane
normal to the common normal. This is measured as
the angle between the axes, Zi and Zi + 1, about axis Xi
+ 1 to be taken positive when rotation is made counter
clockwise.
Zi+1
Z’’’i
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Revolute Joint
• DH@Z (Variable)
• DH@X (Const.)
– Joint offset (b)
– Joint angle ()
– Link length (a)
– Twist angle ()
Zi+1
Z’’’i
Fig. 5.28
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Mathematically
• Translation along Zi
Tb =
1
0

0

0
0
0
1
0
0
1
0
0
0
0 
bi 

1
. . . (5.60a)
• Rotation about Zi
T =
Cθ i
 Sθ
 i
 0

 0
 S θi
Cθ i
0
0
0
0
1
0
0
0
0

1
. . . (5.60b)
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• Translation along Xi+1
1
0
Ta = 
0

0
0
1
0
0
0
0
1
0
ai 
0 
0

1
. . . (5.60c)
• Rotation about Xi+1
T =
1
0

0

0
0
Cαi
0
 S αi
Sαi
0
Cαi
0
0
0
0

1
. . . (5.60d)
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• Total transformation from Frame i to Frame i+1
Ti = TbTTaT
. . . (5.61a)
Do it yourself!
 Sθ i Cα i
Sθ i Sα i
Rotation
Cθ Cα
 Cθ Sα
Matrix
Sα
Cα
i
i
i
i
0
i
0
i
ai C i 
ai Sθ i 
bi 

1 
Position
Cθ i
 Sθ
Ti =  i
 0

 0
. . . (5.61b)
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Spherical-type Arm
• DH-parameters
i
bi
1
0
2
bFill-up
0 /2
the
DH parameters
2
2 (JV)
3
b3
(JV)
ai
i
Link
1 (JV) 0 /2
0
0
0
Fig. 5.32 A spherical arm
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PUMA 560
Fig. 5.35 PUMA 560 and its frames
i
Variable
DH
bi
i
Constant
DH
ai
i
1
0
1
0
-/2
2
0
2
a2
0
3
B3
3
a3
-/2
4
b4
4
0
/2
5
0
5
0
-/2
6
0
6
0
0
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Lecture 4 (SIT Sem. Rm.)
Forward and Inverse Kinematics
(Ch. 6)
by
S.K. Saha
Aug. 12, 2015 (W)@JRL301(Robotics Tech.)
PROPRIETARY MATERIAL. © 2014, 2008 The McGraw-Hill Companies, Inc. All rights reserved. No part of this PowerPoint slide may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this PowerPoint slide, you are using it without
permission.
33
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Forward and Inverse Kinematics
Forward: One soln.
Multiply
+ Add
Inverse: 1st soln.
.
Inverse: nth soln.
Solve
Non-lin. eqns.
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Three-link Planar Arm
• DH-parameters
Link
bi
1
2
3
0
0
0
i
ai
i
1 (JV) a1
the DH
a2
Fill-up
(JV)
2parameters
3 (JV) a3
0
0
0
• Frame transformations
(Homogeneous)
Ti =
Cθ i  Sθ i 0 a i C i 
with the elements 
 SθFill-upCθ
0 a i Sθ i  ,
i
 i
 0
0
1
0 


0
0
0
1


Fig. 5.29 A three-link planar arm
for i=1,2,3
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DH Parameters of Articulated Arm
Link bi
i
ai
i
0
 π/2
1
0 1 (JV)
2
0 2 (JV) a2
0
3
0 3 (JV) a3
0
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Matrices for Articulated Arm
c1
s
T1   1
0

0
c1 c 23
s c
1 23

T
  s 23

 0
 s1 0 
c1 0 
1 0 0 

0
0 1
c3
s
T3   3
0

0
c2
s
T2   2
0

0
0
0
 s3
c3
0
0
0
0
1
0
- c1 s 23
 s1 s 23
 s1
c1
 c 23
0
0
0
 s2
c2
0
0
0
0
1
0
a2 c2 
a2 s2 
0 

1 
a3 c3 
a3 s3 
0 

1 
c1 (a 2 c 2  a 3 c 23 )
s1 (a 2 c 2  a 3 c 23 )
… (6.11)
 (a 2 s 2  a 3 s 23 ) 

1

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Inverse Kinematics
• Unlike Forward Kinematics, general solutions
are not possible.
• Several architectures are to be solved
differently.
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Two-link Arm
p y  a1s1  a2 s12
c2 
p x2  p y2  a12  a22
2 a1 a2
s2   1  c22
2 = atan2 (s2, c2)
s1 
py
Y3
2
Y2
a1
1
X3
RoboAnalyzer
p x  a1c1  a2 c12
Y1
a2  X
2
2
1
px
X1
(a1  a2 c2 ) p y  a2 s2 p x
c1 
2
2
2
2
Δ

a

a

2
a
a
c

p

p
1
2
1 2 2
x
y
Δ
(a1  a2 c2 ) p x  a2 s2 p y
Δ
1 = atan2 (s1, c1)
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Inverse Kinematics of 3-DOF RRR Arm
φ  θ1  θ2  θ3 … (6.18a)
p x  a1c1  a2 c12  a3 c123
… (6.18b)
p y  a1 s1  a 2 s12  a3 s123
… (6.18c)
wx  p x  a3 c φ  a1 c1  a2 c12 … (6.19a)
w y  p y  a3 s φ  a1 s1  a2 s12 … (6.19b)
40
w2x + w2y = a12+ a22 + 2 a1a2c2
… (6.20a)
w12  w22  a12  a 22
s2   1  c22
c2 
2 a1 a 2
2 = atan2 (s2, c2)
wy  (a1  a2 c2 )s1  a2 c1 s2
s1 
Δ
… (6.20b,c)
. . . (6.21)
wx  ( a1  a2c2 )c1  a2 s1s2
(a1  a2 c2 ) wy  a2 s2 wx
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c1 
… (6.22a)
… (6.22b)
(a1  a 2 c 2 ) wx  a 2 s 2 w y
Δ  a12  a22  2a1a2c2  wx2  wy2
Δ
… (6.23a,b)
1 = atan2 (s1, c1)
. . . (6.23c)
3 =  - 1  2
. . . (6.24)
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Numerical Example
• An RRR planar arm (Example 6.15). Input




T 




1
2
3
2
0
0
3

2
1
2
0
0
Rotation0
Matrix 0
1
0
5
Origin
3 
2
of end
effector
3
1
frame
2

0 

1 
4.23
1.86
0
where  = 60o, and a1 = a2 = 2 units, and a3 = 1 unit.
Do it yourself  Verify using RoboAnalyzer
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Using eqs. (6.13b-c),
c2 = 0.866, and s2 = 0.5,
2 = 30o
Next, from eqs. (6.16a-b),
s1 = 0, and c1= 0.866.
1 = 0o.
Finally, from eq. (6.17) ,
Therefore
3 = 30o.
1 = 0o 2 = 30o, and 3 = 30
…(6.30b)
The positive values of s2 was used in evaluating 2 = 30o.
The use of negative value would result in :
1 = 30o 2 = -30o, and 3 = 60o
…(6.30c)
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Watch
• Forward and Inverse Kinematics: Watch 3/3 of
IGNOU Lectures [29min]
https://www.youtube.com/watch?v=duKD8cvtBTI
• For more clarity: Watch 12 of Addis Ababa
Lectures [77 min]
[https://www.youtube.com/watch?v=NXWzk1toze4
• Robotics (13 of Addis Ababa Lectures): Inverse
Kinematics [82 min]
https://www.youtube.com/watch?v=ulP3YiJLiEM
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Velocity Analysis
Jacobian maps joint rates into end-effector’s velocities. It
depends on the manipulator configuration.
ω e 
twistof end - effector : t e    ;
 ve 
t e  Jθ
1 
 
Joint rates : θ   
 n 
 
where J   j1
 e1
J
 e1  a 1e
jn  and
j2
e2

e 2  a 2e


e n  a ne 
en
. . (6.86)
 0 
 ei 
, if Joint i is prismatic
ji  
, if Joint i is revolute ji  


e i  a ie 
ei  aie 
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Jacobian of a 2-link Planar Arm
J  e1  a1e
e 2  a 2e 
where e1  e2  [0 0 1]T
a1e  a1  a 2
 [a1c1  a2c12
a1s1  a2 s12
0]T
a2e  a2
 [a2c12
a2 s12
0]T
  a1 s1  a2 s12
Hence, J  
 a1 c1  a2 c12
 a2 s12 
a2 c12 
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Example: Singularity of 2-link RR Arm
 a1 s1  a 2 s12
J
 a1 c1  a 2 c12
 a 2 s12 

a 2 c12 
2 = 0 or 
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Lecture 4 (SIT Sem. Rm.)
Forward and Inverse Kinematics
(Ch. 6)
by
S.K. Saha
Aug. 12, 2015 (W)@JRL301(Robotics Tech.)
PROPRIETARY MATERIAL. © 2014, 2008 The McGraw-Hill Companies, Inc. All rights reserved. No part of this PowerPoint slide may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this PowerPoint slide, you are using it without
permission.
48
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Statics and Manipulator
Design (Ch. 7)
PROPRIETARY MATERIAL. © 2014, 2008 The McGraw-Hill Companies, Inc. All rights reserved. No part of this PowerPoint slide may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers
and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this PowerPoint slide, you are using it without
permission.
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Principle of Virtual Work
T
w e x
 τ θ
T
… (7.28)
• Relation between two virtual displacements
(Can be derived from velocity expression)
x  Jθ
T
w e Jθ
… (7.29)
w J  τ … (7.31)
T
e
 τ θ
T
τ  J we
T
T
… (7.32)
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Example: 2-link RR Planar Arm
τ1  [e ] [n01 ]1
T
1 1
 a1 f x sθ2  (a2  a1cθ2 )f y
τ 2  [e ] [n12 ]2  a2 f y
T
2 2
τJ f
T
 τ1  T a1sθ 2 a1cθ2  a2
τ  J 
0
a
τ
2

 2
 fx 
0
f 
f

y


0
 0 
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Two Jacobian Matrices
• From
Statics
 a1 sθ 2

J  a1cθ 2  a2

0
0

a2 
0 
• From
 a1 s1  a 2 s12
Kinematics J  
 a1 c1  a 2 c12
 a 2 s12 

a 2 c12 
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Jacobian from Statics in Frame 1
0
c1  s1 0 c 2  s 2 0  a1sθ 2
[J ]1   s1 c1 0  s 2 c 2 0 a1cθ2  a2 a2 
 0
0
1  0
0
1 
0
0 
 a1sθ1  a2 sθ12  a2 sθ12 
  a1cθ1  a2 cθ12
a2 cθ12 

0
0 
… (7.34)
• Without the last row, it is the same as
the one from kinematics  Should be!
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Manipulator Design
• High investment in robot usage  low
technological level of mechanical structure
• Functional Requirements
• Kinetostatic Measures
• Structural Design and Dynamics
• Economics
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Functional Requirements of a
Robot
• Payload
• Mobility
• Configuration
• Speed, Accuracy and Repeatability
• Actuators and Sensors
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bmin  b  bmax, for 0o    360o
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Dexterity and Manipulability
• Dexterity  wd  det(J)
• Manipulability  wm 
… (7.44)
T
det(JJ )
• Nonredundant manipulator  square
Jacobian
wm  det(J )
wd  wm
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Motor Selection (Thumb Rule)
• Rapid movement with high torques (>
3.5 kW): Hydraulic actuator
• < 1.5 kW (no fire hazard): Electric
motors
• 1-5 kW: Availability or cost will
determine the choice
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Simple Calculation
@ McGraw-Hill Education
2 m robot arm to lift 25 kg mass at 10
rpm
• Force = 25 x 9.81 = 245.25 N
• Torque = 245.25 x 2 = 490.5 Nm
• Speed = 2 x 10/60 = 1.047 rad/sec
• Power = Torque x Speed = 0.513 kW
• Simple but sufficient for approximation
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Practical Application
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Trapezoidal
Trajectory
Subscript l for load; m for motor;
G = l/m (< 1); : Motor + Gear box efficiency
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Accelerations & Torques
Ang. accn. during t1:
Ang. accn. during t2: Zero (Const. Vel.)
Ang. accn. during t3:
Torque during t1: T1 =
Torque during t2: T2 =
Torque during t3: T3 =
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RMS Value
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Motor Performance
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Final Selection
• Peak speed and peak torque
requirements , where TPeak is max of
(magnitudes) T1, T2, and T3
• Use individual torque and RMS values
+ Performance curves provided by the
manufacturer.
• Check heat generation + natural
frequency of the drive.
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Dynamics and Control
Measures
• Rule of Thumb
1
 n  r
2
… (7.51)
n
: closed-loop natural frequency
r
: lowest structural resonant frequency
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Manipulator Stiffness
1
1
1
 2 
ke  k1 k2
… (7.48)
ke  equivalent stiffness
  gear ratio
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Link Material Selection
• Mild (low carbon) steel:
Sy = 350 Mpa; Su = 420 Mpa
• High alloyed steel
Sy = 1750-1900 Mpa; Su = 2000-2300
Mpa
• Aluminum
• Sy = 150-500 Mpa; Su = 165-580 Mpa
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Driver Selection
• Driver of a DC motor: A hardware unit
which generates the necessary current
to energize the windings of the motor
• Commercial motors come with
matching drive systems
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Summary
• Forward Kinematics
• Inverse kinematics
– A spatial 6-DOF wrist-portioned has 8
solutions
• Velocity and Jacobian
• Mechanical Design
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Thank You
saha@mech.iitd.ac.in
http://sksaha.com