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1 @ McGraw-Hill Education Lecture 2 Robot Kinematics S.K. Saha Dynamics of Multibody... · 2016....
Transcript of 1 @ McGraw-Hill Education Lecture 2 Robot Kinematics S.K. Saha Dynamics of Multibody... · 2016....
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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 teachersand educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this PowerPoint slide, you are using it withoutpermission.
Lecture 2 Robot Kinematics
byS.K. Saha
Aug. 03’16 (W)@JRL301 (Robotics Tech.)
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Outline• Links and Joints
• Kinematic chains
• Degrees-of-freedom (DOF)
• Pose (≡ Configuration)
• Homogeneous transformation
• Denavit-Hartenberg Parameters
• Conclusions
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Link
Link
Joint
Joint
TAL Robot
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Transformations
• Robot Architecture– Links: A rigid body with 6-DOF– Joints: Couples 2 bodies. Provide
restrictions• Relationship between joint motion (input)
and end-effector motion (output) – Transformations between different
coordinate frames are required
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Joints or Kinematic Pairs• Lower Pair
– Surface contact: Hinge joint of a door
• Higher pair– Line or point contact: Roller or ball rolling
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Lower Pair: Revolute Joint
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Lower Pair: Prismatic Joint
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Lower Pair: Helical Joint
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Lower Pair: Cylindrical Joint
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Lower Pair: Spherical Joint
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Closed Kinematic Chain
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Open Kinematic Chain
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Degrees-of-Freedom (DOF)• Number of independent (or minimum)
coordinates required to fully describe pose or configuration (position + rotation)– A rigid body in 3D space has 6-DOF
• DOF = Coordinates - Constraints– Grubler formula (1917) for planar
mechanisms, DOF = 3 (r-1) – 2p– Kutzbach formula (1929) for spatial
systems, DOF = 6 (r-1) – 5p
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A 4-bar Mechanism
n = 3 (r − 1) − 2p= 3(4-1) − 2×4= 1
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A Robot Manipulator
n = 6 (r − 1) − 5p= 6(7-1) − 5×6 = 6
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n = 3 (r − 1) − 2p= 3(5-1) − 2×6 = 0
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@ McGraw-Hill Education 17
F
p΄
o
p
U
MOM
V
P
W
O
X
Z
Y
Pose ≡ Position + Orientation
Translation: 3Rotation: 3
Total: 6
A moving body Pose or Configuration
Position (noun) or Translation (verb):
Easy (unique)
Orientation (noun) or Rotation
(verb): Difficult (non-unique)
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@ McGraw-Hill Education 18
[ ,
[ ] ,
[ ]
0
0001
u]
v
w
F
F
F
CαSα
SαCα
⎡ ⎤⎢ ⎥
≡ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥
≡ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥
≡ ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
−
. . . (5.20)
Example 5.6 Elementary Rotations @ Z [5.13(a)]
Fig. 5.13
ClueCoordinate
transformation of Class XII
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⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡ααα−α
≡10000
CSSC
ZQ . . . (5.21)
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−≡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−≡
γγγγ
ββ
ββ
CSSC
CS
SC
XY
00
001;
0010
0QQ
. . . (5.22)
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Z Y X
C C C S S S C C S C S SS C S S S C C S S C C S
S C S C C
α β α β γ α γ α β γ α γα β α β γ α γ α β γ α γ
β β γ β γ
≡ =
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
− ++ −
−
Q Q Q Q
Rotations about Z Y (new) X (new) axes
ZYX-Euler angles: 12 sets
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Non-commutative Property: Geometrically
Fig. 5.20
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Non-commutative Property …
Fig. 5.21
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W.R.T. fixed frame: QZY = QYQZ =⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
010001100
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−≡
001010100
9009001090090
Yoo
oo
CS
SCQ
But, QYZ = QZQY = ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−
001100010
Non-commutative Property
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡ −=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡ −≡
100001010
1000909009090
Zoo
oo
CSSC
Q
Hence, QZY ≠ QYZ
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Orientation Description
• Euler angles representation
• Direction cosine representation
• Euler parameters representation, etc.
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Homogeneous Transformation
F
p΄
o
p
U
MOM
V
P
W
O
X
Z
Y
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 . . . (5.46)
⎥⎦
⎤⎢⎣
⎡ ′⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡1][
1][
1][
TF MF poQp
0. . . (5.47)
MF ][][ pTp ′= . . . (5.48)
Homogenous Transformation
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TTT ≠ 1 or T−1 ≠ TT . . . (5.49)
⎥⎥⎦
⎤
⎢⎢⎣
⎡ −=−
1][
T
TT1
0oQQT F . . . (5.50)
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
≡
1000110020100001
T
. . . (5.51)
Example 5.12 Pure Translation
T: Homogenous transformation matrix (4 × 4)
Fig. 5.24 (a)
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rt TTT ≡ . . . (5.53)30 30 0 230 30 0 10 0 1 00 0 0 1
3 1 0 22 21 3 0 12 20 0 1 00 0 0 1
T
o o
o o
C SS C
⎡ ⎤−⎢ ⎥⎢ ⎥≡⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦⎡ ⎤
−⎢ ⎥⎢ ⎥⎢ ⎥
= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
. . . (5.54)
Example 5.14 General Motion
Fig. 5.24 (c)
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@ McGraw-Hill Education 29
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 teachersand educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this PowerPoint slide, you are using it withoutpermission.
Lecture 3 Robot Kinematics
byS.K. Saha
Aug. 17’16 (W)@JRL301 (Robotics Tech.)
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Summary of Last Class
• Kinematic chain: Links and joints• DOF: Parameters-constraints• Position: Simple (like good friend in the
hostel)• Orientation: Confusing and SERIOUS
attention to be paid
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Denavit and Hartenberg (DH) Parameters—Frame Allotment
• Serial chain - Two links connected
by revolute joint, or- Two links connected
by prismatic joint
Fig. 5.27
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• Joint axis i: Link i-1 + link i• Link i: Fixed to frame i+1 (Saha) / frame i (Craig)
DH Variablesbi and θi
[Screw@Z]
Constantsai and αi
[Screw@X]Saha XiXi+1@Zi ZiZi+1@Xi+1
Craig Xi-1Xi@Zi ZiZi+1@Xi
Z’’’i
Zi+1
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Revolute Joint
Fig. 5.28
• DH@Z (Variable)– Joint offset (b)– Joint angle (θ)
• DH@X (Const.)– Link length (a)– Twist angle (α)
Z’’’i
Zi+1
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Tb =⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
1000100
00100001
ib. . . (5.49a)
Tθ =⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡ −
100001000000
ii
ii
CθSθθSCθ
. . . (5.49b)
Mathematically• Translation along Zi
• Rotation about Zi
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⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡−
100000000001
ii
ii
CαSααSCαTα = . . . (5.49d)
Ta =⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
100001000010
001 ia
. . . (5.49c)
• Translation along Xi+1
• Rotation about Xi+1
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Ti = TbTθTaTα . . . (5.50a)
Ti =⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡−
−
10000 iii
iiiiiii
iiiiiii
bCαSαSθaSαCθCαCθSθCaSαSθCαSθCθ θ
. . . (5.50b)
• Total transformation from Frame i to Frame i+1
Rotation Matrix
Position
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Three-link Planar Arm
Ti =⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡ θ−
10000100
00
iiii
iiii
SθaCθSθCaSθCθ
• DH-parameters
, for i=1,2,3
Link bi θi ai αi
1 0 θ1 (JV) a1 02 0 θ2 (JV) a2 03 0 θ3 (JV) a3 0
• Frame transformations(Homogeneous)
Fill-up the DH parameters
Fill-up with the elements
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Conclusions
• DOF and Constraints• Rotation representations• DH Parameters• Configuration and Homogeneous
Transformation• RoboAnalyzer software• Examples
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@ McGraw-Hill Education 40
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 teachersand educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this PowerPoint slide, you are using it withoutpermission.
Lecture 4 (SIT Sem. Rm.)Forward and Inverse Kinematics
(Ch. 6)by
S.K. SahaAug. ??, 2016 (?)@JRL301(Robotics Tech.)
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Recap• Orientation representations
– Non-commutative
• Direction cosines: Has disadv. of 9 param.
• Fixed-axes (RPY) rotations (12 sets)
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Forward and Inverse Kinematics
Inverse: 1st soln..Inverse: nth soln.
Forward: One soln.S
olve N
on-lin. eqns.M
ultiply + A
dd
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Three-link Planar Arm
Ti =⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡ θ−
10000100
00
iiii
iiii
SθaCθSθCaSθCθ
• DH-parameters
, for i=1,2,3
Link bi θi ai αi
1 0 θ1 (JV) a1 02 0 θ2 (JV) a2 03 0 θ3 (JV) a3 0
• Frame transformations(Homogeneous)
Fill-up the DH parameters
Fill-up with the elements
Fig. 5.29 A three-link planar arm
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DH Parameters of Articulated ArmLink bi θi ai αi
1 0 θ1 (JV) 0 − π/2
2 0 θ2 (JV) a2 0
3 0 θ3 (JV) a3 0
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Matrices for Articulated Arm1 1
1 11 0 1 0 0
0 0 0 1
c 0 s 0s 0 c 0
−⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥−⎢ ⎥⎣ ⎦
T
2 2 2 2
2 2 2 22
c s 0 a cs c 0 a s0 0 1 00 0 0 1
−⎡ ⎤⎢ ⎥⎢ ⎥≡⎢ ⎥⎢ ⎥⎣ ⎦
T
3 3 3 3
3 3 3 33
c s 0 a cs c 0 a s0 0 1 00 0 0 1
−⎡ ⎤⎢ ⎥⎢ ⎥≡⎢ ⎥⎢ ⎥⎣ ⎦
T
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
+−−−+−+−
≡
1000sasa0cs
)cac(ascsscs)cac(acssc-cc
233222323
2332211231231
2332211231231
)(T … (6.11)
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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
X2θ2a2
a1
X1
X3
Y2
Y1 Y3
θ1px
py
12211
12211
sasapcacap
y
x
+=+=
21
22
21
22
2 2 aaaapp
c yx −−+=
222 1 cs −±=
θ2 = atan2 (s2, c2)
Δpsap)ca(a
s xy 222211
−+= 22
22122
21 2 yx ppcaaaaΔ +=++≡
Δpsap)ca(a
c yx 222211
++= θ1 = atan2 (s1, c1)
θ1
θ2
RoboA
nalyzer
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Inverse Kinematics of 3-DOF RRR Arm
321 θθθφ ++=123312211 cacacapx ++=
123312211 sasasap y ++=
122113 cacac φ apw xx +=−=122113 sasas φ apw yy +=−=
… (6.18a)
… (6.18b)
… (6.18c)
… (6.19a)… (6.19b)
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w2x + w2
y = a12+ a2
2 + 2 a1a2c2
21
22
21
22
2 2 aaaawwc 21 −−+
= 222 1 cs −±=
θ2 = atan2 (s2, c2) . . . (6.21)
2121221 ssa)ccaa(wx −+=
2121221y sca)sca(aw ++=
Δwsaw)ca(a
s xy 222211
−+=
Δwsaw)ca(a
c yx 222211
++=
22221
22
21 2 yx wwcaaaaΔ +=++≡
θ1 = atan2 (s1, c1) . . . (6.23c)
θ3 = ϕ - θ1 − θ2 . . . (6.24)
… (6.22a)… (6.22b)
… (6.20a)
… (6.20b,c)
… (6.23a,b)
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Numerical Example
3 52 2
3 3 12 2
T
⎡ ⎤− +⎢ ⎥
⎢ ⎥⎢ ⎥
≡ +⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
1 0 32
1 02
0 0 1 00 0 0 1
• An RRR planar arm (Example 6.15). Input
where φ = 60o, and a1 = a2 = 2 units, and a3 = 1 unit.
Rotation Matrix
Origin of end-effectorframe
4.23
1.86
0
Do it yourself Verify using RoboAnalyzer
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Using eqs. (6.13b-c), c2 = 0.866, and s2 = 0.5,
Next, from eqs. (6.16a-b), s1 = 0, and c1= 0.866.
Finally, from eq. (6.17) ,
Therefore …(6.30b)
The positive values of s2 was used in evaluating θ2 = 30o.
The use of negative value would result in :
…(6.30c)
θ2 = 30o
θ1 = 0o.
θ3 = 30o.
θ1 = 0o θ2 = 30o, and θ3 = 30
θ1 = 30o θ2 = -30o, and θ3 = 60o
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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
[ ]1 2where andJ j j j n=
ii
i ie
, if Joint is revolutei ⎡ ⎤
≡ ⎢ ⎥×⎣ ⎦
ej
e aprismatic isointJif, i
ieii ⎥
⎦
⎤⎢⎣
⎡×
≡ae
0j
et Jθ=
1
twistof end - effector : ; Joint rates : ee
en
θ
θ
⎡ ⎤⎡ ⎤ ⎢ ⎥≡ =⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦
ωt θ
v
Jacobian maps joint rates into end-effector’s velocities. It depends on the manipulator configuration.
⎥⎦
⎤⎢⎣
⎡×××
=nee1e aeaeae
eeeJ n2
n221
1. . (6.86)
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Jacobian of a 2-link Planar Arm
[ ]ee 2211 aeaeJ ××=
1 1 2 12 2 12
1 1 2 12 2 12
Hence, Ja s a s a s
a c a c a c− − −⎡ ⎤
= ⎢ ⎥+⎣ ⎦
1 2where [0 0 1]e e T≡ ≡
1 1 2
1 1 2 12 1 1 2 12[ 0]
a a a
e
Ta c a c a s a s
≡ +
≡ + +
2 2
2 12 2 12[ 0]
a a
e
Ta c a s
≡
≡
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Example: Singularity of 2-link RR Arm
⎥⎦
⎤⎢⎣
⎡+
−−−≡
12212211
12212211
cacacasasasa
J θ2 = 0 or π
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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
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 = 2π 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
Subscript l for load; m for motor;G = ωl/ωm (< 1); η: Motor + Gear box efficiency
Trapezoidal Trajectory
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Accelerations & Torques
Ang. accn. during t1:
Ang. accn. during t3:
Ang. accn. during t2: Zero (Const. Vel.)
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
12n rω ω≤
• Rule of Thumb
nω
rω
: closed-loop natural frequency
: lowest structural resonant frequency
… (7.51)
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Manipulator Stiffness
21 2
1 1 1
ek k kη= + ek ≡
η ≡
equivalent stiffness
gear ratio … (7.48)
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Link Material Selection
• Mild (low carbon) steel: Sy = 350 Mpa; Su = 420 Mpa
• High alloyed steelSy = 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