[2022-S2] 02 Robot Kinematics Part 1 new

FACULTY OF ELECTRICAL ENGINEERING & TECHNOLOGY UNIVERSITI MALAYSIA PERLIS EMJ37303 ROBOTIC SYSTEMS KINEMATICS (PART I) “SPATIAL DESCRIPTIONS AND TRANSFORMATIONS” EMJ37303 ROBOTIC SYSTEMS Introduction 2 EMJ37303 ROBOTIC SYSTEMS Introduction 3 EMJ37303 ROBOTIC SYSTEMS Kinematics Problem (𝑥, 𝑦, 𝑧) • How to make sure the robot can follows required trajectory path? • What are the angles needed to be set to obtain the required position? 4 EMJ37303 ROBOTIC SYSTEMS Reference Frames 𝚺𝐞 • Σ𝑤 : World reference frame • Σ𝑒 : End-effector reference frame • Σ𝑐 : Cylinder reference frame • Σ𝑏 : Box reference frame 𝚺𝐛 𝚺𝐰 𝚺𝐜 All joints and object can be described by the position and orientation with the aid of reference frames. 5 EMJ37303 ROBOTIC SYSTEMS Coordinate Systems Left Handed Right Handed 𝑧 𝑧 𝑦 𝑥 𝑦 𝑥 6 Forward Kinematics EMJ37303 ROBOTIC SYSTEMS Robot Kinematics VS Inverse Kinematics What the differences between both approaches? How to solve both? Which one should be used? 7 (𝒙, 𝒚) 𝜽2 Forward Kinematics: EMJ37303 ROBOTIC SYSTEMS Forward Kinematics VS Inverse Kinematics To determine the end-effector position based on joint’s angle (𝜽1 , 𝜽2 ) 𝒀 𝜽1 Forward Kinematics (𝒙, 𝒚) Inverse Kinematics: 𝑿 To determine the joint’s angle based on end-effector position (𝒙, 𝒚) Inverse Kinematics (𝜽1 , 𝜽2 ) 8 • Vector representation EMJ37303 ROBOTIC SYSTEMS Vector Calculus • Transpose • Magnitude 9 EMJ37303 ROBOTIC SYSTEMS Locating Object How to locate an object? • Position 𝑧 P 𝐴 𝑷BORG 𝑦 𝑎𝑥 𝑃 = 𝑏𝑦 = 𝑎𝑥 𝑏𝑦 𝑐𝑧 𝑎𝑥 𝐴 𝑷BORG = 𝑏𝑦 = 𝑎𝑥 𝑐𝑧 𝑐𝑧 𝑏𝑦 𝑇 𝑐𝑧 𝑇 𝑥 10 EMJ37303 ROBOTIC SYSTEMS Locating Object How to locate an object? • Position • Orientation P 𝒙𝐵 ∙ 𝒙𝐴 𝐴 𝒙 𝐵 ∙ 𝒚𝐴 𝐵𝑅 = 𝒙𝐵 ∙ 𝒛𝐴 {B} 𝐴 𝑷BORG 𝒙𝐴 𝜃 𝑍መ𝐵 𝒙𝐵 𝒚𝐵 ∙ 𝒙𝐴 𝒚𝐵 ∙ 𝒚𝐴 𝒚𝐵 ∙ 𝒛𝐴 𝒛𝐵 ∙ 𝒙𝐴 𝒛𝐵 ∙ 𝒚𝐴 𝒛𝐵 ∙ 𝒛𝐴 Dot Product 𝒙𝐴 ∙ 𝒙𝐵 = 𝒙𝐴 𝒙𝑩 cos (𝜃) Since the magnitude of unit vector is 1, then 𝒙𝐴 ∙ 𝒙𝐵 = cos (𝜃) 11 EMJ37303 ROBOTIC SYSTEMS Locating Object (General) 𝐴 𝐵 P 𝐴 𝑷 𝐴 𝐵 𝑷 𝑷BORG = 𝑋𝐵𝑂𝐴 𝑷 = 𝑋𝐵 𝑌𝐵 𝒙𝐵 ∙ 𝒙𝐴 𝐴 𝒙 𝐵 ∙ 𝒚𝐴 𝐵𝑅 = 𝒙𝐵 ∙ 𝒛𝐴 𝑌𝐵𝑂𝐴 𝑍𝐵 𝐶𝐵𝑂𝐴 𝑇 𝑇 𝒚𝐵 ∙ 𝒙𝐴 𝒚𝐵 ∙ 𝒚𝐴 𝒚𝐵 ∙ 𝒛𝐴 𝒛𝐵 ∙ 𝒙𝐴 𝒛𝐵 ∙ 𝒚𝐴 𝒛𝐵 ∙ 𝒛𝐴 𝑷BORG General Form: 𝐴 𝑷 = 𝐵𝐴𝑅 𝐵𝑷 + 𝐴𝑷BORG 𝒙𝐵 ∙ 𝒙𝐴 𝑋𝐴 𝑌𝐴 = 𝒙𝐵 ∙ 𝒚𝐴 𝒙𝐵 ∙ 𝒛𝐴 𝑍𝐴 𝒚𝐵 ∙ 𝒙𝐴 𝒚𝐵 ∙ 𝒚𝐴 𝒚𝐵 ∙ 𝒛𝐴 𝒛𝐵 ∙ 𝒙𝐴 𝒛𝐵 ∙ 𝒚𝐴 𝒛𝐵 ∙ 𝒛𝐴 𝑋𝐵𝑂𝐴 𝑋𝐵 𝑌𝐵 + 𝑌𝐵𝑂𝐴 𝑍𝐵 𝑍𝐵𝑂𝐴 12 EMJ37303 ROBOTIC SYSTEMS Locating Object (General) 𝐴 General Form: Conceptual Form: 𝑷 = 𝐵𝐴𝑅 𝐵𝑷 + 𝐴𝑷BORG 𝐴 𝑷 = 𝐵𝐴𝑇 𝐵𝑷 𝐴 𝐴 𝑷 = 𝐵𝑅 1 0 0 0 𝒙𝐵 ∙ 𝒙𝐴 𝑋𝐴 𝒙 ∙𝒚 𝑌𝐴 = 𝐵 𝐴 𝑍𝐴 𝒙𝐵 ∙ 𝒛𝐴 1 0 𝒚𝐵 ∙ 𝒙𝐴 𝒚𝐵 ∙ 𝒚𝐴 𝒚𝐵 ∙ 𝒛𝐴 0 𝐴 𝑷BORG 1 𝒛𝐵 ∙ 𝒙𝐴 𝒛𝐵 ∙ 𝒚𝐴 𝒛𝐵 ∙ 𝒛𝐴 0 𝐵 𝑷 1 𝑋𝐵𝑂𝐴 𝑌𝐵𝑂𝐴 𝑍𝐵𝑂𝐴 1 𝑋𝐵 𝑌𝐵 𝑍𝐵 1 Homogenous Transformation Matrix 𝐴 𝐵𝑻 Homogeneous Transformation Matrix can describe both orientation and position by a single square matrix. Why? • It is much easier to calculate the inverse of square matrices. • To multiply two matrices, their dimensions must match. 13 EMJ37303 ROBOTIC SYSTEMS Transformation • Fixed frame: Fixed or reference frame is referring to a frame that usually used as global reference frame. The frame is static and fixed. • Moving frame Moving or current frame is referring to the frame that is currently moving or operated Moving Frame 𝑧𝐵 𝑦𝐵 𝑧𝐴 𝐴 𝑷𝐵𝑂𝑅𝐺 𝑦𝐴 𝑥𝐵 𝑥𝐴 Fixed Frame 14 EMJ37303 ROBOTIC SYSTEMS Pure Translation and Rotation Assume that initial position and orientation of frame B is similar to frame A Pure Translation: 𝐴 𝑃 = 𝐵𝑃 + 𝐴𝑃𝐵𝑂𝑅𝐺 Pure Rotation: 𝑧𝐴 𝐴 𝑃 = 𝐵𝐴𝑅 𝐵𝑃 𝑧𝐴 𝑧𝐵 𝑧𝐵 𝑦𝐵 𝑦𝐴 𝑦𝐴 𝑥𝐴 𝑦𝐵 𝑥𝐵 1 0 𝐴 𝐵𝑻 = 0 0 0 1 0 0 0 𝑋𝐵𝑂𝐴 0 𝑌𝐵𝑂𝐴 1 𝑍𝐵𝑂𝐴 0 1 𝑥𝐴 𝑥𝐵 𝒙𝐵 ∙ 𝒙𝐴 𝒙 𝐵 ∙ 𝒚𝐴 𝐴 𝐵𝑻 = 𝒙𝐵 ∙ 𝒛𝐴 0 𝒚𝐵 ∙ 𝒙𝐴 𝒚𝐵 ∙ 𝒚𝐴 𝒚𝐵 ∙ 𝒛𝐴 0 𝒛𝐵 ∙ 𝒙𝐴 𝒛𝐵 ∙ 𝒚𝐴 𝒛𝐵 ∙ 𝒛𝐴 0 0 0 0 1 15 No rotation, so this part is equal to Identity Matrix EMJ37303 ROBOTIC SYSTEMS Pure Translation Represents the frame before translation This column represents the translation 16 EMJ37303 ROBOTIC SYSTEMS Pure Rotation 𝑅11 𝑅 𝑇 = 21 𝑅31 0 𝑅12 𝑅22 𝑅32 0 𝑅13 𝑅23 𝑅33 0 0 0 0 1 17 𝒙𝐵 ∙ 𝒙𝐴 𝐴 𝒙 𝐵 ∙ 𝒚𝐴 𝐵𝑅 = 𝒙𝐵 ∙ 𝒛𝐴 𝒚𝐵 ∙ 𝒙𝐴 𝒚𝐵 ∙ 𝒚𝐴 𝒚𝐵 ∙ 𝒛𝐴 𝒛𝐵 ∙ 𝒙𝐴 𝒛𝐵 ∙ 𝒚𝐴 𝒛𝐵 ∙ 𝒛𝐴 Alternative: Solving based on rotation angle Solve based on current information of orientation EMJ37303 ROBOTIC SYSTEMS Rotation Matrix Alibi Transformation Alias Transformation 18 EMJ37303 ROBOTIC SYSTEMS Rotation Matrix Rotation about 𝒛-axis: 𝐑𝐨𝐭 𝑧, 𝛾 𝑦 𝑃1 (𝑥1 , 𝑦1 , 𝑧1 ) 𝛾 𝑟 𝜃 𝑃0 (𝑥0 , 𝑦0 , 𝑧0 ) 𝑥 Try solving the rotation matrix around 𝑥 and 𝑦 axis. 19 • By assuming the rotation around 𝑥-, 𝑦-, and 𝑧-axis as 𝛼, 𝛽 and 𝛾, the respective rotation matrix are: EMJ37303 ROBOTIC SYSTEMS Rotation Matrix 20 Relative to Fixed Frame EMJ37303 ROBOTIC SYSTEMS Combined Transformations 21 Relative to Fixed Frame 𝑇1 = rot(𝑧, 90°) 𝑇2 = rot(𝑦, 90°) 𝑇 = 𝑇3 𝑇2 𝑇1 EMJ37303 ROBOTIC SYSTEMS Combined Transformations 𝑇3 = trans(4, −3,7) 22 Relative to Fixed Frame EMJ37303 ROBOTIC SYSTEMS Combined Transformations 23 Relative to Fixed Frame 𝑇1 = rot(𝑧, 90°) 𝑇2 = trans(4, −3,7) 𝑇 = 𝑇3 𝑇2 𝑇1 EMJ37303 ROBOTIC SYSTEMS Combined Transformations 𝑇3 = rot(𝑦, 90°) 24 Relative to Moving Frame EMJ37303 ROBOTIC SYSTEMS Combined Transformations 25 Relative to Moving Frame 𝑇1 = rot(𝑎, 90°) 𝑇2 = trans(4, −3,7) 𝑇 = 𝑇1 𝑇2 𝑇3 EMJ37303 ROBOTIC SYSTEMS Combined Transformations 𝑇3 = rot(𝑜, 90°) 26 Relative to Moving & Fixed Frame EMJ37303 ROBOTIC SYSTEMS Combined Transformations 27 Relative to Moving & Fixed Frame EMJ37303 ROBOTIC SYSTEMS Combined Transformations 28 Assume for transformation 𝑇1 , 𝑇2 , 𝑇3 • Transformations relative to Fixed Frame 𝑇 = 𝑇3 𝑇2 𝑇1 • Transformations relative to Moving Frame 𝑇 = 𝑇1 𝑇2 𝑇3 EMJ37303 ROBOTIC SYSTEMS Summary: Combined Transformation Assume for transformation 𝑇1 , 𝑇2 , 𝑇3 , 𝑇4 . Where, 𝑇1 and 𝑇3 are relative to Fixed Frame, while 𝑇2 and 𝑇4 are relative to Moving Frame. 𝑇 = 𝑇3 𝑇1 𝑇2 𝑇4 29 EMJ37303 ROBOTIC SYSTEMS Inverse Transformation 𝑧𝐵 𝑧𝐴 𝑦𝐵 𝐵 𝑷𝑨𝑂𝑅𝐺 𝐴 𝑥𝐴 𝑦𝐴 𝐴 𝐵𝑇 𝐴 −1 𝐵𝑇 𝑷𝐵𝑂𝑅𝐺 = 𝐵 𝐴𝑇 = = 𝐴 𝐴 𝐵𝑅 0 0 0 𝐴 𝑇 𝐵𝑅 0 0 0 𝑥𝐵 𝐵 𝑷BORG 1 𝑷AORG − 𝐵𝐴𝑅𝑇 𝐴𝑷BORG 1 𝑅 −1 = 𝑅𝑇 , 𝑇 −1 ≠ 𝑇 𝑇 𝐵 𝐴𝑅 30 EMJ37303 ROBOTIC SYSTEMS Compound Transformations 𝑈 𝐸𝑇 = 𝑈𝑅𝑇 𝐻𝑅𝑇 𝐻𝐸𝑇= 𝑈𝑃𝑇 𝐸𝑃𝑇 𝑅 𝐻𝑇 =? 𝑈 −1 𝑈 𝑅 𝐻 𝐻 −1 𝑈 −1 𝑈 𝑃 𝐻 −1 𝑇 𝑇 𝑇 𝑇 𝑇 = 𝑅 𝑅 𝐻 𝐸 𝐸 𝑅𝑇 𝑃𝑇 𝐸 𝑇 𝐸 𝑇 𝑅 𝐻𝑇 = 𝐴 𝐵 −1 𝐵 𝑇 𝐴𝑇 =I 𝑈 −1 𝑈 𝑃 𝐻 −1 𝑅𝑇 𝑃𝑇 𝐸 𝑇 𝐸 𝑇 31 EMJ37303 ROBOTIC SYSTEMS Compound Transformations 𝐴 𝑃 = 𝐵𝐴𝑇 𝐵𝑃 𝐵 𝑃 𝐵 𝐴 𝑃= 𝐴 𝑃= 𝐴 𝐵 𝐵𝑇 𝑃 𝐴 𝐵 𝐶 𝐵𝑇 𝐶 𝑇 𝑃 ∴ 𝐵𝐴𝑇 𝐵𝐶𝑇 = 𝐴𝐶𝑇 𝐴 𝐶𝑇 = 𝐴 𝐵 𝐵𝑅 𝑷CORG 𝐴 𝐵 𝐵𝑅 𝐶 𝑅 0 0 𝑃 = 𝐵𝐶𝑇 𝐶𝑃 0 + 𝐴𝑷BORG 1 32 EMJ37303 ROBOTIC SYSTEMS Compound Transformations {C} B C T C D T {B} {D} A B T C E T {A} D E T E A T T CBT CDT DAT = I A B T = CDT DET = CBT −1 ABT −1 EAT −1 C E C B T BAT EAT {E} 33 {C} T = CDT DET EAT ABT EMJ37303 ROBOTIC SYSTEMS Compound Transformations C B C B T C D T {B} {D} A B T {A} D E T E A T {E} T = DET −1 CDT −1 CBT ABT −1 E A I = EAT −1 DET −1 CDT −1 CBT ABT −1 34 EMJ37303 ROBOTIC SYSTEMS Exercise Exercise 1 Solution: 𝑧𝐵 Position: 𝑦𝐵 𝑧𝐴 𝐴 𝑷𝐵𝑂𝑅𝐺 𝑦𝐴 𝑥𝐵 𝑥𝐴 Frame B is located at 3,4,5 unit from frame A, with yB axis is parallel to yA axis. Its xB and zB at 30 deg relative to xA and zA due to a 30 deg rotation around yB axis. Describe Frame B with respect to frame A. 𝐴 𝑷𝐵𝑂𝑅𝐺 = 3 4 5 Orientation: cos 30° 𝐴 𝐵𝑅 = cos 90° cos 120° 𝑇 cos 90° cos 0° cos 90° cos 60° cos 90° cos 30° Thus, frame B can be described with respect to frame A by √3 2 𝐴 𝐵𝑇 = 0 −12 0 0 1 0 0 1 2 0 √3 2 0 3 4 5 1 35 EMJ37303 ROBOTIC SYSTEMS Transformation: Summary {B} Description of a frame A  A BR T =  B 0 0 0 {A} A PBORG {B} B {A} A P A P PBORG A PBORG   1  Transform Mapping A P = ABT B P  A P   BA R  =  1  0 0 0 A PBORG   B P    1  1  Transform Operator A {A} A P 2 Q A P1 P2 = T A P2 1  A P2  0   = 0  1    0 0 0 qx  1 0 q y   A P1    0 1 qz   1   0 0 1 A P2 = A P2 + Q (vector operation) 36

[2022-S2] 02 Robot Kinematics Part 1 new

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