ROBOTICS: ADVANCED CONCEPTS & ANALYSIS

ROBOTICS: ADVANCED CONCEPTS & ANALYSIS MODULE 4 KINEMATICS OF PARALLEL ROBOTS Ashitava Ghosal 1 1 Department of Mechanical Engineering & Centre for Product Design and Manufacture Indian Institute of Science Bangalore 560 012, India Email: asitava@mechengiiscernetin NPTEL, 2010 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 1 / 80

1 CONTENTS 2 LECTURE 1 Introduction Loop-closure Constraint Equations 3 LECTURE 2 Direct Kinematics of Parallel Manipulators 4 LECTURE 3 Mobility of Parallel Manipulators 5 LECTURE 4 Inverse Kinematics of Parallel Manipulators 6 LECTURE 5 Direct Kinematics of Stewart Platform Manipulators 7 ADDITIONAL MATERIAL Problems, References and Suggested Reading ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 2 / 80

OUTLINE 1 CONTENTS 2 LECTURE 1 Introduction Loop-closure Constraint Equations 3 LECTURE 2 Direct Kinematics of Parallel Manipulators 4 LECTURE 3 Mobility of Parallel Manipulators 5 LECTURE 4 Inverse Kinematics of Parallel Manipulators 6 LECTURE 5 Direct Kinematics of Stewart Platform Manipulators 7 ADDITIONAL MATERIAL Problems, References and Suggested Reading ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 3 / 80

INTRODUCTION Parallel manipulators: One or more loops No first or last link No natural choice of end-effector or output link Output link must be chosen Number of joints is more than the degree-of-freedom Several joints are not actuated Un-actuated or passive joints can be multi-degree-of-freedom joints Two main problems: Direct Kinematics and Inverse Kinematics ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 4 / 80

EXAMPLES OF PARALLEL ROBOTS Link 2 l 2 Ŷ L {L} φ 2 Link 1 Link 3 l 3 l 1 Ŷ R {R} θ 1 ˆX L l 0 φ 1 O R ˆX R O L Figure 1: Planar 4-bar Mechanism One-degree-of-freedom mechanism with 4 joints Very well known Link 2 is called coupler and is the typical output link ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 5 / 80

EXAMPLES OF PARALLEL ROBOTS p(x, y, z) Moving Platform Ẑ S3 S2 l3 θ3 {Base} S1 θ2 l1 l2 Axis of R2 9 joints only three P joints actuated Top (moving) platform is the output link O Base Platform θ1 Ŷ Multi-degree-of-freedom spherical(s) joints are passive Axis of R3 ˆX Axis of R1 Figure 2: Three-degree-of-freedom Parallel Manipulator ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 6 / 80

EXAMPLES OF PARALLEL ROBOTS Figure 3: Original Stewart platform (1965) ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 7 / 80

EXAMPLES OF PARALLEL ROBOTS θ 1 b 1 Z d d h θ 3 l 11 l 12 b 3 ψ 1 θ 2 l b 31 2 l ψ 21 φ ψ 1 2 l 32 l l 13 22 23 X Figure 4: Model of a three-fingered hand 3 φ φ 3 2 p 1 Y s s l 33 l s p 3 p 2 Three fingers modeled a R-R-R chain Fingers gripping an object with point contact and no slip Point contact modeled with S joint Object (output link) is an equilateral triangle Three DOF, 12 joints ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 8 / 80

APPLICATIONS OF PARALLEL ROBOTS Modern tyre testing machine Micro-positioning Industrial manufacturing Physik Instrumetente http://wwwphysikinstrumentecom Robotic surgery Figure 5: Precise alignment of mirror Some uses of Gough-Stewart platform ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 9 / 80

DEGREES OF FREEDOM (DOF) Grübler-Kutzbach s criterion DOF = λ(n J 1) + J i=1 F i (1) N Total number of links including the fixed link (or base), J Total number of joints connecting only two links (if joint connects three links then it must be counted as two joints), F i Degrees of freedom at the i th joint, and λ = 6 for spatial, 3 for planar manipulators and mechanisms 4-bar mechanism N = 4, J = 4, J i=1 F i = 1 + 1 + 1 + 1 = 4, λ = 3 DOF = 1 3-RPS manipulator N = 8, J = 9, J i=1 F i = 6 1 + 3 3 = 15, λ = 6 DOF = 3 Three-fingered hand N = 11, J = 12, J i=1 F i = 9 + 9 = 18, λ = 6 DOF = 6 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 10 / 80

DEGREES OF FREEDOM (CONTD) DOF The number of independent actuators In parallel manipulators, J > DOF J DOF joints are passive Example: 4-bar mechanism, J = 4 and DOF = 1 Only one joint is actuated and three are passive Example: 3-RPS manipulator, J = 9 and DOF = 3 6 joints are passive Passive joints can be multi-degree-of-freedom joints In 3-RPS manipulator, three-degree-of-freedom spherical (S) joints are passive In a Stewart platform, the S and U joints are passive Configuration space q = (θ,ϕ) θ are actuated joints & θ R n (n = DOF ) ϕ is the set of passive joints & ϕ R m All passive joints / ϕ (n + m) J ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 11 / 80

LOOP-CLOSURE EQUATIONS CONSTRAINT m passive joint variables m Independent equations required to solve for ϕ for given n actuated variable, θ i, i = 1,2,,n General approach to derive m loop-closure constraint equations 1 Break parallel manipulator into 2 or more serial manipulators, 2 Determine D-H parameters for serial chains and obtain position and orientation of the Break for each chain, 3 Use joint constraint (see Module 2, Lecture 2) at the Break(s) to re-join (close) the parallel manipulator Trick is to break such that 1 The number of passive variables m is least, and 2 Minimum number of constraint equations, η i (q) = 0,i = 1,,m are used ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 13 / 80

CONSTRAINT EQUATIONS 4-BAR EXAMPLE φ3 O2 Link 2 ˆX2 O3 l2 ˆX3 ŶL {L} l1 Link 1 φ2 Link 3 l3 ŶR {R} ˆX1 φ1 θ1 ˆXL l0 OTool, OR ˆXR OL, O1 ˆXTool Figure 6: The four-bar mechanism One loop Fixed frames {L} and {R}, {R} is translated by l 0 along the X axis {1}, {2}, {3}, and {Tool} are as shown Note only ˆX shown for convenience The sequence O L -O 1 -O 2 -O 3 -O Tool can be thought of as a planar 3R manipulator ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 14 / 80

CONSTRAINT EQUATIONS 4-BAR EXAMPLE D-H parameters of the planar 3R manipulator are i α i 1 a i 1 d i θ i 1 0 0 0 θ 1 2 0 l 1 0 ϕ 2 3 0 l 2 0 ϕ 3 From D-H table find 0 3 [T ] (See Slide # 51, Lecture 3, Module 2) For planar 3R and tool of length l 3, find 3 Tool [T ] Tool R [T ] is given Tool R [T ] = cosϕ 1 sinϕ 1 0 0 sinϕ 1 cosϕ 1 0 0 0 0 1 0 0 0 0 1 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 15 / 80

CONSTRAINT EQUATIONS 4-BAR EXAMPLE The loop-closure equations for the four-bar mechanism is L 1[T ] 1 2[T ] 2 3[T ] 3 Tool [T ]Tool R [T ] = L R [T ] Planar loop Only 3 independent equations l 1 cosθ 1 + l 2 cos(θ 1 + ϕ 2 ) + l 3 cos(θ 1 + ϕ 2 + ϕ 3 ) = l 0 l 1 sinθ 1 + l 2 sin(θ 1 + ϕ 2 ) + l 3 sin(θ 1 + ϕ 2 + ϕ 3 ) = 0 θ 1 + ϕ 2 + ϕ 3 + (π ϕ 1 ) = 4π (2) Loop-closure equations: all four joint variables present q = (θ 1,ϕ 1,ϕ 2,ϕ 3 ) The actuated joint θ = θ 1 The passive joints ϕ = (ϕ 1,ϕ 2,ϕ 3 ) In this approach n = 1, m = 3 and J = 4 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 16 / 80

CONSTRAINT EQUATIONS (CONTD) Difficulties in multiplying 4 4 matrices and obtaining constraint equations: 1 Presence of multi-degree-of-freedom spherical (S) and Hooke (U) joints in a loop 2 Obtaining independent loops in the presence of several loops Represent multi-degree-of-freedom joint by two or more one-degree-of-freedom joints and obtain an equivalent 4 4 transformation matrix Obtaining independent loops not easy in this way! ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 17 / 80

B4 P4 CONSTRAINT EQUATIONS (CONTD) P3 B5 Extensible Leg Each leg is U-P-S chain, λ = 6, N = 14, J = 18, J i=1 F i = 36 DOF = 6 6 P joints actuated 30 passive variables P5 {P0} {B0} B3 B6 P2 P6 Fixed Base Top Platform P1 Spherical Joint B2 Prismatic Joint B1 U Joint Many loops For example, 5 of the form B i P i P i+1 B i+1 B i, i = 1,,5, 4 of the form B i P i P i+2 B i+2 B i, i = 1,,4, and 3 of the form B i P i P i+3 B i+3 B i, i = 1,2,3 Each of the 12 loops can have (potentially) 6 independent equations Which 30 equations to choose?! Figure 7: The Stewart-Gough platform ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 18 / 80

4-BAR EXAMPLE REVISITED Breaking at Joint 3 O2 Link 2 l2 φ2 ŶL Link 1 {L} l1 Link 3 l3 ŶR {R} φ1 θ1 ˆXL l0 OL, O1 Fig a ˆXR OR Break Link 2 (x,y) φ3 a b φ2 ŶL Link 1 {L} L l1 p R p Link 3 l3 ŶR {R} φ1 θ1 ˆXL l0 ˆXR OL, O1 Fig b OR Link 2 l2 R p ŶL L p {L} l1 Link 1 Link 3 l3 ŶR {R} φ1 θ1 ˆXL l0 ˆXR OL, O1 Fig c OR Figure 8: The four-bar mechanism broken in different ways ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 19 / 80

4-BAR EXAMPLE REVISITED Alternate way: break loop at third joint (figure 8(a)) One planar 2R manipulator + one planar 1R manipulator Obtain D-H tables for both (see Slide # 62, Lecture 3, Module 2) Easy to obtain L 1 [T ], 1 2 [T ] & R 1 [T ] Using l 2 and l 3, obtain L Tool [T ] and R Tool [T ] From L Tool [T ] extract X and Y components of L p x = l 1 cosθ 1 + l 2 cos(θ 1 + ϕ 2 ), y = l 1 sinθ 1 + l 2 sin(θ 1 + ϕ 2 ) From R Tool [T ], extract vector R p to get x = l 3 cosϕ 1, y = l 3 sinϕ 1 Use constraint for R joint (Slide # 30, Lecture 2, Module 2) x = l 1 cosθ 1 + l 2 cos(θ 1 + ϕ 2 ) = l 0 + l 3 cosϕ 1 y = l 1 sinθ 1 + l 2 sin(θ 1 + ϕ 2 ) = l 3 sinϕ 1 (3) l 0 is the distance along the X axis between {L} and {R} In this case only two constraint equation: q = (θ 1,ϕ 1,ϕ 2 ) ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 20 / 80

4-BAR EXAMPLE REVISITED Another way is to break the second link (see figure 8(b)) Two planar 2R manipulators Obtain the X and Y components of L p as x = l 1 cosθ 1 +a cos(θ 1 +ϕ 2 ), y = l 1 sinθ 1 +a sin(θ 1 +ϕ 2 ) Likewise X and Y components of R p are x = l 3 cosϕ 1 +b cos(ϕ 1 +ϕ 3 ), y = l 3 sinϕ 1 +b sin(ϕ 1 +ϕ 3 ) where l 2 = a +b and the angle ϕ 3 is as shown in figure 8(b) Impose the constraint that the broken link is actually rigid x = l 1 cosθ 1 + a cos(θ 1 + ϕ 2 ) = l 0 + l 3 cosϕ 1 + b cos(ϕ 1 + ϕ 3 ) y = l 1 sinθ 1 + a sin(θ 1 + ϕ 2 ) = l 3 sinϕ 1 + b sin(ϕ 1 + ϕ 3 ) θ 1 + ϕ 2 = ϕ 1 + ϕ 3 + π (4) Similar to equation (2) n = 1, m = 3 and J = 4 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 21 / 80

4-BAR EXAMPLE REVISITED Yet another way to break loop is shown in figure 8(c) Obtain L p and R p as L p = (l 1 cosθ 1,l 1 sinθ 1 ) T, R p = (l 3 cosϕ 1,l 3 sinϕ 1 ) T Enforce the constraint of constant length l 2 to obtain η 1 (θ 1,ϕ 1 ) = (l 1 cosθ 1 l 0 l 3 cosϕ 1 ) 2 + (l 1 sinθ 1 l 3 sinϕ 1 ) 2 l 2 2 = 0 (5) This constraint is analogue of S S pair constraint (see Slide # 34, Lecture 2, Module 2) for planar R R pair Only one constraint equation 1 q = (θ 1,ϕ 1 ), n = m = 1 & J = 4 1 In the four-bar kinematics this is the well known Freudenstein s equation (see Freudenstein, 1954) ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 22 / 80

TWO PROBLEMS IN KINEMATICS OF PARALLEL MANIPULATORS Direct Kinematics Problem: Two-part problem statement Step 1: Given the geometry of the manipulator and the actuated joint variables, obtain passive joint variables Step 2: Obtain position and orientation of a chosen output link Much harder than DK problem for a serial manipulator Leads to the notion of mobility and assemble-ability of a parallel manipulator or a closed-loop mechanism Inverse Kinematics Problem: Given the geometry of the manipulator and the position and orientation of the chosen end-effector or output link, obtain the actuated and passive joint variables Simpler than direct kinematics problem Generally simpler than IK of serial manipulators Often done in parallel One of the origins for the term parallel in parallel manipulators ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 23 / 80

SUMMARY Parallel manipulators: one or more loops & and no natural choice of end-effector Parallel manipulator Number of actuated joints less than total number of joints Degree-of-freedom is less than total number of joints Configuration space of parallel manipulator q = (θ,ϕ) Dimension of q chosen as small as possible Actuated variables θ R n, Passive variables ϕr m Need to derive m constraint equations Two problems Direct kinematics and inverse kinematics ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 24 / 80

DIRECT KINEMATICS OF PARALLEL MANIPULATORS The link dimensions and other geometrical parameters are known The values of the n actuated joints are known First obtain m passive joint variables Obtain (minimal) m loop-closure constraint equations in m passive and n active joint variables Use elimination theory/sylvester s dialytic method/bézout s method (see Module 3, Lecture 4) Solve set of m non-linear equations, if possible, in closed-form for the passive joint variables ϕ i, i = 1,,m Obtain position and orientation of chosen output link from known θ and ϕ Recall no natural end-effector and hence have to be chosen! No general method as compared to the direct kinematics of serial manipulator Approach illustrated with three examples ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 26 / 80

PLANAR 4-BAR MECHANISM φ3 O2 ˆX2 Link 2 O3 l2 ˆX3 ŶL {L} l1 Link 1 φ2 Link 3 l3 ŶR {R} ˆX1 φ1 θ1 ˆXL l0 OTool, OR ˆXR OL, O1 ˆXTool Figure 9: The four-bar mechanism - revisited Simplest possible closed-loop mechanism and studied extensively (see, for example Uicker et al, 2003) A good example to illustrate all steps in kinematics of parallel manipulators! Simple loop-closure equations All steps can be by hand! ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 27 / 80

4-BAR LOOP-CLOSURE EQUATIONS From loop-closure equations (4) (see Figure 8(b)), x l 0 = l 3 cosϕ 1 b cos(θ 1 +ϕ 2 ), y = l 3 sinϕ 1 b sin(θ 1 +ϕ 2 ) Denote δ = θ 1 + ϕ 2, squaring and adding A 1 cosδ + B 1 sinδ + C 1 = 0 (6) where A 1 = x l 0, B 1 = y, C 1 = (1/2b)[(x l 0 ) 2 + y 2 + b 2 l 2 3 ] From the first part of two equation (4) x = l 1 cosθ 1 + acos(θ 1 + ϕ 2 ), y = l 1 sinθ 1 + asin(θ 1 + ϕ 2 ) Squaring, adding, and after simplification gives A 2 cosδ + B 2 sinδ + C 2 = 0 (7) where A 2 = x, B 2 = y, C 2 = (1/2a)[l 2 1 a2 x 2 y 2 ] ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 28 / 80

4-BAR MECHANISM ELIMINATION Convert equations (6) and (7) to quadratics by tangent half-angle substitutions (see Module 3, Lecture 4) Following Sylvester s dialytic elimination method (see Module 3, Lecture 4), det[sm] = 0 gives (A 1 B 2 A 2 B 1 ) 2 = (A 1 C 2 A 2 C 1 ) 2 + (B 1 C 2 B 2 C 1 ) 2 ( ) and δ = 2tan 1 A 1 C 2 A 2 C 1 (B 1 C 2 B 2 C 1 )+(A 1 B 2 A 2 B 1 ) det[sm] = 0, after some simplification, gives 4a 2 b 2 l 0 2 y 2 = [b(x l 0 )(l 2 1 a 2 x 2 y 2 ) ax{(x l 0 ) 2 + y 2 + b 2 l 2 3 }] 2 + (8) y 2 [b(l 2 1 a 2 x 2 y 2 ) a{(x l 0 ) 2 + y 2 + b 2 l 2 3 }] 2 Above sixth-degree curve is the coupler curve 2 2 The coupler curve is extensively studied in kinematics of mechanisms For a more general form of the coupler curve and its interesting properties, see Chapter 6 of Hartenberg and Denavit (1964) ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 29 / 80

4-BAR SOLUTION FOR PASSIVE JOINT VARIABLES The elimination procedure gives δ as a function of (x,y) and the link lengths Since θ 1 is given, ( ) ϕ 2 = δ θ 1 = 2tan 1 A 1 C 2 A 2 C 1 θ 1 (B 1 C 2 B 2 C 1 ) + (A 1 B 2 A 2 B 1 ) (9) The angle ϕ 1 can be obtained from equation (5) l 2 0 + l 2 1 + l 2 3 l 2 2 = cosϕ 1 (2l 1 l 3 cosθ 1 2l 0 l 3 ) + sinϕ 1 (2l 1 l 3 ) (10) Finally, ϕ 3 can be solved from the third equation in equation (4) ϕ 3 = θ 1 + ϕ 2 ϕ 1 π (11) ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 30 / 80

3 4-BAR NUMERICAL EXAMPLE l 0 = 50, l 1 = 10, l 2 = 30, and l 3 = 40 The input link rotates fully (Grashof s criteria) Figure 10(a) shows plot of ϕ 1 vs θ 1 Both set of values plotted From ϕ 1 obtain ϕ 2 and ϕ 3 Two coupler curves shown 25 2 2 15 1 1 05 φ 1 0 y 0 1 05 1 2 15 2 3 0 1 2 3 4 5 6 7 θ 1 (a) ϕ 1 vs θ 1 for 4-bar mechanism 25 0 02 04 06 08 1 12 14 16 18 x (b) Coupler curves for 4-bar mechanism Figure 10: Numerical example for a 4-bar ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 31 / 80

A THREE DOF PARALLEL MANIPULATOR S3 p(x, y, z) Ẑ Moving Platform S2 S1 l2 D-H Table for a R-P-S leg (see Module 2, Lecture 2, Slide # 64) l3 θ3 {Base} O l1 θ2 Axis of R2 i α i 1 a i 1 d i θ i 1 0 0 0 θ 1 2 π/2 0 l 1 0 Axis of R3 Base Platform ˆX θ1 Ŷ Axis of R1 Figure 11: The 3-RPS parallel manipulator Revisited All legs are same θ 1, i = 1,2,3 are passive variables l i, i = 1,2,3 are actuated variables ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 32 / 80

3-DOF EXAMPLE LOOP-CLOSURE EQUATIONS Position vectors of three S joints (see Module 2, Lecture 2, Slide # 65) Base S 1 = (b l 1 cosθ 1,0,l 1 sinθ 1 ) T (12) Base S 2 = ( b 2 + 1 3 3 2 l 2 cosθ 2, 2 b 2 l 2 cosθ 2,l 2 sinθ 2 ) T Base S 3 = ( b 2 + 1 3 3 2 l 3 cosθ 3, 2 b + 2 l 3 cosθ 3,l 3 sinθ 3 ) T Base an equilateral triangle circumscribed by circle of radius b Impose S S pair constraint (see Module 2, Lecture 2, Slide # 34) η 1 (l 1,θ 1,l 2,θ 2 ) = ( Base S 1 Base S 2 ) 2 = k 2 12 η 2 (l 2,θ 2,l 3,θ 3 ) = ( Base S 2 Base S 3 ) 2 = k23 2 η 3 (l 3,θ 3,l 1,θ 1 ) = ( Base S 3 Base S 1 ) 2 = k31 2 (13) S joint variables do not appear Due to S S pair equations! ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 33 / 80

3-DOF EXAMPLE ELIMINATION Assume b = 1 and k 12 = k 23 = k 31 = 3a Eliminate using Sylvester s dialytic method (see Module 3, Lecture 4), θ 1 from η 1 ( ) = 0 and η 3 ( ) = 0 where η 4 (l 1,l 2,l 3,θ 2,θ 3 ) = (A 1 C 2 A 2 C 1 ) 2 + (B 1 C 2 B 2 C 1 ) 2 (A 1 B 2 A 2 B 1 ) 2 = 0 C 1 = 3 3a 2 + l 2 1 + l 2 2 3l 2c 2, A 1 = l 1 l 2 c 2 3l 1, B 1 = 2l 1 l 2 s 2 C 2 = 3 3a 2 + l 2 1 + l 2 3 3l 3c 3, A 2 = l 1 l 3 c 3 3l 1, B 2 = 2l 1 l 3 s 3 Eliminate θ 2 from η 4 ( ) = 0 and η 2 ( ) = 0, with x 3 = tan(θ 3 /2) q 8 (x 2 3 ) 8 + q 7 (x 2 3 ) 7 + + q 1 (x 2 3 ) + q 0 = 0 (14) An eight degree polynomial in x 2 3 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 34 / 80

3-DOF EXAMPLE ELIMINATION Expressions for q i obtained using symbolic algebra software, MAPLE R, are very large Two smaller ones are q 8 = (p 0 a 4 + p 1 a 3 + p 2 a 2 + p 3 a + p 4 ) 2 (p 0 a 4 p 1 a 3 + p 2 a 2 p 3 a + p 4 ) 2 q 0 = (r 0 a 4 + r 1 a 3 + r 2 a 2 + r 3 a + r 4 ) 2 (r 0 a 4 r 1 a 3 + r 2 a 2 r 3 a + r 4 ) 2 where r 0 = p 0 = 9, r 1 = 12(l 3 3), p 1 = 12(l 3 + 3), r 2 = 3(l 2 1 + l 2 2 l 3(l 3 10) 15), p 2 = 3(l 2 1 + l 2 2 l 3(l 3 + 10) 15), r 3 = 2(l 3 3)(l 2 1 + l 2 2 + l 2 3 3), p 3 = 2(l 3 + 3)(l 2 1 + l 2 2 + l 2 3 3), r 4 = l 4 3 8l 3 3 + 3l 2 2 + 18l 2 3 2l 3(l 2 2 + 6) l 2 1 (l 2 2 + 2l 3 3), and p 4 = l 4 3 + 8l 3 3 + 3l 2 2 + 18l 2 3 + 2l 3(l 2 2 + 6) + l 2 1 (l 2 2 + 2l 3 3) 8 possible values of θ 3 for given a and actuated variables (l 1,l 2,l 3 ) T Once θ 3 is obtained, θ 2 obtained from η 2 ( ) = 0 and θ 1 from η 3 ( ) = 0 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 35 / 80

3-DOF EXAMPLE (CONTD) A natural output link is the moving platform Position and orientation of the moving platform: Centroid of moving platform, Base p = 1 3 (Base S 1 + Base S 2 + Base S 3 ) (15) Orientation of moving platform or Base Top [R] is [ Base Base S Top [R] = 1 Base S 2 Base S 1 Base S 2 Ŷ (Base S 1 Base S 2 ) ( Base S 1 Base S 3 ) ( Base S 1 Base S 2 ) ( Base S 1 Base S 3 ) (16) where Ŷ is obtained from the cross-product of the third and first columns Once l i,θ i i = 1,2,3 are known Base p and Base Top [R] can be found Key step was the elimination of passive variables and obtaining a single equation in one passive variable! ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 36 / 80 ]

3-DOF EXAMPLE NUMERICAL EXAMPLE Polynomial in equation (14) is eight degree in (tanθ 3 /2) 2 Not possible to obtain closed-form expressions for θ 1, θ 2, and θ 3 Numerical solution using Matlab R For a = 1/2, and for l 1 = 2/3, l 2 = 3/5 and l 3 = 3/4 Two sets values θ 3 = ±08111, ±08028 radians For the positive values of θ 3, θ 2 = 04809, 02851 radians and θ 1 = 07471, 07593 radians respectively For the set (07471, 04809, 08111), Base p = (00117, 00044,04248) T, and The rotation matrix Base Top [R] is given by Base Top [R] = 08602 05069 00564 04681 08285 03074 02026 02380 09499 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 37 / 80

θ 1 b 6-DOF EXAMPLE D-H PARAMETERS 1 d Z h θ 3 l 11 b 3 d ψ 1 θ 2 l b 31 2 l 12 l ψ 21 φ ψ 1 2 l 32 l l 13 22 23 φ φ 3 2 p 1 Y s Figure 12: 3-RRRS parallel manipulator Revisited 3 X s l 33 l s p 3 p 2 D-H parameters for R-R-R-S chain (see Module 2, Lecture 2, Slide # 67) i α i 1 a i 1 d i θ i 1 0 0 0 θ 1 2 π/2 l 11 0 ψ 1 3 0 l 12 0 ϕ 1 D-H parameters for fingers in {F i }, i = 1,2,3 identical 6DOF parallel manipulator Only 6 out of 12 θ i, ψ i, ϕ i are actuated ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 38 / 80

6-DOF EXAMPLE LOOP-CLOSURE EQUATIONS Position vector of spherical joint i cosθ i (l i1 + l i2 cosψ i + l i3 cos(ψ i + ϕ i )) F i p i = sinθ i (l i1 + l i2 cosψ i + l i3 cos(ψ i + ϕ i )) l i2 sinψ i + l i3 sin(ψ i + ϕ i ) With respect to {Base}, the locations of {F i }, i = 1,2,3, are known and constant Base b 1 = (0, d,h) T, Base b 2 = (0,d,h) T, Base b 3 = (0,0,0) T Orientation of {F i }, i = 1,2,3, with respect to {Base} are also known - {F 1 } and {F 2 } are parallel to {Base} and {F 3 } is rotated by γ about the Ŷ The transformation matrices Base p i [T ] is Base F 1 [T ] 0 1 [T ]1 2 [T ]2 3 [T ]3 p 1 [T ] Last transformation includes l 13 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 39 / 80

6-DOF EXAMPLE LOOP-CLOSURE EQUATIONS Extract position vector Base p 1 from last column of Base F 1 [T ] Base p 1 = Base b 1 + F 1 p 1 = cosθ 1 (l 11 + l 12 cosψ 1 + l 13 cos(ψ 1 + ϕ 1 )) d + sinθ 1 (l 11 + l 12 cosψ 1 + l 13 cos(ψ 1 + ϕ 1 )) h + l 12 sinψ 1 + l 13 sin(ψ 1 + ϕ 1 ) Similarly for second leg cosθ 2 (l 21 + l 22 cosψ 2 + l 23 cos(ψ 2 + ϕ 2 )) Base p 2 = d + sinθ 2 (l 21 + l 22 cosψ 2 + l 23 cos(ψ 2 + ϕ 2 )) h + l 22 sinψ 2 + l 23 sin(ψ 2 + ϕ 2 ) For third leg Base p 3 = cosθ 3 (l 31 + l 32 cosψ 3 + l 33 cos(ψ 3 + ϕ 3 )) [R(Ŷ,γ)] sinθ 3 (l 31 + l 32 cosψ 3 + l 33 cos(ψ 3 + ϕ 3 )) l 32 sinψ 3 + l 33 sin(ψ 3 + ϕ 3 ) ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 40 / 80

6-DOF EXAMPLE LOOP-CLOSURE EQUATIONS Use S S pair constraint to get 3 loop-closure equations η 1 (θ 1,ψ 1,ϕ 1,θ 2,ψ 2,ϕ 2 ) = Base p 1 Base p 2 2 = k 2 12 η 2 (θ 2,ψ 2,ϕ 2,θ 3,ψ 3,ϕ 3 ) = Base p 2 Base p 3 2 = k 2 23 (17) η 3 (θ 3,ψ 3,ϕ 3,θ 1,ψ 1,ϕ 1 ) = Base p 3 Base p 1 2 = k 2 31 where k 12, k 23 and k 31 are constants Actuated: θ 1,ψ 1, θ 2,ψ 2, θ 3, ψ 3 & Passive: ϕ 1, ϕ 2, ϕ 3 Obtain expressions for passive variables using elimination Eliminate ϕ 1 from first and third equation (17) η 4 (ϕ 2,ϕ 3,, ) = 0 Eliminate ϕ 2 from η 4 (ϕ 2,ϕ 3,, ) = 0 and second equation (17) Single equation in ϕ 3 Final equation is 16 th degree polynomial in tan(ϕ 3 /2) Obtained using symbolic algebra software MAPLE R Expressions for the coefficients of the polynomial very long! Numerical example shown next ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 41 / 80

6-DOF EXAMPLE NUMERICAL RESULTS Assume d = 1/2, h = 3/2, l i1 = 1, l i2 = 1/2, l i3 = 1/4 (i = 1,2,3), γ = π/4 and k 12 = k 23 = k 13 = 3/2 For the actuated joint variables, choose θ 1 = 01, ψ 1 = 10, θ 2 = 01, ψ 2 = 12, θ 3 = 03, ψ 3 = 10 radians The sixteenth degree polynomial is obtained as 000012t3 16 000182t3 15 3 005230t13 3 + 013148t12 3 024391t3 11 3 040965t9 3 + 038696t8 3 029811t3 7 + 018502t6 3 009104t5 3 + 003433t4 3 000968t3 3 + 000201t2 3 000037t 3 + 000006 = 0 where t 3 = tan(ϕ 3 /2) Numerical solution gives two real values of ϕ 3 as (08831, 18239) radians Corresponding values of ϕ 1 and ϕ 2 are (03679,01146) radians and (14548, 10448) radians, respectively ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 42 / 80

6-DOF EXAMPLE NUMERICAL RESULTS The position vector of centroid, computed as in the 3-RPS example, using the first set of θ i, ψ i, ϕ i is Base p = 1 3 (Base p 1 + Base p 2 + Base p 3 ) = (13768,02624,01401) T The rotation matrix Base Object [R], computed similar to the 3-RPS example, is 00306 02099 09773 Base Object [R] = 09811 01806 00695 01910 09609 02004 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 43 / 80

MOBILITY OF PARALLEL MANIPULATORS Concept of workspace in serial manipulators All (x,y,z;[r]) such that real solutions for the inverse kinematics exists In parallel manipulators two concepts: mobility and workspace Workspace dependent on the choice of output link Mobility: range of possible motion of the actuated joints in a parallel manipulator Mobility is more important in parallel manipulators! Mobility is determined by geometry/linkage dimensions Loop-closure constraint equations Mobility is related to the ability to assemble a parallel manipulator at a configuration ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 45 / 80

MOBILITY OF PARALLEL MANIPULATORS Mobility: All values of actuated variables such that real value(s) of passive variables exists Determined by direct kinematics No real value of passive variable Cannot be assembled Mobility Obtain conditions for existence of real solutions for the polynomial in one passive variable obtained after elimination Very few parallel manipulators where the direct kinematics can be reduced to the solution of a univariate polynomial of degree 4 or less In most cases mobility determined numerically using search In 4-bar mechanism, mobility can be obtained in closed-form ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 46 / 80

MOBILITY OF 4-BAR MECHANISM Loop-closure constraint equation of a 4-bar η 1 (θ 1,ϕ 1 ) = (l 1 cosθ 1 l 0 l 3 cosϕ 1 ) 2 +(l 1 sinθ 1 l 3 sinϕ 1 ) 2 l 2 2 = 0 On simplification η 1 becomes where P, Q, and R are given by P cosϕ 1 + Q sinϕ 1 + R = 0 (18) P = 2l 0 l 3 2l 1 l 3 c 1, Q = 2l 1 l 3 s 1 R = l 2 0 + l 2 1 + l 2 3 l 2 2 2l 0 l 1 c 1 l 0, l 1, l 2, and l 3 are the link lengths (see figure 6), and c 1, s 1 are the sine and cosine of θ 1, respectively Using tangent half-angle substitutions (see Module 3, Lecture 3) ( Q ± P ϕ 1 = 2tan 1 2 + Q 2 R 2 ) (19) R P ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 47 / 80

MOBILITY OF 4-BAR MECHANISM For real ϕ 1, P 2 + Q 2 R 2 0 Limiting case: P 2 + Q 2 R 2 = 0 Two ϕ 1 s coinciding In the limiting case, the bounds on θ 1 are c 1 = l 2 0 + l 2 1 l 2 3 l 2 2 ± 2l 3l 2 2l 0 l 1 (20) For full rotatability of θ 1 (0 θ 1 2π), θ 1 cannot have any bounds For θ 1 to have full rotatability there cannot be a solution to equation (20)! For full rotatability of θ 1, c 1 > 1 or c 1 < 1 in equation (20) ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 48 / 80

MOBILITY OF 4-BAR MECHANISM For full rotatability/mobility of θ 1, first ϕ 1 be real and then θ 1 be imaginary > Note the order of ϕ 1 and θ 1 The condition c 1 > 1 and c 1 < 1 leads to (l 0 l 1 ) 2 > (l 3 l 2 ) 2 (21) and (l 0 + l 1 ) < (l 3 + l 2 ) (22) Two additional conditions from c 1 > 1, c 1 < 1 lead to l 3 + l 2 + l 1 < l 0 and l 0 + l 1 + l 2 < l 3 Violates triangle inequality Equation (21) gives rise to four inequalities l 0 l 1 > l 3 l 2 l 0 l 1 > l 2 l 3 l 1 l 0 > l 3 l 2 (23) l 1 l 0 > l 2 l 3 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 49 / 80

MOBILITY OF 4-BAR MECHANISM For the case of l 1 < l 0 l 0 + l 2 > l 1 + l 3 l 0 + l 3 > l 1 + l 2 (24) Equations (22) and (24) imply that l 0, l 2 and l 3 are all larger than l 1 Equations (22) and (24) l + s < p + q s, l are the shortest and largest links and p, q are intermediate links Likewise, for l 1 > l 0 l 1 + l 2 > l 0 + l 3 l 1 + l 3 > l 0 + l 2 (25) and again l 0 is the shortest link Concisely represent equations (22) and (25) as l +s < p +q Same as the Grashof s criterion for 4-bar linkages ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 50 / 80

3-DOF PARALLEL MANIPULATOR Three-DOF parallel (3-RPS) manipulator Polynomial is eight degree in x 2 3 a = 05 and (l 1,l 2,l 3 ) [05, 15] Points marked as No real and positive values of x 2 3 Finer search More accurate mobility region l 3 15 1 05 15 1 l 2 Figure 13: Values of (l 1,l 2,l 3 ) for imaginary θ 3 (marked by ) 05 05 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 51 / 80 l 1 1 15

SUMMARY Mobility in parallel manipulators is analogous to workspace 3 in serial manipulators Actuated joint motion can be restricted and not due to joint limits! Mobility of actuated joints determines if an parallel manipulator/mechanism can be assembled in a configuration If no real solution to direct kinematics problem Not possible to assemble Analytical solution for mobility of a 4-bar mechanism yields the well-known Grashof criterion Difficult to find mobility analytically for other manipulators/mechanisms Numerical search based approach can be used 3 Some authors use mobilty in the same sense as degree-of-freedom! ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 52 / 80

INVERSE KINEMATICS OF PARALLEL MANIPULATORS Problem statement: given geometry and link parameters, position and orientation of a chosen output link with respect to a fixed frame, Find the joint (actuated and passive) joint variables Simpler than the direct kinematics problem since no need to worry about the multiple loops or the loop-closure constraint equations Key idea is to break the mechanism into serial chains and obtain the joint angles of each chain in parallel Break parallel manipulators into chains such that no chain is redundant Worst case: Solution of inverse kinematics of a general 6R serial manipulator (See Module 3, Lecture 4) ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 54 / 80

PLANAR 4-BAR MECHANISM ŶL {L} l1 φ2 Link 1 L p θ 1 OL, O1 (x,y) a ˆXL φ l0 (x,y) φ 3 b R p Figure 14: Inverse kinematics of a four-bar mechanism Link 3 l3 O R ŶR {R} φ1 ˆXR Coupler is the chosen output link Given the position of a point L p and the rotation matrix L 2 [R] of the coupler link Planar case x,y coordinates and the orientation angle ϕ given Lengths l 0, l 1, l 2 = a + b, a, b and l 3 are known ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 55 / 80

PLANAR 4-BAR MECHANISM We have x = l 1 cosθ 1 + acos(θ 1 + ϕ 2 ), y = l 1 sinθ 1 + asin(θ 1 + ϕ 2 ) where x and y are known The angle ϕ denoting the orientation of link 2 is given by Solve for θ 1 and ϕ 2 as ϕ = θ 1 + ϕ 2 2π θ 1 = atan2(y asinϕ, x acosϕ), ϕ 2 = ϕ θ 1 In a similar manner, considering the equations x = l 0 + l 3 cosϕ 1 + b cos(ϕ 1 + ϕ 3 ), y = l 3 sinϕ 1 + b sin(ϕ 1 + ϕ 3 ) ϕ = ϕ 1 + ϕ 3 π solve for ϕ 1 and ϕ 3 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 56 / 80

PLANAR 4-BAR MECHANISM ϕ obtained as θ 1 + ϕ 2 2π and as ϕ 1 + ϕ 3 π must be same The four-bar mechanism is a one- degree-of-freedom mechanism and only one of (x,y,ϕ) can be independent x and y are related through the sixth-degree coupler curve (see equation (8)) ϕ must satisfy x cosϕ + y sinϕ = (1/2a)(x 2 + y 2 a 2 l 2 1 ) The constraints on the given position and orientation of the chosen output link, x,y,ϕ, are analogous to the case of the inverse kinematics of serial manipulators when n < 6 (see Module 3, Lecture 3) The inverse kinematics of a four-bar mechanism can be solved when the given position and orientation is consistent ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 57 / 80

A 6-DOF PARALLEL MANIPULATOR Ẑ d d Figure shows one finger as an RRRS chain θ1 h l11 {Base} Ẑ ψ1 Ŷ ˆX l12 Base p φ1 l13 S1 {Object} S2 S3 Given the position and orientation of the gripped object with respect to {Base} Obtain the rotations at the nine joints in the three fingers {F3} γ ˆX Figure 15: Inverse kinematics of sixdegree-of-freedom parallel manipulator ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 58 / 80

INVERSE KINEMATICS OF 6-DOF PARALLEL MANIPULATOR Vector Base p locates the centroid of the gripped object Base Object [R]is also available In {Object}, the location of S 1, Object S 1, is known Hence, (x,y,z) T = Base S 1 = Base Object [R]Object S 1 + Base p Object is known From above (x,y,z) T = cosθ 1 (l 11 + l 12 cosψ 1 + l 13 cos(ψ 1 + ϕ 1 )) d + sinθ 1 (l 11 + l 12 cosψ 1 + l 13 cos(ψ 1 + ϕ 1 )) h + l 12 sinψ 1 + l 13 sin(ψ 1 + ϕ 1 ) Equation (26) can be solved for θ 1, ψ 1 and ϕ 1 using elimination (see Module 3, Lecture 4) from known (x,y,z) T (26) ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 59 / 80

6-DOF PARALLEL MANIPULATOR (CONTD) From equation (26), we get x 2 + (y + d) 2 + (z h) 2 = l 2 11 + l 2 12 + l 2 13 + 2l 11 l 12 cosψ 1 +2l 12 l 13 cosϕ 1 + 2l 11 l 13 cos(ψ 1 + ϕ 1 ) (27) Equation (27) and last equation in (26) can be written as where A i cosψ 1 + B i sinψ 1 + C i = 0, i = 1,2 (28) A 1 = 2l 11 l 12 + 2l 11 l 13 cosϕ 1, A 2 = l 13 sinϕ 1 C 1 = l 2 11 + l 2 12 + l 2 13 + 2l 12 l 13 cosϕ 1 x 2 (y + d) 2 (z h) 2 B 1 = 2l 11 l 13 sinϕ 1, B 2 = l 12 + l 13 cosϕ 1, C 2 = h z ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 60 / 80

6-DOF PARALLEL MANIPULATOR (CONTD) Following Sylvester s dialytic method, eliminate ψ 1 to get 4l 2 11(l 2 12 + l 2 13 + 2l 12 l 13 cosϕ 1 ) = C 2 1 + 4l 2 11(h z) 2 Using tangent half-angle formulas for cosϕ 1 and sinϕ 1, we get a quartic equation a 4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0 = 0 (29) where x = tan(ϕ 1 /2) Solve for ϕ 1 from the quartic and obtain ψ 1 as ( ) ψ 1 = 2tan 1 A 1 C 2 A 2 C 1 (B 1 C 2 B 2 C 1 ) + (A 1 B 2 A 2 B 1 ) Finally, θ 1 is obtained from (30) θ 1 = atan2(y + d,x) (31) The joint variables for the other two fingers can be obtained in same way! ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 61 / 80

IK OF GOUGH-STEWART PLATFORM {P 0 } B0 t P0 p i B0 b i l i P i S Joint P Joint Ẑ Ŷ B i ˆX From Figure 16, an arbitrary platform point P i can be written in {B 0 } as B 0 p i = B 0 P 0 [R] P 0 p i + B 0 t (32) The P 0 p i is a known constant vector in {P 0 } The location of the base connection points B 0 b i are known {B 0 } Figure 16: U Joint A leg of a Stewart platform ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 62 / 80

IK OF GOUGH-STEWART PLATFORM From known B 0 P 0 [R] and translation vector B 0 t, obtain B 0 p 1 [R(Ẑ,γ i )] T ((x,y,z) T B 0 b 1 ) = [R(Ŷ,ϕ i )][R(ˆX,ψ i )](0,0,l i ) T sinϕ 1 cosψ 1 = l 1 sinψ 1 cosϕ 1 cosψ 1 (33) where B 0 p 1 is denoted by (x,y,z) T Three non-linear equations in l 1, ψ 1, ϕ 1 solution l 1 = ± [(x,y,z) T B 0 b1 ] 2 ψ 1 = atan2( Y,± X 2 + Z 2 ) (34) ϕ 1 = atan2(x /cosψ 1,Z/cosψ 1 ) where X,Y,Z are the components of [R(Ẑ,γ i)] T ((x,y,z) T B 0 b 1 ) Perform for each leg to obtain l i, ψ i and ϕ i for i = 1,,6 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 63 / 80

SUMMARY Inverse kinematics involve obtaining actuated joint variables given chosen end-effector position and orientation Key concept is to break the parallel manipulator into simple serial chains Inverse kinematics problem can be solved by considering each serial chain in parallel Inverse kinematics of Gough-Stewart platform much simpler than direct kinematics In general, inverse kinematics problem simpler for parallel manipulator! ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 64 / 80

GOUGH-STEWART MANIPULATORS PLATFORM Gough-Stewart platform Six-DOF parallel manipulator Extensively used in flight simulators, machine tools, force-torque sensors, orienting device etc (Merlet, 2001) Moving Platform P3 P2 Top Platform Moving Platform P2 {P0} P2 P1 P4 P1 Spherical Joint P3 P0 l5 P1 l3 P3 P0 l2 l1 Spherical Joint P5 P6 Prismatic Joint l2 l6 l3 B2 l4 l1 Spherical Joint l4 B3 l5 l6 B2 Prismatic Joint B5 {B0} B6 B1 B0 Prismatic Joint B4 B0 B1 B4 U Joint B3 B1 B0 U Joint Fixed Base Note: Each Base Joint is an U Joint Fixed Base B5 B6 Note: Each Base Joint is an U Joint U Joint Extensible Leg B3 Fixed Base B2 (a) 3 3 Stewart platform (b) 6 3 Stewart platform (c) 6 6 Stewart platform Figure 17: Three configurations of Stewart platform manipulator ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 66 / 80

GEOMETRY OF A LEG {P 0} P0 p i P i B0 t B0 b i l i S Joint P Joint Ẑ Ŷ B i ˆX Hooke ( U ) joint modeled as 2 intersecting R joint Each leg RRPS chain Hooke joint equivalent to successive Euler rotations (see Module 2, Lecture 2, Lecture 2) ϕ i about Ŷ i and ψ i about ˆX i {B 0} Figure 18: -revisited U Joint A leg of a Stewart platform ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 67 / 80

GEOMETRY OF A LEG (CONTD) The vector B 0 p i locating the spherical joint can be written as B 0 p i = B 0 b i + [R(Ẑ,γ i )][R(Ŷ,ϕ i )][R(ˆX,ψ i )](0,0,l i ) T cosγ i sinϕ i cosψ i + sinγ i sinψ i = B 0 b i + l i sinγ i sinϕ i cosψ i cosγ i sinψ i cosϕ i cosψ i (35) Constant vector B 0b i locates the origin O i {i} at the Hooke joint i, Constant angle γ i determines the orientation of {i} with respect to {B 0 }, and l i is the translation of the prismatic (P) joint in leg i B 0 p i is a function of two passive joint variables, ϕ i and ψ i, and the actuated joint variable l i ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 68 / 80

DK OF 3 3 CONFIGURATION 6 legs are B 1 P 1, B 1 P 3, B 2 P 1, B 2 P 2, B 3 P 2 and B 3 P 3 (see Figure 17(a)) 6 actuated and 12 passive variables 12 constraint equations needed Three constraints: Distances between P 1, P 2 and P 3 are constant (similar to 3-RPS) Point P 1 reached in two ways: 3 vector equations or 9 scalar equations B 0 b 1 + B 1 P 1 = B 0 b 2 + B 2 P 1 B 0 b 2 + B 2 P 2 = B 0 b 3 + B 3 P 2 B 0 b 3 + B 3 P 3 = B 0 b 1 + B 1 P 3 16 th degree polynomial in tangent half-angle obtained after elimination (Nanua, Waldron, Murthy, 1990) ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 69 / 80

DK OF 6 3 CONFIGURATION Direct kinematics similar to 3 3 configurations (see Figure 17(b)) 6 legs are B 1 P 1, B 2 P 1, B 3 P 2, B 4 P 2, B 5 P 3 and B 6 P 3 6 actuated and 12 passive variables 12 constraint equations needed Three constraints: Distances between P 1, P 2 and P 3 are constant (similar to 3-RPS) P 1, P 2 and P 3 reached in two ways 9 scalar equations B 0 b 1 + B 1 P 1 = B 0 b 2 + B 2 P 1 B 0 b 3 + B 3 P 2 = B 0 b 4 + B 4 P 2 B 0 b 5 + B 5 P 3 = B 0 b 6 + B 6 P 3 16 th degree polynomial in tangent half-angle obtained after elimination ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 70 / 80

DK OF 6 6 CONFIGURATION IN JOINT SPACE 6 distinct points in the fixed base and moving platform (see Figure 17(c)) Hooke joint modeled as 2 intersecting rotary (R) joint 6 actuated and 12 passive variables Need 12 constraint equations! B 0 p i revisited B 0 p i = B 0 b i + [R(Ẑ,γ i)][r(ŷ,ϕ i )][R(ˆX,ψ i )](0,0,l i ) T cosγ i sinϕ i cosψ i + sinγ i sinψ i = B 0 b i + l i sinγ i sinϕ i cosψ i cosγ i sinψ i cosϕ i cosψ i (36) 6 constraint equations from S S pair constraints (see Module 2, Lecture 2) ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 71 / 80

DK OF 6 6 CONFIGURATION IN JOINT SPACE 6 S S pair constraints η 1 (q) = B 0 p 1 B 0 p 2 2 d12 2 = 0 η 2 (q) = B 0 p 2 B 0 p 3 2 d23 2 = 0 η 3 (q) = B 0 p 3 B 0 p 4 2 d34 2 = 0 η 4 (q) = B 0 p 4 B 0 p 5 2 d45 2 = 0 (37) η 5 (q) = B 0 p 5 B 0 p 6 2 d56 2 = 0 η 6 (q) = B 0 p 6 B 0 p 1 2 d61 2 = 0 Need another 6 independent constraint equations ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 72 / 80

DK OF 6 6 CONFIGURATION IN JOINT SPACE Distance between point B 0 p 1 and B 0 p 3, B 0 p 4 and B 0 p 5 must be constant η 7 (q) = B 0 p 1 B 0 p 3 2 d 2 13 = 0 η 8 (q) = B 0 p 1 B 0 p 4 2 d 2 14 = 0 η 9 (q) = B 0 p 1 B 0 p 5 2 d 2 15 = 0 (38) All six points P i, i = 1,,6 must lie on a plane η 10 (q) = ( B 0 p 1 B 0 p 3 ) ( B 0 p 1 B 0 p 4 ) ( B 0 p 1 B 0 p 2 ) = 0 η 11 (q) = ( B 0 p 1 B 0 p 4 ) ( B 0 p 1 B 0 p 5 ) ( B 0 p 1 B 0 p 3 ) = 0 η 12 (q) = ( B 0 p 1 B 0 p 5 ) ( B 0 p 1 B 0 p 6 ) ( B 0 p 1 B 0 p 4 ) = 0 (39) d ij is the known distance between the spherical joints S i and S j on the top platform ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 73 / 80

DK OF 6 6 CONFIGURATION IN JOINT SPACE 12 non-linear equations in twelve passive variables ϕ i,ψ i, i = 1,,6, and six actuated joint variables l i, i = 1,,6 All equations do not contain all passive variables First equation in (37) is a function of only ϕ 1, ψ 1, l 1, ϕ 2, ψ 2, and l 2 12 equations are not unique and one can have other combinations For direct kinematics, eliminate 11 passive variables from these 12 equations Very hard and not yet done! Direct kinematics of Gough-Stewart platform easier with task space variables ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 74 / 80

DK OF 6 6 CONFIGURATION IN TASK SPACE {B 0} B0 t {P 0} P0 p i B0 b i U Joint S Joint Ẑ P Joint Figure 19: A leg of a Stewart platform -revisited l i P i Ŷ B i ˆX The point P i in {B 0 } B 0 p i = B 0 P 0 [R] P 0 p i + B 0 t (40) where P 0 p i = (p ix,p iy,0) T Denoting point B i by B 0 B i, the leg vector B 0 S i is B 0 S i = B 0 P 0 [R] P 0 p i + B 0 t B 0 b i (41) where B 0 b i = (b ix,b iy,0) T ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 75 / 80

DK OF 6 6 CONFIGURATION IN TASK SPACE The magnitude of the leg vector is l 2 i = (r 11 p ix + r 12 p iy + t x b ix ) 2 + (r 21 p ix + r 22 p iy + t y b iy ) 2 +(r 31 p ix + r 32 p iy + t z b iz ) 2 (42) Using properties of the elements r ij, get (t 2 x + t 2 y + t 2 z ) + 2p ix (r 11 t x + r 21 t y + r 31 t z ) + 2p iy (r 12 t x + r 22 t y + r 32 t z ) 2b ix (t x + p ix r 11 + p iy r 12 ) 2b iy (t y + p ix r 21 + p iy r 22 ) +b 2 i x + b 2 i y + p 2 i x + p 2 i y l 2 i = 0 (43) For six legs, i = 1,,6, six equations of type shown above Additional 3 constraints r 2 11 + r 2 21 + r 2 31 = 1 r 2 12 + r 2 22 + r 2 32 = 1 (44) r 11 r 12 + r 21 r 22 + r 31 r 32 = 0 ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 76 / 80

DK OF 6 6 CONFIGURATION IN TASK SPACE Equations (43) and (44) are nine quadratic equations in nine unknowns, t x, t y, t z, r 11, r 12, r 21, r 22, r 31, and r 32 (see Dasgupta and Mruthyunjaya, 1994) All quadratic terms in equation (43) are square of the magnitude of the translation vector (tx 2 + ty 2 + tz 2 ), and as X and Y component of the vector B 0 t, (r 11 t x + r 21 t y + r 31 t z ) and (r 12 t x + r 22 t y + r 32 t z ), respectively Reduce 9 quadratics to 6 quadratic and 3 linear equations in nine unknowns Starting point of elimination Very hard to eliminate 8 variables from 9 equations to arrive at a univariate polynomial in one unknown Univariate polynomial widely accepted to be of 40th degree (Raghavan, 1993 & Husty, 1996) Continuing attempts to obtain simplest explicit expressions for co-efficients of 40th-degree polynomial ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 77 / 80

SUMMARY Gough-Stewart platform Most important parallel manipulator (see also Module 10, Lecture 2) Most often a symmetric version (also called Semi-Regular Stewart Platform Manipulator SRSPM) is used Extensively used and studied Direct kinematics of 3 3 and 6 3 well understood 6 6 configuration still being studied for simplest direct kinematics equations ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 78 / 80

ADDITIONAL MATERIAL Exercise Problems References & Suggested Reading To view above links Copy link and paste in a New Window/Tab by right click Close new Window/Tab after viewing ASHITAVA GHOSAL (IISC) ROBOTICS: ADVANCED CONCEPTS & ANALYSIS NPTEL, 2010 80 / 80