DYNAMICS OF PARALLEL MANIPULATOR

DYNAMICS OF PARALLEL MANIPULATOR

PARALLEL MANIPULATORS 6-degree of Freedom Flight Simulator

BACKGROUND Platform-type parallel mechanisms 6-DOF MANIPULATORS

INTRODUCTION Under alternative robotic mechanical systems we understand here: Parallel Robots; Multifingered hands; Walking Machines; Rolling Robots. Parallel manipulators are composed of kinematic chains with closed subchains.

KINEMATICS OF PARALLEL MANIPULATORS General six-dof parallel manipulator

KINEMATICS OF PARALLEL MANIPULATORS We can distinguish two platforms: One fixed to the ground B One capable of moving arbitrarily within its workspace M. The moving platform is connected to the fixed platform through six legs, each being regarded as a six-axis serial manipulator whose base is B and whose end-effector is M. The whole leg is composed of six links coupled through six revolutes.

KINEMATICS OF PARALLEL MANIPULATORS Six-dof flight simulator: (a) general layout; (b) geometry of its two platforms

KINEMATICS OF PARALLEL MANIPULATORS In this figure, the fixed platform B is a regular hexagon, while the moving platform M is an equilateral triangle. Moreover, B is connected to M by means of six serial chains, each comprising five revolutes and one prismatic pair. Three of the revolutes bear concurrent axes, and hence, constitute a spherical joint, while two more have axes intersecting at right angles, thus constituting a universal joint. It is to be noted that although each leg of the manipulator has a spherical joint at only one end and a universal joint at the other end.

KINEMATICS OF PARALLEL MANIPULATORS A layout of a leg of the manipulator

KINEMATICS OF PARALLEL MANIPULATORS We analyze the inverse kinematics of one leg of the manipulator. The Denavit-Hartenberg parameters of the leg shown in this figure are given in table: i ai bi αi 1 0 0 90 2 0 0 90 3 0 b 3 0 4 0 b 4 (const) 90 5 0 0 90 6 0 b 6 (const) 0

KINEMATICS OF PARALLEL MANIPULATORS It is apparent that the leg under study is a decoupled manipulator. In view of the DH parameters of this manipulator, we have: Q 1 Q 2 a 3 + a 4 = c where c denotes the position vector of the center C of the spherical wrist and, since frames F 3 and F 4 of the DH notation are related by a pure translation, Q 3 = 1. Upon equating the squares of the Euclidean norms of both sides of the foregoing equation, we obtain: a 3 + a 4 2 = c 2

KINEMATICS OF PARALLEL MANIPULATORS Where, by virtue of the DH parameters of Table: a 3 + a 4 2 = b 3 + b 4 2 Now, since both b 3 and b 4 are positive by construction, we have: b 3 = c b 4 > 0 Note that the remaining five joint variables of the leg under study are not needed for purposes of inverse kinematics, and hence, their calculation could be skipped.

KINEMATICS OF PARALLEL MANIPULATORS We derive three scalar equations in two unknowns, θ 1 and θ 2, namely, b 3 + b 4 = x c c 1 + y c s 1 (b 3 + b 4 = z c 0 = x c s 1 y c c 1 in which c i and s i stand for cosθi and sin θi, respectively θ 1 is derived as: θ 1 = tan y c x c which yields a unique value of θ 1 rather than the two lying π radians apart, for the two coordinates x C determine the quadrant in which θ 1 lies. and y C

KINEMATICS OF PARALLEL MANIPULATORS Once θ 1 is known, θ 2 is derived uniquely from the remaining two equations through its cosine and sine functions, i.e., z c c 2 = s b 3 + b 2 = x cc 1 + y c s 1 4 b 3 + b 4 With the first three joint variables of this leg known, the remaining ones, are calculated as described in foregoing chapter. Therefore, the inverse kinematics of each leg admits two solutions, one for the first three variables and two for the last three.

KINEMATICS OF PARALLEL MANIPULATORS Moreover, since the only actuated joint is one of the first three, which of the two wrist solutions is chosen does not affect the value of b 3, and hence, each manipulator leg admits only one inverse kinematics solution. While the inverse kinematics of this leg is quite straightforward, its direct kinematics is not. Below we give an outline of the solution procedure for the manipulator under study

KINEMATICS OF PARALLEL MANIPULATORS Equivalent simplified mechanism

KINEMATICS OF PARALLEL MANIPULATORS We can replace the pair of legs by a single leg of length li connected to the base plate B by a revolute joint with its axis along A i B i. The resulting simplified structure is kinematically equivalent to the original structure. Now we introduce the coordinate frame F i with origin at the attachment point O i of the ith leg with the base plate B and the notation below: For i=1,2,3 X i is directed from A i to B i Y i is chosen such that Z i is perpendicular to the plane of the hexagonal base and points upwards. O i is set at the intersection of X i and Y i and is designated the center of the revolute joint;

KINEMATICS OF PARALLEL MANIPULATORS Next, we locate the three vertices S 1, S 2 and S 3 of the triangular plate with position vectors stemming from the center O of the hexagon. Furthermore, we need to determine l i and O i. Referring to Figures and letting a i and b i denote the position vectors of points A i and B i respectively, we have: d i = b i a i

KINEMATICS OF PARALLEL MANIPULATORS r i = d i 2 + q ia 2 q ib 2 2d i l i = q ia 2 r i 2 u i = b i a i d i for i = 1,2,3, and hence u i is the unit vector directed from A i to B i.

KINEMATICS OF PARALLEL MANIPULATORS Moreover, the position of the origin O i is given by vector o i, as indicated below: o i = a i + r i u i For i= 1,2,3 Furthermore, let s i be the position vector of S i in frame F i (O i, X i, Y i, Z i ), then: s i = 0 l i cosφ i l i sinφ i For i= 1,2,3

KINEMATICS OF PARALLEL MANIPULATORS Now a frame F 0 (O, X, Y, Z) is defined with origin at O and axes X and Y in the plane of the base hexagon, and related to X i and Y i as depicted in figure. When expressed in frame F 0, s i takes on the form: s i 0 = o i 0 + R i 0 s i For i=1,2,3 Where [R i ] 0 is the matrix that rotate F 0 to frame F i, expressed in and given as: cosα i sinα i 0 R i 0 = sinα i cosα i 0 0 0 1

KINEMATICS OF PARALLEL MANIPULATORS Referring to Figure: cosα i = u i i = u ix sin α i = u i j = u iy

KINEMATICS OF PARALLEL MANIPULATORS After substitution we obtain: s i 0 = o i 0 + l i u ix cosφ i u iy cosφ i sin φ i For i=1,2,3 Since the distances between the three vertices of the triangular plate are fixed, the position vectors s 1, s 2, and s 3 must satisfy the constraints below: s 2 s 1 2 = a 1 2 s 3 s 2 2 = a 2 2 s 1 s 3 2 = a 3 2

KINEMATICS OF PARALLEL MANIPULATORS After expansion, equations take the forms: D 1 cφ 1 + D 2 cφ 2 + D 3 cφ 1 cφ 2 + D 4 sφ 1 sφ 2 + D 5 = 0 (1) E 1 cφ 2 + E 2 cφ 3 + E 3 cφ 2 cφ 3 + E 4 sφ 2 sφ 3 + E 5 (2) F 1 cφ 1 + F 2 cφ 3 + F 3 cφ 1 cφ 3 + F 4 sφ 1 sφ 3 + F 5 = 0 (3) where c( ) and s( ) stand for cos( ) and sin( ), respectively, while coefficients D i, E i, F 5 i 1 functions of the data only and bear the forms shown below: D 1 = 2l 1 o 2 o T 1 Eu 1 D 2 = 2l 2 o 2 o T 1 Eu 2 D 3 = 2l 1 l 2 u T 2 u 1

KINEMATICS OF PARALLEL MANIPULATORS D 4 = 2l 1 l 2 D 5 = o 2 2 + o 2 1 2o T 1 o 2 + l 2 1 + l 2 2 2 a 1 E 1 = 2l 2 o 3 o T 2 Eu 2 E 2 = 2l 3 o 3 o T 2 Eu 3 E 3 = 2l 2 l 3 u T 3 u 2 E 4 = 2l 2 l 3 E 5 = o 2 3 + o 2 1 2o T 3 o 2 + l 2 2 + l 2 2 3 a 2 F 1 = 2l 1 o 1 o T 3 Eu 1 F 2 = 2l 3 o 1 o T 3 Eu 3 F 3 = 2l 1 l 3 u T 3 u 1 F 4 = 2l 1 l 3 F 5 = o 2 3 + o 2 1 2o T 3 o 1 + l 2 1 + l 2 2 3 a 3

KINEMATICS OF PARALLEL MANIPULATORS Our next step is to reduce the foregoing system of three equations in three unknowns to two equations in two unknowns, and hence, obtain two contours in the plane of two of the three unknowns, the desired solutions being determined as the intersections of the two contours. Since the first equation is already free of Φ 3, all we have to do is eliminate it from the other equations. The resulting equations take on the forms: k 1 τ 3 2 + k 2 τ 3 + k 3 = 0 m 1 τ 3 2 + m 2 τ 3 + m 3 = 0

KINEMATICS OF PARALLEL MANIPULATORS Where k 1, k 2, and k 3 are linear combinations of s 2, c 2, and 1. Likewise, m 1, m 2, and m 3 are linear combinations of s 1, c 1 and 1, namely, k 1 = E 1 cφ 2 E 2 E 3 cφ 2 + E 5 k 2 = 2E 4 sφ 2 k 3 = E 1 cφ 2 + E 2 + E 3 cφ 2 + E 5 m 1 = F 1 cφ 1 F 2 F 3 cφ 1 + F 5 m 2 = 2F 4 sφ 1 m 3 = F 1 cφ 1 + F 2 + F 3 cφ 1 + F 5

KINEMATICS OF PARALLEL MANIPULATORS We proceed now by multiplying each of the above equations by τ 3 to obtain two more equations, namely. k 1 τ 3 2 + k 2 τ 3 2 + k 3 τ 3 = 0 m 1 τ 3 2 + m 2 τ 3 2 + m 3 τ 3 = 0 Further, we write equations in homogeneous form as: Φτ 3 = 0 with the 4 x 4 matrix Φ and the 4-dimensional vector τ 3 defined as: Φ = k 1 k 2 k 3 0 m 1 m 2 m 3 0 0 k 1 k 2 k 3 0 m 1 m 2 m 3 τ 3 = τ 3 3 τ 3 2 τ 31

KINEMATICS OF PARALLEL MANIPULATORS Moreover, in view of the form of vector τ 3, we are interested only in nontrivial solutions, which exist only if det(φ) vanishes. We thus have the condition det Φ = 0 (4) Equations (1) and (4) form a system of two equations in two unknowns 1 and 2. Moreover for φ 3 : E 4 sφ 2 E 2 + E 3 cφ 2 sφ 3 = E 1cφ 2 E 5 F 4 sφ 1 F 2 + F 3 cφ 1 cφ 3 F 1 cφ 1 F 5 Knowing the angles 1, 2 and 3 allows us to determine the position vectors of the three vertices of the mobile plate s 1, s 2 and s 3. Since three points define a plane, the pose of the EE is uniquely determined by the positions of its three vertices

CONTOUR-INTERSECTION APPROACH We derive the direct kinematics of a manipulator analyzed by Nanua et al. (1990). This is a platform manipulator whose base plate has six vertices with coordinates expressed with respect to the fixed reference frame F 0 as given below, with all data given in meters: A 1 = (-2.9, -0.9), B 1 =(-1.2, 3.0) A 2 = ( 2.5, 4.1), B 2 =( 3.2, 1.0) A 3 = ( 1.3, -2.3), B 3 = (-1.2,-3.7) The dimensions of the movable triangular plate are, in turn, a 1 = 2.0, a 2 =2.0, a 3 = 3.0

CONTOUR-INTERSECTION APPROACH Determie all possible poses of the moving plate for the six leg-lenghts given as: q 1a = 5.0, q 2a =5.5, q 3a = 5.7 q 1b = 4.5, q 2b =5.0, q 3b = 5.5 SOLUTION: After substitution of the given numerical values: 61.848 36.9561c 1 47.2376c 2 + 33.603c 1 c 2 41.6822s 1 s 2 = 0 28.5721 + 48.6806c 1 20.7097c 1 2 + 68.7942c 2 100.811c 1 c 2 + 35.9634c 1 2 c 2 41.4096c 2 2 + 50.8539c 1 c 2 2 15.613c 1 2 c 2 2 52.9786s 1 2 + 67.6522c 2 s 1 2 13.2765c 2 2 s 1 2 + 74.1623s 1 s 2 25.6617c 1 s 1 s 2 67.953c 2 s 1 s 2 + 33.9241c 1 c 2 s 1 s 2 13.202s 2 2 3.75189c 1 s 2 2 + 6.13542c 1 2 s 2 2 = 0

CONTOUR-INTERSECTION APPROACH

THE UNIVARIATE POLYNOMIAL APPROACH Reduce the two equations above, to a single monovariate polynomial equation. SOLUTION: We obtain two polynomial equations in τ 1, namely, G 1 τ 1 4 + G 2 τ 1 3 + G 3 τ 1 2 + G 4 τ 1 + G 5 = 0 H 1 τ 1 2 + H 2 τ 1 + H 3 = 0 No. 1 (rad) 2 (rad) 3 (rad) 1 0.8335 0.5399 0.8528 2 1.5344 0.5107 0.2712 3-0.8335-0.5399-0.8528 4-0.5107-0.5107-0.2712

THE UNIVARIATE POLYNOMIAL APPROACH Where: G 1 = K 1 τ 2 4 + K 2 τ 2 2 + K 3 G 2 = K 4 τ 2 3 + K 5 τ 2 G 3 = K 6 τ 2 4 + K 7 τ 2 2 + K 8 G 4 = K 9 τ 2 3 + K 10 τ 2 G 5 = K 11 τ 2 4 + K 12 τ 2 2 + K 13 H 1 = L 1 τ 2 2 + L 2 H 2 = L 3 τ 2 H 3 = L 4 τ 2 2 + L 5

THE UNIVARIATE POLYNOMIAL APPROACH Multiplying the first equation by H 1 and the second by G 1 τ 1 and subtract the two equations thus resulting, which leads to a cubic equation in τ 1, namely, G 2 H 1 G 2 H 2 τ 1 3 + G 3 H 1 G 1 H 3 τ 1 2 + G 4 H 1 τ 1 + G 5 H 1 = 0 Likewise multiplying the first by H 1 τ 1 and the second one by G 1 τ 1 3 +G 2 τ 12 and the equations thus resulting are subtracted from each other, one more cubic equation in τ 1 is obtaines : G 1 H 3 G 3 H 1 τ 1 3 + G 4 H 1 + G 3 H 2 G 2 H 3 τ 1 2 + G 5 H 1 + G 4 H 2 τ 1 + G 5 H 2 = 0

THE UNIVARIATE POLYNOMIAL APPROACH Moreover, multiplying the second equation by τ 1, a third cubic equation in τ 1 can be derived: H 1 τ 1 3 + H 2 τ 1 2 + H 3 τ 1 = 0 Now the foregoing equations constitute a homogeneous linea system of four equations in the firt four powers of τ 1 which can be cast in the form: Hτ 1 = 0 Where τ 1 = τ 1 3 τ 1 2 τ 1 1

H = THE UNIVARIATE POLYNOMIAL APPROACH G 2 H 1 G 2 H 2 G 3 H 1 G 1 H 3 G 4 H 1 G 5 H 1 G 3 H 1 G 1 H 3 G 3 H 2 G 2 H 3 + G 4 H 1 G 4 H 2 + G 5 H 1 G 5 H 2 H 1 H 2 H 3 0 0 H 1 H 2 H 3 To admit a nontrivial solution, the determinant of its coefficient matrix must vanish: det (H)=0 and this can be expanded in the form: det(h) = H 1 1 + H 2 2 + H 3 3

Where: THE UNIVARIATE POLYNOMIAL APPROACH Δ 2 = det Δ 2 = det Δ 2 = det G 2 H 1 G 2 H 2 G 4 H 1 G 5 H 1 G 3 H 1 G 1 H 3 G 4 H 2 + G 5 H 1 G 5 H 2 H 1 H 2 0 G 2 H 1 G 2 H 2 G 3 H 1 G 1 H 3 G 5 H 1 G 3 H 1 G 1 H 3 G 3 H 2 G 2 H 3 + G 4 H 1 G 5 H 2 H 1 H 2 0 G 2 H 1 G 1 H 2 G 3 H 1 G 1 H 3 G 4 H 1 G 3 H 1 G 1 H 3 G 3 H 2 G 2 H 3 + G 4 H 1 G 4 H 2 + G 5 H 1 H 1 H 2 H 3

VELOCITY ANALYSES OF PARALLEL MANIPULATORS The inverse velocity analysis of this manipulator consists in 6 determining the six rates of the active joints, b k 1 given the twist t of the moving platform. The velocity analysis of a typical leg leads to a relation of the form of equation namely, J J θ J = t J J= I,II,, IV Where J J is the Jacobian of the Jth leg, is the 6-d joint-rate vector θ J of the same leg and t j is the twist moving platform M with its operation point defined as the point C J

VELOCITY ANALYSES OF PARALLEL MANIPULATORS We thus have: e J J = 1 e 2 0 e 4 e 5 e 6 b 34 e 1 e 3 b 34 e 2 e 3 e 3 0 0 0 t J = ω c J, b 34 = b 3 + b 4 it is apparent that c J = p ω r J with r J defined as the vector directed from C J to the operation point P of the moving platform.

VELOCITY ANALYSES OF PARALLEL MANIPULATORS Multiplying the left side of the equation by I J T Where: I J T = [ 0 T e 3 T ] We obtain: I J T J J θ J = On the other hand we have: I T T J t J = e J c J Upon equating the right-hand sides of the foregoing T equations we have: b J = e J c J b J

VELOCITY ANALYSES OF PARALLEL MANIPULATORS We obtain the relations between the actuated joint rates and the twist of the moving platform, namely b j = (e J t j ) T T ω e J c = k J T t We have the desired relation between the vector of actuated joint rates and the twist of the moving platform, namely. b = Kt With the 6-d vectors b and t defined as the vector of joint variables and the twist of the platform at the operation point, respectively. Moreover, the 6 x 6 matrix K is the Jacobian of the manipulator at hand.

VELOCITY ANALYSES OF PARALLEL MANIPULATORS These quantities are displayed below: b = b I b II b IV t = ω p K = e I r I T e II r II T e VI r VI T T e I T e II T e VI

ACCELLERATION ANALYSES OF PARALLEL MANIPULATORS The acceleration analysis of the same leg is straightforward. Indeed, upon differentiation of both sides of equation with respect to time, one obtains b = K t + Kt Where K takes the form: K = u I T T u II T u VI e I T T e II T e VI

ACCELLERATION ANALYSES OF PARALLEL MANIPULATORS u J is defined as: u J = e J r J Therefore: u J = e J r J + e J Now since the vector r j are fixed to the moving platform, their time-derivatives are simply given by: r J = ω r J On the other hand, vector ej is directed along the log axis, and so, its time-derivative is given by: ω J = ω J e J with ω J defined as the angular velocity of the third leg link: ω J = θ 1 e 1 + θ 2 e 2 J r J

ACCELLERATION ANALYSES OF PARALLEL MANIPULATORS In order to simplify the notation, we start by defining f J = e 1 J g J = e 2 J Now we write the second vector of the equation J J θ J = t J using the foregoing definitions, which yields: ( θ 1 ) J f J b J + b 4 e j + ( θ 2 ) J g J b J + b 4 e j + b J e j = where b 4 is the same for all legs, since all have identical architectures. Now we can eliminate ( θ 2 ) J from the foregoing equation by dot-multiplying its two sides by g j, thereby producing c J

ACCELLERATION ANALYSES OF PARALLEL MANIPULATORS ( θ 1 ) J g J f J b J + b 4 e j + g T T J e j e J c J = g T c J Now it is a simple matter to solve for ( θ 1 ) J from the above equation, which yields: ( θ 1 ) J = g J T T 1 e J e J Where: J = b J + b 4 e j f J g J J c J

ACCELLERATION ANALYSES OF PARALLEL MANIPULATORS Moreover, we can obtain the above expression for ( θ 1 ) J in terms of the platform twist by recalling equation which is reproduced below in a more suitable form for quick reference: c J = C J t where t is the twist of the platform, the 3 x 6 matrix C j being defined as: C J = [R j 1] In which R J is the cross-product matrix of r J

ACCELLERATION ANALYSES OF PARALLEL MANIPULATORS Therefore the expressione sought for ( θ 1 ) J takes the form: ( θ 1 ) J = 1 g T J 1 e J e T J C J t J A similar procedure can be followed to find ( θ 2 ) J the final result being ( θ 2 ) J = 1 f T J 1 e J e T J C J t J

PLANAR AND SPHERICAL MANIPULATORS The velocity analysis of the planar and spherical parallel manipulators of figures are outlined below

PLANAR MANIPULATORS J J θ J = t J J= I,II, III Where J J is the Jacobian matrix of this leg while 3-D vectore of joint rates of this leg: θ J is the J J = 1 1 1 Er J1 Er J2 Er J3 θ J = θ J1 θ J2 θ J3 We multiply both sides of the said equation by a 3-D vector n j perpendicular to the second and the third columns of J J.

PLANAR MANIPULATORS This vector can be most easily determined as the cross product of those two columns, namely, as n = j J2 j J3 = r J2 T Er J3 r J2 r J3 Upon multiplication of both side of the equation by n jt, we obtain: r J2 Er j3 + r J2 r J3 T ErJ1 And hence we can solve directly reriving: θ J1 = r J2 Er J3 ω + r J2 r J3 T c θ J1 = r J2Er J3 ω+ r J2 r J3 T c r J2 Er j3 + r J2 r J3 T ErJ1

PLANAR MANIPULATORS The foregoing equation can be written in the form: With j J and k J defined: j J θ J1 = k J T t J=I,II, III T j J = r J2 r J3 ErJ1 r J2 Er j3 k J = [r T J2 Er J3 T T r J2 r J3 ] If we further define: θ = [ θ I1 θ II2 θ III3 ] And assemble all three foregoing joint-rate-twist relation we obtain: J θ = Kt

PLANAR MANIPULATORS J θ = Kt Where J and K are the two manipulator Jacobians defined as: J = diag j I, j II, j III, K = T k I T k II T k III Expressions for the joint accelerations can be readily derived by differentiation of the foregoing expressions with respect to time.

SPHERICAL MANIPULATOR The velocity analysis of the spherical parallel manipulator can be accomplished similarly. Thus, the velocity relations of the Jth leg take on the form: J J θ J = ω J=I,II,III Where the Jacobian of the Jth leg J J is defined as: J J = e J1 e J2 e J3 While the joint-rate vector of the Jth θ J leg is defined exactly as in the planar case analyzed above. An obvious definition of this vector is, then: n J = e J2 e J3

SPHERICAL MANIPULATOR The desired joint-rate relation is thus readily derived as: j J θ J1 = k J T ω where j J and k J are now defined as: j J = e J1 e J2 e J3 k J = e J2 e J3 The accelerations of the actuated joints can be derived, again, by differentiation of the foregoing expressions. We can then say that in general, parallel manipulators, as opposed to serial ones, have two Jacobian matrices.