Kinematic Isotropy of the H4 Class of Parallel Manipulators

Kinematic Isotropy of the H4 Class of Parallel Manipulators Benoit Rousseau 1, Luc Baron 1 Département de génie mécanique, École Polytechnique de Montréal, benoit.rousseau@polymtl.ca Département de génie mécanique, École Polytechnique de Montréal, luc.baron@polymtl.ca Abstract his paper presents the isotropic conditions for the topological class of H4 parallel manipulators with articulated travelling plate which has four degrees of freedom. First, a generic kinematic model of this class of manipulators is developed, then we impose isotropic conditions on the Jacobian matrix. From the newly obtained equations, design constraints and a design procedure allowing the determination of all isotropic geometries are obtained. he proposed procedure allows the successive choice and calculus of each and all geometrical parameters of an isotropic manipulator of the H4 class. Keywords: isotropy, parallel manipulator, traveling plate, kinematics. Isotropie des manipulateurs parallèles de la classe H4 Résumé Cet article présente les conditions d isotropie de la classe topologique des manipulateurs parallèles à nacelle articulée H4 possédant quatre degrés de liberté. Un modèle cinématique générique de cette classe de manipulateurs est d abord développé, puis on impose une condition isotrope à la matrice jacobienne. Des équations obtenues on trouve les contraintes et une procédure de conception permettant de déterminer toutes les géométries isotropes. La procédure de design proposée permet de choisir et de calculer successivement chacun des paramètres géométriques d un manipulateur isotrope de classe H4. Mots-clé: isotropie, manipulateur parallèle, nacelle articulée, cinématique. 009 CCoMM M 3 Symposium 1

1 INRODUCION We wish to determine the isotropic conditions of parallel manipulators belonging to the H4 topologic class of manipulators. As defined in [1] : a topologic class is the group of mechanisms having the same topology regardless of the geometry. he topology describes the arrangement of the joints of

In order to identify the isotropic conditions, we must search the conditions to be applied to the Jacobian matrices. Since the Jacobian matrices depend both on the posture and the geometry, it is difficult to have a geometry which is isotropic at all postures. herefore, only some particular geometries can reach isotropy, and only for one or a few postures. We are thus searching for the conditions that will allow us to obtain isotropic geometries, not only isotropic postures of a specific geometry. B 1 B û 1 û q q 1 A 1 C 1 θ C A D 1,4 P D,3 A 4 C 4 C 3 A 3 q 4 q 3 û 4 B 4 B 3 û 3 MODELLING OF HE H4 CLASS Figure 3. Schematic diagram of the H4 manipulator Referring to Figure 3, the points {A i } 4 1 are fixed and attached to the base while the point P is attached to the end-effector and is therefore mobile. he joints at points A i and D i are revolute while the joints at points B i and C i are spherical. he links between points A i and B i rotate about the axis û i by an angle q i. he end-effector at P rotates by an angle θ about an axis normal to the plane generated by the relative movement of points C i. Without loss of generality, this axis is chosen as being the unitary vector ˆk in the global frame. Links r i between the spherical joints at points B i and C i form a Π joint which propagates the orientation of axis û i from point A i to point C i. he bars linking the points C i always keep their orientation. When they move one about another, the bars always remain in the same plane so that the rotation of the end-effector located at point P is always about the axis of ˆk which always maintains the same orientation. he position of point A i belonging to the base and the position of the end-effector P are both known relatively to a global frame. Knowing the angle q i of the motorized revolute joint located at A i, it is possible to find the position of B i. From the position and angle θ of the end-effector P, we can find the position of C i see Figure 4). Since points C i and B i are connected by a rigid link, these two points are mathematically related by a rigidity condition. Indeed, the norm of vector r i going from point B i to point C i is constant and equal to r i. 009 CCoMM M 3 Symposium 3

.1 Rigidity Condition he distance r i between points B i and C i belongs to the same rigid link, and hence, is constant, i.e., r i r i = r i 1) Derivating Eq. 1) with respect to time, the rigidity condition expressed in terms of speed is given as : ḃ i r i = ċ i r i ) where ḃi and c i are respectively the speed of points B i and C i expressed in the same reference frame. his expression represent the equiprojectivity of the speeds ḃi and c i on the axis r i linking points B i and C i belonging to the same rigid link. Reference frame B i p i r i Base A i fq i) Closure eq. for leg i P Effector fx, y, z, θ) A i q i C i B i r i = r i C i û i D i s i t i θ P Figure 4. Schematic diagram of the closure equation for each leg Figure 5. An isolated leg. Closure Equations From Figure 5, it is possible to write the closure equation of the kinematic loop associated to leg i : b i = a i + p i q i ) 3) c i = p t i θ) s i 4) r i = c i b i 5) where a i, b i, c i and p are respectively the position vectors of points A i, B i, C i and P. he vector p i is function of q i and the vector t i is function of θ. Substituting Eq. 3) and Eq. 4) in Eq. 5), we obtain : hen substituting Eq. 6) in Eq. 1), we have : r i = p t i θ) s i a i p i q i ) 6) r i = r i p t i θ) s i a i p i q i )) 7) Derivating Eq. 7) with respect to time : r i ṗ + t i ˆk ) ) θ pi û i ) q i = 0 8) 009 CCoMM M 3 Symposium 4

where ˆk is recalled to be a unitary vector parallel to the axis of rotation of the platform. Using the distributivity property of the dot product, Eq. 8) can be rewritten separating the terms ṗ, θ and q i : ˆk) r iṗ + r i t i θ = ri p i û i ) q i 9) Equation 9) links the speed of the motorized joints q i located at the base of the legs to the speeds of the end-effector ṗ and θ. Writing down the equation 9) for all the four legs, separating ṗ and θ, we have the following system of equations : where A Aẋ = B q, ẋ [ ṗ θ ] [, q q1 r 1 r r 3 r 4 q q 3 ] q 4 10) r 1 t 1 ˆk ) r t ˆk ) r 1 p 1 û 1 ) 0 0 r 3 t 3 ˆk ), B 0... 0 11) t 4 ˆk ) 0 0 r 4 p 4 û 4 ) r 4 and ẋ is the velocity of the end-effector, q the velocity vector of the motorized joints, A is of dimension 4 4 and B is diagonal and also 4 4. 3 PROBLEM FORMULAION We wish to determine the isotropic conditions on the Jacobian matrices of Eq. 10), but before this we need to render them adimensional. 3.1 Adimensionalisation A manipulator that can both position and orient itself in space has dimensionally non homogenous Jacobian matrices because they involve dimensional lengths and adimensional angles. he non-homogeneity of the Jacobian matrices is eliminated by introducing a characteristic length [5, 6]. In order to obtain dimensionally homogenous matrices, we need to divide both sides of the Eq. 10) by a characteristic length in order to render them adimensional. Using L as the unit of length and as the unit of time, both sides of Eq. 10) have L / as unit. A further analysis reveals that the Jacobian matrix A has components with dimensions L and L, while the matrix B has L components. In an adimensional form, when λ is taken as the natural length, the Jacobian matrices A and B can be rewritten as : r 1 /λ r 1 t 1 ˆk ) /λ r A /λ r t ˆk ) r 1 p 1 û 1 ) /λ 0 0 λ r 3 /λ r 3 t 3 ˆk ) /λ B 0... 0 r 4 /λ r 4 t 4 ˆk ) r 4 p 4 û 4 ) /λ 0 0 λ 009 CCoMM M 3 Symposium 5

3. Isotropic Condition he isotropic condition on the Jacobian matrices is expressed as : ẋ ẋ = 1 1) It constrains the velocity of the end-effector in all directions to a velocity of unit magnitude, i.e. a unit sphere of dimension m, where m is the total number of degrees of freedom of the manipulator. A Jacobian matrix is said isotropic when it transforms a unit sphere from the m-dimensional space of the end-effector to a n-dimensional sphere in the joint space scaled by a factor α. Substituting Eq. 10) in Eq. 1), we obtain : q A 1 B) A 1 B q = 1 which represents a velocity ellipsoid in the adimensional joint space. he matrix A 1 B is therefore isotropic if its singular values are all identical and different from zero : where C B 1 A 3.3 Isotropy of the Matrix C Matrix C can be expressed as : λ r 1 /g 1 h 1 /g 1 C λ r /g h /g λ r 3 /g 3 h 3 /g 3 λ r 4 /g 4 h 4 /g 4 where g i r i p i û i ) and h i r i t i ˆk). C C = α 1 13) he isotropic conditions of matrix C given in Eq. 13) are orthogonality conditions. he rows of an orthogonal matrix form an orthonormal basis. All rows have the same norm and are mutually orthogonal. It is automatically the same for the columns which also form an orthonormal basis. 3.3.1 Orthogonality Conditions he dot product between two rows or columns) of C must vanish, i.e., [ λ ri /g i h i /g i ][ λ rj /g j h j /g j ] = 0 When developed : ) ) λ ri λ rj + g i g j hi ) ) hj = 0 g i g j 009 CCoMM M 3 Symposium 6

Simplifying g i and developing h i results in : then dividing by r i r j we obtain : λ r i r j + r i t i ˆk))r j t j ˆk)) = 0 λ ˆr j ˆr i t i ˆk))ˆr j t j ˆk)) = 0 14) Vectors r i and r j of Eq. 14) becomes unitary as and ˆr j. Using the definitions of the dot product and the cross product, while fixing without loss of generality) sin t i ˆk = 1 which appears in the cross products : λ cos ˆr j = t i cos t i ˆk t j cos t j ˆk ˆr j 15) We can rewrite Eq. 15) for all pairs of rows of C as σ 1, = t 1,4 λ µ 1 σ,3 = t,3 λ µ t,3 where µ i cos t i ˆk t,3 λ µ σ 1,3 = t 1,4 λ µ 1 λ µ 3 σ 1,4 = t 1,4 λ µ 1 λ µ 4 t,3 λ µ 3 σ,4 = t,3 λ µ t 1,4 λ µ 4 σ 3,4 = t,3 λ µ t 1,4 3 λ µ 4 16) ˆt i ˆk) and σ i,j cos ˆr j ˆr j. From the orthogonality conditions, we have six equations from the non diagonal elements of the identity matrix. he t i appear divided by λ, they are therefore normalized. Because t 1 and t 4 represent the same vector, they are written as t 1,4. he same applies for t and t 3 which are written as t,3. Defining β i t i λ µ i, Eq. 16) becomes : σ 1, = β 1 β σ 1,3 = β 1 β 3 σ 1,4 = β 1 β 4 σ,3 = β β 3 σ,4 = β β 4 σ 3,4 = β 3 β 4 17) From which we find the following constraint : 3.3. Normality conditions All the rows or columns) of C must have the same norm : t 1,4 σ 1, σ 3,4 = σ 1,3 σ,4 = σ 1,4 σ,3 18) [ λ ri /g i h i /g i ][ λ ri /g i h i /g i ] = α When developed : ) ) λ ri λ ri + g i g i hi g i ) ) hi = α By moving the g i on the other side of the equality, we obtain : λ r i r i + r i t i ˆk) ) = α r i p i û i ) ) g i 009 CCoMM M 3 Symposium 7

then dividing by r i results in : λ + t i ˆk) ) ˆr = α i p i û i ) ) 19) Vectors r i of Eq. 19) become unitary. Using the definition of the dot product and the cross product, while fixing without loss of generality) sin t i k i = 1 and sin u i p i = 1 which appear in the cross products : λ + t i cos t i ˆk We can rewrite Eq. 0) for all rows of C as : = α p i cos p i û i 0) 1 + t 1,4 µ λ 1 = α p 1 λ η 1 1 + t,3 µ λ = α p λ η 1 + t,3 µ λ 3 = α p 3 λ η 3 1 + t 1,4 µ λ 4 = α p 4 λ η 4 1) where µ i cos t i ˆk ˆt i ˆk) and η i cos p i û i ˆp i û i ). From the normality conditions we have four equations from the diagonal elements of the identity matrix. Moreover t i and p i appear divided by λ, they are therefore normalized. 3.3.3 Relation Between the σ i,j When the orthogonality condition for the is written in a scalar form, the angles between the appear in the form of σ i,j. A strong relation exists between these angles which must be considered. It is sufficient to know five of these angles to be able to determine the absolute value of the 6th angle using the law of spherical cosines. he relationship that the σ i,j must conform to is : ) ) ) σ1, σ 1,4 σ,4 σ1,3 σ 1,4 σ 3,4 σ,3 σ,4 σ 3,4 arccos ± arccos ± arccos = 0 κ 1,4 κ,4 κ 1,4 κ 3,4 κ,4 κ 3,4 ) where σ i,j cos ˆr j and κ i,j sin ˆr j. Moreover, Eq. ) has real solutions when the following constraints are satisfied : 1 σ 1, σ 1,4 σ,4 κ 1,4 κ,4 1 1 σ 1,3 σ 1,4 σ 3,4 κ 1,4 κ 3,4 1 1 σ,3 σ,4 σ 3,4 κ,4 κ 3,4 1 4 ISOROPIC SOLUIONS In summary, there are 11 equations, see Eqs. 16)1)), associated with the isotropic condition applied to matrix C. It is possible to reduce this system of equation to only one equation by substituing Eq. 16) and Eq. 1) in Eq. ). he choice of the parameters appearing in these equations is made easier when considering the system as a whole rather than as a single equation. 3) 009 CCoMM M 3 Symposium 8

4.1 Design procedure It is possible to obtain an isotropic manipulator by choosing and calculating successively each and all of the parameters appearing in the isotropy equations as we propose in the following procedure : Choice of σ 1,4 : We first choose the value of σ 1,4. Since σ i,j are cosines, σ 1,4 must be chosen within the interval [ 1, 1]. Choice of σ,4 : Considering the constraints Eq. 3) and the extrema of a σ i,j which are 1 and 1, we therefore have : κ 1,4 κ,4 +1) σ 1,4 σ,4 κ 1,4 κ,4 for σ 1, = +1 κ 1,4 κ,4 1) σ 1,4 σ,4 κ 1,4 κ,4 for σ 1, = 1 Hence, with some manipulations, σ,4 must be chosen within the smallest interval between [ 1, +1] and [ 1 σ 1,4, + 1 σ 1,4 ]. Choice of σ 1, : From 1 = Eq. 17), we knowd[-8.587k)-1.87468n)-1.c095a).3554f-948.588:)-3.05095

Calculus of σ,3 and σ 3,4 : Knowing all other σ i,j, we can easily calculate σ,3 and σ 3,4 with Eq. 4). Calculus of β i : σ i,j : From the definition β i and Eq. 17), we can write β 1 as a function of the β 1 = ± σ 1, σ 1,3 σ,3 6) he sign of β 1 can be chosen freely. We can then calculate the other β i easily as : β = σ 1, β 1, β 3 = σ 1,3 β 1, β 4 = σ 1,4 β 1. 7) Calculus of µ i and : Since β 1 and β 4 share the same t 1,4 and that the same applies for β and β 3 which share the same t,3, then we can combine them into the following two relationships : β 1 = β 4 β = β 3 8) µ 1 µ 4 µ µ 3 So, if we combine Eq. 8) to the definitions of β i see Eq. 17)) and µ i see Eq. 16)) : ˆr i ˆr j = β i β j µ i = ˆr i ˆt i ˆk) We then have a system of equations in terms of, β i, µ i and ˆt i ˆk. We are looking for a solution to this system. Without loss of generality, we choose ˆt i ˆk as being along the y axis. In order to have only one possible solution, we choose to restrain ˆr 1 to be in the positive quadrant of the y z plane and to restraint ˆr to have a positive component along the x-axis. We can then numerically find the and the µ i that satisfy these conditions. All the rotations of the vectors about the y-axis and their reflexions about the x, y et z axis are also solutions. Calculus of t i /λ : Having found µ i, we can determine the values of t 1,4 /λ and t,3 /λ from the definition of the β i, see Eq. 17): t 1,4 /λ = β 1 /µ 1 t,3 /λ = β /µ 9) Choice of α : he amplification factor α can be freely chosen as positive non-zero value. Choice of η i : he η i are cosines, we can choose them freely in the interval [ 1, 1]. Calculus of p i /λ : From the Eq. 1), we can the find p i /λ as : 1 + t 1,4 /λ) µ 1 p 1 λ = p 3 λ = α η 1 1 + t,3 /λ) µ 3 α η 3 p λ = p 4 λ = 1 + t,3 /λ) µ α η 1 + t 1,4 /λ) µ 4 α η 4 30) 009 CCoMM M 3 Symposium 10

Other parameters : All the parameters in the equations associated to the isotropic condition have been determined. he other parameters in the kinematics equations can then be determined. hese parameters are the components of the vectors s i, the norm of the vectors r i and the orientation of the vectors p i about the vectors r i. hey can be freely chosen because they have no impact on the local isotropy. Obviously, other criteria can be used to choose them such as workspace or joint motion range, for example. 5 NUMERICAL EXAMPLE Using our design procedure we can now generate isotropic manipulators belonging to the H4 class. Chosing arbitrary values along the design process, knowing that the design procedure is guaranteed to always give an isotropic solution, we obtain the following isotropic geometry shown in Figure 6 :

he vectors describing the positions of points and the orientation of axes are shown as : A 1 = [ 0.1691, 1.719, 0.10] A = [ 0.7137, 1.196, 0.6353] A 3 = [ 0.6083, 0.48, 0.590] A 4 = [ 0.4456, 0.5505, 0.149] B 1 = [ 1.0415, 1.6361, 0.7301] B = [ 1.611, 0.5531, 0.89] B 3 = [ 1.3655, 0.189, 0.019] B 4 = [ 0.9967, 1.4140, 0.8614] C 1 = [ 1.910, 1.0000, 0.0000] C = [ 0.7746, 1.0000, 0.0000] C 3 = [ 0.7746, 1.0000, 0.0000] C 4 = [ 1.910, 1.0000, 0.0000] D 1,4 = [ 1.910, 0.0000, 0.0000] D,3 = [ 0.7746, 0.0000, 0.0000] û 1 = [ 0.6990, 0.054, 0.7147] û 3 = [ 0.3585, 0.6650, 0.6551] P = [ 0.0000, 0.0000, 0.0000] θ = 0.0000 ˆk = [ 0.0000, 0.0000, 1.0000] û = [ 0.5033, 0.86, 0.647] û 4 = [ 0.1514, 0.5837, 0.7977] 6 CONCLUSION here are infinitely many isotropic geometries for the parallel manipulators of the topological class H4. he isotropic conditions have been formulated and a design procedure for the selection and the calculus of the parameters of the manipulator has been proposed. his procedure allows choosing and calculating successively all geometrical parameters of any isotropic manipulators of the H4 class. REFERENCES [1] Jee-Hwan Ryu, editor. Parallel Manipulators : New Developments, chapter 4. I-ech Education and Publishing, Vienna, Austria, 008. [] L. Baron and G. Bernier. he design of parallel manipulators of star topology under isotropic constraint. ASME DEC, Pittsburgh, USA, 001. [3] L Baron, X. Wang, and G. Cloutier. he Isotropic Conditions of Parallel Manipulators of Delta opology. ARK, Caldes de Malavalla, Spain, 00. [4] O. Company, F. Marquet, and F. Pierrot. A new high-speed 4-dof parallel robot synthesis and modeling issues. IEEE ransactions on Robotics and Automation, 193):411 40, 003. [5] O. Ma and J. Angeles. Optimum architecture design of platform manipulators. Advanced Robotics, 1991. Robots in Unstructured Environments, 91 ICAR., Fifth International Conference on, pages 1130 1135 vol., Jun 1991. [6] Jorge Angeles. he design of isotropic manipulator architectures in the presence of redundancies. he International Journal of Robotics Research, 113):196 01, 199. 009 CCoMM M 3 Symposium 1