PR 5 Robot Dynamics & Control /8/7 PR 5: Robot Dynamics & Control Robot Inverse Kinematics Asanga Ratnaweera Department of Mechanical Engieering The Inverse Kinematics The determination of all possible and feasible sets of joint variables, which would achieve the specified position and orientation of the manipulator s end-effector with respect to the base frame. In practice, a robot manipulator control requires knowledge of the endeffector position and orientation for the instantaneous location of each joint as well as knowledge of the joint displacements required to place the endeffector in a new location.
PR 5 Robot Dynamics & Control /8/7 Solvability of Inverse Kinematic Model Inverse Kinematics is complex because the solution is to be found for nonlinear simultaneous equations: Involves harmonic functions (sine and cosine) Number of equations are generally more than the number of unknowns Therefore, some equations are mutually dependent These lead to the possibility of multiple solutions or nonexistence of any solution for a given end-effector position and orientation. Existence of the solution The conditions for existence of solutions to the inverse kinematic problem are : The desired point should lie inside the (restricted) work-envelop The wrist should be a -DOF joint in order to realize all the possible orientations. Consider the general form of kinematics model for a robot manipulator: r r r r4 r r r r 4 T = n r r r r 4 4
PR 5 Robot Dynamics & Control /8/7 Existence of the solution This can yield maximum of equations 9 equations arise from rotational matrix (x) equations arise from the displacement vector Out of 9 equations from rotational matrix involves only three unknowns corresponding to the orientation of the end effector Three Euler angles or RPY angles Therefore, there exists only six independent constraints in n unknowns Thus, in order for a manipulator to have all the possible solutions The number of DOF n should be less than or equal to 6 necessary condition for the existence of a solution 5 Existence of the solution In addition to these 6 independent constraint equations, the tool position and orientation must be such that the limits on the joint motions are not violated. 6
PR 5 Robot Dynamics & Control /8/7 Existence of the solution For manipulators with less than or more than 6 DOF, the solutions are more complex: When DOF are less than 6, it cannot attain the general goal position and orientation in D space- mathematically over-determined. When DOF are more than 6, mathematically under-determined, so that all the unknowns cannot be found. 7 Multiple Solutions Multiple solutions can arise: If more than two joint axes are parallel. If trigonometric functions are existed in the equations. Ex: sinθ = sine(π+ θ) 8 4
PR 5 Robot Dynamics & Control /8/7 Multiple Solutions Multiple solutions depends on: Number of non-zero joint link parameters. In General number of ways to reach a certain goal is directly related to the number of non-zero link parameters. Number of degree of freedom. Manipulator with more than 6 DOF may have infinitely many solutions. 9 Solution techniques There are two approaches to solve inverse kinematics equations Closed form solution Numerical solution 5
PR 5 Robot Dynamics & Control /8/7 Closed form solution General guideline. Solve equations involving only one joint variables first.. Look for pairs or set of equations, which could be reduced to one equation in one joint variable by application of algebraic and trigonometrically identities.. Use tan - functions instead of cos - and sin - 4. Solution is terms of elements of the position vector components are more efficient. tan - function returns the accurate angle in the range of π and π by eliminating the sign of both y and x and detecting whether either x or y is non zero. Closed form solution General guideline 5. Consider the general form of the kinematics equation: T = T n T T 4... n T n = T Each i- T i is a function of only one joint variable, q i. Therefore, pre-multiplying both sides by the inverse: [ T ] T n Tn = T T4... T n = Thus the matrix elements of the right hand side are, zero, constant, or function of the joint variable q. 6
PR 5 Robot Dynamics & Control /8/7 L L T = T = T T T CC S C = S -C S - S S C -S - C C ( L C S ( L C L S + L C + L C + L S ) ) S = sin (θ + θ ) 4 7
PR 5 Robot Dynamics & Control /8/7 Consider the DOF manipulator r r r r 4 r r r r 4 T = T = n r r r r 4 The kinematric equation can be obtained using DH method: CC -CS -S C ( LC + LC ) S C - S S - C S ( L C + L C ) T = T = n S C L S + L S 5 Therefore, CC SC S -C S - S S C -S - C C ( L C S ( L C L S + L C + L C + LS ) r ) = r r r r r r r r r4 r4 r 4 Thus, non trivial equations for three unknowns θ, θ, θ 6 8
PR 5 Robot Dynamics & Control /8/7 From step, θ can be found from the element of the row. -sin θ = r This is not preferred as a correct quadrant of the angle cannot be found. (step ) Alternatively applying step, θ can be isolated by dividing elements 4 of the row and row. C ( L C S ( L C + L C + L C ) = r ) = r 4 4 θ = tan - (r 4 /r 4 ) 7 The other two unknowns cannot be obtained directly: Therefore applying step 5 in the guidelines and post multiplying by the inverse of T : T T 4 = T [ T ] C S T = -S L C C L S [ T ] C - S = S C -L 8 9
PR 5 Robot Dynamics & Control /8/7 Only function of θ Considering the elements of the position vector 9
PR 5 Robot Dynamics & Control /8/7 Example: DOF Spherical Robot The position and the orientation of the end-effector of the robot manipulator shown below is given by T. Determine all the values of joint variables of the robot manipulator..54 -.6 T =.77.866.5.54 -.6.77.6 -.84. The joint displacements allowed are: - <θ < ; -<θ <7 ;.5m <d <.5m Example: DOF Spherical Robot Coordinate frames
PR 5 Robot Dynamics & Control /8/7 Example: DOF Spherical Robot DH Parameter table Joint transformation matrices Example: DOF Spherical Robot Overall transformation matrix Therefore,.54 -.6 T =.77.866.5.54 -.6.77.6 -.84. End-effector position and orientation 4
PR 5 Robot Dynamics & Control /8/7 d d Example: DOF Spherical Robot Equating the elements of the 4 th column CS = SS = d C =.6.84. [() + () ]/() () () () tan θ ()/() ±.84 +.6 =..84 tanθ =.6 [() + () + () ] d =.84 +.6 +. 5 Example: 4 DOF RPPR Robot The joint-link transformation matrices for a 4 DOF RPPR manipulator is shown in the next slide. If the tool configuration matrix at a given instant is as shown below. Determine the joint variables. 6
PR 5 Robot Dynamics & Control /8/7 Example: 4 DOF RPPR Robot Joint transformation matrices 7 4