If any of these solutions suffer from inaccuracies

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Unformatted text preview: CRIPTION. 1. Equate the general transformation matrix to the manipulator transformation matrix. If a particular, rather than a general solution is required a matrix describing the desired position and orientation of the end of the manipulator is equated to the manipulator transformation matrix. 2. Look at both matrices for: (a) elements which contain only one joint variable (b) (b) pairs of elements which will produce expression in only one joint variable when divided (c) elements, or combinations of elements, that can be simplified using trigonometric identities. 3. Having selected an element, equate it to the equate corresponding element in the other matrix to produce produce an equation. Solve this equation to find a description of one fi joint variable in terms of the elements of the general transformation matrix. 4. Repeat step 3 until all the elements identified in step 2 have been used 5. If any of these solutions suffer from inaccuracies, undefined results, or redundant results, set them aside and look for better solutions. Solutions in terms of the elements of the p vector may lead to more efficient Solutions than solutions in terms of the elements of the x, y, or z vectors. 6. If there are more joint angles to be found, premultiply both sides of the matrix equation by the inverse of the A matrix for the first link to produce th th fi a new set of equivalent matrix elements. 7. Repeat steps 2-6 until either solution to all the joint variables have been found, or you have run out of A matrices to pre-multiply. 8. If a suitable solution cannot be found for a joint variable, choose one of those discarded in step 5, taking notes of regions where problems may occur. 9. If a solution cannot be found for a joint variable in terms of the elements of the manipulator transform, it may be that the manipulator cannot achieve the specified position and orientation; Also, theoretical solutions may not be physically attainable because of mechanical limits on the range of joint variable. ⎡1 ⎢0 ⎢ ⎢0 ⎢ 0 ⎣ 0 0 1⎤ ⎥ 1 0 0⎥ 0 1 0⎥ ⎥ 0 0 0⎦ inverse kinematics The robot is actuated in joint space and that is what we can control Once the problem is solved in Cartesian space constraints, need to map these constraints into the robot’s joint space using inverse kinematics E.g. if a straight line trajectory is needed, then need to break the trajectory into a set of joint space values over time so that robot follow the line. inverse kinematics The inverse kinematics mapping is typically one to many Has to choose one solution from a number of valid solutions Th There are degenerate cases with infinite number of solutions (singularities) Some solutions of the inverse mapping may not be physically realizable due to physical joint limits of manipulators. There may not be a closed form solution to the inverse problem at all for some manipulators. However, most of the manipulators use a 3 DOF wrist that has intersecting axes. This allows the separation of the inverse problem into a 3 DOF problem for finding the endpoint of wrist and a 3 DOF problem for finding the orientation of wrist. Numerical methods can be used to find a solution to the inverse problem if a closed form solution does not exist. A redundant robot has an extra DOF which can be useful to reach around obstacles and avoiding collisions with other objects in eth workspace. Inverse kinematics can be solved by various methods : geometric, trigonometric and algebraic. Once solution for a joint variable has been found, the manipulator becomes a reduced th DOF mechanism – i.e. with one joint less. Example Solving the inverse kinematics of Stanford arm ⎡ nx ⎢n T06 = ⎢ y ⎢ nz ⎢ ⎣0 sx ax sy sz 0 ay az 0 px ⎤ py ⎥ ⎥ = T01T12T23T34T45T56 pz ⎥ ⎥ 1⎦ (T01 ) −1T06 = T12T23T34T45T56 = T16 ⎡X ⎢X T16 = ⎢ ⎢X ⎢ ⎣0 X X X X X X 0 0 Cθ1 p x + Sθ1 p y ⎤ ⎡ X ⎥ ⎢X − pz ⎥ =⎢ − Sθ1 p x + Cθ1 p y ⎥ ⎢ X ⎥⎢ 1 ⎦ ⎣0 X X X X X X 0 0 Sθ 2 ⋅ d 3 ⎤ − Cθ 2 ⋅ d 3 ⎥ ⎥ 0.1 ⎥ ⎥ 1 ⎦ 68...
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