ROBOTICS (5 cfu) 09/02/2016. Last and first name Matricola Graduating

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1 ROBOTICS (5 cfu) 09/02/2016 Last and first name Matricola Graduating PART 1 - QUIZ (8 marks) 1. For a 3x3 matrix to be a rotation matrix, it should hold - each row vector has module 1 T F - the product of any two different column vectors is 1 T F - the determinant of the matrix is 1 T F TF (is zero)t 2. A continuous trajectory - is obtained from sampling a wanted movement T F - reproduces only positions and not velocity T F - requires heavy computations T F TFF 3. The iterative solutions of the inverse kinematics for arms - are more precise than the closed form solutions, in every case T F - require a computation time shorter than the closed form solutions T F - have been obtained for anthropomorphic arms T F FFT 4. The Cartesian controller in position and velocity - receives in input the position and velocity of a single joint T F - is effective in reducing the error in the Cartesian space T F - uses position sensors on the joints T F FTT 5. Paths and trajectories for a mobile robot - are the same if the robot is reduced to a point T F - paths are usually computed from maps T F - trajectories could be generated from artificial potentials methods T F FTT 6. The localization problem - is a typical problem of mobile robots T F - the odometer alone is enough to solve it with great accuracy T F - is difficult because there is an unknown uncertainty in the actions of the robot T F TFT 7. The controller in a wheeled mobile robot - can use the cinematic model of the robot T F - uses the solution of the inverse kinematics to drive the robot to a given destination T F - is always working in closed loop T F TTF 8. A purely reactive system - can use hardware and software solutions to implement fast sensing-acting cycles T F - can make the robot able to survive in dynamic environments T F - can be incorporated in the low level of a hierarchical or behaviour architecture T F TTT 9. The image segmentation - is done directly on the acquired image T F - as a possible output can give back a set of segments T F - may be used to find specific lines in the image T F FTT 10. A robot with synchronous drive - has 3 actuated wheels T F - cannot rotate in place T F - is a non holonomic robot T F TFT

2 PART 2- OPEN QUESTIONS - (10 marks) 1. Illustrate and compare the use of velocity in controlling manipulators and mobile robots. 1. Closed loop controller of velocity and position for manipulators. The input is velocity and position of the end effector in Cartesian space; through inverse kinematics and inverse Jacobian go to joint space, move the arm, get new joints values, and transform back to Cartesian through direct kinematics and Jacobian to close the loop. Computationally expensive, so controllers in position that use a predefined time of destination are used in practice. 2. In WMR, usually an open loop controller in position, where the input to the controller is the final destination at a time. Using inverse kinematics the Cartesian position is transformed into the control variables (for instance the velocities of the 2 actuated wheels) and between the possible infinite number of trajectories the one obtained with easy heuristics is built. 2. Illustrate and schematize the path planning methods on the basis of obstacles considered, computational complexity, and completeness of solutions. A first subdivision in two classes is between precise (visibility graphs) or approximated obstacles quadtrees, Voronoi, grids) For the precise description, there is the need to approximate the extended obstacles in many layers, so the computation may be expensive for building the map. For the approximated description it depends on the approximation. For searching the path, the complexity of A* is O( E ), so in worst case quadratic in the number of nodes. About the completeness of the methods, only visibility graph in 2D is a complete method; everything else is not complete method visibility graph quadtree Voronoi grid obstacles precise approximate approximate approximate complexity search O(n 2 ) O(n 2 ) O(n 2 ) O(n) completeness Yes in 2D No guarantee No guarantee No guarantee shortest path Yes in 2D no no no mapping O(n 3 ) for 3D O(nlogn) O(n 2 )

3 3. Define the behaviour-based paradigm for robotics, and give examples of architectures that implement it. See the subsumption architecture notes 4. Describe the possible use of a vision system to detect obstacles for a mobile robot. Indicate the main hardware and software requirements. To detect colour or shape of objects under given angular view it is possible to use 1 camera. For manipulators the camera can be fixed in the scene and looking from top. For detecting the position in 3D we have to compute the distance. Usually a stereo system is needed. Another possibility is to use an integration of a vision system and a distance sensor (see the kinect).

4 PART 3- EXERCISE- (15 marks) 1. Consider the 3 dof RRR robot illustrated. a. define the D-H model of this arm b. describe the working space c. define the Cartesian coordinates that can be solved by the inverse kinematics d. develop a method (algebraic or geometric) to solve the inverse kinematics Solution The working space is part of a sphere Direct kinematix: T matrix Inverse solution: Given the (x, y, z) vector of the end effector, use PAUL (see lesson). A geometric solution is possible considering that the robot is planar. See the solution of the Minimover.

5 2. Consider this map, with 2 rooms and an interconnecting door. The robot is a triangle, illustrated in the initial and final positions. Black figures are obstacles. Use the Voronoi diagram and the visibility graph to obtain a path. Discuss the differences. Voronoi from many random uniform points: many small regions, check if they are free, connect the free ones using A*. Voronoi from obstacles: a generator point in the obstacles, the borders are the path. It is difficult to define points in the walls. Visibility graph: Define the reference point on the robot, for instance the top vertex. Enlarge all obstacles and create one map with this orientation (the same in start and goal). Build the visibility graph and make the path. To traverse the door observe that the wall too has been enlarged. Conclusion: the visibility graph is the best way to traverse the door in case the dimension of the robot is adequate. For random points Voronoi we can miss a free region in the door opening.

6 3. Consider the following programming problem: We have two cubes of equal size and one prism as in figure. Using them we want to build an arch in a given position. - Define the world model, using parameters when possible; - Compute the grasping points from the given positions; - Write the assembly code using any number (1 to 3) of robot arms. begin valori di default di approach e departure frame cubo1, cubo1grasp, cubo1fin, cubo2, cubo2grasp, cubo2fin, triangolo, triangolograsp, triangolofin; scalar lato, altezzatriangolo, lunghezzatriangolo; event primo, secondo; lato = 2*inches; altezzatriangolo = 3*inches; lunghezzatriangolo = 7*inches; cubo1 = FRAME (NILROTN, VECTOR (10, 20, 0) * inches; cubo1fin = FRAME (NILROTN, VECTOR (30, 20, 0) * inches; cubo2 = FRAME (NILROTN, VECTOR (10, 15, 0) * inches; cubo2fin = FRAME (NILROTN, VECTOR (30, 29, 0) * inches; triangolo = FRAME (NILROTN, VECTOR (10, 5, 0) * inches; triangolofin = FRAME (NILROTN, VECTOR (30, 29, 0) * inches; AFFIX cubo1grasp TO cubo1 AT TRANS (ROT(xhat, 180*degrees), VECTOR (lato/2. lato/2, lato/2)*inches) RIGIDLY; AFFIX cubo2grasp TO cubo2 AT TRANS (ROT(xhat, 180*degrees), VECTOR (lato/2. lato/2, lato/2)*inches) RIGIDLY; AFFIX triangolograsp TO triangolo AT TRANS (ROT (xhat, 180*degrees), VECTOR (lato/2, lunghezzatriangolo/2, altezzatriangolo/2)*inches) RIGIDLY; cobegin begin muovi cubo1 move barm to bpark; open bhand; move barm to cubo1grasp; center barm; affix cubo1 to barm rigidly; move cubo1 to cubo1fin; open bhand; unfix cubo1 from barm; signal primo; move barm to bpark; end;

7 begin muovi cubo2 lo stesso con yarm e signal secondo end; coend end begin muoviamo il triangolo move rarm to rpark; move rarm to triangolograsp; center rarm; affix triangolo to rarm rigidly; wait primo; wait secondo; move triangolo to triangolofin; open rhand; unfix triangolo from rarm; move rarm to rpark; end