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OSU ECE 5463 - Manipulator-Kinematics-2

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Manipulator Kinematics-2Read Chapter 3Practice of the Hexapod Model = 0 00 000001 00 1= 0 2.50 0 1 0000 00 1= 0 12.80 000001 00 1=  12.82.5  12.82.5001  12.80 1VI. Specify Body and Foot Transformation• Usually the robot has additional frames fixed to the base (link 0) and the terminal link• In a manipulator:– {S} frame attached to the station–: indicates where the manipulator is within the workstation– {T}: frame attached to the tool (gripper)– For the OSU hexapod:Station: the bodyTool: footSpecify Body and Foot Transformation-continued 1 0 0 0 1 0 00001 0 1 10.980 10.98 1710.9810.980 10.98 1710.980010 0  1.440 1VII. Forward Kinematics for Manipulator• Compute the position/orientation of the tool relative to the station frame• Forward kinematics: Given = , i= 1, …n, find . •••  ArmToolManipulator locationOSU Hexapod LegZ1Z0Z2Z3ZFXFYFX2X3X1X0#1#2#312 + 9/16 2 + 1/2 2 + 7/1617Z3Z2X3X21 + 7/16 X0X1Z1Z0l4l1l2l3l5#0.How did we get ?First, what is ? 10.9810.98010.9810.98000110 000101 0=10.980 10.9810.980 10.9801 0Rotate about z for 10.98Rotate about x for 90RX(10 00c s0s cRZ(c s 0s c 0001Then what is = 1710.981710.981.44Then we have as  = 0001= 10.98010.98 1710.9810.98010.98 1710.980010 0  1.440


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OSU ECE 5463 - Manipulator-Kinematics-2

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