Day 2.4 - 2 y 2 Forward Kinematics 1/17/2011 19 without...

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Forward Kinematics 1/17/2011 16 notice that link 2 moves in a circle centered on frame 1 Forward Kinematics q 2 1 a 1 a 2 ( x , y ) ? x 0 y 0 ( a 1 cos 1 , a 1 sin 1 ) 1 x 1 y 1 ( a 2 cos ( 1 + 2 ), a 2 sin ( 1 + 2 ) )
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Forward Kinematics 1/17/2011 17 because the base frame and frame 1 have the same orientation, we can sum the coordinates to find the position of the end effector in the base frame Forward Kinematics q 2 1 a 1 a 2 x 0 y 0 ( a 1 cos 1 , a 1 sin 1 ) 1 x 1 y 1 ( a 2 cos ( 1 + 2 ), a 2 sin ( 1 + 2 ) ) ( a 1 cos 1 + a 2 cos ( 1 + 2 ), a 1 sin 1 + a 2 sin ( 1 + 2 ) )
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Forward Kinematics 1/17/2011 18 we also want the orientation of frame 2 with respect to the base frame x 2 and y 2 expressed in terms of x 0 and y 0 Forward Kinematics q 2 1 a 1 a 2 x 0 y 0 1 x
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Unformatted text preview: 2 y 2 Forward Kinematics 1/17/2011 19 without proof I claim: Forward Kinematics q 2 1 a 1 a 2 x y 1 x 2 = (cos ( 1 + 2 ), sin ( 1 + 2 ) ) y 2 = (-sin ( 1 + 2 ), cos ( 1 + 2 ) ) x 2 y 2 Inverse Kinematics 1/17/2011 20 given the position (and possibly the orientation) of the end effector, and the dimensions of the links, what are the joint variables? Inverse Kinematics q 2 ? 1 ? a 1 a 2 x y x 2 y 2 ( x , y )...
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This note was uploaded on 02/13/2012 for the course CSE 4421 taught by Professor Burton during the Winter '11 term at York University.

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Day 2.4 - 2 y 2 Forward Kinematics 1/17/2011 19 without...

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