# Day 06 - Forward Kinematics 2 1 a 1 a 2 x y a 1 cos 1 a 1...

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Day 06 Forward Kinematics 1/25/2012 1

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Transform Equations 1/25/2012 2 {0} {1} {2} {3} {4} 0 1 T 1 2 T 0 3 T 3 4 T 4 2 T
Transform Equations 1/25/2012 3 give expressions for: 0 2 T 3 4 T

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Transform Equations 1/25/2012 4 {0} {1} 0 1 T 0 2 T {2} {3} 2 3 T
Transform Equations 1/25/2012 5 how can you find 0 1 T 2 3 T 0 2 T 1 3 T

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Forward Kinematics 1/25/2012 7 given the joint variables and dimensions of the links what is the position and orientation of the end effector? Forward Kinematics 2 1 a 1 a 2 x 0 y 0 p 0 ?

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Forward Kinematics 1/25/2012 8 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

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Unformatted text preview: Forward Kinematics 2 1 a 1 a 2 x y ( 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 ) ) Forward Kinematics 1/25/2012 9 from Day 02 Forward Kinematics 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 p = ( a 1 cos 1 + a 2 cos ( 1 + 2 ), a 1 sin 1 + a 2 sin ( 1 + 2 ) ) Frames 1/25/2012 10 Forward Kinematics 2 1 a 1 a 2 x y x 1 y 1 x 2 y 2 Forward Kinematics 1/25/2012 11 using transformation matrices 1 1 , , 1 a x z D R T 2 2 , , 1 2 a x z D R T 1 2 1 2 T T T...
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Day 06 - Forward Kinematics 2 1 a 1 a 2 x y a 1 cos 1 a 1...

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