Transcript Document

Inverse Kinematics
Given the tool configuration (orientation Rw and
position pw) in the world coordinate within the work
envelope, find the joint variables, (q).
Formulation:
R0n(q) | p0n(q)
T0n(q) =
Rw | pw
------------- = ---------
000 | 1
000|1
==> 12 equations, n unknowns;
r1(q) r2(q) r3(q) | p0n(q)
Re-writing, T0n(q) = - - - - - - - - - - - - - - - - - - - - 0 0
0 | 1
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Indirect Kinematics
Recall: r1(q), r2(q), r3(q) are orthonormal vectors,
==> r1(q) r2(q) = 0; r2(q) r3(q) = 0; r1(q) r3(q) = 0; and
|| r1(q) || = 1; || r2(q) || = 1; || r3(q) || = 1.
==> 6 independent constraints
==> solution exists for q if n >= 6
Note: Solution is NOT unique and may not exist.
To reduce the number of equations, a tool configuration vector,
W is defined.
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Inverse Kinematics - Tool Configuration Vector
Definition: The tool configuration vector, W, in R6 is defined as
p0n(q)
W(q) = - - - - - - - - - - - - - - - - exp( qn/p ) r3(q)
where p0n(q) is the position of the tool frame;
qn is the roll angle of the tool (qn); and
r3(q) is the unit vector giving the direction of the approach
vector (zn ).
Thus, the number of equations becomes 6.
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Formulation of Direct & Inverse Kinematics
Given a robot arm:
Step 0: Determine the vector of joint variables, q.
Step 1: Derive the link coordinates of the arm based upon the D-H
representation and algorithm;
Step 2: Derive kinematics parameters ( q , d , a , a );
Step 3: Derive the arm equation, i.e. the transformation matrix,
R0n(q) | p0n(q)
T0n(q) = - - - - - - - - - - - - 000 | 1
Step 4: Direct Kinematics: Given q, compute T0n(q) ;
Step 5: Inverse Kinematics: Given the tool configuration vector, W,
compute the vector of joint variables, q.
Example: 5-axis spherical robot (details in class notes)
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