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Unformatted text preview: CRIPTION. 1. Equate the general transformation matrix
to the manipulator transformation matrix.
If a particular, rather than a general solution
is required a matrix describing the desired
position and orientation of the end of the
manipulator is equated to the manipulator
transformation matrix. 2. Look at both matrices for: (a) elements which contain only one joint variable
(b)
(b) pairs of elements which will produce
expression in only one joint variable when
divided
(c) elements, or combinations of elements, that can
be simplified using trigonometric identities. 3. Having selected an element, equate it to the
equate corresponding element in the other matrix to
produce
produce an equation.
Solve this equation to find a description of one
fi
joint variable in terms of the elements of the
general transformation matrix. 4. Repeat step 3 until all the elements identified in
step 2 have been used
5.
If any of these solutions suffer from
inaccuracies, undefined results, or redundant
results, set them aside and look for better
solutions. Solutions in terms of the elements of
the p vector may lead to more efficient
Solutions than solutions in terms of the elements
of the x, y, or z vectors. 6. If there are more joint angles to be found, premultiply both sides of the matrix equation by the
inverse of the A matrix for the first link to produce
th
th fi
a new set of equivalent matrix elements.
7. Repeat steps 26 until either solution to all the
joint variables have been found, or you have run
out of A matrices to premultiply. 8. If a suitable solution cannot be found for a joint
variable, choose one of those discarded in step 5,
taking notes of regions where problems may
occur. 9. If a solution cannot be found for a joint
variable in terms of the elements of the
manipulator transform, it may be that the
manipulator cannot achieve the specified position
and orientation; Also, theoretical solutions may
not be physically attainable because of
mechanical limits on the range of joint variable. ⎡1
⎢0
⎢
⎢0
⎢
0
⎣ 0 0 1⎤
⎥
1 0 0⎥
0 1 0⎥
⎥
0 0 0⎦ inverse kinematics
The robot is actuated in joint space and that is
what we can control
Once the problem is solved in Cartesian space
constraints, need to map these constraints into the
robot’s joint space using inverse kinematics
E.g. if a straight line trajectory is needed, then
need to break the trajectory into a set of joint
space values over time so that robot follow the
line. inverse kinematics
The inverse kinematics mapping is typically one
to many
Has to choose one solution from a number of
valid solutions
Th
There are degenerate cases with infinite number
of solutions (singularities)
Some solutions of the inverse mapping may not
be physically realizable due to physical joint
limits of manipulators. There may not be a
closed form solution
to
the
inverse
problem at all for
some manipulators.
However, most of
the
manipulators
use a 3 DOF wrist
that has intersecting
axes. This allows the
separation of the
inverse
problem
into a 3 DOF
problem for finding
the endpoint of
wrist and a 3 DOF
problem for finding
the orientation of
wrist. Numerical methods can be used to find a
solution to the inverse problem if a
closed form solution does not exist. A redundant robot has an extra DOF which
can be useful to reach around obstacles and
avoiding collisions with other objects in eth
workspace. Inverse kinematics can be solved by various
methods :
geometric, trigonometric and algebraic.
Once solution for a joint variable has been
found, the manipulator becomes a reduced
th
DOF mechanism – i.e. with one joint less. Example
Solving the inverse kinematics of Stanford arm
⎡ nx
⎢n
T06 = ⎢ y
⎢ nz
⎢
⎣0 sx ax sy
sz
0 ay
az
0 px ⎤
py ⎥
⎥ = T01T12T23T34T45T56
pz ⎥
⎥
1⎦ (T01 ) −1T06 = T12T23T34T45T56 = T16
⎡X
⎢X
T16 = ⎢
⎢X
⎢
⎣0 X X X
X X
X 0 0 Cθ1 p x + Sθ1 p y ⎤ ⎡ X
⎥ ⎢X
− pz
⎥ =⎢
− Sθ1 p x + Cθ1 p y ⎥ ⎢ X
⎥⎢
1
⎦ ⎣0 X X X
X X
X 0 0 Sθ 2 ⋅ d 3 ⎤
− Cθ 2 ⋅ d 3 ⎥
⎥
0.1 ⎥
⎥
1
⎦
68...
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 Spring '14
 Mechatronics

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