Download - Intelligent Robotic CHAPTER 4
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Between two frames we have a kinematic relationship -basically a translation and a rotation.
This relationship is mathematically represented by a 4 4Homogeneous Transformation Matrix.
z
y
x
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x
y 3 1 Translation
z
r1 r2 r3
r4 r5 r6
r7 r8 r9
3
3 RotationalMatrix
0 0 0
1 3 Perspective
1 Global Scale
Homogeneous transformations:
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4
Base
Frame
Tool Frame
x
Goal Frame
Link Frame
Camera Frame
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We are interested in two kinematics topics:
Forward Kinematics -angles to position (See Chapters 2 & 3)What you are given: The length of each link AND the angle of each joint
What you can find: The position of any point (i.e. its (x, y, z) coordinates)OR
(Given the joint angles for the robot, what is the orientation and position of the endeffector)
Inverse Kinematics - position to angles
What you are given: The length of each link and the position of some point on the robotWhat you can find: The angles of each joint needed to obtain that position.
OR
(Given a desired end effector position what are the joint angles to achieve this)
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For a robot system the inverse kinematic problem isone of the most difficult to solve.
The robot controller must solve a set of non-linearsimultaneous equations.
The problems can be summarised as:
The existence of multiple solutions.
The possible non-existence of a solution.
Singularities.
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Goal
This two link planar manipulator has two possible solutions.This problem gets worse with more Degrees of Freedom.
Redundancy of movement.
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Goal
A goal outside the workspace of the robot has no
solution.
An unreachable point can also be within the
workspace of the manipulator - physical constraints.
A singularity is a place of acceleration - trajectory
tracking.
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2
1
(x , y)
l2
l1
Given: l1, l2,x , y
Find:1, 2
(x , y)
Inverse Kinematics of a Two Link Manipulator
Redundancy: A unique solution to this problem doesnot exist. Notice, that using the givens two solutions are
possible. Sometimes no solution is possible.
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The Geometric Solution
l1
l2
(x , y)
2
1
Using the Law of Cosines:
21
2
2
2
1
221
2
21
2
2
2
1
22
21
2
2
2
1
22
222
2cos
2)cos(
)cos()180cos(
)180cos(2)(cos2
ll
llyx
ll
llyx
llllyxCabbac
2
22
2
Redundant since 2 could be in the first or
fourth quadrant.
xy
yxl
l
cClbBcC
bB
1
22
2211
1
11
22
2
22
2
2
1
22221
tan)sin(sin;xytan
;yx
)sin(
yx
)sin(180sin
yx;180;;;sinsin
Redundancy caused since 2
has two
possible values
Using the Law of Cosines:
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12
21
2
2
2
1
221
2
221
2
2
2
1
21121121
2
2
2
1
21121
2
21
2
2
2
1
2
121121
2
21
2
2
2
1
2
1
2222
2
yx
cos
c2
)(ss)(cc2)(ss2)(ss)(cc2)(cc
yx)()(
ll
ll
llll
llllllllllll
iii
The Algebraic Solution
21
21211
21211
2121
11
)(
ssy)(
ccx)(
)cos(c
cosc
iii
llii
lli
))(sin(cos))(sin(cos)sin(
))(sin(sin))(cos(cos)cos(
:
abbaba
bababa
Note
l1
l2
(x , y)
1
2
Only unknown
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)c(s)s(c
cscss
sinsy
)()c(c
ccc
ccx
2211221
12221211
21211
2212211
21221211
21211
lll
lll
ll
slsll
sslll
ll
We know what 2 is from the previous slide. We
need to solve for 1 . Now we have two equations
and two unknowns (sin 1 and cos 1 )
22
222211
2212
22
1122221
221122
221
221
221
2211
yx
x)c(ys
)c2(sx)c(
1
)c(s)s()c(
)(xy
)c(
)(xc
slll
llllslll
lllll
sls
ll
sls
Substituting for c1 and simplifying
many times
Notice this is the law of cosinesand can be replaced by x2+ y2
22
222211
yx
x)c(yarcsin
slll
))(sin(cos))(sin(cos)sin(
))(sin(sin))(cos(cos)cos(
:
abbaba
bababa
Note