4. Why Robot Manipulation ?
Robot Manipulation is a core robot technology
A. Basic fundamentals of robotics lies in robot manipulation
- kinematics & dynamics
- motion planning & control
- higher mathematics and AI
B. Direct application to industry
- robot manipulators are still a growing market
- technology is now being applied beyond conventional areas
- new research on interactive robots is centered around it
11. Manipulator Kinematics
Position Kinematics:
The mathematics behind making the robot move to a
desired position.
Typically used in:
a. Pick & Place operations
b. Assembly operations
c. Stacking and loading
Velocity Kinematics:
The mathematics behind making the robot move with
desired velocities.
Typically used in:
a. Cutting / Machining
b. Painting Operations
c. Scanning areas
12. Forward & Inverse Kinematics
Forward Kinematics:
I know: The position / velocity of each robot joint.
I need to find out: The end position / velocity of the robot
Inverse Kinematics:
I know: The desired position / velocity of the end point where the robot must reach
I need to find out: The position / velocity of each robot joint
13. Forward & Inverse Kinematics
Forward Kinematics:
!!
SY
I know: The position / velocity of each robot joint.
I need to find out: The end position / velocity of the robot
EA
!!
EX
Inverse Kinematics:
PL
I know: The desired position / velocity of the end point where the robot must reach
M
I need to find out: The position / velocity of each robot joint
O
C
14. Forward Kinematics
Forward Kinematics:
Joint 2 Angle
What is the end point location (X, Y) if the joint Y
angles are: End Point
Joint 1 = 30º and Joint 2 = 60º
Given the 3 link lengths (a1, a2 and a3)
are 10mm each
Joint 1 Angle
X
15. Forward Kinematics
Forward Kinematics:
Joint 2 Angle
What is the end point location (X, Y) if the joint Y
angles are: End Point
Joint 1 = 30º and Joint 2 = 60º
Given the 3 link lengths (a1, a2 and a3)
are 10mm each
Joint 1 Angle
X
a1 = a2 = a3 = 10mm
x = a2*sin (j1) + a3*sin (j1 + j2) = 10 * 0.5 + 10 * 1 = 15mm
y = a2*cos (j1) + a3*cos (j1 + j2) = 10 * 0.866 + 10 * 0 = 8.66mm
16. Inverse Kinematics
Inverse Kinematics:
Joint 2 Angle
Lets say that the end point is at (15, 8.66) mm Y
End Point
Given the 3 link lengths (a1, a2 and a3)
are 10mm each
What are the joint angles?
Joint 1 Angle
X
Is it only 30 & 60 ?
Solve: Hint Cos(A+B) = CosA.CosB – SinA.SinB
17. Inverse Kinematics
From the FK equations:
x = a2*sin (j1) + a3*sin (j1 + j2) Joint 2 Angle
y = a2*cos (j1) + a3*cos (j1 + j2) Y
End Point
Simplifying the notations
x = a2.s1 + a3.s12
y = a2.c1 + a3.c12
If we square and add, we get Joint 1 Angle
x2 + y2 = a22 + a32 + 2a2a3 [ s1.s12 + c1. c12 ]
X
which is to say
x2 + y2 = a22 + a32 + 2a2a3 c2
Thus c2 = (x2 + y2 – a22 – a32) / 2(a2.a3)
Remember: cos(-x) = cos(x)
18. Inverse Kinematics
Now that we have solved joint 2 = j2
Joint 2 Angle
x = a2.s1 + a3.s12 Y
y = a2.c1 + a3.c12
End Point
x = a2.s1 + a3s1c2 + a3s2c1
y = a2.c1 + a3c1c2 – a3s1s2
x = s1 [ a2 + a3 c2 ] + a3s2 c1
y = c1 [ a2 + a3 c2 ] – a3s2 s1
Joint 1 Angle
let k1 = a2 + a3 c2 and k2 = a3s2
x = k1 s1 + k2 c1 X
y = k1 c1 + k2 s1
Substituting: k1 = r cos (t) and k2 = r sin (t)
we will get
t + j1 = Atan2 ( y, x) where t = Atan2( k2, k1)
19. Inverse Kinematics
Final Solution
Solution 1 = [ 30, 60 ] ..... obviously
Solution 2 = [90, -60 ]
In principle:
Inverse Kinematics will almost always give you
more than one solution!
20. Summarizing
Robot Manipulators cover the core concepts of robot design & control
Manipulators are necessarily kinematic chains with links and joints (drives)
The crux of robot arm control is based on the theory of kinematic
Kinematics are used in both Position and Velocity domain
Kinematics are of two kinds – Forward and Inverse Kinematics
Forward Kinematics is easy and is used mainly for Analysis
Inverse Kinematics is complex and is used for robot control and operations
Classical Industrial Robots have structures to easy out complex calculations
Modern day robots are equipped with powerful computers and hence there is no restriction on
the structure