Page 185 - Mechatronics with Experiments

MECHANISMS FOR MOTION TRANSMISSION 171
3. If there is a load with weight W at the tip, determine the necessary torques at joints
1 and 2 in order to balance the load.

In this example, we are asked to determine the following relations,

x = f( ) (3.241)
̇
̇ x = J( )( ) (3.242)
T
Torque = J ( )(Force) (3.243)
Let us attach three coordinate frames to the manipulator. Coordinate frame 0 is attached to
the base and fixed, coordinate frame 1 is attached at joint 1 to link 1 (moves with link 1),
coordinate frame 2 is attached at joint 2 to link 2 (moves with link 2). The position vector of
T
the tip with respect to coordinate frame 2 is simple, r 2P = [ l 001] . The transformation
2
matrices between the three coordinate frames are functions of and as follows,
1
2
1
1
⎡ cos( ) −sin( )0.00.0 ⎤
sin( ) cos( ) 0.00.0
⎢ ⎥
1
1
T 01 = ⎢ ⎥ (3.244)
⎢ 0.0 0.0 1.00.0 ⎥
⎢ ⎥
⎣ 0 0 0 1 ⎦
⎡ cos( ) −sin( )0.0 l ⎤
2
1
2
sin( ) cos( ) 0.00.0
⎢ ⎥
2
2
T 12 = ⎢ ⎥ (3.245)
⎢ 0.0 0.0 1.00.0 ⎥
⎢ ⎥
⎣ 0 0 0 1 ⎦
Hence, the tip vector description in base coordinates,
T
r = [ x y 01] = T ⋅ T ⋅ r (3.246)
0P 0P 0P 01 12 2P
The vector components of r 0P can be expressed as individual functions for more clarity,
x 0P = x ( , ; l , l ) = l ⋅ cos( ) + l ⋅ cos( + ) (3.247)
1 2
2
2
1
2
1
1
0P
1
y 0P = y ( , ; l , l ) = l ⋅ sin( ) + l ⋅ sin( + ) (3.248)
1
0P
1 2
1
1
2
2
2
1
z = 0.0 (3.249)
0P
The Jacobian matrix for this case can simply be determined by taking the derivative
of the forward kinematic relations with respect to time and expressing the equation in the
matrix form to find the Jacobian matrix.
d
̇
̇ x = (x ( , ; l , l )) = J + J ̇ (3.250)
0P 0P 1 2 1 2 11 1 12 2
dt
d
̇
̇ y 0P = (y ( , ; l , l )) = J + J ̇ (3.251)
0P
22 2
21 1
1
1 2
2
dt
where it is easy to show that the elements of the Jacobian matrix J
[ ]
J 11 J 12
J = (3.252)
J J
21 22
[ ]
−l ⋅ sin( ) − l ⋅ sin( + ) −l ⋅ sin( + )
1
2
1
1
2
2
2
1
= (3.253)
l ⋅ cos( ) + l ⋅ cos( + ) l ⋅ cos( + )
1 1 2 1 2 2 1 2