MMAE-540 Introduction to Robotics - Fall 2009 Homework 3 Solutions

MMAE-540 Introduction to Robotics - Fall 2009 Homework 3 Solutions

1. Consider the three-link planar manipulator shown below. Derive the forward kinematics equations using the DH convention. The DH parameters are given as:

x3

y3

l3 q 3 x2

y2

l2

y1

q 2

x1

y0

l1

q 1 x0

Figure 1: Manipulator for Problem 1

Link ai i di i

1 l1 0 0 1 2 l2 0 0 2 3 l3 0 0 3

The transformation matrices are given as:

c1 -s1 0 l1s1

0T1

=

s1 0

c1 0

0 l1s1

1

0

(1)

0 001

c2 -s2 0 l2s2

1T2

=

s2 0

c2 0

0 l2s2

1

0

(2)

0 001

c3 -s3 0 l3s3

2T3

=

s3 0

c3 0

0 l3s3

1

0

(3)

0 001

c123 -s123 0 l1c1 + l2c12 + l3c123

0T3

=

0T1

1T2

2T3

=

s123 0

c123 0

0 l1s1 + l2s12 + l3s123

1

0

(4)

0 00

1

1

The forward kinematics are given in the last column of 0T3:

x = l1c1 + l2c12 + l3c123 y = l1s1 + l2s12 + l3s123

2. Consider the two-link planar manipulator show below with a revolute joint and a prismatic joint. Derive the forward kinematic equations using the DH convention.

x2

a1 y0

y2 y1

a2 x1

q 1 x0

Figure 2: Manipulator for Problem 2

The DH parameters are given as:

Link ai i di i

1 a1 0 0 1 2 a2 0 0 /2

The forward kinematics are the last column of the transformation matrix:

-s1 -c1 0 a1c1 - a2s1

0T2

=

c1 0

-s1 0 a2c1 + a1s1

01

0

(5)

0 00

1

such that:

x = a1c1 - a2s1 y = a2c1 + a1s1

3. Consider the 3D manipulator shown below. Draw a schematic of the manipulator, label the diagram with the appropriate coordinate frames and derive the forward kinematic equations using the DH convention.

There are several configurations you could have chosen as the "base" configuration of the robot. This one was easy to keep the X-axes pointing in the same direction, which makes it easy to visualize the link twists. The axes I choose are shown in Fig 4:

The DH parameters are:

Link ai i di i

1 0 90 d1 1 2 a2 90 0 2 3 0 0 d3 0

The forward kinematics of the manipulator are:

c1 0 s1 0

0T1

=

s1 0

0 1

-c1 0

0

d1

00 0 1

2

Figure 3: Manipulator for Problem 3.

Joint 3 Link 2 X2

Z2 Link 3

X3

Z3

Y1 Joint 2

X1 Link 1

Joint 1 Z0

X0 Link 0 is the base

Figure 4: Solution for manipulator for problem 3

c2 0 s2 a2c2

1T2

=

s2 0

0 1

-c2 0

a2s2

0

00 0 1

1 0 0 0

2T3

=

0 0

1 0

0 1

0

d3

000 1

c1c2 s1 c1s2 a2c1c2 + d3c1s2

0T3 = 0T1 1T2 2T3

=

s1c2 s2

-c1 0

s1s2 -c2

a2c2s1 + d3s1s2

a2s2 - d3c2 + d1

000

1

4. Consider the 3D manipulator shown below. Draw a schematic of the manipulator, label the diagram with the appropriate coordinate frames and derive the forward kinematic equations using the DH convention.

The DH parameters are:

3

Figure 5: Manipulator for Problem 4.

Link ai i di i 1 0 -90 0 1 2 0 -90 d2 0 3 0 0 d3 3

5. Consider the 3D Cartesian manipulator shown below. Solve the inverse position kinematics.

x1

z1

y1

z0

x2

x3

z2

y3

y2 y4 x4 z3

z4

y0 x0

Figure 6: Reference frames for cartesian manipulator for Problem 5.

In this problem its helpful to add in an additional reference frame.

Link ai i di i

1 0 -90 d1 0 2 0 0 d2 -90 3 0 -90 0 0 4 0 0 d3 0

The resulting DH Matrix is given as:

0 0 1 d3

0T3

=

0 1

-1 0

0 0

d2

d1

(6)

0 0 01

As you can see, the inverse problem is very simple for this example since

x d3

y = d2

(7)

z

d1

4

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