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CS 5310/6310 & ME 5220/6220 Exam 2 Solution November 2, 2012

1. (60 pts) Consider the rotary 3-link robot with the following DH parameters:

i ai di i i1 0 d1 /2 2 0 /2 /23 a3 0 0

(a) (5pts) Draw the robot in the zero-angle position.

x

z

x

O

O

0

0

0

1

22

2

3

3

z

x

z

x

z1

2d

d1

O1

3

O3

a

(b) (10pts) Given 0d03, describe which variables affect which coordinates (x,y,z) of the endpointby filling in the table below. Put a cross for each coordinate affected by a particular variable.

Based on this table, which variable would be the best to solve for first?

Variable x0 y0 z01 X X

d2 X X

3 X X X

(c) (15 pts) Find the joint variable identified above.

The endpoint position is:

0d03 = d10z0 + d2

0z1 + a30x3

Project onto the direction z0, which happens to equal x2.

0d03 0z0 = d1 + a3

0x3 0x2

= d1 + a3c3

c3 =0d03

0z0 d1a3

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Use the half-angle tangent formula to find 3.

(d) (30 pts) Find the remaining two joint variables.

Joints 2 and 3 form a planar manipulator, and cast a shadow line in the x0,y0 plane whoseorientation depends only on 1. Hence

1 = tan1

0d03x0d03y

d2 is simply the difference between the radial line in the x0,y0 plane and the contribution of3.

d2 =

0d203x +

0d203y a3s3

2. (40 or 70 pts) Consider the rotary 3-link robot with the following DH parameters:

i ai di i i0 a0 d0 /2 /2

1 0 0 02 a2 0 /2 3 a3 0 /2

(a) (5 pts) Draw the robot in the zero-angle position.

z1 z3

x2 3xx1

z0

x0

O OO1 2 3

O0

a3

a2

d1

z2

z

x 1

1

O1

d0

a0

(b) (20 pts) Given 1d1,3, find

0d03.

From the figure, and noting that x0 = y1,

1d1,3 = d0

1z1 + a0

1x0 +1d03

1d03 =1d

1,3 d01z

1 + a01y

1

The rotation matrix 1R0 can be found by inspection or computation:

1R0 =

0 0 11 0 0

0 1 0

2

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Hence 0d03 = (1R0)

T1d03.

(c) (15 pts) Given 0d03, one of the joint variables can be found independent of the other two. Find

this joint variable.

This is another shadow line problem, except that 2 is the one changing the orientation of the

shadow line.

2 = tan1

0d03y0d03x

(d) (30pts) This problem is for graduate students. Find the remaining two joint variables.

The position equation is:

0d03 = d10z0 + a2

0x2 + a30x3

3 affects the x0,y0 directions, but not d3. Project onto x0, which is the same as x1.

0d03 0x0 = a2

0x2 0x1 + a3

0x3 0x1

= a2c2 + a30x3

0x1

x1 and x3 are separated by a coordinate system, so evaluate midframe by expressing both in

terms of axes 2.

x1 = c2x2 + s2z2

x3 = c3x2 + s3y2

Hence x1 x3 = c2c3. Substituting,

0d03 0x0 = a2c2 + a3c2c3

c3 =0d03

0x0 a2c2a3c2

Once again, find 3 using the half-angle tangent formula. Finally, d1 is simply the difference inthe z0 coordinate once the contribution of3 is subtracted out.

d1 =0d03z a3s3

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