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CS 5310/6310 & ME 5220/6220 Exam 3 Solution November 28, 2012

1. (50pts) This question concerns how to splice together two polynomial trajectories in the joint variable

q. Call the first polynomial q1(t), the second q2(t).

Polynomial q1(t) is valid in the time range 0 t t1, and has an initial position, velocity and

acceleration of zero.

Polynomial q2(t) is valid in the time range t1 t t2, and has a final position ofq2 and a finalvelocity of zero.

At time t = t1, the two polynomials have the same position, velocity and acceleration (unspeci-fied).

(a) (16pts) Write the constraint relations.

Polynomial 1 Polynomial 2 Matching constraints

q1(0) = 0 q2(t2) = q2 q1(t1) = q2(t1)q1(0) = 0 q2(t2) = 0 q1(t1) = q2(t1)

q1(0) = 0 q1(t1) = q2(t1)

(b) (10pts) Write the general form of the two polynomials q1(t) and q2(t) that are of minimal degree.

There are a total of 8 constraints. Polynomial 1 is quartic, polynomial 2 is a quadratic.

q1(t) = a0 + a1t + a2t2 + a3t

3 + a4t4

q2(t) = b0 + b1(t t1) + b2(t t1)2

(c) (10pts) Solve for the coefficients of polynomial 1 which are directly found from the constraints

(no simultaneous equations).

Coefficients a0, a1 and a2 are found immediately.

q1(0) = 0 = a0

q1(0) = 0 = a1

q1(0) = 0 = 2a2

(d) (10pts) Focusing on the constraints for polynomial 2 alone, find the relationships among the

coefficients. You are not solving for the coefficients, only the relationships.

The two constraints for polynomial 2 yield:

q2 = b0 + b1(t2 t1) + b2(t2 t1)2

0 = b1 + 2b2(t2 t1)

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Combining,

b1 = 2b2(t2 t1)

q2 = b0 b2(t2 t1)2

(e) (10pts) Focusing on the matching constraints, find a relationship among the coefficients of poly-

nomial 2 alone. You are not solving for the coefficients, only the relationship, and are not

incoroporating the results from part (d).

For polynomial 1 at time t1,

q2(t1) = b0

q2(t1) = b1

q2(t1) = 2b2

For polynomial 2 at time t1,

q1(t1) = a3t3

1+ a4t

4

1

q1(t1) = 3a3t2

1+ 4a4t

3

1

q1(t1) = 6a3t1 + 12a4t2

1

Equating,

b0 = a3t3

1 + a4t4

1

b1 = 3a3t2

1 + 4a4t3

1

b2 = 3a3t1 + 4a4t2

1

Eliminate coefficients a3 and a4.

b1t1 3b0 = a4t4

1

b2t1 b1 = 2a4t3

1

3b1t1 6b0 = b2t2

1

2

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2. (50pts) Consider the rotary 3-link robot.

i ai di i i1 a1 d1 0 2 0 0 0

3 a3 0 0

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

z0

x0

O

O0

a

d1

x3

z3

x2 O3

z2

z1

x1

O2

1

a3

d2

1

(b) (15pts) Determine the velocity Jacobian. Do not resolve the cross products at this point, except

to eliminate zero cross product terms.

Jv =

0z0

0d030z1

0z2 0d23

0z0 00z2

=

0z0 (d1

0z0 + a10x1 + d2

0z1 + a30x3)

0z10z2 a3

0x30z0 0

0z2

=

0z0 (a1

0x1 + a30x3)

0z10z2 a3

0x30z0 0

0z2

since all the 0zi axes are parallel.

(c) (30pts) Given the linear velocity 0d03 of the endpoint, determine the joint rates.

From the velocity Jacobian, the linear velocity is:

0d03 = 10z0 (a1

0x1 + a30x3) + d2

0z1 + 30z2 a3

0x3

3

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Solve immediately for d2 by projecting onto z1.

0d03 0z1 = d2

The joint angle rate 1 can be found by noticing that 3 cannot cause motion in the x3 direction.

0d03 0x3 = 1

0z0 (a10x1 + a3

0x3) 0x3

= 1(0x3

0z3) a10x1

= 10y3 a1

0x1

= 1a1 cos(/2 + 3)

= 1a1s3

Collect all the known terms on the left and call it vector v.

v = 0d03 10z0 (a1

0x1 + a30x3) d2

0z1

= 30z2 a3

0x3

Project onto x1 to find 3.

v 0x1 = 30z2 a3

0x3 0x1

= 3(0x1

0z1) a30x3

= 30y1) a3

0x3

= 3a3 cos(/2 3)

= 3a3s3

3. (30pts) This problem is for graduate students. Consider the robot of problem 2. Derive the endpoint

acceleration 0d03. Please resolve any cross products and simplify.

Regroup the linear velocity expression and resolve the cross products.

0d03 = 10z0 a1

0x1 + d20z1 + (1 + 3)

0z2 a30x3

= 1a10y1 + d2

0z1 + (1 + 3)a30y3

4

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Differentiating,

0d03 = 1a10y1 + 1a1(

0z11 0y1) + d2

0z1 + (1 + 3)a30y3 + (1 + 3)(

0z01 +0z23) a3

0y

= 1a1 0y1 21a1 0x1 + d2 0z1 + (1 + 3)a3 0y3 (1 + 3)2a3 0x3

where note is taken that all zi axes are constant.

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