### MATLAB Assignment 6

iteration = 1

normOfCorrection = 0.5771

iteration = 2

normOfCorrection = 0.0850

iteration = 3

normOfCorrection = 0.0013

iteration = 4

normOfCorrection = 6.2976e-007

iteration = 5

normOfCorrection = 2.9448e-013

Here's the results that i got for the Newton-Raphson part of the assignment:

iteration = 1

normOfCorrection = 0.5771

iteration = 2

normOfCorrection = 0.0850

iteration = 3

normOfCorrection = 0.0013

iteration = 4

normOfCorrection = 6.2976e-007

iteration = 5

normOfCorrection = 2.9448e-013

iteration = 1

normOfCorrection = 0.5771

iteration = 2

normOfCorrection = 0.0850

iteration = 3

normOfCorrection = 0.0013

iteration = 4

normOfCorrection = 6.2976e-007

iteration = 5

normOfCorrection = 2.9448e-013

I think there's an error in Problems 1 & 2 of the Matlab assignment:

1. The AbsoluteX constraint needs to retain "xPground" and only omit "yPground"

2. The AbsoluteY constraint needs to retain "yPground" and only omit "xPground"

1. The AbsoluteX constraint needs to retain "xPground" and only omit "yPground"

2. The AbsoluteY constraint needs to retain "yPground" and only omit "xPground"

F11Wendy wrote:I think there's an error in Problems 1 & 2 of the Matlab assignment:

1. The AbsoluteX constraint needs to retain "xPground" and only omit "yPground"

2. The AbsoluteY constraint needs to retain "yPground" and only omit "xPground"

Wendy, for AbsoluteX, the xPground will be rolled into the CmotionFunction. That's why CmotionFunction can never assume the value NONE. Say you want to have the x of point P always be 4.532. Then include this 4.532 value into the CmotionFunction. If you want the difference between x of point P and the point on ground with x=4.532 be some function of time (say 0.25*t + sin(t)), than define your motion to read CmotionFunction: 4.532+ 0.25*t + sin(t)

I hope this makes sense.

BTW, the same point holds for AbsoluteY and the yPground value.

Have a good weekend,

Dan

My solution for Question 3 c)

t=.37

Phi=[0 0 0]'

Phiq=[[1 0 0];[0 1 0];[0.685 -0.73 1]]

t=.37

Phi=[0 0 0]'

Phiq=[[1 0 0];[0 1 0];[0.685 -0.73 1]]

In Problem 3B qdot 0 = [-2pi; 0; 0]T. How can the velocity theta dot be zero given the prescribed motion theta(t) = pi/2 + 2pi*t ?

My solution for Question 3.b:

phi = [0;0;0]

phi_q=

[1 0 -1

0 0 0

0 0 1]

nu = [0; 0; 2pi]

gamma = [0; 0; 0]

phi = [0;0;0]

phi_q=

[1 0 -1

0 0 0

0 0 1]

nu = [0; 0; 2pi]

gamma = [0; 0; 0]

F11Jonathan wrote:My solution for Question 3 c)

t=.37

Phi=[0 0 0]'

Phiq=[[1 0 0];[0 1 0];[0.685 -0.73 1]]

My solution for 3C agrees with this.

F11Wendy wrote:In Problem 3B qdot 0 = [-2pi; 0; 0]T. How can the velocity theta dot be zero given the prescribed motion theta(t) = pi/2 + 2pi*t ?

Wendy - you are right, the velocity should have been 2pi.

The PDF has been changed to reflect this correction.

Thank you for catching this.

Dan

P.S. If you worked with the old value is ok, no need to redo anything. In general, the value that you prescribe at the beginning of the analysis is irrelevant for Kinematics (qdot_zero gets computed by simEngine2D for you). They would be relevant though for Dynamics.

My results for Problem 4:

Iteration 1: 5.77e-1

Iteration 2: 8.50e-2

Iteration 3: 1.33e-3

Iteration 4: 6.30e-7

Iteration 5: 2.94e-13

Iteration 1: 5.77e-1

Iteration 2: 8.50e-2

Iteration 3: 1.33e-3

Iteration 4: 6.30e-7

Iteration 5: 2.94e-13

Problem 3.b t=0

Phi=[0;0;0]; Jacobian=[1 0 -1;0 1 0;0 0 1];

nu=[0;0;2*pi];

gamma=[0;(2*pi)^2;0];

Problem 3.c t=0.37

Phi=[0;0;0]; Jacobian=[1 0 0.6845;0 1 -0.7290;0 0 1];

also my q at this time is [1.7290;1.6845;3.8956];

Phi=[0;0;0]; Jacobian=[1 0 -1;0 1 0;0 0 1];

nu=[0;0;2*pi];

gamma=[0;(2*pi)^2;0];

Problem 3.c t=0.37

Phi=[0;0;0]; Jacobian=[1 0 0.6845;0 1 -0.7290;0 0 1];

also my q at this time is [1.7290;1.6845;3.8956];

Results for problem 3:

t =

0.3700

Phi =

0

0

0

Phi_q =

1.0000 0 0.6845

0 1.0000 -0.7290

0 0 1.0000

Results for problem 4:

iterationCounter =

1

normCorrection =

0.5771

iterationCounter =

2

normCorrection =

0.0850

iterationCounter =

3

normCorrection =

0.0013

iterationCounter =

4

normCorrection =

6.2976e-007

iterationCounter =

5

normCorrection =

2.9448e-013

q =

1.0000

1.5708

t =

0.3700

Phi =

0

0

0

Phi_q =

1.0000 0 0.6845

0 1.0000 -0.7290

0 0 1.0000

Results for problem 4:

iterationCounter =

1

normCorrection =

0.5771

iterationCounter =

2

normCorrection =

0.0850

iterationCounter =

3

normCorrection =

0.0013

iterationCounter =

4

normCorrection =

6.2976e-007

iterationCounter =

5

normCorrection =

2.9448e-013

q =

1.0000

1.5708

iteration = 1

normOfCorrection = 0.5771

iteration = 2

normOfCorrection = 0.0850

iteration = 3

normOfCorrection = 0.0013

Final Values

q=[1 1.5708]'

normOfCorrection = 0.5771

iteration = 2

normOfCorrection = 0.0850

iteration = 3

normOfCorrection = 0.0013

Final Values

q=[1 1.5708]'

Iteration1

normCorrection =

0.5771

Iteration2

normCorrection =

0.0850

Iteration3

normCorrection =

0.0013

Iteration4

normCorrection =

6.2976e-007

Iteration5

normCorrection =

2.9448e-013

Here is the value of q:

q =

1.0000

1.5708

normCorrection =

0.5771

Iteration2

normCorrection =

0.0850

Iteration3

normCorrection =

0.0013

Iteration4

normCorrection =

6.2976e-007

Iteration5

normCorrection =

2.9448e-013

Here is the value of q:

q =

1.0000

1.5708

My results for problem 3.c:

t = .37

phi = [0;0;0]

phi_q = [1, 0, .685; 0, 1, -.729; 0, 0, 1]

My results for problem 4 (iteration, norm of correction):

(1, .5771)

(2, .0850)

(3, .0013)

(4, 6.2976e-7)

(5, 2.9448e-13)

t = .37

phi = [0;0;0]

phi_q = [1, 0, .685; 0, 1, -.729; 0, 0, 1]

My results for problem 4 (iteration, norm of correction):

(1, .5771)

(2, .0850)

(3, .0013)

(4, 6.2976e-7)

(5, 2.9448e-13)

Problem 3c

At time 37

Phi_37 =

[ 0]

[ 0]

[-0.0628]

Phi_q_37 =

1.0000 0 0.6367

0 1.0000 -0.7711

0 0 1.0000

Problem 4

iteration= 1

normCorrection = 0.5771

iteration= 2

normCorrection = 0.0850

iteration= 3

normCorrection = 0.0013

iteration=4

normCorrection =6.2976e-007

iteration=5

normCorrection =2.9448e-013

Here is the value of q:

q =

1.0000

1.5708

At time 37

Phi_37 =

[ 0]

[ 0]

[-0.0628]

Phi_q_37 =

1.0000 0 0.6367

0 1.0000 -0.7711

0 0 1.0000

Problem 4

iteration= 1

normCorrection = 0.5771

iteration= 2

normCorrection = 0.0850

iteration= 3

normCorrection = 0.0013

iteration=4

normCorrection =6.2976e-007

iteration=5

normCorrection =2.9448e-013

Here is the value of q:

q =

1.0000

1.5708

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