Question

For HW#14, problem 3, it says that we are not concerning ourselves with orientation. Does this affect our end effector configuration, or is it still: ge(a) = g1(a)g2(a)g3(a)g4 ?

Answer 1

It's still that. You just take the position only, don't worry about the orientation.

Answer 2

So, we need Jbody(a) in order to calculate the psuedo inverse. Do we do the same procedure from last HW to find Jbody(a)?

Answer 3

You don't need Jbody this time, just J. The question doesn't as you to use the body velocity or anything to that effect.

Answer 4

How do we find xi from the (x(t), y(t)) given?

Answer 5

What is xi?

Answer 6

I thought we would calculate alphadot = (psuedo invese) * (xi)

Answer 7

Yes but what do you think is xi?

Answer 8

I would imagine it is some sort of velocity. Would we take the derivative of the (x(t),y(t))?

Answer 9

Yes

Answer 10

I have a 3x1 vector for my psuedo inverse. How do I multiply that by xi = (.5sin(t), .49cos(t) )?

Answer 11

You pseudoinverse should not be 3x1. What formula are you using?

Answer 12

Jsharp = Jbody^T * (Jbody*Jbody^T)^-1
I used the Jbody that we solved for in the last homework and just plugged in our new alpha values

Answer 13

I told you above that you only need J. The manipulator in this homework is different from the one last week and therefore has a different J. Since we are only considering the position and not the orientation, the jacobian should be 2x3. When you apply the formula above you should get (2x3)*(3x3) = 2x3

Answer 14

I have to go

Answer 15

I don't understand how I calculate a 2x3 Jacobian from the information given. Also, I don't know how xi = ( .5sin(t) , .49cos(t) ) can give me a 3x3 matrix.

Answer 16

I have figured out the 2x3 Jacobian. Would my 3x3 xi be :
[.49cos(t), -.5sin(t), 0; .5sin(t), .49cos(t), 0; 0, 0 , 1] ?

Answer 17

Your xi would be a 2x1. You may be using the wrong form or the pseudo-inverse. You want your J# to be a 3x2.

Answer 18

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