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 ?

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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 ?

GT ECE 4560 - Intro to Automation & Robotics
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It's still that. You just take the position only, don't worry about the orientation.
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)?
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.

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How do we find xi from the (x(t), y(t)) given?
What is xi?
I thought we would calculate alphadot = (psuedo invese) * (xi)
Yes but what do you think is xi?
I would imagine it is some sort of velocity. Would we take the derivative of the (x(t),y(t))?
Yes
I have a 3x1 vector for my psuedo inverse. How do I multiply that by xi = (.5sin(t), .49cos(t) )?
You pseudoinverse should not be 3x1. What formula are you using?
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
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
I have to go
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.
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] ?
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.
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