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 2 years ago
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 ?
 2 years ago
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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Chipper10
 2 years ago
Best ResponseYou've already chosen the best response.0It's still that. You just take the position only, don't worry about the orientation.

bbillingsley3
 2 years ago
Best ResponseYou've already chosen the best response.0So, we need Jbody(a) in order to calculate the psuedo inverse. Do we do the same procedure from last HW to find Jbody(a)?

Chipper10
 2 years ago
Best ResponseYou've already chosen the best response.0You don't need Jbody this time, just J. The question doesn't as you to use the body velocity or anything to that effect.

bbillingsley3
 2 years ago
Best ResponseYou've already chosen the best response.0How do we find xi from the (x(t), y(t)) given?

bbillingsley3
 2 years ago
Best ResponseYou've already chosen the best response.0I thought we would calculate alphadot = (psuedo invese) * (xi)

Chipper10
 2 years ago
Best ResponseYou've already chosen the best response.0Yes but what do you think is xi?

bbillingsley3
 2 years ago
Best ResponseYou've already chosen the best response.0I would imagine it is some sort of velocity. Would we take the derivative of the (x(t),y(t))?

bbillingsley3
 2 years ago
Best ResponseYou've already chosen the best response.0I have a 3x1 vector for my psuedo inverse. How do I multiply that by xi = (.5sin(t), .49cos(t) )?

Chipper10
 2 years ago
Best ResponseYou've already chosen the best response.0You pseudoinverse should not be 3x1. What formula are you using?

bbillingsley3
 2 years ago
Best ResponseYou've already chosen the best response.0Jsharp = 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

Chipper10
 2 years ago
Best ResponseYou've already chosen the best response.0I 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

bbillingsley3
 2 years ago
Best ResponseYou've already chosen the best response.0I 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.

bbillingsley3
 2 years ago
Best ResponseYou've already chosen the best response.0I have figured out the 2x3 Jacobian. Would my 3x3 xi be : [.49cos(t), .5sin(t), 0; .5sin(t), .49cos(t), 0; 0, 0 , 1] ?

Chipper10
 2 years ago
Best ResponseYou've already chosen the best response.0Your xi would be a 2x1. You may be using the wrong form or the pseudoinverse. You want your J# to be a 3x2.
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