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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 ?
 one year ago
 one year 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 ?
 one year ago
 one year ago

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Chipper10Best ResponseYou've already chosen the best response.0
It's still that. You just take the position only, don't worry about the orientation.
 one year ago

bbillingsley3Best ResponseYou've already chosen the best response.0
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)?
 one year ago

Chipper10Best ResponseYou've already chosen the best response.0
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.
 one year ago

bbillingsley3Best ResponseYou've already chosen the best response.0
How do we find xi from the (x(t), y(t)) given?
 one year ago

bbillingsley3Best ResponseYou've already chosen the best response.0
I thought we would calculate alphadot = (psuedo invese) * (xi)
 one year ago

Chipper10Best ResponseYou've already chosen the best response.0
Yes but what do you think is xi?
 one year ago

bbillingsley3Best ResponseYou've already chosen the best response.0
I would imagine it is some sort of velocity. Would we take the derivative of the (x(t),y(t))?
 one year ago

bbillingsley3Best ResponseYou've already chosen the best response.0
I have a 3x1 vector for my psuedo inverse. How do I multiply that by xi = (.5sin(t), .49cos(t) )?
 one year ago

Chipper10Best ResponseYou've already chosen the best response.0
You pseudoinverse should not be 3x1. What formula are you using?
 one year ago

bbillingsley3Best ResponseYou've already chosen the best response.0
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
 one year ago

Chipper10Best ResponseYou've already chosen the best response.0
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
 one year ago

bbillingsley3Best ResponseYou've already chosen the best response.0
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.
 one year ago

bbillingsley3Best ResponseYou've already chosen the best response.0
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] ?
 one year ago

Chipper10Best ResponseYou've already chosen the best response.0
Your xi would be a 2x1. You may be using the wrong form or the pseudoinverse. You want your J# to be a 3x2.
 one year ago
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