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bbillingsley3 Group Title

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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  1. Chipper10 Group Title
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    It's still that. You just take the position only, don't worry about the orientation.

    • one year ago
  2. bbillingsley3 Group Title
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    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
  3. Chipper10 Group Title
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    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
  4. bbillingsley3 Group Title
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    How do we find xi from the (x(t), y(t)) given?

    • one year ago
  5. Chipper10 Group Title
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    What is xi?

    • one year ago
  6. bbillingsley3 Group Title
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    I thought we would calculate alphadot = (psuedo invese) * (xi)

    • one year ago
  7. Chipper10 Group Title
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    Yes but what do you think is xi?

    • one year ago
  8. bbillingsley3 Group Title
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    I would imagine it is some sort of velocity. Would we take the derivative of the (x(t),y(t))?

    • one year ago
  9. Chipper10 Group Title
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    Yes

    • one year ago
  10. bbillingsley3 Group Title
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    I have a 3x1 vector for my psuedo inverse. How do I multiply that by xi = (.5sin(t), .49cos(t) )?

    • one year ago
  11. Chipper10 Group Title
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    You pseudoinverse should not be 3x1. What formula are you using?

    • one year ago
  12. bbillingsley3 Group Title
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    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
  13. Chipper10 Group Title
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    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
  14. Chipper10 Group Title
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    I have to go

    • one year ago
  15. bbillingsley3 Group Title
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    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
  16. bbillingsley3 Group Title
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    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
  17. Chipper10 Group Title
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    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.

    • one year ago
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