Question

I am confused on 1a. I think we are supposed to use ge(dot)(t) in order to find the body velocity. Is this my ge(dot)(t):

(1/T)*[xe(T)-xe(0); ye(T)-ye(0); theta(T)-theta(0)]

Answer 1

I think our ge from the last homework was:

{1.52577; -.547367; -.785398} + (t/T)*{.173704; .542958; 1.309}

If I take the derivative of this with respect to t, I get:

ge(dot) = (1/T)*{.173704; .542958; 1.309}

Am I doing this right?

Answer 2

Yes that looks right for gedot. You just plugged in the values from last week correct?

Answer 3

The numbers I used came from partb of problem 1 of last week's hw. The solutions say: "one canidate trajectory connecting the two end effector configurations is:

{xe*(t); ye*(t); theta*(t)} = {1.52577; -.547367; -.785398} + (t/T)*{.173704; .542958; 1.309}

Is this correct?

Answer 4

Yes you seem to have taken the derivative properly.

Answer 5

So, if xi = g^-1(alpha) * (dg(alpha)/dalpha)

Then I would simply multiply the inverse of my first ge value by the derivative that I calculated?

Answer 6

I'm not sure what you mean by your first ge value. You multiply it by the inverse of ge which varies as a function of time. It should look like this:

\[\xi_b = \left[\begin{matrix}R^{-1}(\theta_e(t)) & 0 \\ 0 & 1\end{matrix}\right]*gdot_e\]

Answer 7

What is my ge?

Answer 8

I can't take the inverse of ge = {1.52577; -.547367; -.785398} + (t/T)*{.173704; .542958; 1.309}

Answer 9

You don't need to take the inverse of anything. The formula above is the body velocity - it varies as a function of time.

Answer 10

The theta(t) comes from the trajectory - for any given time t, you can figure out theta and thus the body velocity.

Answer 11

what is R^-1(theta(t))?

Answer 12

At this point, I'm not sure if the orginal gedot or ge I posted is correct

Answer 13

Do you know how to make a rotation matrix \[R(\theta)?\]

Answer 14

R = [cos(theta), -sin(theta; sin(theta) , cos(theta)]

Answer 15

Yes so R^-1 is the inverse of that.

Answer 16

R^-1 = [-cos(theta), -sin(theta); sin(theta), -cos(theta)]

Answer 17

\[R^{-1}(\theta) = R^T(\theta) = R(-\theta)\]

Answer 18

So xi = R(-theta) * [(1/T)*[xe(T)-xe(0); ye(T)-ye(0); theta(T)-theta(0)]]

Answer 19

that would be a 2x2 times a 3x1. That doesn't work. It's what I posted above.

Answer 20

If gdot = (1/T)*[xe(T)-xe(0); ye(T)-ye(0); theta(T)-theta(0)], how do I get something other than a 3x1 matrix?

Answer 21

You don't. But you multiply it on the left by \[\left[\begin{matrix}R^{-1}(\theta(t)) & 0 \\ 0 & 1\end{matrix}\right]\]

Answer 22

Not just R(-theta)

Answer 23

So, xi = [cos(-theta(t)), -sin(-theta(t)), 0; sin(-theta(t)), cos(-theta(t)), 0; 0 , 0, 1] * [(1/T)*[xe(T)-xe(0); ye(T)-ye(0); theta(T)-theta(0)]]

Answer 24

Yes, that xi_body. I gotta go. Good luck

Answer 25

Thanks for the help.

Answer 26

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