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solutions.tex
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solutions.tex
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Forward Kinematics:
(1)
1. Write out the entries of a rotation matrix $^A_BR$.
Solution:
\hat{X}_B\cdot\hat{X}_A \hat{Y}_B\cdot\hat{X}_A \hat{Z}_B\cdot\hat{X}_A
\hat{X}_B\cdot\hat{Y}_A \hat{Y}_B\cdot\hat{Y}_A \hat{Z}_B\cdot\hat{Y}_A
\hat{X}_B\cdot\hat{Z}_A \hat{Y}_B\cdot\hat{Z}_A \hat{Z}_B\cdot\hat{Z}_A
2. Express \hat{X}_B=[0, 1, 0]^T in frame \{A\}.
Solution:
\hat{Y}_B\cdot\hat{X}_A
\hat{Y}_B\cdot\hat{Y}_A
\hat{Y}_B\cdot\hat{Z}_A
3. Now calculate ^B_AR:
Solution:
\hat{X}_B\cdot\hat{X}_A \hat{X}_B\cdot\hat{Y}_A \hat{X}_B\cdot\hat{Z}_A
\hat{Y}_B\cdot\hat{X}_A \hat{Y}_B\cdot\hat{Y}_A \hat{Y}_B\cdot\hat{Z}_A
\hat{Z}_B\cdot\hat{X}_A \hat{Z}_B\cdot\hat{Y}_A \hat{Z}_B\cdot\hat{Z}_A
4. Verify that the solution from (2) indeed reduces to [0, 1, 0]^T when expressed in coordinate system B. Tip: Express multiplications as scalar products and remember that the scalar product of orthogonal vectors is 0 and that of parallel vectors is 1.
Solution:
The solution consists of a 1x3 vector that is obtained by multiplying the 1x3 vector from (2) with each row of ^B_AR. The first entry is:
[\hat{X}_B\cdot\hat{X}_A [\hat{Y}_B\cdot\hat{X}_A+
\hat{X}_B\cdot\hat{Y}_A . \hat{Y}_B\cdot\hat{Y}_A+
\hat{X}_B\cdot\hat{Z}_A] \hat{Y}_B\cdot\hat{Z}_A]
The first vector is ^A\hat{X}_B and the second vector is ^A\hat{Y}_B. As they are orthogonal, their scalar product is zero.
The second entry is:
[\hat{Y}_B\cdot\hat{X}_A [\hat{Y}_B\cdot\hat{X}_A
\hat{Y}_B\cdot\hat{Y}_A . \hat{Y}_B\cdot\hat{Y}_A
\hat{Y}_B\cdot\hat{Z}_A] \hat{Y}_B\cdot\hat{Z}_A]
Both vectors are identical and their scalar product is one.
The third entry is:
[\hat{Z}_B\cdot\hat{X}_A [\hat{Y}_B\cdot\hat{X}_A
\hat{Z}_B\cdot\hat{Y}_A . \hat{Y}_B\cdot\hat{Y}_A
\hat{Z}_B\cdot\hat{Z}_A] \hat{Y}_B\cdot\hat{Z}_A]
The first vector is ^A\hat{Z}_B and the second vector is ^A\hat{Y}_B. As they are orthogonal, the scalar product is zero.
(2)
Consider two coordinate systems \{B\} and \{C\}, whose orientation is given by the rotation matrix $^C_BR$ and have distance $^CP$. Provide the homogenous transform ^C_BT and its inverse ^B_CT.
Solution:
[
^C_BR ^CP
0 0 0 1]