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You are told that markers 3-5 do form the TCS. Marker 4 is used as the origin. The x-axis is directed from marker 4 towards marker 3. The y-axis is directed anteriorly and is normal to the x-axis and the vector from marker 4 towards marker 5. The z-axis is normal to both the x-axis and the y- axis such that the axes form a right-handed basis. Using the values for the location the markers in the Global Coordinate System (GCS) shown in Figure 1, please derive the 4x4 homogeneous transformation matrix that expresses the
with respect to the GCS (Mathinline body --uriencoded--TCS_%7Bshank%7D
) for the conditions shown in Figure 1.Mathinline body --uriencoded--%5e%7BGCS%7D T %5e %7BTCS_%7Bshank%7D%7D You are once again correct and are told that markers 1,2,6, and 7 do form the ACS (at this point, you’re feeling good about the interview). You are told the following: “The origin for the shank ACS is the midpoint of markers 1 and 2. The z-axis is directed from the origin towards the midpoint of markers 6 and 7. The y-axis is directed anteriorly and is normal to the z-axis and a temp vector from marker 1 to marker 2. The x-axis normal to both the y-axis and z-axis and is directed medially.” Using the values for the location the markers in the Global Coordinate System (GCS) shown in Figure 1, please derive the homogeneous transformation matrix that expresses the
with respect to the GCS (Mathinline body --uriencoded--ACS_%7Bshank%7D
) for the conditions shown in Figure 1.Mathinline body --uriencoded--%5e%7BGCS%7D T %5e %7BACS_%7Bshank%7D%7D Using your solutions to parts 1 and 2, please derive the transformation matrix that expresses the
with respect toMathinline body --uriencoded--ACS_%7Bshank%7D
(Mathinline body --uriencoded--TCS_%7Bshank%7D
).Mathinline body --uriencoded--%5e%7BTCS_%7Bshank%7D%7D T %5e %7BACS_%7Bshank%7D%7D Which of the following transformation matrices are constant for the duration of an experiment (and therefore only need to be calculated once), and which vary with time?
Mathinline body --uriencoded--%5e%7BGCS%7D T %5e %7BTCS_%7Bshank%7D%7D Mathinline body --uriencoded--%5e%7BGCS%7D T %5e %7BACS_%7Bshank%7D%7D Mathinline body --uriencoded--%5e%7BTCS_%7Bshank%7D%7D T %5e %7BACS_%7Bshank%7D%7D Mathinline body --uriencoded--%5e%7BTCS_%7Bthigh%7D%7D T %5e %7BTCS_%7Bshank%7D%7D Mathinline body --uriencoded--%5e%7BGCS%7D T %5e %7BTCS_%7Bthigh%7D%7D Mathinline body --uriencoded--%5e%7BGCS%7D T %5e %7BACS_%7Bthigh%7D%7D Mathinline body --uriencoded--%5e%7BTCS_%7Bthigh%7D%7D T %5e %7BACS_%7Bthigh%7D%7D Mathinline body --uriencoded--%5e%7BACS_%7Bthigh%7D%7D T %5e %7BACS_%7Bshank%7D%7D You know that the markers used to establish the ACS are frequently removed from a subject during a motion analysis study. As a final test during your job interview, you are asked to find the knee flexion angle ( from
from question 7.2 above), the knee varus/valgus angle (Mathinline body \alpha
), and the internal rotation angle (Mathinline body \beta
) using a X-Y-Z Euler angle convention for some future instance in time (assume the thigh is the fixed frame “A” and that the shank is the moving frame “B”). Please provide your answer in degrees. To solve for the angles, please use your previous answers and the following additional information that was computed at timeMathinline body \gamma
:Mathinline body t = t_1
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^{GCS}T^{TCS_{thigh}} = \left[ \begin{array}{ccc} -0.4542 & 0.7576 & -0.4687 & 5.1436 \\ -0.7900 & -0.0994 & 0.6049 & -3.2432 \\ 0.4117 & 0.6451 & 0.6437 & 63.9444 \\ 0 & 0 & 0 & 1 \end{array} \right] |
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