Finally, Part 6 asked us to solve the forward kinematics problem for the planar Stewart Platform so that p2 provided only two poses. The value of p2 chosen was 4. These poses can be seen below for when \(\theta\)= 1.3316 and 1.7775.
\(\theta\) = 1.3316, x = 4.8907, y = 1.0400
\(\theta\) = 1.7775, x = 4.8991, y = 0.9991
