Kinematics of the Stewart Platform part 6

Kathleen McLane, Tim Reid, and Paul McNulty

Part 6 asked us to solve the forward kinematics problem for the planar Stewart Platform specified by x1 = 5, (x2,y2) = (0,6), L1 = L3 = 3, L2= 3*sqrt(2), \(\gamma\) = \(\pi\)/4, p1 = 5, p3 = 3, and some value of p2 that would give two zeros for f(\(\theta\)), which we found to be p2 = 4. The methodology we used was the same as in part 4, so we have included the diary from when we went through to find the zeros using the bisection method: p6 diary. The plot of f(\(\theta\)) vs. \(\theta\) can be seen below, and the \(\theta\) values at which the zeros occurred are: \(\theta\) = 1.3316 and 1.7775.

Again, x and y values for the first vertex were given by Project1xy.m, and the other vertices and strut lengths were given and checked respectively using vertices.m.

\(\theta\) = 1.3316, x = 4.8907, y = 1.0400

\(\theta\) = 1.7775, x = 4.8991, y = 0.9991