Kinematics of the Stewart Platform part 5

Kathleen McLane, Tim Reid, and Paul McNulty

Part 5 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 not equal to 7 but close to it. We used p2 = 6.99. 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: p5 diary. The plot of f(\(\theta\)) vs. \(\theta\) can be seen below, and the \(\theta\) values at which the zeros occurred are: \(\theta\) = -0.6969, -0.3035, 0.0139, 0.4560, 0.9780, and 2.5128.

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.

\(\theta\) = -0.6969, x = -4.2554, y = 2.6252

\(\theta\) = -0.3035, x = -4.8446, y = 1.2368

\(\theta\) = 0.0139, x = -4.9487, y = 0.7143

\(\theta\) = 0.4560, x = -0.8155, y = 4.9327

\(\theta\) = 0.9780, x = 2.3196, y = 4.4292

\(\theta\) = 2.5128, x = 3.2243, y = 3.8216