Kinematics of the Stewart Platform part 4

Kathleen McLane, Tim Reid, and Paul McNulty

Part 4 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 = p2 = 5, p3 = 3. In order to do this, we first plotted f(\(\theta\)) vs. \(\theta\) for these parameters, and used the plot to get the intervals in which f(\(\theta\)) crossed the x-axis, which we then used in the bisection script. The plot of f(\(\theta\)) vs. \(\theta\) can be seen below, and the \(\theta\) values at which the zeros occurred are: \(\theta\) = -0.7208, -0.3310, 1.1437, and 2.1159.

We then ran these \(\theta\) values through the Project1xy.m script to get the x and y values of the first vertex and therefore the associated positions. A plot of each position can be seen below. The strut lengths were checked to ensure that p1 = p2 = 5 and p3 = 3 using the vertices.m script, which also gave the positions of the other two vertices in each triangle.

\(\theta\) = -0.7208, x = -1.3784, y = 4.8062

\(\theta\) = -0.3310, x = -0.9147, y = 4.9156

\(\theta\) = 1.1437, x = 4.4817, y = 2.2167

\(\theta\) = 2.1159, x = 4.5718, y = 2.0244