This page describes a solution to the forward kinematics problem which models a two-dimension version of the Stewart Platform. This problem is posed on page 67 of Numerical Analysis, Second Edition by Timothy Sauer. The two dimensional model consists of a triangular platform in a fixed plain controlled by three struts. It is assumed that the length of all three struts is known as is the length of the sides of the triangles. Also known is the three points of the plane to which the struts connect. The forward kinematics problem solves for the placement of one of the vertices of the triangle and the angle to the x-axis from the side of the triangle given the angle of the triangle associated with the vertex. The solutions and associated figures were coded and plotted in Matlab. This project was done by Anna-Rose Wolff and Chris Mazzullo. The website design, f.m, and half of the plots were created by Anna-Rose Wolff. The bisection method and a script to plot the struts were written by Chris Mazzullo.

The equation that specifies valid poses for the platform is f(θ) = 0, where f(θ) is the function given in this MATLAB file: f.m

Here is a graph of f(θ) against θ where θ at f(θ) = 0 specifies an angle value for the platform that is valid. Notice the roots at θ = -π / 4 and π / 4.

Here is what Stewart Platforms using those valid angles with the given strut lengths look like:

(a)	(b)

Here is a graph for f(θ) of a somewhat more complicated platform with more valid angles:

The four roots of the function were pinpointed with the bisection method. These four roots correspond to four valid values of θ, which in turn correspond to these four poses of the Stewart Platform:

θ = -.7208	θ = -.3310
θ = 1.1437	θ = 2.1159

Slightly increasing the righmost strut to a length of 7.01 leads to a platform with yet more roots:

Here is the graph for f(θ)

Here are all six possible poses:

θ = 2.5000	θ = 0.9774
θ = 0.4617	θ = 0.0576
θ = -0.4098	θ = -0.6415

Finally, by plotting f(θ) with a rightmost strut length of 4, a situation with only two valid angles can be found.

Here is the graph for f(θ), notice the two roots:

Here are the two poses:

θ = 1.3316	θ = 1.7775