Robert Eisele
Engineer, Systems Architect and DBA

# Constructing Hexagonal Baseplate for a Stewart Platform

A Stewart platform is a nice way to learn lots of motion dynamics. I'm in the process of creating an own prototype myself, so stay tuned. For now I want to focus on constructing a typical baseplate, which can be 3D printed or cut out off acrylic glass or aluminium.

I wanted to construct the typical hexagonal baseplate in a manner to be able to parametrize it freely. As I wanted a circular base, I decided to define the hexagon based on a circle. After some scribbles, I saw a really nice pattern: You can construct the typical look with two equilateral triangles and their inscribed circle. All we then have to do is calculating the six intersection points of the two triangles:

On the right you get the six coordinates generated with the described algorithm. To allow a drawing and not to destroy my website layout, I limited the radii to 10-80. I think that should be enough for most use-cases. If you still need more freedom, I created a simple OpenSCAD sketch to generate the baseplate, which can be downloaded here.

## Derivation

When it comes to the derivation, the side length of an equilateral triangle is $$a=2r\sqrt{3}$$, since every angle in the triangle has 60°, drawing the angle bisector results in 3 median lines, joining together in the centroid of the triangle. The height of the big triangle is thus the radius $$r$$ plus a portion $$s$$. In general the height of an equilateral triangle is $$\frac{1}{2}\sqrt{3}a$$, as such $$\text{height}=r+s\Rightarrow s=\frac{1}{2}\sqrt{3}a-r$$. Calculating the side $$a$$ is now easy, since we know the small rectangular triangle has a hypothenuse of length $$r$$ and the catheti of length $$s$$ and $$\frac{a}{2}$$. Using pythagorean theorem we can solve for $$a$$:

$r^2 = \frac{a^2}{4} + s^2 \Leftrightarrow r^2=\frac{a^2}{4} + (\frac{1}{2}\sqrt{3}a-r)^2 \Leftrightarrow a= 2r\sqrt{3}$

From that, we can draw three points with $$360°/3=120°$$ apart. The radius $$R$$ from the middle of the circle to the edge of the triangle which has the inscribed circle is then

$\begin{array}{rl}R &= \sqrt{r^2 + (a / 2)^2}\\&= \sqrt{r^2 + (r \sqrt{3})^2}\\&= 2r\end{array}$

With that, a triangle can be drawn based on the radius of the inner circle. We now create a second triangle the same way and rotate it 180°. Then the intersections of these two triangles are the desired points.

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