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.


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.

You might also be interested in the following


Sorry, comments are closed for this article. Contact me if you want to leave a note.