# TENSEGRITY

TENSEGRITIES

## Tetrahedron

In fact it is not a real tetrahedron but a truncated one, because the four edges of the tetrahedron are cut off leaving four equilateral triangles.

To solve the problem we look at the tensegrity from a specfic angle, as can be seen on this picture.

Vertical to this picture the strut and the tendon in the middle lie parallel to eachother. They both lie in the same plane as the picture itself, therefor the z-axes is not relevant for this problem. In fact, the dimensioning of the tetrahedron is now reduced to a two dimensional problem.

s = the length of a strut.

b = the length of the 12 tendons that form the three equilateral triangles.

c = the length of the 6 tendons that connect the four triangles.

φ = the angle that describe the twist between the equilateral triangles. For normal truncated tetrahedrons this angle is 0, but in the case of a tensegrity it is not.

ω = the angle between two axes of a tetrahedron. This angle is 109,47° (see for instance wikipedia). Imagine four lines from the four corners of a tetrahedron to the heart of the tetrahedron. The angle ω is the angle between each pair of these lines. In this case half the angle is even more interesting: 54,74°, and in particular cos(½ω) = √(1/3).

The big circle in the picture describes all the possible ends of tendon c. Remember in this picture tendon c and strut s lie flat in the same plane as the picture..

The small ellipse describes all possible end points of the triangle formed bij tendons b. In fact these points lie in a circle but from this view they form an ellipse. The "flatness" of this ellipse is determined bij the angle ω.

The two illustrations above are nearly the same. Their difference lies in a different angle φ. The pictures form two examples of the possible positions of the tendons b and c and the strut s.

The radius of the big circle depends only on the size of tendon c. In fact the radius is exactly ½ c. The radius of the ellipse is fixed bij the length of tendon b. The relation between radius and tendon is:

 rb = b / √3 (17)

The angle in the ellipse formed by the point where the strut s ends, the haeart of the ellipse and the point where tendon c ends is exactly 120° (equilateral triangle).

Given, tendons b and c the question is: Which angle φ results in a maximum strutlength s?

There are many "clever" solutions to find the correct angle φ, but the computers at hand one can also easily calculate the strutlength for a few hundred different angles φ. The longest strut s is automatically the right answer (because apparantly this is the longest strut that fits in this tendon-web).

The length of the strut can be calculated as follows:
 (½s)2 = (x1)2 + (y1 + y2 + y3)2 (18)

In which:

 x1 = (b/√3) * sin(φ+120) (19)
 y1 = √((½c)2- ((b/√3) * sin(φ))2) (20)
 y2 = (b/√3) * cos(φ) * cos(½ω) (21)
 y3 = - (b/√3) * cos(φ+120) * cos(½ω) (22)

As already mentioned: cos(½ω) = √(1/3)

For the dimensioning of the tetrahedron the following steps should be made:

Choose the tendon lengths b en c. Use these values in equations (19) till (22) en try different values of the angle φ.

Every different angle φ will result in a different value of half the strut length ½ s. The largest ½ s is your answer. The strut of course is twice as long.

You can turn a 3D image of a tetrahedron on page 3D images

Marcelo Pars