# TENSEGRITY

TENSEGRITIES

## Dissimilar

One can not expect great mathematical problems with three strut tensegrities, but nevertheless this page will show that simple calculations can be annoyingly hard to solve. That is probably the general meaning of this page: to show you that simple matters can be difficult.

Maybe this "complexity of simplicity" is one of the reasons why tensegrities always remained at an appropriate distance to the general public.

The way the problem is solved below is just one way. There is at least one other technique that I might demonstrate in another chapter. The beauty of this approach is that you only need to know the The Pythagorean theorem.

First, the simple side of the tensegrity: In the picture one can see a three strut tensegrity with two short struts and one large one. The six vertices of the tensegrity are marked with the lettres A to F. The diagram below shows all the connections between the vertices of the tensegrity.

 A B C D E F A - B tendon - C tendon tendon - D strut tendon - E strut tendon tendon - F tendon strut tendon tendon -

The position of the six verices A to F define the entire construction. Every vertice has three coordinates corresponding with the X-, Y- and Z-axes. For instance vertice E has the coördinates Ex, Ey en Ez.

So, in total there are 6 x 3 = 18 values to be determined.

The 9 tendons describe the following 9 equations:

 (Bx - Ax)2 + (By - Ay)2 + (Bz - Az)2 = tA-B2 (23)
 (Cx - Ax)2 + (Cy - Ay)2 + (Cz - Az)2 = tA-C2 (24)
 (Cx - Bx)2 + (Cy - By)2 + (Cz - Bz)2 = tB-C2 (25)
 (Fx - Ax)2 + (Fy - Ay)2 + (Fz - Az)2 = tA-F2 (26)
 (Dx - Bx)2 + (Dy - By)2 + (Dz - Bz)2 = tB-D2 (27)
 (Ex - Cx)2 + (Ey - Cy)2 + (Ez - Cz)2 = tC-E2 (28)
 (Ex - Dx)2 + (Ey - Dy)2 + (Ez - Dz)2 = tD-E2 (29)
 (Fx - Dx)2 + (Fy - Dy)2 + (Fz - Dz)2 = tD-F2 (30)
 (Fx - Ex)2 + (Fy - Ey)2 + (Fz - Ez)2 = tF-E2 (31)

The 3 struts describe the next 3 equations:

 (Dx - Ax)2 + (Dy - Ay)2 + (Dz - Az)2 = sA-D2 (32)
 (Ex - Bx)2 + (Ey - By)2 + (Ez - Bz)2 = sB-E2 (33)
 (Fx - Cx)2 + (Fy - Cy)2 + (Fz - Cz)2 = sC-F2 (34)

We are free to choose the coordinate system in which we place our construction. Let's place vertex A in the heart of the system and vertex B on the X-axis. Finally we chose A, B en C, in a plane horizontal surface. The coordinates are then as follows:

 Ax = 0 (35)
 Ay = 0 (36)
 Az = 0 (37)
 By = 0 (38)
 Bz = 0 (39)
 Cz = 0 (40)

In total we have 18 equations and 18 variables. This means we should be able to solve the problem.

Now we arrived at the hard part of the puzzle: The determination of 17 coordinates as a function of one variable that we are free to choose.

The first 6 coordinates are given in equations (35) to (40).
With equations (23) and (35) to (39):

 Bx = ± tA-B (41)

Here we choose Bx = + tA-B

With equations (24), (25) and (35) to (40):

 Cx = (Bx2 + tA-C2 - tB-C2) / (2*Bx) (42)

and:

 Cy = ± √(tA-C2 - Cx2) (43)

Also in this case we choose the positive value.

With equations (27) and (32):

 Dx = ( Bx2 + sA-D2 - tB-D2) / (2*Bx) (44)

The value Dy is our only variable we are free choose.

The following mathematics is the determination of all other vertex coordinates as a function of Dy.

With (32) and (35) to (37):

 Dz = ± √(sA-D2 - Dy2 - Dx2) (45)

Also here we choose the positive value.

With (28) and (33):

 Ey = N + W * Ex (46)

in which:

 N = ( sB-E2 - tC-E2 - Bx2 + Cx2 + Cy2 ) / (2*Cy) (47)

and

 W = ( Bx - Cx ) / Dy (48)

With equations (29) and (33):

 Ez = M + V * Ex (49)

in which:

 M = ( sB-E2 - tD-E2 + Dx2 + Dy2 + Dz2 - Bx2 - 2*N*Dy ) / ( 2*Dz ) (50)

and

 V = ( Bx - Dx - W*Dy) / Dz (51)

Maybe it's not very clear in this picture, but the strut between point C and F is larger than the other two struts. The short struts have the same length.

With (46) and (49) in (33):

 Ex = ( - b ± √ ( b2 - 4 * a * c )) / (2 * a) (52)

in which, in this case, a, b and c are equal to:

 a = 1 + W2 + V2 (53)
 b = 2*N*W + 2*M*V - 2*Bx (54)
 c = Bx2 + M2 + N2 - sB-E2 (55)

With (26), (30), (31) and (35) t/m (37):

 Fy = G + P * Fx (56)

in which:

 G = {Ez * ( tD-F2 - tA-F2 - Dx2 - Dy2 - Dz2 ) + Dz * ( tA-F2 - tE-F2 + Ex2 + Ey2 + Ez2) } / (2*Ey*Dz - 2*Dy*Ez) (57)

and

 P = ( Dx*Ez - Dz*Ex ) / ( Dz*Ey - Dy*Ez ) (58)

And with the same equations:

 Fz = H + Q * Fx (59)

in which:

 H = ( tA-F2 - tD-F2 + Dx2 + Dy2 + Dz2 - 2*G*Dy) / (2*Dz ) (60)

and

 Q = ( - Dx - Dy*P ) / ( Dz ) (61)

With equations (26) en (35) t/m (37):

 Fx = ( - b ± √ ( b2 - 4 * a * c )) / (2 * a) (63)

in which, in this case, a, b and c are equal to

 a = 1 + P2 + Q2 (64)
 b = 2*G*P + 2*H*Q (65)
 c = G2 + H2 - tA-F2 (66)

Now, all coordinates are described as function of Dy.

We go back to the tensegrity in the picture in which all tendons are 20,44 cm and the two short struts are 26.0 cm.

The diagram looks now as follows:

 A B C D E F A - B 20,44 - C 20,44 20,44 - D 26,0 20,44 - E 26,0 20,44 20,44 - F 20,44 ??? 20,44 20,44 -

With all equations above we can calculate all the coordinates a function of Dy. In the three diagrams below the values of all coordinates are shown for Dy = -1,8, Dy = -3,8 en Dy = -5,8. Next to each diagram the distance between the points C and F is calculated. It appears that Dy = -3,8 gives the maximum distance between C and F: 36,3 cm. This means the strut length that "compensates" the other two short struts must be 36,3 cm.

 - X Y Z A 0 0 0 B 20,44 0 0 C 10,22 17,70 0 D 16,54 -1,8 -19,98 E -0,77 6,85 -13,39 F 0,39 -13,46 -15,38
Strut length C - F =
√{(-0,39-10,22)2 +
(-13,46-17,70)2 +
(-15,38-0)2} =
36,1 cm
 - X Y Z A 0 0 0 B 20,44 0 0 C 10,22 17,70 0 D 16,54 -3,8 -19,70 E 0,24 7,43 -14,59 F -1,34 -12,91 -15,79
Strut length C - F =
√{(-1,34-10,22)2 +
(-12,91-17,70)2 +
(-15,79-0)2} =
36,3 cm
 - X Y Z A 0 0 0 B 20,44 0 0 C 10,22 17,70 0 D 16,54 -5,8 -19,21 E 2,31 8,63 -16,52 F -3,07 -11,09 -16,89
Strut length C - F =
√{(-3,07-10,22)2 +
(-11,09-17,70)2 +
(-16,89-0)2} =
35,9 cm

In this example all tendons have the same length. But the equations to calculate the coordinates are general equations, so they can be used to calculate all sorts of dissimilar tensegrities, as long these tensegrities have three struts.

Marcelo Pars