]> A Practical Guide to Tensegrity Design: 8 Analyzing Clearances in Tensegrities

Chapter 8

Analyzing Clearances in Tensegrities

8.1 Clearance Analysis: Introduction

A practical factor which must be taken into account in analyzing the outcome of a tensegrity mathematical programming problem is the clearance between (in other words, the distance separating) one member and another. Especially with truss tensegrities, solutions to a problem can very easily have clearances which would result in one member inadvertently intersecting another if the model were implemented. If this is the case, adjustments must be made using length constraints, objective function weights and/or model configuration until satisfactory clearances are obtained.

8.2 Clearance Analysis: Formulas

Two members can be modeled mathematically as two line segments in space. When neither end point of the two line segments coincides, the parameter of interest is the minimum distance between the two line segments. The position of the points on the two segments where this minimum is attained may also be of interest. When the two segments coincide at one end point, the angle between the two segments may be of concern.

8.2.1 Clearance Formulas: Distance Between Two Line Segments

Let $A ⁢B‾$ and $C ⁢D‾$ be two line segments. An arbitrary point, call it $P A⁢B$, on the line obtained by extending the segment $A ⁢B‾$ can be generated as a function of a scalar multiplier, call it $λ A⁢B$, using the formula

$P A⁢B ≡ A+ λ A⁢B ⁢B-A$.

If $λ A⁢B$ is between 0 and 1, this point will lie on the segment $A ⁢B‾$. Similarly, a point on the line coinciding with $C ⁢D‾$ can be generated using the formula

$P C⁢D ≡ C+ λ C⁢D ⁢D-C$.

To find the minimum distance between these two lines (which is not necessarily the distance between the two line segments), values for $λ A⁢B$ and $λ C⁢D$ can be found which minimize the distance between $P A⁢B$ and $P C⁢D$. Thus, the following unconstrained programming problem is arrived at:

$minimize PA⁢B - PC⁢D 2≡ PA⁢ B - PC⁢D ⋅ PA⁢ B - PC⁢D λ A⁢B, λ C⁢D$

This problem can be solved by differentiating the objective function with respect to $λ A⁢B$ and $λ C⁢D$, setting the two resulting equations equal to zero and solving the implied system for $λ A⁢B$ and $λ C⁢D$.

Substituting using

$PA⁢ B - PC⁢D = A+ λ A⁢B ⁢B-A - C+ λ C⁢D ⁢D-C$
$PA⁢ B - PC⁢D = A-C+ λ A⁢B ⁢B-A - λ C⁢D ⁢D-C$

and differentiating results in the system:

$2⁢λ A⁢B ⁢ A ⁢B‾ 2 - 2⁢λ C⁢D ⁢ B-A⋅ D-C +2⁢ A-C⋅ B-A =0$
$- 2⁢λ A⁢B ⁢ B-A⋅ D-C + 2⁢λ C⁢D ⁢ C ⁢D‾ 2 -2⁢ A-C⋅ D-C =0$

Since

$B-A ⋅ PA⁢ B - PC⁢D = λ A⁢B ⁢ A ⁢B‾ 2 - λ C⁢D ⁢ B-A⋅ D-C + B-A⋅ A-C$

and

$D-C ⋅ PA⁢ B - PC⁢D = -λ C⁢D ⁢ C ⁢D‾ 2 + λ A⁢B ⁢ D-C⋅ B-A + D-C⋅ A-C$

this system implies

$B-A ⋅ PA⁢ B - PC⁢D =0$
$D-C ⋅ PA⁢ B - PC⁢D =0$

In other words, the line segment connecting the two closest points on the lines is orthogonal to both lines.

This system can be solved to find values for $λ A⁢B$ and $λ C⁢D$. If either of these values is outside the range $0.01.0$, then this solution is not valid as the distance between the two line segments and boundary solutions must be searched.

The first sort of boundary solution which can be investigated is one in which the minimum distance is attained at one end point of one of the segments with the other minimum point being an interior point of the other segment. Calculating this distance involves another minimization problem. For example, to calculate the distance between $A$ and $C ⁢D‾$ the following minimization problem would need to be solved:

$minimize A- PC⁢D 2≡ A - PC⁢D ⋅ A - PC⁢D λ C⁢D$

This problem can be solved by differentiating the objective function with respect to $λ C⁢D$, setting the resulting equation equal to zero and solving the implied system for $λ C⁢D$.

Substituting using

$A- PC⁢D = A- C+ λ C⁢D ⁢D-C$
$A- PC⁢D = A-C - λ C⁢D ⁢D-C$

and differentiating results in the equation

$2⁢λ C⁢D ⁢ C ⁢D‾ 2 -2⁢ A-C⋅ D-C =0$

or

$λ C⁢D = A-C⋅ D-C C ⁢D‾ 2$

Since

$D-C ⋅ A- PC⁢D = -λ C⁢D ⁢ C ⁢D‾ 2 + D-C⋅ A-C$

the solution equation implies

$D-C ⋅ A- PC⁢D =0$

In other words, the line segment connecting $A$ with the closest point on $C ⁢D‾$ is orthogonal to $C ⁢D‾$.

If $λ C⁢D$ is outside the range $0.01.0$, then again the value is not valid as a minimum distance to the line segment from the point, and the minimum of $A ⁢C‾$ and $A ⁢D‾$ should be selected as the value. $A ⁢C‾$ is calculated using the Pythagorean distance formula $A ⁢C‾ = A-C ⋅ A-C$ and $A ⁢D‾$ is computed similarly.

In searching for a boundary value for the minimum length between the two segments, all four boundary possibilities should be examined ($A$ and $C ⁢D‾$, $B$ and $C ⁢D‾$, $A ⁢B‾$ and $C$, $A ⁢B‾$ and $D$) and the minimum of these taken to be the solution.

8.2.2 Clearance Formulas: Angle Between Two Line Segments

In some situations, the geometry and position of a hub may dictate that problems will ensue if the angle between two of the members the hub connects is too small. In these cases, it is advisable to compute the angle between the centerlines of the relevant members and see if it is greater than the necessary threshold.

The formula for the angle between two line segments is derived from the Schwarz inequality.1 For example, the angle between the two line segments $A ⁢B‾$ and $C ⁢D‾$ is equal to

$arccos A-B ⋅ C-D A ⁢B‾ ⁢ C ⁢D‾$

8.2.3 Clearance Formulas: A Sample Application

The line segment formulas were used to look at the clearances of the struts and interlayer tendons of the 4ν t-octahedron spherical truss of Section 5.3. The planned realization in mind was a structure at a scale of 1 model unit = 90 mm using 8-mm-diameter wooden dowels for struts and fishing line for tendons. The clearance goal was one strut diameter between the outer surfaces of any two members. This reduced to 2 strut diameters between two strut center lines, 1.5 strut diameters between a strut center line and a tendon center line, and 1 strut diameter between two tendon center lines. The diameter of the tendon was regarded as negligible. In model units, these thresholds were $2⋅890 =0.18$, $1.5⋅890 =0.13$ and $1⋅890 =0.09$ respectively.

It was found strut member #3 and a transformed2 version of tendon member #5 had a poor clearance of 0.081 model units. In addition, at 0.17 model units the clearance between the two strut members #1 and #3 was marginally a problem. Increasing the constrained length of the highly-tensioned tendon member #28 from 1 to 1.4 model units increased the first clearance to 0.16 model units and the second clearance to 0.20 model units without creating clearance problems between other members. Table 8.1 shows the values for the lengths and relative member forces of the revised model. Table 8.2 shows the revised values for the coordinates of the basic points.

 Member # Length Relative Force 1 3.000000 -12.309 2 3.000000 -11.701 3 3.000000 -11.604 4 3.000000 -11.265 5 2.357656 4.715 6 2.389582 4.779 7 2.437046 4.874 8 2.365863 4.732 9 2.030353 4.061 10 2.047569 4.095 11 2.040178 4.080 12 1.640244 3.280 13 1.000000 3.950 14 1.000000 5.618 15 1.000000 4.330 16 1.000000 5.030 17 2.498276 0.999 18 2.732085 1.093 19 2.745418 1.098 20 2.924998 1.170 21 1.382591 1.383 22 1.427206 1.427 23 1.500065 1.500 24 0.943181 0.943 25 1.000000 6.055 26 1.000000 6.400 27 1.000000 5.535 28 1.000000 6.260

Table 8.1: 4ν T-Octahedron: Revised Member Lengths and Forces

 Point
Coordinates
 $x$ $y$ $z$
 $P1$ $P2$ $P3$ $P4$
 1.0923 -0.280953 2.03525 -0.128883 0.310054 2.30168 0.820711 1.13782 1.63086 -1.02937 1.2451 1.77746
 $P1'$ $P2'$ $P3'$ $P4'$
 1.57876 0.504378 3.38267 0.618364 0.662396 3.61217 1.16044 1.39379 3.1984 -1.67512 2.09062 2.49157

Table 8.2: 4ν T-Octahedron: Revised Basic Point Coordinates

8.2.4 Clearance Formulas: Another Sample Application

In the double-layer tensegrities considered in Chapters 5 and 6, when lower-frequency designs are considered, the angle between a secondary interlayer tendon and the strut it connects with on the outer layer may be very small and allow the tendon to rub against projections on the hub or strut at that point. This is because of the shallowness of the corresponding secondary tripod. At high-enough frequencies, the situation approximates the planar situation illustrated in Figure A.2 and no special consideration needs to be taken of the secondary interlayer tendons since the angles there will be adequate if the angles for the primary interlayer tendons are. The cure for a deficiency in either of these angles is to shrink the radius of the inner tendon network.

Therefore, as an example of the use of member-angle computations, the angles between the secondary interlayer tendons and the struts they are attached to on the outer layer will be computed for the 4ν t-octahedron spherical truss solution obtained in Section 8.2.3.

 Strut # Primary Tendon # Angle Secondary  Tendon # Angle 1 5 16.5692 9 5.67875 2 6 17.0124 10 7.04962 3 7 17.5822 11 6.50225 4 8 16.6881 12 8.61599

Table 8.3: 4ν T-Octahedron: Strut/Interlayer-Tendon Angles (in Degrees)

Table 8.3 summarizes the results. The angles at the feet of the outer-pointing t-tripods are two to three times as large as the angles at the feet of the inner-pointing t-tripods. At high-enough frequencies, they would approach equality.

1 See Lang71, p. 22. The Schwarz inequality is also known as the Cauchy-Schwarz inequality or the Cauchy-Buniakovskii-Schwarz inequality.

2 The transformation was $xyz ⇒ -x -yz$.