Deployment of a Class 2 Tensegrity BoomJean-Paul Pinauda, Soren Solaria, and Robert E. SkeltonaaUniversity of California, San Diego9500 Gilman Drive, La Jolla, California 92093, U.S.A.AbstractTensegrity structures are special truss structures composed of bars in compression and cables in tension. Mosttensegrity structures under investigation, to date, have been of Class 1, where bars do not touch. In this article,however, we demonstrate the hardware implementation of a 2 stage symmetric Class 2 tensegrity structure,where bars do connect to each other at a pivot. The open loop control law for tendon lengths to accomplishthe desired geometric reconfiguration are computed analytically. The velocity of the structure’s height is chosenand reconfiguration is accomplished in a quasi-static manner, ignoring dynamic effects. The main goal of thisresearch was to design, build, and test the capabilities of the Class 2 structure for deployment concepts and tofurther explore the possibilities of multiple stage structures using the same design and components.Keywords: Tensegrity structures, deployment, reconfiguration, stable unit1. INTRODUCTIONTensegrity structures are built of bars and tendons attached to the ends of the bars.1The bars can resistcompressive force and the strings cannot. Most bar-string configurations which one might conceive are not inequilibrium, and if actually constructed will collapse to a different shape. Only bar-string configurations in astable equilibrium will be called tensegrity structures.2–5If well designed, the application of forces to a tensegrity structure will deform it into a slightly different shapein a way which supp orts the applied forces. Tensegrity structures are very special cases of trusses, where membersare assigned special functions. Some members are always in tension and others are always in compression. Wewill adopt the words “tendons” for the tensile members, and “bars” for compressive members.6A tensegritystructure’s bars cannot be attached to each other through joints that impart torques. The end of a bar can beattached to tendons or ball jointed to other bars.Tensegrity structures are natural candidates to be actively controlled structures since the control system canbe embedded in the structure directly; for example tendons can act as actuators and/or sensors.7–9Shape controlof the tensegrity structure can be accomplished by moving along its equilibrium manifold. The tensegrity unitshown in Fig. 1 is the simplest three–dimensional tensegrity unit which comprises three bars held together inspace by strings so as to form a tensegrity unit. A tensegrity unit comprising of three bars will be called a3–bar tensegrity. A 3–bar tensegrity is constructed by using three bars in each stage which are twisted either inclockwise or in anti–clockwise direction. The top strings connecting the top of each bar support the next stagein which the bars are twisted in a direction opposite to the bars in the previous stage. In this way any numberof stages can be constructed which will have an alternating clockwise and anti–clockwise rotation of the bars ineach successive stage. This is the type of structure in Snelson’s Needle Tower.Contact info:For animations/movies of figures contact authors.J.P.Pinaud: Email: [email protected]. Solari: Email: [email protected] 1. Simplest tensegrity unit: three-bar unit.In contrast to the Class 1 structure,10, 11a Class 2 structure can also be constructed with the same 3–bartensegrity unit. Assembly of two units, with clockwise and anti–clockwise sense, can be “stacked” directly oneach other, resulting in bars connecting shown in Fig. 2a. The following section addresses the analysis andconstruction of a prototype Class 2 tensegrity structure that verifies the deployment methodology used.2. TWO STAGE CLASS 2 TENSEGRITY STRUCTUREDesign of a two-stage Class 2 tensegrity structure begins with the design of the base. The allowable twist angle,α, of a two-stage Class 2 structure with fixed base nodes isπ2−πn, where n is the number of bars in each stage(n = 3 in this paper). The addition of a Reinforcing tendon, to be described in the next section, increases theadmissable twist angle range to (π6,π2).12–14In addition, from these papers, by Masic, we conclude the forcesin the three bar tensegrity unit are realisable and can be scaled with respect to each other. We now turnour attention to geometry of the structure and writing the equations that describe the reconfiguration of thestructure. These equations are vital in deriving the motor command signals that actuate the structure. Choosingthe coordinate axes as shown in Fig. 2 we write the coordinates of each node as followsFirst StageNodal Number Fixed Node? Coordinates1 yes (rbase, 0, 0)2 no (r cos α, r sin α, h)3 yes (rbasecos(β + α), rbasesin(β + α), 0)4 no (r cos(β + 2α), r sin(β + 2α), h)5 yes (rbasecos(2β + 2α), rbasesin(2β + 2α), 0)6 no (r cos(2β + 3α), r sin(2β + 3α), h)Second Stage7 top platform (rbase, 0, 2h)8 top platform (rbasecos(β + α), rbasesin(β + α), 2h)9 top platform (rbasecos(2β + 2α), rbasesin(2β + 2α), 2h)Table 1. Table of nodal coordinates shown in Fig. 2.where the chosen parameters to describe the geometry of the structure are r, h, and α. From symmetry, theangle β is definedβ + α =2π3.(a) VRML Simulation (b) Top ViewFigure 2. Views of 2 stage tensegrity indicating no dal positions and coordinate axes.The saddle tendon length can be computed by calculating the distance between nodes 4 and 2 asS =µ[r cos(2π3+ α) − r cos α ]2+ [r sin(2π3+ α) − r sin α]2¶12= r√3. (1)Similarly, the vertical tendon length can be computed by calculating the distance between nodes 2 and 1 asV =¡[r cos α − rbase]2+ r2sin2α + h2¢12, (2)and the reinforcing tendon length can be computed from the distance between nodes 2 and 3 asR =µ[r cos α − rbasecos(2π3)]2+ [r sin α − rbasesin(2π3)]2+ h2¶12. (3)The parameter r can be shown to depend on h and α, therefore, only two independent parameters are neededto describe the structure completely. Figure 3 can be used to derive the dependence as follows: Apply the lawof cosines to 4ABC(l2bar− h2) = r2base+ r2− 2rrbasecos(2π3+ α), (4)and solving for r via the quadratic formula yieldsr = rbasecos(2π3+ α) ± (r2basecos2(2π3+ α) + l2bar− h2− r2base)12. (5)(a) Top View (b) Side View of a single barFigure 3. Geometric relations