Computational Algorithm for Determining the Generic Mobility of Floating Planar and Spherical Linkages Offer Shai Mechanical Engineering School, Tel-Aviv University Tel-Aviv, Israel Andreas Müller Institute of Mechatronics, Chemnitz, Germany,

Outline of the talk -Constraint Graphs: Bar-joint, Body-bar and Mixed graphs. -The main idea underlying pebble game. -Pebble game decomposes any mechanisms into Assur graphs. -Choosing different drivers/input when using pebble game results in different decompositions. -Pebble game reveals redundancy -Pebble game identifies BJ-BB Assur Graphs. -How Pebble game works.

Body-Bar Constraint Graph G=(V,E) is a body-bar graph iff: - A vertex stands for a rigid body. Thus, a vertex possesses three DOFs (in 2D) or six DOFs in (3D). - Edges stand for kinematic pairs (higher or lower pairs). - An edge is between exactly two bodies. - Between two vertices there can be several edges. Body-bar constraint graph

Bar- Joint Constraint Graph G=(V,E) is a bar-joint graph iff: - A vertex represents a point where binary links interconnect through only lower kinematic pairs (two constraints). - Since the vertex stands for a point it possesses two DOFs (in 2D) or three DOFs (in 3D) - Between two vertices there can be at most one edge. (a) (b) Bar-Joint constraint graph

The main problem: in mechanisms several rigid bodies can be connected through one joint (called “multiple joint”). For mechanisms with multiple joints there is no unique body-bar graph causing difficulties.

A B C B1B1 B2B2 B3B3 B4B4 B5B5 0 B1B1 B2B2 B3B3 B4B4 B5B5 0 Kinematic pairs: (B 2,B 3 ), (B 3,B 4 ) Kinematic pairs: (B 2,B 4 ), (B 3,B 4 ) The problem of Multiple Joints and Body-Bar Graphs Different Body-Bar Graphs for the same mechanism Different structural representations

Different Body-Bar graphs might result with even wrong conclusions

Mixed Constraint Graph (new) G=(V B  V J, E) is a mixed graph iff: - Every vertex can stand for a body or a point. - If the vertex corresponds to a body, v  V B, it possesses three DOFs (in 2D) or six DOFs (in 3D). - If the vertex corresponds to a point (i.e. the location of a joint as in the bar-joint), v  V J, it possesses two DOFs (in 2D) or three DOFs (in 3D).

Easy to represent multiple revolute joint (in 2d): A multiple joint connecting m bodies stands for m-1 revolute joints with collinear axis A B 6 C A B C 9 D D (a) (b) 5

Pebble Game – Combinatorial Algorithm for Generic (topological) Mobility Determination The algorithm determines the correct generic, i.e., topological, mobility of the mechanisms. It is applied upon the constraint graph of the mechanism. The constraint graphs possess only topological properties thus redundancies due to special geometries are excluded.

The main idea underlying Pebble Game Pebbles are assigned to each vertex, equal to the DOF of the physical object represented by that vertex when considered unconstraint. Activating the constraints, represented by edges, by coordinated relocation of pebbles.

Choosing different inputs/drivers when applying Pebble game Next to each input/driver we locate a free pebble. Theorem: After locating the free pebbles next to the drivers there is a unique decomposition into Assur Graphs, and pebble game finds this decomposition. For different inputs/drivers there exist different decompositions into Assur Graphs. B A C B A C D B A C D B A C D A 3 7 C B D A 3 7 C B B A C D B A C B A D C B A C D A 3 7 C B D A 3 7 C B

B A C B A C B A C A B C B 1. Bar-joint Constraint Graph 2. Initialization – two pebbles to each vertex. 3. Output of the Pebble Game Alg. – all the admissible edges are directed and free pebbles. 4. The DOF of the mechanism – total number of free pebbles.5. The DOF of an object – maximum free pebbles can be move to the corresponding vertex. 6. Input links/drivers - free pebbles are located at the corresponding vertices. 7. Assur decomposition – each directed cutset defines an Assur Graph 8. Order of the Assur Decomposition – the order of the directed cutsets.

Choosing different inputs/drivers (edges with free pebbles at the end) - causes pebble game to decompose mechanisms into different Assur Graphs. This theorem was proved to be valid also in 3d.

D A 3 7 C B D B A C D B A C 1. Bar-joint Constraint Graph 2. Initialization – two pebbles to each vertex. D B A C D B A C 3. Output of the Pebble Game Alg. – all the admissible edges are directed and free pebbles. 4. The DOF of the mechanism – total number of free pebbles. D B A C 5. The DOF of an object – maximum free pebbles can be move to the corresponding vertex. 6. Input links/drivers - free pebbles are located at the corresponding vertices. 7. Assur decomposition – each directed cutset defines an Assur Graph 8. Order of the Assur Decomposition – the order of the directed cutsets.

Pebble Game Reveals Redundancy An edge is admissible (not redundant) IFF we can bring 4 free pebbles next to its two end vertices. Edge e=(x,y) can be oriented if p(x) + p(y) >= 4 Where p(z) is the number of free pebbles next to vertex z. Pebble game reveals redundancy in mechanisms. An edge/constraint is redundant IFF it is impossible to bring four free pebbles next to its end vertices. This property assures that pebble game determines the correct generic/topological mobility of the mechanism.

A F B A C D E F B A C D E F B A C D E F B A C D E F C A D B E F Edge (E,C) is inadmissible since there are only 3 free pebbles next to its end vertices. One pebble is missing to vertex E. The algorithm searches for a free pebble. Every vertex that is visited during the search belongs to the over constrained region. No free pebble was found – region {B,E,C,D} is over constrained.Bar-joint Constraint GraphInitialization – two pebbles to each vertex. Output of the Pebble Game Alg. – all the admissible edges are directed and free pebbles. B E D C

Pebble game Identifies Assur Graphs Main theorem : G is an Assur Graph IFF after applying pebble game the constraint graph is strongly connected (there exists a directed path from each vertex to any other vertex) IFF there is no directed cut-set (set of edges that disconnetcs the graph). This theorem is valid for Bar-joint and Body-bar (Baranov Trusses) Assur Graphs and proved to be correct both in 2D and 3D.

C A D J F G I H B E C A D J F G I H B E (A) (B) G H F E I J D C A B G H F E I J D C A B (A) (B) Which one is an Assur Graph? The directed constraint graph is an Assur Graph it is strongly connected there is no directed cut-set. The output of the pebble game

Body – bar Assur Graph (Baranov Truss) After applying pebble game on the BB constraint graph we move 3 free pebbles to any vertex (grounding it). The remaining graph should be strongly connected (does not contain a directed cut-set).

Constraint graph is a Body-bar Assur Graph after grounding any vertex (moving 3 free pebbles next to it) the remaing graph is strongly connected. A structural representation of a Body-bar Assur Graph Constraint graph Initialization – 3 pebbles to each vertex. Output of the pebble, body 1 is grounded. Different bodies can be grounded by moving the 3 free pebbles. Grounding body 2 instead of body 1.

The algorithm arranges the BB and BJ matrices in a block triangular form

Synthesis Understanding properties of the Body- Bar Assur Graphs enables to create them in 2D and 3D.

E.E. Peisakh: An algorithmic description of the structural synthesis of planar Assur groups, Journal of Machinery Manufacture and Reliability,Journal of Machinery Manufacture and Reliability 2007, vol. 36, No. 6, vol. 36, No. 6

The Pebble Game Algorithm : Edge is admissible if the total of free pebbles next to its end vertices is at least four. 1.Initialization – The graph is undirected, i.e., initially all constraints are inactive and all objects/vertices are unconstrained. Assign k(v) pebbles to each vertex v, k=2 for joints and k=3 for bodies. 2.While there exist admissible edges DO: Orientation move – if edge (u,v) admissible then remove a pebble from one of the end vertices, let it be u, the edge is directed and u the tail vertex. u v u v

3. While there are free pebbles left Do: Reorientation move – let (u,v) an undirected edge and make it admissible by bringing free pebbles to its end vertices. If peb(v)<2 (v stands for a joint) or peb(v) < 3 (v stands for a body) then search for a vertex ‘z’, s.t., peb(z)>0 and there is a directed path from v to z. Then, move one pebble from vertex z to v by reversing the direction of all the edges, and peb(z) := peb(z)-1, peb(u):=peb(u) + 1. uv z u v z peb(v)- number of pebbles at vertex v.

Thank you !!!!!