B.5.MPC:5.2 - Use two-colouring to answer a gear question

A designer asks whether every gear in a connected arrangement can rotate while all specified meshes remain engaged. The supplied idealized account has parallel, fixed axes, external gear contacts and positive pitch radii. View every rotation from the same side. At each external mesh, the two gears have opposite signs of angular velocity. Contact geometry and tooth compatibility are additional physical conditions; this first question concerns the consistency of rotation directions.

Construct a finite undirected graph. A vertex represents one gear and an edge represents one of the stipulated external meshes. Assign colour 0 to one rotation direction and colour 1 to the other. An edge requires different colours at its endpoints. This expression makes the physical question a mathematical one: can the vertices be assigned two colours while satisfying every edge?

The graph records actual contact, not geometric closeness on the page. A drawing with crossing lines does not add a mesh between the gears at that crossing. If an actual contact is absent from the list, the result answers only the incomplete list.

Obtain either a colouring or a witness of failure by the following procedure.

1. Choose an uncoloured vertex as a root. Give it colour 0, depth 0 and no parent; place it on a waiting list.
2. Remove the first vertex from the list and inspect its neighbours. Give each uncoloured neighbour the opposite colour, record the removed vertex as its parent, give the neighbour depth one greater and append it to the list.
3. For an already coloured neighbour, compare colours. If they agree, use that edge and the two parent paths to construct the odd-cycle witness explained below, and return it.
4. Continue until the list is empty. Start another root if an uncoloured vertex remains. If every edge joins different colours, return the colouring.

A vertex is added to the waiting list only once. Each recorded parent has smaller depth, so parent paths reach their root. A finite graph therefore yields a result after all relevant vertices and edges have been examined, or after a contradiction has been found. With adjacency lists, constant-cost access to a vertex’s recorded data, and constant-cost insertion and removal in the FIFO waiting queue, the work is proportional to the number of vertices plus edges.

Work a square contact cycle with gears A, B, C and D and contacts AB, BC, CD and DA. Starting at A gives A colour 0; B and D colour 1; and C colour 0. Every contact joins opposite colours. The groups {A,C} and {B,D} therefore give compatible direction choices. Reversing both groups gives the other choice. Four equal pitch circles can be placed with centres at the corners of a square of side twice the pitch radius to obtain those contact incidences; tooth engagement and motion under load still require the corresponding mechanical design.

Now work three gears with contacts AB, BC and CA. Starting at A gives B and C colour 1. Edge BC then joins equal colours. Its endpoints’ parent paths B–A and C–A, together with BC, give the three-edge closed path B–A–C–B. Alternating directions around three external contacts asks B to have both directions at once.

The same reasoning constructs a failure witness in a larger graph. For a same-colour edge, follow its endpoints’ parent paths until their nearest common vertex. Discard the shared path beyond that vertex. The retained paths have even total length because the endpoints have the same depth parity. The extra edge closes a simple odd cycle. Alternating two colours cannot satisfy every edge of an odd cycle. Conversely, if the procedure finishes without such an edge, its returned colours satisfy every listed constraint.

The physical interpretation can be made quantitative without hiding the direction question. Let rᵢ be a gear’s pitch radius and ωᵢ its signed angular velocity. Ideal external meshing with fixed axes gives:

rᵢωᵢ + rⱼωⱼ = 0 at each mesh
qᵢ = rᵢωᵢ, so qⱼ = −qᵢ

For a two-colouring, choose any nonzero value q for one colour and −q for the other, then set ωᵢ = qᵢ/rᵢ. This satisfies those kinematic equations. For an odd cycle, following the equations around the cycle gives q = −q and hence q = 0. In a connected graph all gears then have zero angular velocity. Thus the odd cycle excludes the requested nonzero coupled rotation under the stipulated contact model; a stationary arrangement remains compatible with the equations.

Return the cycle as the particular set of physical contacts responsible for the obstruction. Removing one actual contact from the triangle leaves a chain and permits alternating directions. Deleting only its graph edge leaves the physical obstruction in place. An internal gear contact or a moving carrier changes the contact rule and requires a revised physical account before the colouring test is reused.

A compatible direction assignment answers this bounded question. It does not establish adequate torque, tooth phasing, freedom from interference or motion under the intended load. For those questions, retain the graph result and obtain the missing mechanical contribution. The common Method supplies the contact interpretation, the connection to the computational witness and the return to the design. The gear account and graph argument supply the substantive rules that make that return possible.