Consider planar cubic bipartite graphs. The graph has a 3-edge coloring due to the 4-coloring theorem. By that and its planarity the vertices have an induced orientation. Now traverse the graph's (conjectured) Hamilton cycle. Going with/against the local orientation or the vertices, alternates along a Hamilton cycle, was proven here.
Now, I found the definition of a Petrie Polygon, saying:
A Petrie polygon is a polygon found in a polyhedron or other regular map by travelling along its edges, turning sharp left and sharp right at alternate vertices.
Does that mean that a Hamilton Cycle is a Petrie Polygon with respect to the local orientation of the vertices? Was that ever used somehow?
References welcome...
The more precise definition of a Petrie polygon in $3$ dimensions requires that no $3$ consecutive edges belong to the same facet. (In $n$ dimensions, no $n$ consecutive edges can belong to the same facet.)
In the case of a planar cubic graph (bipartite or otherwise), this uniquely determines the Petrie polygon once you pick two starting edges. Most of the time, you will get a cycle which is not Hamiltonian.
For example, if you try this in the cube, the Petrie polygons will all be identical-looking $6$-cycles (one is shown on the left, below). All Hamiltonian cycles in the cube also look identical (one is shown on the right, below) and include four places where they follow edges along the same face for $3$ consecutive steps, violating the Petrie condition.