Is there anything known about the maximum number of simple-polygonal Hamilton cycles that a straight-line drawing of a Hamiltonian graph can have?

Put differently, if the vertices of a Hamilton graph are mapped to points of the euclidean plane and the edges to straight-line segments connecting the images of their adjacent vertices, how many of the graph's Hamilton cycles are then maximally be mapped to simple polygons.

Take for example $K_5$: if the images of all five vertices are in convex configuration, then exactly one Hamilton cycle is mapped to a simple polygon; if only four vertices are in convex configuration, then four Hamilton cycles are mapped to simple polygons, and if finally only three points are in convex configuration, then six of the Hamiltion cycles are mapped to simple polygons - in no case is it possible to place the points in such a way, that all twelve Hamilton cycles are mapped to simple polygons. So, in case of $K5$ the maximal number I am looking for, would be $6$, provided, I counted right.

Remarks.

I am especially interested in $K_5$ and $K_{3,3}$, mainly, because due to Kuratowski's characterization of planar graphs, one can be sure that permutations of $5$ (resp. $6$) cities, i.e. those permutations which correspond to a Hamilton circuit, can't appear in that order in the optimal tour through all $n$ cities; with $K_4$, this is not so.

Letting $p$ denote the maximum number of simple polygonal tours and $h$ the number of Hamiltonian tours, $h-p$ permutations of a subset of $m$ cities can be excluded from the optimal tour through all $n$ cities.

non-planar graphs ($K^5$ and $K^{3,3}$). Why? These donothave any planar embedding. Would you please clarify? $\endgroup$ – Peter Heinig Aug 20 '17 at 6:03tensionbetween what the unusual term "simply-polygonal tour" might be trying to say, and what logically is the case in a planar graph: even if the shape of the circuit in a specifiedembeddingmatters to you, it is a fact that: 'simple-polygonal tour in the embedding'='circuit in the embedding'. In this sense, I do not see what the use of 'simple-polygonal tour' is meant to achieve. $\endgroup$ – Peter Heinig Aug 20 '17 at 6:57at leastfor graphs which arein additionassumed to be vertex-$3$-connected (which I guess most of the graphs you care about are anyway), this maximization is trivial, in that by a theorem usually (yet perhaps quite ahistorically) attributed to H. Whitney, there is essentially onlyoneembedding, so the maximization is over a one-element set. Would you please clarify? $\endgroup$ – Peter Heinig Aug 20 '17 at 7:06