The cube blend has an edge cut of size $d$ separating half its vertices

For $d\ge2$, the polytope blendPoly d has a set $W$ of $2^d-1$ vertices (one of its two cube lobes) such that at most $d$ edges of the polytope leave $W$. Hence its edge expansion is at most $d/(2^d-1)$, exponentially small in $d$, although the polytope is simple with only $n=3d$ facets. (Balinski's theorem gives the matching lower bound $d$, so the expansion is exactly $d/(2^d-1)$.) Consequently no universal bound of the form $h(G(P))\ge 1/\mathrm{poly}(n+d)$ holds for graphs of bounded H-polytopes, and unweighted expansion cannot be the mechanism behind a polynomial diameter bound.

Proof idea. Same-sign vertices $v_\sigma(S),v_\sigma(T)$ are adjacent iff $|S\triangle T|=1$; the only edges between the two signs are the $d$ bridges $v_+(\{i\})v_-(\{i\})$.