A planar graph $G$ without $4$-cycles or $5$-cycles that $mad(G)\geq\frac{10}{3}$ The well-known Theorem about the girth nunber $g(G)$ and $mad(G)$ in planar graphs  that $mad(G)\leq\frac{2g(G)}{g(G)-2}$. If a planar graph $G$ with $g(G)=5$, then $mad(G)\geq\frac{10}{3}$. Is there an example that a planar graph without $4$-cycles or $5$-cycles $G$ with $mad(G)\geq\frac{10}{3}$ ? If it is fale, than  a planar graph without $4$-cycles or $5$-cycles $G$ is a subset of the set of planar graphs without $3$-cycles or $4$-cycles.

The best known bounds on the strong oriented chromatic number of planar graphs with girth $5, 6$ and $12$ are obtained via the maximum average degree. Therefore, to get bounds on the strong oriented chromatic number of planar graphs without cycles of lengths $4$ to $i, i \geq 4$, it is natural to determine the maximum average degree of these classes. The following two lemmas give tight bounds on the maximum average degree of planar graphs without cycles of lengths $4$ to $i$ for all $i \geq 4$.

$(1)$ If $G$ is a planar graph without cycles of length $4$, then $\operatorname{mad}(G)<\frac{30}{7}$.
$(2)$ For all $\epsilon >0$, there exists a planar graph $G$ without cycles of lengths $4$ to $i$ such that $\operatorname{mad}(G) >3+\frac{3}{i-2}-\epsilon$.
$(3)$ For all $i \geq 5$, if $G$ is a planar graph without cycles of lengths $4$ to $i$, then $\operatorname{mad}(G) < 3+ \frac{3}{i-2}$.
$(4)$ For all $i\geq 5$, for all $\epsilon >0$, there exists a planar graph $G$ without cycles of length $4$ to $i$ such that $\operatorname{mad}(G) >3+\frac{3}{i-2}-\epsilon$.
   Thus by $(3)$, we have, that every planar graph $G$ without cycles of length $4$ to $11$ has $\operatorname{mad}(G)<3+\frac{3}{11-2} =\frac{10}{3}$.   

EDIT: These cases have been discussed in the paper Strong oriented chromatic number of planar graphs without short cycles in page $4$. Hope it helps you in a small way.
