A graph $G=(V,E)$ has thickness $2$ if $E$ can be written as a disjoint union $E=E_1\cup E_2$ so that $G_1:=(V,E_1),G_2:=(V,E_2)$ are planar graphs. For instance, $K_5$ has thickness $2$. It is known that to have thickness $2$, the graph must satisfy $v\geq 5$, and there is an easy upper bound to the number of edges: $$ e\leq 6v-12. $$

Note also that, if $v\geq 5$, $e=6v-12$ implies that $$ v\geq 11. $$

My question is: is the sharp upper bound on the number of edges for a graph with thickness $2$ and number of vertices $v$ known for every $v$? In particular, are there graphs $(V,E)$ with thickness two such that $e=6v-12$ for any finite set $v$ of size $v\geq 11$?

I thought about this question reading this post. Be aware though that in that post they use a different notion of thickness.

Example for $v=12$

An example of such a graph with thickness $2$ with $|V|=12$ is the following.

$$ V=\{N,1,2,3,4,5,1’,2’,3’,4’,5’,S\}, $$ $$ E=\{(N,i), (N,i’),(S,i),(S,i’),(i,j’)\,|\, 1\leq i,j\leq 5, \,i\neq j\}. $$ We note that $|E|=60=6|V|-12$, which is the theoretical upper bound for thickness $2$ graphs. We can decompose $E$ into $$ E=E_1\cup E_2, $$ $$ E_1=\{ (N,i)|1\leq i \leq 5\}\cup \{ (S,i’)|1\leq i \leq 5\}\cup \{ (i,j’)\,|\,1\leq i,j \leq 5,\,j-i=1\text{ or }2\mod 5\}, $$ $$ E_2=\{ (N,i’)|1\leq i \leq 5\}\cup \{ (S,i)|1\leq i \leq 5\}\cup \{ (i,j’)\,|\,1\leq i,j \leq 5,\,j-i=3\text{ or }4\mod 5\}. $$ Then, $G_1=(V,E_1),G_2=(V,E_2)$ are both planar graphs, as they councide with the vertex-edge scheme of an icosahedron. To see that, you can think of $N,S$ as the north and south vertices, while $A:=\{1,2,3,4,5\},B:=\{1’,2’,3’,4’,5’\}$ are the two sets of vertices with same latitude, except that in $G_1$, the set $A$ is in the north emisphere and $B$ is in the southern emisphere, and in $G_2$ it’s the opposite.

So, since $G_1$ and $G_2$ are planar, the graph has clearly thickness two.

The basic idea is: take a ball, draw $12$ vertices on it, and connect them with $30$ red edges to form an icosahedron (so that red edges do not intersect). Then, if you are careful enough, you can draw another $30$ blue edges, so that all edges (red and blue) connect different pairs of points, and such that blue edges do not cross each other.