in Figure \[fig:I\].

\[I&gt;1\] If $\overrightarrow{D} \in \mathcal{I}$, then $obt(\overrightarrow{D}) > 1$.

Assume there exists a $1$-page oriented book embedding of $\overrightarrow{D}$. Since $\overrightarrow{D}$ contains dicycle, by Lemma \[directed cycle\], each vertex of $\overrightarrow{D}_n$ is tight in the embedding, except the endpoints of the loose arc. However $\phi(i)$ and $\phi(j)$ are not adjacent, thus at least one of $\{\phi(i),\phi(j)\}$, say $\phi(i)$, is covered by the loose arc of $\overrightarrow{D_n}$; let $u_i$ be the other endpoint of the $a_i$. Since $u_i$ must be located in the spine above all vertices of the dicycle, or below all vertices of the dicycle, $a_i$ crosses the loose arc of the dicycle, a contradiction to the planarity rule.

![Members of $\mathcal{I}$ (based on a $4$-dicycle)[]{data-label="fig:I"}](I_small.eps)

For members of the class $\mathcal{I}$, the length of the directed cycle is required to be at least four. Therefore, we define the class $\mathcal{T}$ to be the class of all oriented graphs $\overrightarrow{D}$ which contain a $3$-dicycle and such that there exist three arcs $a_1,a_2,a_3 \in A(\overrightarrow{D})$ having one endpoint of degree one and other endpoint $\phi(1), \phi(2), \phi(3)$, respectively, in $\overrightarrow{D_n}$, with $\overrightarrow{D} = \overrightarrow{D_n} \cup \{a_1,a_2,a_3 \}$. As with $\mathcal{I}$, the direction of each arc not contained 