TITLE: Cone of movable curves
QUESTION [9 upvotes]: Let $X$ be a smooth complex projective variety of dimension $n$.
Under the duality between $N_1(X)$ and $N^1(X)$ we know that closure of cone of effective curves $\overline{NE}(X)$ is dual to closure of ample cone $\overline{Amp}(X)$.
It was proved in 2004 that the closure of cone of effective divisors $\overline{Eff}(X)$ is dual to the closure of cone of movable curves $\overline{Mov}(X)$. A movable curve by definition is a curve class $C \in N_1(X)$ such that $C=\pi_*(H_1 H_2 \cdots H_{n-1})$, where $\pi: X' \rightarrow X$ is a birational morphism and $H_i$'s are ample classes on $X'$. 
My question: Let $Q(X)$ be the cone obtained by curve classes $H_1 H_2 \cdots H_{n-1}$ where $H_i$ are ample divisors on $X$ itself. Is it true/false that $\overline{Q}(X)=\overline{Mov}(X)$? i.e. as long as I am interested only in the closure of these cones; do I really miss some curve class if I only restrict my self to intersection of ample classes on $X$ itself.
Can any body give an example where  $\overline{Q}(X) \neq \overline{Mov}(X)$?
Meanwhile, I am only interested in $n=3$ case.

REPLY [3 votes]: The definition you give for movable is actually the definition for strongly movable. A movable curve on X is one that lives in a family that covers a dense open subset of X, or the closure of the convex span of this concept. That the definitions are equivalent is a strong result.
I am mentioning this because I needed the other definition, and Google brought me here.
Cheers.