Let $\Phi$ be an irreducible root system in a Euclidean vector space $V$. Let $W$ denote its Weyl group. Choose a base $\Delta=\{\alpha_1,...,\alpha_r\}$ for $\Phi$. Then $\Delta$ is a basis for $V$. Let $$F_\Delta=\{\omega_1^\vee, ...,\omega_r^\vee\}\subset V^*$$ denote the dual basis. This is the set of fundamental coweights associated to $\Delta$. Now observe that for every $w\in W$, the set $w.\Delta$ is also a base for $\Phi$ with corresponding fundamental coweights $w.F_\Delta$. Moreover, $$ \bigcup_{w\in W} w.\Delta = \Phi. $$
Question: What does the set $\displaystyle X:=\bigcup_{w\in W} w.F_\Delta$ look like? E.g. how many elements does it have? Is there a direct way to obtain $X$ from $\Phi$ without choosing a base?
Added in edit: Note that each element $x\in X\subset V^*$ defines a hyperplane in $V$; namely, $\mathrm{ker}(x)$. I am also interested in the hyperplane arrangement defined by $X$. Is there anything known about this?
