Let $X$ be a topological space. A good cover on $X$ is an open cover such that all finite non-empty intersections are contractible. It is a theorem of Hironaka that (complex) algebraic sets admit triangulations. As a consequence, all algebraic sets admit a good cover - simply take as an open cover the collection of open stars of the vertices in the triangulation. In case $X$ is an algebraic curve with nodes, there is a way of constructing an open cover geometrically. First pass to the resolution of singularities $\pi:\widetilde{X}\rightarrow X$ - this is a compact Riemann surface obtained by blowing up the singular points of $X$. The singular locus of $X$ is a finite collection of nodes, $S$. The set $E = \pi^{-1}(S)$ is finite, and $\pi|_{\widetilde{X}\backslash E}: \widetilde{X}\backslash E\rightarrow X\backslash S$ is a biholomorphism. Note that a node is an example of a normal crossings singularity - one which is locally isomorphic to a finite union of transverse hyperplanes. Since $\widetilde{X}$ is, in particular, a smooth (paracompact) manifold, it admits a good cover by differential geometry (choose a metric on $\widetilde{X}$, then cover it by sufficiently geodesically convex balls). We can construct a cover by convex balls - call it $(U_{\alpha})$ - in which each point of $E$ is contained in exactly one $U_{\alpha}$. The sets $\pi(U_{\alpha})$ then form a good open cover of $X$. Away from $S$, the $\pi(U_{\alpha})$ are homeomorphic to open discs $\mathbb{D}\subset \mathbb{C}$. The $\pi(U_{\alpha})$ which contain a singular point are homeomorphic to $\mathbb{D}\vee \mathbb{D}$, where the two discs are joined at the origin. My question is: >Does this construction generalize to higher dimensional varieties with normal crossings singularities? More precisely, does there exist a good (?) open cover $(U_{\alpha})$ of $\widetilde{X}$ such that $\pi(U_{\alpha})$ is a good open cover of $X$? If so, is there an analogously simple description of the open sets $\pi(U_{\alpha})$ which happen to contain singular points?