I think this should be called the *relatively ample cone* and its closure the *relatively nef cone*. Yes, this already exists and is sensible. This dissertation might be somewhat useful (although things are done for smooth $X$ instead of singular $X$). [UMich dissertation][1] **EDIT:** With regards to your two conditions (and the comments above), I think this condition is basically factoriality for the singularity. Suppose $x \in X$ is an isolated singularity as you describe and suppose that we are working with only one singular point. **Claim:** Suppose that $X$ is factorial, this means that every divisor through $x$ on $X$ is going to be linearly equivalent to a divisor not passing through $x$. *Proof of claim:* Suppose that $D > 0$ is an effective divisor passing through $x \in X$ (the effectivity assumption is harmless since we can write $D = D'-D''$ for effective $D'$ and $D''$). Choose a very ample and sufficiently ample Cartier divisor $A$ such that $\Gamma(X, O_X(A - D))$ is globally generated. Choose a global section $s$ generating the stalk $O_{X,x}(A-D)$ (this is possible since $D$ is Cartier at $x$, since $O_{X,x}$ is a UFD). Consider the inclusion $$\Gamma(X, O_X(A-D)) \subseteq \Gamma(X, O_X(A))$$ and so view $s$ as a global section of $O_X(A)$. It follows that $s$ determines a divisor $S \sim A$. Furthermore, $S$ and $D$ coincide in a neighborhood of $x \in X$ by construction. Further choose $T \sim A$ to be any divisor not passing through $x$ (this is possible since $A$ is very ample). Now then $D - S + T \sim D$, but $D - S + T$ is zero near $x \in X$. This proves the claim. It follows from the claim that there exists a set of generating divisors of the type you want. On the other hand, if you have divisors on $\hat{X}$ that satisfies those two conditions, then the images (pushforwards) of those divisors also generate the divisor class group of $X$. In particular, the divisor class group of $X$ is generated by divisors not passing through $x$. In particular, it is generated by Cartier divisors (since $x \in X$ was the only singular point). This implies that $X$ is factorial. [1]: http://deepblue.lib.umich.edu/bitstream/2027.42/60794/1/ofelguei_1.pdf