Suppose I have a projective plane curve $C = V(F)$ defined over $\mathbb{C}$, where $F$ is some homogeneous polynomial in three variables. For the sake of simplicity, let's assume that $C$ is irreducible, and that $C$ contains a rational point $P$.

If $C$ is smooth of degree 2 (so genus $0$), then we can obtain a universal conformal covering map $\mathbb P^1(\mathbb C) \rightarrow C$ by identifying each point in $\mathbb P^1$ with a line $\ell$ through $P$ in the usual way, and outputting the other intersection point of $\ell$ with $C$.

If $C$ is smooth of degree 3 (so genus $1$), then we can obtain a universal conformal covering map $\mathbb C \rightarrow C$ by applying a change of variables to convert $C$ to Weierstrass form $$ F(x,y,z) = y^2 z - x^3 + g_2 xz^2 - g_3z^3, $$ and then using the Weierstrass $\wp$-function for the associated lattice: $$ z \longmapsto [\wp(z) : \wp'(z) : 1], $$ where we send the poles of $\wp$ to the limiting points at infinity.

If $C$ is singular of geometric genus $\le 1$, then we can resolve the singularities via a normalization in order to obtain a smooth curve, followed by parameterizing as above.

Consider now the case where $C$ has genus $g\ge 2$. We then obtain a Riemann surface of genus $g$ punctured at the singular points. By the uniformization theorem, there is a conformal map $\varphi_C:\mathbb H\rightarrow C$, where $\mathbb H$ is the upper half-plane, corresponding to the universal covering of $C$. However, I am not aware of any explicit formula for this map, even for a specific instance of $C$.

Is there any known way to explicitly parameterize these higher genus curves by a conformal map?