TITLE: Seeking concrete examples of "generic" elliptic fibrations of K3 surfaces
QUESTION [8 upvotes]: For me a K3 surface will be a smooth complex projective variety of dimension 2 that is simply-connected and has trivial canonical bundle.  Given a K3 surface $X$, an elliptic fibration $\pi \colon X \to \mathbb{C}P^1$ is a proper morphism with connected fibers such that all but finitely many fibers are smooth curves of genus 1.  I've learned a little about these from here:

Daniel Huybrechts, Lectures on K3 surfaces, Chapter 11: Elliptic K3 surfaces.

Generically a K3 surface admits no elliptic fibration, but among those that do, generically the fiber of $\pi$ is a smooth curve of genus 1 at all but 24 points, where the fiber is a rational curve with a single double point.
Huybrecht also catalogues the less generic cases where $\pi$ has fewer singular (i.e. non-smooth) fibers, but with correspondingly worse singularities.  On  MathOverflow there's a nice easy example: the Fermat quartic surface admits an elliptic fibration with 6 singular fibers, each of which has 4 double points.
But I'd like to see concrete easy examples of elliptic fibrations with 24 singular fibers.
By 'concrete easy examples' I mean that ideally I would like there to be simple explicit formulas for the K3 surface $X$, the hyper-Kähler structure on $X$, the elliptic fibration $\pi$, the 24 points on $\mathbb{C}\mathrm{P}^1$ with singular fibers, the double points on these fibers, and also the points of $X$ where $d\pi$ is not injective.   But of course I'll settle for whatever I can get!

REPLY [2 votes]: One can get examples by taking a double cover of the Segre-Hirzebruch surface $\mathbb F_4$. Write $C^-$ for the negative section and $C^+$ for a zero section and $F$ for a fiber of the projection $\mathbb F_4\to \mathbb P^1$, so that $C^+$ is linearly equivalent to $C^-+4F$. Set $L=2C^-+6F$ and choose $D$ in the linear system $|3C^+|$. Then there is a double cover $f\colon X \to \mathbb F_4$ branched on $C^-+D$ and such that $f_*\mathcal O_X=\mathcal O_{\mathbb F_4}\oplus L^{-1}$; if $D$ is smooth, the surface $X$ is also smooth. Standard formulae for double covers give $h^1(\mathcal O_Y)=0$ and $K_Y=0$, so $Y$ is a K3.
The map $X\to \mathbb F_4\to \mathbb P^1$ is an elliptic fibration.
The  singular points of the elliptic fibers lie above the points where $D$ is tangent to the ruling of $\mathbb F_4$. If the tangency is simple the singularity of the elliptic fiber is a node. A dimension count shows that for a general choice of $D$ all the tangency points of $D$ to a ruling are simple.  I think that using computer algebra it should not be hard to find explicit examples of $D$ where this condition holds.