# The sparsest planar net that captures every unit segment

Let $\cal C = \lbrace C_i \rbrace$ be a collection of rectifiable curves in the plane with the property that every unit-length segment meets at least one curve in at least one point. Call such a collection $\cal C$ a needle net: any unit-length "needle" is captured by the net.

I would like to find the sparsest needle net, sparse in the sense that the curves have minimum length per unit area. That is, the limit of $L/A$ of the ratio of the length $L$ of the curves within a region to that region's area $A$, as the region grows large, is as small as possible.

For example, a regular grid of orthogonal parallel lines separated by $\sqrt{2}/2$ is a needle net: the diagonal of each square cell of the grid has length $1$. If I've calculated correctly, the length of its curves (lines) within each unit area region $L/A$ is $2 \sqrt{2}$. See left below, where a unit-length diagonal is highlighted in red, and the region of the plane I used to compute $L/A$ is marked. Again if I've calculated correctly, the equilateral-triangle tiling of the plane obtained from three sets of parallel lines is less efficient, and the packing arrangement of unit-diameter circles shown right above is less efficient still.

Is the square-grid the sparsest needle net? This feels like a question that has been addressed before, perhaps in another guise. If so, a pointer would be welcomed. Thanks!

Update. Roland Bacher's more efficient needle net: Is this the optimal net?

• I've given up trying to work out what it gives you, but you can move the circles in the right hand figure further apart for a small saving. It doesn't look like it will beat the square lattice though. – Ben Barber Nov 16 '12 at 14:30

Without error of my part, a paving with regular hexagons with sides of length $1/2$ gives $L/A=4\sqrt{3}/3\sim 2.3094$. This could very well be the optimal candidate.