Density of particles in hexagonal lattice I need to calculate, in a 2D hexagonal lattice of point particles in which the nearest neighbours are a distance apart $a$, what's the density of particles. What I really need is, if $\rho$ is the number of particles per squared centimeter, what's the number $\rho a^2$.
I've tried using closepacking densities but this obviously doesn't give the correct result.
I don't know how to calculate it and hints (or solutions) would be appreciated.
 A: I think your result is right but your procedure of calculation maybe wrong.
Following your image：

You may find that the area of this parallelogram is $V=(N-1)^2a^2cos30$ because the length of each edge should be $(N-1)a$.
Then we should consider that some points on the edges are shared by multiple parallelograms. For example, the red points are shared by 2 parallelograms and can only be considered as half points. Only the gray points completely inside the parallelogram can be viewed as full points.
Finally, we will have:
point number = $(N-2)^2$ (full points) + $4*1/2*(N-2)$ (half points) + $1$ (all corner points), which equals to $N^2-4N+4+2N-4+1=(N-1)^2$.
Since $V=(N-1)^2a^2cos30$, the $\rho$ will still equals to $2/(\sqrt3*a^2)$.
A: Ok, finally solved it in a very simple geometrical way.
IF we take a square slanted lattice in the hexagonal lattice, like in the image, which is $N$ particles along each side. Then the number of particles inside is $N^2$. The volume of that area is just $V=(Na)^2\cos(30)$, and so:
$$\rho a^2=\frac{2}{\sqrt 3}$$.
I'm sorry for asking, I was frustrating myself with this.

