# Determining a distance from a hexagonal close packing system

My friend and I have been feverishly arguing about something and I really want to know who's right. The question is as follows (bear with me, it's a little hard to define the problem)

An HCP structure, if you are unfamiliar, looks like this picture http://users-phys.au.dk/philip/pictures/solid_crystalstructures/closepack.gif. The first picture represents the base, then it builds in the $\hat{z}$ direction.

It is arranged in such a way that the volume of free space is minimized and the volume contained by the spheres is maximized. Now, the next part might not seem like it makes sense, but I don't know how else to phrase the question. Define regular unit vectors $\hat{x}, \hat{y}, \hat{z}$. Usually we wouldn't want to do this when considering this structure, since the middle of two rows of spheres are not solely in one of these three directions, however, hopefully towards the end this makes more sense.

Looking at the picture, place the origin in the center of the upper left hand sphere. Our coordinate system points in the usual direction, where $\hat{x}$ is pointing straight down on the page (so moving one unit looks like it goes from the center of the first sphere to directly between the two spheres below), $\hat{y}$ is pointing to the right (one unit taking you from the middle of the first sphere to the middle of the next sphere) and finally the $\hat{z}$ direction pointing out of the page (moving one unit would end in the same result as the $\hat{x}$, just in the $\hat{z}$ direction)

Now my question is, I want to find magnitude of the $\hat{x}$ component of the distance between one sphere on the bottom row and one sphere touching it a row directly above it (in the $\hat{z}$ direction).

Let me know if the question is not clear! I know that the total distance between the two spheres of interest is $2R$ where $R$ is the radius of each sphere, but I can't seem to determine the distance of interest. I hope someone can help because I'm on the side of thinking this is possible =p. Thanks =)

EDIT: From the answer to the problem, the $\hat{y}$ component of the distance I'm talking about is $\sqrt{\frac{2}{3}}R$. I'm guessing from rotational symmetry this is the same as the $\hat{x}$ component, but I'm still wondering how to get it.

A sphere in top row is directly above a point equidistant $P$ from and coplanar with centers of the three spheres touching it in the lower row. Point $P$'s distance from center of any of these three spheres is $2R/\sqrt{3}$. In your coordinate system, its x-component is $R/\sqrt{3}$.
• Hmm, this seems correct, but from the rotational symmetry of the problem I would think that the $\hat{y}$ component of the same distance should be equal to $\hat{x}$ component. In the answer to this problem, they say the distance for $\hat{y}$ is $\sqrt{\frac{2}{3}}R$ so I'm not sure this is correct... – Spaderdabomb Oct 9 '13 at 17:26