# How do I calculate the distance between the center points of ellipses in a face centered cubic lattice?

I am currently writing a mathematical essay concerning the close packing of various 3D shapes. One example I am investigating is the shape of an oblate spheroid (ellipsoid) with a minor axis of 5 mm and a major axis of 15 mm. I am aiming to find the maximum packing density of such a spheroid in a face-centered cubic lattice. In order to calculate this, I have taken a look at a cubic section from such an infinite lattice. This cube contains half a spheroid on each face and an additional 8 eighths of a spheroid in each corner. It was fairly simple calculating the dimensions of the top and bottom surfaces of the cube, but the sides have been more of an issue.

To make it short: I need to find the distance between the center points of two ellipses represented by the red arrow in the following graph. The graph shows the cube from one of the side faces.

The graph can be infinitely stacked in all direction to produce a 2D representation of the 3D lattice I am investigating. It is not to scale, nor are the ratios graphed correctly, but the numbers are correct. Any help is greatly appreciated.

• The outer rectangle cannot be a square. – Aretino Dec 17 '17 at 18:42
• It is not meant to be. Sorry for the bad quality of the graph, I have not gotten around using graphing software yet. – Alex vW Dec 17 '17 at 19:43
• In that case you must add some other detail. In the sketch, the horizontal and vertical distances between the four outer ellipses seem to be equal: is that the case? – Aretino Dec 17 '17 at 20:16
• Imagine vertiaclly squashing the spherical fcc face slice by the ratio of your ellipse axis. Isn't the center to center distance you want just the half diagonal of the cell face rectangle? and the rectangle sides are in the same ratio as the ellipse major/minor axis? – f5r5e5d Dec 17 '17 at 21:03
• @Aretino That is another problem with the graph. The horizontal distance between the ellipses is actually smaller than the vertical one. The inaccuracy of the graph is starting to annoy me quite a bit, but as of now I am still looking for a simple easy to use graphing software. Sketching this problem would take quite some time though I would think. Sorry for the inconvenience. – Alex vW Dec 17 '17 at 21:35

Here's a picture of what happens. In any case, the distance between centers is half the rectangle diagonal. But the rectangle horizontal side $X$ and vertical side $Y$ must satisfy the following equation: $$Y=10\sqrt{1-{X^2\over900}}.$$
• The first position should be the one with the two upper ellipses touching. In that case the rectangle width $X$ is the same as the major axis of the central ellipse: inserting $X=15$ into the above equation one gets $$Y=10\sqrt{1-{15^2\over900}}=5\sqrt3.$$. I made the animation with GeoGebra, a great (and free) piece of software. – Aretino Dec 18 '17 at 15:29
• $X$ and $Y$ are simply four times the coordinates $(x,y)$ of a tangency point, which in turn obey the ellipse equation $${x^2\over7.5^2}+{y^2\over2.5^2}=1.$$ Hence: $${X^2\over30^2}+{Y^2\over10^2}=1.$$ – Aretino Dec 18 '17 at 16:29