Find the height of a square "pyramid" *General Form*

Question

This is the general shape of a pyramid stacked using squares. For example, the first layer would have 1 sphere, the second layer would have 4, and each layer would have $n^2$ spheres. I want to know a general formula of the height with $n$ layers expressed with $r$ as a function of $h_n(r)$.

I've made to $n=2$ so far, but I can't figure out the height beyond $2$ or even the general form.

My progress:

Obviously, we know that the height if it's 1 layer would be $2r$, or $h_1(r)=2r$.

It is also relatively easy if we have $2$ layers. We can construct a pyramid by connecting the $5$ centers. Using Pythagorean theorem, we can know that the middle portion is $\sqrt{2}r$. Therefore, the total height would be $h_2(r)=r(2+\sqrt{2})$

I start to struggle with $3$ layers. Please help me solving for that as well as a general form. Hopefully this version is more detailed and well-written question.

Answer 1

Using a model with $r=1$ for simplicity, let $A$ be the center of the upper circle and $B,C,D,E$ be the centers of the underlying four circles. These points represent a pyramid with $|AB|=|AC|=|AD|=|AE|=|BC|=|CD|=|DE|=|EB|=2$.

The base $BCDE$ of the pyramid is a $2\times2$ square, all the distances from $B,C,D,E$ to the center of the square are equal to $\sqrt2$, and obviously, the height $|AO|=\sqrt2$. Accounting for the radii of the balls on the top and the bottom, the height of two layers of spheres in such an arrangement must be $|AO|+2=2+\sqrt2$. Adding one layer below, we just need to add $\sqrt2$, so the total height of the pyramid of 14 balls is $2+2\sqrt2$, and for the balls of radius $r$ it would be $2(1+\sqrt2)\,r$.

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Edit

The total height of the pyramid of $n$ layers of balls with $n\times n$ balls on the bottom layer would then be

\begin{align} h_n(r) &= (2+\sqrt2 (n-1))\,r ,\\ h_1&=2 ,\\ h_2&=(2+\sqrt2)\,r ,\\ h_3&=(2+2\sqrt2)\,r ,\\ h_4&=(2+3\sqrt2)\,r ,\\ h_5&=(2+4\sqrt2)\,r ,\\ &\cdots \end{align}

This is a diagonal cross-section of the 14-balls pyramid:

Answer 2

Presumably the balls in each layer are arranged in a square lattice and neighbours are touching. If four balls are at the corners of a square, figure out how high a ball must be over the centre of the square so that the distance from the four balls is twice the radius.

Answer 3

Spheres stacked in this way can be considered as part of a HCP (Hexagonal Closed Packed) crystal structure. This kind of sphere packing has well known properties, in particular the fact that successive parallel planes passing through the centers of spheres of a given level are distant one from the other by distance

$$d=\sqrt{2}r \ \ \ \text{where r is the common radii of spheres}$$

(see this)

Therefore, adding an initial $r$ and a final $r$ for the bottom and for the top, the answer is:

$$2d+2r=2r\left(1+\sqrt{2}\right)$$

Remark: A well-written document about these issues.