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.enter image description here 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.

  • $\begingroup$ Well done @CellSecret. Your question now merits more attention, and has improved a lot from its previous state. See, making sure that every question of yours is like this will be an extremely beneficial attitude to have on the site : it leads to greater attention, better focus on the question, and better answers. I am sure your next question will be better. $\endgroup$ – Teresa Lisbon May 7 at 5:24
  • $\begingroup$ @TeresaLisbon Thanks! I will try my best on every question. :) But please put your suggestions if my questions need to be improved. For this one, I just got randomly downvoted until someone said that I should revise it. $\endgroup$ – Interstigation May 7 at 5:32
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    $\begingroup$ Good point, @CellSecret. I admit I did not explain my downvote, and I'm sorry about that. The reason is for that is because I visit about 175 questions like this everyday, trying to help improve each of them. Sometimes I'm able to leave a comment, but sometimes I can only downvote, sometimes only close. I wish I could do more, but I have time constraints, and very few people do my job for me. So thanks very much for taking the message from the answer below and improving your question, and I promise I'll give you suggestions on your next questions if and when I see them. (+1 to the question) $\endgroup$ – Teresa Lisbon May 7 at 5:37
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    $\begingroup$ Thanks :) The process of writing the problem actually helped me into thinking it deeper too! $\endgroup$ – Interstigation May 7 at 5:41
  • $\begingroup$ Great. That would be one of the advantage points of having you think through your problem and frame it better. Have a good day! $\endgroup$ – Teresa Lisbon May 7 at 5:45

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 $|AB|=|AC|=|AD|=|AE|=|BC|=|CD|=|DE|=|EB|=2$.

enter image description here

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 $2(1+\sqrt2)\,r$.


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:

enter image description here

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    $\begingroup$ It would be super-duper wonderful if you consider reading the Enforcement of Quality Standards on meta.math.se. Have a wonderful day! $\endgroup$ – amWhy May 6 at 21:40
  • $\begingroup$ @g.kov I beg your pardon, you were right, I was wrong. I erase my erroneous remark. I thought it was an FCC "crystal structure" instead of a HCP structure... Besides [+1}, nice graphics. $\endgroup$ – Jean Marie May 7 at 12:51
  • $\begingroup$ @Jean Marie: Thanks, it's OK. The images were prepared using Asymptote. $\endgroup$ – g.kov May 7 at 13:19
  • $\begingroup$ Mine is with Matlab... less easy to manipulate. $\endgroup$ – Jean Marie May 7 at 13:20
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    $\begingroup$ I looked at these 14 balls in interactive 3d webgl from all possible angles, trying to figure out how it would be better to illustrate it. $\endgroup$ – g.kov May 7 at 16:34

enter image description here

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:


Remark: A well-written document about these issues.

  • $\begingroup$ So $2r(1+\sqrt{2})$ is just the case with $3$ layers? And we have to add $d$ for every layer added right? Therefore, $h(n)=(n-1)d+2r$? $\endgroup$ – Interstigation May 7 at 15:17
  • $\begingroup$ That's write. I forgot to say it. $\endgroup$ – Jean Marie May 7 at 15:18
  • $\begingroup$ This diagram is a bit hard to see than the one above. Still great explanation though! (+1) $\endgroup$ – Interstigation May 7 at 15:24
  • $\begingroup$ I have attempted to play on transparency (initial pyramid in grey)... but I should choose different colors. $\endgroup$ – Jean Marie May 7 at 15:31

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.


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