# If you accumulate 100 identical spheres how many will end up on the surface?

So far I have asked this question in physics, engineering and math sections but have not gotten an answer in three years. If you have 100 identical spheres and group them into A sphere shape, how many will end up on the surface? They can be packed tight or loose in a hexagon pattern or cubicle pattern. They can be symmetrically or evenly spaced. I realize the answer can be off by one or two depending on how you group them, But in general it seems like something that can be answered. Is there a formula that could be applied to any number of spheres? For example what if you have 120 identical spheres, how many will be on the surface then? Imagine an atom with 120 nuclei in the nucleus. Thanks for your help.

EDIT:After seeing some of the responses it reminded me of my thought process a few years ago when I began to think about this. How do you define surface, what type of packing, etc. etc. I too tried the plastic bags and ended up using different colored magnetic balls. Experimenting has worked just fine and I asked the question because I truly thought a simple formula would be available. As for what the surface is I have found another question that actually comes closer to what I'm interested in. No matter how many spheres are accumulated into a spherical shape you can separate them into two categories. Spheres in the (interior) and spheres on the (surface or touching a sphere on the surface). This seems confusing at first but it actually simplifies the surface question by literally filling the gaps. So really the question is how many spheres are not on the surface or touch a sphere on the Surface? Even still it can be difficult counting spheres when you transfer between hexagonal, cubical or symmetrically spherical shaped. That is why I have hoped for quicker solution.

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 24, 2020 at 5:09
• Colleagues, it would be a shame if this question gets closed. Sep 25, 2020 at 21:24
• Too short for an answer, but this might be helpful : en.wikipedia.org/wiki/Waterman_polyhedron Sep 29, 2020 at 21:43

### Experimental approach

Here are 100 marbles in a polyethylene bag: If a marble touches the bag, it is counted as being on the surface. I did not count, and the bag was not perfectly round. A balloon might be better.

### Theoretical approach

This works for large $$N,$$ and the result is as Ivan predicted (from scaling arguments in 2D and 3D, I would imagine). We neglect the packing defects in the bulk, and treat the surface as having no curvature and no defects. The maximal packing density of spheres is well-known by chemists,

$$P_\mathrm{3D} = \frac{π \cdot \sqrt{2}}{6} = 0.74046\ldots\tag{1}$$

The maximal packing density of circles is less well-known on this site, but you can look it up on Wikipedia:

$$P_\mathrm{2D} = \frac{π \cdot \sqrt{3}}{6} = 0.90687\ldots\tag{2}$$

If you have $$N$$ marbles with radius $$r$$, you can estimate the radius $$R$$ of the enclosing sphere, get its area ($$4 \pi R^2$$), and calculate how many circles (of area $$\pi r^2$$) fit in that area. The resulting expression scales with $$N$$ to the power of two-thirds, as Ivan said:

$$N_\mathrm{surface} = 4 \cdot P_{\mathrm{2D}} \cdot \left(\frac{N}{P_{\mathrm{3D}}}\right)^\frac{2}{3}\tag{3}$$

For example, for a million spheres, about four percent are on the surface. For a single sphere, all are on the surface. The expression for $$N = 1$$ yields $$4.4$$, pretty bad. For $$N = 100,$$ it yields $$95.2$$ That is a not a bad estimate; from the experimental approach, I see more than $$30$$ marbles on the surface, and the picture only shows one side.

### Approximating a sphere with a cube

If you have 125 spheres that pack in a primitive cubic fashion, you would have a cube with 5 spheres in each direction. The top and bottom, right and left, front and back layer would be on the surface, leaving a 3 by 3 by 3 cube in the center. So 27 spheres would be on the inside and the remaining 98 would be on the surface. When we take those same 125 and pack them tighter and such that they fit in a minimal sphere, the surface area would get smaller, and the packing on the surface would get denser, leading to a similar fraction of spheres on the surface.

• @Ivan Thanks for 80% of the answer.
– Karsten
Sep 25, 2020 at 2:24
• @Poutnik The bag is better than glue because it also helps to define which spheres are on the surface. Also, you can vary the number of spheres more easily, and adjust the packing.
– Karsten
Sep 25, 2020 at 2:26
• The plastic bag seems to work fine. Last night I was using plastic wrap but the bag seems to work better. Thanks everyone for your input, this is very helpful. Sep 25, 2020 at 2:45
• @Karsten Theis I did mention the bag/wrapping as well. I mentioned the glue to experimentally evaluate the particular sphere packing by counting, what is not available by bag wrapping. Sep 25, 2020 at 8:48
• If you multiply the top and bottom of your $P_{3D}$ formula by $\sqrt{2}$ then it will look more like the $P_{2D}$ one. Sep 26, 2020 at 8:12

Note that the classical idea of an atomic nucleus as a set of nucleon balls is very wrong, similarly as another classical idea of an atom as a planetary system of orbiting and spinning electron balls.

I suppose you have had bad luck until now as this task is in opinion of my non-math brain resistant to both general theoretical ( geometry, topology ) and experimental analysis.

I doubt there is a general formula ( or very complex at the best ), aside of the asymptotic formula for the group radius $$R$$ much bigger then the sphere radius $$r$$, wher $$N_\mathrm{surf} = a \cdot N^{2/3}$$.

The problem for small N as 100 and the related effective diameter is, there is collision between maintaining the given sphere arrangement and following the spherical shape.

Are there supposed centripetal forces corrupting sphere arrangement in favour of more spherical shape ?

If there is given the space filling factor $$f$$ ( a fraction of volume filed by spheres) then for $$R \gg r$$ the sphere volume

$$V \simeq \frac {4N}{3f} \cdot \pi \cdot r^3 \tag{1}$$

and

$$R=r \cdot {\left(\frac Nf\right)}^{1/3} \tag{2}$$

The outer layer of thickness $$2r$$ has volume $$V_\mathrm{s} = \frac 43 \cdot \pi \cdot \left( R^3 - \left(R-2r\right)^3 \right)= \frac 43 \cdot \pi \cdot \left( 3R^2.2r - 3R.4r^2 + 8r^3 \right) \\= \frac 43 \cdot \pi \cdot r^3 \left( 6{(\frac Rr)}^2 - 12 \frac Rr + 8 \right) \\= \frac 43 \cdot \pi \cdot r^3 \left( 6 \cdot {(\frac Nf)}^{2/3} - 12 \cdot {(\frac Nf)}^{1/3} + 8 \right)\tag{3}$$

The volume of spheres in this layer is approximately $$V_\mathrm{s1} \simeq V_\mathrm{s} \cdot f = \frac 43 \cdot \pi \cdot r^3 \cdot f \cdot \left( 6 \cdot {(\frac Nf)}^{2/3} - 12 \cdot {(\frac Nf)}^{1/3} + 8 \right) \tag{4}$$

The number of the surface spheres is then approximately

$$N_\mathrm{surf} \simeq \frac {V_\mathrm{s1}}{V_\mathrm{1 sphere}}= \frac {\frac 43 \cdot \pi \cdot r^3 \cdot f \cdot \left( 6 \cdot {(\frac Nf)}^{2/3} - 12 \cdot {(\frac Nf)}^{1/3} + 8 \right) }{\frac 43 \cdot \pi \cdot r^3} \\ =f \cdot \left( 6 \cdot {(\frac Nf)}^{2/3} - 12 \cdot {(\frac Nf)}^{1/3} + 8 \right) \tag{5}$$

For a very big N, we receive the previously posted:

$$N_\mathrm{surf} \simeq 6 \cdot f^{1/3} \cdot N^{2/3} \tag{5a}$$

Note that this asymptotic solution for big N considers surface spheres as those touched by the thought wrapper foil around the object.

For the tightest arrangent is $$f \simeq 0.74$$, then

$$N_\mathrm{surf} \simeq 0.74 \cdot \left( 6 \cdot {(\frac {100}{0.74})}^{2/3} - 12 \cdot {(\frac {100}{0.74})}^{1/3} + 8 \right) \simeq 77 \tag{6}$$

The function $$N_\mathrm{surf} = f(N,f)$$ is not much sensitive to the space filling factor for small N. $$N_\mathrm{surf} = f(100,0.6) \simeq 74$$.

If we consider instead of that a loose rectangular arrangement as $$\ce{NaCl(s)}$$ crystal structure $$5 \cdot 5 \cdot 4=100$$, then the surface sphere number would be $$100 - 3 \cdot 3 \cdot 2=82$$.

For more information, see the table below.

$$\begin{array}{r|rrrr} \hline &f&&& \\ N&0.74&0.7&0.65&0.6 \\ \hline 10&10&10&10&10 \\ 20&19&19&19&19 \\ 50&43&43&43&42 \\ 100&77&76&75&74 \\ 200&134&132&130&128 \\ 500&270&266&261&256 \\ 1000&450&444&435&425 \\ 2000&744&732&717&701 \\ 5000&1425&1402&1371&1339 \\ 10000&2313&2275&2224&2170 \\ 20000&3738&3674&3590&3502 \\ 50000&7010&6887&6727&6559 \\ 100000&11242&11044&10785&10511 \\ 200000&17992&17672&17254&16813 \\ 500000&33415&32815&32032&31207 \\ 1000000&53294&52334&51079&49757 \\ \hline \end{array}$$

• As posed I really don't think you can get any better than the dimensional argument. To go any further the packing has to be much better described, for instance in terms of mathematical restrictions on the form the radial distribution function can take, and what constitutes "being on the surface" needs to be defined. For instance consider 1m balls and a 1 light year sphere. Is just 4 as a tetrahedron on the surface OK, with the rest tightly packed in the centre? Or can all 100 be on the 1 light year surface? From the vague description I don't think so, but I'm not sure. Sep 23, 2020 at 11:53
• @Ian Bush As surface I understand spheres touched by a foil if wrapped by it. Sep 23, 2020 at 15:41
• As surface I understand spheres touching other spheres for a ill defined "down, or below"... Sep 24, 2020 at 16:40
• The whole task is ill defined and rather self-purposed. I guess for big N I can assume the overall convex arrangement. Sep 24, 2020 at 16:45
• I tried evaluating $N_\mathrm{surf} \simeq \frac {V_\mathrm{s1}}{V_\mathrm{1 sphere}}$ but didn't come up with your result. What is the expression for $V_\mathrm{1 sphere}$? Sep 25, 2020 at 12:44

Here I used a numerical approach roughly equivalent to the analytic approach in Poutnik's answer. The procedure is to generate the smallest enclosing sphere containing within its radius 100 hexagonally close-packed beads, and then determine the number of beads in a second smaller concentric enclosing sphere. If r=1 is the radius of a bead, R=5.074 is the radius of the sphere enclosing 100 close-packed beads. Then a concentric enclosing sphere that is smaller by r/2 contains 69 beads, one that is smaller by r contains 52 beads, and one that is smaller by 2r contains 24 beads. Note that in order to place exactly 100 beads into a perfect sphere, the arrangement of beads is not symmetrical.

There are alternate (and better) ways of distinguishing between interior and surface beads, for instance to count the number of contacts. In an hcp arrangement this is a maximum of 12, with surface beads having less. Using such counting reveals that only 25 beads are not on the surface and make the max possible contacts. Like the previous estimation method, contact counting one does not check whether there are gaps etc. These clearly exist as shown in the following (interior beads are colored blue, surface ones black): Since N=100 is not very big, some of the approximations used in the analytic approaches proposed in other answers are bound to fail. Consider my answer: the number of interior beads predicted by a numerical method is nearly the same as the number of surface beads according to the large N approximation. In fact the number of surface beads is more nearly that expected for a cube (see Poutnik's answer) rather than an ideal (large N) sphere. This problem was made to be solved by experiment (numerical or otherwise).