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Sphere packing problem: Pyramid design

Sphere packing problems are a maths problems which have been considered over many centuries – they concern the optimal way of packing spheres so that the wasted space is minimised.  You can achieve an average packing density of around 74% when you stack many spheres together, but today I want to explore the packing density of 4 spheres (pictured above) enclosed in a pyramid.

Considering 2 dimensions

First I’m going to consider the 2D cross section of the base 3 spheres.  Each sphere will have a radius of 1.  I will choose A so that it is at the origin.  Using some basic Pythagoras this will give the following coordinates:

Finding the centre

Next I will stack my single sphere on top of these 3, with the centre of this sphere directly in the middle.  Therefore I need to find the coordinate of D.  I can use the fact that ABC is an equilateral triangle and so:

3D coordinates

Next I can convert my 2D coordinates into 3D coordinates.  I define the centre of the 3 base circles to have 0 height, therefore I can add z coordinates of 0.  E will be the coordinate point with the same x and y coordinates as D, but with a height, a, which I don’t yet know:

In order to find I do a quick sketch, seen below:

Here I can see that I can find the length AD using trig, and then the height DE (which is my a value) using Pythagoras:

Drawing spheres

The general equation for spheres with centre coordinate (a,b,c) and radius 1 is:

Therefore the equation of my spheres are:

Plotting these on Geogebra gives:

Drawing a pyramid

Next I want to try to draw a pyramid such that it encloses the spheres.  This is quite difficult to do algebraically – so I’ll use some technology and a bit of trial and error.

First I look at creating a base for my pyramid.  I’ll try and construct an equilateral triangle which is a tangent to the spheres:

This gives me an equilateral triangle with lengths 5.54. I can then find the coordinate points of F,G,H and plot them in 3D.  I’ll choose point E so that it remains in the middle of the shape, and also has a height of 5.54 from the base. This gives the following:

As we can see, this pyramid does not enclose the spheres fully.  So, let’s try again, this time making the base a little bit larger than the 3 spheres:

This gives me an equilateral triangle with lengths 6.6.  Taking the height of the pyramid to also be 6.6 gives the following shape:

This time we can see that it fully encloses the spheres.  So, let’s find the density of this packing.  We have:

Therefore this gives:

and we also have:

Therefore the density of our packaging is:

Given our diagram this looks about right – we are only filling less than half of the available volume with our spheres.

Comparison with real data

We can see that this task has been attempted before using computational power – the table above shows the average density for a variety of 2D and 3D shapes.  The pyramid here was found to have a density of 46% – so our result of 44% looks pretty close to what we should be able to achieve.  We could tweak our measurements to see if we could improve this density.

So, a nice mixture of geometry, graphical software, and trial and error gives us a nice result.  You could explore the densities for other 2D and 3D shapes and see how close you get to the results in the table.

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Tetrahedral Numbers – Stacking Cannonballs

This is one of those deceptively simple topics which actually contains a lot of mathematics – and it involves how spheres can be stacked, and how they can be stacked most efficiently.  Starting off with the basics we can explore the sequence:

1, 4, 10, 20, 35, 56….

These are the total number of cannons in a stack as the stack gets higher.  From the diagram we can see that this sequence is in fact a sum of the triangular numbers:

S1 = 1

S2 1+3

S3 1+3+6

S4 1+3+6+10

So we can sum the first n triangular numbers to get the general term of the tetrahedral numbers. Now, the general term of the triangular numbers is 0.5n2 + 0.5n therefore we can think of tetrahedral numbers as the summation:

$\bf \sum_{k=1}^{n}0.5k+0.5k^2 = \sum_{k=1}^{n}0.5k+\sum_{k=1}^{n}0.5k^2$

But we have known results for the 2 summations on the right hand side:

$\bf \sum_{k=1}^{n}0.5k =\frac{n(n+1)}{4}$

and

$\bf \huge \sum_{k=1}^{n}0.5k^2 = \frac{n(n+1)(2n+1)}{12}$

and when we add these two together (with a bit of algebraic manipulation!) we get:

$\bf S_n= \frac{n(n+1)(n+2)}{6}$

This is the general formula for the total number of cannonballs in a stack n rows high. We can notice that this is also the same as the binomial coefficient:

$\bf S_n={n+2\choose3}$

Therefore we also can find the tetrahedral numbers in Pascals’ triangle (4th diagonal column above).

The classic maths puzzle (called the cannonball problem), which asks which tetrahedral number is also a square number was proved in 1878. It turns out there are only 3 possible answers. The first square number (1) is also a tetrahedral number, as is the second square number (4), as is the 140th square number (19,600).

We can also look at something called the generating function of the sequence. This is a polynomial whose coefficients give the sequence terms. In this case the generating function is:

$\bf \frac{x}{(x-1)^4} = x + 4x^2 + 10x^3 + 20x^4 ...$

Having looked at some of the basic ideas behind the maths of stacking spheres we can look at a much more complicated mathematical problem. This is called Kepler’s Conjecture – and was posed 400 years ago. Kepler was a 17th century mathematician who in 1611 conjectured that there was no way to pack spheres to make better use of the given space than the stack above. The spheres pictured above fill about 74% of the given space. This was thought to be intuitively true – but unproven. It was chosen by Hilbert in the 18th century as one of his famous 23 unsolved problems. Despite much mathematical efforts it was only finally proved in 1998.

If you like this post you might also like:

The Poincare Conjecture – the search for a solution to one of mathematics greatest problems.

IB Revision

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