Thursday, 6 January 2011

Mathematical Problems 4: Close packed circles

While I was at university, I remember studying close-packed spheres and how they form in crystalline structures.  We were given the figure that close-packed spheres occupy the largest proportion of space available - more than they'd fill if they were arranged in squares, for example.

To explain what close-packed circles look like, here's a diagram.

Sometimes, they're described as hexagonally close-packed; as you can see from the diagram, the circles form hexagonal arrangements.  This makes it easier to calculate how much of the area the circles are filling... especially if we break the hexagon down into equilateral triangles...


Now, the triangle marked in bold contains three sixths of a circle - half the circle, in other words, which has an area of A = 0.5 x pi r2

The triangle is an equilateral triangle (all sides are equal, all angles are equal at 60 degrees).  I don't know any shortcuts for working out the area of an equilateral triangle, so I'll do it long hand:

Area = base x height x 0.5

The base = the diameter of a circle = 2 r
The height is found through trigonometry:  tan 60 = h / r   therefore h = r tan 60

Area = 2 r x r tan 60 x 0.5
Area of triangle = r2 x tan 60

Area of half a circle (contained within the triangle) = 0.5 x pi x r2

The proportion of the triangle's area which is covered by the semi circle = area semicircle / area triangle

Proportion = 0.5 x pi x r2  / r2 x tan 60  and the r2 cancel, so proportion = 0.5 pi / tan 60

pi / 2 =1.5707
tan 60 = 1.732

Proportion = 1.5707 / 1.732 = 0.9069 which is 91%.

Now I accept I haven't worked it out for packing by spheres in space, but I thought I'd start simple and work from there...  maybe next time!






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