Block Fractals

We estimate the area of the limiting shape by finding the area of each stage in the construction, and finding the limit.

The initial stage consists of four cubes, each of side length 1.

Each cube has 6 faces, each of area 1.

Interior faces do not contribute to the surface area, and three pairs of cubes share two faces, making them interior.

So the area is

A0 = 4*6*1 - 6*1

The first stage of the construction consists of 16 cubes, each of side length 1/2.

Each cube has 6 faces, each of area 1/4.

This shape consists of four scaled copies of the initial shape, each with six interior faces.

In addition, the joining of these copies forms six interior faces.

So the area is

A1 = 4*(4*6*(1/4) - 6*(1/4)) - 6*(1/4)

Continuing in this way, the second stage of the construction consists of 64 cubes, each with side length 1/4.

Each cube has 6 faces, each of area (1/4)2.

So the area is

A2 = 4*(4*(4*6*(1/4)2 - 6*(1/4)2) - 6*(1/4)2) - 6*(1/4)2

= 6*43*(1/4)2 - 6*(1/4)2*(1 + 4 + 42)

Following this pattern, we see

An = 6*4n+1*(1/4)n - 6*(1/4)n*(1 + 4 + 42 + ... + 4n)

= 6*4n+1*(1/4)n - 6*(1/4)n*((1 - 4n+1)/(1 - 4))

= 6*4 + (6/3)(1/4)n - (6/3)(1/4)n4n+1

= 24 + 2*(1/4)n - 8

So An -> 16 as n -> infinity.

This is an upper bound for the area of the limiting shape. Unfortunately, we cannot conclude that the area of the limiting shape is 16, because of a well-known example.

Return to block fractal computations.