The Koch Curve or the von Koch snowflake was discovered by Helge von Koch (1870-1924) in 1904. It is a closed fractal curve of infinite length within a finite region of space, enclosing a finite area. The construction of such a curve and some of its unique properties is explained in this post, with a connection to porous media.
Take an equilateral triangle as shown below in Figure 1, with sides of, say, 1 unit each. This is the “zeroth” iteration of the Koch curve.


Figure 1: Iteration zero (and) Figure 2: Iteration one
For the first iteration we split each side of the equilateral triangle into 3 equal parts, with the middle 1/3rd being replaced essentially by another smaller equilateral triangle made of side of 1/3 unit. The resulting figure is shown below.
What we have done in the first iteration is to replace each side made of 1 unit stick by four, 1/3 unit sticks, arranged such that their end points remain fixed between the original 1 unit length of iteration zero. This procedure automatically fixes the angles of the protruding equilateral triangle in Figure 2, and thus increases the original area confined by the equilateral triangle of Figure 1, to that of Figure 2.
By following the previous procedure, in the second iteration, we could now generate Figure 3 out of Figure 2, with each of the sides of Figure 2 now being replaced in their middle thirds with even smaller equilateral triangles of sides made of 1/9 unit sticks.


Figure 3: Iteration two (and) Figure 4: Iteration three
Results of iteration three, four and five are shown in Figure 4, 5 and 6 respectively.


Figure 5: Iteration four (and) Figure 6: Iteration five
The procedure could be repeated many times, to reach the von Koch Snowflake, when the iteration tends to infinity. There is an applet available at [http://www.efg2.com/lab], for you to try out more iterations and other curves, starting from other basic geometrical figures.
Intuitively we could see that this resulting figure would have “infinite” length for its sides (if we could possibly draw it) while enclosing a “finite area” as shown in the above figures. How much would be the area when we actually do the infinite iterations?
To answer this let us put this procedure in a mathematical footing. It only involves simple algebra.
The derivation for calculating the enclosed area done here follows the explanation given in When Least is Best by Paul J. Nahin. Similar explanations are available in the Wikipedia page for the Koch Curve.
Let after n iterations (n > or = 0)
- Nn = number of sides
- ln = length of each side
- Ln = length of perimeter = Nn x ln
For instance, if l0 = 1, N0 = 3, then L0 = 3. Since we increase in every iteration the number of sides by a factor of four (compare Figure 1 and Figure 2) and since N0 = 3, then it means
Nn = 3 x 4n, where n = 0, 1, 2, ……
Further, for each iteration, the length of a side decreases by a factor of 3. for instance we used a stick of length unit 1 to form the side of Figure 1, while in Figure 2 each side of Figure 1 is replaced by 4 sticks of length unit 1/3 of the original. Since l0 = 1, we could write then,
The perimeter after the n’th iteration becomes
which, in other words, means
So, the length of this closed curve does actually goes to infinity. Now, for the area it encloses.
We know from our high school trigonometry that the area of a triangle can be calculated using the Heron’s formula
where ln is the length of a side, with s = 1/2 x (ln + ln + ln) = 3/2 x ln.
Using the above formula, for Figure 1, the zeroth iteration of the Koch curve, we could write
By comparing Figure 1 and Figure 2 we could see that we are increasing the area by adding 3 x 4 n-1 equilateral triangles (a triangle for each side) with ln = 1/3n. For example, with n = 1, we increase from A0 to A1, by adding three equilateral triangles, each with l1 = 1/3.The area of each of these triangles is obviously,
and so for the figure after n iterations
For clarity, if we write explicitly for the first few n values
Therefore, in general we could write
The expression inside the braces of the above equation can be seen to be in a geometric progression, which then could be simplified as
So, the area enclosed by the closed curve of infinite length is actually only 60 percent more than that of the original area of the equilateral triangle we started in Figure 1. A remarkable property indeed.
Now to give all of this a porous medium perspective, we could think of each successive figure from 1 through 6 to be of finer and finer “resolution level”, depicting however, the same porous medium. The ‘gray color region’ inside of each figure being one constituent and the ‘outside’ of this figure being the other constituent (of course, bounded by a outer boundary, which is not seen in these pictures), with the Koch curve forming the “identifiable interface” between these two constituents.
The visual resolution going from Figure 1 to 2, the next level of the Koch curve ‘porous medium’, is 3. Because, each original length of the earlier iteration is divided into four parts of each 1/3 of the original length. So, indirectly, the visual resolution is represented by the interface length in a porous medium.
We could see now why visual resolution is important to define a porous medium, as we kept saying the previous posts on this topic. Even the interface length is increasing in the above example of a porous medium, if we resolve things finer and finer, an effect which is bound to influence interface phenomena, like heat transfer across the constituents of a porous medium.




