The Koch Curve and Visual Resolution

August 28, 2006 · 13 Comments

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.

KochCurveIteration0

KochCurveIteration1

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.

KochCurveIteration2

KochCurveIteration3

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.

KochCurveIteration4

KochCurveIteration5

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)

N_n = number of sides
l_n = length of each side
L_n = length of perimeter = N_n x l_n

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

N_n = 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,

$l_n = 1\cdot\left(\frac{1}{3}\right)^n = \frac{1}{3^n}, n=0,1,2,\cdots$

The perimeter after the n’th iteration becomes

$L_n = N_nl_n = 3 \cdot 4^n\cdot \frac{1}{3^n} = 3 \cdot (\frac{4}{3})^n$

which, in other words, means

$lim _{n \rightarrow \infty }[L_n] = lim _{n \rightarrow \infty } \left[ 3 \cdot (\frac{4}{3})^n \right] = \infty$

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

$\sqrt {s(s-l_n)(s-l_n)(s-l_n)} = \sqrt { \frac {3}{2}l_n \times \frac{1}{2}l_n \times \frac{1}{2}l_n \times \frac{1}{2}l_n } = \frac {\sqrt {3}}{4}(l_n)^2$

where l_n is the length of a side, with s = 1/2 x (l_n + l_n + l_n) = 3/2 x l_n.

Using the above formula, for Figure 1, the zeroth iteration of the Koch curve, we could write

$A_0 = \frac{\sqrt{3}}{4}$

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 l_n = 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,

$\frac{\sqrt{3}}{4}\left(\frac{1}{3^n}\right)^2 = \frac{\sqrt{3}}{4}\cdot\frac{1}{9^n}$

and so for the figure after n iterations

$A_n = A_{n-1} + 3 \times 4^{n-1} \times \frac {\sqrt {3}}{4} \times \frac {1}{9^n}$

$= A_{n-1} + A_0 \times \frac{1}{3} \times (\frac {4}{9})^{n-1}$

For clarity, if we write explicitly for the first few n values

$n = 1: A_1 = A_0 + A_0 \times \frac{1}{3}\left(\frac{4}{9}\right) ^0 = A_0\Big[1+\frac{1}{3}\left(\frac{4}{9}\right) ^0\Big]$

$n = 2: A_2 = A_1 + A_0 \times \frac{1}{3}\left(\frac{4}{9}\right) ^1 = A_0\Big[1+\frac{1}{3}\left(\frac{4}{9}\right) ^0 + \frac{1}{3}\left(\frac{4}{9}\right)\Big]$

$n = 3: A_3 = A_2 + A_0 \times \frac{1}{3}\left(\frac{4}{9}\right) ^2 = A_0\Big[1+\frac{1}{3}\left(\frac{4}{9}\right) ^0 + \frac{1}{3}\left(\frac{4}{9}\right) ^1 + \frac{1}{3}\left(\frac{4}{9}\right) ^2 \Big]$

Therefore, in general we could write

$\lim_{n\to\infty} A_n = A_0 \Big[1+\frac{1}{3}\left(\frac{4}{9}\right) ^0 + \frac{1}{3}\left(\frac{4}{9}\right) ^1 + \frac{1}{3}\left(\frac{4}{9}\right) ^2 + \cdots \Big]$

$= A_0 \Big[1 + \frac{1}{3} \Big( 1 + \frac{4}{9} + \left(\frac{4}{9}\right) ^2 + \cdots \Big) \Big]$

The expression inside the braces of the above equation can be seen to be in a geometric progression, which then could be simplified as

$\lim_{n\to\infty} A_n = A_0 \Big[1+\frac{1}{3} \cdot \frac{1}{1 - \frac {4}{9}} \Big] = \frac {8}{5} A_0$

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.

Categories: Maths
Tagged: extremal problem, fractal porous medium, fractals, koch curve, poruos medium, Science, von koch, von koch snowflake

13 responses so far ↓

Kaoslantida // July 31, 2010 at 7:48 pm | Reply

Sama krivulja nikada ne prelazi granicu te krunice. Beskona?no duga?ka krivulja zarobljena je kona?no duga?kom i jo k tome savrenom krivuljom: krunicom. Da se ?ovjek zamisli Reference Fractal math patterns Wolfram Mathworld Maths.org Nonscience.info Kaoslantida
Venkat // August 28, 2006 at 3:25 am | Reply

Looks like the definition of Porous media has been beaten to death, and I seem to be extremely late for the discussion.

Carrying on the discussion that we were having in class – I shall try and summarize what came out of it

Koch curve is an instance where there is a finite increase in area (~ volume) and an infinite increase in the perimeter (~ interface area) and we had agreed that there is a possibility of modifying the Koch curve to keep the interface area constant and an an increase in volume, or another posibility is to decrease the interface area and shrink it to zero (?) and increase the volume (?). These look like mathematical possibilites to me, and I have tried looking at pages on the net to prove/disprove this, but I have not been lucky.

On a slightly tangential note, it seems worthwhile to mention something that I was thinking about. It’s regarding the Jordan-curve theorem. For those who are unaware of this,

It states that “A simple closed curve divides the plane into two components – the one that is inside and the one that is outside.” And loosely, a simple closed curve is one of finite arc length which doesn’t cross itself.

I remember wondering in first semester when this was first taught to me if the finiteness of arc length is that important after all. Seems like Koch curve is an example of curve with an infinite arc length and it does have a region inside and outside. I am not sure if there is a more general version of the Jordan-curve theorem. I am not aware of it, but it looks like an extension of the Jordan curve theorem should be present. I am sorry for imposing a mathematical twist to the whole thing.
Selvakumar.A // August 28, 2006 at 12:25 pm | Reply

In the class it was explained that the koch curve is continuous ane “nowhere ” differentiable. I think this statement is true only as “n” tends to infiniy.
gaddeswarup // August 28, 2006 at 1:03 pm | Reply

A simple closed curve is just the homeomorphic image ( in this case continuous, one-to-one) image of a circle. It need not be rectifiable. Jordan-Brouwer Theorem is valid with that definition.
The proofs are still difficult and the modern ones use some algebraic invariants like fundamental group or homology ( I do not know Brouwer’s proof) and are based on Alexander’s arguments. See Munkres’s Topology or Armstrong’s “Basic Topology” both available in Indian editions.
Selvakumar.A // August 28, 2006 at 1:09 pm | Reply

Does it mean that while doing an analysis, We must take a resolution level corresponding to an area 1.6 times the actual value within the specified limits, (only) then the results will be comparable with experimental values ?
eganesan // August 28, 2006 at 5:59 pm | Reply

Added a link to Nonoscience.
scan man // August 29, 2006 at 11:10 pm | Reply

‘simple algebra’ indeed!!!

All I can say is that the last figure looks beautiful :)

Can’t say the same about the squiggles and the numbers in the equation though :(

If you do this in your spare time, I can’t imagine what your work is like!!
Arunn // August 30, 2006 at 10:12 am | Reply

Scan Man: My work is as enjoyable as these squiggles and associated thoughts…
eganesan: thanks for the link.
Swarup: Thanks. Very encouraging to have you around commenting at my blog with your expertise. You are welcome to contribute more at this blog, from your rich experience and knowledge. I am not a trained mathematician like you, but certainly would love to know more new things in that subject, from willing and generous sources like you.
Selvakumar: Yes, the nowhere differentiable tag is valid for the Koch curve only when n tends to infinity. Actually, only when n tends to infinity, it is a Koch curve and behaves as a fractal – exhibiting self similar structure.
To answer your other question, we need to know a concept called the representative elemental volume (REV) for a porous medium. We will do it in a future post.
Venkat: Thanks for your thoughts. Swarup has shared his thoughts on yours.
Kiranjeet // November 20, 2006 at 2:43 am | Reply

Thank you, this is by far the most helpful website I have found to help me understand the Koch snowflake.
Arunn // November 20, 2006 at 2:47 am | Reply

Kiranjeet: Good to know the post was helpful to you. Keep visiting and discussing.
Math Carnival at Nonoscience // February 24, 2007 at 10:28 am | Reply

[...] of mathematics has been published by Mark at Good Math Bad Math. A good collection for the weekend. The Koch Curve and Visual Resolution from this blog is also featured in the carnival.[...]
bharath // February 25, 2007 at 5:20 am | Reply

nice interesting topic.

To show area remains bounded it is sufficient to show the object at no point spills over the initial square drawn around it.

This can be shown using 2 observations:
1. Each edge at any step of the way is aligned in one of the 3 directions.
2. Replacing the middle by an equilateral triangle does not spill beyond the square for which the initial edge is the diagonal.

try it. the proof will need no computation.

And it is *not* necessary to have fractals to get curves of infinite length enclosing finite area. If you take a square of length 1 on each side. pick the top side and introduce simple nested Cauchy intervals. From right end each interval, drop down by 0.5, go horizontally below the left end of the next interval and jump up 0.5 to the left of the next interval. Continue this to form a closed curve.

The area will be bounded by 1 at all times and length will increase by 0.5(2)^k every step. which is faster increase in perimeter than Koch curve.
Arunn // February 26, 2007 at 8:22 am | Reply

bharath: thanks for the informative comments.

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