Von Koch Curve
Watch the video below to see the first six stages of the infinite process for generating the Von Koch curve.
The original equilateral triangle has side $1$ unit. Work out the length of this curve in the first few stages and the length of the fractal curve formed when the process goes on indefinitely.
Now suppose you made a poster for your classroom with coloured paper by drawing an equilateral triangle for Stage 0 and then cutting smaller equilateral triangles and sticking them on the edge. What is the total area of all the triangles you would stick on one edge if you could continue the process indefinitely to make the Von Koch curve? So what is the area inside the Von Koch curve?
Find the dimension of the Von Koch curve using the formula $n=m^d$, where where $n$ is the number of self similar pieces in the generator and $m$ is the magnification factor.
See First Forward for a ten part series giving an introduction to Logo programming for beginners. Can you program the Von Koch curve?
Thank you Shaun from Nottingham High School and Andrei from Tudor Vianu National College, Bucharest, Romania for your solutions. Both Shaun and Andrei used essentially the same method.
To solve this problem I made a table with the first $3$ stages and then I determined analytical expressions for the quantities in the table as a function on $n$:
Stage | Number of sides | Length of curve | Area contained inside the curve |
$0$ | $3$ | $3$ | ${\sqrt 3\over 4}$ |
$1$ | $3\times 4=12$ | $3\times{4\over 3}$ | ${\sqrt 3\over 4}\left[1+3\left({1\over 9}\right)\right]$ |
$2$ | $3\times 4^2=48$ | $3\times\left({4\over 3}\right)^2$ | ${\sqrt 3\over 4}\left[1+3\left({1\over 9}\right)+12\left({1\over 9}\right)^2\right]$ |
Now I made the following observations:
- The length of each edge at the $(n+1)$th stage will be one third of the length of the edgeat the $n$th stage.
- The number of edges at the ($n+1)$th stage is $4$ times greater then the number of edges at the $n$th stage.
- From the last two observations the length of the curve is increased by a factor ${4\over 3}$ at each stage.
- The area of each additional small triangle included inside the curve at the $(n+1)$th stage will be one ninth of the area of the additional triangles included at the $n$th stage.
- After stage $2$ the extra area included within the curve at the $(n+1)$th stage is ${4\over 9}$ths of the extra area included at the $n$th stage.
Stage | Number of edges | Length | Area contained inside the curve |
$n$ | $3\times 4^n$ | $3\times\left({4\over 3}\right)^n$ |
${\sqrt 3\over 4}\left [1+3\left({1\over 9}\right) \left(1 +
\left({4\over 9}\right) + \left({4\over 9}\right)^2 + \ldots +
\left({4\over 9}\right)^{n-1} \right)\right]$
|
$n \to \infty$ | $\infty$ | $\infty$ | ${2\sqrt 3\over 5}$ |
To find the area contained inside the fractal curve we need the infinite sum of the geometric series: $$1 + \left({4\over 9}\right) + \left({4\over 9}\right)^2 + \left({4\over 9}\right)^3 + \ldots$$ which is $${1\over \left(1-{4\over 9}\right)} = {9\over 5}.$$ Hence the area inside the fractal curve is ${2\sqrt 3\over 5}$.
In order to calculate the dimension $d$ in the formula $n=m^d$, where $n$ is the number of self similar pieces and $m$ is the magnification factor, I see that for this problem $n =4$ and $m = 3$.
So, the dimension of the Von Koch Curve is $$d = {\log 4\over \log 3}= \log_3 4 \approx 1.262$$.
This is a Logo program to draw the Von Koch curve.
You can download a free copy of the MSW Logo software from http://www.softronix.com/logo.html
to allvonkoch
;draws 6 stages on same screen
cs pu fd 100 lt 90 fd 300 rt 90 pd
vonkoch2 200 0
pu rt 90 fd 250 lt 90 pd
vonkoch2 200 1
pu rt 90 fd 250 lt 90 pd
vonkoch2 200 2
pu bk 300 lt 90 fd 500 rt 90 pd
vonkoch2 200 3
pu rt 90 fd 250 lt 90 pd
vonkoch2 200 4
pu rt 90 fd 250 lt 90 pd
vonkoch2 200 5 end
to side :x :y
if :y=0 [fd :x stop]
side :x/3 :y-1
lt 60 side :x/3 :y-1
rt 120 side :x/3 :y-1
lt 60 side :x/3 :y-1
end
to vonkoch1
;superimpose 6 stages
cs pu bk 300 lt 90 fd 200 rt 90 pd
vonkoch2 500 0
vonkoch2 500 1
vonkoch2 500 2
vonkoch2 500 3
vonkoch2 500 4
vonkoch2 500 5
end
to vonkoch2 :x :y
;draws single curve size :x stage :y at current cursor position
repeat 3 [side :x :y rt 120]
end
to vonkoch3
;draws sixth stage
cs pu bk 300 lt 90 fd 200 rt 90 pd
vonkoch2 500 5
end