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# The Koch Snowflake

Extracts from this document...

Introduction

Constança Cassinelli

The Koch Snowflake – Portfolio SL Type I

Stage 0         Stage 1          Stage 2        Stage 3        Stage 4

1. Using an initial side length of 1, the table below was created to show the values  of Nn, ln, Pn, and An for n=0, 1, 2 and 3.
 n Nn Ln Pn An 0 3 1 3 0.433 1 12 1/3 4 0.577 2 48 1/9 5.33 0.642 3 192 1/27 7.11 0.670

It can be seen that each successive terms of values for Nn are multiplied by 4 : -

3x4=12,   12x4 =48,   48x4=192 , etc;

The relationship between each values of Ln is that each of these are multiplied by 1/3 to equal the successive value :-

1x1/3 = 1/3,    1/3 x 1/3 = 1/9,    1/9 x 1/3 = 1/27 , etc;

The perimeter was found by multiplying the

Middle

1. The graph for Number of sides supports the generalisation that to calculate the proceding value of number of sides, the preceding value must be multiplied by 4.

It can be deduced from the graph of Side length that 1/3n can be applied to all values of n to calculate the length of the sides of the fractal. Verifying it with the values of side length in the table :

n = 0 : 1/30 = 1

n = 1 : 1/31 = 1/3

n = 2 : 1/3² = 1/9

n = 3 : 1/3³ = 1/27

The perimeter is found using the formula Nn x ln and it is clear from the graph that as the values of n increase, the perimeter increases almost constantly.

Conclusion

 N Pn 7 22.475 8 29.966 9 39.955 10 53.273

As the value of n gets larger and larger, the perimeter keeps growing infinitely, e.g :-

The area however, keeps growing approaching a number but never exceeding it : -

 N An (for greater accuracy 4 d.p. was used) 7 0.6912 8 0.6917 9 0.6919 10 0.6920

Therefore as n -> ∞, the difference between the values of area decreases more and more.

Hence a general formula for area can be found by using the same equation used to find the areas, however adding all of the values up:

√3/4 x l²(1 + ∑nn=1 3x4n-1/9n )

The area is limited while the perimeter is infinite.

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