Koch Snowflake Analysis
Classic Koch Snowflake and a Variation of the Koch Snowflake
Jarred Sareault
Introduction:
In this project, we need to find the area and perimeter of both the Classic and Variation Koch Snowflake for the first five levels. Also we need to create and implement general forms for the area and perimeter of the Classic/Variation Snowflakes to find the total area and perimeter of the final snowflake for each. For both the Classic and Variation Koch Snowflake, an equilateral triangle is used to start. We will be using basic arithmetic based on what we know about triangles to find the perimeter and area of the starting triangle. The information gathered here is crucial in our implementation of a sequence for finding the perimeter of the first five/final …show more content…
Using (3.0^(1/2)/4) as a constant, we multiply it by the Sides[k-1], which equals 3*(4)^k-1. 3*(4)^k-1 equals (3/4)*(4)^k and we multiply this by Length[k]^2. Length[k]^2 is (1/3)^k * (1/3)^k.
Therefore, the general formula of area for the Classic Koch Snowflake is:
Using the general formula for area, we can evaluate the area for the first five iterations of the Classic Koch Snowflake as being 0.5773502693, 0.6415002993, 0.6700114237, 0.6826830345, 0.6883148616 respectively. We can treat the area formula as a geometric series, where r = (4/9) and since (4/9)<1, the area for the final snowflake converges to 0.6928203232. The Variation Koch Snowflake starts with the same equilateral triangle as the Classic Koch snowflake. This means finding the area and the perimeter of the initial triangle is the same as the classic. However, by looking at each iteration after the initial equilateral triangle, it is clear that there will be discrepancies in either the general formula for the area or …show more content…
However, The Classic and Variation Koch Snowflakes differ in their general formula for area because the Variation Koch Snowflake is an alternating series. The area of the final snowflake for the Classic and the Variation both converge to a number, but the Variation Koch Snowflake is going to have a smaller final area because it loses area for every odd iteration. As shown in this project, it is important to realize how an alternating series can affect the area of fractal iterations but an alternating series for area has zero effect on the perimeter for each iteration of a fractal all the way to the final
Jarred Sareault
Introduction:
In this project, we need to find the area and perimeter of both the Classic and Variation Koch Snowflake for the first five levels. Also we need to create and implement general forms for the area and perimeter of the Classic/Variation Snowflakes to find the total area and perimeter of the final snowflake for each. For both the Classic and Variation Koch Snowflake, an equilateral triangle is used to start. We will be using basic arithmetic based on what we know about triangles to find the perimeter and area of the starting triangle. The information gathered here is crucial in our implementation of a sequence for finding the perimeter of the first five/final …show more content…
Using (3.0^(1/2)/4) as a constant, we multiply it by the Sides[k-1], which equals 3*(4)^k-1. 3*(4)^k-1 equals (3/4)*(4)^k and we multiply this by Length[k]^2. Length[k]^2 is (1/3)^k * (1/3)^k.
Therefore, the general formula of area for the Classic Koch Snowflake is:
Using the general formula for area, we can evaluate the area for the first five iterations of the Classic Koch Snowflake as being 0.5773502693, 0.6415002993, 0.6700114237, 0.6826830345, 0.6883148616 respectively. We can treat the area formula as a geometric series, where r = (4/9) and since (4/9)<1, the area for the final snowflake converges to 0.6928203232. The Variation Koch Snowflake starts with the same equilateral triangle as the Classic Koch snowflake. This means finding the area and the perimeter of the initial triangle is the same as the classic. However, by looking at each iteration after the initial equilateral triangle, it is clear that there will be discrepancies in either the general formula for the area or …show more content…
However, The Classic and Variation Koch Snowflakes differ in their general formula for area because the Variation Koch Snowflake is an alternating series. The area of the final snowflake for the Classic and the Variation both converge to a number, but the Variation Koch Snowflake is going to have a smaller final area because it loses area for every odd iteration. As shown in this project, it is important to realize how an alternating series can affect the area of fractal iterations but an alternating series for area has zero effect on the perimeter for each iteration of a fractal all the way to the final