The Koch Snowflake
The Koch Snowflake
The snowflake model was created in 1904 by Helen von Koch. This snowflake appeared to be one of the earliest fractal curves. The fractal is built by starting with an equilateral triangle. One must remove the inner third of each side and replace it with another equilateral triangle. The process is repeated indefinitely. The length of each side is one which will help you determine the perimeter of each triangle. With having the perimeter of each triangle, the height can be determined so the area can be defined. The height must be determined because to use the formula A=12bh to find the area of the traingle, the height must be known. After repeating the process for the triangles, the graph below displays the number of sides (Nn) for each snowflake, the length of a single side (In), the length of the perimeter (Pn) and the area (An). In 0 1 2 3 1 1/3 1/9 1/27 Nn 3 12 48 192 Pn 3 4 16/3 64/9 An 3/4 3/4(1+13) 3/4(1+13+427)
3/4(1+13+427+16243)
The behavior of the graph above proves that each time a new snowflake is formed, the perimeter increases by 49. So, by simply multiplying 49to the area prior to, you will generate the area for the next sequence. When A4 occurs, these are the following results;
In 4 181
Nn 192
Pn 25627
An 3/4(1+13+427+16243+642187)
When A6 occurs, these are the following results; In 6 1729 Nn 3072 Pn 1024243 ) An 3/4(1+13+427+16243+642187+1024177147
The pattern that is emerged after experimenthing further into the snowflakes is An+1=An+(3*4n9n)A0 and n1. Theoretically the area gets too small to see so it is a limit is when N gets to 7. So, N could never reach with the area being noticable.
The snowflake model was created in 1904 by Helen von Koch. This snowflake appeared to be one of the earliest fractal curves. The fractal is built by starting with an equilateral triangle. One must remove the inner third of each side and replace it with another equilateral triangle. The process is repeated indefinitely. The length of each side is one which will help you determine the perimeter of each triangle. With having the perimeter of each triangle, the height can be determined so the area can be defined. The height must be determined because to use the formula A=12bh to find the area of the traingle, the height must be known. After repeating the process for the triangles, the graph below displays the number of sides (Nn) for each snowflake, the length of a single side (In), the length of the perimeter (Pn) and the area (An). In 0 1 2 3 1 1/3 1/9 1/27 Nn 3 12 48 192 Pn 3 4 16/3 64/9 An 3/4 3/4(1+13) 3/4(1+13+427)
3/4(1+13+427+16243)
The behavior of the graph above proves that each time a new snowflake is formed, the perimeter increases by 49. So, by simply multiplying 49to the area prior to, you will generate the area for the next sequence. When A4 occurs, these are the following results;
In 4 181
Nn 192
Pn 25627
An 3/4(1+13+427+16243+642187)
When A6 occurs, these are the following results; In 6 1729 Nn 3072 Pn 1024243 ) An 3/4(1+13+427+16243+642187+1024177147
The pattern that is emerged after experimenthing further into the snowflakes is An+1=An+(3*4n9n)A0 and n1. Theoretically the area gets too small to see so it is a limit is when N gets to 7. So, N could never reach with the area being noticable.