Perimeter of Koch Snowflake?

Question

What is the perimeter of a Koch Snowflake if the lengths are infinite? Also if a Koch Curve is in its second stage or itineration would the length be 1/9, and the third would be 1/27?

Answer 1

The "point" of the Koch snowflake, and fractals in general, is that the perimeter increases with each iteration. You know, the coastline of Britain gets longer and longer the more you pay attention to small variations.

Yes, if the first iteration is a triangle with each side length 1/3 (perimeter 1), then the second iteration has side length 1/9 and perimeter 12/9 ~ 1.33. The third iteration has side length 1/27 and perimeter 48/27 ~ 1.78. For good measure, the fourth iteration has 192 edges of length 1/81 each, perimeter 2.37

If the lengths are infinite, the perimeter is infinite. I'm guessing what you mean to ask is what happens as the number of ieterations goes to infinity. At each step, the number of edges gets multiplied by 4 and the length of each side is divided by 3. It's not hard to work out that the perimeter at step n is (4/3)^(n-1). As n increases, that goes to infinity.

Answer 2

"The Koch snowflake is an example of a shape with an infinite perimeter which bounds a finite area." (from the ref.)
Each stage increases the number of line segments by a factor of 4 and reduces the line segment length to 1/3 that of the previous stage. If the first stage is an equilateral triangle with 3 sides of length 1, the second stage has a total of 12 line segments of length 1/3 and the third stage has a total of 48 line segments of length 1/9.

Answer 3

Suppose your original straight line is of length L. Then at step 0, the perimeter is L. At step 1, you will have four straight lines. Each of these lines is 1/3 of the length of the line in step 0. Thus, the perimeter will be 4 * (1/3) * L = (4/3)L At step 2, you will have four times of the four straight lines in step 1. Each of this line is 1/3 of the length of the lines in step 1. Thus, the perimeter will be 4 * (1/3) * (4/3)L = (16/9)L This pattern will go on, so for step 3, the perimeter will be 4 * (1/3) * (16/9)L = (64/27)L, and for step 4, the perimeter will be 4 * (1/3) * (64/27)L = (256/81)L In short for step n, the perimeter will be (4/3)^n * L, where L is the length of your original straight line in step 0.