I technically did not have to render the big center snowflake because it will be ‘drawn’ by the edges of the surrounding snowflake. So I start out by finding the center the big middle snowflake, and surround it with 6 of the smaller ones. If you know the two sizes of Koch snowflakes, all you need to do is place them in the right places. I think I may have just been reluctant to create a JSTOR account to read it, I’m not sure.īut anyway, the construction methods described are for if you are using solid shapes, but I really wanted to draw this tiling as paths to achieve the same effect. I honestly don’t know why I didn’t see this the first time, this would have made my job easier. …using the two methods of construction shown above we can tile the plane with two sizes of tile, their diameters being in ratio sqrt(3):1 It describes various additive and subtractive ways to create the tessellation, but the most interesting part is Going back to the Wikipedia article on the Koch snowflake, it cites a paper describing the tessellation 3. To start, I reimplemented the same approach, but the inaccuracy became very obvious, especially for SVGs, because you can really blow it up and see that the snowflakes were not flush to one another. It was also really slow, and if you recursed too many times, it could cause the browser window to hang. I never really talked about that first implementation because I knew it was a hack. You see, my hacky implementation has been haunting me all this while. Approach 2: actually read literature on the thingįour years later in 2019 I decided to reimplement the Koch tessellation to create an SVG version so I could use a plotter to draw it. The little circles were there for debugging but I liked them so much that I just left them in. Given the side length of an equilateral triangle, l, it’s area, A can be written asĤ iterations of tessellations. But I was lazy and just went with the formula to find the area of an equilateral triangle. There are ways to estimate the surface area of a Koch snowflake 2. With this information, I can use a formula approximating the surface area of a snowflake given its edge length to reverse engineer the length of a snowflake with a third of the area. I can’t find the source for this for the life of me, so you’ll just have to take my word for it 1. In 2015, I read somewhere that the smaller snowflake will have a third area of the larger one. But which two sizes? Approach 1: estimating from area We know that it’s not possible to do with only one size – you need two sizes of the snowflake. There are also a lot of articles on the Koch snowflake out there so I’m not gonna do too much into it.īut did you know that the Koch snowflake shape (not just the curve) can be recursed upon? I found out about this from the Wikipedia page, but couldn’t really find any implementations in code. I talk about it in my 2016 !!Con talk on L-Systems, another concept I’m fond of.
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