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## Main Question or Discussion Point

According to the link below, fractal dimension is an exponent of some sort:

http://www.vanderbilt.edu/AnS/psychology/cogsci/chaos/workshop/Fractals.html

The Hausdorff Dimension (aka fractal dimension) is denoted as

If we were to look at any image and use Hausdorff Dimension approximation methods such as the box counting method (http://classes.yale.edu/fractals/fracanddim/boxdim/BoxDim.html) for approximating the Hausdorff Dimension which is

The reason I ask is because in the case of the Koch Snowflake, we know the initiator and generator (refer to first link if you're not familiar with these two terms) because it is something created by man; in other words, we already know its

http://www.vanderbilt.edu/AnS/psychology/cogsci/chaos/workshop/Fractals.html

The Hausdorff Dimension (aka fractal dimension) is denoted as

*D*in the website above. And*r*is the base number.If we were to look at any image and use Hausdorff Dimension approximation methods such as the box counting method (http://classes.yale.edu/fractals/fracanddim/boxdim/BoxDim.html) for approximating the Hausdorff Dimension which is

*D*in*N=r^D*. The link describes how to find*D*using the box-counting method, but it doesn't explain how to derive at*r*. Is there a way in how we get*r*using the box counting method or any other Hausdorff Dimension approximation methods?The reason I ask is because in the case of the Koch Snowflake, we know the initiator and generator (refer to first link if you're not familiar with these two terms) because it is something created by man; in other words, we already know its

*D*and*r*because these values are chosen by man (aka man-made). However, if we were to take a picture of a real tree in my backyard for example (trees in general have a wonderful fractal dimensional branching pattern), we can use the box counting method to approximate at*D*without knowing*r*. So I wanted to know if we can derive at*r*using the box counting method or any other Hausdorff Dimension approximation methods.