Fractal Dimensions

Date: 01/10/98 at 22:48:06
From: James
Subject: Fractals and fractal dimensions

How do you calculate the fractal dimension of a fractal?

Thank you.

Date: 01/11/98 at 14:38:30
From: Doctor Anthony
Subject: Re: Fractals and fractal dimensions

Dimensions of Fractals.

If we divide a line segment into equal parts, each 1/nth of the 
whole, we will need n^1 parts to reassemble the original line segment.  
The dimension of the object is equal to the exponent.

If we divide a square into smaller squares, each with side 1/n of the 
whole, we shall require n^2 parts to reassemble the original square.  
Again the mathematical dimension is equal to the exponent.

Similarly, if we divide a cube into smaller cubes, each with edge 
1/nth the original cube, we shall require n^3 parts to reassemble the 
original cube.  Again the dimension is equal to the exponent.

In all three cases - the line segment, the square, and the cube, we 
can piece together n^d smaller parts with edge scaled down by 1/n from 
the original, and rebuild the original.

Now the SIMILARITY DIMENSION 'd' is defined as the ratio:

               log(number of parts)
       d =  ---------------------------
             log(1/linear magnification)

Thus for the line segment

               ln(n^1)
           d = -------   =  1
               ln(n)

For square      ln(n^2)      2.ln(n)
            d = -------  =  --------   = 2
                 ln(n)        ln(n) 

For cube        ln(n^3)      3.ln(n)
            d = --------  = --------   =  3
                 ln(n)        ln(n)

The similarity dimension is a basic tool in studying fractals.

If you think of the Koch snowflake, each side of the triangle develops 
into 4 smaller parts; scaled down to 1/3 size, the triangles are self-
similar. Hence the dimension

                               ln(4)
             snow flake   d = -------  =  1.261...
                               ln(3)     

                              ln(3)
The Sierpinski triangle   d = -----  = 1.584.....
                              ln(2}

                               ln(8)
The Sierpinski carpet     d = -------  =  1.892....
                               ln(3)   

                               ln(20)
The Sierpinski sponge     d =  ------  =  2.726....
                               ln(3)

                               ln(2)
The Cantor set            d = -------  =  0.630....
                               ln(3)   

The large dark areas of the Mandelbrot set are two-dimensional but the 
entire set, being not exactly self-similar, does not have a similarity 
dimension. On the other hand, the similarity dimension is only one of 
many that yield fractal dimensions. The boundary of the Mandelbrot set 
wiggles so violently as to be considered two-dimensional in the 
Hausdorff-Besicovich dimension.
 
-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/