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 Pages:    Go Down       Author Topic: fractal Dimension of a square and a filled circle(disk)  (Read 1616 times) Description: 0 Members and 1 Guest are viewing this topic.
Fearings
Forums Newbie Posts: 2 « on: September 02, 2015, 04:56:11 PM »

Hello folks,

as much as i would like to understand fractals, i still can't figure out the following problem:

If a filled circle(disk) does have a fractal Dimension of 2, what fractal Dimension would a filled square have?

I did look up fractal dimension of nanoparticles, they are often calculated with a boxcounting-method. Using the formula: D = -lim (x-->0)  ln (N(x))/ln(x) you would get a fractal dimension of 2 if you apply it to a square. I attached a figure to display the problem.

Help would be appreciated « Last Edit: September 15, 2015, 08:31:41 AM by Fearings » Logged
hobold
Fractal Bachius Posts: 573 « Reply #1 on: September 02, 2015, 05:38:20 PM »

they are often calculated with a boxcounting-method. Using the formula: D = -lim (x-->0)  ln (N(x))/ln(x) you would get a fractal dimension of 2 if you apply it to a square.

A filled square is a classical two dimensional object (with finite circumfence and positive surface area). The box counting method correctly reflects this.

I am guessing you were expecting unit circle and unit square to have different fractal dimension, because they are not equal "size"? In that case, the misunderstanding is that fractal dimension is not a measurement of "size". Instead, dimension determines what exact kind of "size" is appropriate: either length, or surface area, or volume, and so on.

The trouble with fractal dimension is that we end up with infinitely many kinds of "size", and that breaks our intuitive assumptions based on classical geometry. Logged
lkmitch
Fractal Lover  Posts: 238  « Reply #2 on: September 02, 2015, 05:43:12 PM »

The filled-in square has a fractal (box counting) dimension of 2, like the disc.  Another way of looking at fractal dimension is how much space (relatively) does the object occupy?  A filled-in square or disc occupies all of the space (area) within its boundary, but no more, and has a fractal dimension equal to its topological (regular) dimension, 2.  A straight line segment has a fractal dimension of 1, because it occupies just enough space (length) for its topological dimension (1).  A fractal segment, like the Koch snowflake, occupies more space than a straight line and has a higher fractal dimension.  The Hilbert curve, while essentially a line, occupies so much space (the same as a filled-in square) that its fractal dimension is 2. Logged
Fearings
Forums Newbie Posts: 2 « Reply #3 on: September 03, 2015, 09:20:22 AM »

Thank you very much for your help, i think i do undestand the problem much better now.

My dispute still exists, well at least still in my head:
- if I calculate the fractal Dimension of a square (lets say Unit square 1x1) with the formula i posted, a filled square will have a fractal Dimension of 2
- if i do the same for a filled unit circle (disk) and decrease the size of my boxes (lets say take the right most picture with a box size of 1/8), x (or in the picture e) will be 1/8, just as for the square. But obviously it will not contain all 64 boxes, thus have a smaller dimension then 2.

Is this just a problem of the method? Cause i found literature would use the box counting-method to determine the fractal dimension of particles (they are generally spherical), but as i see it, a filled disk will never reach a fractal dimension of 2 using THIS method. Logged
hobold
Fractal Bachius Posts: 573 « Reply #4 on: September 03, 2015, 11:09:27 AM »

if i do the same for a filled unit circle (disk) and decrease the size of my boxes (lets say take the right most picture with a box size of 1/8), x (or in the picture e) will be 1/8, just as for the square. But obviously it will not contain all 64 boxes, thus have a smaller dimension then 2.

The box counting algorithm only cares for the limit after infinitely many iterations. The problem you describe is a boundary error; i.e. something that happens at the rim of the circle, not deeper inside the disk. But as you keep halving box size, the number of boxes overlapping the circle rim is dwarfed by the number of boxes inside the disk. Eventually, what is happening inside the disk will dominate the ultimate outcome, and boundary effects become irrelevant.

Another way to see that circle and box really behave the same way is this: imagine the unit square is first rotated by some oblique angle, say, 31 degrees; but keep the grid for box counting upright. Then the square case will suffer from boundary errors, too. But obviously the nature of the square has not changed, so you still expect the same result.

(In fact, practical implementations of the box counting algorithm must apply a little extra effort to negate boundary artifacts. One way to do that is to randomly move the boxes' coordinate origin in every iteration, and even to rotate the box grid randomly in every iteration.) Logged
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