Posted by
Gabriel Landini on
Jun 20, 2016; 10:27am
URL: http://imagej.273.s1.nabble.com/Convert-macro-from-NIH-Image-to-ImageJ-tp5015304p5016694.html
On Monday 20 Jun 2016 18:15:30 Dini Nurfiani wrote:
> Thank you for your explanation. I am a bit confused since you explained
> about dilation dimension. Are you referring to my first&previous question
> about dilation macro? Because my recent&new question is about Euclidean
> Distance Map macro.
I commented on the figures you included in your email.
The EDM is somewhat equivalent to the nested dilations, but the fractal
dimension is not computed as the figure seemed to suggest: log of distance
transform vs log of pixels, but the the way I outlined in my previous email.
It is 1- slope of (log(diameter) vs log(length[diameter]) ).
> But from your explanation about dilation dimension, I tried it as well. I
> calculated the perimeter (length) first by dividing Area(epsilon)/epsilon
> and also plotted the log radius vs log perimeter as you mentioned. I found
> the fractal dimension for Koch Snowflake using that method was 1,2283,
> since the slope is - 0,2283. That value is quite far from the theoretical
> value, which is 1,2685.
1. You need to make sure that the largest dilated version of the curve fits
*completely* within the frame of the image. So the curve has to have lots of
empty space around. In the figure you pasted before, the EDM was truncated by
the image borders. That will underestimate the number of pixels of the large
dilation discs.
2. Do not use diameters which are too small. At small scales the image you
used is made of short straight lines (with D=1).
Hope it helps
Gabriel
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