Saturated and desaturated 3D shapes

Saturated

Here are some more examples of saturated shape (see last post), this time 3D. I've used the inversion sets table as a starting point.

These saturated surfaces all have fractal dimension 2, so they are 2D surfaces. However their surface areas are infinite. A measurable area needs to be in m^2 log^n m where n is larger for the lumpier shapes:

These are also saturated surfaces, but this time saturated with spheres rather than lumps or dents:

Neither are fractals and neither have finite surface area. 

These curves all have fractal dimension 1, they are non-fractal (branching) curves, however their total length is infinite. You need to measure their size in m log^n m for some n representing how saturated they are:

(ignore the sphere in the centres of the last one, an artifact of too few iterations)

This set is disparate points, or dust. It's fractal dimension is 0 like its topological dimension, so it is not a fractal. But the number of points is infinite, you need to 'count' the points in log^n m:

So saturated shapes are relatively easy to create in 3D. 

All of these shapes are nowhere differentiable. They are more lumpy than C(1) fractals like the Hevea project shapes which are differentiable and also have finite length/area.

Desaturated

This is a desaturated tree surface fractal. Its fractal dimension is something like 2.5, but the roughness grows at the finer details, such that its area in m^2.5 is zero. 

Here is an equivalent with the cluster-tree fractal:

 

In both cases the desaturated shapes look sharp compared to the rough fractals, and the saturated shapes at the top look blunt.

Here's a desaturated shell-shell:

The equivalent for the void-tree again has most of the geometry at the finer scales:

Colour

Because desaturated shapes have a proportionally larger light-absorbing surface content in the smaller scales, the reflected light at smaller wavelengths is less, so they should be tinted brown on average. 

Conversely the saturated shapes should be tinted blue, which is conveniently compatible with analogy of being saturated with water.

saturated   (blunt features)                desaturated   (sharp features)

In general the tint from brown to grey to blue is based on a signed n, where desaturated use negative values, larger |n| having a stronger colour. I expect the tint to be mild, so the slight gradient from red to blue ought to be more like a dusky brown and a dusky turquoise rather than red and blue.