fimplicit3: colormapping (3DContourPlot)

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Niklas Kurz
Niklas Kurz 2021-1-30
回答： Scott Ronquist 2021-1-30
This is totally optional and off pure interest:
If I plot a manifold with this powerful function:
f = @(x,y,z) x.^2 + y.^2 - z.^2;
fimplicit3(f)
The shape is exactly as I want. Just if: how do I change colorgradient?
I'm not well versed with terms like Edgecolor etc.

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Scott Ronquist
Scott Ronquist 2021-1-30
Hi Niklas,
The color gradient in this figure can be changed by using the colormap function. Predefined colormaps are listed here, but any custom color map can be used by inputting a m x 3 numeric matrix into colormap. See example below:
>> f = @(x,y,z) x.^2 + y.^2 - z.^2;
>> fimplicit3(f)
>> colormap("cool")
Hope this helps!

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