fimplicit3: colormapping (3DContourPlot)
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Niklas Kurz
le 30 Jan 2021
Réponse apportée : Scott Ronquist
le 30 Jan 2021
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
le 30 Jan 2021
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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