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The follow example created by software grasshopper3d, I cann't reproduce it using ColorFunction or Image3DProjection or other way.

enter image description here

Limited to my ability, I can only provide very little useful information.

f2d[x_, y_] := Cos[x] + Cos[y];

g3d[x_, y_] := Sinc[x]*Sinc[y];

p1 = ContourPlot[f2d[x, y], {x, -5, 5}, {y, -5, 5}]

p2 = Plot3D[g3d[x, y], {x, -5, 5}, {y, -5, 5}]

(Grasshopper3d)enter image description here

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  • $\begingroup$ Thanks to @David G. Stork ! There are two new problem: (1) "Texture" have a different graphical logic from "ColorFunction", when export into a cad format, "ColorFunction" can export the color correctly but "Texture" without color. (2) For the Answer using "Texture" method, when the texture is 2d plot instead of JPG, "Export" Function will give an Error: Export::noscaled. $\endgroup$ Commented Jul 20, 2024 at 13:06
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    $\begingroup$ As to the new questions in the comment, they are different enough from the original one, you may consider posting a new question, if existing posts e.g. mathematica.stackexchange.com/search?q=texture+export don't answer your question. $\endgroup$ Commented Aug 28, 2024 at 8:43
  • $\begingroup$ @xzczd, I will delete the question, and I think this hyperlink you have given should be added with a tag of 'projection'. Some people may not know much about the concept of 'texture'. $\endgroup$ Commented Aug 28, 2024 at 23:26
  • $\begingroup$ You don't need to. (And you're not able to, I believe, because there's a upvoted answer there :) ) Just take it easy, "marking as duplicate" is not a punishment. Your post will be kept in the site as a road sign. $\endgroup$ Commented Aug 29, 2024 at 0:11

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p1 = ContourPlot[f2d[x, y], {x, -5, 5}, {y, -5, 5}, Frame -> None]; Plot3D[Sinc[x] Sinc[y], {x, -5, 5}, {y, -5, 5}, Mesh -> None, PlotStyle -> Texture[p1]]

enter image description here

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