I would like to texture and deform a sphere with with images using various mapping functions between x,y and phi,theta. I figure I should start with $f(phi,theta)=(x,y)$ as a test case. However, I have run into a problem.

When I run:

lena = ExampleData[{"TestImage", "Lena"}];
SphericalPlot3D[1, {ϕ, 1, π}, {θ, 0, 2 Pi}, 
  ColorFunction -> Function[{x, y, z, ϕ, θ, r}, RGBColor[ImageValue[lena, {ϕ, θ}]]]

I receive a number of error messages most notably:

ImageValue::imgrng: The specified argument
{(1. -Graphics`SphericalPlot3DDump`slotfourmin)/(Graphics`SphericalPlot3DDump`slotfourmax-Graphics`SphericalPlot3DDump`slotfourmin),
should be an image, a graphics object, or a list of coordinates.

It seems like SphericalPlot3D is trying to get a range on the color values my color function generates and for some reason it does this with the symbols like Graphics`SphericalPlot3DDump`slotfivemin instead of numerical values.

I can try to get around this like so:

cf[x_, y_, z_, ϕ_NumericQ, θ_NumericQ, r_] :=RGBColor[ImageValue[lena, {ϕ *10, θ *10}]]

SphericalPlot3D[1, {ϕ, 0, π}, {θ, 0, 2 Pi}, ColorFunction -> cf]

But that just produces an ordinary sphere with some weird triangular artifacts.

This maybe an XY problem, and I would be interested in other solutions to this color mapping and deforming problem, maybe using PlotStyle->Texture["imag"]

  • $\begingroup$ This scheme works perfectly with Plot3D: Plot3D[Max[ImageValue[lena, {x, y}]], {x, 0, 500}, {y, 0, 500}, ColorFunction -> Function[{x, y, z}, RGBColor[ImageValue[lena, {x*500, y*500}]]], PlotRange -> All, PlotPoints -> 120, Mesh -> None] $\endgroup$ – alessandro Jan 16 '19 at 4:00

Won't comment on reasons or alternative solutions now but this seems to do the trick:

lena = ExampleData[{"TestImage", "Lena"}];

img[ϕ_?NumericQ, θ_?NumericQ] :=  RGBColor[ImageValue[lena, 512 {ϕ, θ}]]

SphericalPlot3D[1, {ϕ, 1, π}, {θ, 0, 2 Pi}, 
 ColorFunction ->  Function[{x, y, z, ϕ, θ, r}, img[ϕ, θ]], 
 PlotPoints -> 200]

enter image description here

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.