Skip to main content

How can I use multiple Textures in Graphics3D


I am creating a Graphics3D object for a box using 6 Polygons (1 per side). I can use one single Texture on one or more sides but I can not find a way to use different Textures for different sides of the box. Is there any way to achieve this?



Answer




You can set Texture before each polygon


t = ImageResize[ExampleData@#, {100, 100}] & /@ 
ExampleData["ColorTexture"][[;; 6]];

vtc = {{0, 0}, {1, 0}, {1, 1}, {0, 1}};
coords = {{{0, 0, 0}, {0, 1, 0}, {1, 1, 0}, {1, 0, 0}}, {{0, 0,
0}, {1, 0, 0}, {1, 0, 1}, {0, 0, 1}}, {{1, 0, 0}, {1, 1, 0}, {1,
1, 1}, {1, 0, 1}}, {{1, 1, 0}, {0, 1, 0}, {0, 1, 1}, {1, 1,
1}}, {{0, 1, 0}, {0, 0, 0}, {0, 0, 1}, {0, 1, 1}}, {{1, 0,
1}, {1, 1, 1}, {0, 1, 1}, {0, 0, 1}}};


Graphics3D[{Table[{Texture@t[[i]],
Polygon[coords[[i]], VertexTextureCoordinates -> vtc]}, {i, 6}]}]

enter image description here


Or you can use one texture with proper texture coordinates like in cube maps


enter image description here


cubemap = ImageResize[#, Scaled[0.5]] &@ Import@"http://i.stack.imgur.com/hSCfz.jpg";

vtccm = {{##}, {# + 1, #2}, {# + 1, #2 + 1}, {#, #2 + 1}} & @@@ {{1,

1}, {0, 2}, {1, 2}, {2, 2}, {3, 2}, {1, 3}}/4;

α = 0.;
Graphics3D[{EdgeForm[], Texture[cubemap],
Polygon[coords, VertexTextureCoordinates -> vtccm]},
Lighting -> {{"Ambient", White}}, Boxed -> False, ViewAngle -> 1,
ViewCenter -> {1, 1, 1}/2,
ViewVector -> {1, 1, 1}/2 + {Cos[α], Sin[α], 0}/8,
RotationAction -> "Clip", ImageSize -> 500,
ViewVertical -> {0, 0, 1}]


enter image description here


There is no visible joints!


Related: Implementing a first person view of 3D objects in a scene


Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]