Gradient Fill using Graphics in Mathematica

Question

How could I create the following using Rectangle[] in Graphics[]?

enter image description here

score 7 · Accepted Answer · answered Nov 15 '11 at 19:49

7

Using Polygon, you can

Graphics[{EdgeForm[Black], 
  Polygon[{{0, 0}, {3, 0}, {3, 1}, {0, 1}}, 
   VertexColors -> {White, Red, Red, White}]}]

answered Nov 15 '11 at 19:49

Szabolcs

3

And just for fun (requires a webcam to work): pts = {{0, 0}, {1, 0}, {1, 1}, {0, 1}}; Graphics[{Texture[CurrentImage[]], Polygon[pts, VertexTextureCoordinates -> pts]}] – Arnoud Buzing Nov 16 '11 at 00:31
CurrentImage[] - an interesting example of a function that works so much better in the online documentation than it does on my machine... :) And how do you switch it off... – cormullion Nov 16 '11 at 14:37
@cormullion It works pretty well here. Try `Dynamic@CurrentImage[]`. The very first frame it captures might be black (it needs time to adjust), and if there are several `Dynamic` ones, it gets a bit unreliable. `ImageCapture[]` opens an UI that lets you switch it off. – Szabolcs Nov 16 '11 at 14:44

score 5 · Answer 2 · answered Nov 16 '11 at 12:16

5

Also:

Graphics[Raster[{Range[100]/100}, ColorFunction -> (Blend[{White, Red}, #] &)], 
 AspectRatio -> .3, 
 Frame       -> True, 
 FrameTicks  -> None]

enter image description here

answered Nov 16 '11 at 12:16

Dr. belisarius

2 Answers2