Please consider :
Manipulate[
Row[{
Graphics[Disk[]],
Graphics[{
Polygon[{{0, 0}, {3, 0}, {3, 1}, {0, 1}},
VertexColors -> {White, Blend[{White, Blue}],
Blend[{White, Blue}], White}],
Black, Thick,
Line[{{i, 0}, {i, 1}}]}, ImageSize -> 300]}],
{i, 0, 3}]
Using Szabolcs`s solution on Gradient Filling
How could I color the disk with the color located underneath the Black Line ?
Here is one solution which works because the color on the left is White
and the gradient is linear.
With[{max = 3, color = Blend[{White, Blue}]},
Manipulate[
Row[{Graphics[{Opacity[i/max], color, Disk[]}],
Graphics[{Polygon[{{0, 0}, {max, 0}, {max, 1}, {0, 1}},
VertexColors -> {White, color, color, White}], Black, Thick,
Line[{{i, 0}, {i, 1}}]}, ImageSize -> 300]}], {i, 0, max}]]
If you had two different colors for each end (i.e., something other than White
), the Opacity
approach won't work. Instead, you can use the optional blending fraction argument to Blend
the colors in the desired proportion. Here's an example:
With[{max = 3, color1 = Red, color2 = Green},
Manipulate[
Row[{Graphics[{Blend[{color1, color2}, i/max], Disk[]}],
Graphics[{Polygon[{{0, 0}, {max, 0}, {max, 1}, {0, 1}},
VertexColors -> {color1, color2, color2, color1}], Black,
Thick, Line[{{i, 0}, {i, 1}}]}, ImageSize -> 300]}], {i, 0,
max}]]
If you need to do this for a blend of colours other than something and white, Opacity
won't be suitable. You could instead stay closer to Szabolcs' original solution using the second argument to Blend
like so:
skyBlue = Blend[{White,Blue}];
Manipulate[ Row[{ Graphics[{Blend[{White,skyBlue},i/3], Disk[]}],
Graphics[{ Polygon[{{0, 0}, {3, 0}, {3, 1}, {0, 1}},
VertexColors -> {White, skyBlue,
skyBlue, White}], Black, Thick,
Line[{{i, 0}, {i, 1}}]}, ImageSize -> 300]}], {i, 0, 3}]
I have divided i
by 3 because that parameter is meant to vary between 0
and 1
.