Gradient with HSV rather than RGB in OpenGL - opengl

OpenGL can colour a rectangle with a gradient of colours from 1 side to the other. I'm using the following code for that in C++
glBegin(GL_QUADS);
{
glColor3d(simulationSettings->hotColour.redF(), simulationSettings->hotColour.greenF(), simulationSettings->hotColour.blueF());
glVertex2d(keyPosX - keyWidth/2, keyPosY + keyHight/2);
glColor3d(simulationSettings->coldColour.redF(), simulationSettings->coldColour.greenF(), simulationSettings->coldColour.blueF());
glVertex2d(keyPosX - keyWidth/2, keyPosY - keyHight/2);
glColor3d(simulationSettings->coldColour.redF(), simulationSettings->coldColour.greenF(), simulationSettings->coldColour.blueF());
glVertex2d(keyPosX + keyWidth/2, keyPosY - keyHight/2);
glColor3d(simulationSettings->hotColour.redF(), simulationSettings->hotColour.greenF(), simulationSettings->hotColour.blueF());
glVertex2d(keyPosX + keyWidth/2, keyPosY + keyHight/2);
}
I'm using some Qt libraries to do the conversions between HSV and RGB. As you can see from the code, I'm drawing a rectangle with colour gradient from what I call hotColour to coldColour.
Why am I doing this? The program I made draws 3D Vectors in space and indicates their length by their colour. The user is offered to choose the hot (high value) and cold (low value) colours, and the program will automatically do the gradient using HSV scaling.
Why HSV scaling? because HSV is single valued along the colour map I'm using, and creating gradients with it linearly is a very easy task. For the user to select the colours, I offer him a QColourDialog colour map
http://qt-project.org/doc/qt-4.8/qcolordialog.html
On this colour map, you can see that red is available on the right and left side, making it impossible to have a linear scale for this colour-map with RGB. But with HSV, the linear scale is very easily achievable, where I just have to use a linear scale between 0 and 360 for Hue values.
With this paradigm, we can see that hot and cold colours define the direction of the gradient, so for example, if I choose hue to be 0 for cold and 359 for hot, HSV will give me a gradient between 0 and 359, and will include the whole spectrum of colours in the gradient; whilst, in OpenGL, it will basically go from red to red, which is no gradient!!!!!!
How can I force OpenGL to use an HSV gradient rather than RGB? The only idea that occurs to me is slicing the rectangle I wanna colour and do many gradients over smaller rectangles, but I think this isn't the most efficient way to do it.
Any ideas?

How can I force OpenGL to use an HSV gradient rather than RGB?
I wouldn't call it "forcing", but "teaching". The default way of OpenGL to interpolate vertex attributes vectors is by barycentric interpolation of the single vector elements based on the NDC coordinates of the fragment.
You must tell OpenGL how to turn those barycentric interpolated HSV values into RGB.
For this we introduce a fragment shader that assumes the color vertex attribute not being RGB but HSV.
#version 120
varying vec3 vertex_hsv; /* set this in appropriate vertex shader to the vertex attribute data*/
vec3 hsv2rgb(vec3 hsv)
{
float h = hsv.x * 6.; /* H in 0°=0 ... 1=360° */
float s = hsv.y;
float v = hsv.z;
float c = v * s;
vec2 cx = vec2(v*s, c * ( 1 - abs(mod(h, 2.)-1.) ));
vec3 rgb = vec3(0., 0., 0.);
if( h < 1. ) {
rgb.rg = cx;
} else if( h < 2. ) {
rgb.gr = cx;
} else if( h < 3. ) {
rgb.gb = cx;
} else if( h < 4. ) {
rgb.bg = cx;
} else if( h < 5. ) {
rgb.br = cx;
} else {
rgb.rb = cx;
}
return rgb + vec3(v-cx.y);
}
void main()
{
gl_FragColor = hsv2rgb(vertex_hsv);
}

You can do this with a fragment shader. You draw a quad and apply your fragment shader which does the coloring you want to the quad. The way I would do this is to set the colors of the corners to the HSV values that you want, then in the fragment shader convert the interpolated color values from HSV back to RGB. For more information on fragment shaders see the docs.

Related

fwidth(uv) giving strange results in glsl

I checked the result of the filter-width GLSL function by coloring it in red on a plane around the camera.
The result is a bizarre pattern. I thought that it would be a circular gradient on the plane extending around the camera relative to distance. The further pixels uniformly represent more distant UV coordinates between pixels at further distances.
Why isn't fwidth(UV) a simple gradient as a function of distance from the camera? I don't understand how it would work properly if it isn't, because I want to anti-alias pixels as a function of amplitude of the UV coordinates between them.
float width = fwidth(i.uv)*.2;
return float4(width,0,0,1)*(2*i.color);
UVs that are close = black, and far = red.
Result:
the above pattern from fwidth is axis aligned, and has 1 axis of symmetry. it couldnt anti-alias 2 axis checkerboard or an n-axis texture of perlin noise or a radial checkerboard:
float2 xy0 = float2(i.uv.x , i.uv.z) + float2(-0.5, -0.5);
float c0 = length(xy0); //sqrt of xx+yy, polar coordinate radius math
float r0 = atan2(i.uv.x-.5,i.uv.z-.5);//angle polar coordinate
float ww =round(sin(c0* freq) *sin(r0* 50)*.5+.5) ;
Axis independent aliasing pattern:
The mipmaping and filtering parameters are determined by the partial derivatives of the texture coordinates in screen space, not the distance (actually as soon as the fragment stage kicks in, there's no such thing as distance anymore).
I suggest you replace the fwidth visualization with a procedurally generated checkerboard (i.e. (mod(uv.s * k, 1) > 0.5)*(mod(uv.t * k, 1) < 0.5)), where k is a scaling parameter) you'll see that the "density" of the checkerboard (and the aliasing artifacts) is the highst, where you've got the most red in your picture.

How to render color gradients that start from one side to other side in opengl?

I want to render color gradient that start from one side to another. In the figure as you can see, the color gradients start from minimum bounding box to maximum bounding box. The code is below.
Point c = (VERTICES[i] - min) / (max - min);
p_COLOR[i].setRGB(c.red, c.green, c.blue);
Here the issue is, the color gradient is not following any direction (for example down to up: down side has two colors red and bluish purple). But I need to apply gradient that start from -x to x OR -y to y OR -z to z, means if red color start from -y then only red color covers -y side area then by moving gradient from -y to y, increment the color with red after blue then green then yellow etc. There should not be red and bluish purple at down side, there must be only red then moving upward to blue then green and so on.
how can I do that ? what would be the equation ?
Your solution is very simple, yet very limited. You should only use one component of your VERTICES struct (I assume VERTICES[i] holds position of the current vertex) if you want the gradient to apply across only one axis. Split your code into two parts:
The gradient function
vec3 Gradient (float param)
{
param = clamp(param, 0, 1);
// we are treating these colors as HSL now:
vec3 ColorA (0, 0.5, 0.5);
vec3 ColorB (1, 0.5, 0.5);
// so now we need to convert them to plain old RGB
// note how the blending equation remains the same
return HSL2RGB(param * ColorA + (1-param) * ColorB); // simple blending equation
}
The coloring function
float c = ((VERTICES[i] - min) / (max - min)).x; // or y,z
p_COLOR[i].setRGB(Gradient(c));
The shaders are in normal GLSL, so please excuse my ignorance in ES.

GLSL Distance Field transparency

I am after smooth texture based outline effect in OpenGL. So far I tried mostly all kinds of edge detection algorithms which result mostly in crude and jagged outlines. Then I read about Distance Field. I found an example which does pretty nice distance field. Here is the GLSL code:
#version 420
layout(binding=0) uniform sampler2D colorMap;
flat in vec4 diffuseOut;
in vec2 uvsOut;
out vec4 outputColor;
const float ALPHA_THRESHOLD = 0.9;
const float NUM_SPOKES = 36.0; // Number of radiating lines to check in.
const float ANGULAR_STEP =360.0 / NUM_SPOKES;
const int ZERO_VALUE =128; // Color channel containing 0 => -128, 128 => 0, 255 => +127
int in_StepSize=15; // Distance to check each time (larger steps will be faster, but less accurate).
int in_MaxDistance=30; // Maximum distance to search out to. Cannot be more than 127!
vec4 distField(){
vec2 pixel_size = 1.0 / vec2(textureSize(colorMap, 0));
vec2 screenTexCoords = gl_FragCoord.xy * pixel_size;
int distance;
if(texture(colorMap, screenTexCoords).a == 0.0)
{
// Texel is transparent, search for nearest opaque.
distance = ZERO_VALUE + 1;
for(int i = in_StepSize; i < in_MaxDistance; i += in_StepSize)
{
if(find_alpha_at_distance(screenTexCoords, float(i) * pixel_size, 1.0))
{
i = in_MaxDistance + 1; // BREAK!
}
else
{
distance = ZERO_VALUE + 1 + i;
}
}
}
else
{
// Texel is opaque, search for nearest transparent.
distance = ZERO_VALUE;
for(int i = in_StepSize; i <= in_MaxDistance; i += in_StepSize)
{
if(find_alpha_at_distance(screenTexCoords, float(i) * pixel_size, 0.0))
{
i = in_MaxDistance + 1; // BREAK!
}
else
{
distance = ZERO_VALUE - i;
}
}
}
return vec4(vec3(float(distance) / 255.0) * diffuseOut.rgb, 1.0 - texture(colorMap, screenTexCoords).a);
}
void main()
{
outputColor= distField();
}
The result of this shader covers the whole screen using the diffuse color for filling the screen area outside the Distance Field outline.Here is how it looks like :
What I need is to leave all the area which has the solid red fill outside the distance field as transparent.
I came to the solution by using Distance Field gray scale 8 bit alpha map.Stefan Gustavson
describes in detail how to do it.Basically one needs to generate the distance field version of the original texture.Then this texture is rendered with the primitive normally in the first pass into an FBO.In the second pass the alpha blending mode should be on.The texture from the first pass in used with the screen quad.At this stage the the fragment shader samples the alpha from that texture.This results in both smooth edges and alpha transparency around the edges.
Here is the result:
Based on the screenshot I'm assuming you're rendering a fullscreen quad? If that's the case Tim just provided the answer, try:
glEnable( GL_BLEND );
glBlendFunc(GL_SRC_ALPHA, GL_ONE_MINUS_SRC_ALPHA);
Before you render the quad. Obviously if you're going to render non-transparent stuff too, I advise you to render those first so you won't get depth buffer problems. When you're done drawing the transparent stuff, call:
glDisable( GL_BLEND );
To turn alphablending off again.

x-coordinate modulo 2 == 1.0 needs different color

I need to write a shader where the color of the pixel are black when the following equation is true:
(x-coordinate of pixel) mod 2 == 1
If it is false, the pixel should be white. Therefore I searched the web but it did not work.
More information:
I've an OpenGL scene with 800 x 600 resolution and the teapot in it. The teapot is red. Now I need to create that zebra look.
Here is some code I've wrote, but it didn'T work:
FragmentShader:
void main(){
if (mod(gl_FragCoord[0].x * 800.0 , 2.0) == 0){
gl_FragColor = vec4(1.0,1.0,1.0,1.0);
}else{
gl_FragColor = vec4(0.0,0.0,0.0,1.0);
}
}
VertexShader:
void main(void)
{
gl_Position = ftransform();
gl_TexCoord[0] = gl_MultiTexCoord0;
}
As far as I know, gl_FragCood.x is in range(0,1) therefore I need to multiply with width.
Interesting you mention the need to multiply with the width, have you tried without the * 800.0 in there? The range of gl_FragCoord is such that the distance between adjacent pixels is 1.0, for example [0.0, 800.0] or possibly [0.5, 800.5].
Remove the width multiplication and see if it works.
Instead of comparing directly to 0, try doing a test against 1.0, e.g.
void main(){
if (mod(gl_FragCoord[0].x , 2.0) >= 1.0){
gl_FragColor = vec4(1.0,1.0,1.0,1.0);
}else{
gl_FragColor = vec4(0.0,0.0,0.0,1.0);
}
}
That'll avoid precision errors and the cost of rounding.
As emackey points out, gl_FragCoord is specified in window coordinates, which:
... result from scaling and translating Normalized
Device Coordinates by the viewport. The parameters to glViewport() and
glDepthRange() control this transformation. With the viewport, you can
map the Normalized Device Coordinate cube to any location in your
window and depth buffer.
So you also don't actually want to multiply by 800 — the incoming coordinates are already in pixels.

Gradient "miter" in OpenGL shows seams at the join

I am doing some really basic experiments around some 2D work in GL. I'm trying to draw a "picture frame" around an rectangular area. I'd like for the frame to have a consistent gradient all the way around, and so I'm constructing it with geometry that looks like four quads, one on each side of the frame, tapered in to make trapezoids that effectively have miter joins.
The vert coords are the same on the "inner" and "outer" rectangles, and the colors are the same for all inner and all outer as well, so I'd expect to see perfect blending at the edges.
But notice in the image below how there appears to be a "seam" in the corner of the join that's lighter than it should be.
I feel like I'm missing something conceptually in the math that explains this. Is this artifact somehow a result of the gradient slope? If I change all the colors to opaque blue (say), I get a perfect solid blue frame as expected.
Update: Code added below. Sorry kinda verbose. Using 2-triangle fans for the trapezoids instead of quads.
Thanks!
glClearColor(1.0, 1.0, 1.0, 1.0);
glClear(GL_COLOR_BUFFER_BIT);
glEnableClientState(GL_VERTEX_ARRAY);
glEnableClientState(GL_COLOR_ARRAY);
// Prep the color array. This is the same for all trapezoids.
// 4 verts * 4 components/color = 16 values.
GLfloat colors[16];
colors[0] = 0.0;
colors[1] = 0.0;
colors[2] = 1.0;
colors[3] = 1.0;
colors[4] = 0.0;
colors[5] = 0.0;
colors[6] = 1.0;
colors[7] = 1.0;
colors[8] = 1.0;
colors[9] = 1.0;
colors[10] = 1.0;
colors[11] = 1.0;
colors[12] = 1.0;
colors[13] = 1.0;
colors[14] = 1.0;
colors[15] = 1.0;
// Draw the trapezoidal frame areas. Each one is two triangle fans.
// Fan of 2 triangles = 4 verts = 8 values
GLfloat vertices[8];
float insetOffset = 100;
float frameMaxDimension = 1000;
// Bottom
vertices[0] = 0;
vertices[1] = 0;
vertices[2] = frameMaxDimension;
vertices[3] = 0;
vertices[4] = frameMaxDimension - insetOffset;
vertices[5] = 0 + insetOffset;
vertices[6] = 0 + insetOffset;
vertices[7] = 0 + insetOffset;
glVertexPointer(2, GL_FLOAT , 0, vertices);
glColorPointer(4, GL_FLOAT, 0, colors);
glDrawArrays(GL_TRIANGLE_FAN, 0, 4);
// Left
vertices[0] = 0;
vertices[1] = frameMaxDimension;
vertices[2] = 0;
vertices[3] = 0;
vertices[4] = 0 + insetOffset;
vertices[5] = 0 + insetOffset;
vertices[6] = 0 + insetOffset;
vertices[7] = frameMaxDimension - inset;
glVertexPointer(2, GL_FLOAT , 0, vertices);
glColorPointer(4, GL_FLOAT, 0, colors);
glDrawArrays(GL_TRIANGLE_FAN, 0, 4);
/* top & right would be as expected... */
glDisableClientState(GL_COLOR_ARRAY);
glDisableClientState(GL_VERTEX_ARRAY);
As #Newbie posted in the comments,
#quixoto: open your image in Paint program, click with fill tool somewhere in the seam, and you see it makes 90 degree angle line there... means theres only 1 color, no brighter anywhere in the "seam". its just an illusion.
True. While I'm not familiar with this part of math under OpenGL, I believe this is the implicit result of how the interpolation of colors between the triangle vertices is performed... I'm positive that it's called "Bilinear interpolation".
So what to do to solve that? One possibility is to use a texture and just draw a textured quad (or several textured quads).
However, it should be easy to generate such a border in a fragment shader.
A nice solution using a GLSL shader...
Assume you're drawing a rectangle with the bottom-left corner having texture coords equal to (0,0), and the top-right corner with (1,1).
Then generating the "miter" procedurally in a fragment shader would look like this, if I'm correct:
varying vec2 coord;
uniform vec2 insetWidth; // width of the border in %, max would be 0.5
void main() {
vec3 borderColor = vec3(0,0,1);
vec3 backgroundColor = vec3(1,1,1);
// x and y inset, 0..1, 1 means border, 0 means centre
vec2 insets = max(-coord + insetWidth, vec2(0,0)) / insetWidth;
If I'm correct so far, then now for every pixel the value of insets.x has a value in the range [0..1]
determining how deep a given point is into the border horizontally,
and insets.y has the similar value for vertical depth.
The left vertical bar has insets.y == 0,
the bottom horizontal bar has insets.x = 0,, and the lower-left corner has the pair (insets.x, insets.y) covering the whole 2D range from (0,0) to (1,1). See the pic for clarity:
Now we want a transformation which for a given (x,y) pair will give us ONE value [0..1] determining how to mix background and foreground color. 1 means 100% border, 0 means 0% border. And this can be done in several ways!
The function should obey the requirements:
0 if x==0 and y==0
1 if either x==1 or y==1
smooth values in between.
Assume such function:
float bias = max(insets.x,insets.y);
It satisfies those requirements. Actually, I'm pretty sure that this function would give you the same "sharp" edge as you have above. Try to calculate it on a paper for a selection of coordinates inside that bottom-left rectangle.
If we want to have a smooth, round miter there, we just need another function here. I think that something like this would be sufficient:
float bias = min( length(insets) , 1 );
The length() function here is just sqrt(insets.x*insets.x + insets.y*insets.y). What's important: This translates to: "the farther away (in terms of Euclidean distance) we are from the border, the more visible the border should be", and the min() is just to make the result not greater than 1 (= 100%).
Note that our original function adheres to exactly the same definition - but the distance is calculated according to the Chessboard (Chebyshev) metric, not the Euclidean metric.
This implies that using, for example, Manhattan metric instead, you'd have a third possible miter shape! It would be defined like this:
float bias = min(insets.x+insets.y, 1);
I predict that this one would also have a visible "diagonal line", but the diagonal would be in the other direction ("\").
OK, so for the rest of the code, when we have the bias [0..1], we just need to mix the background and foreground color:
vec3 finalColor = mix(borderColor, backgroundColor, bias);
gl_FragColor = vec4(finalColor, 1); // return the calculated RGB, and set alpha to 1
}
And that's it! Using GLSL with OpenGL makes life simpler. Hope that helps!
I think that what you're seeing is a Mach band. Your visual system is very sensitive to changes in the 1st derivative of brightness. To get rid of this effect, you need to blur your intensities. If you plot intensity along a scanline which passes through this region, you'll see that there are two lines which meet at a sharp corner. To keep your visual system from highlighting this area, you'll need to round this join over. You can do this with either a post processing blur or by adding some more small triangles in the corner which ease the transition.
I had that in the past, and it's very sensitive to geometry. For example, if you draw them separately as triangles, in separate operations, instead of as a triangle fan, the problem is less severe (or, at least, it was in my case, which was similar but slightly different).
One thing I also tried is to draw the triangles separately, slightly overlapping onto one another, with a right composition mode (or OpenGL blending) so you don't get the effect. I worked, but I didn't end up using that because it was only a tiny part of the final product, and not worth it.
I'm sorry that I have no idea what is the root cause of this effect, however :(