Is it possible to have different scissor test paramenters for each instance drawn using glDrawElementsInstanced? If so, how? If not, what would be the cheapest way to implement the scissor test in the fragment shader and are there any performance considerations in doing so?
No, you can't change the scissor rectangle during a draw call. A couple of options come to mind:
Specify the clip rectangle as a per instance vertex attribute, pass it through to the fragment shader, check for fragments to be within the rectangle, and discard them otherwise.
Apply user clip planes.
The second option should be more efficient, since it allows you to clip away the geometry outside the clip rectangle before it enters the fragment shader.
To apply user clip planes with modern OpenGL, you assign values to elements of the predefined gl_ClipDistance array in the vertex shader, and enable GL_CLIP_DISTANCEi. Here are the main steps to apply this in your use case:
Introduce a per-instance vertex attribute that defines the clip rectangle. You can nicely fit the rectangle description into an attribute of type vec4. For example, define the attribute to contain the values (xMin, xMax, yMin, yMax) for the rectangle you want to clip against.
In the vertex shader, define the attribute:
in vec4 ClipRect;
and set the clip distances based on your vertex coordinates coord:
gl_ClipDistance[0] = coord.x - ClipRect[0];
gl_ClipDistance[1] = ClipRect[1] - coord.x;
gl_ClipDistance[2] = coord.y - ClipRect[2];
gl_ClipDistance[3] = ClipRect[3] - coord.y;
Note that the assigned distance must be positive for points you want to keep, negative for points you want to clip. You will also have to make sure that the coordinates you use here are in the same coordinate system that the rectangle boundaries are specified in.
In the client code, before starting to draw, enable the clip planes:
glEnable(GL_CLIP_DISTANCE0);
glEnable(GL_CLIP_DISTANCE1);
glEnable(GL_CLIP_DISTANCE2);
glEnable(GL_CLIP_DISTANCE3);
Related
I am making a retro-style game with OpenGL, and I want to draw my own cubemaps for it. Here is an example of one:
As you can tell, there is no perspective warping anywhere; each face is fully equiangular. When using this as a cubemap, the result is this:
As you can see, it looks box-y, and not spherical at all. I know of a solution to this, which is to remap each point on the cubemap to a a sphere position. I have done this manually by creating a sphere mesh and mapping the cubemap texture onto it (and then rendering that to an environment map), but this is time-consuming and complicated.
I seek a different solution: in my fragment shader, I hope to remap the sampling ray to a sphere position, instead of a cube position. Here is my original fragment shader, without any changes:
#version 400 core
in vec3 cube_edge;
out vec3 color;
uniform samplerCube skybox_sampler;
void main(void) {
color = texture(skybox_sampler, cube_edge).rgb;
}
I can get a ray that maps to the sphere by just normalizing cube_edge, but that doesn't change anything, for some reason. After messing around a bit, I tried this mapping, which almost works, but not quite:
vec3 sphere_edge = vec3(cube_edge.x, normalize(cube_edge).y, cube_edge.z);
As you can see, some faces become spherical in nature, whereas the top face warps inwards, instead of outwards.
I also tried the results from this site: http://mathproofs.blogspot.com/2005/07/mapping-cube-to-sphere.html, but the faces were not curved outwards enough.
I have been stuck on this for so long now - if you know how I can change my cube to sphere mapping in my fragment shader, or if that's even possible, please let me know!
As you can tell, there is no perspective warping anywhere; each face is fully equiangular.
This premise is incorrect. You hand-drew some images; this doesn't make them equiangular.
'Equiangular cubemap' (EAC) specifically means a cubemap remapped by this formula (section 2.4):
u = 4/pi * atan(u)
v = 4/pi * atan(v)
Let's recognize first that the term is misleading, because even though EAC aims at reducing the variation in sampling rate, the sampling rate is not constant. In fact no 2d projection of any part of a sphere can truly be equi-angular; this is a mathematical fact.
Nonetheless, we can try to apply this correction. Implemented in GLSL fragment shader as:
d /= max(abs(d.x), max(abs(d.y), abs(d.z));
d = atan(d)/atan(1);
gives the following result:
Compare it with the uncorrected d:
As you can see the EAC projection shrinks the pixels in the middle by a little bit, and expands them near the corners, so that they cover more equal area.
Instead, it appears that you want a cylindrical projection around the horizon. It can be implemented like so:
d /= length(d.xy);
d.xy /= max(abs(d.x), abs(d.y));
d.xy = atan(d.xy)/atan(1);
Which gives the following result:
However there's no artifact-free way to fit the top/bottom square faces of the cube onto the circular faces of the cylinder -- which is why you see the artifacts there.
Bottom-line: you cannot fit the image that you drew onto a sphere in a visually pleasing way. You should instead re-focus your effort on alternative ways of authoring your environment map. I recommend you try using an equidistant cylindrical projection for the horizon, cap it with solid colors above/below a fixed latitude, and use billboards for objects that cannot be represented in that projection.
Your problem is that the size of the geometry on which the environment is placed is too small. You are not looking at the environment but at the inside of a small cube in which you are sitting. The environment map should behave as if you are always in the center of the map and the environment is infinitely far away. I suggest to draw the environment map on the far plane of the viewing frustum. You can do this by setting the z-component of the clip space position equal to the w-component in the vertex shader. If you set z to w, you guarantee that the final z value of the position will be 1.0. This is the z value of the far plane. (You can do that with Swizzling gl_Position = clipPos.xyww). It is quite sufficient to draw a cube and wrap the environment by looking up the map with the interpolated vertices of the cube. In the case of a samplerCube, the 3-dimensional texture coordinate is treated as a direction vector. You can use the vertex coordinate of the cube to look up the texture.
Vertex shader:
cube_edge = inVertex.xyz;
vec4 clipPos = projection * view * vec4(inVertex.xyz, 1.0);
gl_Position = clipPos.xyww;
Fragment shader:
color = texture(skybox_sampler, cube_edge).rgb;
The solution is also explained in detail at LearnOpenGL - Cubemap.
I have a cube that can be rotated using mouse navigation in PyOpenGL. I want to create small sections on each face of the cube and render the different sections with different colors/illumination. It is like having a certain light source and the cube is considered as a room being illuminated with the light source. How do I set my desired values for each section ? Is it possible to do so ?
Extend your fragment shader to expect some kind of interpolated value (Either just add "face coordinates" to your vertices that goes from 0 to 1 across the face, or by transforming your texture uv coordinates if you have them). Then you can just use if-else inside the fragment shader.
i.e.
if (coords.x < 0.5 && coords.y < 0.5) // one quarter of the face
Lightning=...
FragColor=...
else if ...
You could use a geometry shader and some exponential function to distribute the colors in each face. If i undertand correcltly you want something like this:
In this case you will pass the colors you want as vertex attributes for the face you want and in the geometry shader you will compute the distance from each face-vertex to the face's center. Then you will pass that distance to the fragment shader. Passing the distance to the Fshader will interpolate the value for each fragment.
As far as I understand, location of a point/pixel cannot be a fraction, at least on a raster graphics system where hardwares use pixels to display images.
Then, why and how does OpenGL use fractional values for plotting pixels?
For example, how is it possible: glVertex2f(0.15f, 0.51f); ?
This command does not plot any pixels. It merely defines the location of a point in 3D space (you'll notice that there are 3 coordinates, while for a pixel on the screen you'd only need 2). This is the starting point for the OpenGL pipeline. This point then goes through a lot of transformations before it ends up on the screen.
Also, the coordinates are unitless. For example, you can say that your viewport is between 0.0f and 1.0f, then these coordinates make a lot of sense. Basically you have to think of these point in terms of mathematics, not pixels.
I would suggest some reading on how OpenGL transformations work, for example here, here or the tutorial here.
The vectors you pass into OpenGL are not viewport positions but arbitrary numbers in some vector space. Only after a chain of transformations these numbers are mapped into viewport pixel positions. With the old fixed function pipeline this could be anything that can be represented by a vector–matrix multiplication.
These days, where everything is programmable (shaders) the mapping can very well be any kind of function you can think of. For example the values you pass into glVertex (immediate mode call, but available to shaders with OpenGL-2.1) may be interpreted as polar coordinates in the vertex shader:
This is a perfectly valid OpenGL-2.1 vertex shader that interprets the vertex position to be in polar coordinates. Note that due to triangles and lines being straight edges and polar coordinates being curvilinear this gives good visual results only for points or highly tesselated primitives.
#version 110
void main() {
gl_Position =
gl_ModelViewProjectionMatrix
* vec4( gl_Vertex.y*vec2(sin(gl_Vertex.x),cos(gl_Vertex.x)) , 0, 1);
}
As you can see here the valus passed to glVertex are actually arbitrary, unitless components of vectors in some vector space. Only by applying some transformation to the viewport space these vectors gain meaning. Hence it makes no way to impose a certain value range onto the values that go into the vertex attribute.
Vertex and pixel are very different things.
It's quite possible to have all your vertices within one pixel (although in this case you probably need help with LODing).
You might want to start here...
http://www.glprogramming.com/blue/ch01.html
Specifically...
Primitives are defined by a group of one or more vertices. A vertex defines a point, an endpoint of a line, or a corner of a polygon where two edges meet. Data (consisting of vertex coordinates, colors, normals, texture coordinates, and edge flags) is associated with a vertex, and each vertex and its associated data are processed independently, in order, and in the same way.
And...
Rasterization produces a series of frame buffer addresses and associated values using a two-dimensional description of a point, line segment, or polygon. Each fragment so produced is fed into the last stage, per-fragment operations, which performs the final operations on the data before it's stored as pixels in the frame buffer.
For your example, before glVertex2f(0.15f, 0.51f) is on the screen, there are many transforms to be done. Making complex thing crudely simpler, after moving your vertex to view space (applying camera position and direction), the magic here is (1) projection matrix, and (2) viewport setting.
Internally, OpenGL "screen coordinates" are in a cube (-1, -1, -1) - (1, 1, 1), :
http://www.matrix44.net/cms/wp-content/uploads/2011/03/ogl_coord_object_space_cube.png
Projection matrix 'squeezes' the frustum in this cube (which you do in vertex shader), assuming you have perspective transform - if projection is orthogonal, the projection is just a tube, limited by near and far values (and like in both cases, scaling factors):
http://www.songho.ca/opengl/files/gl_projectionmatrix01.png
EDIT: Maybe better example here:
http://www.opengl-tutorial.org/beginners-tutorials/tutorial-3-matrices/#The_Projection_matrix
(EDIT: The Z-coordinate is used as depth value) When fragments are finally transferred to pixels on texture/framebuffer/screen, these are multiplied with viewport settings:
https://www3.ntu.edu.sg/home/ehchua/programming/opengl/images/GL_2DViewportAspectRatio.png
Hope this helps!
I'm creating game that uses orthogonal view(2D). I'm trying to understand the value of gl_Position in vertex shader.
From what I understand x and y coordinates translate to screen position in range of -1 to 1, but I'm quite confused with role of the z and w, I only know that the w value should be set to 1.0
For the moment I just use gl_Position.xyw = vec3(Position, 1.0);, where Position is 2D vertex position
I use OpenGL 3.2.
Remember that openGL must also work for 3D and it's easier to expose the 3D details than to create a new interface for just 2D.
The Z component is to set the depth of the vertex, points outside -1,1 (after perspective divide) will not be drawn and for the values between -1,1 it will be checked against a depth buffer to see if the fragment is behind some previously drawn triangle and not draw it if it should be hidden.
The w component is for a perspective divide and allowing the GPU to interpolate the values in a perspective correct way. Otherwise the textures looks weird.
So I'm supposed to Texture Map a specific model I've loaded into a scene (with a Framebuffer and a Planar Pinhole Camera), however I'm not allowed to use OpenGL and I have no idea how to do it otherwise (we do use glDrawPixels for other functionality, but that's the only function we can use).
Is anyone here able enough to give me a run-through on how to texture map without OpenGL functionality?
I'm supposed to use these slides: https://www.cs.purdue.edu/cgvlab/courses/334/Fall_2014/Lectures/TMapping.pdf
But they make very little sense to me.
What I've gathered so far is the following:
You iterate over a model, and assign each triangle "texture coordinates" (which I'm not sure what those are), and then use "model space interpolation" (again, I don't understand what that is) to apply the texture with the right perspective.
I currently have my program doing the following:
TL;DR:
1. What is model space interpolation/how do I do it?
2. What explicitly are texture coordinates?
3. How, on a high level (in layman's terms) do I texture map a model without using OpenGL.
OK, let's start by making sure we're both on the same page about how the color interpolation works. Lines 125 through 143 set up three vectors redABC, greenABC and blueABC that are used to interpolate the colors across the triangle. They work one color component at a time, and each of the three vectors helps interpolate one color component.
By convention, s,t coordinates are in source texture space. As provided in the mesh data, they specify the position within the texture of that particular vertex of the triangle. The crucial thing to understand is that s,t coordinates need to be interpolated across the triangle just like colors.
So, what you want to do is set up two more ABC vectors: sABC and tABC, exactly duplicating the logic used to set up redABC, but instead of using the color components of each vertex, you just use the s,t coordinates of each vertex. Then for each pixel, instead of computing ssiRed etc. as unsigned int values, you compute ssis and ssit as floats, they should be in the range 0.0f through 1.0f assuming your source s,t values are well behaved.
Now that you have an interpolated s,t coordinate, multiply ssis by the texel width of the texture, and ssit by the texel height, and use those coordinates to fetch the texel. Then just put that on the screen.
Since you are not using OpenGL I assume you wrote your own software renderer to render that teapot?
A texture is simply an image. A texture coordinate is a 2D position in the texture. So (0,0) is bottom-left and (1,1) is top-right. For every vertex of your 3D model you should store a 2D position (u,v) in the texture. That means that at that vertex, you should use the colour the texture has at that point.
To know the UV texture coordinate of a pixel in between vertices you need to interpolate the texture coordinates of the vertices around it. Then you can use that UV to look up the colour in the texture.