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!
Related
I read the Khronos wiki on this, but I don't really understand what it is saying. What exactly does textureGrad do?
I think it samples multiple mipmap levels and computes some color mixing using the explicit derivative vectors given to it, but I am not sure.
When you sample a texture, you need the specific texture coordinates to sample the texture data at. For sake of simplicity, I'm going to assume a 2D texture, so the texture coordinates are a 2D vector (s,t). (The explanation is analogous for other dimensionalities).
If you want to texture-map a triangle, one typically uses one of two strategies to get to the texture coordinates:
The texture coordinates are part of the model. Every vertex contains the 2D texture coordinates as a vertex attribute. During rasterization, those texture coordinates are interpolated across the primitive.
You specify a mathematic mapping. For example, you could define some function mapping the 3D object coordinates to some 2D texture coordinates. You can for example define some projection, and project the texture onto a surface, just like a real projector would project an image onto some real-world objects.
In either case, each fragment generated when rasterizing the typically gets different texture coordinates, so each drawn pixel on the screen will get a different part of the texture.
The key point is this: each fragment has 2D pixel coordinates (x,y) as well as 2D texture coordinates (s,t), so we can basically interpret this relationship as a mathematical function:
(s,t) = T(x,y)
Since this is a vector function in the 2D pixel position vector (x,y), we can also build the partial derivatives along x direction (to the right), and y direction (upwards), which are telling use the rate of change of the texture coordinates along those directions.
And the dTdx and dTdy in textureGrad are just that.
So what does the GPU need this for?
When you want to actually filter the texture (in contrast to simple point sampling), you need to know the pixel footprint in texture space. Each single fragment represents the area of one pixel on the screen, and you are going to use a single color value from the texture to represent the whole pixel (multisampling aside). The pixel footprint now represent the actual area the pixel would have in texture space. We could calculate it by interpolating the texcoords not for the pixel center, but for the 4 pixel corners. The resulting texcoords would form a trapezoid in texture space.
When you minify the texture, several texels are mapped to the same pixel (so the pixel footprint is large in texture space). When you maginify it, each pixel will represent only a fraction of the corresponding texel (so the footprint is quiete small).
The texture footprint tells you:
if the texture is minified or magnified (GL has different filter settings for each case)
how many texels would be mapped to each pixel, so which mipmap level would be appropriate
how much anisotropy there is in the pixel footprint. Each pixel on the screen and each texel in texture space is basically a square, but the pixel footprint might significantly deviate from than, and can be much taller than wide or the over way around (especially in situations with high perspective distortion). Classic bilinear or trilinear texture filters always use a square filter footprint, but the anisotropic texture filter will uses this information to
actually generate a filter footprint which more closely matches that of the actual pixel footprint (to avoid to mix in texel data which shouldn't really belong to the pixel).
Instead of calculating the texture coordinates at all pixel corners, we are going to use the partial derivatives at the fragment center as an approximation for the pixel footprint.
The following diagram shows the geometric relationship:
This represents the footprint of four neighboring pixels (2x2) in texture space, so the uniform grid are the texels, and the 4 trapezoids represent the 4 pixel footprints.
Now calculating the actual derivatives would imply that we have some more or less explicit formula T(x,y) as described above. GPUs usually use another approximation:
the just look at the actual texcoords the the neighboring fragments (which are going to be calculated anyway) in each 2x2 pixel block, and just approximate the footprint by finite differencing - the just subtracting the actual texcoords for neighboring fragments from each other.
The result is shown as the dotted parallelogram in the diagram.
In hardware, this is implemented so that always 2x2 pixel quads are shaded in parallel in the same warp/wavefront/SIMD-Group. The GLSL derivative functions like dFdx and dFdy simply work by subtracting the actual values of the neighboring fragments. And the standard texture function just internally uses this mechanism on the texture coordinate argument. The textureGrad functions bypass that and allow you to specify your own values, which means you control the what pixel footprint the GPU assumes when doing the actual filtering / mipmap level selection.
I am drawing a stack of decals on a quad. Same geometry, different textures. Z-fighting is the obvious result. I cannot control the rendering order or use glPolygonoffset due to batched rendering. So I adjust depth values inside the vertex shader.
gl_Position = uMVPMatrix * pos;
gl_Position.z += aDepthLayer * uMinStep * gl_Position.w;
gl_Position holds clip coordinates. That means a change in z will move a vertex along its view ray and bring it to the front or push it to the back. For normalized device coordinates the clip coords get divided by gl_Position.w (=-Zclip). As a result the depth buffer does not have linear distribution and has higher resolution towards the near plane. By premultiplying gl_Position.w that should be fixed and I should be able to apply a flat amount (uMinStep) to the NDC.
That minimum step should be something like 1/(2^GL_DEPTH_BITS -1). Or, since NDC space goes from -1.0 to 1.0, it might have to be twice that amount. However it does not work with these values. The minStep is roughly 0.00000006 but it does not bring a texture to the front. Neither when I double that value. If I drop a zero (scale by 10), it works. (Yay, thats something!)
But it does not work evenly along the frustum. A value that brings a texture in front of another while the quad is close to the near plane does not necessarily do the same when the quad is close to the far plane. The same effect happens when I make the frustum deeper. I would expect that behaviour if I was changing eye coordinates, because of the nonlinear z-Buffer distribution. But it seems that premultiplying gl_Position.w is not enough to counter that.
Am I missing some part of the transformations that happen to clip coords? Do I need to use a different formula in general? Do I have to include the depth range [0,1] somehow?
Could the different behaviour along the frustum be a result of nonlinear floating point precision instead of nonlinear z-Buffer distribution? So maybe the calculation is correct, but the minStep just cannot be handled correctly by floats at some point in the pipeline?
The general question: How do I calculate a z-Shift for gl_Position (clip coordinates) that will create a fixed change in the depth buffer later? How can I make sure that the z-Shift will bring one texture in front of another no matter where in the frustum the quad is placed?
Some material:
OpenGL depth buffer faq
https://www.opengl.org/archives/resources/faq/technical/depthbuffer.htm
Same with better readable formulas (but some typos, be careful)
https://www.opengl.org/wiki/Depth_Buffer_Precision
Calculation from eye coords to z-buffer. Most of that happens already when I multiply the projection matrix.
http://www.sjbaker.org/steve/omniv/love_your_z_buffer.html
Explanation about the elements in the projection matrix that turn into the A and B parts in most depth buffer calculation formulas.
http://www.songho.ca/opengl/gl_projectionmatrix.html
So when drawing a rectangle on OpenGL, if you give the corners of the rectangle texture coordinates of (0,0), (1,0), (1,1) and (0, 1), you'll get the standard rectangle.
However, if you turn it into something that's not rectangular, you'll get a weird stretching effect. Just like the following:
I saw from this page below that this can be fixed, but the solution given is only for trapezoidal values only. Also, I have to be doing this over many rectangles.
And so, the questions is, what is the proper way, and most efficient way to get the right "4D" texture coordinates for drawing stretched quads?
Implementations are allowed to decompose quads into two triangles and if you visualize this as two triangles you can immediately see why it interpolates texture coordinates the way it does. That texture mapping is correct ... for two independent triangles.
That diagonal seam coincides with the edge of two independently interpolated triangles.
Projective texturing can help as you already know, but ultimately the real problem here is simply interpolation across two triangles instead of a single quad. You will find that while modifying the Q coordinate may help with mapping a texture onto your quadrilateral, interpolating other attributes such as colors will still have serious issues.
If you have access to fragment shaders and instanced vertex arrays (probably rules out OpenGL ES), there is a full implementation of quadrilateral vertex attribute interpolation here. (You can modify the shader to work without "instanced arrays", but it will require either 4x as much data in your vertex array or a geometry shader).
Incidentally, texture coordinates in OpenGL are always "4D". It just happens that if you use something like glTexCoord2f (s, t) that r is assigned 0.0 and q is assigned 1.0. That behavior applies to all vertex attributes; vertex attributes are all 4D whether you explicitly define all 4 of the coordinates or not.
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.
Im having a bit of trouble with getting a depth value that I'm storing in a Float texture (or rather i don't understand the values). Essentially I am creating a deffered renderer, and in one of the passes I am storing the depth in the alpha component of a floating point render target. The code for that shader looks something like this
Define the clip position as a varying
varying vec4 clipPos;
...
In the vertex shader assign the position
clipPos = gl_Position;
Now in the fragment shader I store the depth:
gl_FragColor.w = clipPos.z / clipPos.w;
This by and large works. When I access this render target in any subsequent shaders I can get the depth. I.e something like this:
float depth = depthMap.w;
Am i right to assume that 0.0 is right in front of the camera and 1 is in the distance? Because I am doing some fog calculations based on this but they don't seem to be correct.
fogFactor = smoothstep( fogNear, fogFar, depth );
fogNear and fogFar are uniforms I send to the shader. When the fogNear is set to 0, I would have thought I get a smooth transition of fog from right in front of the camera to its draw distance. However this is what I see:
When I set the fogNear to 0.995, then I get something more like what Im expecting:
Is that correct, it just doesn't seem right to me? (The scale of the geometry is not really small / too large and neither is the camera near and far too large. All the values are pretty reasonable)
There are two issues with your approach:
You assume the depth is in the range of [0,1], buit what you use is clipPos.z / clipPos.w, which is NDC z coord in the range [-1,1]. You might be better of by directly writing the window space z coord to your depth texture, which is in [0,1] and will simply be gl_FragCoord.z.
The more serious issue that you assume a linear depth mapping. However, that is not the case. The NDC and window space z value is not a linear representation of the distance to the camera plane. It is not surprisinng that anything you see in the screenshot is very closely to 1. Typical, fog calculations are done in eye space. However, since you only need the z coord here, you simply could store the clip space w coordinate - since typically, that is just -z_eye (look at the last row of your projection matrix). However, the resulting value will be not in any normailized range, but in [near,far] that you use in your projection matrix - but specifying fog distances in eye space units (which normally are indentical to world space units) is more intuitive anyway.