Is there a way using OpenGL or GLUT to project a point from the model-view matrix into an associated texture matrix? If not, is there a commonly used library that achieves this? I want to modify the texture of an object according to a ray cast in 3D space.
The simplest case would be:
A ray is cast which intersects a quad, mapped with a single texture.
The point of intersection is converted to a value in texture space clamped between [0.0,1.0] in the x and y axis.
A 3x3 patch of pixels centered around the rounded value of the resulting texture point is set to an alpha value of 0.( or another RGBA value which is convenient, for the desired effect).
To illustrate here is a more complex version of the question using a sphere, the pink box shows the replaced pixels.
I just specify texture points for mapping in OpenGL, I don't actually know how the pixels are projected onto the sphere. Basically I need to to the inverse of that projection, but I don't quite know how to do that math, especially on more complex shapes like a sphere or an arbitrary convex hull. I assume that you can somehow find a planar polygon that makes up the shape, which the ray is intersecting, and from there the inverse projection of a quad or triangle would be trivial.
Some equations, articles and/or example code would be nice.
There are a few ways you could accomplish what you're trying to do:
Project a world coordinate point into normalized device coordinates (NDCs) by doing the model-view and projection transformation matrix multiplications by yourself (or if you're using old-style OpenGL, call gluProject), and perform the perspective division step. If you use a depth coordinate of zero, this would correspond to intersecting your ray at the imaging plane. The only other correction you'd need to do map from NDCs (which are in the range [-1,1] in x and y) into texture space by dividing the resulting coordinate by two, and then shifting by .5.
Skip the ray tracing all together, and bind your texture as a framebuffer attachment to a framebuffer object, and then render a big point (or sprite) that modifies the colors in the neighborhood of the intersection as you want. You could use the same model-view and projection matrices, and will (probably) only need to update the viewport to match the texture resolution.
So I found a solution that is a little complicated, but does the trick.
For complex geometry you must determine which quad or triangle was intersected, and use this as the plane. The quad must be planar(obviously).
Draw a plane in the identity matrix with dimensions 1x1x0, map the texture on points identical to the model geometry.
Transform the plane, and store the inverse of each transform matrix in a stack
Find the point at which the the plane is intersected
Transform this point using the inverse matrix stack until it returns to identity matrix(it should have no depth(
Convert this point from 1x1 space into pixel space by multiplying the point by the number of pixels and rounding. Or start your 2D combining logic here.
Related
OpenGL spec:
It says: However, depth values for polygons must be interpolated by (14.10).
Why? Are the z coordinates depth values in camera space? If so, we should use perspective correctly barycentric coordinates to interpolate them, isn't it?(like equation 14.9)
Update:
So the z coordinates are NDC coordinates(which already divided by w). I have a small demo which implement a rasterizer. When I use linear interpolation of the NDC z coordinates, the result is a bit unusual(image below). While I use perspective correctly interpolation of camera z coordinates, the result is ok.
This is the perspective projection matrix I use:
Why? Are the z coordinates depth values in camera space? If so, we should use perspective correctly barycentric coordinates to interpolate them, isn't it?
No, they are not. They are in window space, meaning they already have been divided by w. It is correct that if you wanted to interpolate camrea space z, you would have to apply perspective correction. But for NDC and window space Z this would be wrong - after all, the perspective transformation (as achieved by perspective projection matrix and perspective divide) still maps straight lines to straight lines, and flat trinagles to flat triangles. That's why we use the hyperbolically distorted Z values as depth in the first place. This is also a property that is exploited for the hierarchical depth test optimization. Have a look at my answer here for some more details, including a few diagrams.
Im making an editor in which I want to build a terrain map. I want to use the mouse to increase/decrease terrain altitude to create mountains and lakes.
Technically I have a heightmap I want to modify at a certain texcoord that I pick out with my mouse. To do this I first go from screen coordinates to world position - I have done that. The next step, going from world position to picking the right texture coordinate puzzles me though. How do I do that?
If you are using a simple hightmap, that you use as a displacement map in lets say the y direction. The base mesh lays in the xz plain (y=0).
You can discard the y coordinate from world coordinate that you have calculated and you get the point on the base mesh. From there you can map it to texture space the way, you map your texture.
I would not implement it that way.
I would render the scene to a framebuffer and instead of rendering a texture the the mesh, colorcode the texture coordinate onto the mesh.
If i click somewhere in screen space, i can simple read the pixel value from the framebuffer and get the texture coordinate directly.
The rendering to the framebuffer should be very inexpensive anyway.
Assuming your terrain is a simple rectangle you first calculate the vector between the mouse world position and the origin of your terrain. (The vertex of your terrain quad where the top left corner of your height map is mapped to). E.g. mouse (50,25) - origin(-100,-100) = (150,125).
Now divide the x and y coordinates by the world space width and height of your terrain quad.
150 / 200 = 0.75 and 125 / 200 = 0.625. This gives you the texture coordinates, if you need them as pixel coordinates instead simply multiply with the size of your texture.
I assume the following:
The world coordinates you computed are those of the mouse pointer within the view frustrum. I name them mouseCoord
We also have the camera coordinates, camCoord
The world consists of triangles
Each triangle point has texture coordiantes, those are interpolated by barycentric coordinates
If so, the solution goes like this:
use camCoord as origin. Compute the direction of a ray as mouseCoord - camCoord.
Compute the point of intersection with a triangle. Naive variant is to check for every triangle if it is intersected, more sophisticated would be to rule out several triangles first by some other algorithm, like parting the world in cubes, trace the ray along the cubes and only look at the triangles that have overlappings with the cube. Intersection with a triangle can be computed like on this website: http://www.lighthouse3d.com/tutorials/maths/ray-triangle-intersection/
Compute the intersection points barycentric coordinates with respect to that triangle, like that: https://www.scratchapixel.com/lessons/3d-basic-rendering/ray-tracing-rendering-a-triangle/barycentric-coordinates
Use the barycentric coordinates as weights for the texture coordinates of the corresponding triangle points. The result are the texture coordinates of the intersection point, aka what you want.
If I misunderstood what you wanted, please edit your question with additional information.
Another variant specific for a height map:
Assumed that the assumptions are changed like that:
The world has ground tiles over x and y
The ground tiles have height values in their corners
For a point within the tile, the height value is interpolated somehow, like by bilinear interpolation.
The texture is interpolated in the same way, again with given texture coordinates for the corners
A feasible algorithm for that (approximative):
Again, compute origin and direction.
Wlog, we assume that the direction has a higher change in x-direction. If not, exchange x and y in the algorithm.
Trace the ray in a given step length for x, that is, in each step, the x-coordinate changes by that step length. (take the direction, multiply it with step size divided by it's x value, add that new direction to the current position starting at the origin)
For your current coordinate, check whether it's z value is below the current height (aka has just collided with the ground)
If so, either finish or decrease step size and do a finer search in that vicinity, going backwards until you are above the height again, then maybe go forwards in even finer steps again et cetera. The result are the current x and y coordinates
Compute the relative position of your x and y coordinates within the current tile. Use that for weights for the corner texture coordinates.
This algorithm can theoretically jump over very thin tops. Choose a small enough step size to counter that. I cannot give an exact algorithm without knowing what type of interpolation the height map uses. Might be not the worst idea to create triangles anyway, out of bilinear interpolated coordinates maybe? In any case, the algorithm is good to find the tile in which it collides.
Another variant would be to trace the ray over the points at which it's x-y-coordinates cross the tile grid and then look if the z coordinate went below the height map. Then we know that it collides in this tile. This could produce a false negative if the height can be bigger inside the tile than at it's edges, as certain forms of interpolation can produce, especially those that consider the neighbour tiles. Works just fine with bilinear interpolation, though.
In bilinear interpolation, the exact intersection can be found like that: Take the two (x,y) coordinates at which the grid is crossed by the ray. Compute the height of those to retrieve two (x,y,z) coordinates. Create a line out of them. Compute the intersection of that line with the ray. The intersection of those is that of the intersection with the tile's height map.
Simplest way is to render the mesh as a pre-pass with the uvs as the colour. No screen to world needed. The uv is the value at the mouse position. Just be careful though with mips/filtering etv
How can one mix ortographic and perspective projection in openGL?
Some 2d elements have to be drawn in screen space (no scaling, rotation, etc..)
These 2d elements have a z position, they have to appear in front/behind of other 3d elements.
So i set up orographic projection, draw all 2d elements, then setup perspective projection and draw all 3d elements.
The result is that all 2d elements are drawn on top. It seems that the z values from the orto projection and the z values from the perspective projection are not compatible (GL_DEPTH_TEST).
Separately all 2d and all 3d elements work fine, the problem is when i try to mix them.
Does the prespective projection changes the z values? In what way?
Is it possible to use z values from orto projection mixed with z values from perspective projection for depth test, or this whole concept is flawed?
Bare opengl1.5
It seems that the z values from the orto projection and the z values from the perspective projection are not compatible (GL_DEPTH_TEST).
That is indeed the case. Perspective transformation maps the Z values nonlinear to the depth buffer values. The usual way to address this problem is to copy the depth buffer after the perspective pass into a depth texture and use that as an additional input in the fragment shader of the orthographic drawn stuff, reverse the nonlinearity in the depth input and compare the incoming Z coordinate with that; then discard appropriately.
It's also possible to emit linear depth values in the perspective drawn geometry fragment shaders, however the depth nonlinearity of perspective projection has its purpose; without it you loose depth precision where it matters most, close to the point of view.
How can i render a textured plane at some z-pos to be visible towards infinity?
I could achieve this by drawing really huge plane, but if i move my camera off the ground to higher altitude, then i would start to see the plane edges, which i want to avoid being seen.
If this is even possible, i would prefer non-shader method.
Edit: i tried with the 4d coordinate system as suggested, but: it works horribly bad. my textures will get distorted even at camera position 100, so i would have to draw multiple textured quads anyways. perhaps i could do that, and draw the farthest quads with the 4d coordinate system? any better ideas?
Edit2: for those who dont have a clue what opengl texture distortion is, here's example from the tests i did with 4d vertex coords:
(in case image not visible: http://img828.imageshack.us/img828/469/texturedistort.jpg )
note that it only happens when camera gets far enough, in this case its only 100.0 units away from middle! (middle = (0,0) where my 4 triangles starts to go towards infinity). usually this happens around at 100000.0 or something. but with 4d vertices it seems to happen earlier for some reason.
You cannot render an object of infinite size.
You are more than likely confusing the concept of projection with rendering objects of infinite size. A 4D homogeneous coordinate who's W is 0 represents a 3D position that is at infinity relative to the projection. But that doesn't mean a point infinitely far from the camera; it means a point infinitely close to the camera. That is, it represents a point who's Z coordinate (before multiplication with the perspective projection matrix) was equal to the camera position (in camera space, this is 0).
See under perspective projection, a point that is in the same plane as the camera is infinitely far away on the X and Y axes. That is the nature of the perspective projection. 4D homogeneous coordinates allow you to give them all finite numbers, and therefore you can do useful mathematics to them (like clipping).
4D homogeneous coordinates do not allow you to represent an infinitely large surface.
Drawing an infinitely large plane is easy - all you need is to compute the horizon line in screen coordinates. To do so, you have to simply take two non-collinear 4D directions (say, [1, 0, 0, 0] and [0, 0, 1, 0]), then compute their position on the screen (by multiplying manually with the view-matrix and the projection matrix, and then clipping into viewport coordinates. When you have these two points, you can compute a 2D line through the screen and clip it against it. There, you have your infinity plane (the lower polygon). However, it is difficult to display a texture on this plane, because it would be infinitely large. But if your texture is simple (e.g. a grid), then you can compute it yourself with 4D coordinates, using the same schema like above - computing points and their corresponding vanishing point and connecting them.
I have a very general question. I wish to determine the boundary points of a number of objects (comprising 30-50 closed polygons (z) each having around 300 points(x,y,z)). I am working with a fixed viewport which is rotated about x,y and z-axes (alpha, beta, gamma) wrt origin of coordinate system for polygons.
As I see it there are two possibilities: perspective projection or raytracing. Perspective projection would seem to requires a large number of matrix operations for each point to determine its position is within or without the viewport.
Or given the large number of points would I better to raytrace the viewport pixels to object?
i.e. determine whether there is an intersection and then whether intersection occurs within or without object(s).
In either case I will write this result as 0 (outside) or 1 (inside) to 200x200 an integer matrix representing the viewport
Thank you in anticipation
Perspective projection (and then scan-converting the polygons in image coordinates) is going to be a lot faster.
The matrix transform that is required in the case of perspective projection (essentially the world-to-camera matrix) is required in exactly the same way when raytracing. However, with perspective projection, you're only transforming the corner points, whereas with raytracing, you're transforming all the points in the image.
You should be able to use perspective projection and a perspective projection matrix to compute the position of the vertices in screen space? It's hard to understand what you want to do really. If you want to create an image of that 3D scene then with only few polygons it would be hard to see any difference anyway between ray tracing and rasterisation if your code is optimised (you will still need to use an acceleration structure for the ray tracing approach), however yes rasterisation is likely to be faster anyway.
Now if you need to compute the distance from between the eye (the camera's origin) and the geometry visible through the camera's view, the I don't see why you can't use the depth value of any sample for any pixel in the image and use the inverse of the perspective projection matrix to find its distance in camera space.
Why is speed an issue in your problem? Otherwise use RT indeed.
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