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.
Related
I heard clipping should be done in clipping coordinate system.
The book suggests a situation that a line is laid from behind camera to in viewing volume. (We define this line as PQ, P is behind camera point)
I cannot understand why it can be a problem.
(The book says after finishing normalizing transformation, the P will be laid in front of camera.)
I think before making clipping coordinate system, the camera is on original point (0, 0, 0, 1) because we did viewing transformation.
However, in NDCS, I cannot think about camera's location.
And I have second question.
In vertex shader, we do model-view transformation and then projection transformation. Finally, we output these vertices to rasterizer.
(some vertices's w is not equal to '1')
Here, I have curiosity. The rendering pipeline will automatically do division procedure (by w)? after finishing clipping.
Sometimes not all the model can be seen on screen, mostly because some objects of it lie behind the camera (or "point of view"). Those objects are clipped out. If just a part of the object can't be seen, then just that part must be clipped leaving the rest as seen.
OpenGL clips
OpenGL does this clipping in Clipping Coordinate Space (CCS). This is a cube of size 2w x 2w x 2w where 'w' is the fourth coordinate resulting of (4x4) x (4x1) matrix and point multiplication. A mere comparison of coordinates is enough to tell if the point is clipped or not. If the point passes the test then its coordinates are divided by 'w' (so called "perspective division"). Notice that for ortogonal projections 'w' is always 1, and with perspective it's generally not 1.
CPU clips
If the model is too big perhaps you want to save GPU resources or improve the frame rate. So you decide to skip those objects that are going to get clipped anyhow. Then you do the maths on your own (on CPU) and only send to the GPU the vertices that passed the test. Be aware that some objects may have some vertices clipped while other vertices of this same object may not.
Perhaps you do send them to GPU and let it handle these special cases.
You have a volume defined where only objects inside are seen. This volume is defined by six planes. Let's put ourselves in the camera and look at this volume: If your projection is perspective the six planes build a "fustrum", a sort of truncated pyramid. If your projection is orthogonal, the planes form a parallelepiped.
In order to clip or not to clip a vertex you must use the distance form the vertex to each of these six planes. You need a signed distance, this means that the sign tells you what side of the plane is seen form the vertex. If any of the six distance signs is not the right one, the vertex is discarded, clipped.
If a plane is defined by equation Ax+By+Cz+D=0 then the signed distance from p1,p2,p3 is (Ap1+Bp2+Cp3+D)/sqrt(AA+BB+C*C). You only need the sign, so don't bother to calculate the denominator.
Now you have all tools. If you know your planes on "camera view" you can calculate the six distances and clip or not the vertex. But perhaps this is an expensive operation considering that you must transform the vertex coodinates from model to camera (view) spaces, a ViewModel matrix calculation. With the same cost you use your precalculated ProjectionViewModel matrix instead and obtain CCS coordinates, which are much easier to compare to '2w', the size of the CCS cube.
Sometimes you want to skip some vertices not due to they are clipped, but because their depth breaks a criteria you are using. In this case CCS is not the right space to work with, because Z-coordinate is transformed into [-w, w] range, depth is somehow "lost". Instead, you do your clip test in "view space".
I'm working with openGL but this is basically a math question.
I'm trying to calculate the projection matrix, I have a point on the view plane R(x,y,z) and the Normal vector of that plane N(n1,n2,n3).
I also know that the eye is at (0,0,0) which I guess in technical terms its the Perspective Reference Point.
How can I arrive the perspective projection from this data? I know how to do it the regular way where you get the FOV, aspect ration and near and far planes.
I think you created a bit of confusion by putting this question under the "opengl" tag. The problem is that in computer graphics, the term projection is not understood in a strictly mathematical sense.
In maths, a projection is defined (and the following is not the exact mathematical definiton, but just my own paraphrasing) as something which doesn't further change the results when applied twice. Think about it. When you project a point in 3d space to a 2d plane (which is still in that 3d space), each point's projection will end up on that plane. But points which already are on this plane aren't moving at all any more, so you can apply this as many times as you want without changing the outcome any further.
The classic "projection" matrices in computer graphics don't do this. They transfrom the space in a way that a general frustum is mapped to a cube (or cuboid). For that, you basically need all the parameters to describe the frustum, which typically is aspect ratio, field of view angle, and distances to near and far plane, as well as the projection direction and the center point (the latter two are typically implicitely defined by convention). For the general case, there are also the horizontal and vertical asymmetries components (think of it like "lens shift" with projectors). And all of that is what the typical projection matrix in computer graphics represents.
To construct such a matrix from the paramters you have given is not really possible, because you are lacking lots of parameters. Also - and I think this is kind of revealing - you have given a view plane. But the projection matrices discussed so far do not define a view plane - any plane parallel to the near or far plane and in front of the camera can be imagined as the viewing plane (behind the camere would also work, but the image would be mirrored), if you should need one. But in the strict sense, it would only be a "view plane" if all of the projected points would also end up on that plane - which the computer graphics perspective matrix explicitely does'nt do. It instead keeps their 3d distance information - which also means that the operation is invertible, while a classical mathematical projection typically isn't.
From all of that, I simply guess that what you are looking for is a perspective projection from 3D space onto a 2D plane, as opposed to a perspective transformation used for computer graphics. And all parameters you need for that are just the view point and a plane. Note that this is exactly what you have givent: The projection center shall be the origin and R and N define the plane.
Such a projection can also be expressed in terms of a 4x4 homogenous matrix. There is one thing that is not defined in your question: the orientation of the normal. I'm assuming standard maths convention again and assume that the view plane is defined as <N,x> + d = 0. From using R in that equation, we can get d = -N_x*R_x - N_y*R_y - N_z*R_z. So the projection matrix is just
( 1 0 0 0 )
( 0 1 0 0 )
( 0 0 1 0 )
(-N_x/d -N_y/d -N_z/d 0 )
There are a few properties of this matrix. There is a zero column, so it is not invertible. Also note that for every point (s*x, s*y, s*z, 1) you apply this to, the result (after division by resulting w, of course) is just the same no matter what s is - so every point on a line between the origin and (x,y,z) will result in the same projected point - which is what a perspective projection is supposed to do. And finally note that w=(N_x*x + N_y*y + N_z*z)/-d, so for every point fulfilling the above plane equation, w= -d/-d = 1 will result. In combination with the identity transform for the other dimensions, which just means that such a point is unchanged.
Projection matrix must be at (0,0,0) and viewing in Z+ or Z- direction
this is a must because many things in OpenGL depends on it like FOG,lighting ... So if your direction or position is different then you need to move this to camera matrix. Let assume your focal point is (0,0,0) as you stated and the normal vector is (0,0,+/-1)
Z near
is the distance between focal point and projection plane so znear is perpendicular distance of plane and (0,0,0). If assumption is correct then
znear=R.z
otherwise you need to compute that. I think you got everything you need for it
cast line from R with direction N
find closest point to focal point (0,0,0)
and then the z near is the distance of that point to R
Z far
is determined by the depth buffer bit width and z near
zfar=znear*(1<<(cDepthBits-1))
this is the maximal usable zfar (for mine purposes) if you need more precision then lower it a bit do not forget precision is higher near znear and much much worse near zfar. The zfar is usually set to the max view distance and znear computed from it or set to min focus range.
view angle
I use mostly 60 degree view. zang=60.0 [deg]
Common males in my region can see up to 90 degrees but that is peripherial view included the 60 degree view is more comfortable to view.
Females have a bit wider view ... but I did not heard any complains from them on 60 degree views ever so let assume its comfortable for them too...
Aspect
aspect ratio is determined by your OpenGL window dimensions xs,ys
aspect=(xs/ys)
This is how I set the projection matrix:
glMatrixMode(GL_PROJECTION);
glLoadIdentity();
gluPerspective(zang/aspect,aspect,znear,zfar);
// gluPerspective has inacurate tangens so correct perspective matrix like this:
double perspective[16];
glGetDoublev(GL_PROJECTION_MATRIX,perspective);
perspective[ 0]= 1.0/tan(0.5*zang*deg);
perspective[ 5]=aspect/tan(0.5*zang*deg);
glLoadMatrixd(perspective);
deg = M_PI/180.0
perspective is projection matrix copy I use it for mouse position conversions etc ...
If you do not correct the matrix then you will be off when using advanced things like overlapping more frustrum to get high precision depth range. I use this to obtain <0.1m,1000AU> frustrum with 24bit depth buffer and the inaccuracy would cause the images will not fit perfectly ...
[Notes]
if the focal point is not really (0,0,0) or you are not viewing in Z axis (like you do not have camera matrix but instead use projection matrix for that) then on basic scenes/techniques you will see no problem. They starts with use of advanced graphics. If you use GLSL then you can handle this without problems but fixed OpenGL function can not handle this properly. This is also called PROJECTION_MATRIX abuse
[edit1] few links
If your view is standard frustrum then write the matrix your self gluPerspective otherwise look here Projections for some ideas how to construct it
[edit2]
From your comment I see it like this:
f is your viewing point (axises are the global world axises)
f' is viewing point if R would be the center of screen
so create projection matrix for f' position (as explained above), create transform matrix to transform f' to f. The transformed f must have Z axis the same as in f' the other axises can be obtained by cross product and use that as camera or multiply booth together and use as abused Projection matrix
How to construct the matrix is explained in the Understanding transform matrices link from my earlier comments
I am confused about the position of objects in opengl .The eye position is 0,0,0 , the projection plane is at z = -1 . At this point , will the objects be in between the eye position and and the plane (Z =(0 to -1)) ? or its behind the projection plane ? and also if there is any particular reason for being so?
First of all, there is no eye in modern OpenGL. There is also no camera. There is no projection plane. You define these concepts by yourself; the graphics library does not give them to you. It is your job to transform your object from your coordinate system into clip space in your vertex shader.
I think you are thinking about projection wrong. Projection doesn't move the objects in the same sense that a translation or rotation matrix might. If you take a look at the link above, you can see that in order to render a perspective projection, you calculate the x and y components of the projected coordinate with R = V(ez/pz), where ez is the depth of the projection plane, pz is the depth of the object, V is the coordinate vector, and R is the projection. Almost always you will use ez=1, which makes that equation into R = V/pz, allowing you to place pz in the w coordinate allowing OpenGL to do the "perspective divide" for you. Assuming you have your eye and plane in the correct places, projecting a coordinate is almost as simple as dividing by its z coordinate. Your objects can be anywhere in 3D space (even behind the eye), and you can project them onto your plane so long as you don't divide by zero or invalidate your z coordinate that you use for depth testing.
There is no "projection plane" at z=-1. I don't know where you got this from. The classic GL perspective matrix assumes an eye space where the camera is located at origin and looking into -z direction.
However, there is the near plane at z<0 and eveything in front of the near plane is going to be clipped. You cannot put the near plane at z=0, because then, you would end up with a division by zero when trying to project points on that plane. So there is one reasin that the viewing volume isn't a pyramid with they eye point at the top but a pyramid frustum.
This is btw. also true for real-world eyes or cameras. The projection center lies behind the lense, so no object can get infinitely close to the optical center in either case.
The other reason why you want a big near clipping distance is the precision of the depth buffer. The whole depth range between the front and the near plane has to be mapped to some depth value with a limited amount of bits, typically 24. So you want to keep the far plane as close as possible, and shift away the near plane as far as possible. The non-linear mapping of the screen-space z coordinate makes this even more important, as that the precision is non-uniformely distributed over that range.
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.
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.
Most of this information can be found on www.scratchapixel.com