Variable gl_Position output from a GLSL vertex shader must have 4 coordinates. In OpenGL, it seems w coordinate is used to scale the vector, by dividing the other coordinates by it. What is the purpose of w in Vulkan?
Shaders and projections in Vulkan behave exactly the same as in OpenGL. There are small differences in depth ranges ([-1, 1] in OpenGL, [0, 1] in Vulkan) or in the origin of the coordinate system (lower-left in OpenGL, upper-left in Vulkan), but the principles are exactly the same. The hardware is still the same and it performs calculations in the same way both in OpenGL and in Vulkan.
4-component vectors serve multiple purposes:
Different transformations (translation, rotation, scaling) can be
represented in the same way, with 4x4 matrices.
Projection can also be represented with a 4x4 matrix.
Multiple transformations can be combined into one 4x4 matrix.
The .w component You mention is used during perspective projection.
All this we can do with 4x4 matrices and thus we need 4-component vectors (so they can be multiplied by 4x4 matrices). Again, I write about this because the above rules apply both to OpenGL and to Vulkan.
So for purpose of the .w component of the gl_Position variable - it is exactly the same in Vulkan. It is used to scale the position vector - during perspective calculations (projection matrix multiplication) original depth is modified by the original .w component and stored in the .z component of the gl_Position variable. And additionally, original depth is also stored in the .w component. After that (as a fixed-function step) hardware performs perspective division and divides position stored in the gl_Position variable by its .w component.
In orthographic projection steps performed by the hardware are exactly the same, but values used for calculations are different. So the perspective division step is still performed by the hardware but it does nothing (position is dived by 1.0).
gl_Position is a Homogeneous coordinates. The w component plays a role at perspective projection.
The projection matrix describes the mapping from 3D points of the view on a scene, to 2D points on the viewport. It transforms from eye space to the clip space, and the coordinates in the clip space are transformed to the normalized device coordinates (NDC) by dividing with the w component of the clip coordinates (Perspective divide).
At Perspective Projection the projection matrix describes the mapping from 3D points in the world as they are seen from of a pinhole camera, to 2D points of the viewport. The eye space coordinates in the camera frustum (a truncated pyramid) are mapped to a cube (the normalized device coordinates).
Perspective Projection Matrix:
r = right, l = left, b = bottom, t = top, n = near, f = far
2*n/(r-l) 0 0 0
0 2*n/(t-b) 0 0
(r+l)/(r-l) (t+b)/(t-b) -(f+n)/(f-n) -1
0 0 -2*f*n/(f-n) 0
When a Cartesian coordinate in view space is transformed by the perspective projection matrix, then the the result is a Homogeneous coordinates. The w component grows with the distance to the point of view. This cause that the objects become smaller after the Perspective divide, if they are further away.
In computer graphics, transformations are represented with matrices. If you want something to rotate, you multiply all its vertices (a vector) by a rotation matrix. Want it to move? Multiply by translation matrix, etc.
tl;dr: You can't describe translation along the z-axis with 3D matrices and vectors. You need at least 1 more dimension, so they just added a dummy dimension w. But things break if it's not 1, so keep it at 1 :P.
Anyway, now we begin with a quick review on matrix multiplication:
You basically put x above a, y above b, z above c. Multiply the whole column by the variable you just moved, and sum up everything in the row.
So if you were to translate a vector, you'd want something like:
See how x and y is now translated by az and bz? That's pretty awkward though:
You'd have to account for how big z is whenever you move things (what if z was negative? You'd have to move in opposite directions. That's cumbersome as hell if you just want to move something an inch over...)
You can't move along the z axis. You'll never be able to fly or go underground
But, if you can make sure z = 1 at all times:
Now it's much clearer that this matrix allows you to move in the x-y plane by a, and b amounts. Only problem is that you're conceptually levitating all the time, and you still can't go up or down. You can only move in 2D.
But you see a pattern here? With 3D matrices and 3D vectors, you can describe all the fundamental movements in 2D. So what if we added a 4th dimension?
Looks familiar. If we keep w = 1 at all times:
There we go, now you get translation along all 3 axis. This is what's called homogeneous coordinates.
But what if you were doing some big & complicated transformation, resulting in w != 1, and there's no way around it? OpenGL (and basically any other CG system I think) will do what's called normalization: divide the resultant vector by the w component. I don't know enough to say exactly why ('cause scaling is a linear transformation?), but it has favorable implications (can be used in perspective transforms). Anyway, the translation matrix would actually look like:
And there you go, see how each component is shrunken by w, then it's translated? That's why w controls scaling.
Related
I was quite confused on how the projection matrix worked so I researched and I discovered a few other things but after researching a few days, I just wanted to confirm my understanding is correct. I might use a few wrong terms but my brain was exhausted after writing this. A few topics I just researched briefly like screen coordinates and window transform so I didn’t write much about it and my knowledge might be incorrect. Is everything I’ve written here correct or mostly correct? Correct me on anything if I’m wrong.
What does the projection matrix do?
So the perspective projection matrix defines a frustum that is a truncated pyramid. Anything outside of that frustum/frustum range will be clipped. I'll get more on that later. The perspective projection matrix also adds perspective. To make the vertices follow the rules of perspective, the perspective projection matrix manipulates the vertex's w component (the homogenous component) depending on how far the vertex is from the viewer (the farther the vertex is, the higher the w coordinate will increase).
Why and how does the w component make the world look perceptive?
The w component makes the world look perceptive because in the perspective division (perspective division happens in the vertex post processing stage), when the x, y and z is divided by the w component, the vertex coordinate will be scaled smaller depending on how big the w component is. So essentially, the w component scales the object smaller the farther the object is.
Example:
Vertex position (1, 1, 2, 2).
Here, the vertex is 2 away from the viewer. In perspective division the x, y, and z will be divided by 2 because 2 is the w component.
(1/2, 1/2, 2/2) = (0.5, 0.5, 1).
As shown here, the vertex coordinate has been scaled by half.
How does the projection matrix decide what will be clipped?
The near and far plane are the limits of where the viewer can see (anything beyond the far plane and before the near plane will be clipped). Any coordinate will also have to go through a clipping check to see if it has to be clipped. The clipping check is checking whether the vertex coordinate is within a frustum range of -w to w. If it is outside of that range, it will be clipped.
Let's say I have a vertex with a position of (2, 130, 90, 90).
x value is 2
y value is 130
z value is 90
w value is 90
This vertex must be within the range of -90 to 90. The x and z value is within the range but the y value goes beyond the range thus the vertex will be clipped.
So after the vertex shader is finished, the next step is vertex post processing. In vertex post processing the clipping happens and also perspective division happens where clip space is converted into NDC (normalized device coordinates). Also, viewport transform happens where NDC is converted to window space.
What does perspective division do?
Perspective division essentially divides the x, y, and z component of a vertex with the w component. Doing this actually does two things, converts the clip space to Normalized device coordinates and also add perspective by scaling the vertices.
What is Normalized Device Coordinates?
Normalized Device Coordinates is the coordinate system where all coordinates are condensed into an NDC box where each axis is in the range of -1 to +1.
After NDC is occurred, viewport transform happens where all the NDC coordinates are converted screen coordinates. NDC space will become window space.
If an NDC coordinate is (0.5, 0.5, 0.3), it will be mapped onto the window based on what the programmer provided in the function glViewport. If the viewport is 400x300, the NDC coordinate will be placed at pixel 200 on x axis and 150 on y axis.
The perspective projection matrix does not decide what is clipped. After transforming a world coordinate with the projection, you get a clipspace coordinate. This is a Homogeneous coordinates. Base on this coordinate the Rendering Pipeline clips the scene. The clipping rule is -w < x, y, z < w. In the following process of the rendering pipeline, the clip space coordinates is transformed into the normalized device space by the perspective divide (x, y, z)' = (x/w, y/w, z/w). This division by the w component gives the perspective effect. (See also What exactly are eye space coordinates? and Transform the modelMatrix)
In OpenGL, I have read that a vertex should be represented by (x,y,z,w), where w = z. This is to enable perspective divide, whereby (x,y,z) are divided by w in order to determine their screen position due to the perspective effect. And if they were just divided by the original z value, then the z would be 1 everywhere.
My question is: Why do you need to divide the z component by w at all? Why can you not just divide the x and y components by z, such that the screen coordinates have the perspective effect applied, and then just used the original unmodified z component for depth testing? In this way, you would not have to use the w component at all....
Obviously I am missing something!
3D computer graphics is typically handled with homogeneous coordinates and in a projective vector space. The math behind this is a bit more than "just divide by w".
Using 4D homogeneous vectors and 4x4 matrices has the nice advantage that all sorts of affine transformations (and this includes especially the translation, which also relies on w) and projective transforms can be represented by simple matrix multiplications.
In OpenGL, I have read that a vertex should be represented by
(x, y, z, w), where w = z.
That is not true. A vertex should be represented by (x, y, z, w), where w is just w. In your typical case, the input w is actually 1, so it is usually not stored in the vertex data, but added on demand in the shaders etc.
Your typical projection matrix will set w_clip = -z_eye. But that is a different thing. This means that you just project along the -z direction in eye space. You could also put w_clip=2 *x_eye -3*y_eye + 4 * z_eye there, and your axis of projection would have the direction (2, -3, 4, 0).
My question is: Why do you need to divide the z component by w at all? Why can you not just divide the x and y components by z, such that the screen coordinates have the perspective effect applied, and then just used the original unmodified z component for depth testing?
Conceptually, the space is distorted along all 3 dimensions, not just in x and y. Furthermore, in the beginning, GPUs had just 16 bit or 24 bit integer precision for the depth buffer. In such a scenario, you definitively want to have a denser representation near the camera, and a sparse one far away.
Nowadays, with programmable vertex shaders and floating-point depth buffer formats, you can basically just store the z_eye value in the depth buffer, and use this for depth testing. However, this is typically referred to as W buffering, because the (clip space) w component is used.
There is another conceptional issue if you would divide by z: you wouldn't be able to use an orthogonal projection, you always would force some kind of perspective. Now one might argue that the division by z doesn't have to happen automatically, but one could apply it when needed in the vertex shader. But this won't work either. You must not apply the perspective divide in the vertex shader, because that would project points which lie behind the camera in front of the camera. As the vertex shader does not work on whole primitives, this would completely screw up any primitive there if at least one vertex lies behind the camera and another one lies in front of it. To deal with that situation, the clipping has to be applied before the divide - hence the name clip space.
In this way, you would not have to use the w component at all.
That is also not true. The w component is further used down the pipeline. It is essential for perspective-correct attribute interpolation.
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.
I started to read lesson 1 in learningwebgl blog, and I noticed this part:
var pMatrix = mat4.create();
mat4.perspective(45, gl.viewportWidth / gl.viewportHeight, 0.1, 100.0, pMatrix);
I roughly understand how matrices (translation/rotation/multiple) works, but I have no idea what mat4.perspective(...) means. What is it used for? What is the result, if I multiply a vector with this matrix?
The perspective matrix is used to scale, and possibly translate or flip the coordinate system in preparation for the perspective divide. Since the the perspective projection operation involves a divide, it cannot be represented by a linear matrix transformation alone.
In a programmable graphics pipeline (see pixel shaders) you cannot see the divide operation - it is still one of the fixed-function parts. The programmer controls it by tweaking the variables involved in the operation. In the case of the perspective divide it is the projection matrix that gives you this control.
The projection matrix is used to convert world-coordinates to screen coordinates.
The positions in your three-dimensional virtual world are triplets of x, y and z coordinates. When you want to draw something (or rather tell OpenGL to draw something) it needs to calculation where these coordinates are on the users screen.
This calculation is implemented with matrix multiplication.
A vector consisting of x, y and z (and a fourth value of 1 which is necessary to allow the matrix to do some operations like scaling) is multiplied with a matrix to receive a new set of x, y and z coordinates (4th value is discarded) which represent where this point is on the users screen (the z-coordinate is required to determine which objects are in front of others).
The function mat4.perspective generates a projection matrix which generates a matrix which does exactly that. The arguments are:
The field-of-view in degree (45)
the aspect ratio of the field of view (the aspect ratio of the viewport)
the minimal distance from the viewer which is still drawn (0.1 world units)
the maximum distance from the viewer which is still drawn (100.0 world units)
the array in which the generated matrix is stored (pMatrix)
When a point is multiplied with this matrix, the result are the screen coordinates where this point has to be drawn.