I am trying to orient a 3d object at the world origin such that it doesn't change its position wrt camera when I move the camera OR change its field of view. I tried doing this
Object Transform = Inverse(CameraProjectionMatrix)
How do I undo the perspective divide because when I change the fov, the object is affected by it
In detail it looks like
origin(0.0, 0.0, 0.0, 1.0f);
projViewInverse = Camera.projViewMatrix().inverse();
projectionMatrix = Camera.projViewMatrix();
projectedOrigin = projectionMatrix * origin;
topRight(0.5f, 0.5f, 0.f);
scaleFactor = 1.0/projectedOrigin.z();
scale(scaleFactor,scaleFactor,scaleFactor);
finalMatrix = projViewInverse * Scaling(w) * Translation(topRight);
if you use gfx pipeline where positions (w=1.0) and vectors (w=0.0) are transformed to NDC like this:
(x',y',z',w') = M*(x,y,z,w) // applying transforms
(x'',y'') = (x',y')/w' // perspective divide
where M are all your 4x4 homogenyuous transform matrices multiplied in their order together. If you want to go back to the original (x,y,z) you need to know w' which can be computed from z. The equation depends on your projection. In such case you can do this:
w' = f(z') // z' is usually the value encoded in depth buffer and can obtained
(x',y') = (x'',y'')*w' // screen -> camera
(x,y) = Inverse(M)*(x',y',z',w') // camera -> world
However this can be used only if you know the z' and can derive w' from it. So what is usually done (if we can not) is to cast ray from camera focal point through the (x'',y'') and stop at wanted perpendicular distance to camera. For perspective projection you can look at it as triangle similarity:
So for each vertex you want to transform you need its projected x'',y'' position on the znear plane (screen) and then just scale the x'',y'' by the ratio between distances to camera focal point (*z1/z0). Now all we need is the focal length z0. That one dependss on the kind of projection matrix you use. I usually encounter 2 versions when you are in camera coordinate system then point (0,0,0) is either the focal point or znear plane. However the projection matrix can be any hence also the focal point position can vary ...
Now when you have to deal with aspect ratio then the first method deals with it internally as its inside the M. The second method needs to apply inverse of aspect ratio correction before conversion. So apply it directly on x'',y''
Related
I have fov angle = 60, width = 640 and height = 480 of window, near = 0.01 and far = 100 planes and I get projection matrix using glm::perspective()
glm::perspective(glm::radians(fov),
width / height,
zNear,
zFar);
It works well.
Then I want to change projection type to orthogonal, but I don't knhow how to compute input parameters of glm::ortho() properly.
I've tried many ways, but problem is after switching to orthographic projection size of model object become another.
Let I have a cube with center in (0.5, 0.5, 0.5) and length size 1, and camera with mEye in (0.5, 0.5, 3), mTarget in (0.5, 0.5, 0.5) and mUp (0, 1, 0). View matrix is glm::lookAt(mEye, mTarget, mUp)
With perspective projection it works well. With glm::ortho(-width, width, -height, height, zNear, zFar) my cube became a small pixel in the center of window.
Also I've tried implement this variant How to switch between Perspective and Orthographic cameras keeping size of desired object
but result is (almost) same as before.
So, first question is how to compute ortho parameters for saving original view size of object/position of camera?
Also, zooming with
auto distance = glm::length(mTarget - mEye)
mEye = mTarget - glm::normalize(mTarget - mEye) * distance;
have no effect with ortho. Thus second question is how to implement zooming in case of ortho projection?
P.s.
I assume I understand ortho correctly. Proportions of model doesn't depends on depth, but nevertheless I still can decide where camera is for setting size of model properly and using zoom. Also I assume it is simple and trivial task, for example, when developing a 3D-viewer/editor/etc. Correct me if it is not.
how to compute ortho parameters for saving original view size of object/position of camera?
At orthographic projection the 3 dimensional scene is parallel projection to the 2 dimensional viewport.
This means that the objects, which are projected on the viewport always have the same size, independent of their depth (distance to the camera).
The perspective projection describes the mapping from 3D points in the world as they are seen from of a pinhole camera, to 2D points of the viewport.
This means an object which is projected on the viewport becomes smaller, by its depth.
If you switch form perspective to orthographic projection only the objects in 1 plane, which is planar (parallel) to the viepwort, and keeps its depth. Note, a plane is 2 dimensional and has no "depth". This cause that a 3 dimensional object never can "look" the same, when the projection is switched. But a 2 dimensional billboard can keep it's size.
The ration of depth an size at perspective projection is linear and can be calculated. It depends on the field of view angle only:
float ratio_size_per_depth = atan(glm::radians(fov / 2.0f) * 2.0f;
If you want to set up an orthographic projection, which keeps the size for a certain distance (depth) then you have to define the depth first:
e.g. Distance to the target point:
auto distance = glm::length(mTarget - mEye);
the projection can be set up like this:
float aspect = width / height
float size_y = ratio_size_per_depth * distance;
float size_x = ratio_size_per_depth * distance * aspect;
glm::mat4 orthProject = glm::ortho(-size_x, size_x, -size_y, size_y, 0.0f, 2.0f*distance);
how to implement zooming in case of ortho projection?
Scale the XY components of the orthographic projection:
glm::mat4 orthProject = glm::ortho(-size_x, size_x, -size_y, size_y, 0.0f, 2.0f*distance);
float orthScale = 2.0f;
orthProject = glm::scale(orthProject, glm::vec3(orthScale, orthScale, 1.0f));
Set a value for orthScale which is > 1.0 for zoom in and a value which is < 1.0 for zoom out.
Can someone tell me how to make triangle vertices collide with edges of the screen?
For math library I am using GLM and for window creation and keyboard/mouse input I am using GLFW.
I created perspective matrix and simple array of triangle vertices.
Then I multiplied all this in vertex shader like:
gl_Position = projection * view * model * vec4(pos, 1.0);
Projection matrix is defined as:
glm::mat4 projection = glm::perspective(
45.0f, (GLfloat)screenWidth / (GLfloat)screenHeight, 0.1f, 100.0f);
I have fully working camera and projection. I can move around my "world" and see triangle standing there. The problem I have is I want to make sure that triangle collide with edges of the screen.
What I did was disable camera and only enable keyboard movement. Then I initialized translation matrix as glm::translate(model, glm::vec3(xMove, yMove, -2.5f)); and scale matrix to scale by 0.4.
Now all of that is working fine. When I press RIGHT triangle moves to the right when I press UP triangle moves up etc... The problem is I have no idea how to make it stop moving then it hits edges.
This is what I have tried:
triangleRightVertex.x is glm::vec3 object.
0.4 is scaling value that I used in scaling matrix.
if(((xMove + triangleRightVertex.x) * 0.4f) >= 1.0f)
{
cout << "Right side collision detected!" << endl;
}
When I move triangle to the right it does detect collision when x of the third vertex(bottom right corner of triangle) collides with right side but it goes little bit beyond before it detects. But when I tried moving up it detected collision when half of the triangle was up.
I have no idea what to do here can someone explain me this please?
Each of the vertex coordinates of the triangle is transformed by the model matrix form model space to world space, by the view matrix from world space to view space and by the projection matrix from view space to clip space. gl_Position is the Homogeneous coordinate in clip space and further transformed by a Perspective divide from clip space to normalized device space. The normalized device space is a cube, with right, bottom, front of (-1, -1, -1) and a left, top, back of (1, 1, 1).
All the geometry which is in this (volume) cube is "visible" on the viewport.
In clip space the clipping of the scene is performed.
A point is in clip space if the x, y and z components are in the range defined by the inverted w component and the w component of the homogeneous coordinates of the point:
-w <= x, y, z <= w
What you want to do is to check if a vertex x coordinate of the triangle is clipped. SO you have to check if the x component of the clip space coordinate is in the view volume.
Calculate the clip space position of the vertices on the CPU, as it does the vertex shader.
The glm library is very suitable for things like that:
glm::vec3 triangleVertex = ... ; // new model coordinate of the triangle
glm::vec4 h_pos = projection * view * model * vec4(triangleVertex, 1.0);
bool x_is_clipped = h_pos.x < -h_pos.w || h_pos.x > h_pos.w;
If you don't know how the orientation of the triangle is transformed by the model matrix and view matrix, then you have to do this for all the 3 vertex coordinates of the triangle-
The target is to draw a shape, lets say a triangle, pixel-perfect (vertices shall be specified in pixels) and be able to transform it in the 3rd dimension.
I've tried it with a orthogonal projection matrix and everything works fine, but the shape doesn't have any depth - if I rotate it around the Y axis it looks like I would just scale it around the X axis. (because a orthogonal projection obviously behaves like this). Now I want to try it with a perspective projection. But with this projection, the coordinate system changes completely, and due to this I can't specify my triangles verticies with pixels. Also if the size of my window changes, the size of shape changes too (because of the changed coordinate system).
Is there any way to change the coordinate system of the perspective projection so that I can specify my vertices like if I would use the orthogonal projection? Or do anyone have a Idea how to achieve the target described in the first sentence?
The projection matrix describes the mapping from 3D points of a scene, to 2D points of 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. The NDC are in range (-1,-1,-1) to (1,1,1).
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
where:
aspect = w / h
tanFov = tan( fov_y * 0.5 );
prjMat[0][0] = 2*n/(r-l) = 1.0 / (tanFov * aspect)
prjMat[1][1] = 2*n/(t-b) = 1.0 / tanFov
I assume that the view matrix is the identity matrix, and thus the view space coordinates are equal to the world coordinates.
If you want to draw a polygon, where the vertex coordinates are translated 1:1 into pixels, then you have to draw the polygon in parallel plane to the viewport. This means all points have to be draw with the same depth. The depth has to choose that way, that the transformation of a point in normalized device coordinates, by the inverse projection matrix gives the vertex coordinates in pixel. Note, the homogeneous coordinates given by the transformation with the inverse projection matrix, have to be divided by the w component of the homogeneous coordinates, to get cartesian coordinates.
This means, that the depth of the plane depends on the field of view angle of the projection:
Assuming you set up a perspective projection like this:
float vp_w = .... // width of the viewport in pixel
float vp_h = .... // height of the viewport in pixel
float fov_y = ..... // field of view angle (y axis) of the view port in degrees < 180°
gluPerspective( fov_y, vp_w / vp_h, 1.0, vp_h*2.0f );
Then the depthZ of the plane with a 1:1 relation of vertex coordinates and pixels, will be calculated like this:
float angRad = fov_y * PI / 180.0;
float depthZ = -vp_h / (2.0 * tan( angRad / 2.0 ));
Note, the center point of the projection to the view port is (0,0), so the bottom left corner point of the plane is (-vp_w/2, -vp_h/2, depthZ) and the top right corner point is (vp_w/2, vp_h/2, depthZ). Ensure, that the near plane of the perspective projetion is less than -depthZ and the far plane is greater than -depthZ.
See further:
Both depth buffer and triangle face orientation are reversed in OpenGL
Transform the modelMatrix
I want to convert 2D screen coordinates to 3D world coordinates. I have searched a lot but I did not get any satisfying result.
Note: I am not using OpenGL nor any other graphics library.
Data which I have:
Screen X
Screen Y
Screen Height
Screen Width
Aspect Ratio
If you have the Camera world Matrix and Projection Matrix this is pretty simple.
If you don't have the world Matrix you can compute it from it's position and rotation.
worldMatrix = Translate(x, y, z) * RotateZ(z_angle) * RotateY(y_angle) * RotateX(x_angle);
Where translate returns the the 4x4 translation matrices and Rotate returns the 4x4 rotation matrices around the given axis.
The projection matrix can be calculated from the aspect ratio, field of view angle, and near and far planes.
This blog has a good explanation of how to calculate the projection matrix.
You can unproject the screen coordinates by doing:
mat = worldMatrix * inverse(ProjectionMatrix)
dir = transpose(mat) * <x_screen, y_screen, 0.5, 1>
dir /= mat[3] + mat[7] + mat[11] + mat[15]
dir -= camera.position
Your ray will point from the camera in the direction dir.
This should work, but it's not a super concreate example on how to do this.
Basically you just need to do the following steps:
calculate camera's worldMatrix
calculate camera's projection matrix
multiply worldMatrix with inverse projection matrix.
create a point <Screen_X_Value, Screen_Y_Value, SOME_POSITIVE_Z_VALUE, 1>
apply this "inverse" projection to your point.
then subtract the cameras position form this point.
The resulting vector is the direction from the camera. Any point along that ray are the 3D coordinates corresponding to your 2D screen coordinate.
Hey guys I am trying to do a Camera class that uses lookAt from glm library. I have 4 points, the first one is eye, that is the camera position in space, the second one is the look, that is the point where the camera is looking at, the third one is the upp, that set the orientation of camera, and the forth is side, that is a cross product of look - eye and upp -eye.
So in the end I got a base of 3 vectors all of them with origin in the eye point. I got a coordinate system of the camera.
In my camera class, I want to be able to rotate about the coordinate system of the camera, not the coordinate system of the world. So what I am doing is rotate about one of the axis of the coordinate system of the camera.
I construct the class with initial values like this:
void Observer::initialize(glm::vec3 eye, glm::vec3 look, glm::vec3 upp, glm::vec3 side)
{
this->eye = eye; // (0.0, 0.0, 0.0)
this->look = look; // (0.0, 0.0, -1.0)
this->upp = upp; // (0.0, 1.0, 0.0)
this->side = side; // (1.0, 0.0, 0.0)
}
When I want to rotate the coordinate system about the x axis for example I call the function from glm like this:
void Observer::pitch(GLfloat pitch)
{
glm::mat4 rotate(1.0f);
rotate = glm::rotate(rotate, pitch, side - eye);
look = glm::vec3(rotate * glm::vec4(look, 1.0f));
upp = glm::vec3(rotate * glm::vec4(upp, 1.0f));
}
So far, I am understanding that all my points still form a coordinate system for the camera and all vector are perpendicular between each other.
But then I use these points I got with the lookAt function, to position the camera in the world.
glm::mat4 view = glm::lookAt(eye, look, upp);
And multiply this matrix with the modelview matrix from OpenGL
If I start to rotate a lot, the camera after a few rotations "reflect" the rotation, like I was rotating in the other way (I don't know how to describe what is really happening in a better way =s).
I really don't understand what is happening. I should normalize the vectors after I apply the rotation? Am I having a problem with gimbal lock (I don't know a lot about gimbal lock)?
When you do incremental rotations on vectors as you have done, numerical errors mount. When error causes the up and look-at vectors to point nearly in the same direction or opposite each other, the camera transformation calculation is unstable and whacky things can happen. Gimbal lock. Length changes cause different problems.
A solution that has worked for me is to re-orthogonalize the up and look-at vectors after each rotation. To do this, compute their cross-product L, then adjust (really replace) the up vector by crossing L with the lookAt. After all this re-normalize both up and look-at to unit length.
Though the orthogonalization-normalization is a fast operation, you don't really have to do it with every camera motion.
Note that when you correct the vectors like this you are actually doing part of the lookAt matrix calculation, so consider implementing your own to avoid an unnecessary cross product. See e.g. this prior SO article on this topic.