I am trying to render 3D point cloud from the depth data which I saved from opengl framebuffer. Basically, I took different depth samples from different n viewpoints (which are already known) for the rendered model centered at (0, 0, 0). I successfully saved the depth maps but now I want to extract x, y, z coordinated from these depth maps. For this, I am back projecting point from image to world. To get world coordinates I use the following equation P = K_inv [R|t]_inv * p. to calculate the world coordinates.
To calculate the image intrinsics matrix I used information from the opengl camera matrix, glm::perspective(fov, aspect, near_plane, far_plane). The intrinsic matrix K is calculated as
where
If I transform the coordinates in camera origin (i.e., no extrinsic transformation [R|t]), I get a 3D model for a single Image. To fuse multiple depths maps, I also need extrinsic transformation which I am calculating as from the OpenGL lookat matrix glm::lookat(eye=n_viewpoint_coorinates, center=(0, 0, 0), up=(0, 1, 0)). The extrisnics matrix is calculated as below (ref: http://ksimek.github.io/2012/08/22/extrinsic/
But when I fuse two depth images they are misaligned. I think the extrinsic matrix is not correct. I also tried to use glm::lookat matrix directly but that does not work as well. The fused model snapshot is shown below
Can someone suggest, what is wrong with my approach. Is it the extrinsic matrix that is wrong (which I am damn sure of)?
Finally, I managed to solve this by myself. Instead of doing transformation inside the OpenGL, I did the transformation outside of the OpenGL. Basically, I kept the camera constant and at some distance from the model and did rotation transformation on the model, and then finally render the model without lookat matrix (or just 4x4 identity matrix). I don't know why using lookat matrix does not gave me the result or maybe it is due something I was missing. To backproject the model into world coordinates I would just take the inverse of the exact transformation I did initially before feeding the model to OpenGL.
Related
I'm trying to use the GLM mathematics library to warp the projection matrix such that multiple cameras (e.g. a left eye and a right eye) are viewing the same view, but the camera frustum is distorted.
This is what I am trying to do with my cameras using GLM:
image
Note: One of the challenges I have for my camera, is that I only have in my OpenGL Graphics API a read-only view matrix and projection matrix. So even once I have the view and projection matrix, I can only move my camera using x,y,z for eye position, and then yaw,pitch,roll for view direction. Then I have standard OpenGL calls, but I am not sure how to shift or warp the projection matrix as shown below.
I think this second picture also accurately shows what I am trying to achieve with OpenGL.
Any tips on how to approach this problem? Do I need to implement a special lookAt function, or is there something built into GLM that can help me compute the right view?
A possible way to do this is:
Assume that X is the position of the "nose" (the half point between the 2 eyes).
Let V be the direction orthogonal to the left eye - right eye line.
The function X + t * V describes the "line of sight" in your last
picture.
Take t > 0 and set that as your look at point P/
Create 2 cameras with centers at the eyes, both with the exact same up direction, then use glm::lookat to look at P.
I am currently facing an issue with my Structure from Motion program based on OpenCv.
I'm gonna try to depict you what it does, and what it is supposed to do.
This program lies on the classic "structure from motion" method.
The basic idea is to take a pair of images, detect their keypoints and compute the descriptors of those keypoints. Then, the keypoints matching is done, with a certain number of tests to insure the result is good. That part works perfectly.
Once this is done, the following computations are performed : fundamental matrix, essential matrix, SVD decomposition of the essential matrix, camera matrix computation and finally, triangulation.
The result for a pair of images is a set of 3D coordinates, giving us points to be drawn in a 3D viewer. This works perfectly, for a pair.
Indeed, here is my problem : for a pair of images, the 3D points coordinates are calculated in the coordinate system of the first image of the image pair, taken as the reference image. When working with more than two images, which is the objective of my program, I have to reproject the 3D points computed in the coordinate system of the very first image, in order to get a consistent result.
My question is : How do I reproject 3D points coordinate given in a camera related system, into an other camera related system ? With the camera matrices ?
My idea was to take the 3D point coordinates, and to multiply them by the inverse of each camera matrix before.
I clarify :
Suppose I am working on the third and fourth image (hence, the third pair of images, because I am working like 1-2 / 2-3 / 3-4 and so on).
I get my 3D point coordinates in the coordinate system of the third image, how do I do to reproject them properly in the very first image coordinate system ?
I would have done the following :
Get the 3D points coordinates matrix, apply the inverse of the camera matrix for image 2 to 3, and then apply the inverse of the camera matrix for image 1 to 2.
Is that even correct ?
Because those camera matrices are non square matrices, and I can't inverse them.
I am surely mistaking somewhere, and I would be grateful if someone could enlighten me, I am pretty sure this is a relative easy one, but I am obviously missing something...
Thanks a lot for reading :)
Let us say you have a 3 * 4 extrinsic parameter matrix called P. To match the notations of OpenCV documentation, this is [R|t].
This matrix P describes the projection from world space coordinates to the camera space coordinates. To quote the documentation:
[R|t] translates coordinates of a point (X, Y, Z) to a coordinate system, fixed with respect to the camera.
You are wondering why this matrix is non-square. That is because in the usual context of OpenCV, you are not expecting homogeneous coordinates as output. Therefore, to make it square, just add a fourth row containing (0,0,0,1). Let's call this new square matrix Q.
You have one such matrix for each pair of cameras, that is you have one Qk matrix for each pair of images {k,k+1} that describes the projection from the coordinate space of camera k to that of camera k+1. Those matrices are inversible because they describe isometries in homogeneous coordinates.
To go from the coordinate space of camera 3 to that of camera 1, just apply to your points the inverse of Q2 and then the inverse of Q1.
I am having profound issues regarding understanding the transformations involved in VTK. OpenGL has fairly good documentation and I was of the impression that VTK is verym similar to OpenGL (it is, in many ways). But when it comes to transformations, it seems to be an entirely different story.
This is a good OpenGL documentation about transforms involved:
http://www.songho.ca/opengl/gl_transform.html
The perspective projection matrix in OpenGL is:
I wanted to see if this formula applied in VTK will give me the projection matrix of VTK (by cross-checking with VTK projection matrix).
Relevant Camera and Renderer Parameters:
camera->SetPosition(0,0,20);
camera->SetFocalPoint(0,0,0);
double crSet[2] = {10, 1000};
renderer->GetActiveCamera()->SetClippingRange(crSet);
double windowSize[2];
renderWindow->SetSize(1280,720);
renderWindowInteractor->GetSize(windowSize);
proj = renderer->GetActiveCamera()->GetProjectionTransformMatrix(windowSize[0]/windowSize[1], crSet[0], crSet[1]);
The projection transform matrix I got for this configuration is:
The (3,3) and (3,4) values of the projection matrix (lets say it is indexed 1 to 4 for rows and columns) should be - (f+n)/(f-n) and -2*f*n/(f-n) respectively. In my VTK camera settings, the nearz is 10 and farz is 1000 and hence I should get -1.020 and -20.20 respectively in the (3,3) and (3,4) locations of the matrix. But it is -1010 and -10000.
I have changed my clipping range values to see the changes and the (3,3) position is always nearz+farz which makes no sense to me. Also, it would be great if someone can explain why it is 3.7320 in the (1,1) and (2,2) positions. And this value DOES NOT change when I change the window size of the renderer window. Quite perplexing to me.
I see in VTKCamera class reference that GetProjectionTransformMatrix() returns the transformation matrix that maps from camera coordinates to viewport coordinates.
VTK Camera Class Reference
This is a nice depiction of the transforms involved in OpenGL rendering:
OpenGL Projection Matrix is the matrix that maps from eye coordinates to clip coordinates. It is beyond doubt that eye coordinates in OpenGL is the same as camera coordinates in VTK. But is the clip coordinates in OpenGL same as viewport coordinates of VTK?
My aim is to simulate a real webcam camera (already calibrated) in VTK to render a 3D model.
Well, the documentation you linked to actually explains this (emphasis mine):
vtkCamera::GetProjectionTransformMatrix:
Return the projection transform matrix, which converts from camera
coordinates to viewport coordinates. This method computes the aspect,
nearz and farz, then calls the more specific signature of
GetCompositeProjectionTransformMatrix
with:
vtkCamera::GetCompositeProjectionTransformMatrix:
Return the concatenation of the ViewTransform and the
ProjectionTransform. This transform will convert world coordinates to
viewport coordinates. The 'aspect' is the width/height for the
viewport, and the nearz and farz are the Z-buffer values that map to
the near and far clipping planes. The viewport coordinates of a point located inside the frustum are in the range
([-1,+1],[-1,+1], [nearz,farz]).
Note that this neither matches OpenGL's window space nor normalized device space. If find the term "viewport coordinates" for this aa poor choice, but be it as it may. What bugs me more with this is that the matrix actually does not transform to that "viewport space", but to some clip-space equivalent. Only after the perspective divide, the coordinates will be in the range as given for the above definition of the "viewport space".
But is the clip coordinates in OpenGL same as viewport coordinates of
VTK?
So that answer is a clear no. But it is close. Basically, that projection matrix is just a scaled and shiftet along the z dimension, and it is easy to convert between those two. Basically, you can simply take znear and zfar out of VTK's matrix, and put it into that OpenGL projection matrix formula you linked above, replacing just those two matrix elements.
I am trying to create a dataset of images of objects at different poses, where each image is annotated with camera pose (or object pose).
If, for example, I have a world coordinate system and I place the object of interest at the origin and place the camera at a known position (x,y,z) and make it face the origin. Given this information, how can I calculate the pose (rotation matrix) for the camera or for the object.
I had one idea, which was to have a reference coordinate i.e. (0,0,z') where I can define the rotation of the object. i.e. its tilt, pitch and yaw. Then I can calculate the rotation from (0,0,z') and (x,y,z) to give me a rotation matrix. The problem is, how to now combine the two rotation matrices?
BTW, I know the world position of the camera as I am rendering these with OpenGL from a CAD model as opposed to physically moving a camera around.
The homography matrix maps between homogeneous screen coordinates (i,j) to homogeneous world coordinates (x,y,z).
homogeneous coordinates are normal coordinates with a 1 appended. So (3,4) in screen coordinates is (3,4,1) as homogeneous screen coordinates.
If you have a set of homogeneous screen coordinates, S and their associated homogeneous world locations, W. The 4x4 homography matrix satisfies
S * H = transpose(W)
So it boils down to finding several features in world coordinates you can also identify the i,j position in screen coordinates, then doing a "best fit" homography matrix (openCV has a function findHomography)
Whilst knowing the camera's xyz provides helpful info, its not enough to fully constrain the equation and you will have to generate more screen-world pairs anyway. Thus I don't think its worth your time integrating the cameras position into the mix.
I have done a similar experiment here: http://edinburghhacklab.com/2012/05/optical-localization-to-0-1mm-no-problemo/
I've been writing a 2D basic game engine in OpenGL/C++ and learning everything as I go along. I'm still rather confused about defining vertices and their "position". That is, I'm still trying to understand the vertex-to-pixels conversion mechanism of OpenGL. Can it be explained briefly or can someone point to an article or something that'll explain this. Thanks!
This is rather basic knowledge that your favourite OpenGL learning resource should teach you as one of the first things. But anyway the standard OpenGL pipeline is as follows:
The vertex position is transformed from object-space (local to some object) into world-space (in respect to some global coordinate system). This transformation specifies where your object (to which the vertices belong) is located in the world
Now the world-space position is transformed into camera/view-space. This transformation is determined by the position and orientation of the virtual camera by which you see the scene. In OpenGL these two transformations are actually combined into one, the modelview matrix, which directly transforms your vertices from object-space to view-space.
Next the projection transformation is applied. Whereas the modelview transformation should consist only of affine transformations (rotation, translation, scaling), the projection transformation can be a perspective one, which basically distorts the objects to realize a real perspective view (with farther away objects being smaller). But in your case of a 2D view it will probably be an orthographic projection, that does nothing more than a translation and scaling. This transformation is represented in OpenGL by the projection matrix.
After these 3 (or 2) transformations (and then following perspective division by the w component, which actually realizes the perspective distortion, if any) what you have are normalized device coordinates. This means after these transformations the coordinates of the visible objects should be in the range [-1,1]. Everything outside this range is clipped away.
In a final step the viewport transformation is applied and the coordinates are transformed from the [-1,1] range into the [0,w]x[0,h]x[0,1] cube (assuming a glViewport(0, w, 0, h) call), which are the vertex' final positions in the framebuffer and therefore its pixel coordinates.
When using a vertex shader, steps 1 to 3 are actually done in the shader and can therefore be done in any way you like, but usually one conforms to this standard modelview -> projection pipeline, too.
The main thing to keep in mind is, that after the modelview and projection transforms every vertex with coordinates outside the [-1,1] range will be clipped away. So the [-1,1]-box determines your visible scene after these two transformations.
So from your question I assume you want to use a 2D coordinate system with units of pixels for your vertex coordinates and transformations? In this case this is best done by using glOrtho(0.0, w, 0.0, h, -1.0, 1.0) with w and h being the dimensions of your viewport. This basically counters the viewport transformation and therefore transforms your vertices from the [0,w]x[0,h]x[-1,1]-box into the [-1,1]-box, which the viewport transformation then transforms back to the [0,w]x[0,h]x[0,1]-box.
These have been quite general explanations without mentioning that the actual transformations are done by matrix-vector-multiplications and without talking about homogenous coordinates, but they should have explained the essentials. This documentation of gluProject might also give you some insight, as it actually models the transformation pipeline for a single vertex. But in this documentation they actually forgot to mention the division by the w component (v" = v' / v'(3)) after the v' = P x M x v step.
EDIT: Don't forget to look at the first link in epatel's answer, which explains the transformation pipeline a bit more practical and detailed.
It is called transformation.
Vertices are set in 3D coordinates which is transformed into a viewport coordinates (into your window view). This transformation can be set in various ways. Orthogonal transformation can be easiest to understand as a starter.
http://www.songho.ca/opengl/gl_transform.html
http://www.opengl.org/wiki/Vertex_Transformation
http://www.falloutsoftware.com/tutorials/gl/gl5.htm
Firstly be aware that OpenGL not uses standard pixel coordinates. I mean by that for particular resolution, ie. 800x600 you dont have horizontal coordinates in range 0-799 or 1-800 stepped by one. You rather have coordinates ranged from -1 to 1 later send to graphic card rasterizing unit and after that matched to particular resolution.
I ommited one step here - before all that you have an ModelViewProjection matrix (or viewProjection matrix in some simple cases) which before all that will cast coordinates you use to an projection plane. Default use of that is to implement a camera which converts 3D space of world (View for placing an camera into right position and Projection for casting 3d coordinates into screen plane. In ModelViewProjection it's also step of placing a model into right place in world).
Another case (and you can use Projection matrix this way to achieve what you want) is to use these matrixes to convert one range of resolutions to another.
And there's a trick you will need. You should read about modelViewProjection matrix and camera in openGL if you want to go serious. But for now I will tell you that with proper matrix you can just cast your own coordinate system (and ie. use ranges 0-799 horizontaly and 0-599 verticaly) to standarized -1:1 range. That way you will not see that underlying openGL api uses his own -1 to 1 system.
The easiest way to achieve this is glOrtho function. Here's the link to documentation:
http://www.opengl.org/sdk/docs/man/xhtml/glOrtho.xml
This is example of proper usage:
glMatrixMode (GL_PROJECTION)
glLoadIdentity ();
glOrtho (0, 800, 600, 0, 0, 1)
glMatrixMode (GL_MODELVIEW)
Now you can use own modelView matrix ie. for translation (moving) objects but don't touch your projection example. This code should be executed before any drawing commands. (Can be after initializing opengl in fact if you wont use 3d graphics).
And here's working example: http://nehe.gamedev.net/tutorial/2d_texture_font/18002/
Just draw your figures instead of drawing text. And there is another thing - glPushMatrix and glPopMatrix for choosen matrix (in this example projection matrix) - you wont use that until you combining 3d with 2d rendering.
And you can still use model matrix (ie. for placing tiles somewhere in world) and view matrix (in example for zooming view, or scrolling through world - in this case your world can be larger than resolution and you could crop view by simple translations)
After looking at my answer I see it's a little chaotic but If you confused - just read about Model, View, and Projection matixes and try example with glOrtho. If you're still confused feel free to ask.
MSDN has a great explanation. It may be in terms of DirectX but OpenGL is more-or-less the same.
Google for "opengl rendering pipeline". The first five articles all provide good expositions.
The key transition from vertices to pixels (actually, fragments, but you won't be too far off if you think "pixels") is in the rasterization stage, which occurs after all vertices have been transformed from world-coordinates to screen coordinates and clipped.