Once that I calibrated my camera and I know horizontal and vertical fields of view (through OpenCV's calibrationMatrixValues function), is it possible to know the X and Y angles (from the center) of a particular pixel?
Suppose I have a camera with fields of view of 100° (h) and 80° (v) and a resolution of 500x400. The angle/pixel ratio is (0.2°, 0.2°). So the central pixel will be (0°, 0°), its left neighbor (-0.2°, 0), the topmost central pixel (0°, +40°) and so on.
Is actually this relation constant through all the image or there is a formula to perform this calculation? Is the obtained information reliable?
This is going to be the first step for triangulating objects in a multi camera environment.
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I have two images (I and I’) from a camera (intrinsics K is known). The y axis of the camera is always vertical w.r.t. the real world. The second image has been captured w.r.t. the first by rotating the camera about the camera center by known rotation matrix R. There is no overlap between the images. However the images I and I' are in continuum ie. the right boundary of I coincides with the left boundary of I' (ie captured from same location in real world). If (x,y) and (x',y') are corresponding locations in images I and I' the homography can be written as (K and R are known)
My question is given a location in image I, how do I find the location in image I'. The homography equation has a factor s. How do I determine this factor s? Also is this factor constant for all locations in the image?
If this factor s is constant, is it correct to use the above equation for all pairs of neighboring pixels of I and I' at the boundary to find s.
Two cameras , Calibration is done between them and both intrinsic and extrinsic matrices are obtained , I am able to get (U,V) of the first camera , How could i get (U,V) of the second camera ? What is the kind of transformation could be made ?
Positions of cameras is fixed
Homography is the way a two 2D planes could be related
Since these cameras are paralel to each other(i.e. stereo), y axis of a point(x,y) in the first image will remain the same in the second image, i.e . y' = y. Only x will change. ( y is vertical axis, x is horizontal).
There are some techniques to find x'. The easiest one is normalized cross correlation. Choose a window around the points, do normalized cross correlation. The result will be an array of width of the image.
Unless you are searching for a point in a smooth region, maximum value in your array (peak) is expected to be your matching point.
Alternatively, you can try SIFT/SURF feature but I am not expert on those. I only know there are functions you can use in Matlab (such as detectSURFfeatures).
Note that if you are using two different cameras, you have to calibrate both of them.
I do some image processing task in 3D and I have a problem.
I use a simulator which provides me an special kind of cameras which can tell the distance between the position of camera and any arbitrary point, using the pixels of that point in the image of camera. For example I can get the distance between camera and the object which is placed in pixel 21:34.
Now I need to calculate the real distance between two arbitrary pixels in the image of camera.
It is easy when camera is vertical and placed on the above of the field and all objects are on the ground but when camera is horizontal the depth of objects in image is different.
So, how should I do?
Simple 3D reconstruction will accomplish this. The distance from camera to points in 3D is along optical axis that is Z, which you already have. You will need X, Y as well:
X = u*Z/f;
Y = v*Z/f,
where f is camera focal length in pixels, Z your distance in mm or meters and u,v is an image centered coordinates: u = column-width/2, v = height/2-row. Note the asymmetry due to the fact that rows go down while Y and v go up. As soon as you get your X, Y, Z the distance in 3D is given by Euclidean formula:
dist = sqrt((X1-X2)2+(Y1-Y2)2+(Z1-Z2)2)
I am searching lots of resources on internet for many days but i couldnt solve the problem.
I have a project in which i am supposed to detect the position of a circular object on a plane. Since on a plane, all i need is x and y position (not z) For this purpose i have chosen to go with image processing. The camera(single view, not stereo) position and orientation is fixed with respect to a reference coordinate system on the plane and are known
I have detected the image pixel coordinates of the centers of circles by using opencv. All i need is now to convert the coord. to real world.
http://www.packtpub.com/article/opencv-estimating-projective-relations-images
in this site and other sites as well, an homographic transformation is named as:
p = C[R|T]P; where P is real world coordinates and p is the pixel coord(in homographic coord). C is the camera matrix representing the intrinsic parameters, R is rotation matrix and T is the translational matrix. I have followed a tutorial on calibrating the camera on opencv(applied the cameraCalibration source file), i have 9 fine chessbordimages, and as an output i have the intrinsic camera matrix, and translational and rotational params of each of the image.
I have the 3x3 intrinsic camera matrix(focal lengths , and center pixels), and an 3x4 extrinsic matrix [R|T], in which R is the left 3x3 and T is the rigth 3x1. According to p = C[R|T]P formula, i assume that by multiplying these parameter matrices to the P(world) we get p(pixel). But what i need is to project the p(pixel) coord to P(world coordinates) on the ground plane.
I am studying electrical and electronics engineering. I did not take image processing or advanced linear algebra classes. As I remember from linear algebra course we can manipulate a transformation as P=[R|T]-1*C-1*p. However this is in euclidian coord system. I dont know such a thing is possible in hompographic. moreover 3x4 [R|T] Vector is not invertible. Moreover i dont know it is the correct way to go.
Intrinsic and extrinsic parameters are know, All i need is the real world project coordinate on the ground plane. Since point is on a plane, coordinates will be 2 dimensions(depth is not important, as an argument opposed single view geometry).Camera is fixed(position,orientation).How should i find real world coordinate of the point on an image captured by a camera(single view)?
EDIT
I have been reading "learning opencv" from Gary Bradski & Adrian Kaehler. On page 386 under Calibration->Homography section it is written: q = sMWQ where M is camera intrinsic matrix, W is 3x4 [R|T], S is an "up to" scale factor i assume related with homography concept, i dont know clearly.q is pixel cooord and Q is real coord. It is said in order to get real world coordinate(on the chessboard plane) of the coord of an object detected on image plane; Z=0 then also third column in W=0(axis rotation i assume), trimming these unnecessary parts; W is an 3x3 matrix. H=MW is an 3x3 homography matrix.Now we can invert homography matrix and left multiply with q to get Q=[X Y 1], where Z coord was trimmed.
I applied the mentioned algorithm. and I got some results that can not be in between the image corners(the image plane was parallel to the camera plane just in front of ~30 cm the camera, and i got results like 3000)(chessboard square sizes were entered in milimeters, so i assume outputted real world coordinates are again in milimeters). Anyway i am still trying stuff. By the way the results are previosuly very very large, but i divide all values in Q by third component of the Q to get (X,Y,1)
FINAL EDIT
I could not accomplish camera calibration methods. Anyway, I should have started with perspective projection and transform. This way i made very well estimations with a perspective transform between image plane and physical plane(having generated the transform by 4 pairs of corresponding coplanar points on the both planes). Then simply applied the transform on the image pixel points.
You said "i have the intrinsic camera matrix, and translational and rotational params of each of the image.” but these are translation and rotation from your camera to your chessboard. These have nothing to do with your circle. However if you really have translation and rotation matrices then getting 3D point is really easy.
Apply the inverse intrinsic matrix to your screen points in homogeneous notation: C-1*[u, v, 1], where u=col-w/2 and v=h/2-row, where col, row are image column and row and w, h are image width and height. As a result you will obtain 3d point with so-called camera normalized coordinates p = [x, y, z]T. All you need to do now is to subtract the translation and apply a transposed rotation: P=RT(p-T). The order of operations is inverse to the original that was rotate and then translate; note that transposed rotation does the inverse operation to original rotation but is much faster to calculate than R-1.
Is it possible to obtain front-parallel view of image or camera position by 2d-3d points relation using OpenCV?
For this I have intrinsic and extrinsic parameters. I have also 3d coordinates of set of control points (which lies in one plane) on image (relation 2d-3d).
In fact I need location and orientation of camera, but it is not difficult to find it if I can convert image to fronto-parallel view.
If it is not possible to do with OpenCV, are the other libraries which can solve this task?
Solution is based on the formulas in the OpenCV documentation Camera Calibration and 3D Reconstruction
Let's consider numerical form without distortion coefficient (in contrast with matrix form).
We have u and v.
It is easy to calculate x' and y'.
But x and y can not be calculate because we can choose any non-zero z.
Line in 3d corresponds to one point in 2d image.
To solve this we take two points for z=1 and z=2.
Then we find 2 points in 3d space which specify line (x1,y1,z1) and (x2,y2,z2).
Then we can apply R-1 to (x1,y1,z1) and (x2,y2,z2) which results in line determined by two points (X1, Y1, Z1) and (X1, Y1, Z1).
Since our control points lie in one plane (let plane is Z=0 for simplicity) we can find corresponding X and Y point which is a point in 3d.
After applying normalization from mm to pixels we obtain fronto-parallel image.
(If we have input image distorted we should undistort it first)