I'd like to get homography matrix to Bird's eye view and I know the projection Matrix of the camera. Is there any relation between them?
Thanks.
A projection matrix is defined as a product of camera's intrinsic (e.g. focal length, principal points, etc.) and extrinsic (rotation and translation) matrices. The question is w.r.t what your rotation and translation are? For example, I can imagine another camera or an object in 3D with respect to which you have these rotations and translations. Otherwise your projection is just an intrinsic matrix.
Think first about the pieces of information you need to know to obtain a bird’s eye view: you need to know at least how your camera is oriented w.r.t ground surface. If you also know camera elevation you can create a metric reconstruction. But since you mentioned a homography, I assume that you consider a bird’s eye view of a flat surface since a homography maps the points on two flat surfaces, in your case the points on a flat ground to the points on your flat sensor.
Let’s consider a pinhole camera equation. It basically says that
[u, v, 1]T ~ A*[R|t][x, y, z, 1]T, where A is a camera intrinsic matrix. Now since you deal with a ground plane, you can align a new coordinate system with it by setting z=0; R|t are rotation and translation matrices from this coordinate system into your camera-aligned system;
Next, note that your R|t is a 3x4 matrix and it looses one dimension since z=0; it becomes 3x3 or Homography which is equal now to H=A*R’|t; Ok all we did was proving that a homography mapping existed between the ground and your sensor;
Now, you want another kind of homography that happens during pure camera rotations and zooms between points on sensor before and after rotations/zoom; that is you want to rotate the camera down and possibly zoom out. Again, think in terms of a pinhole camera equation: originally you had H1=A ( here I threw out R|T as irrelevant for now) and then you rotated your camera and you have H2=AR; in other words, H1 is how you make your image now and H2 is how you want your image look like.
The relations between two is what you want to find, H12, and it is also a homography since the Homography is a family of transformations (use this simple heuristics: what happens in a family stays in the family). Since the same surface can generate images either with H1 or H2 we can assemble H12 by undoing H1 (back to the ground plane) and applying H2 (from the ground to a sensor bird’s eye view); in a way this resembles operations with vectors, you just have to respect the order of matrix application from the right to the left:
H12 = H2*H1-1=A*R*A-1=P*A-1 , where we substituted the expressions for H1, H2 and finally for a projection matrix (in case you do have it)
This is your answer, and if the rotation R is unknown it can be guessed from the camera orientation w.r.t. the ground or calculated using solvePnP() from the opeCV library. Finally, when I do this on a cell phone I just use its accelerometer readings as a good approximation since when a cell phone is not accelerated the readings represent a gravity vector which gives the rotation w.r.t. flat horizontal ground.
When you plot your bird’s eye view as an image you will notice that its boundaries turned from rectangular into some kind of a trapezoid (due to a camera frustum shape) and there are some holes at the distant locations (due to the insufficient sampling rate). You can interpolate inside the holes using wrapPerspective()
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I am using opencv to calibrate my webcam. So, what I have done is fixed my webcam to a rig, so that it stays static and I have used a chessboard calibration pattern and moved it in front of the camera and used the detected points to compute the calibration. So, this is as we can find in many opencv examples (https://docs.opencv.org/3.1.0/dc/dbb/tutorial_py_calibration.html)
Now, this gives me the camera intrinsic matrix and a rotation and translation component for mapping each of these chessboard views from the chessboard space to world space.
However, what I am interested in is the global extrinsic matrix i.e. once I have removed the checkerboard, I want to be able to specify a point in the image scene i.e. x, y and its height and it gives me the position in the world space. As far as I understand, I need both the intrinsic and extrinsic matrix for this. How should one proceed to compute the extrinsic matrix from here? Can I use the measurements that I have already gathered from the chessboard calibration step to compute the extrinsic matrix as well?
Let me place some context. Consider the following picture, (from https://docs.opencv.org/2.4/modules/calib3d/doc/camera_calibration_and_3d_reconstruction.html):
The camera has "attached" a rigid reference frame (Xc,Yc,Zc). The intrinsic calibration that you successfully performed allows you to convert a point (Xc,Yc,Zc) into its projection on the image (u,v), and a point (u,v) in the image to a ray in (Xc,Yc,Zc) (you can only get it up to a scaling factor).
In practice, you want to place the camera in an external "world" reference frame, let's call it (X,Y,Z). Then there is a rigid transformation, represented by a rotation matrix, R, and a translation vector T, such that:
|Xc| |X|
|Yc|= R |Y| + T
|Zc| |Z|
That's the extrinsic calibration (which can be written also as a 4x4 matrix, that's what you call the extrinsic matrix).
Now, the answer. To obtain R and T, you can do the following:
Fix your world reference frame, for example the ground can be the (x,y) plane, and choose an origin for it.
Set some points with known coordinates in this reference frame, for example, points in a square grid in the floor.
Take a picture and get the corresponding 2D image coordinates.
Use solvePnP to obtain the rotation and translation, with the following parameters:
objectPoints: the 3D points in the world reference frame.
imagePoints: the corresponding 2D points in the image in the same order as objectPoints.
cameraMatris: the intrinsic matrix you already have.
distCoeffs: the distortion coefficients you already have.
rvec, tvec: these will be the outputs.
useExtrinsicGuess: false
flags: you can use CV_ITERATIVE
Finally, get R from rvec with the Rodrigues function.
You will need at least 3 non-collinear points with corresponding 3D-2D coordinates for solvePnP to work (link), but more is better. To have good quality points, you could print a big chessboard pattern, put it flat in the floor, and use it as a grid. What's important is that the pattern is not too small in the image (the larger, the more stable your calibration will be).
And, very important: for the intrinsic calibration, you used a chess pattern with squares of a certain size, but you told the algorithm (which does kind of solvePnPs for each pattern), that the size of each square is 1. This is not explicit, but is done in line 10 of the sample code, where the grid is built with coordinates 0,1,2,...:
objp[:,:2] = np.mgrid[0:7,0:6].T.reshape(-1,2)
And the scale of the world for the extrinsic calibration must match this, so you have several possibilities:
Use the same scale, for example by using the same grid or by measuring the coordinates of your "world" plane in the same scale. In this case, you "world" won't be at the right scale.
Recommended: redo the intrinsic calibration with the right scale, something like:
objp[:,:2] = (size_of_a_square*np.mgrid[0:7,0:6]).T.reshape(-1,2)
Where size_of_a_square is the real size of a square.
(Haven't done this, but is theoretically possible, do it if you can't do 2) Reuse the intrinsic calibration by scaling fx and fy. This is possible because the camera sees everything up to a scale factor, and the declared size of a square only changes fx and fy (and the T in the pose for each square, but that's another story). If the actual size of a square is L, then replace fx and fy Lfx and Lfy before calling solvePnP.
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.
I have used openCV to calculate the homography relating to views of the same plane by using features and matching them. Is there any way to recover the plane itsself or the plane normal from this homography? (I am looking for an equation where H is the input and the normal n is the output.)
If you have the calibration of the cameras, you can extract the normal of the plane, but not the distance to the plane (i.e. the transformation that you obtain is up to scale), as Wikipedia explains. I don't know any implementation to do it, but here you are a couple of papers that deal with that problem (I warn you it is not straightforward): Faugeras & Lustman 1988, Vargas & Malis 2005.
You can recover the real translation of the transformation (i.e. the distance to the plane) if you have at least a real distance between two points on the plane. If that is the case, the easiest way to go with OpenCV is to first calculate the homography, then obtain four points on the plane with their 2D coordinates and the real 3D ones (you should be able to obtain them if you have a real measurement on the plane), and using PnP finally. PnP will give you a real transformation.
Rectifying an image is defined as making epipolar lines horizontal and lying in the same row in both images. From your description I get that you simply want to warp the plane such that it is parallel to the camera sensor or the image plane. This has nothing to do with rectification - I’d rather call it an obtaining a bird’s-eye view or a top view.
I see the source of confusion though. Rectification of images usually involves multiplication of each image with a homography matrix. In your case though each point in sensor plane b:
Xb = Hab * Xa = (Hb * Ha^-1) * Xa, where Ha is homography from the plane in the world to the sensor a; Ha and intrinsic camera matrix will give you a plane orientation but I don’t see an easy way to decompose Hab into Ha and Hb.
A classic (and hard) way is to find a Fundamental matrix, restore the Essential matrix from it, decompose the Essential matrix into camera rotation and translation (up to scale), rectify both images, perform a dense stereo, then fit a plane equation into 3d points you reconstruct.
If you interested in the ground plane and you operate an embedded device though, you don’t even need two frames - a top view can be easily recovered from a single photo, camera elevation from the ground (H) and a gyroscope (or orientation vector) readings. A simple diagram below explains the process in 2D case: first picture shows how to restore Z (depth) coordinate to every point on the ground plane; the second picture shows a plot of the top view with vertical axis being z and horizontal axis x = (img.col-w/2)*Z/focal; Here is img.col is image column, w - image width, and focal is camera focal length. Note that a camera frustum looks like a trapezoid in a birds eye view.
I know that in the general case, making this conversion is impossible since depth information is lost going from 3d to 2d.
However, I have a fixed camera and I know its camera matrix. I also have a planar calibration pattern of known dimensions - let's say that in world coordinates it has corners (0,0,0) (2,0,0) (2,1,0) (0,1,0). Using opencv I can estimate the pattern's pose, giving the translation and rotation matrices needed to project a point on the object to a pixel in the image.
Now: this 3d to image projection is easy, but how about the other way? If I pick a pixel in the image that I know is part of the calibration pattern, how can I get the corresponding 3d point?
I could iteratively choose some random 3d point on the calibration pattern, project to 2d, and refine the 3d point based on the error. But this seems pretty horrible.
Given that this unknown point has world coordinates something like (x,y,0) -- since it must lie on the z=0 plane -- it seems like there should be some transformation that I can apply, instead of doing the iterative nonsense. My maths isn't very good though - can someone work out this transformation and explain how you derive it?
Here is a closed form solution that I hope can help someone. Using the conventions in the image from your comment above, you can use centered-normalized pixel coordinates (usually after distortion correction) u and v, and extrinsic calibration data, like this:
|Tx| |r11 r21 r31| |-t1|
|Ty| = |r12 r22 r32|.|-t2|
|Tz| |r13 r23 r33| |-t3|
|dx| |r11 r21 r31| |u|
|dy| = |r12 r22 r32|.|v|
|dz| |r13 r23 r33| |1|
With these intermediate values, the coordinates you want are:
X = (-Tz/dz)*dx + Tx
Y = (-Tz/dz)*dy + Ty
Explanation:
The vector [t1, t2, t3]t is the position of the origin of the world coordinate system (the (0,0) of your calibration pattern) with respect to the camera optical center; by reversing signs and inversing the rotation transformation we obtain vector T = [Tx, Ty, Tz]t, which is the position of the camera center in the world reference frame.
Similarly, [u, v, 1]t is the vector in which lies the observed point in the camera reference frame (starting from camera center). By inversing the rotation transformation we obtain vector d = [dx, dy, dz]t, which represents the same direction in world reference frame.
To inverse the rotation transformation we take advantage of the fact that the inverse of a rotation matrix is its transpose (link).
Now we have a line with direction vector d starting from point T, the intersection of this line with plane Z=0 is given by the second set of equations. Note that it would be similarly easy to find the intersection with the X=0 or Y=0 planes or with any plane parallel to them.
Yes, you can. If you have a transformation matrix that maps a point in the 3d world to the image plane, you can just use the inverse of this transformation matrix to map a image plane point to the 3d world point. If you already know that z = 0 for the 3d world point, this will result in one solution for the point. There will be no need to iteratively choose some random 3d point. I had a similar problem where I had a camera mounted on a vehicle with a known position and camera calibration matrix. I needed to know the real world location of a lane marking captured on the image place of the camera.
If you have Z=0 for you points in world coordinates (which should be true for planar calibration pattern), instead of inversing rotation transformation, you can calculate homography for your image from camera and calibration pattern.
When you have homography you can select point on image and then get its location in world coordinates using inverse homography.
This is true as long as the point in world coordinates is on the same plane as the points used for calculating this homography (in this case Z=0)
This approach to this problem was also discussed below this question on SO: Transforming 2D image coordinates to 3D world coordinates with z = 0
With FeatureDetector I get features on two images with the same element and match this features with BruteForceMatcher.
Then I'm using OpenCv function findHomography to get homography matrix
H = findHomography( src2Dfeatures, dst2Dfeatures, outlierMask, RANSAC, 3);
and getting H matrix, then align image with
warpPerspective(img1,alignedSrcImage,H,img2.size(),INTER_LINEAR,BORDER_CONSTANT);
I need to know rotation angle, scale, displacement of detected element. Is there any simple way to get this than some big equations? Some evaluated formulas just to put data in?
Homography would match projections of your elements lying on a plane or lying arbitrary in 3D if the camera goes through a pure rotation or zoom and no translation. So here are the cases we are talking about with indication of what is the input to our calculations:
- planar target, pure rotation, intra-frame homography
- planar target, rotation and translation, target to frame homography
- 3D target, pure rotation, frame to frame mapping (constrained by a fundamental matrix)
In case of the planar target, a pure rotation is easy to calculate through your frame-to-frame Homography (H12):
given intrinsic camera matrix A, plane to image homographies for frame H1, and H2 that can be expressed as H1=A, H2=A*R, H12 = H2*H1-1=ARA-1 and thus R=A-1H12*A
In case of elements lying on a plane, rotation with translation of the camera (up to unknown scale) can be calculated through decomposition of target-to-frame homography. Note that the target can be just one of the views. Assuming you have your original planar target as an image (taken at some reference orientation) your task is to decompose the homography between images H12 which can be done through SVD. The first two columns of H represent the first two columns of the rotation martrix and be be recovered through H=ULVT, [r1 r2] = UDVT where D is 3x2 Identity matrix with the last row being all 0. The third column of a rotation matrix is just a vector product of the first two columns. The last column of the Homography is a translation vector times some constant.
Finally for arbitrary configuration of points in 3D and pure camera rotation, the rotation is calculated using the essential matrix decomposition rather than homography, see this
cv::decomposeProjectionMatrix();
and
cv::RQDecomp3x3();
are both similar to what you want to achive.
None of them is perfect. The theory behind them and why you cannot extract all params from a 3x3 matrix is a bit cumbersome. But the short answer is that a 3x3 proj matrix is a simplification from the complete 4x4 one, based on the fact that all points stay in the same plane.
You can try to use levenberg marquardt optimalization, where parameters will be translation and rotation, equations will represent by computed distances between features from two images(use only inliers from ransac homography).
Here is C++ implementation of LM http://www.ics.forth.gr/~lourakis/levmar/