I have an image of a chessboard taken at some angle. Now I want to warp perspective so the chessboard image look again as if was taken directly from above.
I know that I can try to use 'findHomography' between matched points but I wanted to avoid it and use e.g. rotation data from mobile sensors to build homography matrix on my own. I calibrated my camera to get intrinsic parameters. Then lets say the following image has been taken at ~60degrees angle around x-axis. I thought that all I have to do is to multiply camera matrix with rotation matrix to obtain homography matrix. I tried to use the following code but looks like I'm not understanding something correctly because it doesn't work as expected (result image completely black or white.
import cv2
import numpy as np
import math
camera_matrix = np.array([[ 5.7415988502105745e+02, 0., 2.3986181527877352e+02],
[0., 5.7473682183375217e+02, 3.1723734404756237e+02],
[0., 0., 1.]])
distortion_coefficients = np.array([ 1.8662919398453856e-01, -7.9649812697463640e-01,
1.8178068172317731e-03, -2.4296638847737923e-03,
7.0519002388825025e-01 ])
theta = math.radians(60)
rotx = np.array([[1, 0, 0],
[0, math.cos(theta), -math.sin(theta)],
[0, math.sin(theta), math.cos(theta)]])
homography = np.dot(camera_matrix, rotx)
im = cv2.imread('data/chess1.jpg')
gray = cv2.cvtColor(im,cv2.COLOR_BGR2GRAY)
im_warped = cv2.warpPerspective(gray, homography, (480, 640), flags=cv2.WARP_INVERSE_MAP)
cv2.imshow('image', im_warped)
cv2.waitKey()
pass
I also have distortion_coefficients after calibration. How can those be incorporated into the code to improve results?
This answer is awfully late by several years, but here it is ...
(Disclaimer: my use of terminology in this answer may be imprecise or incorrect. Please do look up on this topic from other more credible sources.)
Remember:
Because you only have one image (view), you can only compute 2D homography (perspective correspondence between one 2D view and another 2D view), not the full 3D homography.
Because of that, the nice intuitive understanding of the 3D homography (rotation matrix, translation matrix, focal distance, etc.) are not available to you.
What we say is that with 2D homography you cannot factorize the 3x3 matrix into those nice intuitive components like 3D homography does.
You have one matrix - (which is the product of several matrices unknown to you) - and that is it.
However,
OpenCV provides a getPerspectiveTransform function which solves the 3x3 perspective matrix (using homogenous coordinate system) for a 2D homography between two planar quadrilaterals.
Link to documentation
To use this function,
Find the four corners of the chessboard on the image. These will be your source coordinates.
Supply four rectangle corners of your choice. These will be your destination coordinates.
Pass the source coordinates and destination coordinates into the getPerspectiveTransform to generate a 3x3 matrix that is able to dewarp your chessboard to an upright rectangle.
Notes to remember:
Mind the ordering of the four corners.
If the source coordinates are picked in clockwise order, the destination also needs to be picked in clockwise order.
Likewise, if counter-clockwise order is used, do it consistently.
Likewise, if z-order (top left, top right, bottom left, bottom right) is used, do it consistently.
Failure to order the corners consistently will generate a matrix that executes the point-to-point correspondence exactly (mathematically speaking), but will not generate a usable output image.
The aspect ratio of the destination rectangle can be chosen arbitrarily. In fact, it is not possible to deduce the "original aspect ratio" of the object in world coordinates, because "this is 2D homography, not 3D".
One problem is that to multiply by a camera matrix you need some concept of a z coordinate. You should start by getting basic image warping given Euler angles to work before you think about distortion coefficients. Have a look at this answer for a slightly more detailed explanation and try to duplicate my result. The idea of moving your image down the z axis and then projecting it with your camera matrix can be confusing, let me know if any part of it does not make sense.
You do not need to calibrate the camera nor estimate the camera orientation (the latter, however, in this case would be very easy: just find the vanishing points of those orthogonal bundles of lines, and take their cross product to find the normal to the plane, see Hartley & Zisserman's bible for details).
The only thing you need to do is estimate the homography that maps the checkers to squares, then apply it to the image.
Related
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.
My problem statement is very simple. But I am unable to get the opencv calibration work for me. I am using the code from here : source code.
I have to take images parallel to the camera at a fixed distance. I tried taking test images (about 20 of them) only parallel to the camera as well as at different planes. Also I changed the size and the no of squares.
What would be the best way to calibrate in this scenario?
The undistorted image is cropped later, that's why it looks smaller.
After going through the images closely, the pincushion distortion seems to have been corrected. But the "trapezoidal" distortion still remains. Since the camera is mounted in a closed box, the planes at which I can take images is limited.
To simplify what Vlad already said: It is theoretically impossible to calibrate your camera with test images only parallel to the camera. You have to change your calibration board's orientation. In fact, you should have different orientation in each test image.
Check out the first two images in the link below to see how the calibration board should be slanted (or tilted):
http://www.vision.caltech.edu/bouguetj/calib_doc/
think about calibration problem as finding a projection matrix P:
image_points = P * 3d_points, where P = intrinsic * extrinsic
Now just bear with me:
You basically are interested in intrinsic part but the calibration algorithm has to find both intrinsic and extrinsic. Now, each column of projection matrix can be obtained if you select a 3D point at infinity, for example xInf = [1, 0, 0, 0]. This point is at infinity because when you transform it from homogeneous coordinates to Cartesian you get
[1/0, 0, 0]. If you multiply a projection matrix with a point at infinity you will get its corresponding column (1st for Xinf, 2nd for yInf, 3rd for zInf and 4th for camera center).
Thus the conclusion is simple - to get a projection matrix (that is a successful calibration) you have to clearly see points at infinity or vanishing points from the converging extensions of lines in your chessboard rig (aka end of the railroad tracks at the horizon). Your images don’t make it easy to detect vanishing points since you don’t slant your chessboard, nor rotate nor scale it by stepping back. Thus your calibration will always fail.
I am doing stereo calibration of two cameras (let's name them L and R) with opencv. I use 20 pairs of checkerboard images and compute the transformation of R with respect to L. What I want to do is use a new pair of images, compute the 2d checkerboard corners in image L, transform those points according to my calibration and draw the corresponding transformed points on image R with the hope that they will match the corners of the checkerboard in that image.
I tried the naive way of transforming the 2d points from [x,y] to [x,y,1], multiply by the 3x3 rotation matrix, add the rotation vector and then divide by z, but the result is wrong, so I guess it's not that simple (?)
Edit (to clarify some things):
The reason I want to do this is basically because I want to validate the stereo calibration on a new pair of images. So, I don't actually want to get a new 2d transformation between the two images, I want to check if the 3d transformation I have found is correct.
This is my setup:
I have the rotation and translation relating the two cameras (E), but I don't have rotations and translations of the object in relation to each camera (E_R, E_L).
Ideally what I would like to do:
Choose the 2d corners in image from camera L (in pixels e.g. [100,200] etc).
Do some kind of transformation on the 2d points based on matrix E that I have found.
Get the corresponding 2d points in image from camera R, draw them, and hopefully they match the actual corners!
The more I think about it though, the more I am convinced that this is wrong/can't be done.
What I am probably trying now:
Using the intrinsic parameters of the cameras (let's say I_R and I_L), solve 2 least squares systems to find E_R and E_L
Choose 2d corners in image from camera L.
Project those corners to their corresponding 3d points (3d_points_L).
Do: 3d_points_R = (E_L).inverse * E * E_R * 3d_points_L
Get the 2d_points_R from 3d_points_R and draw them.
I will update when I have something new
It is actually easy to do that but what you're making several mistakes. Remember after stereo calibration R and L relate the position and orientation of the second camera to the first camera in the first camera's 3D coordinate system. And also remember to find the 3D position of a point by a pair of cameras you need to triangulate the position. By setting the z component to 1 you're making two mistakes. First, most likely you have used the common OpenCV stereo calibration code and have given the distance between the corners of the checker board in cm. Hence, z=1 means 1 cm away from the center of camera, that's super close to the camera. Second, by setting the same z for all the points you are saying the checker board is perpendicular to the principal axis (aka optical axis, or principal ray), while most likely in your image that's not the case. So you're transforming some virtual 3D points first to the second camera's coordinate system and then projecting them onto the image plane.
If you want to transform just planar points then you can find the homography between the two cameras (OpenCV has the function) and use that.
I have a small cube with n (you can assume that n = 4) distinguished points on its surface. These points are numbered (1-n) and form a coordinate space, where point #1 is the origin.
Now I'm using a tracking camera to get the coordinates of those points, relative to the camera's coordinate space. That means that I now have n vectors p_i pointing from the origin of the camera to the cube's surface.
With that information, I'm trying to compute the affine transformation matrix (rotation + translation) that represents the transformation between those two coordinate spaces. The translation part is fairly trivial, but I'm struggling with the computation of the rotation matrix.
Is there any build-in functionality in OpenCV that might help me solve this problem?
Sounds like cvGetPerspectiveTransform is what you're looking for; cvFindHomograpy might also be helpful.
solvePnP should give you the rotation matrix and the translation vector. Try it with CV_EPNP or CV_ITERATIVE.
Edit: Or perhaps you're looking for RQ decomposition.
Look at the Stereo Camera tutorial for OpenCV. OpenCV uses a planar chessboard for all the computation and sets its Z-dimension to 0 to build its list of 3D points. You already have 3D points so change the code in the tutorial to reflect your list of 3D points. Then you can compute the transformation.