What is the Project Tango lens distortion model? - opencv

The Project Tango C API documentation says that the TANGO_CALIBRATION_POLYNOMIAL_3_PARAMETERS lens distortion is modeled as:
x_corr_px = x_px (1 + k1 * r2 + k2 * r4 + k3 * r6) y_corr_px = y_px (1
+ k1 * r2 + k2 * r4 + k3 * r6)
That is, the undistorted coordinates are a power series function of the distorted coordinates. There is another definition in the Java API, but that description isn't detailed enough to tell which direction the function maps.
I've had a lot of trouble getting things to register properly, and I suspect that the mapping may actually go in the opposite direction, i.e. the distorted coordinates are a power series of the undistorted coordinates. If the camera calibration was produced using OpenCV, then the cause of the problem may be that the OpenCV documentation contradicts itself. The easiest description to find and understand is the OpenCV camera calibration tutorial, which does agree with the Project Tango docs:
But on the other hand, the OpenCV API documentation specifies that the mapping goes the other way:
My experiments with OpenCV show that its API documentation appears correct and the tutorial is wrong. A positive k1 (with all other distortion parameters set to zero) means pincushion distortion, and a negative k1 means barrel distortion. This matches what Wikipedia says about the Brown-Conrady model and will be opposite from the Tsai model. Note that distortion can be modeled either way depending on what makes the math more convenient. I opened a bug against OpenCV for this mismatch.
So my question: Is the Project Tango lens distortion model the same as the one implemented in OpenCV (documentation notwithstanding)?
Here's an image I captured from the color camera (slight pincushioning is visible):
And here's the camera calibration reported by the Tango service:
distortion = {double[5]#3402}
[0] = 0.23019999265670776
[1] = -0.6723999977111816
[2] = 0.6520439982414246
[3] = 0.0
[4] = 0.0
calibrationType = 3
cx = 638.603
cy = 354.906
fx = 1043.08
fy = 1043.1
cameraId = 0
height = 720
width = 1280
Here's how to undistort with OpenCV in python:
>>> import cv2
>>> src = cv2.imread('tango00042.png')
>>> d = numpy.array([0.2302, -0.6724, 0, 0, 0.652044])
>>> m = numpy.array([[1043.08, 0, 638.603], [0, 1043.1, 354.906], [0, 0, 1]])
>>> h,w = src.shape[:2]
>>> mDst, roi = cv2.getOptimalNewCameraMatrix(m, d, (w,h), 1, (w,h))
>>> dst = cv2.undistort(src, m, d, None, mDst)
>>> cv2.imwrite('foo.png', dst)
And that produces this, which is maybe a bit overcorrected at the top edge but much better than my attempts with the reverse model:

The Tango C-API Docs state that (x_corr_px, y_corr_px) is the "corrected output position". This corrected output position needs to then be scaled by focal length and offset by center of projection to correspond to a distorted pixel coordinates.
So, to project a point onto an image, you would have to:
Transform the 3D point so that it is in the frame of the camera
Convert the point into normalized image coordinates (x, y)
Calculate r2, r4, r6 for the normalized image coordinates (r2 = x*x + y*y)
Compute (x_corr_px, y_corr_px) based on the mentioned equations:
x_corr_px = x (1 + k1 * r2 + k2 * r4 + k3 * r6)
y_corr_px = y (1 + k1 * r2 + k2 * r4 + k3 * r6)
Compute distorted coordinates
x_dist_px = x_corr_px * fx + cx
y_dist_px = y_corr_px * fy + cy
Draw (x_dist_px, y_dist_px) on the original, distorted image buffer.
This also means that the corrected coordinates are the normalized coordinates scaled by a power series of the normalized image coordinates' magnitude. (this is the opposite of what the question suggests)
Looking at the implementation of cvProjectPoints2 in OpenCV (see [opencv]/modules/calib3d/src/calibration.cpp), the "Poly3" distortion in OpenCV is being applied the same direction as in Tango. All 3 versions (Tango Docs, OpenCV Tutorials, OpenCV API) are consistent and correct.
Good luck, and hopefully this helps!
(Update: Taking a closer look at a the code, it looks like the corrected coordinates and distorted coordinates are not the same. I've removed the incorrect parts of my response, and the remaining parts of this answer are still correct.)

Maybe it's not the right place to post, but I really want to share the readable version of code used in OpenCV to actually correct the distortion.
I'm sure that I'm not the only one who needs x_corrected and y_corrected and fails to find an easy and understandable formula.
I've rewritten the essential part of cv2.undistortPoints in Python and you may notice that the correction is performed iteratively. This is important, because the solution for polynom of 9-th power does not exist and all we can do is to apply its the reveresed version several times to get the numerical solution.
def myUndistortPoint((x0, y0), CM, DC):
[[k1, k2, p1, p2, k3, k4, k5, k6]] = DC
fx, _, cx = CM[0]
_, fy, cy = CM[1]
x = x_src = (x0 - cx) / fx
y = y_src = (y0 - cy) / fy
for _ in range(5):
r2 = x**2 + y**2
r4 = r2**2
r6 = r2 * r4
rad_dist = (1 + k4*r2 + k5*r4 + k6*r6) / (1 + k1*r2 + k2*r4 + k3*r6)
tang_dist_x = 2*p1 * x*y + p2*(r2 + 2*x**2)
tang_dist_y = 2*p2 * x*y + p1*(r2 + 2*y**2)
x = (x_src - tang_dist_x) * rad_dist
y = (y_src - tang_dist_y) * rad_dist
x = x * fx + cx
y = y * fy + cy
return x, y
To speed up, you can use only three iterations, on most cameras this will give enough precision to fit the pixels.

Related

Error in radial distortion based on (zoomed in) OpenCV-style camera parameters

Our AR device is based on a camera with pretty strong optical zoom. We measure the distortion of this camera using classical camera-calibration tools (checkerboards), both through OpenCV and the GML Camera Calibration tools.
At higher zoom levels (I'll use 249 out of 255 as an example) we measure the following camera parameters at full HD resolution (1920x1080):
fx = 24545.4316
fy = 24628.5469
cx = 924.3162
cy = 440.2694
For the radial and tangential distortion we measured 4 values:
k1 = 5.423406
k2 = -2964.24243
p1 = 0.004201721
p2 = 0.0162647516
We are not sure how to interpret (read: implement) those extremely large values for k1 and k2. Using OpenCV's classic "undistort" operation to rectify the image using these values seems to work well. Unfortunately this is (much) too slow for realtime usage.
The thumbnails below look similar, clicking them will display the full size images where you can spot the difference:
Camera footage
Undistorted using OpenCV
That's why we want to take the opposite aproach: leave the camera footage be distorted and apply a similar distortion to our 3D scene using shaders. Following the OpenCV documentation and this accepted answer in particular, the distorted position for a corner point (0, 0) would be
// To relative coordinates
double x = (point.X - cx) / fx; // -960 / 24545 = -0.03911
double y = (point.Y - cy) / fy; // -540 / 24628 = -0.02193
double r2 = x*x + y*y; // 0.002010
// Radial distortion
// -0.03911 * (1 + 5.423406 * 0.002010 + -2964.24243 * 0.002010 * 0.002010) = -0.039067
double xDistort = x * (1 + k1 * r2 + k2 * r2 * r2);
// -0.02193 * (1 + 5.423406 * 0.002010 + -2964.24243 * 0.002010 * 0.002010) = -0.021906
double yDistort = y * (1 + k1 * r2 + k2 * r2 * r2);
// Tangential distortion
... left out for brevity
// Back to absolute coordinates.
xDistort = xDistort * fx + cx; // -0.039067 * 24545.4316 + 924.3162 = -34.6002 !!!
yDistort = yDistort * fy + cy; // -0.021906 * 24628.5469 + 440.2694 = = -99.2435 !!!
These large pixel displacements (34 and 100 pixels at the upper left corner) seem overly warped and do not correspond with the undistorted image OpenCV generates.
So the specific question is: what is wrong with the way we interpreted the values we measured, and what should the correct code for distortion be?

Computing 3D coordinates of keypoints in multiple images

I have multiple images of an object taken by the same calibrated camera. Let's say calibrated means both intrinsic and extrinsic parameters (I can put a checkerboard next to the object, so all parameters can be retrieved). On these images I can find matching keypoints using SIFT or SURF, and some matching algorithm, this is basic OpenCV. But how do I do the 3D reconstruction of these points from multiple images? This is not a classic stereo arrangement, so there are more than 2 images with the same object points on them, and I want to use as many as possible for increased accuracy.
Are there any built-in OpenCV functions that do this?
(Note that this is done off-line, the solution does not need to be fast, but robust)
I guess you are looking for so-called Structur from motion approaches. They are using multiple images from different viewpoints and return a 3D reconstruction (e.g. a pointcloud). It looks like OpenCV has a SfM module in the contrib package, but I have no experiences with it.
However, I used to work with bundler. It was quite uncomplicated and returns the entire information (camera calibration and point positions) as text file and you can view the point cloud with Meshlab. Please note that it uses SIFT keypoints and descriptors for correspondence establishment.
I think I have found a solution for this. Structure from motion algorithms deal with the case where the cameras are not calibrated, but in this case all intrinsic and extrinsic parameters are known.
The problem degrades into a linear least squares problem:
We have to compute the coordinates for a single object point:
X = [x, y, z, 1]'
C = [x, y, z]'
X = [[C], [1]]
We are given n images, which have these transformation matrices:
Pi = Ki * [Ri|ti]
These matrices are already known. The object point is projected on the images at
U = [ui, vi]
We can write in homogeneous coordinates (the operator * represents both matrix multiplication, dot product and scalar multiplication):
[ui * wi, vi * wi, wi]' = Pi * X
Pi = [[p11i, p12i, p13i, p14i],
[p21i, p22i, p23i, p24i],
[p31i, p32i, p33i, p34i]]
Let's define the following:
p1i = [p11i, p12i, p13i] (the first row of Pi missing the last element)
p2i = [p21i, p22i, p23i] (the second row of Pi missing the last element)
p3i = [p31i, p32i, p33i] (the third row of Pi missing the last element)
a1i = p14i
a2i = p24i
a3i = p34i
Then we can write:
Q = [x, y, z]
wi = p3i * Q + a3i
ui = (p1i * Q + a1i) / wi =
= (p1i * Q + a1i) / (p3i * Q + a3i)
ui * p3i * Q + ui * a3i - p1i * Q - a1i = 0
(ui * p3i - p1i) * Q = a1i - a3i
Similarly for vi:
(vi * p3i - p2i) * Q = a2i - a3i
And this holds for i = 1..n. We can write this in matrix form:
G * Q = b
G = [[u1 * p31 - p11],
[v1 * p31 - p21],
[u2 * p32 - p12],
[v2 * p32 - p22],
...
[un * p3n - p1n],
[vn * p3n - p2n]]
b = [[a11 - a31 * u1],
[a21 - a31 * v1],
[a12 - a32 * u2],
[a22 - a32 * v2],
...
[a1n - a3n * un],
[a2n - a3n * vn]]
Since G and b are known from the Pi matrices, and the image points [ui, vi], we can compute the pseudoinverse of G (call it G_), and compute:
Q = G_ * b

Obtain sigma of gaussian blur between two images

Suppose I have an image A, I applied Gaussian Blur on it with Sigam=3 So I got another Image B. Is there a way to know the applied sigma if A,B is given?
Further clarification:
Image A:
Image B:
I want to write a function that take A,B and return Sigma:
double get_sigma(cv::Mat const& A,cv::Mat const& B);
Any suggestions?
EDIT1: The suggested approach doesn't work in practice in its original form(i.e. using only 9 equations for a 3 x 3 kernel), and I realized this later. See EDIT1 below for an explanation and EDIT2 for a method that works.
EDIT2: As suggested by Humam, I used the Least Squares Estimate (LSE) to find the coefficients.
I think you can estimate the filter kernel by solving a linear system of equations in this case. A linear filter weighs the pixels in a window by its coefficients, then take their sum and assign this value to the center pixel of the window in the result image. So, for a 3 x 3 filter like
the resulting pixel value in the filtered image
result_pix_value = h11 * a(y, x) + h12 * a(y, x+1) + h13 * a(y, x+2) +
h21 * a(y+1, x) + h22 * a(y+1, x+1) + h23 * a(y+1, x+2) +
h31 * a(y+2, x) + h32 * a(y+2, x+1) + h33 * a(y+2, x+2)
where a's are the pixel values within the window in the original image. Here, for the 3 x 3 filter you have 9 unknowns, so you need 9 equations. You can obtain those 9 equations using 9 pixels in the resulting image. Then you can form an Ax = b system and solve for x to obtain the filter coefficients. With the coefficients available, I think you can find the sigma.
In the following example I'm using non-overlapping windows as shown to obtain the equations.
You don't have to know the size of the filter. If you use a larger size, the coefficients that are not relevant will be close to zero.
Your result image size is different than the input image, so i didn't use that image for following calculation. I use your input image and apply my own filter.
I tested this in Octave. You can quickly run it if you have Octave/Matlab. For Octave, you need to load the image package.
I'm using the following kernel to blur the image:
h =
0.10963 0.11184 0.10963
0.11184 0.11410 0.11184
0.10963 0.11184 0.10963
When I estimate it using a window size 5, I get the following. As I said, the coefficients that are not relevant are close to zero.
g =
9.5787e-015 -3.1508e-014 1.2974e-015 -3.4897e-015 1.2739e-014
-3.7248e-014 1.0963e-001 1.1184e-001 1.0963e-001 1.8418e-015
4.1825e-014 1.1184e-001 1.1410e-001 1.1184e-001 -7.3554e-014
-2.4861e-014 1.0963e-001 1.1184e-001 1.0963e-001 9.7664e-014
1.3692e-014 4.6182e-016 -2.9215e-014 3.1305e-014 -4.4875e-014
EDIT1:
First of all, my apologies.
This approach doesn't really work in the practice. I've used the filt = conv2(a, h, 'same'); in the code. The resulting image data type in this case is double, whereas in the actual image the data type is usually uint8, so there's loss of information, which we can think of as noise. I simulated this with the minor modification filt = floor(conv2(a, h, 'same'));, and then I don't get the expected results.
The sampling approach is not ideal, because it's possible that it results in a degenerated system. Better approach is to use random sampling, avoiding the borders and making sure the entries in the b vector are unique. In the ideal case, as in my code, we are making sure the system Ax = b has a unique solution this way.
One approach would be to reformulate this as Mv = 0 system and try to minimize the squared norm of Mv under the constraint squared-norm v = 1, which we can solve using SVD. I could be wrong here, and I haven't tried this.
Another approach is to use the symmetry of the Gaussian kernel. Then a 3x3 kernel will have only 3 unknowns instead of 9. I think, this way we impose additional constraints on v of the above paragraph.
I'll try these out and post the results, even if I don't get the expected results.
EDIT2:
Using the LSE, we can find the filter coefficients as pinv(A'A)A'b. For completion, I'm adding a simple (and slow) LSE code.
Initial Octave Code:
clear all
im = double(imread('I2vxD.png'));
k = 5;
r = floor(k/2);
a = im(:, :, 1); % take the red channel
h = fspecial('gaussian', [3 3], 5); % filter with a 3x3 gaussian
filt = conv2(a, h, 'same');
% use non-overlapping windows to for the Ax = b syatem
% NOTE: boundry error checking isn't performed in the code below
s = floor(size(a)/2);
y = s(1);
x = s(2);
w = k*k;
y1 = s(1)-floor(w/2) + r;
y2 = s(1)+floor(w/2);
x1 = s(2)-floor(w/2) + r;
x2 = s(2)+floor(w/2);
b = [];
A = [];
for y = y1:k:y2
for x = x1:k:x2
b = [b; filt(y, x)];
f = a(y-r:y+r, x-r:x+r);
A = [A; f(:)'];
end
end
% estimated filter kernel
g = reshape(A\b, k, k)
LSE method:
clear all
im = double(imread('I2vxD.png'));
k = 5;
r = floor(k/2);
a = im(:, :, 1); % take the red channel
h = fspecial('gaussian', [3 3], 5); % filter with a 3x3 gaussian
filt = floor(conv2(a, h, 'same'));
s = size(a);
y1 = r+2; y2 = s(1)-r-2;
x1 = r+2; x2 = s(2)-r-2;
b = [];
A = [];
for y = y1:2:y2
for x = x1:2:x2
b = [b; filt(y, x)];
f = a(y-r:y+r, x-r:x+r);
f = f(:)';
A = [A; f];
end
end
g = reshape(A\b, k, k) % A\b returns the least squares solution
%g = reshape(pinv(A'*A)*A'*b, k, k)

Calculate distance (disparity) OpenCV

-- Update 2 --
The following article is really useful (although it is using Python instead of C++) if you are using a single camera to calculate the distance: Find distance from camera to object/marker using Python and OpenCV
Best link is Stereo Webcam Depth Detection. The implementation of this open source project is really clear.
Below is the original question.
For my project I am using two camera's (stereo vision) to track objects and to calculate the distance. I calibrated them with the sample code of OpenCV and generated a disparity map.
I already implemented a method to track objects based on color (this generates a threshold image).
My question: How can I calculate the distance to the tracked colored objects using the disparity map/ matrix?
Below you can find a code snippet that gets the x,y and z coordinates of each pixel. The question: Is Point.z in cm, pixels, mm?
Can I get the distance to the tracked object with this code?
Thank you in advance!
cvReprojectImageTo3D(disparity, Image3D, _Q);
vector<CvPoint3D32f> PointArray;
CvPoint3D32f Point;
for (int y = 0; y < Image3D->rows; y++) {
float *data = (float *)(Image3D->data.ptr + y * Image3D->step);
for (int x = 0; x < Image3D->cols * 3; x = x + 3)
{
Point.x = data[x];
Point.y = data[x+1];
Point.z = data[x+2];
PointArray.push_back(Point);
//Depth > 10
if(Point.z > 10)
{
printf("%f %f %f", Point.x, Point.y, Point.z);
}
}
}
cvReleaseMat(&Image3D);
--Update 1--
For example I generated this thresholded image (of the left camera). I almost have the same of the right camera.
Besides the above threshold image, the application generates a disparity map. How can I get the Z-coordinates of the pixels of the hand in the disparity map?
I actually want to get all the Z-coordinates of the pixels of the hand to calculate the average Z-value (distance) (using the disparity map).
See this links: OpenCV: How-to calculate distance between camera and object using image?, Finding distance from camera to object of known size, http://answers.opencv.org/question/5188/measure-distance-from-detected-object-using-opencv/
If it won't solve you problem, write more details - why it isn't working, etc.
The math for converting disparity (in pixels or image width percentage) to actual distance is pretty well documented (and not very difficult) but I'll document it here as well.
Below is an example given a disparity image (in pixels) and an input image width of 2K (2048 pixels across) image:
Convergence Distance is determined by the rotation between camera lenses. In this example it will be 5 meters. Convergence distance of 5 (meters) means that the disparity of objects 5 meters away is 0.
CD = 5 (meters)
Inverse of convergence distance is: 1 / CD
IZ = 1/5 = 0.2M
Size of camera's sensor in meters
SS = 0.035 (meters) //35mm camera sensor
The width of a pixel on the sensor in meters
PW = SS/image resolution = 0.035 / 2048(image width) = 0.00001708984
The focal length of your cameras in meters
FL = 0.07 //70mm lens
InterAxial distance: The distance from the center of left lens to the center of right lens
IA = 0.0025 //2.5mm
The combination of the physical parameters of your camera rig
A = FL * IA / PW
Camera Adjusted disparity: (For left view only, right view would use positive [disparity value])
AD = 2 * (-[disparity value] / A)
From here you can compute actual distance using the following equation:
realDistance = 1 / (IZ – AD)
This equation only works for "toe-in" camera systems, parallel camera rigs will use a slightly different equation to avoid infinity values, but I'll leave it at this for now. If you need the parallel stuff just let me know.
if len(puntos) == 2:
x1, y1, w1, h1 = puntos[0]
x2, y2, w2, h2 = puntos[1]
if x1 < x2:
distancia_pixeles = abs(x2 - (x1+w1))
distancia_cm = (distancia_pixeles*29.7)/720
cv2.putText(imagen_A4, "{:.2f} cm".format(distancia_cm), (x1+w1+distancia_pixeles//2, y1-30), 2, 0.8, (0,0,255), 1,
cv2.LINE_AA)
cv2.line(imagen_A4,(x1+w1,y1-20),(x2, y1-20),(0, 0, 255),2)
cv2.line(imagen_A4,(x1+w1,y1-30),(x1+w1, y1-10),(0, 0, 255),2)
cv2.line(imagen_A4,(x2,y1-30),(x2, y1-10),(0, 0, 255),2)
else:
distancia_pixeles = abs(x1 - (x2+w2))
distancia_cm = (distancia_pixeles*29.7)/720
cv2.putText(imagen_A4, "{:.2f} cm".format(distancia_cm), (x2+w2+distancia_pixeles//2, y2-30), 2, 0.8, (0,0,255), 1,
cv2.LINE_AA)
cv2.line(imagen_A4,(x2+w2,y2-20),(x1, y2-20),(0, 0, 255),2)
cv2.line(imagen_A4,(x2+w2,y2-30),(x2+w2, y2-10),(0, 0, 255),2)
cv2.line(imagen_A4,(x1,y2-30),(x1, y2-10),(0, 0, 255),2)
cv2.imshow('imagen_A4',imagen_A4)
cv2.imshow('frame',frame)
k = cv2.waitKey(1) & 0xFF
if k == 27:
break
cap.release()
cv2.destroyAllWindows()
I think this is a good way to measure the distance between two objects

Describing nonlinear transformation between two images, using homography

A one to one point matching has already been established
between the blue dots on the two images.
The image2 is the distorted version of the image1. The distortion model seems to be
eyefish lens distortion. The question is:
Is there any way to compute a transformation matrix which describes this transition.
In fact a matrix which transforms the blue
dots on the first image to their corresponding blue dots on the second image?
The problem here is that we don’t know the focal length(means images are uncalibrated), however we do have
perfect matching between around 200 points on the two images.
the distorted image:
I think what you're trying to do can be treated as a distortion correction problem, without the need of the rest of a classic camera calibration.
A matrix transformation is a linear one and linear transformations map always straight lines into straight lines (http://en.wikipedia.org/wiki/Linear_map). It is apparent from the picture that the transformation is nonlinear so you cannot describe it with a matrix operation.
That said, you can use a lens distortion model like the one used by OpenCV (http://docs.opencv.org/doc/tutorials/calib3d/camera_calibration/camera_calibration.html) and obtaining the coefficients shouldn't be very difficult. Here is what you can do in Matlab:
Call (x, y) the coordinates of an original point (top picture) and (xp, yp) the coordinates of a distorted point (bottom picture), both shifted to the center of the image and divided by a scaling factor (same for x and y) so they lie more or less in the [-1, 1] interval. The distortion model is:
x = ( xp*(1 + k1*r^2 + k2*r^4 + k3*r^6) + 2*p1*xp*yp + p2*(r^2 + 2*xp^2));
y = ( yp*(1 + k1*r^2 + k2*r^4 + k3*r^6) + 2*p2*xp*yp + p1*(r^2 + 2*yp^2));
Where
r = sqrt(x^2 + y^2);
You have 5 parameters: k1, k2, k3, p1, p2 for radial and tangential distortion and 200 pairs of points, so you can solve the nonlinear system.
Be sure the x, y, xp and yp arrays exist in the workspace and declare them global:
global x y xp yp
Write a function to evaluate the mean square error given a set of arbitrary distortion coefficients, say it's called 'dist':
function val = dist(var)
global x y xp yp
val = zeros(size(xp));
k1 = var(1);
k2 = var(2);
k3 = var(3);
p1 = var(4);
p2 = var(5);
r = sqrt(xp.*xp + yp.*yp);
temp1 = x - ( xp.*(1 + k1*r.^2 + k2*r.^4 + k3*r.^6) + 2*p1*xp.*yp + p2*(r.^2 + 2*xp.^2));
temp2 = y - ( yp.*(1 + k1*r.^2 + k2*r.^4 + k3*r.^6) + 2*p2*xp.*yp + p1*(r.^2 + 2*yp.^2));
val = sqrt(temp1.*temp1 + temp2.*temp2);
Solve the system with 'fsolve":
[coef, fval] = fsolve(#dist, zeros(5,1));
The values in 'coef' are the distortion coefficients you're looking for. To correct the distortion of new points (xp, yp) not present in the original set, use the equations:
r = sqrt(xp.*xp + yp.*yp);
x_corr = xp.*(1 + k1*r.^2 + k2*r.^4 + k3*r.^6) + 2*p1*xp.*yp + p2*(r.^2 + 2*xp.^2);
y_corr = yp.*(1 + k1*r.^2 + k2*r.^4 + k3*r.^6) + 2*p2*xp.*yp + p1*(r.^2 + 2*yp.^2);
Results will be shifted to the center of the image and scaled by the factor you used above.
Notes:
Coordinates must be shifted to the center of the image as the distortion is symmetric with respect to it.
It should't be necessary to normalize to the interval [-1, 1] but it is comon to do it so the distortion coefficients obtained are more or less of the same order of magnitude (working with powers 2, 4 and 6 of pixel coordinates would need very small coefficients).
This method doesn't require the points in the image to be in an uniform grid.

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