I have an rendered Image. I want to apply radial and tangential distortion coefficients to my image that I got from opencv. Even though there is undistort function, there is no distort function. How can I distort my images with distortion coefficients?
I was also looking for the same type of functionality. I couldn't find it, so I implemented it myself. Here is the C++ code.
First, you need to normalize the image point using the focal length and centers
rpt(0) = (pt_x - cx) / fx
rpt(1) = (pt_y - cy) / fy
then distort the normalized image point
double x = rpt(0), y = rpt(1);
//determining the radial distortion
double r2 = x*x + y*y;
double icdist = 1 / (1 - ((D.at<double>(4) * r2 + D.at<double>(1))*r2 + D.at<double>(0))*r2);
//determining the tangential distortion
double deltaX = 2 * D.at<double>(2) * x*y + D.at<double>(3) * (r2 + 2 * x*x);
double deltaY = D.at<double>(2) * (r2 + 2 * y*y) + 2 * D.at<double>(3) * x*y;
x = (x + deltaX)*icdist;
y = (y + deltaY)*icdist;
then you can translate and scale the point using the center of projection and focal length:
x = x * fx + cx
y = y * fy + cy
Related
I'm trying to better understand the calibrateCamera and SolvePnP functions in OpenCV, specifically the rotation vectors returned by these functions which I believe is an axis-angle rotation vector (NOT as I had thought initially the yaw,pitch,roll angles). I would like to know the rotation around the x,y and z axis of my checkerboard image. The OpenCV functions return a rotation vector in the form rot = [a,b,c]
Using this answer
as a guide I calculate the angle theta with theta = sqrt(a^2,b^2,c^2) and the rotation axis v = [a/theta, b/theta, c/theta];
Then I take these values and use the Axis-Angle To Euler conversion on euclideanspace.com. shown here:
heading = atan2(y * sin(angle)- x * z * (1 - cos(angle)) , 1 - (y^2 + z^2 ) * (1 - cos(angle)))
attitude = asin(x * y * (1 - cos(angle)) + z * sin(angle))
bank = atan2(x * sin(angle)-y * z * (1 - cos(angle)) , 1 - (x^2 + z^2) * (1 - cos(angle)))
I'm using one of the example OpenCV checkerboard images (Left01.jpg), shown below (note the frame axes in the upper left corner with red = x, green = y, blue = z
Using this image I get a rotation vector from calibrateCamera of [0.166,0.294,0.014]
Running these values through the calculations discussed and converting to degrees I get:
heading = 16.7 deg
attitude = 1.7 deg
bank = 9.3 deg
I believe these correspond to yaw,pitch,roll? The 16.7 degree heading seems high looking at the image, but it's hard to tell. Does this make sense? What would be the correct way to figure out the euler angles (angles around each axis) given the OpenCV rotation vector? Snippets of my code are shown below.
double RMSError = calibrateCamera(
objectPointsArray,
imagePointsArray,
img.size(),
intrinsics,
distortion,
rotation,
translation,
CALIB_ZERO_TANGENT_DIST |
CALIB_FIX_K3 | CALIB_FIX_K4 | CALIB_FIX_K5 |
CALIB_FIX_ASPECT_RATIO);
Mat rvec = rotation.at(0);
//try and get the rotation angles here
//https://stackoverflow.com/questions/12933284/rodrigues-into-eulerangles-and-vice-versa
float theta = sqrt(pow(rvec.at<double>(0),2) + pow(rvec.at<double>(1),2) + pow(rvec.at<double>(2),2));
Mat axis = (Mat_<double>(1, 3) << rvec.at<double>(0) / theta, rvec.at<double>(1) / theta, rvec.at<double>(2) / theta);
float x_ = axis.at<double>(0);
float y_ = axis.at<double>(1);
float z_ = axis.at<double>(2);
//this is yaw,pitch,roll respectively...maybe
float heading = atan2(y_ * sin(theta) - x_ * z_ * (1 - cos(theta)), 1 - (pow(y_,2) + pow(z_,2)) * (1 - static_cast<double>(cos(theta))));
float attitude = asin(x_ * y_ * (1 - cos(theta) + z_ * sin(theta)));
float bank = atan2(x_ * sin(theta) - y_ * z_ * (1 - cos(theta)), 1 - (pow(x_, 2) + pow(z_, 2)) * (1 - static_cast<double>(cos(theta))));
float headingDeg = heading * (180 / 3.14);
float attitudeDeg = attitude * (180 / 3.14);
float bankDeg = bank * (180 / 3.14);
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?
I am new to metal. I want to draw a semi circle. Any ideas or suggestions to draw a semi circle. I tried using tesselation
float u = patch_coord.x;
float v = patch_coord.y;
float w = patch_coord.z;
float x = u * patchIn.control_points[0].position.x + v * patchIn.control_points[1].position.x + w * patchIn.control_points[2].position.x;
float y = u * patchIn.control_points[0].position.y + v * patchIn.control_points[1].position.y + w * patchIn.control_points[2].position.y;
float2 tangent = normalize(float2(x, y)) / 25;
When I normalize always it is in the middle of the screen. I want to draw circle in my finger touch.
I have a fisheye lens:
I would like to undistort it. I apply the FOV model:
rd = 1 / ω * arctan (2 * ru * tan(ω / 2)) //Equation 13
ru = tan(rd * ω) / 2 / tan(ω / 2) //Equation 14
as found in equations (13) and (14) of the INRIA paper "Straight lines have to be straight" https://hal.inria.fr/inria-00267247/document.
The code implementation is the following:
Point2f distortPoint(float w, float h, float cx, float cy, float omega, Point2f input) {
//w = width of the image
//h = height of the image
//cx = center of the lens in the image, aka w/2
//cy = center of the lens in the image, aka h/2
Point2f tmp = new Point2f();
//We normalize the coordinates of the point
tmp.x = input.x / w - cx / w;
tmp.y = input.y / h - cy / h;
//We apply the INRIA key formula (FOV model)
double ru = sqrt(tmp.x * tmp.x + tmp.y * tmp.y);
double rd = 1.0f / omega * atan(2.0f * ru * tan(omega / 2.0f));
tmp.x *= rd / ru;
tmp.y *= rd / ru;
//We "un-normalize" the point
Point2f ret = new Point2f();
ret.x = (tmp.x + cx / w) * w;
ret.y = (tmp.y + cy / h) * h;
return ret;
}
I then used the OpenCV remap function:
//map_x and map_y are computed with distortPoint
remap(img, imgUndistorted, map_x, map_y, INTER_LINEAR, BORDER_CONSTANT, Scalar(0, 0, 0));
I managed to get the distortion model from the lens manufacturer. It's a table of image_height as a function of the field angle:
Field angle(deg) Image Height (mm)
0 0
1 height1
2 height2
3 height3
...
89 height89
90 height90
To give an idea, each height is small and inferior to 2mm.
I found an interesting paper here: https://www.altera.com/content/dam/altera-www/global/en_US/pdfs/literature/wp/wp-01073-flexible-architecture-fisheye-correction-automotive-rear-view-cameras.pdf
How can I modify my pixel-unit undistortion function to factor in the mm-unit table provided by the manufacturer in order to have the most accurate possible undistorted image?
I better explain my problem with an Image
I have a contour and a line which is passing through that contour.
At the intersection point of contour and line I want to draw a perpendicular line at the intersection point of a line and contour up to a particular distance.
I know the intersection point as well as slope of the line.
For reference I am attaching this Image.
If the blue line in your picture goes from point A to point B, and you want to draw the red line at point B, you can do the following:
Get the direction vector going from A to B. This would be:
v.x = B.x - A.x; v.y = B.y - A.y;
Normalize the vector:
mag = sqrt (v.x*v.x + v.y*v.y); v.x = v.x / mag; v.y = v.y / mag;
Rotate the vector 90 degrees by swapping x and y, and inverting one of them. Note about the rotation direction: In OpenCV and image processing in general x and y axis on the image are not oriented in the Euclidian way, in particular the y axis points down and not up. In Euclidian, inverting the final x (initial y) would rotate counterclockwise (standard for euclidean), and inverting y would rotate clockwise. In OpenCV it's the opposite. So, for example to get clockwise rotation in OpenCV: temp = v.x; v.x = -v.y; v.y = temp;
Create a new line at B pointing in the direction of v:
C.x = B.x + v.x * length; C.y = B.y + v.y * length;
(Note that you can make it extend in both directions by creating a point D in the opposite direction by simply negating length.)
This is my version of the function :
def getPerpCoord(aX, aY, bX, bY, length):
vX = bX-aX
vY = bY-aY
#print(str(vX)+" "+str(vY))
if(vX == 0 or vY == 0):
return 0, 0, 0, 0
mag = math.sqrt(vX*vX + vY*vY)
vX = vX / mag
vY = vY / mag
temp = vX
vX = 0-vY
vY = temp
cX = bX + vX * length
cY = bY + vY * length
dX = bX - vX * length
dY = bY - vY * length
return int(cX), int(cY), int(dX), int(dY)