convert an image from Cartesian to Polar - image

I'm trying to convert an image with many circles with the same center, from Cartesian to Polar (so that the new image will be the circles but lines instead of the circles, see the image below), and that's working out just fine using the following code:
[r, c] = size(img);
r=floor(r/2);
c=floor(c/2);
[X, Y] = meshgrid(-c:c-1,-r:r-1);
[theta, rho] = cart2pol(X, Y);
subplot(221), imshow(img), axis on;
hold on;
subplot(221), plot(xCenter,yCenter, 'r+');
subplot(222), warp(theta, rho, zeros(size(theta)), img);
view(2), axis square;
The problem is, I don't understand why does it even work? (obviously it's not my code), I mean, when I use the function cart2pol I don't even use the image, it's just some vectors x and y generated from the meshgrid function..
and another problem is, I want somehow to have a new image (not just to be able to draw it with the wrap function) which is the original image but by the theta and rho coordinates (meaning the same pixels but rearranged)... I'm not even sure how to ask this, in the end I want to have an image which is a matrix so that I can sum each row and turn the matrix into a column vector...

You can think of your image as being a 2D matrix, where each pixel has an X and Y coordinate
[(1,1) (1,2) (1,3) .... (1,c)]
[(2,1) (2,2) (2,3) .... (2,c)]
[(3,1) (3,2) (3,3) .... (3,c)]
[.... .... .... .... .... ]
[(r,1) (r,2) (r,3) .... (r,c)]
In the code that you posted, it maps each of these (X,Y) coordinates to it's equivalent polar coordinate (R, theta) using the center of the image floor(c/2) and floor(r/2) as the reference point.
% Map pixel value at (1,1) to it's polar equivalent
[r,theta] = cart2pol(1 - floor(r/2),1 - floor(c/2));
So whatever pixel value was used for (1,1) should now appear in your new polar coordinate space at (r,theta). It is important to note that to do this conversion, no information about the actual pixel values in the image matters, rather we just want to perform this transformation for each pixel within the image.
So first we figure out where the center of the image is:
[r, c] = size(img);
r = floor(r / 2);
c = floor(c / 2);
Then we figure out the (X,Y) coordinates for every point in the image (after the center has already been subtracted out
[X, Y] = meshgrid(-c:c-1,-r:r-1);
Now convert all of these cartesian points to polar coordinates
[theta, rho] = cart2pol(X, Y);
All that warp now does, is say "display the value of img at (X,Y) at it's corresponding location in (theta, rho)"
warp(theta, rho, zeros(size(theta)), img);
Now it seems that you want a new 2D image where the dimensions are [nTheta, nRho]. To do this, you could use griddata to interpolate your scattered (theta, rho) image (which is displayed by warp above) to a regular grid.
% These is the spacing of your radius axis (columns)
rhoRange = linspace(0, max(rho(:)), 100);
% This is the spacing of your theta axis (rows)
thetaRange = linspace(-pi, pi, 100);
% Generate a grid of all (theta, rho) coordinates in your destination image
[T,R] = meshgrid(thetaRange, rhoRange);
% Now map the values in img to your new image domain
theta_rho_image = griddata(theta, rho, double(img), T, R);
Take a look at all the interpolation methods for griddata to figure out which is most appropriate for your scenario.
There were a couple other issues (like the rounding of the center) which caused the result to be slightly incorrect. A fully working example is provided below
% Create an image of circles
radii = linspace(0, 40, 10);
rows = 100;
cols = 100;
img = zeros(rows, cols);
for k = 1:numel(radii)
t = linspace(0, 2*pi, 1000);
xx = round((cos(t) * radii(k)) + (cols / 2));
yy = round((sin(t) * radii(k)) + (rows / 2));
toremove = xx > cols | xx < 1 | yy > rows | yy < 1;
inds = sub2ind(size(img), xx(~toremove), yy(~toremove));
img(inds) = 1;
end
[r,c] = size(img);
center_row = r / 2;
center_col = c / 2;
[X,Y] = meshgrid((1:c) - center_col, (1:r) - center_row);
[theta, rho] = cart2pol(X, Y);
rhoRange = linspace(0, max(rho(:)), 1000);
thetaRange = linspace(-pi, pi, 1000);
[T, R] = meshgrid(thetaRange, rhoRange);
theta_rho_image = griddata(theta, rho, double(img), T, R);
figure
subplot(1,2,1);
imshow(img);
title('Original Image')
subplot(1,2,2);
imshow(theta_rho_image);
title('Polar Image')
And the result

Related

Interp2 of image with transformed coordinates

I have 2 greyscale images that i am trying to align using scalar scaling 1 , rotation matrix [2,2] and translation vector [2,1]. I can calculate image1's transformed coordinates as
y = s*R*x + t;
Below the resulting images are shown.
The first image is image1 before transformation,
the second image is image1 (red) with attempted interpolation using interp2 shown on top of image2 (green)
The third image is when i manually insert the pixel values from image1 into an empty array (that has the same size as image2) using the transformed coordinates.
From this we can see that the coordinate transformation must have been successful, as the images are aligned although not perfectly (which is to be expected since only 2 coordinates were used in calculating s, R and t) .
How come interp2 is not producing a result more similar to when i manually insert pixel values?
Below the code for doing this is included:
Interpolation code
function [transformed_image] = interpolate_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
% doesn't help if i use get_grid that the other function is using here
[~, grid_xr, grid_yr] = get_ipgrid(im_r);
[x_t, grid_xt, grid_yt] = get_ipgrid(im_t);
y = s*R*x_t + t;
yx = reshape(y(1,:), m,n);
yy = reshape(y(2,:), m,n);
transformed_image = interp2(grid_xr, grid_yr, im_r, yx, yy, 'nearest');
end
function [x, grid_x, grid_y] = get_ipgrid(image)
[m,n] = size(image);
[grid_x,grid_y] = meshgrid(1:n,1:m);
x = [reshape(grid_x, 1, []); reshape(grid_y, 1, [])]; % X is [2xM*N] coordinate pairs
end
The manual code
function [transformed_image] = transform_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
[x_t, grid_xt, grid_yt] = get_grid(im_t);
y = s*R*x_t + t;
ymat = reshape(y',m,n,2);
yx = ymat(:,:,1);
yy = ymat(:,:,2);
transformed_image = zeros(m,n);
for i = 1:m
for j = 1:n
% make sure coordinates are inside
if (yx(i,j) < m & yy(i,j) < n & yx(i,j) > 0.5 & yy(i,j) > 0.5)
transformed_image(round(yx(i,j)),round(yy(i,j))) = im_r(i,j);
end
end
end
end
function [x, grid_x, grid_y] = get_grid(image)
[m,n] = size(image);
[grid_y,grid_x] = meshgrid(1:n,1:m);
x = [grid_x(:) grid_y(:)]'; % X is [2xM*N] coordinate pairs
end
Can anyone see what i'm doing wrong with interp2? I feel like i have tried everything
Turns out i got interpolation all wrong.
In my question i calculate the coordinates of im1 in im2.
However the way interpolation works is that i need to calculate the coordinates of im2 in im1 such that i can map the image as shown below.
This means that i also calculated the wrong s,R and t since they were used to transform im1 -> im2, where as i needed im2 -> im1. (this is also called the inverse transform). Below is the manual code, that is basically the same as interp2 with nearest neighbour interpolation
function [transformed_image] = transform_image(im_r,im_t,s,R,t)
[m,n] = size(im_t);
[x_t, grid_xt, grid_yt] = get_grid(im_t);
y = s*R*x_t + t;
ymat = reshape(y',m,n,2);
yx = ymat(:,:,1);
yy = ymat(:,:,2);
transformed_image = zeros(m,n);
for i = 1:m
for j = 1:n
% make sure coordinates are inside
if (yx(i,j) < m & yy(i,j) < n & yx(i,j) > 0.5 & yy(i,j) > 0.5)
transformed_image(i,j) = im_r(round(yx(i,j)),round(yy(i,j)));
end
end
end
end

Rotate image in Matlab about arbitrary points

I am trying to rotate an image in Matlab about an arbitrary set of points.
So far, I have used imrotate, but it looks like imrotate only rotates about the center.
Is there a nice way of doing this without first padding the image and then using imrotate?
Thank you
The "nice way" is using imwarp
Building the transformation matrix is a little tricky.
I figured out how to build it from a the following question: Matlab image rotation
The transformation supports rotation, translation and zoom.
Parameters:
(x0, y0) is the center point that you rotate around it.
phi is rotation angle.
sx, sy are horizontal and vertical zoom (set to value to 1).
W and H are width and height of input (and output) images.
Building 3x3 transformation matrix:
T = [sx*cos(phi), -sx*sin(phi), 0
sy*sin(phi), sy*cos(phi), 0
(W+1)/2-((sx*x0*cos(phi))+(sy*y0*sin(phi))), (H+1)/2+((sx*x0*sin(phi))-(sy*y0*cos(phi))), 1];
Using imwarp:
tform = affine2d(T);
J = imwarp(I, tform, 'OutputView', imref2d([H, W]), 'Interp', 'cubic');
Here is a complete executable code sample:
I = imresize(imread('peppers.png'), 0.5); %I is the input image
[H, W, ~] = size(I); %Height and Width of I
phi = 120*pi/180; %Rotate 120 degrees
%Zoom coefficients
sx = 1;
sy = 1;
%Center point (the point that the image is rotated around it).
x0 = (W+1)/2 + 50;
y0 = (H+1)/2 + 20;
%Draw white cross at the center of the point of the input image.
I(y0-0.5:y0+0.5, x0-19.5:x0+19.5, :) = 255;
I(y0-19.5:y0+19.5, x0-0.5:x0+0.5, :) = 255;
%Build transformation matrix.
T = [sx*cos(phi), -sx*sin(phi), 0
sy*sin(phi), sy*cos(phi), 0
(W+1)/2-((sx*x0*cos(phi))+(sy*y0*sin(phi))), (H+1)/2+((sx*x0*sin(phi))-(sy*y0*cos(phi))), 1];
tform = affine2d(T);
J = imwarp(I, tform, 'OutputView', imref2d([H, W]), 'Interp', 'cubic');
%Draw black cross at the center of the output image:
J(end/2:end/2+1, end/2-15:end/2+15, :) = 0;
J(end/2-15:end/2+15, end/2:end/2+1, :) = 0;
%Shows that the center of the output image is the point that the image was rotated around it.
figure;imshow(J)
Input image:
Output image:
Note:
Important advantage over other methods (like imrotate after padding), is that the center coordinates don't have to be integer values.
You can rotate around (100.4, 80.7) for example.

Plotting a 2D Moving Image in MatLab [duplicate]

I'm trying to plot small images on a larger plot... Actually its isomap algorithm, I got many points, now each point correspond to some image, I know which image is it... The porblem is how to load that image and plot on the graph?
One more thing I have to plot both image and the points, so, basically the images will overlap the points.
Certainly, the type of image given here
Something like this should get you started. You can use the low-level version of the image function to draw onto a set of axes.
% Define some random data
N = 5;
x = rand(N, 1);
y = rand(N, 1);
% Load an image
rgb = imread('ngc6543a.jpg');
% Draw a scatter plot
scatter(x, y);
axis([0 1 0 1]);
% Offsets of image from associated point
dx = 0.02;
dy = 0.02;
width = 0.1;
height = size(rgb, 1) / size(rgb, 2) * width;
for i = 1:N
image('CData', rgb,...
'XData', [x(i)-dx x(i)-(dx+width)],...
'YData', [y(i)-dy y(i)-(dy+height)]);
end

Remove barrel distortion from an image in MATLAB [duplicate]

BOUNTY STATUS UPDATE:
I discovered how to map a linear lens, from destination coordinates to source coordinates.
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
1). I actually struggle to reverse it, and to map source coordinates to destination coordinates. What is the inverse, in code in the style of the converting functions I posted?
2). I also see that my undistortion is imperfect on some lenses - presumably those that are not strictly linear. What is the equivalent to-and-from source-and-destination coordinates for those lenses? Again, more code than just mathematical formulae please...
Question as originally stated:
I have some points that describe positions in a picture taken with a fisheye lens.
I want to convert these points to rectilinear coordinates. I want to undistort the image.
I've found this description of how to generate a fisheye effect, but not how to reverse it.
There's also a blog post that describes how to use tools to do it; these pictures are from that:
(1) : SOURCE Original photo link
Input : Original image with fish-eye distortion to fix.
(2) : DESTINATION Original photo link
Output : Corrected image (technically also with perspective correction, but that's a separate step).
How do you calculate the radial distance from the centre to go from fisheye to rectilinear?
My function stub looks like this:
Point correct_fisheye(const Point& p,const Size& img) {
// to polar
const Point centre = {img.width/2,img.height/2};
const Point rel = {p.x-centre.x,p.y-centre.y};
const double theta = atan2(rel.y,rel.x);
double R = sqrt((rel.x*rel.x)+(rel.y*rel.y));
// fisheye undistortion in here please
//... change R ...
// back to rectangular
const Point ret = Point(centre.x+R*cos(theta),centre.y+R*sin(theta));
fprintf(stderr,"(%d,%d) in (%d,%d) = %f,%f = (%d,%d)\n",p.x,p.y,img.width,img.height,theta,R,ret.x,ret.y);
return ret;
}
Alternatively, I could somehow convert the image from fisheye to rectilinear before finding the points, but I'm completely befuddled by the OpenCV documentation. Is there a straightforward way to do it in OpenCV, and does it perform well enough to do it to a live video feed?
The description you mention states that the projection by a pin-hole camera (one that does not introduce lens distortion) is modeled by
R_u = f*tan(theta)
and the projection by common fisheye lens cameras (that is, distorted) is modeled by
R_d = 2*f*sin(theta/2)
You already know R_d and theta and if you knew the camera's focal length (represented by f) then correcting the image would amount to computing R_u in terms of R_d and theta. In other words,
R_u = f*tan(2*asin(R_d/(2*f)))
is the formula you're looking for. Estimating the focal length f can be solved by calibrating the camera or other means such as letting the user provide feedback on how well the image is corrected or using knowledge from the original scene.
In order to solve the same problem using OpenCV, you would have to obtain the camera's intrinsic parameters and lens distortion coefficients. See, for example, Chapter 11 of Learning OpenCV (don't forget to check the correction). Then you can use a program such as this one (written with the Python bindings for OpenCV) in order to reverse lens distortion:
#!/usr/bin/python
# ./undistort 0_0000.jpg 1367.451167 1367.451167 0 0 -0.246065 0.193617 -0.002004 -0.002056
import sys
import cv
def main(argv):
if len(argv) < 10:
print 'Usage: %s input-file fx fy cx cy k1 k2 p1 p2 output-file' % argv[0]
sys.exit(-1)
src = argv[1]
fx, fy, cx, cy, k1, k2, p1, p2, output = argv[2:]
intrinsics = cv.CreateMat(3, 3, cv.CV_64FC1)
cv.Zero(intrinsics)
intrinsics[0, 0] = float(fx)
intrinsics[1, 1] = float(fy)
intrinsics[2, 2] = 1.0
intrinsics[0, 2] = float(cx)
intrinsics[1, 2] = float(cy)
dist_coeffs = cv.CreateMat(1, 4, cv.CV_64FC1)
cv.Zero(dist_coeffs)
dist_coeffs[0, 0] = float(k1)
dist_coeffs[0, 1] = float(k2)
dist_coeffs[0, 2] = float(p1)
dist_coeffs[0, 3] = float(p2)
src = cv.LoadImage(src)
dst = cv.CreateImage(cv.GetSize(src), src.depth, src.nChannels)
mapx = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
mapy = cv.CreateImage(cv.GetSize(src), cv.IPL_DEPTH_32F, 1)
cv.InitUndistortMap(intrinsics, dist_coeffs, mapx, mapy)
cv.Remap(src, dst, mapx, mapy, cv.CV_INTER_LINEAR + cv.CV_WARP_FILL_OUTLIERS, cv.ScalarAll(0))
# cv.Undistort2(src, dst, intrinsics, dist_coeffs)
cv.SaveImage(output, dst)
if __name__ == '__main__':
main(sys.argv)
Also note that OpenCV uses a very different lens distortion model to the one in the web page you linked to.
(Original poster, providing an alternative)
The following function maps destination (rectilinear) coordinates to source (fisheye-distorted) coordinates. (I'd appreciate help in reversing it)
I got to this point through trial-and-error: I don't fundamentally grasp why this code is working, explanations and improved accuracy appreciated!
def dist(x,y):
return sqrt(x*x+y*y)
def correct_fisheye(src_size,dest_size,dx,dy,factor):
""" returns a tuple of source coordinates (sx,sy)
(note: values can be out of range)"""
# convert dx,dy to relative coordinates
rx, ry = dx-(dest_size[0]/2), dy-(dest_size[1]/2)
# calc theta
r = dist(rx,ry)/(dist(src_size[0],src_size[1])/factor)
if 0==r:
theta = 1.0
else:
theta = atan(r)/r
# back to absolute coordinates
sx, sy = (src_size[0]/2)+theta*rx, (src_size[1]/2)+theta*ry
# done
return (int(round(sx)),int(round(sy)))
When used with a factor of 3.0, it successfully undistorts the images used as examples (I made no attempt at quality interpolation):
Dead link
(And this is from the blog post, for comparison:)
If you think your formulas are exact, you can comput an exact formula with trig, like so:
Rin = 2 f sin(w/2) -> sin(w/2)= Rin/2f
Rout= f tan(w) -> tan(w)= Rout/f
(Rin/2f)^2 = [sin(w/2)]^2 = (1 - cos(w))/2 -> cos(w) = 1 - 2(Rin/2f)^2
(Rout/f)^2 = [tan(w)]^2 = 1/[cos(w)]^2 - 1
-> (Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
However, as #jmbr says, the actual camera distortion will depend on the lens and the zoom. Rather than rely on a fixed formula, you might want to try a polynomial expansion:
Rout = Rin*(1 + A*Rin^2 + B*Rin^4 + ...)
By tweaking first A, then higher-order coefficients, you can compute any reasonable local function (the form of the expansion takes advantage of the symmetry of the problem). In particular, it should be possible to compute initial coefficients to approximate the theoretical function above.
Also, for good results, you will need to use an interpolation filter to generate your corrected image. As long as the distortion is not too great, you can use the kind of filter you would use to rescale the image linearly without much problem.
Edit: as per your request, the equivalent scaling factor for the above formula:
(Rout/f)^2 = 1/(1-2[Rin/2f]^2)^2 - 1
-> Rout/f = [Rin/f] * sqrt(1-[Rin/f]^2/4)/(1-[Rin/f]^2/2)
If you plot the above formula alongside tan(Rin/f), you can see that they are very similar in shape. Basically, distortion from the tangent becomes severe before sin(w) becomes much different from w.
The inverse formula should be something like:
Rin/f = [Rout/f] / sqrt( sqrt(([Rout/f]^2+1) * (sqrt([Rout/f]^2+1) + 1) / 2 )
I blindly implemented the formulas from here, so I cannot guarantee it would do what you need.
Use auto_zoom to get the value for the zoom parameter.
def dist(x,y):
return sqrt(x*x+y*y)
def fisheye_to_rectilinear(src_size,dest_size,sx,sy,crop_factor,zoom):
""" returns a tuple of dest coordinates (dx,dy)
(note: values can be out of range)
crop_factor is ratio of sphere diameter to diagonal of the source image"""
# convert sx,sy to relative coordinates
rx, ry = sx-(src_size[0]/2), sy-(src_size[1]/2)
r = dist(rx,ry)
# focal distance = radius of the sphere
pi = 3.1415926535
f = dist(src_size[0],src_size[1])*factor/pi
# calc theta 1) linear mapping (older Nikon)
theta = r / f
# calc theta 2) nonlinear mapping
# theta = asin ( r / ( 2 * f ) ) * 2
# calc new radius
nr = tan(theta) * zoom
# back to absolute coordinates
dx, dy = (dest_size[0]/2)+rx/r*nr, (dest_size[1]/2)+ry/r*nr
# done
return (int(round(dx)),int(round(dy)))
def fisheye_auto_zoom(src_size,dest_size,crop_factor):
""" calculate zoom such that left edge of source image matches left edge of dest image """
# Try to see what happens with zoom=1
dx, dy = fisheye_to_rectilinear(src_size, dest_size, 0, src_size[1]/2, crop_factor, 1)
# Calculate zoom so the result is what we wanted
obtained_r = dest_size[0]/2 - dx
required_r = dest_size[0]/2
zoom = required_r / obtained_r
return zoom
I took what JMBR did and basically reversed it. He took the radius of the distorted image (Rd, that is, the distance in pixels from the center of the image) and found a formula for Ru, the radius of the undistorted image.
You want to go the other way. For each pixel in the undistorted (processed image), you want to know what the corresponding pixel is in the distorted image.
In other words, given (xu, yu) --> (xd, yd). You then replace each pixel in the undistorted image with its corresponding pixel from the distorted image.
Starting where JMBR did, I do the reverse, finding Rd as a function of Ru. I get:
Rd = f * sqrt(2) * sqrt( 1 - 1/sqrt(r^2 +1))
where f is the focal length in pixels (I'll explain later), and r = Ru/f.
The focal length for my camera was 2.5 mm. The size of each pixel on my CCD was 6 um square. f was therefore 2500/6 = 417 pixels. This can be found by trial and error.
Finding Rd allows you to find the corresponding pixel in the distorted image using polar coordinates.
The angle of each pixel from the center point is the same:
theta = arctan( (yu-yc)/(xu-xc) ) where xc, yc are the center points.
Then,
xd = Rd * cos(theta) + xc
yd = Rd * sin(theta) + yc
Make sure you know which quadrant you are in.
Here is the C# code I used
public class Analyzer
{
private ArrayList mFisheyeCorrect;
private int mFELimit = 1500;
private double mScaleFESize = 0.9;
public Analyzer()
{
//A lookup table so we don't have to calculate Rdistorted over and over
//The values will be multiplied by focal length in pixels to
//get the Rdistorted
mFisheyeCorrect = new ArrayList(mFELimit);
//i corresponds to Rundist/focalLengthInPixels * 1000 (to get integers)
for (int i = 0; i < mFELimit; i++)
{
double result = Math.Sqrt(1 - 1 / Math.Sqrt(1.0 + (double)i * i / 1000000.0)) * 1.4142136;
mFisheyeCorrect.Add(result);
}
}
public Bitmap RemoveFisheye(ref Bitmap aImage, double aFocalLinPixels)
{
Bitmap correctedImage = new Bitmap(aImage.Width, aImage.Height);
//The center points of the image
double xc = aImage.Width / 2.0;
double yc = aImage.Height / 2.0;
Boolean xpos, ypos;
//Move through the pixels in the corrected image;
//set to corresponding pixels in distorted image
for (int i = 0; i < correctedImage.Width; i++)
{
for (int j = 0; j < correctedImage.Height; j++)
{
//which quadrant are we in?
xpos = i > xc;
ypos = j > yc;
//Find the distance from the center
double xdif = i-xc;
double ydif = j-yc;
//The distance squared
double Rusquare = xdif * xdif + ydif * ydif;
//the angle from the center
double theta = Math.Atan2(ydif, xdif);
//find index for lookup table
int index = (int)(Math.Sqrt(Rusquare) / aFocalLinPixels * 1000);
if (index >= mFELimit) index = mFELimit - 1;
//calculated Rdistorted
double Rd = aFocalLinPixels * (double)mFisheyeCorrect[index]
/mScaleFESize;
//calculate x and y distances
double xdelta = Math.Abs(Rd*Math.Cos(theta));
double ydelta = Math.Abs(Rd * Math.Sin(theta));
//convert to pixel coordinates
int xd = (int)(xc + (xpos ? xdelta : -xdelta));
int yd = (int)(yc + (ypos ? ydelta : -ydelta));
xd = Math.Max(0, Math.Min(xd, aImage.Width-1));
yd = Math.Max(0, Math.Min(yd, aImage.Height-1));
//set the corrected pixel value from the distorted image
correctedImage.SetPixel(i, j, aImage.GetPixel(xd, yd));
}
}
return correctedImage;
}
}
I found this pdf file and I have proved that the maths are correct (except for the line vd = *xd**fv+v0 which should say vd = **yd**+fv+v0).
http://perception.inrialpes.fr/CAVA_Dataset/Site/files/Calibration_OpenCV.pdf
It does not use all of the latest co-efficients that OpenCV has available but I am sure that it could be adapted fairly easily.
double k1 = cameraIntrinsic.distortion[0];
double k2 = cameraIntrinsic.distortion[1];
double p1 = cameraIntrinsic.distortion[2];
double p2 = cameraIntrinsic.distortion[3];
double k3 = cameraIntrinsic.distortion[4];
double fu = cameraIntrinsic.focalLength[0];
double fv = cameraIntrinsic.focalLength[1];
double u0 = cameraIntrinsic.principalPoint[0];
double v0 = cameraIntrinsic.principalPoint[1];
double u, v;
u = thisPoint->x; // the undistorted point
v = thisPoint->y;
double x = ( u - u0 )/fu;
double y = ( v - v0 )/fv;
double r2 = (x*x) + (y*y);
double r4 = r2*r2;
double cDist = 1 + (k1*r2) + (k2*r4);
double xr = x*cDist;
double yr = y*cDist;
double a1 = 2*x*y;
double a2 = r2 + (2*(x*x));
double a3 = r2 + (2*(y*y));
double dx = (a1*p1) + (a2*p2);
double dy = (a3*p1) + (a1*p2);
double xd = xr + dx;
double yd = yr + dy;
double ud = (xd*fu) + u0;
double vd = (yd*fv) + v0;
thisPoint->x = ud; // the distorted point
thisPoint->y = vd;
This can be solved as an optimization problem. Simply draw on curves in images that are supposed to be straight lines. Store the contour points for each of those curves. Now we can solve the fish eye matrix as a minimization problem. Minimize the curve in points and that will give us a fisheye matrix. It works.
It can be done manually by adjusting the fish eye matrix using trackbars! Here is a fish eye GUI code using OpenCV for manual calibration.

matlab plot graph of data over an image

What I would like to do is plot an image of a graph (from say a pdf file or a scanned image). Next, I would like to overlay an axis on the graph in the image, and then plot data on that axis (over the image).
Using imtool, I know the coordinates of the graph in the image (x range = ~52-355 pixels, and y range = 23(top) - 262(bottom) pixels in this case).
This is what I have tried:
I = imread('C:\MATLAB\R2014a\help\images\ref\ftrans2_fig.png');
I = squeeze(uint8(mean(I,3)));
figure, imshow(I)
[rows, cols] = size(I);
x_data = (-1 : .01 : +1)';
y_data = 1 - x_data.^2;
h1 = axes('Position',([52, 23, 355-52, 262-23] ./ [cols, rows, cols, rows] ));
set(h1, 'Color', 'none')
hold on
plot(x_data, y_data, '-rx')
Question: Knowing the pixel coordinates of the graph in the image, how do I determine the proper position of the axis in the figure, (my code fails to account for the actual size of the figure box, the gray border around the image). I have to do this for several images and sets of data, so I would like an automated method, assuming I find the coordinates of the graphs in the image ahead of time.
Thanks for your reply! (1st time posting, please be kind)
You may be able to solve your problem by forcing the image onto the same axis as the plot. Try this:
I = imread('C:\MATLAB\R2014a\help\images\ref\ftrans2_fig.png');
I = squeeze(uint8(mean(I,3)));
[rows, cols] = size(I);
x_data = (-1 : .01 : +1)';
y_data = 1 - x_data.^2;
h1 = axes('Position',([52, 23, 355-52, 262-23] ./ [cols, rows, cols, rows] ));
set(h1, 'Color', 'none')
hold on
image(I, 'Parent', h1);
plot(h1, x_data, y_data, '-rx')
That should at ensure that the plot axis and the image axis have the same origin, as they will be one and the same. You may need to adjust your sizing code. Let me know if that doesn't do it for you.
Good Luck!
I think I have it figured out.
It would have been easier if I could use:
figure, h1=imshow(I)
get(h1,'Position')
but that results in "The name 'Position' is not an accessible property for an instance of class 'image'."
Instead, this appears to work:
I = imread('C:\MATLAB\R2014a\help\images\ref\ftrans2_fig.png');
I = squeeze(uint8(mean(I,3)));
in_mag = 300;
figure, imshow(I, 'Border', 'tight', 'InitialMagnification', in_mag)
[rows, cols] = size(I);
x_data = (-1 : .01 : +1)';
y_data = 1 - x_data.^2;
% Coord of graph in image pixels
x_0 = 50; x_max = 354; y_0 = 262; y_max = 23;
h1 = axes('Position',([x_0, rows-y_0, x_max-x_0, y_0-y_max] ...
./ [cols, rows, cols, rows] ));
set(h1,'Color','none')
hold on
plot(x_data, y_data, '-rx')
ylim([0,1.4])
set(gca,'YColor', [0 0 1], 'XColor', [0 0 1])
However, if anybody has a better idea, I would be very happy to explore it!
Thanks

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