First of all, I have this code which creates a curved surface that looks like a flying carpet.
#include "colors.inc"
camera {
location <5,140,25>
look_at <0, 0, 0>
angle 7}
light_source {
<20, 20, 20>
White
}
#declare Ball =
sphere{<0,0,0>,0.25
texture{
pigment{color rgb<1,0.65,0.0>}
finish {diffuse 0.9 phong 1}
}// end of texture
}// end of sphere
#declare E = 5;
#declare Z = -E; // start value Z
#declare EndZ = E; // end value Z
#declare Step = 0.2;// step value
//------- loop start Z:
#while ( Z < EndZ + Step)
#declare X = -E; // start value X
#declare EndX = E; // end value X
//------ loop start X:
#while ( X < EndX + Step)
object{Ball
translate<X,0.05*X*sin(X-2*Z)
+ 0.1*Z*cos(3*X-Z),Z>}
#declare X = X+Step;//next X value
#end // --------------- loop end X
#declare Z = Z+Step;//next Z value
#end // --------------- loop end Z
The output of this code:
now I want to map an simple image like this
Instead of yellow balls, I want map that simple pattern on the curved surface so that pattern can be wavy kind of looking.
I have tried to map the ball and loop it, but obviously the output was not the single square wave-looking image. (I don't want that picture repeated, just single curved image.)
would that be possible??
Thank you..
Currently you pre-declare a sphere and then translate it. If you apply a texture then it will be translated along with the sphere, so all will get the same texture.
Use the loop to create the sphere with the proper coordinates at once, inside a union. Then apply the texture to that union. Scale and position the texture accordingly.
I was trying to implement the IBVS algorithm (the one explained in the Introduction here) in MATLAB myself, but I am facing the following problem : The algorithm seems to work only for the cases that the camera does not have to change its orientation in respect to the world frame.For example, if I just try to make one vertex of the initial (almost) square go closer to its opposite vertex, the algorithm does not work, as can be seen in the following image
The red x are the desired projections, the blue circles are the initial ones and the green ones are the ones I get from my algorithm.
Also the errors are not exponentially dereasing as they should.
What am I doing wrong? I am attaching my MATLAB code which is fully runable. If anyone could take a look, I would be really grateful. I took out the code that was performing the plotting. I hope it is more readable now. Visual servoing has to be performed with at least 4 target points, because else the problem has no unique solution. If you are willing to help, I would suggest you take a look at the calc_Rotation_matrix() function to check that the rotation matrix is properly calculated, then verify that the line ds = vc; in euler_ode is correct. The camera orientation is expressed in Euler angles according to this convention. Finally, one could check if the interaction matrix L is properly calculated.
function VisualServo()
global A3D B3D C3D D3D A B C D Ad Bd Cd Dd
%coordinates of the 4 points wrt camera frame
A3D = [-0.2633;0.27547;0.8956];
B3D = [0.2863;-0.2749;0.8937];
C3D = [-0.2637;-0.2746;0.8977];
D3D = [0.2866;0.2751;0.8916];
%initial projections (computed here only to show their relation with the desired ones)
A=A3D(1:2)/A3D(3);
B=B3D(1:2)/B3D(3);
C=C3D(1:2)/C3D(3);
D=D3D(1:2)/D3D(3);
%initial camera position and orientation
%orientation is expressed in Euler angles (X-Y-Z around the inertial frame
%of reference)
cam=[0;0;0;0;0;0];
%desired projections
Ad=A+[0.1;0];
Bd=B;
Cd=C+[0.1;0];
Dd=D;
t0 = 0;
tf = 50;
s0 = cam;
%time step
dt=0.01;
t = euler_ode(t0, tf, dt, s0);
end
function ts = euler_ode(t0,tf,dt,s0)
global A3D B3D C3D D3D Ad Bd Cd Dd
s = s0;
ts=[];
for t=t0:dt:tf
ts(end+1)=t;
cam = s;
% rotation matrix R_WCS_CCS
R = calc_Rotation_matrix(cam(4),cam(5),cam(6));
r = cam(1:3);
% 3D coordinates of the 4 points wrt the NEW camera frame
A3D_cam = R'*(A3D-r);
B3D_cam = R'*(B3D-r);
C3D_cam = R'*(C3D-r);
D3D_cam = R'*(D3D-r);
% NEW projections
A=A3D_cam(1:2)/A3D_cam(3);
B=B3D_cam(1:2)/B3D_cam(3);
C=C3D_cam(1:2)/C3D_cam(3);
D=D3D_cam(1:2)/D3D_cam(3);
% computing the L matrices
L1 = L_matrix(A(1),A(2),A3D_cam(3));
L2 = L_matrix(B(1),B(2),B3D_cam(3));
L3 = L_matrix(C(1),C(2),C3D_cam(3));
L4 = L_matrix(D(1),D(2),D3D_cam(3));
L = [L1;L2;L3;L4];
%updating the projection errors
e = [A-Ad;B-Bd;C-Cd;D-Dd];
%compute camera velocity
vc = -0.5*pinv(L)*e;
%change of the camera position and orientation
ds = vc;
%update camera position and orientation
s = s + ds*dt;
end
ts(end+1)=tf+dt;
end
function R = calc_Rotation_matrix(theta_x, theta_y, theta_z)
Rx = [1 0 0; 0 cos(theta_x) -sin(theta_x); 0 sin(theta_x) cos(theta_x)];
Ry = [cos(theta_y) 0 sin(theta_y); 0 1 0; -sin(theta_y) 0 cos(theta_y)];
Rz = [cos(theta_z) -sin(theta_z) 0; sin(theta_z) cos(theta_z) 0; 0 0 1];
R = Rx*Ry*Rz;
end
function L = L_matrix(x,y,z)
L = [-1/z,0,x/z,x*y,-(1+x^2),y;
0,-1/z,y/z,1+y^2,-x*y,-x];
end
Cases that work:
Ad=2*A;
Bd=2*B;
Cd=2*C;
Dd=2*D;
Ad=A+1;
Bd=B+1;
Cd=C+1;
Dd=D+1;
Ad=2*A+1;
Bd=2*B+1;
Cd=2*C+1;
Dd=2*D+1;
Cases that do NOT work:
Rotation by 90 degrees and zoom out (zoom out alone works, but I am doing it here for better visualization)
Ad=2*D;
Bd=2*C;
Cd=2*A;
Dd=2*B;
Your problem comes from the way you move the camera from the resulting visual servoing velocity. Rather than
cam = cam + vc*dt;
you should compute the new camera position using the exponential map
cam = cam*expm(vc*dt)
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.
Hello so I have an original image and I cropped a part of that image(template) and wrote a code with SAD algorithim to detect on the original image the part i cropped (template) where it exists and draw a rectangle on it.
The code doesnt have any errors , but the rectangle draw doesnt match the template , so I guess the 'output ' variable is the problem can you please help me
I2=imread('img.PNG');
I2=rgb2gray(I2);
[r,c]= size(I2);
%padding the image
%padding
B = padarray(I2,[24 24],'replicate' ,'both');
%imshow(B);
%creating template
temp=imread('crop_img.PNG');
temp= rgb2gray(temp);
%imshow(temp)
size(temp)
output = zeros (size(I2));
K = size(temp)
x=1;
y=1;
for i = 25 : r-24
for j = 25: c-24
%dif = temp -I2(i:i+47,j:j+47) ;
K = imabsdiff(temp,B(i-24:i+24,j-24:j+24));
output(i-24, j-24) = sum (K(:));
end
end
%gettting min value in output
min_value = output(1,1)
for i=1 : r
for j=1 :c
if(output(i,j)<min_value)
min_value=output(i,j);
row=i;
col=j;
end
end
end
row
col
output(1,465)
output(6,200)
%draw rectangle on matching area
%Create the shape inserter object.
shapeInserter = vision.ShapeInserter;
%Define the rectangle dimensions as [x y width height].
rectangle = int32([row col 24 24]);
%Draw the rectangle and display the result.
J = step(shapeInserter, I2, rectangle);
imshow(J);
There is an issue with image coordinates vs. matrix coordinates - you need to flip your row and column variables. Also, with shapeInserter, the coordinates are for the corner of the rectangle, so to make something centred on your output, you'd need something along the lines of:
rectangle = uint8([col-12 row-12 24 24]);
In MATLAB, it is usually not required to loop over every pixel of an image - it's much more efficient to work on the whole image at once. For example, you don't need a loop here:
min_value = output(1,1)
for i=1 : r
for j=1 :c
if(output(i,j)<min_value)
min_value=output(i,j);
row=i;
col=j;
end
end
end
This can be replaced by:
min_value = min(output(:));
[row,col] = find(output==min_value,1,'first');
using Hough Transform, how can I detect and get coordinates of (x0,y0) and "a" and "b" of an ellipse in 2D space?
This is ellipse01.bmp:
I = imread('ellipse01.bmp');
[m n] = size(I);
c=0;
for i=1:m
for j=1:n
if I(i,j)==1
c=c+1;
p(c,1)=i;
p(c,2)=j;
end
end
end
Edges=transpose(p);
Size_Ellipse = size(Edges);
B = 1:ceil(Size_Ellipse(1)/2);
Acc = zeros(length(B),1);
a1=0;a2=0;b1=0;b2=0;
Ellipse_Minor=[];Ellipse_Major=[];Ellipse_X0 = [];Ellipse_Y0 = [];
Global_Threshold = ceil(Size_Ellipse(2)/6);%Used for Major Axis Comparison
Local_Threshold = ceil(Size_Ellipse(1)/25);%Used for Minor Axis Comparison
[Y,X]=find(Edges);
Limit=numel(Y);
Thresh = 150;
Para=[];
for Count_01 =1:(Limit-1)
for Count_02 =(Count_01+1):Limit
if ((Count_02>Limit) || (Count_01>Limit))
continue
end
a1=Y(Count_01);b1=X(Count_01);
a2=Y(Count_02);b2=X(Count_02);
Dist_01 = (sqrt((a1-a2)^2+(b1-b2)^2));
if (Dist_01 >Global_Threshold)
Center_X0 = (b1+b2)/2;Center_Y0 = (a1+a2)/2;
Major = Dist_01/2.0;Alpha = atan((a2-a1)/(b2-b1));
if(Alpha == 0)
for Count_03 = 1:Limit
if( (Count_03 ~= Count_01) || (Count_03 ~= Count_02))
a3=Y(Count_03);b3=X(Count_03);
Dist_02 = (sqrt((a3 - Center_Y0)^2+(b3 - Center_X0)^2));
if(Dist_02 > Local_Threshold)
Cos_Tau = ((Major)^2 + (Dist_02)^2 - (a3-a2)^2 - (b3-b2)^2)/(2*Major*Dist_02);
Sin_Tau = 1 - (Cos_Tau)^2;
Minor_Temp = ((Major*Dist_02*Sin_Tau)^2)/(Major^2 - ((Dist_02*Cos_Tau)^2));
if((Minor_Temp>1) && (Minor_Temp<B(end)))
Acc(round(Minor_Temp)) = Acc(round(Minor_Temp))+1;
end
end
end
end
end
Minor = find(Acc == max(Acc(:)));
if(Acc(Minor)>Thresh)
Ellipse_Minor(end+1)=Minor(1);Ellipse_Major(end+1)=Major;
Ellipse_X0(end+1) = Center_X0;Ellipse_Y0(end+1) = Center_Y0;
for Count = 1:numel(X)
Para_X = ((X(Count)-Ellipse_X0(end))^2)/(Ellipse_Major(end)^2);
Para_Y = ((Y(Count)-Ellipse_Y0(end))^2)/(Ellipse_Minor(end)^2);
if (((Para_X + Para_Y)>=-2)&&((Para_X + Para_Y)<=2))
Edges(X(Count),Y(Count))=0;
end
end
end
Acc = zeros(size(Acc));
end
end
end
Although this is an old question, perhaps what I found can help someone.
The main problem of using the normal Hough Transform to detect ellipses is the dimension of the accumulator, since we would need to vote for 5 variables (the equation is explained here):
There is a very nice algorithm where the accumulator can be a simple 1D array, for example, and that runs in . If you wanna see code, you can look at here (the image used to test was that posted above).
If you use circle for rough transform is given as rho = xcos(theta) + ysin(theta)
For ellipse since it is
You could transform the equation as
rho = axcos(theta) + bysin(theta)
Although I am not sure if you use standard Hough Transform, for ellipse-like transforms, you could manipulate the first given function.
If your ellipse is as provided, being a true ellipse and not a noisy sample of points;
the search for the two furthest points gives the ends of the major axis,
the search for the two nearest points gives the ends of the minor axis,
the intersection of these lines (you can check it's a right angle) occurs at the centre.
If you know the 'a' and 'b' of an ellipse then you can rescale the image by factor of a/b in one direction and look for circle. I am still thinking about what to do when a and b are unknown.
If you know that it is circle then use Hough transform for circles. Here is a sample code:
int accomulatorResolution = 1; // for each pixel
int minDistBetweenCircles = 10; // In pixels
int cannyThresh = 20;
int accomulatorThresh = 5*_accT+1;
int minCircleRadius = 0;
int maxCircleRadius = _maxR*10;
cvClearMemStorage(storage);
circles = cvHoughCircles( gryImage, storage,
CV_HOUGH_GRADIENT, accomulatorResolution,
minDistBetweenCircles,
cannyThresh , accomulatorThresh,
minCircleRadius,maxCircleRadius );
// Draw circles
for (int i = 0; i < circles->total; i++){
float* p = (float*)cvGetSeqElem(circles,i);
// Draw center
cvCircle(dstImage, cvPoint(cvRound(p[0]),cvRound(p[1])),
1, CV_RGB(0,255,0), -1, 8, 0 );
// Draw circle
cvCircle(dstImage, cvPoint(cvRound(p[0]),cvRound(p[1])),
cvRound(p[2]),CV_RGB(255,0,0), 1, 8, 0 );
}