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OpenCV detect if points lie along line/plane

开发者 https://www.devze.com 2023-02-23 05:48 出处:网络
I am working on a form of autocalibration for an optics device which is currently performed manually. The first part of the calibration is to determine whether a light beam has illuminated the set of

I am working on a form of autocalibration for an optics device which is currently performed manually. The first part of the calibration is to determine whether a light beam has illuminated the set of 'calibration' points.

I am using OpenCV and have thresholded and cropped the image to leave only the possible relevant points. I know want to determine if these points lie along a stright (horizontal) line; if they a sufficient number do the beam is in the correct position! (The points lie in a straight line but the beam is often bent so hitt开发者_高级运维ing most of the points suffices, there are 21 points which show up as white circles when thresholded).

I have tried using a histogram but on the thresholded image the results are not correct and am now looking at Hough lines, but this detects straight lines from edges wwhere as I want to establish if detected points lie on a line.

This is the threshold code I use:

cvThreshold(output, output, 150, 256, CV_THRESH_BINARY);

The histogram results with anywhere from 1 to 640 bins (image width) is two lines at 0 and about 2/3rds through of near max value. Not the distribution expected or obtained without thresholding.

Some pictures to try to illistrate the point (note the 'noisy' light spots which are a feature of the system setup and cannot be overcome):

12 points in a stright line next to one another (beam in correct position)

The sort of output wanted (for illistration, if the points are on the line this is all I need to know!)

Any help would be greatly appreciated. One thought was to extract the co-ordinates of the points and compare them but I don't know how to do this.


Incase it helps anyone here is a very basic (the first draft) of some simple linaear regression code I used.

    // Calculate the averages of arrays x and y
double xa = 0, ya = 0;

for(int i = 0; i < n; i++) 
{
    xa += x[i];
    ya += y[i];
}   
xa /= n;
ya /= n;

// Summation of all X and Y values
double sumX = 0;
double sumY = 0;
// Summation of all X*Y values
double sumXY = 0;
// Summation of all X^2 and Y^2 values
double sumXs = 0;
double sumYs = 0;

for(int i = 0; i < n; i++)
{
    sumX = sumX + x[i];
    sumY = sumY + y[i];

    sumXY = sumXY + (x[i] * y[i]);

    sumXs = sumXs + (x[i] * x[i]);
    sumYs = sumYs + (y[i] * y[i]);
}
// (X^2) and (Y^2) sqaured
double Xs = sumX * sumX;
double Ys = sumY * sumY;

// Calculate slope, m
slope = (n * sumXY - sumX * sumY) / (n* sumXs - Xs);    

// Calculate intercept
intercept = ceil((sumY - slope * sumX) / n);

// Calculate regression index, r^2
double r_top = (n * sumXY - sumX * sumY); 
double r_bottom = sqrt((n* sumXs - Xs) * (n* sumYs - Ys));
double r = 0;

// Check line is not perfectly vertical or horizontal
if(r_top == 0 || r_bottom == 0)
    r = 0;
else
    r = r_top/ r_bottom;

There are more efficeint ways of doing this (see CodeCogs or AGLIB) but as quick fix this code seems to work.

To detect Circles in OpenCV I dropped the Hough Transform and adapeted codee from this post: Detection of coins (and fit ellipses) on an image

It is then a case of refining the co-ordinates (removing any outliers etc) to determine if the circles lie on a horizontal line from the slope and intercept values of the regression.


Obtain the x,y coordinates of the thresholded points, then perform a linear regression to find a best-fit line. With that line, you can determine the r^2 value which effectively gives you the quality of fit. Based on that fitness measure, you can determine your calibration success.

Here is a good discussion.


you could do something like this, altough it is an aproximation:

var dw = decide a medium dot width in pixels

maxdots = 0;
for each line of the image {
  var dots = 0;
  scan by incrementing x by dw {
     if (color==dotcolor) dots++;
  }
  if (dots>maxdots) maxdots=dots;
}

maxdots would be the best result...

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