How to fitline on my binary image and find distance

Hello,

I have a question about my image processing problem. I have convert the image into binary and now i would like to use fitline on my image. Basically want to smooth the line with the curve. I am a beginner for MATLAB software. Please help me to find the solution. Also i want to find distance of one thin line to thick line (top to bottom).

I am attachng my picture as png. Also my code is there.

Answer 1

[Image Analyst]

Your question is very unclear and ambiguous. Like, exactly what does "want to find distance of one thin line to thick line (top to bottom)" mean??? Where are thin lines and where are the thick lines in your image?

Here is another guess at what you might mean. I just find the topmost and bottommost white pixel in the image for each column, and compute the distance (height) between them:

% By Image Analyst
clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 22;

%--------------------------------------------------------------------------------------------------------
%    READ IN IMAGE
folder = pwd;
baseFileName = 'image.jpeg';
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
% Check if file exists.
if ~exist(fullFileName, 'file')
	% The file doesn't exist -- didn't find it there in that folder.
	% Check the entire search path (other folders) for the file by stripping off the folder.
	fullFileNameOnSearchPath = baseFileName; % No path this time.
	if ~exist(fullFileNameOnSearchPath, 'file')
		% Still didn't find it.  Alert user.
		errorMessage = sprintf('Error: %s does not exist in the search path folders.', fullFileName);
		uiwait(warndlg(errorMessage));
		return;
	end
end
grayImage = imread(fullFileName);

% Get the dimensions of the image.
% numberOfColorChannels should be = 1 for a gray scale image, and 3 for an RGB color image.
[rows, columns, numberOfColorChannels] = size(grayImage);
if numberOfColorChannels > 1
	% It's not really gray scale like we expected - it's color.
	% Use weighted sum of ALL channels to create a gray scale image.
	grayImage = rgb2gray(grayImage);
	% ALTERNATE METHOD: Convert it to gray scale by taking only the green channel,
	% which in a typical snapshot will be the least noisy channel.
	% grayImage = grayImage(:, :, 2); % Take green channel.
end
subplot(2, 2, 1);
imshow(grayImage, []);
impixelinfo;
title('Original Gray Scale Image', 'FontSize', fontSize, 'Interpreter', 'None');
hFig = gcf;
hFig.WindowState = 'maximized'; % May not work in earlier versions of MATLAB.
drawnow;

%--------------------------------------------------------------------------------------------------------
%    SEGMENTATION OF IMAGE
% Get a binary image
binaryImage = imbinarize(grayImage);
% Get rid of big white frame around the border of the image.
binaryImage = imclearborder(binaryImage);
subplot(2, 2, 2);
imshow(binaryImage, []);
impixelinfo;
title('Binary Image', 'FontSize', fontSize, 'Interpreter', 'None');
hFig = gcf;
hFig.WindowState = 'maximized'; % May not work in earlier versions of MATLAB.
drawnow;

% Scan image finding the distance from topmost white pixel to bottommost.
topRows = nan(1, columns);
bottomRows = nan(1, columns);
for col = 1 : columns
	thisColumn = binaryImage(:, col);
	t = find(thisColumn, 1, 'first');
	if ~isempty(t)
		topRows(col) = t;
		bottomRows(col) = find(thisColumn, 1, 'last');
	end
end
% Plot top rows and bottom rows.
subplot(2, 2, 3);
x = 1 : columns;
plot(x, topRows, 'b-', 'LineWidth', 2);
grid on;
hold on;
plot(x, bottomRows, 'r-', 'LineWidth', 2);
title('Top and Bottom Rows', 'FontSize', fontSize);
xlabel('Column', 'FontSize', fontSize);
ylabel('Row', 'FontSize', fontSize);
legend('Top Rows', 'Bottom Rows');

% Plot heights.
heights = bottomRows - topRows;
subplot(2, 2, 4);
x = 1 : columns;
plot(x, heights, 'b-', 'LineWidth', 2);
grid on;
title('Heights', 'FontSize', fontSize);
xlabel('Column', 'FontSize', fontSize);
ylabel('Height in Rows', 'FontSize', fontSize);

Answer 2

[Image Analyst]

Did you try polyfit() to fit a line to the pixels?

% By Image Analyst
clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 22;

%--------------------------------------------------------------------------------------------------------
%    READ IN IMAGE
folder = pwd;
baseFileName = 'image.jpeg';
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
% Check if file exists.
if ~exist(fullFileName, 'file')
	% The file doesn't exist -- didn't find it there in that folder.
	% Check the entire search path (other folders) for the file by stripping off the folder.
	fullFileNameOnSearchPath = baseFileName; % No path this time.
	if ~exist(fullFileNameOnSearchPath, 'file')
		% Still didn't find it.  Alert user.
		errorMessage = sprintf('Error: %s does not exist in the search path folders.', fullFileName);
		uiwait(warndlg(errorMessage));
		return;
	end
end
grayImage = imread(fullFileName);

% Get the dimensions of the image.
% numberOfColorChannels should be = 1 for a gray scale image, and 3 for an RGB color image.
[rows, columns, numberOfColorChannels] = size(grayImage);
if numberOfColorChannels > 1
	% It's not really gray scale like we expected - it's color.
	% Use weighted sum of ALL channels to create a gray scale image.
	grayImage = rgb2gray(grayImage);
	% ALTERNATE METHOD: Convert it to gray scale by taking only the green channel,
	% which in a typical snapshot will be the least noisy channel.
	% grayImage = grayImage(:, :, 2); % Take green channel.
end
subplot(2, 1, 1);
imshow(grayImage, []);
impixelinfo;
title('Original Gray Scale Image', 'FontSize', fontSize, 'Interpreter', 'None');
hFig = gcf;
hFig.WindowState = 'maximized'; % May not work in earlier versions of MATLAB.
drawnow;

%--------------------------------------------------------------------------------------------------------
%    SEGMENTATION OF IMAGE
% Get a binary image
binaryImage = imbinarize(grayImage);
% Get rid of big white frame around the border of the image.
binaryImage = imclearborder(binaryImage);
subplot(2, 1, 2);
imshow(binaryImage, []);
impixelinfo;
title('Binary Image', 'FontSize', fontSize, 'Interpreter', 'None');
hFig = gcf;
hFig.WindowState = 'maximized'; % May not work in earlier versions of MATLAB.
drawnow;

% Get all coordinates
[y, x] = find(binaryImage);
% Fit a line through them
coefficients = polyfit(x, y, 1)
% Get a fitted line
xFit = 1 : columns;
yFit = polyval(coefficients, xFit);
% Plot the line in red over the image.
hold on;
plot(xFit, yFit, 'r-', 'LineWidth', 2);
% Put the equation up
txt = sprintf('y = (%f) * x + (%f)', coefficients(1), coefficients(2));
text(100, 200, txt, 'Color', 'r', 'FontSize', 15, 'FontWeight', 'bold');

[Prashant]

Hello Image Analyst,

Thank you so much for your answer. i am sorry for confusing question. what you did that is fine. but i do not want to remove any detail from my image. I want to measure the distance between two point and also want to make my image boundary a bit smooth so i can get the distance what i want. here i am sending you the foto that which distance i want to measure . See attached file Image.jpg. Also the RGB image in attached name DSC_2637.png .

Also please have a look my code name check.m, here i threshold all plane (RGB) because i want to detect red color in image and then i got the binary result like this in picture. now i want to find the distance as i said to see attach Image.jpg.

thank you in advance.

[Image Analyst]

If you've segmented your image so that the oblong shape is the black region between the laser line and the white, then see my code below. It finds the top line, bottom line, and height of the region.

[Prashant]

Thanks for your answer. would be great if you give me the solution for what i want for my image, i mean would be great for me if i can get the specific code from you.

[Image Analyst]

Here's a start for you:

% By Image Analyst
clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 22;

%--------------------------------------------------------------------------------------------------------
%    READ IN IMAGE
folder = pwd;
baseFileName = 'dsc_2637.png';
% Get the full filename, with path prepended.
fullFileName = fullfile(folder, baseFileName);
% Check if file exists.
if ~exist(fullFileName, 'file')
	% The file doesn't exist -- didn't find it there in that folder.
	% Check the entire search path (other folders) for the file by stripping off the folder.
	fullFileNameOnSearchPath = baseFileName; % No path this time.
	if ~exist(fullFileNameOnSearchPath, 'file')
		% Still didn't find it.  Alert user.
		errorMessage = sprintf('Error: %s does not exist in the search path folders.', fullFileName);
		uiwait(warndlg(errorMessage));
		return;
	end
end
grayImage = imread(fullFileName);

% Get the dimensions of the image.
% numberOfColorChannels should be = 1 for a gray scale image, and 3 for an RGB color image.
[rows, columns, numberOfColorChannels] = size(grayImage);
if numberOfColorChannels > 1
	% It's not really gray scale like we expected - it's color.
	% Use weighted sum of ALL channels to create a gray scale image.
% 	grayImage = rgb2gray(grayImage);
	% ALTERNATE METHOD: Convert it to gray scale by taking only the green channel,
	% which in a typical snapshot will be the least noisy channel.
	grayImage = grayImage(:, :, 1); % Take red channel.
end
% Crop it to the right lines.
grayImage = grayImage(1750:2250, 800:end);
subplot(2, 1, 1);
imshow(grayImage, []);
impixelinfo;
title('Red Channel Image', 'FontSize', fontSize, 'Interpreter', 'None');
hFig = gcf;
hFig.WindowState = 'maximized'; % May not work in earlier versions of MATLAB.
drawnow;

%--------------------------------------------------------------------------------------------------------
%    SEGMENTATION OF IMAGE
% Get a binary image by interactively thresholding using the function at
% https://www.mathworks.com/matlabcentral/fileexchange/29372-thresholding-an-image
% 156 seems good for this image.
lowThreshold = 156;
highThreshold = 255;
[lowThreshold, highThreshold] = threshold(lowThreshold, highThreshold, grayImage);
binaryImage = grayImage > lowThreshold & grayImage < highThreshold;
subplot(2, 1, 2);
imshow(binaryImage, []);
impixelinfo;
title('Binary Image', 'FontSize', fontSize, 'Interpreter', 'None');
hFig = gcf;
hFig.WindowState = 'maximized'; % May not work in earlier versions of MATLAB.
drawnow;

[Prashant]

Thanks,

when i run this code then i am getting this error:

Unrecognized function or variable 'threshold'.

Error in Image_Analyst (line 58)

[lowThreshold, highThreshold] = threshold(lowThreshold, highThreshold, grayImage);

[Image Analyst]

Did you see this line:

% Get a binary image by interactively thresholding using the function at
% https://www.mathworks.com/matlabcentral/fileexchange/29372-thresholding-an-image

so if you're not going to get the interactive thresholding app from my File Exchange, just comment out that line and it will use the initial values for lowThreshold, and highThreshold.

[Prashant]

Okay thats perfectly working. Now how can i find distance as i mentioned earlier in attached file image.jpg?

distance should be in cm.

[Image Analyst]

Well you're going to have to calibrate. See my attached demo.

[Prashant]

I used image that we got from last code. I used the same calibration demo that u gave. It is working for me, now just want to know that how to find that is one pixel =?. is there any standard calculation to find? how to know the value of one pixel?

[Image Analyst]

Prashant, I'm not sure you ran or read my spatial calibration demo yet. You can see that you click endpoints of a line and tell it how many real world units it is. Then it computes and tells you the meters per pixel. Let's say your wall was 10 meters long and you clicked on each side and it turned out to be 1000 pixels. Then the spatial calibration factor is 10/1000 = 0.01 meters per pixel. So one pixel is 0.01 meters. 57 pixels is 57 * 0.01 = 0.57 meters.

Not sure what you mean when you ask "is there any standard calculation to find?" The standard calculation is division. Like 10/1000 -- that is division. Please explain what that question means.

[Prashant]

Thanks for the explaination. I understood exactly now how it works.

I asked this question because i do not know about the measurent of the wall. I just took a photo with my phone and i analysed that image without camera calibration process.

so basically, i have to see that how long is my wall and then i can count or find the distance or?

[Image Analyst]

Yes, get your tape measure out, walk over to the wall, and measure it. Then use something like my program to find the pixel distance between the same points on the wall that you measured the distance between. Then divide those two distances -- (real world distance) / (pixel distance). Then anytime you measure a pixel distance in your program, multiply it by that ratio to get the real world distance.

[Prashant]

Yes understand.. thank you so much... :)

[Prashant]

Hello Image Analyst,

I have one image, i want to measure the angle of intersection between two line as shown in Image. I have trace the boundary that i want, now i want to draw an 'X' at the point of intersection. also want to find an angle.

please help me how to write that code. I also attached my code file name Task.m

the picture i got with boundary name Trace.jpg and the original Image name DSC_2637.png.

[Image Analyst]

Threshold the image then scan it column-by-column to get the top edge. Now you have the line and you can find the top line with polyfit() or fitPolynomicalRANSAC(). Same for bottom line. Then find where the edge deviates a fair amount from the fitted line and fit a slope to that section. Pretty each. Try it. I didn't do it for you but I believe I've shown you before how to do it in other questions of yours, basically

mask = grayImage > someThreshold
topRows = zeros(1, columns)
for col =1 : columns
    t = find(mask(:, col), 1, 'first')
    if ~isempty(t)
        topRows(col) = t;
    end
end
% Fit line to left portion
lefty = topRows(1:someColumn);
x = 1 : someColumn;
leftCoefficients = polyfit(x, lefty, 1)
leftFitted = 
% Fit line to right portion
righty = topRows(someOtherColumn:end);
rightCoefficients = polyfit(x, righty, 1)
rightFitted = 
% Find differences
leftDiff = topRows - leftFitted;
rightDiff = topRows - rightFitted;
col1 = find(leftDiff > someValue, 1, 'first')
col2 = find(leftDiff > someValue, 1, 'first')
ySlanted = topRows(col1:col2)
x = col1:col2;
coefficients = polyfit(x, ySlanted, 1)

etc. As you can see, not hard at all. I have full confidence that you can do it. Go ahead and finish it, and proudly post your achievement.

[Prashant]

Hallo Image analyst,

I have tried my code and i am attaching that image file. Also i attached my code name image.m. please have a look.

As you can see in picture, i am getting green boundary on the Image but my intersection point x is not on that point of an image. I want x point that i mentioned in fig.

[Image Analyst]

Try findchangepts()

[Prashant]

Hello Image Analyst,

I have one more problem in the different image. Could you please have a look. I am attaching the image name carlight.jpg

I want to measure the distance automatically as shown in fig, i want to find the distance from the laser light and the curve or as shown in fig. could you please help me how to write code.

thank you.

[Image Analyst]

Try scanning column by column. If there are exactly two regions, go from the top or center of the first region to the top of the second region. Or you could just measure the height of the black gap between them. It just depends on what your definition of distance is.

clc;    % Clear the command window.
fprintf('Beginning to run %s.m ...\n', mfilename);
close all;  % Close all figures (except those of imtool.)
imtool close all;  % Close all imtool figures.
clear;  % Erase all existing variables.
workspace;  % Make sure the workspace panel is showing
fontSize = 15;

mask = imread('image.bmp');
mask = mask(:, :, 1) > 0;
hFig = figure;
subplot(2, 1, 1);
imshow(mask);
hold on;
hFig.WindowState = 'maximized';
[rows, columns] = size(mask)
distances = zeros(1, columns);
distances2 = zeros(1, columns);
for col = 1 : columns
    thisColumn = mask(:, col);
    props = regionprops(thisColumn, 'Centroid', 'PixelList');
    if length(props) == 2
        % Only two regions which is what we need.
		% Get the center of the thick top line.
        yCentroid = props(1).Centroid(2);
        yTop = props(1).PixelList(1,2);
		% Get the top of the bottom region.
        yTop2 = props(2).PixelList(1,2);
		distances(col) = abs(yTop2 - yCentroid);
		distances2(col) = abs(yTop2 - yTop);
		line([col, col], [yTop, yTop2], 'Color', 'r');
    end
end
subplot(2, 1, 2);
plot(distances, 'b-', 'LineWidth', 2);
hold on;
plot(distances2, 'r-', 'LineWidth', 2);
xlim([1, columns]);
grid on;
xlabel('Column', 'FontSize', fontSize);
ylabel('Distance', 'FontSize', fontSize);
legend('Distance from Centroid', 'Distance between tops');

Answer 3

[Prashant]

Hello Image Analyst,

Thank you for your answer.

I have tried your code but i can not get the result which i have shown in fig. Also by using mask my image is become inaccurate for example the edge of the image surface is blurred.

I want to find the distance in cm or whatever the unit. But i don't know that how can i select that particular region.

Could you please write code for me with the explaination. the image is in attached name binary.jpg.

[Image Analyst]

You need to clear that border with imclearborder. My code looks for exactly 2 regions but if you have that white border all around, then the columns with the two regions are not the places you want. They'll be the placed with no blob and just the places with the white frame. Why is there a white frame around your image anyway?

Answer 4

[Prashant]

As you know that this image is from Headlight. I have made a photo with my camera and it is look like this in attched image name ''DSC_2637.png ''. So i have convert the image in binary then i got the image like attached image called ''binary.jpg ''.

Now i want to see that how can i get that distance of that marked region.

could you please write that code for me?

[Image Analyst]

Explain how "convert the image in binary" was done? Because simply calling imbinarize() or whatever will not give the white rectangular frame around the image. How did that white frame get there?????

[Prashant]

Here is my code that i convert my image into binary.

[Image Analyst]

OK, now this gives a binary image without that white frame around it. I think you must have saved the whole figure, not just the image alone. Anyway, the code gives this:

What's not working about that? And you never did explain what point in the above skinny blob you want to measure from - the top or the middle of it. Which is it? My plots show it both ways.

[Prashant]

Hi ,

Yes it is working, just i was a bit confuse because in the graph it shows two result for me. and i wanted result only from top line (bottom border) to bottom line (top border). So what will be the code for showing only this area?

thank you image Analyst. I am really glad that you help me. I am doing my Masterthesis and for that i have to do some programming and i am not that good with programming. but with your help i am learning and improving my skill. Thank you so much. I would like to be your student. :)

[Image Analyst]

Thanks for the compliments. If the image has "rows" rows, then, at each column, the top skinny blob will start at y1 and end at y2 (which could be the same in the case of a single pixel wide line), and have a centroid at (y1+y2)/2. Let's call the centroid y3. The big bottom blob will start at y4 and go down to y5=rows. So we have several possible distances:

1. y1 to y4 (top of upper skinny blob to top of lower big blob)
2. y2 to y4 (centroid of upper skinny blob to top of lower big blob)
3. y3 to y4 (bottom of upper skinny blob to top of lower big blob)

Which do you want? I believe my code gives you all three distances.

[Prashant]

Thanks for the explaination, I want bottom of upper skinny blob to top of lower big blob.

[Image Analyst]

Then you'd use a y value for the bottom of the skinny blob, as I'm sure you've already figured out by now.

yTop = props(1).PixelList(end,2);

[Prashant]

Yes now i know how to select that line.

thank you again.

[Prashant]

Hello Image Analyst,

I hope you are well. I would like to ask you one more question that this time i did single camera calibration and for that i used checkboard on the wall and did calibration. the square size of the checkboard is 34mm, after i removed that checkboard an with the same distance and same position i made a photo of my image on the same plane. Now i would like to know about the distance between laser line and the the upper surface of the image that is mentioned in image.

I do not know that how can i do further. could you please help me to find real world distance by using camera calibration.

i have attched my code name Versuch.m and the image that i want to measure name IMG_0519.JPG.

The image for calibration i can't send you because the file size is big. If it is possible, could you give me your email or smthng.

thank you in advance

[Prashant]

One of the calibration image that is used that i attched here.

[Rik]

1. Measure the distance in pixels between two corners on your checkerboard image.
2. Measure that same distance in real world units.
3. Measure the disance on the image you want to know.
4. Use the first two to determine your calibration factor and multiply by the third to get the distance in real world units.

This all assumes your camera did not move significantly and you are using the same resolution image (or you know the exact conversion).

PS When you contact people, they tend to prefer you sending them a link to your question. Please do that next time.

[Prashant]

HI Rik,

Thank you for your answer.

But i did not understand that why i need to measure the distance in pixel of two corners of checkboard. I am using camera calibration tool and real world distance of checkerboard square is 34mm that i know. Have you seen my code name Versuch.m i followed the same step like matlab tutorial ''Measuring planar object with camera calibration''

[Rik]

That cameraParams variable probably contains the required information. What did a Google search turn up? This isn't really my field of expertise. I've only had to do this once, which only required a keystone correction I could do manually.

[Prashant]

Thanks Rik,

google did not show me any solution and i saw also in matlab community but could not find my solution.

if you know smthng then pls help me to write code for my problem

[Image Analyst]

Try my attached spatial calibration demo.

[Prashant]

Hi Image Analyst,

I have already used your spatial calibration demo and it work for me perfectly. Now i am doing with camera calibration and want to find distance automatically by using reference image like checkerboard. because with spatial calibration it might not an accurate result for me because in this case i have to draw the line and then this code will give me result and would be posibility that i will draw th line on that contains some other area of pixel that i do not want.

Therefore i am using camera calibration otherwise your code is perfecct working.

[Prashant]

Hi Image Analyst,

https://de.mathworks.com/matlabcentral/answers/555463-how-to-fitline-on-my-binary-image-and-find-distance#comment_940293 in this link we have solved this problem. for more details and understand about my question please see the content in the link.

I would like to ask you that in this problem we found that filled area with red lines, does it shows the number of pixels ? if yes then how can i find the number of pixel in one column? for example i want to know that in coulmn number ([2300, 2395]), how many pixels are there? then how can i find this?

I need this information because i would like to know the distance of filled area with red line but not all the area only for the particular coulmn. for example i have written this code:

is my assumption is correct?

[rows, columns] = size(Iopenned);
distances = zeros(1, columns);
for col = 1 : columns
        thisColumn = Iopenned(:, col);
        props = regionprops(thisColumn, 'Centroid', 'PixelList');
        if length(props) == 2
        % Only two regions which is what we need
%       get the bottom of the top region
        yTop = props(1).PixelList(end,2);
% 		Get the top of the bottom region.
        yTop2 = props(2).PixelList(1,2);
%         distances(col) = abs(yTop2 - yTop); %[2292, 2367]
        distances([2300, 2395]) = abs(yTop2 - yTop); %this column i want
		line([2578,2582], [2292, 2367], [yTop, yTop2], 'Color', 'r'); %[2578,2582], [2292, 2367]
% 		line([col, col], [yTop, yTop2], 'Color', 'r');
        end
end
% After this code i want to find the real distance by using referance image (checkerboard)
%so i used this formula by using number of pixels in this ([2300, 2395])column which i decided and 
%the number of pixels in one square bölock, so at the end i will get a real distance
distances2 = distances([2300, 2395]).* squareSize/pixelsinsquare;

[Image Analyst]

No, that code is no good. For one thing, you look at every column when you said you're only interested in two columns. Then your distance array only assigns those two columns even though the loop iterator is on a different column - makes no sense. Finally you don't need to use regionprops at all when you can simply use find() on those two columns - no loop at all is needed if you want only those two columns:

distances2(1) = find(Iopenned(:, 2300), 1, 'last') - find(Iopenned(:, 2300), 1, 'first');
distances2(2) = find(Iopenned(:, 2395), 1, 'last') - find(Iopenned(:, 2395), 1, 'first');

And again, it seems like you're totally ignoring my advice on how to spatially calibrate more accurately using the bounding box of the entire checkerboard. Why did you ignore that advice and go with your initial, less accurate guess of using a single square instead?

[Prashant]

Hi image analyst,

i did not ignored your advice. I have written in here to just mentioned that what i want to do but i do not want to use only one square block. I will use as you said the whole blocks with finding pixels.

The second thing is that when i found distance between those 2 column then i will get number of pixels in that column with which is filled with red lines?

[Image Analyst]

Well I think your original binarization is no good, so everything after that is questionable. First, if I remember correctly, it looks like your image is over exposed. You should use a lower exposure. You could improve the contrast of the line in several ways. You could use perhaps a black stripe on the wall instead of a laser. Using light on light (a laser) doesn't give much contrast, while I think a black painted line would give more contrast. Or you could use different polarizations for the laser and the headlight. Put polarizers in front of the lights at perpendicular directions and then a rotatable one in front of the camera and take two images, one with the polarizer rotated to see the laser and another rotated to see the headlight. Another way to see the line better through the bright part of the light pattern is to do CLAHE on the image, meaning use adapthisteq() on the image to flatten it and improve contrast. Doing any or all of these should improve your images.

[Prashant]

Thanks for the suggestion. I would like to tell you that i could not use the stripe because the laser is installed inside the headlight and it is there to see the distance from red line to my image surface. The second thing is that we do not have polarizers in our company. so i could not also use that because i am doing my thesis with company and they do not have this kind of stuff. let's see if i can find some solution.

[Prashant]

Hallo Image Analyst,

I would to ask you that in my headlight projection figure, I would like to detect the reference point for example when the horizontal straight line is changing to curve and again line will be staight. in this case when line will change to curve, i would like to put there one reference point with red round mark and I want that will happen automatically for every figure.

An example is shown in figure.

[Image Analyst]

I'd try scanning with a window of, say 51 pixels or something, and fitting a line to it and looking for the slope. Then when the slope exceeds to value, like .3 or whatever, that's the edge. Untested code:

windowSize = 51;
slopes = zeros(1, columns);
for k = ceil(windowSize/2) : columns - ceil(windowSize/2)
    thisSection = binaryImage(k : k + windowSize - 1);
    coefficients = polyfit(1 : windowSize, thisSection, 1);
    slopes(k) = coefficients(1);
end
plot(slopes, 'b-', 'LineWidth', 2);
grid on;
xlabel('Column', 'FontSize', 18);
ylabel('Slope', 'FontSize', 18);
% Find where slope exceeds 0.3
steepColumns = find(slopes > 0.3)
xline(steepColumns, 'Color', 'r')

[Prashant]

Hello Image Analyst,

Thank you for your help. I have run this code and i got error like this:

I attached my code and figure please have a look.

'columns' requires Database Toolbox.

Error in Knickpunkt (line 44)
slopes = zeros(1, columns);

[Image Analyst]

columns is the number of columns in your image. You should have seen from our many prior messages that you do

[rows, columns, numberOfColorChannels] = size(binaryImage);

[Prashant]

after i got this another error:

could you do correction in my code?

Error using xline (line 29)
Value must be nonempty.

Error in Knickpunkt (line 56)
xline(steepColumns, 'Color', 'r')

[Prashant]

Hallo Image Analyst,

Could you please have a look or give me some correction.

Thank you in Advance.

[Image Analyst]

A number of errors on that. Mainly because you didn't use the right window width or have it scan the top rows. You scanned the image instead. Also you should take the largest 4-connected blob. Corrected code is attached.

[Prashant]

How to know that window size is 51?

[Image Analyst]

I just eyeballed it from the plot and that seemed like it would be a good window width. And it worked out well. Feel free to experiment with different widths and choose a different one if you want, but use an odd number.

[Prashant]

Ok I understood.

[Prashant]

I have one more problem that i have one image with vertical lines and i would like to detect only two of them and want to find distance in pixels between that two line. How could i measure that? could you please help me with code, the image is in attached file.

An example of image that i want to find distance in pixel

[Image Analyst]

Use regionprops on the binary image to find the height of the black blobs. Then extract the two tallest vertical lines and find their centroids and the distance between their x component of the centroids. Untested code:

mask = grayImage < 128; % Whatever - somehow you have to get the black lines to be true and background to be false.
% Get rid of small noise specks.
mask = bwareaopen(mask, 10); % Only keep blobs 10 pixels or bigger.
% Label the image
labeledImage = bwlabel(mask);
props = regionprops(labeledImage, 'BoundingBox')
allBB = vertcat(props.BoundingBox);
allHeights = allBB(:, end);
% Sort them to get the two with the tallest height.
[sortedHeight, sortOrder] = sort(allHeights, 'descend');
% Extract the two tallest blobs
twoVertical = ismember(labeledImage, sortOrder(1)) | ismember(labeledImage, sortOrder(2));
% Now get centroids of that image to find the distance between the centroids.
props = regionprops(twoVertical, 'Centroid');
horizontalDistance = abs(props(2).Centroid(1) - props(1).Centroid(1))

[Prashant]

Thanks Image Analyst,

could you please test it also with my figure.

[Image Analyst]

Did it not work? If not, attach your original, unannotated image, or the binary image.

[Prashant]

Here is the figure and its name 'blank.png'

[Image Analyst]

That's not the same as what you originally showed. This one has 4 lines and they go all the way vertically across the image. What distances do you want? Do you want 3 distances at every row/line of the image? I think you can solve this knowing what you've already learned about image processing.

[Prashant]

Hi image analyst,

I need any two vertical line (one near to other) and the rest i do not want. I need the distance between those two line so i can use it for my calibration code which can measure my object with reference to this two line distance measurement. so that is why i need this distance value in pixel. I know the real world distance is 100 cm between two vertical line. for example you gave me a code for counting number of pixels of square block in checkerboard like tha same way i want to find for this image.

this image is original one that i posted.

[Image Analyst]

I'd sum the inverted image vertically to get the horizontal profile, and then look for peaks.

horizontalProfile = sum(255 - grayImage);
plot(horizontalProfile, 'b-', 'LineWidth', 2);
grid on;
[peakValues, peakLocations] = findpeaks(horizontalProfile);
plot(peakLocations, peakValues, 'rv', 'MarkerSize', 15, 'LineWidth', 2);