radius of rod along its length in an image

I have an image (attached) containing a rough rod. I am trying to find the values of radius along the rod's length. The true radius at the edge of the rod is known to me. What would be a good/ elegant way of doing this using matlab image processing tools?

[DGM]

I don't see how you're going to separate the object from the background in this image. Given that it's reflective, this is likely going to be a challenge. You might exploit reflections and creative staging to increase the edge contrast against a dark background, or you might try photographing it against a diffuse backlit surface. I've used a light table for getting profile images of parts before, and I suppose any ad-hoc equivalent would work too.

The bigger problem is that the object obscures itself due to the distance to the camera. The ends of the object are effectively presented at an angle, despite the camera's axis being perpendicular to the part. If you can get the camera further away from the object and use a smaller lens angle, that would reduce the effect. Doing those two things will also reduce the apparent scale distortion caused by perspective.

[MAM09]

Thank you for your suggestions. If the issues you mentioned were not there, how would I actually measure the radius in matlab? I know how to get dimensions from images of simple objects such as circles, but not sure how to do the same for radius along the length of this rod (at 1 mm intervals).

[Image Analyst]

Just threshold to get the rod, then sum the binary image vertically. The sum is the diameter so divide it by 2.

mask = imbinarize(grayImage); % Assuming the mask looks good.....I have not tried it.
horizontalProfile = sum(mask, 1);
radii = horizontalProfile / 2;

Answer 1

[DGM]

I ended up having to do a little manual touchup on the image (attached). It's still not a very good result, and the perspective probably makes the results meaningless, but it should serve as an example.

A = imread('wavyrod.jpg');
A = rgb2gray(A);

% crop it down to save time
A = imcrop(A,[0.5 962.5 4032 572]);

% try to flatten the image and smooth out any noise
Af = imflatfield(A,200);
Af = medfilt2(Af,[10 10]);

%imshow(Af)

% threshold the image and try to clean it up
m = Af>50;
m = bwareafilt(m,1);
m = imfill(m,'holes');
m = imopen(m,strel('disk',30,0));

figure
%imshow(m)

% omit the very end, since it's not a full diameter due to the viewing angle
startx = 3908; 
m = m(:,1:startx);

% object width in pixels
profile = sum(m,1); 

% assuming end diameter is known, that could be used for scaling
enddia = 25; % mm
scale_estimate1 = enddia/mean(profile(end-10:end)) % from the diameter

scale_estimate1 = 0.0596

% or the ruler could be used for scaling
% this is the distance beween the the lower 2 and 23 cm tickmarks on the ruler
% i just measured this manually
rulerspan = [3440 40]; % pixels [x y]
spanlen = 210; % mm
scale_estimate2 = spanlen/norm(rulerspan) % from the ruler

scale_estimate2 = 0.0610

% we now have two estimates of the scaling factor
% for this example, i'm going to assume the second has preferable accuracy
% note that this is reversed, aligning the visible end of the rod with zero
profilemm = profile * enddia/mean(profile(end-10:end));

% create scaled xdata for the profile
xposmm = linspace(scale_estimate2*numel(profilemm),0,numel(profilemm));

plot(xposmm,profilemm)

% now let's say we want to interpolate this xdata every 1mm
% start by getting new xdata
xposmm_new = 0:1:floor(xposmm(1));

% one approach would be to fit a spline to both interpolate and denoise
pp = fit(xposmm',profilemm','smoothingspline','smoothingparam',0.5);
profilemm_new = pp(xposmm_new);

figure
plot(xposmm_new,profilemm_new) % the new data at every mm

Obviously, the radius would be half that.

I don't feel confident to trust the ruler for scaling, since its length varies by about 12% across the image. In a better setup, using the ruler may be more effective.

[Image Analyst]

If you want better accuracy you'd use a telecentric lens or use knowledge of the geometry of the setup to correct for magnification (the light rays are not parallel if you're not telecentric - they're like a fan beam.)

[MAM09]

Thank you so much for all your assistance. I will try and get a better image and perform the steps you mentioned to see if I get the desired results or not. Thanks again

[Image Analyst]

To get the radii in mm:

radii = profilemm / 2;

This is the radius of the rod for every column in the image.

[MAM09]

Hello again, using the method suggested by DGM I got radius values quite close to the actual ones (attached). I Had to modify the code slightly but kept the idea the same. The difference between actual radius values and those extrated from image is about 8X10^-3 inch in most cases which is consistent with the resolution of the camera used. But at some locations at the rod, the error is about 10^-2 inch. I guess the edges were a bit darker at these locations (non-uniform lighting?). Are there ways to improve these local errors?

[DGM]

For the given image (or a similar setup), I would expect that you're right. The accuracy in getting the exact edge isolated varies due to the reflectivity of the object and the texture of the background. Some edges look exactly like the background since they're a reflection of it.

As I think I mentioned, I've resorted to backlighting parts on a light table to get good edge contrast, but I haven't done it with shiny round parts. Reflections might still be an obstacle, and I know my current digital camera does not like to behave with the flicker from the light table, so that might be another source of frustration.

I guess the brute simple way to do it would be to just smoke the part with lampblack and photograph it against a plain white background. That would be a bit of a mess, but it would be way easier.

[MAM09]

Thank you for your suggestions. The reflective nature of the rod is indeed an issue.

Answer 2

[yanqi liu]

yes，sir，may be ocr the number and use the scale information to get result，now we can see the ocr process.

im = imread('https://ww2.mathworks.cn/matlabcentral/answers/uploaded_files/896745/image.jpg');
bw = im2bw(im);
bwt = ~rot90(bw,2);
c = ocr(bwt,'CharacterSet','0123456789');
bts = [];
wts = [];
for i = 1 : length(c.Words)
    if length(c.Words{i}) ~= 2 || c.WordBoundingBoxes(i,4) > size(bw,2)/40
        continue;
    end
    bts(end+1,:) = c.WordBoundingBoxes(i,:);
    wts{end+1} = c.Words{i};
end
[~,ind] = max(bts(:,1));
rect = bts(ind,:); 
word = wts{ind};
figure; imshow(rot90(im,2),[]); hold on;
hold on; rectangle('position', rect, 'EdgeColor', 'r', 'LineWidth', 2)
text(rect(1),rect(2)-rect(4)/2,word,'Color','r','FontSize',20);

[MAM09]

Hello, thanks for the response. But does the above method give me the radius at different points along the length of the rod? If yes, how? Sorry, I could not understand how the radius values of the rod can be found.