How can I find circle diameter using arc ?

I need to find the diameter of a circle. However, the whole circle is not visible in the image, only the arc is shown. How can i find the diameter of the circle using this arc alone ?

Answer 1

[Image Analyst]

See attached demo.

[Syed Sameer Ul Hasan]

Thank you so much Image Analyst. It works perfectly !!!! Thanks alot

[sloppydisk]

Did the radius end up being different from my answer?

[Syed Sameer Ul Hasan]

Yes. Your answer gave the result 314 while this code gave 515

[Syed Sameer Ul Hasan]

Image Analyst, thanks for the code. However is there a way to increase the accuracy of the output even more ?

[sloppydisk]

In fact my code gave 521.6413 pixels.

[Image Analyst]

Yes, use a higher resolution camera, and include more of the circle. Improve your lighting. And, depending on the situation and geometry, using a telecentric lens ( a special type of lens that uses parallel rays of light) might help.

Answer 2

[sloppydisk]

Take three points on the arc, draw lines between them and take the bissectors of those lines, where they meet is the center O. The radius is the distance from any of the points to the center O.

Or if you want to do it algebraically take a look at this link .

[Syed Sameer Ul Hasan]

Thank you for your answer sloppy disk. But how can I take the three points on the arc programmatically ? I Actually want the system to work in real time using a camera so user interference has to be zero.

[sloppydisk]

I would probably do something like this:

rgbImage = imread('002.jpg');
grayImage = rgb2gray(rgbImage);

% take edge
bw = edge(grayImage, 'canny', .2);
% clean up edge
[X, Y] = meshgrid(1:size(bw, 2), flip(1:size(bw, 1)));
bw(Y<50|Y>.8*max(Y)) = 0;
figure(1) 
imshow(bw) % show edge

bw = flip(bw);
[y, x] = find(bw); % take indices of edge

figure(2)

plot(x, y) % plot edge
[z, r] = fitcircle([x y]); % do fit giving center z and radius r

t = linspace(0, 2*pi, 50);
xfit = z(1) + r*cos(t); yfit = z(2) + r*sin(t);
hold on
plot(xfit, yfit) % plot fit of edge
hold off

X = ['Radius is ',num2str(r), ' pixels'];
disp(X)

Using fitcircle from file exchange.

[Syed Sameer Ul Hasan]

Thanks for your quick response. Did you use this fitcircle.m file? https://www.mathworks.com/matlabcentral/fileexchange/15060-fitcircle-m

I used the one above but the radius is not coming out to be correct.

Thanks

[sloppydisk]

Yeah I used that one, I linked to it in my comment. How do you know it's not correct? In my fit plot it fits the data closely.

[Syed Sameer Ul Hasan]

Because i have this circle at hand. I do know it's real diameter. And yes i did mulitiply the radius by calibration factor (mm/pixel)

[sloppydisk]

Did you also multiply by two to go from radius to diameter? Otherwise I don't know what's wrong.

[Syed Sameer Ul Hasan]

Yes it did. Still thanks alot for the help.

Btw, can you help me with this question of mine ?

https://www.mathworks.com/matlabcentral/answers/403903-how-to-remove-frame-boundary-from-object-boundary

Thanks

[John D'Errico]

Note that is a moderately poor way to compute approximate circle radius, especially if the arc is not a perfect circle. You are taking two points as normals off of a bumpy & vaguely circular arc, extending lines. So to find the intersection of two lines that are defined by noisy estimates, will generate a noisy result, but worse, the intersection computation will exaggerate that noise. The bast case would be if you had multiple normal vectors to that curve, with the most useful information probably arising when the lines were orthogonal.

Higher quality tools that compute a circular arc will try to use the entire curve, not just effectively two points form that curve.

[Syed Sameer Ul Hasan]

Thanks for your response John D'Errico. What do you suggest I should do to get the most accurate result ?

[sloppydisk]

@John D'Errico Yes the picture was only meant as an example to show the general principle of getting the center geometrically. In my example code I did use a least squares fit using the entire curve from fitcircle.m from the file exchange.

Answer 3

[Image Analyst]

Threshold, extract largest blob with bwareafilt(), find perimeter with bwboundaries(), then use the FAQ http://matlab.wikia.com/wiki/FAQ#How_can_I_fit_a_circle_to_a_set_of_XY_data.3F

[Syed Sameer Ul Hasan]

Thank you for your reply Image Analyst. I have seen this faq before but didn't understand the working and imputs of this function. If you could please do that it would be very helpful for me

Answer 4

[John D'Errico]

Use the entire curve if possible. Well, all of it that you have. Not just a couple of isolated points.

Extract points from the entire accessible arc, as (x,y) pairs. At this point, it becomes a pure circle fitting problem.

Some years ago, I wrote a circle fitting tool. It tries to produce a fit that does as well as possible, even allowing the use of various solvers like backslash, pinv, or robustfit as alternatives. However, there are many algorithms one might postulate. I'll admit, I'm not in love with the one listed in the FAQ, but it is not that bad either. In some quick tests, it was comparable to the one I wrote. So if your data truly comes from a circle, you probably won't do too much better. And your data should not be too noisy since that arc is quite clear.

Can I do better than that though? Were I to try to do a bit better, I might look to using a maximum likelihood estimation, assuming noise in both x and y on each point. But your data is not that noisy, not that crappy that it is worth the effort.

So where will any "problems" arise? They will come from deviations from pure circularity. Thus, is that arc a perfectly circular arc, with a perfectly fixed radius? I doubt it. In fact, I'll bet that the largest amount of error in your circle parameter estimates come from those deviations from circularity, thus, lack of fit to a true circle, rather than from noise in the fit.

If that is true, then what does it mean to try to do better, fitting a circle perfectly to data that is not perfectly circular? There is a limit beyond which the fit becomes an exercise in absurdity. How can you improve things to best effect? Do a better job of extracting those points on the perimeter of the circle. Improving your data is always a good idea, because that means the algorithms meant to post-process the data will have a far easier time of it.

I would look carefully at the data you have. Compare it to the fit you estimate. Does it have systematic patterns in the residuals between your data and the circle fit. If you see patterns, that means your largest problem lies in lack of fit. And if my conjecture is true, then you won't improve things by making a better circle fit.

Answer 5

[Christine Tobler]

There's a recent blog post by Loren on fitting a sine wave to a picture by eye:

Is This a Sine Wave? - Loren on the Art of MATLAB

This won't give you best accuracy, but could be a nice way of doing the fit by hand, and verifying if the final circle you choose fits the picture well.