How can I find the radius of this object, preferably at multiple x,y points for a consistent output?

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Sanam Pun
Sanam Pun 2024-2-7
编辑： Matt J 2024-2-7
Hi,
Using the grid as a scale and performing spatial calibration on Matlab, I have managed to find the radius using r = s^2+L^2/2*s^2. However this only uses 3 data points. I was looking for a more "robust" method of doing this - i.e fitting multiple points in the image and finding multiple radius/curv values that should ideally all be the same since x^2+y^2 = r^2.
Any advice please?

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Matt J
Matt J 2024-2-7
编辑：Matt J 2024-2-7
You can use circularFit() from this FEX download,
https://www.mathworks.com/matlabcentral/fileexchange/87584-object-oriented-tools-to-fit-plot-conics-and-quadrics
It will allow you to fit the arcs based on more than 3 points. However, it will only fit one object at a time. So, if you have multiple arcs from different objects, you must process them separately.
Also, bear in mind that the circularity of the arcs in your image may be warped depending on the perspective of the camera. The fitting tool doesn't take camera perspective into account.
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Sanam Pun
Sanam Pun 2024-2-7
Does viscircles use a particular unit measurement? Or does it automatically scale? Also, from my understanding, all I need is a center points and radius?
imshow(imread('xyz.jpg'));
hold on;
viscircles(centers,radius);
hold off
Matt J
Matt J 2024-2-7
编辑：Matt J 2024-2-7
viscircles will be in whatever scale the XData and YData of imshow are. You can change the default scale using,
imshow(imread('xyz.jpg') ,XData=[xmin,xmax] , YData=[ymin,ymax]);

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更多回答（1 个）

Image Analyst
Image Analyst 2024-2-7
See the FAQ:
How can I fit a circle to a set of XY data
or my attached function.
  2 个评论
Sanam Pun
Sanam Pun 2024-2-7
Hi, cheers for this! I used your function and it works perfectly! (doubled checked with a circle I already know the values of i.e xcentre, ycentre ).
Matt J
Matt J 2024-2-7
编辑：Matt J 2024-2-7
Ran in:
Comparing the fitting performances of the two methods is a bit subjective, I suppose, seeing as it's such a short arc. But, I like mine better ;-)
load xytrial
cf1=circularFit(xy); %Matt J
[xCenter, yCenter, radius] = circlefit(xy(1,:), xy(2,:));
cf2=circularFit.groundtruth([],[xCenter,yCenter], radius); %Image Analyst
%%Compare
H(1)=plot(cf1);
hold on;
H(2)=plot(cf2,{'Color','g','LineStyle','--'});
hold off
legend(H,'Matt J','Image Analyst','Location','SE')
axis([1.2180 3.0 1.6036 3.3441]*1000); axis equal

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