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

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?

[Dyuman Joshi]

Have you tried searching on the forum for similar questions asked (and answered) before?

[Sanam Pun]

Hi, I have found some similar topics and have looked into it, however I was unable to apply it to my specific scenario...

Answer 1

[Matt J]

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.

[Sanam Pun]

Hi Matt,

Thanks for this! My Matlab skills aren't that great so following your script is a little bit difficult to me. Any chances you could elobrate on how I can utilise the code to generate solutions? Also just to be clear, this code assumes the center points and radius is unkown?

[Matt J]

Yes, everything about the circle is assumed unknown. If you attach a .mat file with a set of xy-points, I can demonstrate its usage.

[Sanam Pun]

Apologies for the delay - here's an example, not sure if they will be enough though.

[Matt J]

load xytrial

cf=circularFit(xy)

cf = 
  circularFit with properties:

    center: [2.7050e+03 1.4496e+03]
    radius: 1.6825e+03

plot(cf)
axis([0,2*cf.center(1), 0, 2.5*cf.center(2)])

[Sanam Pun]

Right! I see. This is great stuff. Thank you again for this :)

[Image Analyst]

You can use viscircles, but it's interesting to note that plot will take such an object directly, and not need just x and y data.

[Sanam Pun]

Just to add, how can I plot a circle in the actual image from the given center points and radius calculation? Technically, if there is no or very little distortion from the camera, it should be a near perfect circle?

[Matt J]

Any two plots can be superimposed, including images using hold, e.g.,

imshow(imread('cameraman.tif')); hold on
plot((1:150),'y'); hold off

or you can use visciircles as @Image Analyst mentioned earlier.

[Sanam Pun]

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]

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]);

Answer 2

[Image Analyst]

See the FAQ:

How can I fit a circle to a set of XY data

or my attached function.

[Sanam Pun]

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]

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