![screenshot.jpg](https://www.mathworks.com/matlabcentral/mlc-downloads/downloads/submissions/34767/versions/3/screenshot.jpg)
How to create a circles (smallest and biggest circle) based on points in an image given, then find the center and radius?
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Jafar Nur Arafah
am 15 Jan. 2019
Kommentiert: Jafar Nur Arafah
am 16 Jan. 2019
Hi,
I have extract some points from Image Scanning, I need to create 2 circle (smallest circle and largest circle) in order to calculate the error of circularity (Rmax - Rmin), in order to do that I need to find the center and radius. Did any have done this kind of case?
Below are some image as reference:
- Points extract from image scanning
![test.GIF](https://www.mathworks.com/matlabcentral/answers/uploaded_files/200746/test.gif)
2. Example of circles need to be done
![example.GIF](https://www.mathworks.com/matlabcentral/answers/uploaded_files/200747/example.gif)
Any help will be appreciated.
Thanks in advance
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Akzeptierte Antwort
Image Analyst
am 15 Jan. 2019
To get the minimal bounding outer circle, try this:
![screenshot.jpg](https://www.mathworks.com/matlabcentral/mlc-downloads/downloads/submissions/34767/versions/3/screenshot.jpg)
I asked John for the largest interior circle, and he said that's a much more difficult problem and he doesn't have code for that.
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Weitere Antworten (1)
KSSV
am 15 Jan. 2019
YOu can try to fit a circle with the coordinates (x,y) you have. Check the below code:
function [xc,yc,R,a] = circfit(x,y)
%CIRCFIT Fits a circle in x,y plane
%
% [XC, YC, R, A] = CIRCFIT(X,Y)
% Result is center point (yc,xc) and radius R. A is an optional
% output describing the circle's equation:
%
% x^2+y^2+a(1)*x+a(2)*y+a(3)=0
% by Bucher izhak 25/oct/1991
n=length(x); xx=x.*x; yy=y.*y; xy=x.*y;
A=[sum(x) sum(y) n;sum(xy) sum(yy) sum(y);sum(xx) sum(xy) sum(x)];
B=[-sum(xx+yy) ; -sum(xx.*y+yy.*y) ; -sum(xx.*x+xy.*y)];
a=A\B;
xc = -.5*a(1);
yc = -.5*a(2);
R = sqrt((a(1)^2+a(2)^2)/4-a(3));
3 Kommentare
KSSV
am 15 Jan. 2019
Ou are running the code in a wrong way...you have to save the above code into a function and call the function.
Image Analyst
am 16 Jan. 2019
Bearbeitet: Image Analyst
am 16 Jan. 2019
This fits a circle through the data, which could be part of the solution, but it does not find the max bounding circle (like my Answer).
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