MATLAB: Circle least squares fit for 3D data

circle fitleast squares

Hi everyone,

I have 6000 x coordinates, y coordinates and z cooridinates that form a circle that does not perfectly occupy one plane. I am looking to fit the data with a circle but I can only find functions online that do it for x and y coordinates and not including z. I would appreciate any help in creating some code for this as I am not sure where to start (I am a beginner in MATLAB!)

Thank you in advance

Ciara

Best Answer

X=csvread('data.csv');
XC = mean(X,1);
Y=X-XC;
[~,~,V]=svd(Y,0);
Q = V(:,[1 2]); % basis of the plane
Y=Y*Q;
xc=Y(:,1);
yc=Y(:,2);
M=[xc.^2+yc.^2,-2*xc,-2*yc];
% Fit ellipse through (xc,yc)
P = M\ones(size(xc));
a=P(1);
P = P/a;
r=sqrt(P(2)^2+P(3)^2+1/a); % radius
xyzc = XC' + Q*P(2:3); % center
theta = linspace(0,2*pi);
c = xyzc + r*Q*[cos(theta); sin(theta)]; % fit circle
close all
plot3(X(:,1),X(:,2),X(:,3),'.');
hold on
plot3(c(1,:),c(2,:),c(3,:),'r','LineWidth', 2);
axis equal

Related Solutions

MATLAB: Find points of intersection circle with line

clc; clear all ;
C = [1.5 1.5] ;   % center of cricle 
r = 1.5 ;         % radius of circle 
% 
x = 0:1:3 ; 
y = 0:1:3 ; 
[X,Y] = meshgrid(x,y) ; 
%
th = linspace(0,2*pi) ; 
xc = C(1)+r*cos(th) ; 
yc = C(2)+r*sin(th) ; 
L1 = [xc;yc] ;
% GEt intersecdtion points 
% along X 
[m,n] = size(X) ; 
for i = 1:m
  L2 = [X(i,:) ; Y(i,:)] ; 
  P{i} = InterX(L1,L2)
end
for i = 1:n
  L2 = [X(:,i) Y(:,i)]' ;
  Q{i} = InterX(L1,L2) ;
end
P = cell2mat([P Q]) ;
plot(xc,yc,'b') ;
hold on
plot(X,Y,'r')
plot(X',Y','r')
plot(P(1,:),P(2,:),'*k')

InterX can be downloaded from here: https://in.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections?focused=5165138&tab=function

MATLAB: Solving two equations with two unknows

You could undoubtedly solve these two equations using 'solve' of the Symbolic Toolbox or 'fsolve' of the Optimization Toolbox. However, this is the classic problem of finding the two intersections of two circles which is well understood, and the following matlab code is very likely more efficient than either of these methods.

Let P1 and P2 be column vectors containing the coordinates of the two circles' centers - [xA;yA] and [xB;yB] in your case - and let r1 and r2 be the two respective radii. Then the two intersections of the circles are points Q1 and Q2 where Q1 lies off to the left and Q2 off to the right as you move from P1 toward P2.

P21 = P2-P1;
d2 = sum((P21).^2);
P0 = (P1+P2)/2+(r1+r2)*(r1-r2)/2/d2*P21;
Q0 = sqrt(((r1+r2)^2-d2)*(d2-(r1-r2)^2))/2/d2*[0,-1;1,0]*P21;
Q1 = P0+Q0;
Q2 = P0-Q0;

Note: For the above Q0 computation to work properly it is essential that P1 and P2 be column vectors, not row vectors. If they are row vectors, the Q0 code should be changed to its transpose:

Q0 = sqrt(((r1+r2)^2-d2)*(d2-(r1-r2)^2))/2/d2*P21*[0,1;-1,0];

Best Answer

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MATLAB: Find points of intersection circle with line

MATLAB: Solving two equations with two unknows

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