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zhhhero

铜虫 (初入文坛)

[求助] matllab中空间圆如何拟合?

求大侠,帮忙,如何在matlab中最小二乘法拟合空间圆,求出圆心,半径。最好有matlab程序,谢谢
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zhhhero

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引用回帖:
4楼: Originally posted by dbb627 at 2013-05-02 10:23:35
空间圆的点必然共面,将空间坐标转到平面坐标下,拟合后再转回去。

有matlab的代码吗
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5楼2013-05-02 10:25:58
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change0618

铁杆木虫 (著名写手)

方丈大师

【答案】应助回帖

★ ★
感谢参与,应助指数 +1
xiegangmai: 金币+2, 谢谢参与! 2013-05-02 23:10:37
可以去matlab App中心下载
CODE:
function   [xc,yc,R,a] = circfit(x,y)
%
%   [xc yx R] = circfit(x,y)
%
%   fits a circle  in x,y plane in a more accurate
%   (less prone to ill condition )
%  procedure than circfit2 but using more memory
%  x,y are column vector where (x(i),y(i)) is a measured point
%
%  result is center point (yc,xc) and radius R
%  an optional output is the vector of coeficient a
% describing the circle's equation
%
%   x^2+y^2+a(1)*x+a(2)*y+a(3)=0
%
%  By:  Izhak bucher 25/oct /1991,
    x=x(:); y=y(:);
   a=[x y ones(size(x))]\[-(x.^2+y.^2)];
   xc = -.5*a(1);
   yc = -.5*a(2);
   R  =  sqrt((a(1)^2+a(2)^2)/4-a(3));

CODE:
%try_circ_fit
%
% IB
%
% revival of a 13 years old code


  % Create data for a circle + noise
  
  th = linspace(0,2*pi,20)';
  R=1.1111111;
  sigma = R/10;
  x = R*cos(th)+randn(size(th))*sigma;
  y = R*sin(th)+randn(size(th))*sigma;
  
   plot(x,y,'o'), title(' measured points')
   pause(1)
   
   % reconstruct circle from data
   [xc,yc,Re,a] = circfit(x,y);
      xe = Re*cos(th)+xc; ye = Re*sin(th)+yc;
   
     plot(x,y,'o',[xe;xe(1)],[ye;ye(1)],'-.',R*cos(th),R*sin(th)),
     title(' measured fitted and true circles')
      legend('measured','fitted','true')
      text(xc-R*0.9,yc,sprintf('center (%g , %g );  R=%g',xc,yc,Re))
     xlabel x, ylabel y
     axis equal
  

CODE:
function [center,rad,v1n,v2nb] = circlefit3d(p1,p2,p3)
% circlefit3d: Compute center and radii of circles in 3d which are defined by three points on the circumference
% usage: [center,rad,v1,v2] = circlefit3d(p1,p2,p3)
%
% arguments: (input)
%  p1, p2, p3 - vectors of points (rowwise, size(p1) = [n 3])
%               describing the three corresponding points on the same circle.
%               p1, p2 and p3 must have the same length n.
%
% arguments: (output)
%  center - (nx3) matrix of center points for each triple of points in p1,  p2, p3
%
%  rad    - (nx1) vector of circle radii.
%           if there have been errors, radii is a negative scalar ( = error code)
%
%  v1, v2 - (nx3) perpendicular vectors inside circle plane
%
% Example usage:
%
%  (1)
%      p1 = rand(10,3);
%      p2 = rand(10,3);
%      p3 = rand(10,3);
%      [center, rad] = circlefit3d(p1,p2,p3);
%      % verification, result should be all (nearly) zero
%      result(:,1)=sqrt(sum((p1-center).^2,2))-rad;
%      result(:,2)=sqrt(sum((p2-center).^2,2))-rad;
%      result(:,3)=sqrt(sum((p3-center).^2,2))-rad;
%      if sum(sum(abs(result))) < 1e-12,
%       disp('All circles have been found correctly.');
%      else,
%       disp('There had been errors.');
%      end
%
%
% (2)
%       p1=rand(4,3);p2=rand(4,3);p3=rand(4,3);
%       [center,rad,v1,v2] = circlefit3d(p1,p2,p3);
%       plot3(p1(:,1),p1(:,2),p1(:,3),'bo');hold on;plot3(p2(:,1),p2(:,2),p2(:,3),'bo');plot3(p3(:,1),p3(:,2),p3(:,3),'bo');
%       for i=1:361,
%           a = i/180*pi;
%           x = center(:,1)+sin(a)*rad.*v1(:,1)+cos(a)*rad.*v2(:,1);
%           y = center(:,2)+sin(a)*rad.*v1(:,2)+cos(a)*rad.*v2(:,2);
%           z = center(:,3)+sin(a)*rad.*v1(:,3)+cos(a)*rad.*v2(:,3);
%           plot3(x,y,z,'r.');
%       end
%       axis equal;grid on;rotate3d on;
%
%
% Author: Johannes Korsawe
% E-mail: johannes.korsawe@volkswagen.de
% Release: 1.0
% Release date: 26/01/2012

% Default values
center = [];rad = 0;v1n=[];v2nb=[];

% check inputs
% check number of inputs
if nargin~=3,
    fprintf('??? Error using ==> cirlefit3d\nThree input matrices are needed.\n');rad = -1;return;
end
% check size of inputs
if size(p1,2)~=3 || size(p2,2)~=3 || size(p3,2)~=3,
    fprintf('??? Error using ==> cirlefit3d\nAll input matrices must have three columns.\n');rad = -2;return;
end
n = size(p1,1);
if size(p2,1)~=n || size(p3,1)~=n,
    fprintf('??? Error using ==> cirlefit3d\nAll input matrices must have the same number or rows.\n');rad = -3;return;
end
% more checks are to follow inside calculation

% Start calculation
% v1, v2 describe the vectors from p1 to p2 and p3, resp.
v1 = p2 - p1;v2 = p3 - p1;
% l1, l2 describe the lengths of those vectors
l1 = sqrt((v1(:,1).*v1(:,1)+v1(:,2).*v1(:,2)+v1(:,3).*v1(:,3)));
l2 = sqrt((v2(:,1).*v2(:,1)+v2(:,2).*v2(:,2)+v2(:,3).*v2(:,3)));
if find(l1==0) | find(l2==0), %#ok<OR2>
    fprintf('??? Error using ==> cirlefit3d\nCorresponding input points must not be identical.\n');rad = -4;return;
end
% v1n, v2n describe the normalized vectors v1 and v2
v1n = v1;for i=1:3, v1n(:,i) = v1n(:,i)./l1;end
v2n = v2;for i=1:3, v2n(:,i) = v2n(:,i)./l2;end
% nv describes the normal vector on the plane of the circle
nv = [v1n(:,2).*v2n(:,3) - v1n(:,3).*v2n(:,2) , v1n(:,3).*v2n(:,1) - v1n(:,1).*v2n(:,3) , v1n(:,1).*v2n(:,2) - v1n(:,2).*v2n(:,1)];
if find(sum(abs(nv),2)<1e-5),
    fprintf('??? Warning using ==> cirlefit3d\nSome corresponding input points are nearly collinear.\n');
end
% v2nb: orthogonalization of v2n against v1n
dotp = v2n(:,1).*v1n(:,1) + v2n(:,2).*v1n(:,2) + v2n(:,3).*v1n(:,3);
v2nb = v2n;for i=1:3,v2nb(:,i) = v2nb(:,i) - dotp.*v1n(:,i);end
% normalize v2nb
l2nb = sqrt((v2nb(:,1).*v2nb(:,1)+v2nb(:,2).*v2nb(:,2)+v2nb(:,3).*v2nb(:,3)));
for i=1:3, v2nb(:,i) = v2nb(:,i)./l2nb;end

% remark: the circle plane will now be discretized as follows
%
% origin: p1                    normal vector on plane: nv
% first coordinate vector: v1n  second coordinate vector: v2nb

% calculate 2d coordinates of points in each plane
% p1_2d = zeros(n,2); % set per construction
% p2_2d = zeros(n,2);p2_2d(:,1) = l1; % set per construction
p3_2d = zeros(n,2); % has to be calculated
for i = 1:3,
    p3_2d(:,1) = p3_2d(:,1) + v2(:,i).*v1n(:,i);
    p3_2d(:,2) = p3_2d(:,2) + v2(:,i).*v2nb(:,i);
end

% calculate the fitting circle
% due to the special construction of the 2d system this boils down to solving
% q1 = [0,0], q2 = [a,0], q3 = [b,c] (points on 2d circle)
% crossing perpendicular bisectors, s and t running indices:
% solve [a/2,s] = [b/2 + c*t, c/2 - b*t]
% solution t = (a-b)/(2*c)

a = l1;b = p3_2d(:,1);c = p3_2d(:,2);
t = 0.5*(a-b)./c;
scale1 = b/2 + c.*t;scale2 = c/2 - b.*t;

% centers
center = zeros(n,3);
for i=1:3,
    center(:,i) = p1(:,i) + scale1.*v1n(:,i) + scale2.*v2nb(:,i);
end

% radii
rad = sqrt((center(:,1)-p1(:,1)).^2+(center(:,2)-p1(:,2)).^2+(center(:,3)-p1(:,3)).^2);

2楼2013-05-02 09:22:48
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zhhhero

铜虫 (初入文坛)

对了,我是有10组空间点三维坐标,现在想拟合空间圆。
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3楼2013-05-02 09:58:24
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dbb627

荣誉版主 (著名写手)

【答案】应助回帖


感谢参与,应助指数 +1
xiegangmai: 金币+1, 谢谢参与! 2013-05-02 23:10:45
空间圆的点必然共面,将空间坐标转到平面坐标下,拟合后再转回去。
The more you learn, the more you know, the more you know, and the more you forget. The more you forget, the less you know. So why bother to learn.
4楼2013-05-02 10:23:35
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