I am attempting to apply a transform matrix by setting the 'Matrix' property of a matlab hgtransform object. The transform matrix is below:
866.0254e-003 500.0000e-003 0.0000e+000 500.0000e-003
500.0000e-003 -866.0254e-003 0.0000e+000 500.0000e-003
0.0000e+000 0.0000e+000 1.0000e+000 0.0000e+000
0.0000e+000 0.0000e+000 0.0000e+000 1.0000e+000
This particular matrix is intended to represent a translation
(0.5, 0.5, 0)
and rotation around the Z axis of pi/6.
When I try to do this:
% make a unit box
sx = 1;
sy = 1;
sz = 1;
shapeData.Vertices = [ -sx/2, -sy/2, -sz/2;
sx/2, -sy/2, -sz/2;
sx/2, sy/2, -sz/2;
-sx/2, sy/2, -sz/2;
-sx/2, -sy/2, sz/2;
sx/2, -sy/2, sz/2;
sx/2, sy/2, sz/2;
-sx/2, sy/2, sz/2; ];
shapeData.Faces = [ 1, 4, 3, 2;
1, 5, 6, 2;
2, 6, 7, 3;
7, 8, 4, 3;
8, 5, 1, 4;
8, 7, 6, 5 ];
figure;
axes;
transformObject = hgtransform (gca);
patchObject = patch (gca, ...
'Faces', shapeData.Faces, ...
'Vertices', shapeData.Vertices, ...
'FaceColor', 'red', ...
'FaceAlpha', 1.0, ...
'EdgeColor', 'none', ...
'FaceLighting', 'gouraud', ...
'AmbientStrength', 0.15, ...
'Parent', transformObject);
M = [ ...
866.0254e-003 500.0000e-003 0.0000e+000 500.0000e-003; ...
500.0000e-003 -866.0254e-003 0.0000e+000 500.0000e-003; ...
0.0000e+000 0.0000e+000 1.0000e+000 0.0000e+000; ...
0.0000e+000 0.0000e+000 0.0000e+000 1.0000e+000; ...
];
set ( transformObject, 'Matrix', M );
I get the error:
Error using matlab.graphics.primitive.Transform/set
Invalid value for Matrix property
Why?
EDIT
The code that generated the transform matrix. First you need the following class which constructs orientation (rotation) matrices:
classdef orientmat
properties (GetAccess = public, SetAccess = protected)
orientationMatrix;
end
methods
function this = orientmat (spectype, spec)
% orentmat constructor
%
% Syntax
%
% om = orientmat (spectype, spec)
%
% Input
%
%
switch spectype
case 'orientation'
this.orientationMatrix = spec;
case 'euler'
this.orientationMatrix = SpinCalc('EA123toDCM', rad2deg (spec), eps (), 1);
case 'euler123'
this.orientationMatrix = SpinCalc('EA123toDCM', rad2deg (spec), eps (), 1);
case 'euler321'
this.orientationMatrix = SpinCalc('EA321toDCM', rad2deg (spec), eps (), 1);
case 'vector'
% axis and angle (angle in rad = norm of matrix)
wcrs = [ 0 spec(3) -spec(2)
-spec(3) 0 spec(1)
spec(2) -spec(1) 0] ;
this.orientationMatrix = expm (wcrs);
case '2vectors'
% normalise the fisr vector
spec.vec1 = this.unit (spec.vec1);
spec.vec2 = this.unit (spec.vec2);
spec.vec3 = cross (spec.vec1, spec.vec2);
spec.vec2 = this.unit (cross (this.unit (spec.vec3), spec.vec1));
switch spec.vec1axis
case 1
X = spec.vec1;
if spec.vec2axis == 2
Y = spec.vec2;
Z = spec.vec3;
elseif spec.vec2axis == 3
Y = spec.vec3;
Z = spec.vec2;
end
case 2
Y = spec.vec1;
if spec.vec2axis == 1
X = spec.vec2;
Z = spec.vec3;
elseif spec.vec2axis == 3
X = spec.vec3;
Z = spec.vec2;
end
case 3
Z = spec.vec1;
if spec.vec2axis == 2
X = spec.vec2;
Y = spec.vec3;
elseif spec.vec2axis == 3
X = spec.vec3;
Y = spec.vec2;
end
end
this.orientationMatrix = [ X, Y, Z ];
end
end
end
% operator overloading
methods
function om = plus (om1, om2)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix + om2.orientationMatrix);
end
function om = minus (om1, om2)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix - om2.orientationMatrix);
end
function om = times (om1, om2)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix .* om2.orientationMatrix);
end
function om = mtimes (om1, om2)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix * om2.orientationMatrix);
end
function om = double (om1)
om = om1.orientationMatrix;
end
function om = uminus (om1)
om = mbdyn.pre.orientmat ('orientation', -om1.orientationMatrix);
end
function om = uplus (om1)
om = mbdyn.pre.orientmat ('orientation', +om1.orientationMatrix);
end
function om = transpose (om1)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix.');
end
function om = ctranspose (om1)
om = mbdyn.pre.orientmat ('orientation', om1.orientationMatrix');
end
end
methods (Access = private)
function out = unit (self, vec)
out = vec ./ norm (vec);
end
end
end
Then do:
om = orientmat ('2vectors', struct ('vec1axis', 1, 'vec1', [cos(pi/6);sin(pi/6);0], 'vec2axis', 3, 'vec2', [0;0;1]));
M = [ om.orientationMatrix, [0.5; 0.5; 0]; 0, 0, 0, 1 ];
Now there may be an issue with the rotation not actually being what I intend, but as far as I can see it is still a valid transformation matrix?
The answer was that Matlab only accepts non-negative scaling terms, see Transforms Supported by hgtransform
Related
I have a problem with the following code. In the evidence part, the value is very small so, in the end, the probabilities are not calculated. I need to normalize it, but in which part should it be?
The code in MATLAB is:
clear all; close all; clc;
randn('seed', 1234);
resistivities = [50 200 2000 1500];
thicknesses = [500 100 200];
Par_real = [resistivities, thicknesses];
dataFreq = logspace(log10(0.001), log10(1000), 100);
[Ydata, phase] = modelMT2(Par_real, dataFreq);
sigma = 0.1;
Yexp = Ydata + sigma*randn(size(Ydata));
plot(dataFreq, Yexp, '.'); hold on; plot(dataFreq, Ydata, '-')
nsamples = 20000;
R1 = 5;
R2 = 2050;
P1 = 25;
P2 = 500;
Resis = R1 + (R2-R1)*rand(nsamples, 7);
Profs = P1 + (P2-P1)*rand(nsamples, 6);
for ii=1:nsamples
par3C = [Resis(ii, 1:3), Profs(ii, 1:2)];
par4C = [Resis(ii, 1:4), Profs(ii, 1:3)];
par5C = [Resis(ii, 1:5), Profs(ii, 1:4)];
par7C = [Resis(ii, 1:7), Profs(ii, 1:6)];
Like_M3C(ii) = log_likelihood(#modelMT2, dataFreq, Yexp, sigma, par3C);
Like_M4C(ii) = log_likelihood(#modelMT2, dataFreq, Yexp, sigma, par4C);
Like_M5C(ii) = log_likelihood(#modelMT2, dataFreq, Yexp, sigma, par5C);
Like_M7C(ii) = log_likelihood(#modelMT2, dataFreq, Yexp, sigma, par7C);
end
figure()
subplot(1, 2, 1)
plot(exp(Like_M5C))
subplot(1, 2, 2)
hist(exp(Like_M5C))
Evidencia(1) = mean(exp(Like_M3C));
Evidencia(2) = mean(exp(Like_M4C));
Evidencia(3) = mean(exp(Like_M5C));
Evidencia(4) = mean(exp(Like_M7C));
Denominador = sum(Evidencia);
PPMM = Evidencia/Denominador;
fprintf('La probabilidad de los modelos : \n');
fprintf('--------------------------------\n');
fprintf('Modelo M3C: %.4f \n', PPMM(1));
fprintf('Modelo M4C: %.4f \n', PPMM(2));
fprintf('Modelo M5C: %.4f \n', PPMM(3));
fprintf('Modelo M7C: %.4f \n', PPMM(4));
figure()
model = [1, 2, 3, 4];
bar(model, PPMM), grid on
ylim([0, 1])
xlabel('Modelo')
ylabel('Probabilidad del modelo')
function [LogPDF_post] = log_likelihood(Mod, xx, data, sigma, oldpar)
erro = (Mod(oldpar, xx) - data)';
LogPDF_post = -0.5 * erro' * 1/sigma^2 * erro;
end
I have tried to normalize the likelihood as follows, but it doesn't work. It gives equal probability in all cases.
function [LogPDF_norma] = log_likelihood(Mod, xx, data, sigma, oldpar)
erro = (Mod(oldpar, xx) - data)';
LogPDF_post = -0.5 * erro' * 1/sigma^2 * erro;
LogPDF_norma = (1/max(LogPDF_post))*LogPDF_post;
end
I am trying to reproduce the colorbar below for my chart. Specifically it is the string colorbar axis title (above and below the colorbar) that I am struggling with, the rest seems fine.
Below is my current code:
time_begin = [1981, 1, 1, 0,0,0];
time_end = [2010,12,31,23,0,0];
years = (time_begin(1):time_end(1))';
nyears = length(years);
TXx1 = randi(100, 288, 192, 30);
lat = rand(192, 1);
lon = rand(288, 1);
time = rand(30,1);
M = numel(lon);
N = numel(lat);
slope1 = zeros(M, N);
intercept1 = zeros(M, N);
T = numel(time);
x = ([ones(T, 1) years]);
for i = 1 : M
for j = 1 : N
y1 = squeeze(TXx1(i, j, :));
c = regress(y1, x);
intercept1(i, j) = c(1);
slope1(i, j) = c(2);
end
end
TXx2 = randi(100, 288, 192, 30);
slope2 = zeros(M, N);
intercept2 = zeros(M, N);
T = numel(time);
x = ([ones(T, 1) years]);
for i = 1 : M
for j = 1 : N
y2 = squeeze(TXx2(i, j, :));
c = regress(y2, x);
intercept2(i, j) = c(1);
slope2(i, j) = c(2);
end
end
figure()
set(gcf, 'color', 'w');
temp = [slope1(:) slope2(:)];
temp(temp == 0) = NaN;
[Q,Qc] = hist3(temp,'Nbins',[100 100],'CDataMode','auto');
Qx = cell2mat(Qc(1));
Qy = cell2mat(Qc(2));
Q = Q';
Q = Q./trapz(Qy,trapz(Qx,Q,2));
surf(Qx,Qy,Q,'FaceColor','interp','EdgeColor','none')
grid off
set(gca, 'Fontsize', 12, 'Fontweight', 'Bold'); %added
cmap = jet(500);
cmap(1,:) = [1,1,1];
colormap(cmap);
h=colorbar;
set(h,'position',[.91 .34 .031 .475]) %[xposition yposition width height].
set(get(h,'ylabel'),'string','Point density');
set(h,'XTickLabel',{'Low',' ',' ',' ',' ','High',});
view(0,90)
Here is my current colourbar:
Replace this line:
set(h,'XTickLabel',{'Low',' ',' ',' ',' ','High',});
with:
h.YTick = h.Limits;
h.YTickLabel = {'Low', 'High'};
This makes two tick-lines and two tick labels. By setting YTicks of the colorbar to its limits, the tick-lines get overlapped by colorbar boundary. So these get hidden and now there is no need to remove them.
However there is this TickLength property which can be used otherwise i.e.
h.TickLength = 0;
Result:
I want to move a red star marker along the spiral trajectory with an equal distance of 5 units between the red star points on its circumference like in the below image.
vertspacing = 10;
horzspacing = 10;
thetamax = 10*pi;
% Calculation of (x,y) - underlying archimedean spiral.
b = vertspacing/2/pi;
theta = 0:0.01:thetamax;
x = b*theta.*cos(theta)+50;
y = b*theta.*sin(theta)+50;
% Calculation of equidistant (xi,yi) points on spiral.
smax = 0.5*b*thetamax.*thetamax;
s = 0:horzspacing:smax;
thetai = sqrt(2*s/b);
xi = b*thetai.*cos(thetai);
yi = b*thetai.*sin(thetai);
plot(x,y,'b-');
hold on
I want to get a figure that looks like the following:
This is my code for the circle trajectory:
% Initialization steps.
format long g;
format compact;
fontSize = 20;
r1 = 50;
r2 = 35;
r3= 20;
xc = 50;
yc = 50;
% Since arclength = radius * (angle in radians),
% (angle in radians) = arclength / radius = 5 / radius.
deltaAngle1 = 5 / r1;
deltaAngle2 = 5 / r2;
deltaAngle3 = 5 / r3;
theta1 = 0 : deltaAngle1 : (2 * pi);
theta2 = 0 : deltaAngle2 : (2 * pi);
theta3 = 0 : deltaAngle3 : (2 * pi);
x1 = r1*cos(theta1) + xc;
y1 = r1*sin(theta1) + yc;
x2 = r2*cos(theta2) + xc;
y2 = r2*sin(theta2) + yc;
x3 = r3*cos(theta3) + xc;
y3 = r3*sin(theta3) + yc;
plot(x1,y1,'color',[1 0.5 0])
hold on
plot(x2,y2,'color',[1 0.5 0])
hold on
plot(x3,y3,'color',[1 0.5 0])
hold on
% Connecting Line:
plot([70 100], [50 50],'color',[1 0.5 0])
% Set up figure properties:
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0, 0, 1, 1]);
drawnow;
axis square;
for i = 1 : length(theta1)
plot(x1(i),y1(i),'r*')
pause(0.1)
end
for i = 1 : length(theta2)
plot(x2(i),y2(i),'r*')
pause(0.1)
end
for i = 1 : length(theta3)
plot(x3(i),y3(i),'r*')
pause(0.1)
end
I can't think of a way to compute distance along a spiral, so I'm approximating it with circles, in hopes that it will still be useful.
My solution relies on the InterX function from FEX, to find the intersection of circles with the spiral. I am providing an animation so it is easier to understand.
The code (tested on R2017a):
function [x,y,xi,yi] = q44916610(doPlot)
%% Input handling:
if nargin < 1 || isempty(doPlot)
doPlot = false;
end
%% Initialization:
origin = [50,50];
vertspacing = 10;
thetamax = 5*(2*pi);
%% Calculation of (x,y) - underlying archimedean spiral.
b = vertspacing/(2*pi);
theta = 0:0.01:thetamax;
x = b*theta.*cos(theta) + origin(1);
y = b*theta.*sin(theta) + origin(2);
%% Calculation of equidistant (xi,yi) points on spiral.
DST = 5; cRes = 360;
numPts = ceil(vertspacing*thetamax); % Preallocation
[xi,yi] = deal(NaN(numPts,1));
if doPlot && isHG2() % Plots are only enabled if the MATLAB version is new enough.
figure(); plot(x,y,'b-'); hold on; axis equal; grid on; grid minor;
hAx = gca; hAx.XLim = [-5 105]; hAx.YLim = [-5 105];
hP = plot(xi,yi,'r*');
else
hP = struct('XData',xi,'YData',yi);
end
hP.XData(1) = origin(1); hP.YData(1) = origin(2);
for ind = 2:numPts
P = InterX([x;y], makeCircle([hP.XData(ind-1),hP.YData(ind-1)],DST/2,cRes));
[~,I] = max(abs(P(1,:)-origin(1)+1i*(P(2,:)-origin(2))));
if doPlot, pause(0.1); end
hP.XData(ind) = P(1,I); hP.YData(ind) = P(2,I);
if doPlot, pause(0.1); delete(hAx.Children(1)); end
end
xi = hP.XData(~isnan(hP.XData)); yi = hP.YData(~isnan(hP.YData));
%% Nested function(s):
function [XY] = makeCircle(cnt, R, nPts)
P = (cnt(1)+1i*cnt(2))+R*exp(linspace(0,1,nPts)*pi*2i);
if doPlot, plot(P,'Color',lines(1)); end
XY = [real(P); imag(P)];
end
end
%% Local function(s):
function tf = isHG2()
try
tf = ~verLessThan('MATLAB', '8.4');
catch
tf = false;
end
end
function P = InterX(L1,varargin)
% DOCUMENTATION REMOVED. For a full version go to:
% https://www.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections
narginchk(1,2);
if nargin == 1
L2 = L1; hF = #lt; %...Avoid the inclusion of common points
else
L2 = varargin{1}; hF = #le;
end
%...Preliminary stuff
x1 = L1(1,:)'; x2 = L2(1,:);
y1 = L1(2,:)'; y2 = L2(2,:);
dx1 = diff(x1); dy1 = diff(y1);
dx2 = diff(x2); dy2 = diff(y2);
%...Determine 'signed distances'
S1 = dx1.*y1(1:end-1) - dy1.*x1(1:end-1);
S2 = dx2.*y2(1:end-1) - dy2.*x2(1:end-1);
C1 = feval(hF,D(bsxfun(#times,dx1,y2)-bsxfun(#times,dy1,x2),S1),0);
C2 = feval(hF,D((bsxfun(#times,y1,dx2)-bsxfun(#times,x1,dy2))',S2'),0)';
%...Obtain the segments where an intersection is expected
[i,j] = find(C1 & C2);
if isempty(i), P = zeros(2,0); return; end
%...Transpose and prepare for output
i=i'; dx2=dx2'; dy2=dy2'; S2 = S2';
L = dy2(j).*dx1(i) - dy1(i).*dx2(j);
i = i(L~=0); j=j(L~=0); L=L(L~=0); %...Avoid divisions by 0
%...Solve system of eqs to get the common points
P = unique([dx2(j).*S1(i) - dx1(i).*S2(j), ...
dy2(j).*S1(i) - dy1(i).*S2(j)]./[L L],'rows')';
function u = D(x,y)
u = bsxfun(#minus,x(:,1:end-1),y).*bsxfun(#minus,x(:,2:end),y);
end
end
Result:
Note that in the animation above, the diameter of the circle (and hence the distance between the red points) is 10 and not 5.
Why doesn't color filter below find green peppers?
The code:
function [ outhsv ] = ColorFilter( hsv, h, s )
%COLORFILTER Summary of this function goes here
% Detailed explanation goes here
if nargin < 2
h = [];
end
if nargin < 3
s = [];
end
if size(h,2)==1
h = padarray(h, [0 1], 1/100, 'post');
end
if size(s,2)==1
s = padarray(s, [0 1], 1/100, 'post');
end
if isempty(h)
v_of_h = ones(size(hsv,1), size(hsv,2));
else
v_of_h = WeightFunction( hsv(:,:,1), h(:,1), h(:,2));
end
if isempty(s)
v_of_s = ones(size(hsv,1), size(hsv,2));
else
v_of_s = WeightFunctionOnce( hsv(:,:,2), s(:,1), s(:,2));
end
outhsv = hsv;
outhsv(:,:,3) = hsv(:,:,3) .* v_of_h .* v_of_s;
function y = WeightFunction( x, mu, sigma )
%y = WeightFunctionOnce(x,mu,sigma) + WeightFunctionOnce(x-1,mu,sigma);
y = 1 - (1-WeightFunctionOnce(x,mu,sigma)) .* (1-WeightFunctionOnce(x-1,mu,sigma));
function y = WeightFunctionOnce( x, mu, sigma )
if nargin<2
mu=0;
elseif nargin<3
sigma=1./100.;
end
if any(size(mu) ~= size(sigma))
error('mu and sigma should be of the same size');
end
y = zeros([size(x) numel(mu)]);
for i=1:numel(mu)
y(:,:,i) = exp(-((x - mu(i)) .^ 2 ./ (2 .* sigma(i) .^ 2)));
end
%y = sum(y,3)/size(y,3);
y = 1-prod(1-y,3);
Display code:
hue = 120;
h = [hue/360 0.05];
s = [];
rgb1 = imread('huescale.png');
%rgb1 = imread('peppers.png');
hsv1 = rgb2hsv(rgb1);
hsv2 = ColorFilter(hsv1, h, s);
rgb2 = hsv2rgb(hsv2);
bitmask = hsv1(:,:,1)>(h(1)-h(2)) & hsv1(:,:,1)<(h(1)+h(2));
figure;
subplot(3,1,1); imshow(rgb1);
subplot(3,1,2); imshow(rgb2);
subplot(3,1,3); imshow(bitmask);
result on scale
(works)
result on peppers:
(does not)
Why?
If you looked closer at the H values, those green peppers are kind of yellowish, so you might want to widen the rule a bit.
I would suggest something in between 0.15 and 0.5. You can also combine with saturation channel, say only consider portions of images that are vibrant, i.e., we want to get rid of the onions. Try the following codes to get a preview.
hsv_dat = rgb2hsv(imread('peppers.png'));
imagesc(hsv_dat(:,:,1) > 0.15 & hsv_dat(:,:,1) < 0.5 & hsv_dat(:,:,2) > 0.3)
colormap(gray)
You should get
I am trying to implement least squares circle fitting following this paper (sorry I can't publish it). The paper states, that we could fit a circle, by calculating the geometric error as the euclidean distance (Xi'') between a specific point (Xi) and the corresponding point on the circle (Xi'). We have three parametres: Xc (a vector of coordinates the center of circle), and R (radius).
I came up with the following MATLAB code (note that I am trying to fit circles, not spheres as it is indicated on the images):
function [ circle ] = fit_circle( X )
% Kör paraméterstruktúra inicializálása
% R - kör sugara
% Xc - kör középpontja
circle.R = NaN;
circle.Xc = [ NaN; NaN ];
% Kezdeti illesztés
% A köz középpontja legyen a súlypont
% A sugara legyen az átlagos négyzetes távolság a középponttól
circle.Xc = mean( X );
d = bsxfun(#minus, X, circle.Xc);
circle.R = mean(bsxfun(#hypot, d(:,1), d(:,2)));
circle.Xc = circle.Xc(1:2)+random('norm', 0, 1, size(circle.Xc));
% Optimalizáció
options = optimset('Jacobian', 'on');
out = lsqnonlin(#ort_error, [circle.Xc(1), circle.Xc(2), circle.R], [], [], options, X);
end
%% Cost function
function [ error, J ] = ort_error( P, X )
%% Calculate error
R = P(3);
a = P(1);
b = P(2);
d = bsxfun(#minus, X, P(1:2)); % X - Xc
n = bsxfun(#hypot, d(:,1), d(:,2)); % || X - Xc ||
res = d - R * bsxfun(#times,d,1./n);
error = zeros(2*size(X,1), 1);
error(1:2:2*size(X,1)) = res(:,1);
error(2:2:2*size(X,1)) = res(:,2);
%% Jacobian
xdR = d(:,1)./n;
ydR = d(:,2)./n;
xdx = bsxfun(#plus,-R./n+(d(:,1).^2*R)./n.^3,1);
ydy = bsxfun(#plus,-R./n+(d(:,2).^2*R)./n.^3,1);
xdy = (d(:,1).*d(:,2)*R)./n.^3;
ydx = xdy;
J = zeros(2*size(X,1), 3);
J(1:2:2*size(X,1),:) = [ xdR, xdx, xdy ];
J(2:2:2*size(X,1),:) = [ ydR, ydx, ydy ];
end
The fitting however is not too good: if I start with the good parameter vector the algorithm terminates at the first step (so there is a local minima where it should be), but if I perturb the starting point (with a noiseless circle) the fitting stops with very large errors. I am sure that I've overlooked something in my implementation.
For what it's worth, I implemented these methods in MATLAB a while ago. However, I did it apparently before I knew about lsqnonlin etc, as it uses a hand-implemented regression. This is probably slow, but may help to compare with your code.
function [x, y, r, sq_error] = circFit ( P )
%# CIRCFIT fits a circle to a set of points using least sqaures
%# P is a 2 x n matrix of points to be fitted
per_error = 0.1/100; % i.e. 0.1%
%# initial estimates
X = mean(P, 2)';
r = sqrt(mean(sum((repmat(X', [1, length(P)]) - P).^2)));
v_cen2points = zeros(size(P));
niter = 0;
%# looping until convergence
while niter < 1 || per_diff > per_error
%# vector from centre to each point
v_cen2points(1, :) = P(1, :) - X(1);
v_cen2points(2, :) = P(2, :) - X(2);
%# distacnes from centre to each point
centre2points = sqrt(sum(v_cen2points.^2));
%# distances from edge of circle to each point
d = centre2points - r;
%# computing 3x3 jacobean matrix J, and solvign matrix eqn.
R = (v_cen2points ./ [centre2points; centre2points])';
J = [ -ones(length(R), 1), -R ];
D_rXY = -J\d';
%# updating centre and radius
r_old = r; X_old = X;
r = r + D_rXY(1);
X = X + D_rXY(2:3)';
%# calculating maximum percentage change in values
per_diff = max(abs( [(r_old - r) / r, (X_old - X) ./ X ])) * 100;
%# prevent endless looping
niter = niter + 1;
if niter > 1000
error('Convergence not met in 1000 iterations!')
end
end
x = X(1);
y = X(2);
sq_error = sum(d.^2);
This is then run with:
X = [1 2 5 7 9 3];
Y = [7 6 8 7 5 7];
[x_centre, y_centre, r] = circFit( [X; Y] )
And plotted with:
[X, Y] = cylinder(r, 100);
scatter(X, Y, 60, '+r'); axis equal
hold on
plot(X(1, :) + x_centre, Y(1, :) + y_centre, '-b', 'LineWidth', 1);
Giving: