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I am trying to draw a cone, connected to the sphere in Matlab. I have the point [x1,y1,z1] outside of the sphere [x2,y2,z2] with R radius and I want it to be the top of the cone, created out of tangents.
On this pictures you can see what I have in mind:
Below you can see what I have already done. I am using it in order to mark the part of the Earth's surface, visible from the satellite position in orbit. Unfortunately, the cone in this picture is approximate, I need to create accurate one, connected with surface. For now, it is not only inaccurate, but also goes under it.
I am creating the sphere with this simple code (I am skipping the part of putting the map on it, it's just an image):
r = 6371.0087714;
[X,Y,Z] = sphere(50);
X2 = X * r;
Y2 = Y * r;
Z2 = Z * r;
surf(X2,Y2,Z2)
props.FaceColor= 'texture';
props.EdgeColor = 'none';
props.FaceLighting = 'phong';
figure();
globe = surface(X2,Y2,Z2,props);
Let's assume that I have the single point in 3D:
plot3(0,0,7000,'o');
How can I create such a cone?
There are two different questions here:
How to calculate cone dimensions?
How to plot lateral faces of a 3D cone?
Calculating Cone Dimensions
Assuming that center of sphere is located on [0 0 0]:
d = sqrt(Ax^2+Ay^2+Az^2);
l = sqrt(d^2-rs^2);
t = asin(rs/d);
h = l * cos(t);
rc = l * sin(t);
Plotting the Cone
The following function returns coordinates of lateral faces of cone with give apex point, axis direction, base radius and height, and the number of lateral faces.
function [X, Y, Z] = cone3(A, V, r, h, n)
% A: apex, [x y z]
% V: axis direction, [x y z]
% r: radius, scalar
% h: height, scalar
% n: number of lateral surfaces, integer
% X, Y, Z: coordinates of lateral points of the cone, all (n+1) by 2. You draw the sphere with surf(X,Y,Z) or mesh(X,Y,Z)
v1 = V./norm(V);
B = h*v1+A;
v23 = null(v1);
th = linspace(0, 2*pi, n+1);
P = r*(v23(:,1)*cos(th)+v23(:,2)*sin(th));
P = bsxfun(#plus, P', B);
zr = zeros(n+1, 1);
X = [A(1)+zr P(:, 1)];
Y = [A(2)+zr P(:, 2)];
Z = [A(3)+zr P(:, 3)];
end
The Results
rs = 6371.0087714; % globe radius
A = rs * 2 * [1 1 1]; % sattelite location
V = -A; % vector from sat to the globe center
% calculating cone dimensions
d = norm(A); % distance from cone apex to sphere center
l = (d^2-rs^2)^.5; % length of generating line of cone
sint = rs/d; % sine of half of apperture
cost = l/d; % cosine of half of apperture
h = l * cost; % cone height
rc = l * sint; % cone radius
% globe surface points
[XS,YS,ZS] = sphere(32);
% cone surface points
[XC, YC, ZC] = cone3(A, V, rc, h, 32);
% plotting results
hold on
surf(XS*rs,YS*rs,ZS*rs, 'facecolor', 'b', 'facealpha', 0.5, 'edgealpha', 0.5)
surf(XC, YC, ZC, 'facecolor', 'r', 'facealpha', 0.5, 'edgealpha', 0.5);
axis equal
grid on
Animating the satellite
The simplest way to animate objects is to clear the whole figure by clf and plot objects again in new positions. But a way more efficient method is to plot all objects once and in each frame, only update positioning data of moving objects:
clc; close all; clc
rs = 6371.0087714; % globe radius
r = rs * 1.2;
n = 121;
t = linspace(0, 2*pi, n)';
% point on orbit
Ai = [r.*cos(t) r.*sin(t) zeros(n, 1)];
[XS,YS,ZS] = sphere(32);
surf(XS*rs,YS*rs,ZS*rs, 'facecolor', 'b', 'facealpha', 0.5, 'edgealpha', 0.5)
hold on
[XC, YC, ZC] = cone3(Ai(1, :), Ai(1, :), 1, 1, 32);
% plot a cone and store handel of surf object
hS = surf(XC, YC, ZC, 'facecolor', 'r', 'facealpha', 0.5, 'edgealpha', 0.5);
for i=1:n
% calculating new point coordinates of cone
A = Ai(i, :);
V = -A;
d = norm(A);
l = (d^2-rs^2)^.5;
sint = rs/d;
cost = l/d;
h = l * cost;
rc = l * sint;
[XC, YC, ZC] = cone3(A, V, rc, h, 32);
% updating surf object
set(hS, 'xdata', XC, 'ydata', YC, 'zdata', ZC);
pause(0.01); % wait 0.01 seconds
drawnow(); % repaint figure
end
Another sample with 3 orbiting satellites:
I'm looking for a simple way for creating a random unit vector constrained by a conical region. The origin is always the [0,0,0].
My solution up to now:
function v = GetRandomVectorInsideCone(coneDir,coneAngleDegree)
coneDir = normc(coneDir);
ang = coneAngleDegree + 1;
while ang > coneAngleDegree
v = [randn(1); randn(1); randn(1)];
v = v + coneDir;
v = normc(v);
ang = atan2(norm(cross(v,coneDir)), dot(v,coneDir))*180/pi;
end
My code loops until the random generated unit vector is inside the defined cone. Is there a better way to do that?
Resultant image from test code bellow
Resultant frequency distribution using Ahmed Fasih code (in comments).
I wonder how to get a rectangular or normal distribution.
c = [1;1;1]; angs = arrayfun(#(i) subspace(c, GetRandomVectorInsideCone(c, 30)), 1:1e5) * 180/pi; figure(); hist(angs, 50);
Testing code:
clearvars; clc; close all;
coneDir = [randn(1); randn(1); randn(1)];
coneDir = [0 0 1]';
coneDir = normc(coneDir);
coneAngle = 45;
N = 1000;
vAngles = zeros(N,1);
vs = zeros(3,N);
for i=1:N
vs(:,i) = GetRandomVectorInsideCone(coneDir,coneAngle);
vAngles(i) = subspace(vs(:,i),coneDir)*180/pi;
end
maxAngle = max(vAngles);
minAngle = min(vAngles);
meanAngle = mean(vAngles);
AngleStd = std(vAngles);
fprintf('v angle\n');
fprintf('Direction: [%.3f %.3f %.3f]^T. Angle: %.2fº\n',coneDir,coneAngle);
fprintf('Min: %.2fº. Max: %.2fº\n',minAngle,maxAngle);
fprintf('Mean: %.2fº\n',meanAngle);
fprintf('Standard Dev: %.2fº\n',AngleStd);
%% Plot
figure;
grid on;
rotate3d on;
axis equal;
axis vis3d;
axis tight;
hold on;
xlabel('X'); ylabel('Y'); zlabel('Z');
% Plot all vectors
p1 = [0 0 0]';
for i=1:N
p2 = vs(:,i);
plot3ex(p1,p2);
end
% Trying to plot the limiting cone, but no success here :(
% k = [0 1];
% [X,Y,Z] = cylinder([0 1 0]');
% testsubject = surf(X,Y,Z);
% set(testsubject,'FaceAlpha',0.5)
% N = 50;
% r = linspace(0, 1, N);
% [X,Y,Z] = cylinder(r, N);
%
% h = surf(X, Y, Z);
%
% rotate(h, [1 1 0], 90);
plot3ex.m:
function p = plot3ex(varargin)
% Plots a line from each p1 to each p2.
% Inputs:
% p1 3xN
% p2 3xN
% args plot3 configuration string
% NOTE: p1 and p2 number of points can range from 1 to N
% but if the number of points are different, one must be 1!
% PVB 2016
p1 = varargin{1};
p2 = varargin{2};
extraArgs = varargin(3:end);
N1 = size(p1,2);
N2 = size(p2,2);
N = N1;
if N1 == 1 && N2 > 1
N = N2;
elseif N1 > 1 && N2 == 1
N = N1
elseif N1 ~= N2
error('if size(p1,2) ~= size(p1,2): size(p1,2) and/or size(p1,2) must be 1 !');
end
for i=1:N
i1 = i;
i2 = i;
if i > N1
i1 = N1;
end
if i > N2
i2 = N2;
end
x = [p1(1,i1) p2(1,i2)];
y = [p1(2,i1) p2(2,i2)];
z = [p1(3,i1) p2(3,i2)];
p = plot3(x,y,z,extraArgs{:});
end
Here’s the solution. It’s based on the wonderful answer at https://math.stackexchange.com/a/205589/81266. I found this answer by googling “random points on spherical cap”, after I learned on Mathworld that a spherical cap is this cut of a 3-sphere with a plane.
Here’s the function:
function r = randSphericalCap(coneAngleDegree, coneDir, N, RNG)
if ~exist('coneDir', 'var') || isempty(coneDir)
coneDir = [0;0;1];
end
if ~exist('N', 'var') || isempty(N)
N = 1;
end
if ~exist('RNG', 'var') || isempty(RNG)
RNG = RandStream.getGlobalStream();
end
coneAngle = coneAngleDegree * pi/180;
% Generate points on the spherical cap around the north pole [1].
% [1] See https://math.stackexchange.com/a/205589/81266
z = RNG.rand(1, N) * (1 - cos(coneAngle)) + cos(coneAngle);
phi = RNG.rand(1, N) * 2 * pi;
x = sqrt(1-z.^2).*cos(phi);
y = sqrt(1-z.^2).*sin(phi);
% If the spherical cap is centered around the north pole, we're done.
if all(coneDir(:) == [0;0;1])
r = [x; y; z];
return;
end
% Find the rotation axis `u` and rotation angle `rot` [1]
u = normc(cross([0;0;1], normc(coneDir)));
rot = acos(dot(normc(coneDir), [0;0;1]));
% Convert rotation axis and angle to 3x3 rotation matrix [2]
% [2] See https://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle
crossMatrix = #(x,y,z) [0 -z y; z 0 -x; -y x 0];
R = cos(rot) * eye(3) + sin(rot) * crossMatrix(u(1), u(2), u(3)) + (1-cos(rot))*(u * u');
% Rotate [x; y; z] from north pole to `coneDir`.
r = R * [x; y; z];
end
function y = normc(x)
y = bsxfun(#rdivide, x, sqrt(sum(x.^2)));
end
This code just implements joriki’s answer on math.stackexchange, filling in all the details that joriki omitted.
Here’s a script that shows how to use it.
clearvars
coneDir = [1;1;1];
coneAngleDegree = 30;
N = 1e4;
sol = randSphericalCap(coneAngleDegree, coneDir, N);
figure;plot3(sol(1,:), sol(2,:), sol(3,:), 'b.', 0,0,0,'rx');
grid
xlabel('x'); ylabel('y'); zlabel('z')
legend('random points','origin','location','best')
title('Final random points on spherical cap')
Here is a 3D plot of 10'000 points from the 30° spherical cap centered around the [1; 1; 1] vector:
Here’s 120° spherical cap:
Now, if you look at the histogram of the angles between these random points at the coneDir = [1;1;1], you will see that the distribution is skewed. Here’s the distribution:
Code to generate this:
normc = #(x) bsxfun(#rdivide, x, sqrt(sum(x.^2)));
mysubspace = #(a,b) real(acos(sum(bsxfun(#times, normc(a), normc(b)))));
angs = arrayfun(#(i) mysubspace(coneDir, sol(:,i)), 1:N) * 180/pi;
nBins = 16;
[n, edges] = histcounts(angs, nBins);
centers = diff(edges(1:2))*[0:(length(n)-1)] + mean(edges(1:2));
figure('color','white');
bar(centers, n);
xlabel('Angle (degrees)')
ylabel('Frequency')
title(sprintf('Histogram of angles between coneDir and random points: %d deg', coneAngleDegree))
Well, this makes sense! If you generate points from the 120° spherical cap around coneDir, of course the 1° cap is going to have very few of those samples, whereas the strip between the 10° and 11° caps will have far more points. So what we want to do is normalize the number of points at a given angle by the surface area of the spherical cap at that angle.
Here’s a function that gives us the surface area of the spherical cap with radius R and angle in radians theta (equation 16 on Mathworld’s spherical cap article):
rThetaToH = #(R, theta) R * (1 - cos(theta));
rThetaToS = #(R, theta) 2 * pi * R * rThetaToH(R, theta);
Then, we can normalize the histogram count for each bin (n above) by the difference in surface area of the spherical caps at the bin’s edges:
figure('color','white');
bar(centers, n ./ diff(rThetaToS(1, edges * pi/180)))
The figure:
This tells us “the number of random vectors divided by the surface area of the spherical segment between histogram bin edges”. This is uniform!
(N.B. If you do this normalized histogram for the vectors generated by your original code, using rejection sampling, the same holds: the normalized histogram is uniform. It’s just that rejection sampling is expensive compared to this.)
(N.B. 2: note that the naive way of picking random points on a sphere—by first generating azimuth/elevation angles and then converting these spherical coordinates to Cartesian coordinates—is no good because it bunches points near the poles: Mathworld, example, example 2. One way to pick points on the entire sphere is sampling from the 3D normal distribution: that way you won’t get bunching near poles. So I believe that your original technique is perfectly appropriate, giving you nice, evenly-distributed points on the sphere without any bunching. This algorithm described above also does the “right thing” and should avoid bunching. Carefully evaluate any proposed algorithms to ensure that the bunching-near-poles problem is avoided.)
it is better to use spherical coordinates and convert it to cartesian coordinates:
coneDirtheta = rand(1) * 2 * pi;
coneDirphi = rand(1) * pi;
coneAngle = 45;
N = 1000;
%perfom transformation preventing concentration of points around the pole
rpolar = acos(cos(coneAngle/2*pi/180) + (1-cos(coneAngle/2*pi/180)) * rand(N, 1));
thetapolar = rand(N,1) * 2 * pi;
x0 = rpolar .* cos(thetapolar);
y0 = rpolar .* sin(thetapolar);
theta = coneDirtheta + x0;
phi = coneDirphi + y0;
r = rand(N, 1);
x = r .* cos(theta) .* sin(phi);
y = r .* sin(theta) .* sin(phi);
z = r .* cos(phi);
scatter3(x,y,z)
if all points should be of length 1 set r = ones(N,1);
Edit:
since intersection of cone with sphere forms a circle first we create random points inside a circle with raduis of (45 / 2) in polar coordinates and as #Ahmed Fasih commented to prevent concentration of points near the pole we should first transform this random points, then convert polar to cartesian 2D coordinates to form x0 and y0
we can use x0 and y0 as phi & theta angle of spherical coordinates and add coneDirtheta & coneDirphi as offsets to these coordinates.
then convert spherical to cartesian 3D coordinates
The below code plots circles in Matlab. How can I specify the MarkerEdgeColor and MarkerFaceColor in it.
function plot_model
exit_agents=csvread('C:\Users\sony\Desktop\latest_mixed_crowds\December\exit_agents.csv');
%scatter(exit_agents(:,2),exit_agents(:,3),pi*.25^2,'filled');
for ii =1:size(exit_agents,1),
circle(exit_agents(ii,2),exit_agents(ii,3),0.25);
end
end
function h = circle(x,y,r)
hold on
th = 0:pi/50:2*pi;
xunit = r * cos(th) + x;
yunit = r * sin(th) + y;
h = plot(xunit, yunit);
hold off
end
Using plot and scatter scales them weirdly when zooming. This is not what I wish for.
There are various options to plot circles. The easiest is, to actually plot a filled rectangle with full curvature:
%// radius
r = 2;
%// center
c = [3 3];
pos = [c-r 2*r 2*r];
r = rectangle('Position',pos,'Curvature',[1 1], 'FaceColor', 'red', 'Edgecolor','none')
axis equal
With the update of the graphics engine with R2014b this is really smooth:
If you have an older version of Matlab than R2014b, you will need to stick with your trigonometric approach, but use fill to get it filled:
%// radius
r = 2;
%// center
c = [3 3];
%// number of points
n = 1000;
%// running variable
t = linspace(0,2*pi,n);
x = c(1) + r*sin(t);
y = c(2) + r*cos(t);
%// draw filled polygon
fill(x,y,[1,1,1],'FaceColor','red','EdgeColor','none')
axis equal
The "resolution" can be freely scaled by the number of points n.
Your function could then look like
function h = circle(x,y,r,MarkerFaceColor,MarkerEdgeColor)
hold on
c = [x y];
pos = [c-r 2*r 2*r];
r = rectangle('Position',pos,'Curvature',[1 1], ...
'FaceColor', MarkerFaceColor, 'Edgecolor',MarkerEdgeColor)
hold off
end
In matlab, if I have a code which draws a circle and generates 100 random points inside it. I want to use k means to cluster these 100 points into 30 clusters with a circle around each cluster to differentiate between the clusters and i want to mark the center if each cluster.this is the code of the circle and the 100 random points inside it . Any help please ?
%// Set parameters
R =250; %// radius
C = [0 0]; %// center [x y]
N = 100; %// number of points inside circle
%// generate circle boundary
t = linspace(0, 2*pi,100);
x = R*cos(t) + C(1);
y = R*sin(t) + C(2);
%// generate random points inside it
th = 2*pi*rand(N,1);
r = R*randnlimit(0, 1, 0.5, 1, [N 1]);
xR = r.*cos(th) + C(1);
yR = r.*sin(th) + C(2);
%// Plot everything
figure(1), clf, hold on
% subplot(1,N0,1);
plot(x,y,'b');
hold on
text(0,0,'C')
plot(xR,yR,'p')
axis equal
zR=cell(N,1);
for i=1:N
zR{i,1}= [xR(i) yR(i)];
end
m=cell2mat(zR);
function clusterCircle()
%// Set parameters
R = 250; %// radius
C = [0 0]; %// center [x y]
N = 100; %// number of points inside circle
k = 30; %// number of clusters
%// generate circle boundary
t = linspace(0, 2*pi, 100);
x = R*cos(t) + C(1);
y = R*sin(t) + C(2);
%// generate random points inside it
th = 2*pi*rand(N,1);
r = R*randnlimit(0, 1, 0.5, 1, [N 1]);
xR = r.*cos(th) + C(1);
yR = r.*sin(th) + C(2);
%// some simple k-means:
% initial centroids:
% -> use different method, if k > N
% -> can be done more reasonable (e.g. run k-Means for different
% seeds, select seeds equidistant, etc.)
xC = xR(1:k)';
yC = yR(1:k)';
% run:
clusters = zeros(N,1);
clusters_old = ones(N,1);
while sum((clusters - clusters_old).^2) > 0
clusters_old = clusters;
[~,clusters] = min((bsxfun(#minus,xR,xC)).^2 + ...
(bsxfun(#minus,yR,yC)).^2 , [] , 2);
for kIdx = 1:k
xC(kIdx) = mean(xR(clusters==kIdx));
yC(kIdx) = mean(yR(clusters==kIdx));
end
end
%// Plot everything
figure(1);
clf;
hold on;
% -> plot circle and center
text(C(1),C(2),'C');
plot(x,y,'k');
% -> plot clusters
co = hsv(k);
for kIdx = 1:k
% plot cluster points
plot(xR(clusters==kIdx),yR(clusters==kIdx),'p','Color',co(kIdx,:));
% plot cluster circle
maxR = sqrt(max((xR(clusters==kIdx)-xC(kIdx)).^2 + ...
(yR(clusters==kIdx)-yC(kIdx)).^2));
x = maxR*cos(t) + xC(kIdx);
y = maxR*sin(t) + yC(kIdx);
plot(x,y,'Color',co(kIdx,:));
% plot cluster center
text(xC(kIdx),yC(kIdx),num2str(kIdx));
end
axis equal
end
%// creates random numbers, not optimized!
function rn = randnlimit(a,b,mu,sigma,sz)
rn = zeros(sz);
for idx = 1:prod(sz)
searchOn = true;
while searchOn
rn_loc = randn(1) * sigma + mu;
if rn_loc >= a && rn_loc <= b
searchOn = false;
end
end
rn(idx) = rn_loc;
end
end
I have a set of 3 datasets which I want to plot in MATLAB, but the 'x' axis, I want to give in the form of a circle instead of of straight bottom line. Any idea on how to do it?
An example plot:
The normal command for plotting in MATLAB is plot(x, data1, x data2, x, data3), in that the x axis is taken as the straight line. I want the x axis taken as a circle. Does anyone know the command for it please.
#Alok asks if you want a polar plot. I tell you that you do want a polar plot ! See the Matlab documentation for the function polar() and its relations, such as cart2pol. Depending on your exact requirements (I haven't followed your link) you may find it relatively easy or quite difficult to produce exactly the plot you want.
The following is a complete example to show how to map the data from a line axis to a circle.
I show two ways of achieving the goal:
one where the three data series are overlapping (i.e all mapped to the same range)
the other option is to draw them superimposed (on different adjacent ranges)
The basic idea: if you have a series D, then map the points to a circle where the radius is equal to the values of the data using:
theta = linspace(0, 2*pi, N); %# divide circle by N points (length of data)
r = data; %# radius
x = r.*cos(theta); %# x-coordinate
y = r.*sin(theta); %# y-coordinate
plot(x, y, '-');
Option 1
%# some random data
K = 3;
N = 30;
data = zeros(K,N);
data(1,:) = 0.2*randn(1,N) + 1;
data(2,:) = 0.2*randn(1,N) + 2;
data(3,:) = 0.2*randn(1,N) + 3;
center = [0 0]; %# center (shift)
radius = [data data(:,1)]; %# added first to last to create closed loop
radius = normalize(radius',1)'+1; %# normalize data to [0,1] range
figure, hold on
%# draw outer circle
theta = linspace(5*pi/2, pi/2, 500)'; %# 'angles
r = max(radius(:)); %# radius
x = r*cos(theta)+center(1);
y = r*sin(theta)+center(2);
plot(x, y, 'k:');
%# draw mid-circles
theta = linspace(5*pi/2, pi/2, 500)'; %# 'angles
num = 5; %# number of circles
rr = linspace(0,2,num+2); %# radiuses
for k=1:num
r = rr(k+1);
x = r*cos(theta)+center(1);
y = r*sin(theta)+center(2);
plot(x, y, 'k:');
end
%# draw labels
theta = linspace(5*pi/2, pi/2, N+1)'; %# 'angles
theta(end) = [];
r = max(radius(:));
r = r + r*0.2; %# shift to outside a bit
x = r*cos(theta)+center(1);
y = r*sin(theta)+center(2);
str = strcat(num2str((1:N)','%d'),{}); %# 'labels
text(x, y, str, 'FontWeight','Bold');
%# draw the actual series
theta = linspace(5*pi/2, pi/2, N+1);
x = bsxfun(#times, radius, cos(theta)+center(1))';
y = bsxfun(#times, radius, sin(theta)+center(2))';
h = zeros(1,K);
clr = hsv(K);
for k=1:K
h(k) = plot(x(:,k), y(:,k), '.-', 'Color', clr(k,:), 'LineWidth', 2);
end
%# legend and fix axes
legend(h, {'M1' 'M2' 'M3'}, 'location', 'SouthOutside', 'orientation','horizontal')
hold off
axis equal, axis([-1 1 -1 1] * r), axis off
Option 2
%# some random data
K = 3;
N = 30;
data = zeros(K,N);
data(1,:) = 0.2*randn(1,N) + 1;
data(2,:) = 0.2*randn(1,N) + 2;
data(3,:) = 0.2*randn(1,N) + 3;
center = [0 0]; %# center (shift)
radius = [data data(:,1)]; %# added first to last to create closed loop
radius = normalize(radius',1)'; %# normalize data to [0,1] range
radius = bsxfun( #plus, radius, (1:2:2*K)' ); %# 'make serieson seperate ranges by addition
figure, hold on
%# draw outer circle
theta = linspace(5*pi/2, pi/2, 500)'; %# 'angles
r = max(radius(:))+1; %# radius
x = r*cos(theta)+center(1);
y = r*sin(theta)+center(2);
plot(x, y, 'k:');
%# draw mid-circles
theta = linspace(5*pi/2, pi/2, 500)'; %# 'angles
r = 1.5; %# radius
for k=1:K
x = r*cos(theta)+center(1);
y = r*sin(theta)+center(2);
plot(x, y, 'k:');
r=r+2; %# increment radius for next circle
end
%# draw labels
theta = linspace(5*pi/2, pi/2, N+1)'; %# 'angles
theta(end) = [];
r = max(radius(:))+1;
r = r + r*0.2; %# shift to outside a bit
x = r*cos(theta)+center(1);
y = r*sin(theta)+center(2);
str = strcat(num2str((1:N)','%d'),{}); %# 'labels
text(x, y, str, 'FontWeight','Bold');
%# draw the actual series
theta = linspace(5*pi/2, pi/2, N+1);
x = bsxfun(#times, radius, cos(theta)+center(1))';
y = bsxfun(#times, radius, sin(theta)+center(2))';
h = zeros(1,K);
clr = hsv(K);
for k=1:K
h(k) = plot(x(:,k), y(:,k), '.-', 'Color', clr(k,:), 'LineWidth', 2);
end
%# legend and fix axes
legend(h, {'M1' 'M2' 'M3'}, 'location', 'SouthOutside', 'orientation','horizontal')
hold off
axis equal, axis([-1 1 -1 1] * r), axis off
I should mention that normalize() is a custom function, it simply performs minmax normalization ((x-min)/(max-min)) defined as:
function newData = normalize(data, type)
[numInst numDim] = size(data);
e = ones(numInst, 1);
minimum = min(data);
maximum = max(data);
range = (maximum - minimum);
if type == 1
%# minmax normalization: (x-min)/(max-min) => x in [0,1]
newData = (data - e*minimum) ./ ( e*(range+(range==0)) );
end
%# (...)
end
You can find here all available MATLAB 2-D and 3-D plot functions.
Sorry, if it may be not a proper answer to your question (you already have plenty). I recently found very powerful tool to plot on circle - CIRCOS: http://mkweb.bcgsc.ca/circos/
Have a look, figures are really amazing. It's not Matlab-based, but Perl, and it's free. May be you will find it useful.