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I am plotting 4 random separate lines in a plot, and I want them to be connected to form a square. This is what I've got so far, any pointers would be really helpful. Also, this is my first time using MATLAB so please don't come for me.
N = 1;
L = 0.2;
x = L +rand(1,N)*(1-2*L);
y = L +rand(1,N)*(1-2*L);
angs = rand(1,N)*360;
xe = x - L*cosd(angs);
ye = y + L*sind(angs);
xb = -x - L*cosd(angs);
yb = y + L*sind(angs);
ax = axes;
plot(ax,[x;xe],[y;ye],[x;xb],[y;yb])
axis square
I need them to be 4 lines of the same length that join to make a square. The above code gives me 2 connecting lines but I really don't know where to go from there.
My attempt of the above code:
The plot function simply plots the x- and y- coordinates that you give it. Giving it the correct coordinates to plot is your job. So how would you figure out the coordinates of a square, given a starting point and the length and direction of one of the sides?
Suppose we have
p0 = rand(1, 2); % p0(1) => x; p0(2) => y
L = 10;
d0 = rand(1, 2); % The direction vector.
First, let's convert the direction into a unit vector.
u_vec = d0 ./ norm(d0);
Next, let's rotate this vector by 90 degrees to get the unit vector for the perpendicular side
u_perp = ([0, -1; 1 0] * u_vec')';
Using vector addition, we know that the second point is p1 = p0 + L * u_vec. The third point is p2 = p1 + L * u_perp, and the final point is p3 = p_0 + L * u_perp. We can stack these vectors just to make it easier to type out the plotting code:
l1 = [p0; p1];
l2 = [p1; p2];
l3 = [p2; p3];
l4 = [p3; p0];
And finally plot them:
figure();
hold on;
plot(l1(:, 1), l1(:, 2));
plot(l2(:, 1), l2(:, 2));
plot(l3(:, 1), l3(:, 2));
plot(l4(:, 1), l4(:, 2));
axis square;
Or, you could stack everything into one array and plot using a single line:
sq = [p0; p1; p2; p3; p0];
figure();
plot(sq(:, 1), sq(:, 2));
axis square;
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
How do you rotate 3D objects around each of the three axes when coordinates are of MATLAB style (X, Y and Z kept in different arrays).
This code is a start. I think I have found the rotation matrix (for pi/2 rotation around x), here it is called rotX90_2. But how should rotX90_2 operate on X, Y, Z?
[X,Y,Z] = cylinder;
% rotX90_1 = makehgtform('xrotate',pi/2) gives
rotX90_1 = ...
[1 0 0 0;
0 0 -1 0;
0 1 0 0;
0 0 0 1];
rotX90_2 = rotX90_1(1:3, 1:3);
% Here rotX90_2 should operate on [X,Y,Z] in order to ...
% rotate it 90 degrees around x, but how is this done?
% == What code should be put here to rotate the cylinder? ==
surf(X,Y,Z);
I have just started using MATLAB. As I understand, the basic ways to manipulate 3D graphics are to either operate on the X, Y, Z, like here or, you can first run graphics routines like h = surf(X, Y, Z); and then operate on graphics objects, using f.ex. hgtransform.
It is convenient to translate and scale using X, Y, Z. - You just add and multiply by scalars. But I ask this question to understand how to rotate.
If you operate on the graphic objects, on the other hand, you can use the function hgtransform. But you must then first create other objects, since hgtransform does not operate directly on the graphic objects, as I understand. (Except functions like rotatex(h, angle). F.ex, I have not found a corresponding "translatex(h, distance)". That surprised me. Maybe I didn't look well enough.)
OK I am new to this. Any simple, practical pointers how to easily rotate, scale and translate MATLAB 3D coordinates/objects (around the coordinate system axes) would be appreciated.
Edit:
According to Prakhar's answer below, which works, the code needed to fill the gap above is the following. Thank you, Prakhar.
[row, col] = size(X);
coordinates = [reshape(X, [row*col, 1]), reshape(Y, [row*col, 1]), reshape(Z, [row*col, 1])];
rC = coordinates * rotX90_2;
X = reshape(rC(:, 1), [row, col]);
Y = reshape(rC(:, 2), [row, col]);
Z = reshape(rC(:, 3), [row, col]);
Let's say R is the appropriate 3x3 rotation matrix.
coordinates = [X Y Z];
rotatedCoordinates = coordinates * R;
(Assuming X, Y, and Z are column vectors of same size)
Now, you can get the new X, Y, and Z coordinates from rotatedCoordinates as rotatedCoordinates(:, 1), rotatedCoordinates(:, 2), and rotatedCoordinates(:, 3), respectively.
EDIT: Another alternative when X, Y, Z are 2D matrices:
[X, Y, Z] = cylinder;
[row, col] = size(X);
coordinates = [reshape(X, [row*col, 1]), reshape(Y, [row*col, 1]), reshape(Z, [row*col, 1])];
rC = coordinates*R;
Xn = reshape(rC(:, 1), [row, col]);
Yn = reshape(rC(:, 2), [row, col]);
Zn = reshape(rC(:, 3), [row, col]);
I am trying to write some code to generate a plot similar to the one below on matlab (taken from here):
I have a set of points on a curve (x_i,y_i,z_i). Each point generates a Gaussian distribution (of mean (x_i,y_i,z_i) and covariance matrix I_3).
What I did is I meshed the space into npoint x npoints x npoints and computed the sum of the probability densities for each of the 'sources' (x_i,y_i,z_i) in each point (x,y,z). Then, if the value I get is big enough (say 95% of the maximum density), I keep the point. otherwise I discard it.
The problem with my code is that it is too slow (many for loops) and the graph I get doesn't look like the one below:
Does anyone know whether there is a package to get a similar plot as the one below?
Using isosurface we can do reasonably well. (Although I'm not honestly sure what you want, I think this is close:
% Create a path
points = zeros(10,3);
for ii = 2:10
points(ii, :) = points(ii-1,:) + [0.8 0.04 0] + 0.5 * randn(1,3);
end
% Create the box we're interested in
x = linspace(-10,10);
y = x;
z = x;
[X,Y,Z] = meshgrid(x,y,z);
% Calculate the sum of the probability densities(ish)
V = zeros(size(X));
for ii = 1:10
V = V + 1/(2*pi)^(3/2) * exp(-0.5 * (((X-points(ii,1)).^2 + (Y-points(ii,2)).^2 + (Z-points(ii,3)).^2)));
end
fv = isosurface(X,Y,Z,V, 1e-4 * 1/(2*pi)^(3/2), 'noshare');
fv2 = isosurface(X,Y,Z,V, 1e-5 * 1/(2*pi)^(3/2), 'noshare');
p = patch('vertices', fv.vertices, 'faces', fv.faces);
set(p,'facecolor', 'none', 'edgecolor', 'blue', 'FaceAlpha', 0.05)
hold on;
p2 = patch('vertices', fv2.vertices, 'faces', fv2.faces);
set(p2,'facecolor', 'none', 'edgecolor', 'red', 'FaceAlpha', 0.1)
scatter3(points(:,1), points(:,2), points(:,3));
How do I draw an ellipse and an ellipsoid using MATLAB?
(x^2/a^2)+(y^2/b^2)=1
n=40;
a=0; b=2*pi;
c=0; d=2*pi;
for i=1:n
u=a+(b-a)*(i-1)/(n-1);
for j=1:m
v=a+(d-c)*(j-1)/(m-1);
x(i,j)=sin(u)*cos(v);
y(i,j)=sin(u)*sin(v);
z(i,j)=cos(u);
end
end
mesh(x,y,z);
But I want the shape?
Ellipse article on Wikipedia had a simple JavaScript code to draw ellipses.
It uses the parametric form:
x(theta) = a0 + ax*sin(theta) + bx*cos(theta)
y(theta) = b0 + ay*sin(theta) + by*cos(theta)
where
(a0,b0) is the center of the ellipse
(ax,ay) vector representing the major axis
(bx,by) vector representing the minor axis
I translated the code into a MATLAB function:
calculateEllipse.m
function [X,Y] = calculateEllipse(x, y, a, b, angle, steps)
%# This functions returns points to draw an ellipse
%#
%# #param x X coordinate
%# #param y Y coordinate
%# #param a Semimajor axis
%# #param b Semiminor axis
%# #param angle Angle of the ellipse (in degrees)
%#
narginchk(5, 6);
if nargin<6, steps = 36; end
beta = -angle * (pi / 180);
sinbeta = sin(beta);
cosbeta = cos(beta);
alpha = linspace(0, 360, steps)' .* (pi / 180);
sinalpha = sin(alpha);
cosalpha = cos(alpha);
X = x + (a * cosalpha * cosbeta - b * sinalpha * sinbeta);
Y = y + (a * cosalpha * sinbeta + b * sinalpha * cosbeta);
if nargout==1, X = [X Y]; end
end
and an example to test it:
%# ellipse centered at (0,0) with axes length
%# major=20, ,minor=10, rotated 50 degrees
%# (drawn using the default N=36 points)
p = calculateEllipse(0, 0, 20, 10, 50);
plot(p(:,1), p(:,2), '.-'), axis equal
The answers from Jacob and Amro are very good examples for computing and plotting points for an ellipse. I'll address some easy ways you can plot an ellipsoid...
First, MATLAB has a built-in function ELLIPSOID which generates a set of mesh points given the ellipsoid center and the semi-axis lengths. The following creates the matrices x, y, and z for an ellipsoid centered at the origin with semi-axis lengths of 4, 2, and 1 for the x, y, and z directions, respectively:
[x, y, z] = ellipsoid(0, 0, 0, 4, 2, 1);
You can then use the function MESH to plot it, returning a handle to the plotted surface object:
hMesh = mesh(x, y, z);
If you want to rotate the plotted ellipsoid, you can use the ROTATE function. The following applies a rotation of 45 degrees around the y-axis:
rotate(hMesh, [0 1 0], 45);
You can then adjust the plot appearance to get the following figure:
axis equal; %# Make tick mark increments on all axes equal
view([-36 18]); %# Change the camera viewpoint
xlabel('x');
ylabel('y');
zlabel('z');
Also, if you want to use the rotated plot points for further calculations, you can get them from the plotted surface object:
xNew = get(hMesh, 'XData'); %# Get the rotated x points
yNew = get(hMesh, 'YData'); %# Get the rotated y points
zNew = get(hMesh, 'ZData'); %# Get the rotated z points
I've adapted this excellent ellipse plotting script from MATLAB Central for your requirement of
function plotEllipse(a,b,C)
% range to plot over
%------------------------------------
N = 50;
theta = 0:1/N:2*pi+1/N;
% Parametric equation of the ellipse
%----------------------------------------
state(1,:) = a*cos(theta);
state(2,:) = b*sin(theta);
% Coordinate transform (since your ellipse is axis aligned)
%----------------------------------------
X = state;
X(1,:) = X(1,:) + C(1);
X(2,:) = X(2,:) + C(2);
% Plot
%----------------------------------------
plot(X(1,:),X(2,:));
hold on;
plot(C(1),C(2),'r*');
axis equal;
grid;
end
Note: change N to define the resolution of your ellipse
Here's an ellipse centered at (10,10) with a = 30 and b = 10
Ellipse article on Wikipedia and Rotation matrix.
Rewrite that functions:
rotate by rotAngle counter-clockwise around (0,0)
Coordinate transform to (cx, cy)
function [X,Y] = calculateEllipse(cx, cy, a, b, rotAngle)
%# This functions returns points to draw an ellipse
%#
%# #param x X coordinate
%# #param y Y coordinate
%# #param a Semimajor axis
%# #param b Semiminor axis
%# #param cx cetner x position
%# #param cy cetner y position
%# #param angle Angle of the ellipse (in degrees)
%#
steps = 30;
angle = linspace(0, 2*pi, steps);
% Parametric equation of the ellipse
X = a * cos(angle);
Y = b * sin(angle);
% rotate by rotAngle counter clockwise around (0,0)
xRot = X*cosd(rotAngle) - Y*sind(rotAngle);
yRot = X*sind(rotAngle) + Y*cosd(rotAngle);
X = xRot;
Y = yRot;
% Coordinate transform
X = X + cx;
Y = Y + cy;
end
and an example to test it:
[X,Y] = calculateEllipse(0, 0, 20, 10, 0);
plot(X, Y, 'b'); hold on; % blue
[X,Y] = calculateEllipse(0, 0, 20, 10, 45);
plot(X, Y, 'r'); hold on; % red
[X,Y] = calculateEllipse(30, 30, 20, 10, 135);
plot(X, Y, 'g'); % green
grid on;
Create two vectors, one of the x-coordinates of the points of the circumference of the ellipsoid, one of the y-coordinates. Make these vectors long enough to satisfy your accuracy requirements. Plot the two vectors as (x,y) pairs joined up. I'd drop the for loops from your code, much clearer if you use vector notation. Also I'd format your question using the SO markup for code to make it all clearer to your audience.
The simplest way might be to use the Matlab function
pdeellip(xc,yc,a,b,phi)
For example:
pdeellip(0,0,1,0.3,pi/4)
However, this is simple solution is good to have a quick glance how the ellipse looks like. If you want to have a nice plot, look at the other solutions.
I don't know in which version of Matlab this was added, but it's available at least from version R2012b on.