I am trying to use the spherical harmonics to represent a perturbation of a spherical object in an acoustic and fluid flow in 3D.
So far, I have been able to use a 2D perturbation on a 3D spherical object, however, I would like to extend that.
The current code represents a 3D sphere with a 2D perturbation, so the perturbation is only in x and y:
clearvars; clc; close all;
Nx = 128;
Ny = 128;
Nz = 128;
Lx =128;
Ly = 128;
Lz = 128;
xi = (0:Nx-1)/Nx*2*pi;
xi_x = 2*pi/Lx;
x = xi/xi_x;
yi = (0:Ny-1)/Ny*2*pi;
yi_y = 2*pi/Ly;
y = yi/yi_y;
zi = (0:Nz-1)/Nz*2*pi;
zi_z = 2*pi/Lz;
z = zi/zi_z;
[X,Y,Z] = meshgrid(x,y,z);
A = 2*pi / Lx;
B = 2*pi / Ly;
C = 2*pi / Lz;
x0 = 64;
y0 = 64;
z0 = 64;
rx0 = 20;
ry0 = 20;
rz0 = 20;
p = 3;
b = 0.1; % pert amplitude
c = 12;
d = 1;
a = 4;
theta = atan2(Y -y0, X-x0) - (pi/c);
p0 = ((X-x0) .* (X-x0)) /(rx0 * rx0) + ((Y-y0) .* (Y-y0))/(ry0 * ry0) + ((Z-z0) .* (Z-z0))/(rz0 * rz0);
Test =d + a * exp((-1. * p0 .* (1 - b .* cos(c * theta))).^p) ;
figure
isosurface(X,Y,Z,Test);
shading flat;
grid on;
Which returns the isosurface:
However, I would like to achieve something similar to this plot (perturbation in z as well):
This is my attempt using spherical harmonics to reproduce the above picture:
clearvars; clc; close all;
%in spherical coord
%calculate r
Nx = 128;
Ny = 128;
Nz = 128;
Lx =128;
Ly = 128;
Lz = 128;
xi = (0:Nx-1)/Nx*2*pi;
xi_x = 2*pi/Lx;
x = xi/xi_x;
yi = (0:Ny-1)/Ny*2*pi;
yi_y = 2*pi/Ly;
y = yi/yi_y;
zi = (0:Nz-1)/Nz*2*pi;
zi_z = 2*pi/Lz;
z = zi/zi_z;
r = sqrt(x.^2 + y.^2 + z.^2);
% Create the grid
delta = pi/127;
%Taking for instance l=1, m=-1 you can generate this harmonic on a (azimuth, elevation) grid like this:
azimuths = 0 : delta : pi;
elevations = 0 : 2*delta : 2*pi;
[R, A, E] = ndgrid(r, azimuths, elevations); %A is phi and E is theta
H = 0.25 * sqrt(3/(2*pi)) .* exp(-1j*A) .* sin(E) .* cos(E);
%transform the grid back to cartesian grid like this:
%can also add some radial distortion to make things look nicer:
%the radial part depends on your domain
X = r .* cos(A) .* sin(E);
Y = r .* sin(A) .* sin(E);
Z = r .* cos(E);
%parameters
x0 = 64;
y0 = 64;
z0 = 64;
rx0 = 20;
ry0 = 20;
rz0 = 20;
p = 3;
b = 0.1; % pert amplitude
%c = 12;
d = 1;
a = 4;
p0 = ((X-x0) .* (X-x0)) /(rx0 * rx0) + ((Y-y0) .* (Y-y0))/(ry0 * ry0) + ((Z-z0) .* (Z-z0))/(rz0 * rz0);
Test1 =d + a * exp((-1. * p0 .*H).^p) ;
figure
isosurface(X,Y,Z,real(Test1)); %ERROR
This gives me the following error:
Error using griddedInterpolant
Grid arrays must have NDGRID structure.
Is the issue in the way I am setting up the spherical harmonics? or the functional form of Test1?? Thanks
I think the problem is that you've created a NDGrid in the second code. In the first code that worked you created a meshgrid.
isosurface expects a grid in meshgrid format and transforms it later into an NDGrid format with permute just for the usage of the griddedInterpolant function. By the input of an NDGrid this operation will fail.
Why did you switch to creating a NDGrid? Did you get the same error when using meshgrid?
EDIT
OK, new theory: I think the problem is the grid itsself. Read the documentation on ndgrid for more information but for short: A NDGrid format is a completely rectangular grid where all nodes are exclusivly surrounded by 90° angles. By spanning up a grid with ndgrid(r, azimuths, elevations) or meshgrid(r, azimuths, elevations) you are getting this rectangular grid but of course this grid is meaningless because r, azimuths and elevations represent spherical coordinates. By later converting R, A and E into the carthesian coordinates X, Y and Z you get a propper spherical grid but the grid isn't a NDGrid structure anymore since it's not rectangular anymore.
So you have to find a workaround to calculate with your spherical grid.
Maybe you can use another function to visualize your data that dosn't take any grid format as input
Or maybe you can try working with a rectangular cartesian grid by setting all values outside the sphere to 0. (You have to refine your grid accordingly to achieve a good estimation of the sphere)
For a given PMF p=f(\theta) for \theta between 0 and 2\pi, i computed the CDF in Matlab as
theta=0:2*pi/n:2*pi
for i=1:n
cdf(i)=trapz(theta(1:i),p(1:i));
end
and the result is verified.
I tried to do the same with cumsum as cdf=cumsum(p)*(2*pi)/n but the result is wrong. why?
How can i compute the CDF if the given PMF is in 2D asp=f(\theta,\phi) ? Can i do it without going into detail as explained here ?
In 1D case you can use cumsum to get the vectorized version of loop (assuming that both theta and p are column vectors):
n = 10;
theta = linspace(0, 2*pi, n).';
p = rand(n,1);
cdf = [0; 0.5 * cumsum((p(1:n-1) + p(2:n)) .* diff(theta(1:n)))];
In 2D case the function cumsum will be applied two times, in vertical and horizontal directions:
nthet = 10;
nphi = 10;
theta = linspace(0, 2*pi, nthet).'; % as column vector
phi = linspace(0, pi, nphi); % as row vector
p = rand(nthet, nphi);
cdf1 = 0.5 * cumsum((p(1:end-1, :) + p(2:end, :)) .* diff(theta), 1);
cdf2 = 0.5 * cumsum((cdf1(:, 1:end-1) + cdf1(:, 2:end)) .* diff(phi), 2);
cdf = zeros(nthet, nphi);
cdf(2:end, 2:end) = cdf2;
I'm trying to surface bumpy spheres and wrinkled spheres in Matlab using
p = 1 + 0,2 * sin(phi * m) * sin(teta * n)
teta = 0:6.23;
phi = 0:3.14;
[teta,phi] = meshgrid(teta, phi);
figure, hold on
for m = 1:12
for n = 1:12
p = 1 + 0.2*sin(m*teta).*sin(n*phi);
surf(teta,phi,p)
pause(0.05)
clf('reset')
end
end
But it's not drawing any sphere just surfaces...what am I doing wrong and what should I change :) Thanks a lot!!!
I don't understand your example so I created a new one:
%We define the spherical coordinates.
theta = linspace(0,2*pi,50);
phi = linspace(0,pi,50);
[x1,x2] = meshgrid(linspace(0,12*pi,50),linspace(0,12*pi,50)); %the variation of rho will produce a bumpy sphere.
rho = 0.1*(sin(x1)+cos(x2))+1;
[theta,phi] = meshgrid(theta,phi);
%we calculate the cartesian coordinates.
x = rho.*cos(theta).*sin(phi);
y = rho.*sin(theta).*sin(phi);
z = rho.*cos(phi);
%plot
surf(x,y,z);
RESULT BUMPY
RESULT WRINKLED
Simply change the rho parameter by:
rho = 0.1*(sin(x1))+1;
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
I am using MATLAB to calculate the height of a ball in case of projectile motion. Here is my code :
v = 10;
teta = 20; % angle of the projectile motion
vx = v*cos(teta); % velocity in x axis
vy = v*sin(teta); % velocity in y axis
x = 0:10;
y = zeros(length(x),1);
for xx=1:length(x)
% here I calculate the height of the ball in y axis
y(xx,:) = vy*(xx/vx)-(0.5*9.81*(xx./vx)^2);
% here I want to break the system when the height of the ball is
% zero (on the ground surface)
if y(xx,:) == 0
break
end
end
plot(x,y, '*')
The problem that when I plot the motion it does not stop when the height is zero
but it continues for negative values of y axis ..
What is the wrong in my if-statement?
You should never compare a float or double number to an integer. When you say if y(xx,:)== 0 you are breaking the loop only when the height is exactly zero, and that will never happen (just see the values stored in y).
If you want to keep your data from been negative, you should try if y(xx,:) < 0. Keep in mind that if you use this approach, you should always delete the last value added to y, because when that statement breaks you loop, you've already stored the first negative value. So, just add y(xx)= 0; before the break line.
<teacher_mode>
What I would do is use constant time increments rather than using constant x increments. This is not only more intuitive, but also provides a much better building block for if you'd want to level up in the future. This approach makes it easier to include additional effects (like airdrag, 3D, bounces, ...) and is a stepping stone towards more advanced methods to do these computations (numerical integration of the vector-valued equations of motion; see ode45 in MATLAB for some examples).
Here is an example of a "bouncy" ball:
v = 100; % initial speed
theta = 20; % initial angle (°) with X-axis
vx = v * cosd(theta); % initial velocity in x axis
vy = v * sind(theta); % initial velocity in y axis
T = 0 : 0.01 : 50; % times to sample
x = zeros(length(T),1);
y = zeros(length(T),1);
ay = -9.80665; % gravity
ax = 0; % ...wind perhaps
ex = 0.9; % elasticity in X-direction
ey = 0.5; % elasticity in Y-direction
for ii = 2:numel(T)
% Update time step
dt = T(ii) -T(ii-1);
% Update speed
vx = vx + ax*dt;
vy = vy + ay*dt;
% Update position
x(ii) = x(ii-1) + [vx ax/2] * dt.^(1:2)';
y(ii) = y(ii-1) + [vy ay/2] * dt.^(1:2)';
% Bounce the ball!
if y(ii) <= 0
y(ii) = 0;
vx = ex* vx;
vy = abs( ey*vy );
end
end
clf, hold on
plot(x,y)
</teacher_mode>
You can avoid for loop at all:
v=10;
theta=20; % angle of the projectile motion
vx=v*cos(theta); % velocity in x axis
vy=v*sin(theta); % velocity in y axis
x=0:0.1:10;
y=zeros(size(x));
y=vy.*(x./vx)-(0.5*9.81*(x./vx).^2); %% Calculate values for all x
x=x(y>=0); %% keep the x values where y>=0
y=y(y>=0); %% keep the y valuse where y>=0
plot(x,y,'*')