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)
This question already has an answer here:
Machine learning - Linear regression using batch gradient descent
(1 answer)
Closed 6 years ago.
I tried normal equation, and the result was correct.
However, when I used gradient descent, the figure turned out to be wrong. I referred to online resources, but I failed to find out what's wrong. I don't think there's anything special in the following code.
clear;
clc;
m = 100; % generate 100 points
noise = randn(m,1); % 100 noise of normal distribution
x = rand(m, 1) * 10; % generate 100 x's ranging from 0 to 10
y = 10 + 2 * x + noise;
plot (x, y, '.');
hold on;
X = [ones(m, 1) x];
theta = [0; 0];
plot (x, X * theta, 'y');
hold on;
% Method 1 gradient descent
alpha = 0.02; % alpha too big will cause going far away from the result
num_iters = 5;
[theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
% Method 2 normal equation
% theta = (pinv(X' * X )) * X' * y
plot (x, X * theta, 'r');
function [theta, J_history] = gradientDescent(X, y, theta, alpha, num_iters)
m = length(y);
J_history = zeros(num_iters, 1);
for iter = 1:num_iters,
theta = theta - alpha * (1/m) * (X' * (X * theta - y));
% plot (X(:, 2), X * theta, 'g');
% hold on;
J_history(iter) = costFunction(X, y, theta);
end
end
function J = costFunction( X, y, theta )
m = length(y);
predictions = X * theta; % prediction on all m examples
sqrErrors = (predictions - y).^2; % Squared errors
J = 1/(2*m) * sum(sqrErrors);
end
Your code is correct. The problem is small number of iterations.
One can take num_iters = 5000; and see that theta converges to right value ([10; 2]).
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'm new with Matlab and Machine Learning and I tried to make a gradient descent function without using matrix.
m is the number of example on my training set
n is the number of feature for each example
The function gradientDescentMulti takes 5 arguments:
X mxn Matrix
y m-dimensional vector
theta : n-dimensional vector
alpha : a real number
nb_iters : a real number
I already have a solution using matrix multiplication
function theta = gradientDescentMulti(X, y, theta, alpha, num_iters)
for iter = 1:num_iters
gradJ = 1/m * (X'*X*theta - X'*y);
theta = theta - alpha * gradJ;
end
end
The result after iterations:
theta =
1.0e+05 *
3.3430
1.0009
0.0367
But now, I tried to do the same without matrix multiplication, this is the function:
function theta = gradientDescentMulti(X, y, theta, alpha, num_iters)
m = length(y); % number of training examples
n = size(X, 2); % number of features
for iter = 1:num_iters
new_theta = zeros(1, n);
%// for each feature, found the new theta
for t = 1:n
S = 0;
for example = 1:m
h = 0;
for example_feature = 1:n
h = h + (theta(example_feature) * X(example, example_feature));
end
S = S + ((h - y(example)) * X(example, n)); %// Sum each feature for this example
end
new_theta(t) = theta(t) - alpha * (1/m) * S; %// Calculate new theta for this example
end
%// only at the end of the function, update all theta simultaneously
theta = new_theta'; %// Transpose new_theta (horizontal vector) to theta (vertical vector)
end
end
The result, all the theta are the same :/
theta =
1.0e+04 *
3.5374
3.5374
3.5374
If you look at the gradient update rule, it may be more efficient to actually compute the hypothesis of all of your training examples first, then subtract this with the ground truth value of each training example and store these into an array or vector. Once you do this, you can then compute the update rule very easily. To me, it doesn't appear that you're doing this in your code.
As such, I rewrote the code, but I have a separate array that stores the difference in the hypothesis of each training example and ground truth value. Once I do this, I compute the update rule for each feature separately:
for iter = 1 : num_iters
%// Compute hypothesis differences with ground truth first
h = zeros(1, m);
for t = 1 : m
%// Compute hypothesis
for tt = 1 : n
h(t) = h(t) + theta(tt)*X(t,tt);
end
%// Compute difference between hypothesis and ground truth
h(t) = h(t) - y(t);
end
%// Now update parameters
new_theta = zeros(1, n);
%// for each feature, find the new theta
for tt = 1 : n
S = 0;
%// For each sample, compute products of hypothesis difference
%// and the right feature of the sample and accumulate
for t = 1 : m
S = S + h(t)*X(t,tt);
end
%// Compute gradient descent step
new_theta(tt) = theta(tt) - (alpha/m)*S;
end
theta = new_theta'; %// Transpose new_theta (horizontal vector) to theta (vertical vector)
end
When I do this, I get the same answers as using the matrix formulation.
I am trying to calculate maximum distance and maximum height of a projectile for an angle theta below.
I assume my way of plotting the graphs of distance against theta and height against theta (on the same graph) are wrong. Any pointers on that will be helpful.
e=100;
m = 1;
g = 9.8;
cd = 0.55;
r = 0.02;
p = 1.21;
a = pi*r^2;
v = sqrt((2*e)/m);
k = (1/2)*cd*a*p;
% For loop to calculate Distance and
for theta = (0:pi/4);
vx = v*cos(theta);
vy = v*sin(theta);
t = sqrt(m/(g*k))*atan(vy*sqrt(k/(m*g)));
x = (m/k)* log((1/vx)+(k/m)*t) - log(1/(vx));
h = (m/k)*(log(cos(atan(vy*sqrt(k/(m*g))-sqrt((g*k)/m)*t))-log(cos((atan(vy*sqrt(k/m*g)))))));
plot (x, theta);
plot (h, theta);
end
You don't need to loop through different values of theta. Try this instead:
theta = (0:0.01:pi/4); % theta = [0 0.01 0.02 0.03 ... pi/4]
vx = v*cos(theta);
vy = v*sin(theta);
t = sqrt(m/(g*k)) * atan(vy .* sqrt(k/(m*g))); % element wise matrix multiplication
x = (m/k)* log((1./vx)+(k/m)*t) - log(1./(vx));
h = (m/k)*(log(cos(atan(vy .* sqrt(k/(m*g))- sqrt((g*k)/m)*t))-log(cos((atan(vy .* sqrt(k/m*g)))))));
plot (x, theta);
plot (h, theta);
x vs theta:
Also try this:
MATLAB: Creating a Function to Plot Projectile with Drag
MATLAB: Numerical approximation of projectile motion with air resistance
Hope this helps!
The variable t is not defined. Also the loop variable should be integer, for example:
N = 10;
for i = 0:N-1
theta = pi/4*i/(N-1)
(...)
end