Multiple 3d random walks

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Hi everyone I have the next code that creates a 3d random walk but i would like to have in the 3d space some random points(seeds) and start from each one of them a random walk at the same time. Can anyone help me?? My code for the random walk is the following:
lamda = 3; %Mean free path
numberOfSteps = 100000; % Totlal number of steps
x(1) = rand; % Initial position (x)
y(1) = rand; % Initial position (y)
z(1) = rand; % Initial position (z)
for i = 1:numberOfSteps
r = -lamda*log(rand()); % Distance Travelled
theta = pi*rand(); % Arbritary angle in between 0 and Pi
phi = 2*pi*rand(); % Arbritary angle in between 0 and 2Pi
dx = r*sin(theta)*cos(phi); % Step Size (x)
dy = r*sin(theta)*sin(phi); % Step Size (y)
dz = r*cos(theta); % Step Size (z)
x(i+1) = x(i) + dx; % Position at the end of the first step (x)
y(i+1) = y(i) + dy; % Position at the end of the second step (y)
z(i+1) = z(i) + dz; % Position at the end of the third step (z)
end
plot3(x, y, z, 'k');

采纳的回答

Torsten
Torsten 2022-12-8
编辑:Torsten 2022-12-8
lambda = 3; %Mean free path
numberOfSteps = 10000; % Total number of steps
numberOfPaths = 3; % Number of paths
x(1,:) = rand(1,numberOfPaths); % Initial position (x)
y(1,:) = rand(1,numberOfPaths); % Initial position (y)
z(1,:) = rand(1,numberOfPaths); % Initial position (z)
for j = 1:numberOfPaths
for i = 1:numberOfSteps
r = -lambda*log(rand()); % Distance Travelled
theta = pi*rand(); % Arbritary angle in between 0 and Pi
phi = 2*pi*rand(); % Arbritary angle in between 0 and 2Pi
dx = r*sin(theta)*cos(phi); % Step Size (x)
dy = r*sin(theta)*sin(phi); % Step Size (y)
dz = r*cos(theta); % Step Size (z)
x(i+1,j) = x(i,j) + dx; % Position at the end of the first step (x)
y(i+1,j) = y(i,j) + dy; % Position at the end of the second step (y)
z(i+1,j) = z(i,j) + dz; % Position at the end of the third step (z)
end
end
plot3(x(:,1), y(:,1), z(:,1), 'k');
hold on
plot3(x(:,2), y(:,2), z(:,2), 'r');
plot3(x(:,3), y(:,3), z(:,3), 'g');
hold off
  3 个评论
Torsten
Torsten 2022-12-8
Sorry, I don't know.
Maybe you should open a separate question for this.
lena kappa
lena kappa 2022-12-8
It's ok thank you very much ☺!

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