Circular vortex with spin vectors

I need help to create a circular vortex with different polarizations like converging, diverging and clockwise etc.,
I attached an image for reference.

2 Kommentare

Dyuman Joshi
Dyuman Joshi am 24 Jul. 2023
Please show what you have attempted yet.
% Parameters
numPoints = 100; % Number of points in the vortex
spinMagnitude = 0.5; % Magnitude of the spin vectors
radius = 1; % Radius of the vortex
% Generate theta values
theta = linspace(0, 2*pi, numPoints);
% Generate x and y coordinates
x = radius * cos(theta);
y = radius * sin(theta);
% Generate spin vectors
spinVectors = spinMagnitude * ones(size(x));
% Plot the vortex
figure;
quiver(x, y, spinVectors.*cos(theta), spinVectors.*sin(theta), 'b');
axis equal;
title('Circular Diverging Vortex');
xlabel('X');
ylabel('Y');

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Antworten (2)

Dyuman Joshi
Dyuman Joshi am 24 Jul. 2023

1 Stimme

% Parameters
numPoints = 50; % Number of points in the vortex
spinMagnitude = 0.5; % Magnitude of the spin vectors
r1 = 1; % Radius of the outer vortex
r2 = 0.5; %Radius of the inner vortex
% Generate theta values
theta = linspace(0, 2*pi, numPoints);
% Generate x and y coordinates
x = cos(theta);
y = sin(theta);
% Generate spin vectors
spinVectors = spinMagnitude * ones(size(x));
%%Radially outward arrows
figure;
quiver(r1*x, r1*y, spinVectors.*x, spinVectors.*y, 'b');
hold on
quiver(r2*x, r2*y, spinVectors.*x, spinVectors.*y, 'b');
axis equal;
xlabel('X');
ylabel('Y');
%%Radially outward arrows leaning in a counter clockwise direction
figure
quiver(r1*(x+y), r1*(y-x), spinVectors.*x, spinVectors.*y, 'b');
hold on
quiver(r2*(x+y), r2*(y-x), spinVectors.*x, spinVectors.*y, 'b');
axis equal;
xlabel('X');
ylabel('Y');

4 Kommentare

In case you want Radially outward arrows which are leaning clockwise, try this -
% Parameters
numPoints = 50; % Number of points in the vortex
spinMagnitude = 0.5; % Magnitude of the spin vectors
r1 = 1; % Radius of the outer vortex
r2 = 0.5; %Radius of the inner vortex
% Generate theta values
theta = linspace(0, 2*pi, numPoints);
% Generate x and y coordinates
x = cos(theta);
y = sin(theta);
% Generate spin vectors
spinVectors = spinMagnitude * ones(size(x));
%%Radially outward arrows leaning in a clockwise direction
figure
quiver(r1*(x-y), r1*(x+y), spinVectors.*x, spinVectors.*y, 'b');
hold on
quiver(r2*(x-y), r2*(x+y), spinVectors.*x, spinVectors.*y, 'b');
axis equal;
xlabel('X');
ylabel('Y');
Sateesh Kandukuri
Sateesh Kandukuri am 28 Jul. 2023
Dear @Dyuman Joshi, thank you so much for your response. Actually, I am looking for a pattern something like as below.
Let's consider a 200-unit dia circular geometry with a unit vector spacing. I want to define my base geometry with normalized spin vectors with different in-plane possible orientations. Could you please help me in this wasy?
Dyuman Joshi
Dyuman Joshi am 31 Jul. 2023
Do you only have these images to work with? or do you have any data or any other piece of information?
Sateesh Kandukuri
Sateesh Kandukuri am 31 Jul. 2023
Actually, the polarization of the system is defined by P = (cosφ, sinφ, 0), where φ = tan-1(y/x) +Ψ and (x,y) are the spatial coordinates in the system plane with the origin at the centre. The angle Ψ determines the polarization orientation. For Ψ = 0, it gives diverging vortex. I hope this piece of information helpful.

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Bruno Luong
Bruno Luong am 31 Jul. 2023
Bearbeitet: Bruno Luong am 31 Jul. 2023

1 Stimme

[x,y] = ndgrid(linspace(-1,1,10));
x = x(:)';
y = y(:)';
xy = [x; y];
r = vecnorm(xy, 2, 1);
r(r > 1) = NaN;
xyn = xy ./ r;
for k=1:12
Psi = 2*pi*rand();
R = [cos(Psi), -sin(Psi);
sin(Psi), cos(Psi)];
V = R * xyn;
vx = V(1,:);
vy = V(2,:);
subplot(3,4,k);
quiver(x, y, vx, vy, 'linewidth', 2);
set(gca, 'visible', 'off')
end

2 Kommentare

Sateesh Kandukuri
Sateesh Kandukuri am 31 Jul. 2023
Dear @Bruno Luong, I want to define these normalized spin vectors on a 200-unit diameter circular geometry with a unit vector spacing, and I need it in a separate image based on the angle Psi.
Bruno Luong
Bruno Luong am 31 Jul. 2023
Bearbeitet: Bruno Luong am 31 Jul. 2023
Feel free to adapt my code to your need.
I gave you a recipe of the cake, if you want strawberry flavor, you need to adapt my recipe and make your own cake.

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Version

R2020b

Gefragt:

am 24 Jul. 2023

Bearbeitet:

am 31 Jul. 2023

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