The trick here is to be able to identify the points on the GRID that are also on your circle. You want all points whose distance from the center of the circle is equal to the radius of the circle.
Your code is going to loop through all the co-ordinates (x,y) / points and wherever the point lies on the circle, energize it.
The distance from the center of a circle between the center point (h, k) and the point on the circle (x, y), where the radius is represented by "r": √((x - h)² + (y - k)²) ) From Pythagoras theorem. I ran the code below in Matlab.
% Define the grid size
grid_rows = 10; % Example size, adjust as needed
grid_cols = 10; % Example size, adjust as needed
% Define the center of the grid
midrow = grid_rows / 2;
midcol = grid_cols / 2;
% Define the radius of the circle
radius2 = 4; % Updated radius
% Create a grid with the same size as your existing grid
volt_grid = zeros(grid_rows, grid_cols);
% Draw the circle (for visualization)
rectangle('Position', [midrow-radius2, midcol-radius2, radius2*2, radius2*2], 'Curvature', [1, 1]);
% Loop through the grid to assign 1 volt to the boundary of the circle
for i = 1:grid_rows
for j = 1:grid_cols
% Calculate the distance from the center of the circle
dist = sqrt((i - midrow)^2 + (j - midcol)^2);
% Check if the point is on the boundary of the circle
if abs(dist - radius2) < 0.5 % Allowing a small margin for boundary points
volt_grid(i, j) = 1; % Assign 1 volt
end
end
end
% Display the grid with voltages
imagesc(volt_grid);
colorbar;
title('Voltage Distribution on Circle Boundary');
