在正方形网格中创建一个圆

Question

我试图通过固定平行板电容器程序来解决以下二维椭圆型PDE静电问题。但我有问题来绘制圆圈区域。我怎样才能画出圆形区域而不是正方形呢？

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% I use following two lines to label the 50V and 100V squares
% (but it should be two circles)
                
     V(pp1-r_circle_small:pp1+r_circle_small,pp1-r_circle_small:pp1+r_circle_small) = 50;
     V(pp2-r_circle_big:pp2+r_circle_big,pp2-r_circle_big:pp2+r_circle_big) = 100;



    % Contour Display for electric potential
    figure(1)
    contour_range_V = -101:0.5:101;
    contour(x,y,V,contour_range_V,'linewidth',0.5);
    axis([min(x) max(x) min(y) max(y)]);
    colorbar('location','eastoutside','fontsize',10);
    xlabel('x-axis in meters','fontsize',10);
    ylabel('y-axis in meters','fontsize',10);
    title('Electric Potential distribution, V(x,y) in volts','fontsize',14);
    h1=gca;
    set(h1,'fontsize',10);
    fh1 = figure(1); 
    set(fh1, 'color', 'white')

    % Contour Display for electric field
    figure(2)
    contour_range_E = -20:0.05:20;
    contour(x,y,E,contour_range_E,'linewidth',0.5);
    axis([min(x) max(x) min(y) max(y)]);
    colorbar('location','eastoutside','fontsize',10);
    xlabel('x-axis in meters','fontsize',10);
    ylabel('y-axis in meters','fontsize',10);
    title('Electric field distribution, E (x,y) in V/m','fontsize',14);
    h2=gca;
    set(h2,'fontsize',10);
    fh2 = figure(2); 
    set(fh2, 'color', 'white')

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Answer 1

由于索引的方式，您正在创建一个正方形(请参阅this post on indexing)。您已经指定了从pp1-r_circle_small到pp1+r_circle_small运行的行，并为列指定了类似的行。考虑到这个Swiss cheese is not an option，您正在创建一个完整的正方形。

从几何上我们知道，从sqrt((X-X0)^2 - (Y-Y0)^2) < R到(X0,Y0)圆心的距离内的所有点(半径为R )都在圆内，其余点在圆周之外。这意味着您可以简单地构建一个掩码：

% Set up your grid
Xsize = 30;  % Your case: 1
Ysize = 30;  % Your case: 1
step = 1;  % Amount of gridpoints; use 0.001 or something
% Build indexing grid for circle search, adapt as necessary
X = 0:step:Xsize;
Y = 0:step:Ysize;
[XX,YY] = meshgrid(X, Y);
V = zeros(numel(X), numel(Y));

% Repeat the below for both circles
R = 10;  % Radius of your circle; your case 0.1 and 0.15
X0 = 11;  % X coordinate of the circle's origin; your case 0.3 and 0.7
Y0 = 15;  % Y coordinate of the circle's origin; your case 0.3 and 0.7

% Logical check to see whether a point is inside or outside
mask = sqrt( (XX - X0).^2 + (YY - Y0).^2) < R;

V(mask) = 50;  % Give your circle the desired value

imagesc(V)  % Just to show you the result
axis equal % Use equal axis to have no image distortion

mask是一个逻辑矩阵，包含1 (点在圆圈内)和0 (点在外部)。然后，可以使用此掩码对潜在的网格V进行逻辑索引，以将其设置为所需的值。

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注意:这显然不会创建一个完美的圆，因为你不能在正方形网格上绘制一个完美的圆。网格越细，就越像你的“圆圈”。这显示了使用step = 0.01的结果

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注2:您需要对X、Y、X0、Y0和R的定义进行一周的调整，以匹配您的值。