我为此编写了一个小脚本。随时提出建议和编辑。它通过检查每个球体的表面是否落在所有其他球体的体积内来工作。
对于球体相交,在sphere() 函数调用中使用更多面会更好(但更慢)。这应该在可视化中给出更密集的结果。对于单独的球体可视化,较小的数字(~50)就足够了。请参阅 cmets 了解如何可视化每个。
close all
clear
clc
% centers : 3 x N matrix of [X;Y;Z] coordinates
% dist : 1 x N vector of sphere radii
%% Plot spheres (fewer faces)
figure, hold on % One figure to rule them all
[x,y,z] = sphere(50); % 50x50-face sphere
for i = 1 : size(centers,2)
h = surfl(dist(i) * x + centers(1,i), dist(i) * y + centers(2,i), dist(i) * z + centers(3,i));
set(h, 'FaceAlpha', 0.15)
shading interp
end
%% Plot intersection (more faces)
% Create a 1000x1000-face sphere (bigger number = better visualization)
[x,y,z] = sphere(1000);
% Allocate space
xt = zeros([size(x), size(centers,2)]);
yt = zeros([size(y), size(centers,2)]);
zt = zeros([size(z), size(centers,2)]);
xm = zeros([size(x), size(centers,2), size(centers,2)]);
ym = zeros([size(y), size(centers,2), size(centers,2)]);
zm = zeros([size(z), size(centers,2), size(centers,2)]);
% Calculate each sphere
for i = 1 : size(centers, 2)
xt(:,:,i) = dist(i) * x + centers(1,i);
yt(:,:,i) = dist(i) * y + centers(2,i);
zt(:,:,i) = dist(i) * z + centers(3,i);
end
% Determine whether the points of each sphere fall within another sphere
% Returns booleans
for i = 1 : size(centers, 2)
[xm(:,:,:,i), ym(:,:,:,i), zm(:,:,:,i)] = insphere(xt, yt, zt, centers(1,i), centers(2,i), centers(3,i), dist(i)+0.001);
end
% Exclude values of x,y,z that don't fall in every sphere
xmsum = sum(xm,4);
ymsum = sum(ym,4);
zmsum = sum(zm,4);
xt(xmsum < size(centers,2)) = 0;
yt(ymsum < size(centers,2)) = 0;
zt(zmsum < size(centers,2)) = 0;
% Plot intersection
for i = 1 : size(centers,2)
xp = xt(:,:,i);
yp = yt(:,:,i);
zp = zt(:,:,i);
zp(~(xp & yp & zp)) = NaN;
surf(xt(:,:,i), yt(:,:,i), zp, 'EdgeColor', 'none');
end
这里是insphere 函数
function [x_new,y_new,z_new] = insphere(x,y,z, x0, y0, z0, r)
x_new = (x - x0).^2 + (y - y0).^2 + (z - z0).^2 <= r^2;
y_new = (x - x0).^2 + (y - y0).^2 + (z - z0).^2 <= r^2;
z_new = (x - x0).^2 + (y - y0).^2 + (z - z0).^2 <= r^2;
end
示例可视化
对于这些示例中使用的 6 个球体,在我的笔记本电脑上运行组合可视化平均需要 1.934 秒。
6 个球体的交点:
实际 6 个球体:
下面,我将两者结合起来,这样您就可以在球体视图中看到交叉点。
对于这些示例:
centers =
-0.0065 -0.3383 -0.1738 -0.2513 -0.2268 -0.3115
1.6521 -5.7721 -1.7783 -3.5578 -2.9894 -5.1412
1.2947 -0.2749 0.6781 0.2438 0.4235 -0.1483
dist =
5.8871 2.5280 2.7109 1.6833 1.9164 2.1231
我希望这可以帮助其他可能希望看到这种效果的人。