【发布时间】：2016-10-22 22:00:16
【问题描述】：
我们假设 3D 空间中有一个半球和三个三角形。半球底部的中心点用 C 表示。半球底部的半径用 R 表示。半球底部的法向量用 n 表示。
第一个三角形由三个点 V1、V2 和 V3 给出。第二个三角形由三个点 V4、V5 和 V6 给出。第三个三角形由三个点 V7、V8 和 V9 给出。点 V1、V2、……、V9 的位置是任意的。现在，我们进一步假设眼睛位于点 E。请注意，三角形可能会阻挡从眼睛到半球表面的视线；因此，肉眼可能看不到半球表面的某些区域。
请开发一种方法来估计肉眼可以看到的半球表面面积？ 这是矩形而不是半球的代码：
function r = month_1()
P1 = [0.918559, 0.750000, -0.918559];
P2 = [0.653394, 0.649519, 1.183724];
P3 = [-0.918559, -0.750000, 0.918559];
P4 = [-0.653394, -0.649519, -1.183724];
V1 = [0.989609, -1.125000, 0.071051];
V2 = [1.377838, -0.808013, -0.317178];
V3 = [1.265766, -0.850481, 0.571351];
V4 = [0.989609, -1.125000, 0.071051];
V5 = [1.265766, -0.850481, 0.571351];
V6 = [0.601381, -1.441987, 0.459279];
V7 = [0.989609, -1.125000, 0.071051];
V8 = [1.377838, -0.808013, -0.317178];
V9 = [0.713453, -1.399519, -0.429250];
E = [1.714054, -1.948557, 0.123064];
C = [0,1,0];
Radius = 2;
n = [0,1,0];
%hold on
patches.vertices(1,:)= P1;
patches.vertices(2,:)= P2;
patches.vertices(3,:)= P3;
patches.vertices(4,:)= P4;
patches.vertices(5,:)= V1;
patches.vertices(6,:)= V2;
patches.vertices(7,:)= V3;
patches.vertices(8,:)= V4;
patches.vertices(9,:)= V5;
patches.vertices(10,:)= V6;
patches.vertices(11,:)= V7;
patches.vertices(12,:)= V8;
patches.vertices(13,:)= V9;
patches.faces(1,:)= [5,6,7];
patches.faces(2,:)= [8,9,10];
patches.faces(3,:)= [11,12,13];
patches.faces(4,:)= [1,2,3];
patches.faces(5,:)= [1,3,4];
patches.facevertexcdata = 1;
patch(patches);
shading faceted; view (3);
% dispatch([1,1,1])
hold on
Num = 20;
Sum = 0;
%Srec = norm(cross(P1 - P4, P3 - P4));
for i = 1:Num
x1 = rand;
x2 = rand;
v1 = P1-P4;
v2 = P3-P4;
Samp = P4+v1*x1+v2*x2;
ABC = fun_f(E, Samp, V1,V2,V3)*fun_f(E,Samp, V4, V5, V6)*fun_f(E,Samp, V7,V8,V9);
Sum = Sum + ABC;
% if ABC ==1
% plot3(Samp(1), Samp(2), Samp(3),'r +'), hold on
% else
% plot3(Samp(1), Samp(2), Samp(3),'b +'), hold on
% end
%............................
[x, y, z]= sphere;
patches = surf2patch(x,y,z,z);
patches.vertices(:,3) = abs(patches.vertices(:,3));
patches.facevertexcdata = 1;
patch(patches)
shading faceted; view(3)
daspect([1 1 1])
%............................
end
%r = Sum/Num;
%view(31, 15)
%end
r = Sum/Num*norm(cross (P1-P4,P3-P4));
disp(sprintf('the integration is: %.5f',r));
disp(sprintf('the accurate result is: %.5f',norm(cross(P1-P4,P3-P4)/4)));
function res = fun_f(LineB, LineE, V1, V2, V3)
O = LineB;
Len = norm(LineE-LineB);
v = (LineE-LineB)/Len;
N = cross(V2-V1, V3-V1)/norm(cross(V2-V1, V3-V1));
if dot(N,v)~=0
tp = dot(N, V1-O)/ dot(N,v);
% if tp >=0 && tp <= (1:3);
P = LineB+tp*v(1:3);
A = V1 - P;
B = V2 - P;
Stri1 = norm(cross(A,B))/2;
A = V1 - P;
B = V3 - P;
Stri2 = norm(cross(A,B))/2;
A = V3 - P;
B = V2 - P;
Stri3 = norm(cross(A,B))/2;
A = V1 - V2;
B = V3 - V2;
Stotal = norm(cross(A,B))/2;
if (Stri1 + Stri2 + Stri3)> (Stotal + 1e-8)
res = 1;
else
res = 0;
end
else
res =1;
end
end
end
【问题讨论】：
-
我有一个想法，但我只能给你数学，而不是 MatLab。你想要吗？
-
@willywonkadailyblah 是的，我想要。请在数学中给出想法，然后我将其转换为matlab。
标签： matlab simulation matlab-figure matlab-cvst geometry-surface