Question

我是Matlab编程的新手。我使用EIDORS和Netgen通过四面体网格创建了两个模型。然后我需要使用Matlab创建一个基于四面体元素交集的地图。所以，为了找到我尝试使用点测试方法的交集。请参阅链接。的 http://steve.hollasch.net/cgindex/geometry/ptintet.html

让我们假设model1和model2。

首先，我从两个模型中提取顶点并创建矩阵。

然后，我使用点测试方法计算模型之间的交集。

对于model1，我有3000 ++四面体元素，而model2我有8000 ++四面体元素。

为了计算交点，我逐个循环以确定两个模型之间的哪些元素具有交集，然后通过将model2的元素编号索引到model1的元素编号来创建矩阵。

但是，对于少数元素来说似乎是零，这是不可能的，因为来自model1的所有元素应至少与来自model2的一些元素相交。

因此，最后我希望得到一个矩阵，包括（model1的元素编号x model2的元素编号，它与model1中的相应元素相交）。可以帮我解决这个问题吗？请参阅我的代码。

function [elemno,deter0,deter1,deter2,deter3,deter4] = checkp(filename1,filename2);

%/* to check whether the vertices of a layered model''s element are inside the
% tetrahedron of the generic model
%after the model is created using netgen and eidors, there will be a struct    type 
%named model_parameters.
%model_parameters.vtx refers to the vertices which consists of (nodes x   vertices %(x,y,z))
%model_parameters.simp refers to the elements which consists (numberofelements x nodes) nodes are linked to the vertices.*/

filename1 = [filename1 '.mat'];
filename2 =[filename2 '.mat'];

first = load(filename1);
second =load(filename2);

vtx = first.model_parameters.vtx;
simp = first.model_parameters.simp;
[simpr,simpc] = size(simp);

vtx2 = second.model_parameters.vtx;
simp2= second.model_parameters.simp;
[simpr2,simpc2] = size(simp2);


%//extracting the vertices of the elements from the simplices(element)
for loop1 = 1 : simpr
    elemx(loop1,1) = vtx(simp(loop1,1),1);      
    elemx(loop1,2) = vtx(simp(loop1,2),1);
    elemx(loop1,3) = vtx(simp(loop1,3),1);
    elemx(loop1,4) = vtx(simp(loop1,4),1);

    elemy(loop1,1) = vtx(simp(loop1,1),2);
    elemy(loop1,2) = vtx(simp(loop1,2),2);
    elemy(loop1,3) = vtx(simp(loop1,3),2);
    elemy(loop1,4) = vtx(simp(loop1,4),2);

    elemz(loop1,1) = vtx(simp(loop1,1),3);
    elemz(loop1,2) = vtx(simp(loop1,2),3);
    elemz(loop1,3) = vtx(simp(loop1,3),3);
    elemz(loop1,4) = vtx(simp(loop1,4),3);

end

for loop2 = 1:simpr2
    elemx2(loop2,1) = vtx2(simp2(loop2,1),1);
    elemx2(loop2,2) = vtx2(simp2(loop2,2),1);
    elemx2(loop2,3) = vtx2(simp2(loop2,3),1);
    elemx2(loop2,4) = vtx2(simp2(loop2,4),1);

    elemy2(loop2,1) = vtx2(simp2(loop2,1),2);
    elemy2(loop2,2) = vtx2(simp2(loop2,2),2);
    elemy2(loop2,3) = vtx2(simp2(loop2,3),2);
    elemy2(loop2,4) = vtx2(simp2(loop2,4),2);

    elemz2(loop2,1) = vtx2(simp2(loop2,1),3);
    elemz2(loop2,2) = vtx2(simp2(loop2,2),3);
    elemz2(loop2,3) = vtx2(simp2(loop2,3),3);
    elemz2(loop2,4) = vtx2(simp2(loop2,4),3);

end
%//point test calculation
r =[1;1;1;1];
for a = 1:simpr
    m=1;
    for b=1:simpr2  
        for n = 1:4
            p = [elemx2(b,n),elemy2(b,n),elemz2(b,n)];
            n1=[elemx(a,1),elemy(a,1),elemz(a,1)];
            n2=[elemx(a,2),elemy(a,2),elemz(a,2)];
            n3=[elemx(a,3),elemy(a,3),elemz(a,3)];
            n4=[elemx(a,4),elemy(a,4),elemz(a,4)];
            d0 =[n1;n2;n3;n4];
            d0 =[d0 r];
            d1 =[p;n2;n3;n4];
            d1 =[d1 r];
            d2 =[n1;p;n3;n4];
            d2 =[d2 r];
            d3 =[n1;n2;p;n4];
            d3 =[d3 r];
            d4 =[n1;n2;n3;p];
            d4 =[d4 r];
            deter0 = sign(det(d0));
            deter1 = sign(det(d1));
            deter2 = sign(det(d2));
            deter3 = sign(det(d3));
            deter4 = sign(det(d4));
            if isequal(deter0,deter1,deter2,deter3,deter4)
                elemno(a,m) = b;   
                m=m+1;
                break;
            else
                continue;
            end
        end
    end
end

Answer 1

Note that all vertices of two tetrahedrons could lie outside each other, but these tetrahedrons do intersect, so point test is not reliable method.

Possible robust and fast approach is the method of separating axes.

I've used 2D version for quick selection of intersecting pairs between two sets containing a lot of (10^4 and 10^6) convex polygons, but 3D version also looks simple enough.

使用点测试找到两个四面体的交点

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