Question

我试图使用MATLAB测试边界体积算法背后的数学（在射线追踪之前）。

到目前为止，我已经成功创建了相对平凡的轴对齐边界体积，我相信我已成功创建了一个边界球体。

我试图创建一个对齐边界体积的对象 - 但是我相信我已经让原理轴正确，因为盒子看起来是合适的形状 - 我无法正确翻译它＃34 ;到＆＃34;形状。

基本上我的问题是 - 在我的算法中我做错了什么？如何将我的边界体积转换为形状。

我一直使用的两个来源是Maths for 3D Games，还有一个blog，它提供了一些关于翻译的指示 - 但似乎并没有很好地发挥作用。< / p>

我已将源代码放在下面 - 非常感谢！

    %//===================Declare vertices and faces===================
vp_vtx = [379.379,684.302,319.752,711.497,215.956,439.237,600,600,732.938,418.084,600,600,747.081;
    225.836,158.305,394.208,337.480,84.834,245.958,212.858,342.251,322.344,330.576,300,416.190,398.885;
    0.933,0.917,0.904,0.878,0.854,0.923,0.914,0.949,0.943,0.908,0.896,0.940,0.933;
    1,1,1,1,1,1,1,1,1,1,1,1,1];
ws_fcs = [2,4,3,6,6,7,7,10,10; %//Faces forming shape
    4,3,1,7,11,9,13,11,13;
    5,5,5,11,10,13,11,13,12];
%//===================Create an AA bounding box===================
x = vp_vtx(1,:);y = vp_vtx(2,:);z = vp_vtx(3,:); %//Seperate vertex coordinates
bb_vtx = [min(x),max(x),min(x),max(x),min(x),max(x),min(x),max(x); %//Take min/max from each
    min(y),min(y),max(y),max(y),min(y),min(y),max(y),max(y); %//To form enclosing box
    min(z),min(z),min(z),min(z),max(z),max(z),max(z),max(z)];

bb_fcs = [1,2,6,1,1,3; 2,4,5,5,2,4;4,8,7,7,6,8; 3,6,8,3,5,7]; %//Allocate faces of box

figure(); grid on; hold on; xlabel('x'); ylabel('y'); zlabel('z');
scatter3(vp_vtx(1,:),vp_vtx(2,:),vp_vtx(3,:),'r'); %//Plot shape
patch('Faces',ws_fcs','Vertices',vp_vtx(1:3,:)', 'Facecolor', 'r','FaceAlpha', 0.1)
patch('Faces',bb_fcs', 'Vertices',bb_vtx','FaceColor','g','FaceAlpha', 0.05);%//Plot enclosing box

mean_point = sum(vp_vtx,2)/length(vp_vtx);
C = zeros(4,4); %//Create 4x4 empty matrix
for i = 1:length(vp_vtx)
    C = C+(vp_vtx(:,i)-mean_point)*(vp_vtx(:,i)-mean_point)'; %//Sum to get covarience matrix
end;
C = C/length(vp_vtx); %//Scale by the number of samples

[y,v] = eig(C(1:3,1:3)) ;%//Get eigenvalues & eigen vectors

R = y(:,1); %//Eigen vectors & values form object aligned axes
S = y(:,2);
T = y(:,3);%//T is principle axis as derived from largest eigenvalue

%//========Create an Object Orientated Bounding Box=========
dot_arr = zeros(size(vp_vtx));
for i = 1:length(vp_vtx)  %//Create array of dot products with each OO axis
    dot_arr(1,i) = dot(vp_vtx(1:3,i),T);
    dot_arr(2,i) = dot(vp_vtx(1:3,i),R);
    dot_arr(3,i) = dot(vp_vtx(1:3,i),S);
end
%//Get min/max variation in each OO axis
a = 0.5*(min(dot_arr(1,:)) + max(dot_arr(1,:)));
b = 0.5*(min(dot_arr(2,:)) + max(dot_arr(2,:)));
c = 0.5*(min(dot_arr(3,:)) + max(dot_arr(3,:)));
%//Centre is point where the 3 planes of the box intersect? (from book)
q = a*T + b*R + c*S;

Tr = vertcat(horzcat(T,S,R,q),[0,0,0,1]); %//Transform & translate original AA box
bb_vtx = Tr*vertcat(bb_vtx, ones(1,length(bb_vtx)));
patch('Faces',bb_fcs', 'Vertices',bb_vtx(1:3,:)','FaceColor','g','FaceAlpha', 0.05);

创建对象边界体积

0 个答案: