Question

我用椭球函数绘制两个椭圆体，并使用旋转手柄。为了计算交叉点体积（即透镜）和椭圆体之间没有旋转的交叉体积的情况，我使用了球体交叉体积的分析代码。但是我被卡住了，如果椭圆体处于某些旋转状态，我将如何计算它们之间的交叉体积（即，间距）。我还需要计算交叉点直径。镜头椭球相对于沿z轴的最大轴具有不同的旋转角度。这是一个代码：

 draw ellipsoid, [x,y,z] = ellipsoid(xc,yc,zc,xr,yr,zr,n);
 (xc,yc,zc)=center; semi-axis lengths = (Fmax,Fmin,Fmin);n
X=[0,2];  two ellipsoid x coordinates  i.e 0 is first ellipsoid
           and 2 is second ellipsoid respectively
Y=[0,2];   two ellipsoid y coordinates
Z=[0,2];   two ellipsoid z coordinates;
ROTATIONANGLE=[90,30];
RMIN=[1,2];  two ellipsoid minor axis radius 
RMAX=[3,5];  two ellipsoid major axis radius.
for a = 1:2  display ellipsoid
        [x, y, z] = ellipsoid(X(a),Y(a),Z(a),RMIN(a),RMAX(a),3.25,30);
S = surfl(x, y, z);
    rotate(S,[0,0,1],ROTATIONANGLE(a))
colormap copper
axis equal
xlabel('X')
ylabel('Y')`enter code here
zlabel('Z')
check the boolean condition 
hold on
end

Answer 1

据我所知，Matlab中没有这样的功能。我会使用Monte Carlo method来找到交叉点的近似体积。

您需要从某个框中生成随机点，并检查它们是否属于交叉点。你可以算出有多少分落入交叉点。知道了盒子的体积，落入交叉点的点数以及生成点的总数，就可以计算出所需的体积。

在图片上你可以看到交叉点内的红点：

找到一些足以解决问题的迭代非常重要。我通过计算一个小椭圆体的体积来验证该方法，这个椭球体放在一个较大的体内。在这种情况下，交点是一个小椭球本身，其体积可以通过分析找到。

您可以在下一张图片上看到相对误差作为迭代次数的函数：

以下是代码：

该代码仅适用于围绕Z轴的旋转

X=[0,2];   %two ellipsoid x coordinates
Y=[0,2];   %two ellipsoid y coordinates
Z=[0,2];   %two ellipsoid z coordinates
ROTATIONANGLE=[90,30];
AXIS_A=[1,2];  %two ellipsoid minor axis radius 
AXIS_B=[3,5];  %two ellipsoid major axis radius
AXIS_C=[3.25,3.25];  %two ellipsoid major axis radius

ranges = zeros(3, 2);

do_plot = 1; %either plot ellipsoids or not

step_number = 100000;

for i = 1:2  %display ellipsoid

    if (do_plot == 1)
        [x, y, z] = ellipsoid(X(i),Y(i),Z(i),AXIS_A(i),AXIS_B(i),AXIS_C(i),20);
        S = surf(x, y, z);
        alpha(.1);
        rotate(S,[0,0,1], ROTATIONANGLE(i), [X(i),Y(i),Z(i)]);
    end

    %calculate the ranges for the simulation box
    ranges(1, 1) = min(ranges(1, 1), X(i) -   max(AXIS_A(i), AXIS_B(i))  );
    ranges(1, 2) = max(ranges(1, 2), X(i) +   max(AXIS_A(i), AXIS_B(i))  );

    ranges(2, 1) = min(ranges(2, 1), Y(i) -   max(AXIS_A(i), AXIS_B(i))  );
    ranges(2, 2) = max(ranges(2, 2), Y(i) +   max(AXIS_A(i), AXIS_B(i))  );

    ranges(3, 1) = min(ranges(3, 1), Z(i) - AXIS_C(i));
    ranges(3, 2) = max(ranges(3, 2), Z(i) + AXIS_C(i));

    if (do_plot == 1)
        hold on;
    end
end

counter = 0; %how many points targeted the intersection

for i = 1:step_number

    R = rand(3, 1).*(ranges(:, 2) - ranges(:, 1)) + ranges(:, 1); %a random point

    n = 1;
    val = calc_ellipsoid( R(1), R(2), R(3), X(n),Y(n),Z(n),AXIS_A(n),AXIS_B(n),AXIS_C(n),ROTATIONANGLE(n)*pi/180);

    if (val <= 1.0)
        n = 2;
        val = calc_ellipsoid( R(1), R(2), R(3), X(n),Y(n),Z(n),AXIS_A(n),AXIS_B(n),AXIS_C(n),ROTATIONANGLE(n)*pi/180);

        if (val <= 1.0)

            if (do_plot == 1)
                plot3(R(1), R(2), R(3), 'or', 'MarkerSize', 1, 'MarkerFaceColor','r');
            end

            counter = counter + 1;
        end
    end

end

cube_vol = 1; %the volume of the simulation box
for i=1:3
    cube_vol = cube_vol * (ranges(i, 2) - ranges(i, 1));
end

%approximated volume of the intersection
section_vol = cube_vol * counter / step_number;

display(['Cube volume: ', num2str(cube_vol)]);
display(['Targeted points: ', num2str(counter), ' from ', num2str(step_number)]);
display(['Section volume: ', num2str(section_vol)]);

if (do_plot == 1)
    hold off;
end

查找点是否属于椭圆体的函数（如果val是＆lt; = 1.0），则执行此操作：

function [ val ] = calc_ellipsoid( x, y, z, x0, y0, z0, a, b, c, theta)

    x_cmp = ((x - x0)*cos(theta) + (y - y0)*sin(theta))^2   /(a^2);
    y_cmp = ((x - x0)*sin(theta) - (y - y0)*cos(theta))^2   /(b^2);
    z_cmp = (z - z0)^2 / (c^2);

    val = x_cmp + y_cmp + z_cmp;
end

<强>更新

为了找到共同身体的最大轴，您可以执行以下操作：

保存所有落入数组交叉点的点（代码中为R_arr）
从结果数组中构造一个凸包：

[K, v] = convhull(R_arr(:, 1), R_arr(:, 2), R_arr(:, 3));

它将为您提供一个近似的体积v和用于构建船体的点的索引K
现在您有一个点的子集，它们位于交叉点体的表面上。数组中的所有点都存在多次。让我们摆脱重复：

K = K(:); K = unique(K, 'stable');
现在阵列非常小（上面的示例大约300点），你可以通过它找到最长的距离。我为它编写了函数findAxis。

[A, B, d] = findAxis(R_arr(K, :));

它会返回最远的点A和B，以及它们之间的距离d。
现在，当您知道这两个点时，您可以定义属于轴的另一个点。使用函数calc_coord：

C = calc_coord(A, B, ranges(3, 1));

D = calc_coord(A, B, ranges(3, 2));

我使用数组ranges中的值来指定轴的z - 坐标。
绘制轴！

对于上面的例子，情节如下所示：

凸包的体积为5.933。蒙特卡罗方法给了我们6.1889，因此结果非常接近。

最长轴的长度为4.3264。

为了获得体积近似更好的直觉，我计算了相对误差：

正如您所看到的，即使经过几次迭代，蒙特卡罗方法也能提供更好的准确性。

以下是具有轴计算功能的更新代码：

X=[0,2];   %two ellipsoid x coordinates
Y=[0,2];   %two ellipsoid y coordinates
Z=[0,2];   %two ellipsoid z coordinates
ROTATIONANGLE=[90,30];
AXIS_A=[1,2];  %two ellipsoid minor axis radius 
AXIS_B=[3,5];  %two ellipsoid major axis radius
AXIS_C=[3.25,13.25];  %two ellipsoid major axis radius

ranges = zeros(3, 2);

do_plot = 1; %either plot ellipsoids or not

step_number = 1000000;

for i = 1:2  %display ellipsoid

    if (do_plot == 1)
        [x, y, z] = ellipsoid(X(i),Y(i),Z(i),AXIS_A(i),AXIS_B(i),AXIS_C(i),20);
        S = surf(x, y, z);
        alpha(.05);
        rotate(S,[0,0,1], ROTATIONANGLE(i), [X(i),Y(i),Z(i)]);
    end

    %calculate the ranges for the simulation box
    ranges(1, 1) = min(ranges(1, 1), X(i) -   max(AXIS_A(i), AXIS_B(i))  );
    ranges(1, 2) = max(ranges(1, 2), X(i) +   max(AXIS_A(i), AXIS_B(i))  );

    ranges(2, 1) = min(ranges(2, 1), Y(i) -   max(AXIS_A(i), AXIS_B(i))  );
    ranges(2, 2) = max(ranges(2, 2), Y(i) +   max(AXIS_A(i), AXIS_B(i))  );

    ranges(3, 1) = min(ranges(3, 1), Z(i) - AXIS_C(i));
    ranges(3, 2) = max(ranges(3, 2), Z(i) + AXIS_C(i));

    if (do_plot == 1)
        hold on;
    end
end

counter = 0; %how many points targeted the intersection

R_arr = [];

for i = 1:step_number

    R = rand(3, 1).*(ranges(:, 2) - ranges(:, 1)) + ranges(:, 1); %a random point

    n = 1;
    val = calc_ellipsoid( R(1), R(2), R(3), X(n),Y(n),Z(n),AXIS_A(n),AXIS_B(n),AXIS_C(n),ROTATIONANGLE(n)*pi/180);

    if (val <= 1.0)
        n = 2;
        val = calc_ellipsoid( R(1), R(2), R(3), X(n),Y(n),Z(n),AXIS_A(n),AXIS_B(n),AXIS_C(n),ROTATIONANGLE(n)*pi/180);

        if (val <= 1.0)

            if (do_plot == 1)
                %plot3(R(1), R(2), R(3), 'or', 'MarkerSize', 1, 'MarkerFaceColor','r');
            end

            counter = counter + 1;

            R_arr = [R_arr; R'];
        end
    end

end

cube_vol = 1; %the volume of the simulation box
for i=1:3
    cube_vol = cube_vol * (ranges(i, 2) - ranges(i, 1));
end

%approximated volume of the intersection
section_vol = cube_vol * counter / step_number;

display(['Cube volume: ', num2str(cube_vol)]);
display(['Targeted points: ', num2str(counter), ' from ', num2str(step_number)]);
display(['Section volume: ', num2str(section_vol)]);

if (counter > 0)

    %hull
    [K, v] = convhull(R_arr(:, 1), R_arr(:, 2), R_arr(:, 3));


    display(['Approximated volume: ', num2str(v)]);
    trisurf(K, R_arr(:, 1), R_arr(:, 2), R_arr(:, 3), 'FaceColor','cyan')
    hold on;
    axis equal

    K = K(:);
    K = unique(K, 'stable');

    [A, B, d] = findAxis(R_arr(K, :));

    plot3(A(1, 1), A(1, 2), A(1, 3), 'or', 'MarkerSize', 10, 'MarkerFaceColor','r');
    plot3(B(1, 1), B(1, 2), B(1, 3), 'or', 'MarkerSize', 10, 'MarkerFaceColor','r');

    %axis
    C = calc_coord(A, B, ranges(3, 1));
    D = calc_coord(A, B, ranges(3, 2));

    p = [C; D];

    plot3(p(:, 1), p(:, 2), p(:, 3), '--g', 'LineWidth', 3);

else
    display('There is no intersection!');
end

hold off;

功能：

function [ C] = calc_coord( A, B, z)
    x1 = A(1, 1);
    x2 = B(1, 1);

    y1 = A(1, 2);
    y2 = B(1, 2);

    z1 = A(1, 3);
    z2 = B(1, 3);

    y = (y2 - y1)*(z - z1)/(z2 - z1) + y1;
    x = (x2 - x1)*(y - y1)/(y2 - y1) + x1;

    C = [x, y, z];
end

function [ A, B, d ] = findAxis( point_arr )

    A_ind = 0; B_ind = 0; d = -1;

    n = size(point_arr, 1);
    i = 1;

    while (i < n)

        for j=(i+1):n

            cur_d = norm(point_arr(i, :) - point_arr(j, :));

            if (cur_d > d)
                d = cur_d;
                A_ind = i;
                B_ind = j;
            end
        end

        i = i + 1;
    end

    A = point_arr(A_ind, :); 
    B = point_arr(B_ind, :);
end

<强>说明

功能calc_ellipsoid

该函数基于椭球方程（椭圆体围绕z轴旋转角度theta）：

该函数定义给定点是否位于椭球内部

%the function calculates a value for some given point, which shows if the
%point is inside of the ellipsoid or not.
%for a point to be inside, the value has to be <=1

function [ val ] = calc_ellipsoid( x, y, z, x0, y0, z0, a, b, c, theta)

    %x, y, z - coordinates of the point to be checked
    %x0, y0, z0 - center of the ellipsoid
    %a, b, c - axes of the ellipsoid
    %theta - angle of the rotation about the Z-axis

    x_cmp = ((x - x0)*cos(theta) + (y - y0)*sin(theta))^2   /(a^2);
    y_cmp = ((x - x0)*sin(theta) - (y - y0)*cos(theta))^2   /(b^2);
    z_cmp = (z - z0)^2 / (c^2);

    val = x_cmp + y_cmp + z_cmp;
end

如果您需要使用其他形状，您需要找到这样的方程式，以检查您的点是否属于形状。椭球方程是最简单的方法之一。

功能findAxis

该函数从点云中搜索两个点，它们之间的距离最长。在这里我们假设交叉点形状的轴经过这两个点（这是一个非常粗略的假设，但它有效）。

%the function is given an array of all points which lay on the constructed
%body. we assume, that the axis of the body is the longest line, connecting
%some two points of this array. the function goes through all the points
%and calculates the euclidian distance between them. 
%the function returns points A and B, and the distance between them 

function [ A, B, d ] = findAxis( point_arr )

    %point_arr - points from the bodies surface
    %A, B - end points of the found exis
    %d - distance between A and B

    A_ind = 0; B_ind = 0; d = -1;

    n = size(point_arr, 1);
    i = 1;

    while (i < n)
        for j=(i+1):n

            cur_d = norm(point_arr(i, :) - point_arr(j, :));

            if (cur_d > d)
                d = cur_d;
                A_ind = i;
                B_ind = j;
            end
        end

        i = i + 1;
    end

    A = point_arr(A_ind, :); 
    B = point_arr(B_ind, :);
end

功能calc_coord

该功能基于3D空间中的线方程：

该等式定义了一条线，该线穿过坐标为(x1, y1, z1)和(x2, y2, z2)的2个点。我们使用这个函数两个在交点外形找到另外两个点，以便绘制轴更具说明性。如果你不打算拍出漂亮的照片，你真的不需要这个功能。

%the function is given two points A and B, which were found in the
%'findAxis' function. we want to find a new point C, which lay on the axis,
%and have z-coordinate equal to 'z'. it is done merely to visualize the
%axis in a better way.

function [C] = calc_coord( A, B, z)
    x1 = A(1, 1);
    x2 = B(1, 1);

    y1 = A(1, 2);
    y2 = B(1, 2);

    z1 = A(1, 3);
    z2 = B(1, 3);

    y = (y2 - y1)*(z - z1)/(z2 - z1) + y1;
    x = (x2 - x1)*(y - y1)/(y2 - y1) + x1;

    C = [x, y, z];
end

椭圆体的交点体积

1 个答案: