从3d矢量绘制3d等高线图 [英] drawing 3d contour plot from 3d vector

问题描述

我想为3D数据绘制轮廓图.

I want to draw a contour plot for 3D data.

我要在x，y，z方向上施加力，因此我想为此绘制轮廓3

I have a force in x,y,z directions I want to plot the contour3 for that

Fy和Fz的Fx的尺寸= 21x21X21

the dimensions of the Fx = 21x21X21 same for Fy and Fz

我发现力= f * vector(x，y，z) 然后

I am finding force = f*vector(x,y,z) Then

Fx(x,y,z) = force(1)
Fy(x,y,z) = force(2)
Fz(x,y,z) = force(3)

我做了以下事情，但是它对我不起作用?为什么以及如何绘制该图

I did the following but it is not working with me ?? why and how can I plot that

FS = sqrt(Fx.^2 + Fy.^2 + Fz.^2);

x = -10:1:10;
[X,Y] = meshgrid(x); 
for i=1:length(FS)

   for j = 1:length(FS)
       for k=1:length(FS)
        contour3(X,Y,FS(i,j,k),10)
        hold on
      end
   end
end

这是我遇到的错误 使用轮廓3时出错(第129行) 当Z是向量时，X和Y也必须是向量.

This is the error I am getting Error using contour3 (line 129) When Z is a vector, X and Y must also be vectors.

推荐答案

我不确定该解决方案是否是您想要的.

I'm not sure if this solution is what you want.

您的问题是contour和contour3是表示2D对象中标量场的图.请注意，球是2D对象-每个点都由角度theta和phi定义-即使它是空间"中的对象，也不是平面"中的对象.

Your problem is that contour and contour3 are plots to represent scalar field in 2D objects. Note that ball is 2D object - every single point is defined by angles theta and phi - even it is an object in "space" not in "plane".

用于表示矢量场的有quiver，quiver3，streamslice和streamline函数.

For representation of vector fields there is quiver, quiver3, streamslice and streamline functions.

如果要使用轮廓图，则必须将数据从矢量场转换为标量场.因此，必须将格式为 F = f(x，y，z)的数据转换为H = f(x，y)的形式.在那种情况下，H是MxN矩阵，x和y分别是Mx1和Nx1向量.然后contour3(x,y,H)将起作用，从而生成所谓的3D图形.

If you want to use contour plot, you have to transform your data from vector field to scalar field. So your data in form F = f(x,y,z) must be transformed to form of H = f(x,y). In that case H is MxN matrix, x and y are Mx1 and Nx1 vectors, respectively. Then contour3(x,y,H) will work resulting in so-called 3D graph.

如果依赖向量字段，则必须指定6个向量/矩阵，它们的大小与x，y，z坐标和Fx，Fy，Fz向量值的大小相同. 在那种情况下，quiver3(x，y，z，Fx，Fy，Fz)将起作用，从而生成6D图形. 明智地使用它！

If you rely on vector field You have to specify 6 vectors/matrices of the same size of corresponding x, y, z coordinates and Fx, Fy, Fz vector values. In that case quiver3(x,y,z,Fx,Fy,Fz) will work resulting in 6D graph. Use it wisely!

当我评论Ander的答案时，您可以使用色彩空间获取更多尺寸，因此您可以创建5D或理论上为6D，因为您有x，y，z坐标用于位置，而R，G，B坐标用于价值观.我建议为5D图形使用静态(x，y，R，G，B)，为6D使用动画(x，y，t，R，G，B). 明智地使用它！

As I comment the Ander's answer, you can use colourspace to get more dimensions, so You can create 5D or, theoretically, 6D, because you have x, y, z coordinates for position and R, G, B coordinates for the values. I'd recommend using static (x,y,R,G,B) for 5D graph and animated (x,y,t,R,G,B) for 6D. Use it wisely!

在示例中，我显示了上面提到的所有方法.我选择了重力场并计算出比重心低0.25个单位的平面.

In the example I show all approaches mentioned above. i chose gravity field and calculate the plane 0.25 units below the centre of gravity.

假设在极坐标中定义的力场为 F =- r /r ^ 3; F = 1/r ^ 2. 这里 x 和 y 都在-1; 1范围内，并且大小相同N. F 是MxMx3矩阵，其中 F (ii，jj)是对应于 x (ii)和 y (jj). 矩阵 H (ii，jj)是 F (ii，jj)和 X ， Y 和 Z 是坐标矩阵.

Assume a force field defined in polar coordinates as F=-r/r^3; F=1/r^2. Here both x and yare in range of -1;1 and same size N. F is the MxMx3 matrix where F(ii,jj) is force vector corresponding to x(ii) and y(jj). Matrix H(ii,jj) is the norm of F(ii,jj) and X, Y and Z are matrices of coordinates.

最后一个命令可确保F值在(-1; 1)范围内. F./2+0.5移动F的值，使其适合RGB范围.颜色含义为:

Last command ensures that F values are in (-1;1) range. The F./2+0.5 moves values of F so they fit into RGB range. The colour meaning will be:

• 黑色代表(-1，-1，-1)
• 红色代表(1，-1，-1)
• 灰色代表(0,0,0)

取消注释要查看的绘图类型.对于颤抖，请使用resolution为0.1，对于其他情况，请使用0.01.

Un-comment the type of plot You want to see. For quiver use resolution of 0.1, for other cases use 0.01.

clear all,close all
% Definition of coordinates 
resolution=0.1;
x=-1:resolution:1;
y=x;
z=-.25;

%definition of matrices
F=zeros([max(size(x))*[1 1],3]);    % matrix of the force
X=zeros(max(size(x))*[1 1]);        % X coordinates for quiver3
Y=X;                                % Y coordinates for quiver3
Z=X+z;                              % Z coordinates for quiver3                              

% Force F in polar coordinates
% F=-1/r^2
% spherical -> cartesian transformation

for ii=1:max(size(x))
  for jj=1:max(size(y))
    % temporary variables for transformations
    xyz=sqrt(x(ii)^2+y(jj)^2+z^2);
    xy= sqrt(x(ii)^2+y(jj)^2);
    sinarc=sin(acos(z/xyz));

    %filling the quiver3 matrices
    X(ii,jj)=x(ii);
    Y(ii,jj)=y(jj);

    F(ii,jj,3)=-z/xyz^2;
    if xy~=0 % 0/0 error for x=y=0
       F(ii,jj,2)=-y(jj)/xyz/xy*sinarc;
       F(ii,jj,1)=-x(ii)/xyz/xy*sinarc;
    end
    H(ii,jj)=sqrt(F(ii,jj,1)^2+F(ii,jj,2)^2+F(ii,jj,3)^2);
  end
end

F=F./max(max(max(F)));
% quiver3(X,Y,Z,F(:,:,1),F(:,:,2),F(:,:,3));
% image(x,y,F./2+0.5),set(gca,'ydir','normal');
% surf(x,y,Z,F./2+.5,'linestyle','none')
% surf(x,y,H,'linestyle','none')
surfc(x,y,H,'linestyle','none')
% contour3(x,y,H,15)

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