如何绘制经过四点的圆（球体）？

2024-5-24 • tag-icon

如何绘制经过四点的圆（球体）？

我正在尝试绘制一个通过四个点的圆（球体），B, C, E, F就像这张图片一样

我尝试过tikz-3dplot我的代码

\documentclass[border=3mm,12pt]{standalone}
\usepackage{fouriernc}
\usepackage{tikz,tikz-3dplot} 
\usepackage{tkz-euclide}
\usetkzobj{all}

 \usetikzlibrary{intersections,calc,backgrounds}

 \begin{document}

 \tdplotsetmaincoords{70}{110}
  %\tdplotsetmaincoords{80}{100}
 \begin{tikzpicture}[tdplot_main_coords,scale=1.5]
 \pgfmathsetmacro\a{3}
 \pgfmathsetmacro\b{4}
 \pgfmathsetmacro\h{5}

 % definitions
 \path
 coordinate(A) at (0,0,0)
coordinate (B) at (\a,0,0)
coordinate (C) at (0,\b,0)                           
coordinate (S) at (0,0,\h)                
coordinate (E) at  ({\a*\h^2/(\a*\a + \h*\h)},0,{(\a*\a*\h)/(\a*\a + \h*\h)})
coordinate (F) at  (0,{(\b*\h*\h)/(\b*\b + \h*\h)},{(\b*\b*\h)/(\b*\b + \h*\h)});
 \draw[dashed,thick]
       (A) -- (B)  (A) -- (C)  (A) -- (E)  (S)--(A)  (F)--(A);
       \draw[thick]
       (S) -- (B) -- (C) -- cycle;
       \draw[thick] 
       (F) -- (B) (C)--(E) (F)--(E);
\tkzMarkRightAngle(S,E,A);
\tkzMarkRightAngle(S,F,A);

 \foreach \point/\position in {A/below,B/left,C/below,S/above,E/left,F/above}
 {
   \fill (\point) circle (.8pt);
   \node[\position=3pt] at (\point) {$\point$};
 }

 \end{tikzpicture}
 \end{document}

并得到

如何绘制一个通过四个点的圆（球体）B, C, E, F？

答案1

圆由 3 个点确定。当然，球体至少需要 4 个边界点才能确定。但是，球体的投影（即圆）不一定会通过这些点的投影。（实际上，如果球体由这些点唯一确定，边界圆（即球体在屏幕坐标上的投影）将永远不会通过这些点的所有投影，因为要做到这一点，这些点需要位于一个平面上，但这样它们就不再唯一地确定圆了。）

这里展示了两种构造经过某些点的圆的方法：

虚线圆穿过F、E和C。它根据此要求固定。因此，它会B略微偏离。
红色虚线圆穿过BC这些点的中点和。它与和的误差F很小E。

\documentclass[border=3mm,12pt]{standalone}
\usepackage{fouriernc}
\usepackage{tikz,tikz-3dplot} 
\usepackage{tkz-euclide}
\usetkzobj{all}
\usetikzlibrary{calc,through}
\tikzset{circle through 3 points/.style n args={3}{%
insert path={let    \p1=($(#1)!0.5!(#2)$),
                    \p2=($(#1)!0.5!(#3)$),
                    \p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
                    \p4=($(#1)!0.5!(#3)!1!90:(#3)$),
                    \p5=(intersection of \p1--\p3 and \p2--\p4)
                    in },
at={(\p5)},
circle through= {(#1)}
}}

 \usetikzlibrary{intersections,calc,backgrounds}

 \begin{document}

 \tdplotsetmaincoords{70}{110}
  %\tdplotsetmaincoords{80}{100}
 \begin{tikzpicture}[tdplot_main_coords,scale=1.5]
 \pgfmathsetmacro\a{3}
 \pgfmathsetmacro\b{4}
 \pgfmathsetmacro\h{5}

 % definitions
 \path
 coordinate(A) at (0,0,0)
coordinate (B) at (\a,0,0)
coordinate (C) at (0,\b,0)                           
coordinate (S) at (0,0,\h)                
coordinate (E) at  ({\a*\h^2/(\a*\a + \h*\h)},0,{(\a*\a*\h)/(\a*\a + \h*\h)})
coordinate (F) at  (0,{(\b*\h*\h)/(\b*\b + \h*\h)},{(\b*\b*\h)/(\b*\b + \h*\h)});
 \draw[dashed,thick]
       (A) -- (B)  (A) -- (C)  (A) -- (E)  (S)--(A)  (F)--(A);
       \draw[thick]
       (S) -- (B) -- (C) -- cycle;
       \draw[thick] 
       (F) -- (B) (C)--(E) (F)--(E);
\tkzMarkRightAngle(S,E,A);
\tkzMarkRightAngle(S,F,A);

 \foreach \point/\position in {A/below,B/left,C/below,S/above,E/left,F/above}
 {
   \fill (\point) circle (.8pt);
   \node[\position=3pt] at (\point) {$\point$};
 }
  \node[circle through 3 points={F}{E}{C},draw=blue,dotted]{};
  \draw[red,dashed] 
  let \p1=($(B)-(C)$), \n1={veclen(\x1,\y1)/2} in ($(B)!0.5!(C)$) circle (\n1);
\end{tikzpicture}
\end{document}

也可以构造球体。但是，如上所述，其边界可能不会经过任何点。

附录：在您的设置中，四个点不能确定一个唯一的球体，因为它们都位于一个平面上。使用 Mathematica，我能够将其表示F为线性组合

 F = x B + y C + z E

在哪里

因此，在这种设置下，不可能绘制出独特的球体。

答案2

非常感谢marmot纠正了这个圆圈BCEF 这里。

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,3dtools}
\begin{document}
    \begin{tikzpicture}[line cap=round,line join=round,
        3d/install view={phi=110,theta=70},
        declare function={a=3;b=4;h=5;},c/.style={circle,fill,inner sep=1pt}]
        \path
        (0,0,0) coordinate (A)  
        (a,0,0) coordinate (B)  
        (0,b,0) coordinate (C)                             
        (0,0,h)  coordinate (S)
        %(a/2,b/2,0)  coordinate (M)
        ;
        
        \path[3d/circumsphere center={A={(A)},B={(B)},C={(C)},D={(S)}}]
        coordinate (I);
        \pgfmathsetmacro{\myR}{sqrt(TD("(I)-(A)o(I)-(A)"))} ;
        \draw[3d/screen coords] (I) circle[radius=\myR];
        \path[3d/circumcircle center={A={(A)},B={(B)},C={(C)}}] coordinate (O);
        \path pic{3d/circle on sphere={R=\myR,C={(I)}, P={(O)}}};  
        \path[3d/line through={(S) and (B) named lSB}];
        \path[3d/line through={(S) and (C) named lSC}];
        \path[3d/project={(A) on lSB}] coordinate (E);
        \path[3d/project={(A) on lSC}] coordinate (F);
        
        %\path[3d/circumcircle center={A={(E)},B={(B)},C={(C)}}] coordinate (M);
        \path pic[draw=none]{3d circle through 3 points={%
                A={(B)},B={(E)},C={(F)},center name=M}};
        \pgfmathsetmacro{\myr}{tddistance("(B)","(M)")}%     
        \pgfmathsetmacro{\myMC}{TD("(C)-(M)")}
        \pgfmathtruncatemacro{\itest}{screendepth(\myMC)<0?0:1}
        \ifnum\itest=0
        \begin{scope}
            \clip[3d/screen coords] (I) circle[radius=\myR];
            \path pic[3d/hidden]{3d circle through 3 points={%
                    A={(B)},B={(E)},C={(F)},center name=M}};
        \end{scope}
        \begin{scope}
            \clip[3d/screen coords,even odd clip] (I) circle[radius=\myR]
            [generous outside path];
            \path pic[3d/visible]{3d circle through 3 points={%
                    A={(B)},B={(E)},C={(F)},center name=M}};
        \end{scope}
        \else
        \tikzset{3d/define orthonormal dreibein={A={(B)},B={(C)},C={(F)}}}
        \begin{scope}[x={(ex)},y={(ey)},z={(ez)},shift={(M)}]
            \draw[3d/hidden] (B) arc[start angle=180,end angle=0,radius=\myr];
            \draw[3d/visible] (B) arc[start angle=180,end angle=360,radius=\myr];
        \end{scope}
        \fi
        \path foreach \p/\g in {A/90,B/-90,C/0,S/90,E/180,F/90,I/0,M/-90}
        {(\p)node[c]{}+(\g:2.5mm) node{$\p$}};
        
        \draw[3d/hidden] (S) -- (A) (S) --(B) (S) -- (C) (A) -- (B) -- (C) -- cycle (A) -- (E) (A) -- (F);
        \end{tikzpicture}
\end{document}

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