一个月前我在群里给出的方案是：

• 把全路径划分为多个子路径，预先保存中间顶点
• “均匀”的添加箭头，只要保证子路径使用middle arrow=0.5的默认值即可
• 利用bengding,ext.arrows,ext.arrows-plus库提供的\arrow**{ext_Centered Stealth[bend]}}来准确的控制箭头的位置和正确的弯曲

\documentclass[tikz,border=3pt]{standalone}
\usepackage{lete-sans-math}
\usetikzlibrary{bending,arrows.meta,decorations.markings,ext.arrows,ext.arrows-plus}
\tikzset{
    middle arrow/.style={
        semithick,postaction={decorate},
        decoration={markings,mark=at position #1 with {\arrow**{ext_Centered Stealth[bend]}}}
    },middle arrow/.default={0.5}
}
\begin{document}
\begin{tikzpicture}[>=Stealth]
    \def\ang{17.5}\def\R{2.5}
    \pgfmathsetmacro{\r}{2*\R*sin(\ang/2)}
    \draw[thick,fill=gray!30]
        ({\R-\r},0) coordinate (A)
        arc (180:{90+\ang/2}:\r) coordinate (B)
        arc (\ang:{90-\ang}:\R) coordinate (C) 
        arc ({-\ang/2}:{-180+\ang/2}:\r) coordinate (D) 
        arc ({90+\ang}:{180-\ang}:\R) coordinate (E) 
        arc ({90-\ang/2}:0:\r) coordinate (F) 
        -- cycle;
    \path[middle arrow=0.4] (F) -- (A);
    \path[middle arrow] (A) arc (180:{90+\ang/2}:\r);
    \path[middle arrow] (B) arc (\ang:{90-\ang}:\R);
    \path[middle arrow=0.6] (C) arc ({-\ang/2}:{-180+\ang/2}:\r);
    \path[middle arrow] (D) arc ({90+\ang}:{180-\ang}:\R);
    \path[middle arrow] (E) arc ({90-\ang/2}:0:\r);
    \fill[radius=2pt] 
        (-\R,0) circle[] node[below] {$(-1,0)$}
        (+\R,0) circle[] node[below] {$(1,0)$}
        (0,+\R) circle[] node[above] {$(0,1)$}
        ;
    \draw[->] (-3,0) -- (3,0) node[right] {$\Re$};
    \draw[->] (0,-.5) node[above left] {$O$} -- (0,3.5) node[above] {$\Im$};
\end{tikzpicture}
\end{document}

不过这样其实似乎涉嫌重复绘制各段路径...

    \path[middle arrow] (A) arc (180:{90+\ang/2}:\r);
    \path[middle arrow] (B) arc (\ang:{90-\ang}:\R);
    \path[middle arrow=0.6] (C) arc ({-\ang/2}:{-180+\ang/2}:\r);
    \path[middle arrow] (D) arc ({90+\ang}:{180-\ang}:\R);
    \path[middle arrow] (E) arc ({90-\ang/2}:0:\r);

放大看看细节：

不知道是否有更好的方案...

这里的难点在于「s0=cos(pi/n)」仅仅是一个特例，如何找到符合要求的小椭圆，是一个看上去复杂得多的数学问题。对于特例：

\documentclass{standalone}
\usepackage{luadraw}
\usepackage[svgnames]{xcolor}
\usepackage{fourier-otf}
\begin{document}
    \begin{luadraw}{name=poncelet_ellipe}
    local ld = luadraw
    local cpx = ld.cpx
    local g = ld.graph:new{window={-5,5,-4,4}}
    local i, Zp, arg = cpx.I, cpx.Zp, cpx.arg
    local sin, cos, pi = math.sin, math.cos, math.pi
    local a, b, t = 4, 2.5, pi/3
    local n = 5
    local s0 = cos(pi/n)
    local U = Zp(1,t)
    local V = ld.rotate(U,360/n)
    g:Savematrix()
    g:Composematrix({0,a,b*i}) --
    g:Dcircle({0,1},"thick");g:Dcircle({0,s0})
    g:Dlabeldot("$P$",U,{pos="NE"})
    for k = 1, n do
        g:Dseg({U,V},1,"line cap=round,magenta,semithick")
        g:Rotate(360/n)
    end
    g:Restorematrix()
    g:Show()
\end{luadraw}
\end{document}

这是一个半成品，但我不打算继续下去了，有以下几个缺陷：

• 借助chemarr提供的\xrightleftharpoons箭头并非良策
• 上下的字符区域和箭头的距离不完全相同，修改很困难，参见这里
• 需要反复借助\rotatebox[origin=c]{90}{...}
• 左右间距调整纯靠随缘\;以及\,并不准确，且难以和\Big\updownarrow做到精确同高
• 红温😡

\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage{graphicx}
\DeclareMathOperator{\Inv}{Inv}
\DeclareMathOperator{\Gal}{Gal}
\usepackage{chemarr}%<---
\usepackage{lua-visual-debug}
\begin{document}

$X \xrightleftharpoons[o]{o} Y$

$X \xrightleftharpoons[{\;ooo}]{{ooo\;}} Y$

$X \xrightleftharpoons[{\rotatebox[origin=c]{-90}{$\Gal$}}]{{\rotatebox[origin=c]{-90}{$\,\Inv\,$}}} Y$

$X \xrightleftharpoons[\quad{\rotatebox[origin=c]{-90}{$\,\Gal\,$}}\quad]{\quad{\rotatebox[origin=c]{-90}{$\,\Inv\,$}}\quad} Y$

\bigskip

$X \rotatebox[origin=c]{90}{\xrightleftharpoons[\quad{\rotatebox[origin=c]{-90}{$\,\Gal\,$}}\quad]{\quad{\rotatebox[origin=c]{-90}{$\,\Inv\,$}}\quad}} Y$

\end{document}

下面是完整的半成品代码：

\documentclass{article}
\usepackage{amsmath,amssymb}
\usepackage{graphicx}
\DeclareMathOperator{\Inv}{Inv}
\DeclareMathOperator{\Gal}{Gal}
\usepackage{chemarr}%<---

\begin{document}

\begin{equation*}
\begin{array}{*{3}{c}}
    \Inv{H_2} & & \Inv{H_1}
    \\[1mm]
    \rotatebox{90}{$=$} & & \rotatebox{90}{$=$}
    \\[1mm]
    E_2 & \supseteq & E_1
    \\[2mm]
    \;\rotatebox[origin=c]{90}{$\xrightleftharpoons[\;\,{\rotatebox[origin=c]{-90}{$\;\Gal\;$}}\,\;]{{\rotatebox[origin=c]{-90}{$\;\Inv\;$}}}$} & & \Big\updownarrow
    \\[2mm]
    H_2 & \subseteq & H_1
    \\[1mm]
    \rotatebox{90}{$=$} & & \rotatebox{90}{$=$}
    \\[1mm]
    \Gal(K/E_2) & & \Gal(K/E_1)
\end{array}
\end{equation*}

\end{document}

最佳实践还应当是tikzcd...有更准确的位置设置和箭头支持:-[

Update 20261013

一个基于gradbars宏包的优雅方案：

\documentclass{article}
\usepackage[ebgaramond]{fontsetup}
\usepackage[landscape]{geometry}
\usepackage{tabularray}
\usepackage{gradbars}
\pagestyle{empty}
\gradbarssetup{
  preset=presentation,
  height=18pt,width=3cm,
  row padding=0.5pt,
  min=-9.5,max=9.5,
  color=green!40,
  negative color=red!40,
  track color=white,
  text color=black,
  precision=1
}
\newcommand{\mygradbar}[1]{%
  \begin{tikzpicture}[baseline=(b.base)]
    \node[inner sep=0pt,outer sep=0pt] (b) {\gradbar[label=none]{#1}};
    \node[inner sep=0pt] at (b.center) {#1};
  \end{tikzpicture}}

\begin{document}
\begin{center}
\begin{tblr}{
  colspec={*{5}{Q[c,m]}},
  hlines,vlines,stretch=0,
  rows={ht=18pt,rowsep=0.5pt},
  row{1}={bg=lightgray,font=\bfseries\large},
  row{2-Z}={cmd=\mygradbar}
}
  Item1 & Item2 & Item3 & Item4 & Item5 \\
  0.5 & 1.5 & 2.5 & 3.5 & 4.5 \\
  -1 & -2 & -3 & -4 & -5 \\
  -6.5 & -5.4 & -4.3 & -3.2 & -2.1 \\
  6.5 & -5.4 & 0 & -3.2 & 2.1 \\
  -0.5 & -1.5 & -8.5 & -3.5 & -4.5 \\
  -5.5 & -9.5 & -2.5 & -7.5 & -6.5 \\
  -1 & 2 & -3 & 4 & -5 \\
  -3.1 & -2.1 & -1.1 & 1 & -1.1 \\
  1 & 4 & 2 & 8 & 5 \\
  5.5 & 6.5 & 7.5 & 8.5 & 9.5 \\
\end{tblr}
\end{center}
\end{document}

Update 20161028

其实上面的代码需要手动实现：

\newcommand{\mygradbar}[1]{%
  \begin{tikzpicture}[baseline=(b.base)]
    \node[inner sep=0pt,outer sep=0pt] (b) {\gradbar[label=none]{#1}};
    \node[inner sep=0pt] at (b.center) {#1};
  \end{tikzpicture}}

一点也不优雅，在v1.1.0中，支持label align=cell，详见这里的报告：

\documentclass{article}
\usepackage[ebgaramond]{fontsetup}
\usepackage[landscape]{geometry}
\usepackage{tabularray}
\usepackage{gradbars}
\pagestyle{empty}
\gradbarssetup{
  preset=presentation,
  height=18pt,width=3cm,
  row padding=0.5pt,
  min=-9.5,max=9.5,
  color=green!40,
  negative color=red!40,
  track color=white,
  text color=black,
  precision=1,
  % label overflow=auto, % default
  label align=cell %<--- new feature in v1.1.0
}

\begin{document}
\begin{center}
\begin{tblr}{
  colspec={*{5}{Q[c,m]}},
  hlines,vlines,stretch=0,
  rows={ht=18pt,rowsep=0.5pt},
  row{1}={bg=lightgray,font=\bfseries\large},
  row{2-Z}={cmd=\gradbar}
}
  Item1 & Item2 & Item3 & Item4 & Item5 \\
  0.5 & 1.5 & 2.5 & 3.5 & 4.5 \\
  -1 & -2 & -3 & -4 & -5 \\
  -6.5 & -5.4 & -4.3 & -3.2 & -2.1 \\
  6.5 & -5.4 & 0 & -3.2 & 2.1 \\
  -0.5 & -1.5 & -8.5 & -3.5 & -4.5 \\
  -5.5 & -9.5 & -2.5 & -7.5 & -6.5 \\
  -1 & 2 & -3 & 4 & -5 \\
  -3.1 & -2.1 & -1.1 & 1 & -1.1 \\
  1 & 4 & 2 & 8 & 5 \\
  5.5 & 6.5 & 7.5 & 8.5 & 9.5 \\
\end{tblr}
\end{center}
\end{document}

有没有更好的方法只画alpha和beta面内的交线,我想可能是裁剪吧,或者计算交线在面内的两个端点.
另外直线的点法式local l={M(0,1,1),vecK}中,每次画直线都是以该点为中心画直线的

这里的交线 local L1,L2 = ld.interPP(P1,Ps),ld.interPP(P2,Ps)获得的直线形式为形如{A,u}，这里的A是交线“a”和“b”的中点，很巧的是，由于我们的刻意构造，在这个视角下，交线的方向恰为vecI所在的方向，同时我们在：

 g:addPlane(P1,{contrast=0,rectangle={vecI,2,4},edge=true,edgecolor="DarkBlue",hidden=true}),
        g:addPlane(P2,{contrast=0,rectangle={vecI,2,4},edge=true,edgecolor="DarkBlue",hidden=true}),

沿着vecI绘制的长度为2个单位，所以前后延长1个单位即可恰好得到两个端点：

g:Dpolyline3d({{O1-vecI,O1+vecI},{O2-vecI,O2+vecI}},"semithick,red")

完整代码：

\documentclass{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}
\usepackage{fourier-otf}
\begin{document}
    \begin{luadraw}{name=intersected_surface_Dscene}
    local ld = luadraw
    local pt3d, cpx = ld.pt3d, ld.cpx
    local M, Mc, Ms = pt3d.M, pt3d.Mc, pt3d.Ms
    local Origin, vecI, vecJ, vecK = pt3d.Origin, pt3d.vecI, pt3d.vecJ, pt3d.vecK
    local g = ld.graph3d:new{
        window3d = {-5,5,-5,5,-5,5},
        window = {-3,3,-2.5,2.5},
        viewdir = {"yz",0.75,40}
    }
    local vecN = ld.rotate3d(vecK,-50,{Origin,vecI})
    local P1,P2 = {0.75*vecK,vecK},{-0.75*vecK,vecK}
    local Ps = {Origin,vecN}
    local L1,L2 = ld.interPP(P1,Ps),ld.interPP(P2,Ps)
    ld.Hiddenlines = true; ld.Hiddenlinestyle="dashed"
    g:Dscene3d(
        g:addPlane(P1,{contrast=0,rectangle={vecI,2,4},edge=true,edgecolor="DarkBlue",hidden=true}),
        g:addPlane(P2,{contrast=0,rectangle={vecI,2,4},edge=true,edgecolor="DarkBlue",hidden=true}),
        g:addPlane(Ps,{contrast=0,rectangle={vecI,2,4.5},edge=true,edgecolor="Orange",hidden=true})
    )
    local O1,_ = table.unpack(L1)
    local O2,_ = table.unpack(L2) -- 
    g:Dpolyline3d({{O1-vecI,O1+vecI},{O2-vecI,O2+vecI}},"semithick,red")
    g:Dlabel3d(
        "$\\gamma$",O2+0.9*vecJ+1.1*vecI-0.6*vecK,{},
        "$\\alpha$",O1-1.1*vecJ+1.1*vecI,{pos="NE",node_options="text=DarkBlue"},
        "$\\beta$",O2-2.25*vecJ+1.1*vecI,{},
        "$a$",O1,{pos="W",node_options="text=red"},
        "$b$",O2,{}
    )
    g:Show()
    \end{luadraw}
\end{document}

需要提供上述代码...不然没有人能回答...

大猜测时代

没有代码其实很难猜测...但要注意的是，对于g:Dline3d方法，并没有scale选项...我在这里提过类似的议题，作者说请改用g:Dseg3d:

\documentclass[]{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}
\usepackage{fourier-otf}
\begin{document}
    \begin{luadraw}{name=Dline_scale}
    local ld = luadraw
    local pt3d, cpx = ld.pt3d, ld.cpx
    local M, Mc, Ms = pt3d.M, pt3d.Mc, pt3d.Ms
    local Origin, vecI, vecJ, vecK = pt3d.Origin, pt3d.vecI, pt3d.vecJ, pt3d.vecK
    local g = ld.graph3d:new{
        window3d = {-5,5,-5,5,-5,5},
        window = {-2,5,-1,4},
        viewdir = {"yz",0.75,40}
    }
    g:Dpolyline3d({{Origin,2*vecI},{Origin,4*vecJ},{Origin,4*vecK}},"semithick,-Stealth")
    g:Dline3d({3*vecK,vecJ},"teal,thick")
    g:Dseg3d({2*vecK-vecJ,2*vecK+vecJ},1,"magenta,thick")
    g:Dseg3d({vecK-vecJ,vecK+vecJ},0.5,"cyan,thick")
    g:Show()
    \end{luadraw}
\end{document}

比较遵照你的方案（基于g:Dscene3d）的方法：

\documentclass{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}
\usepackage{fourier-otf}
\begin{document}
    \begin{luadraw}{name=intersected_surface_Dscene}
    local ld = luadraw
    local pt3d, cpx = ld.pt3d, ld.cpx
    local M, Mc, Ms = pt3d.M, pt3d.Mc, pt3d.Ms
    local Origin, vecI, vecJ, vecK = pt3d.Origin, pt3d.vecI, pt3d.vecJ, pt3d.vecK
    local g = ld.graph3d:new{
        window3d = {-5,5,-5,5,-5,5},
        window = {-3,3,-2,2},
        viewdir = {"yz",0.75,40}
    }
    local vecN = ld.rotate3d(vecK,-70,{Origin,vecI})
    ld.Hiddenlines = true; ld.Hiddenlinestyle="dashed"
    g:Dscene3d(
        g:addPlane({Origin,vecK},{contrast=0,rectangle={vecI,2,4},edge=true,edgecolor="DarkBlue",hidden=true}),
        g:addPlane({Origin,vecN},{contrast=0,rectangle={vecI,2,2.5},edge=true,edgecolor="Orange",hidden=true}),
        g:addDots({Origin}),
        g:addLabel(
            "$\\ell$",2.5*vecI,{pos="SW"},
            "$P$",Origin,{pos="E",dist=0.1},
            "$\\alpha$",vecI-2*vecJ,{pos="NE",dir={vecJ,-vecI},color="DarkBlue",dist=0},
            "$\\beta$",ld.rotate3d(vecI-vecJ,-70,{Origin,vecI}),{pos="E",dir={-vecI,pt3d.prod(vecN,-vecI)},color="Orange"}
        )
    )
    g:Dpolyline3d({-2.5*vecI,2.5*vecI}, "thick")
    g:Show()
    \end{luadraw}
\end{document}

这里的「灰色」的关键在于 contrast 选项（color的默认值就是white，我是能省就省的，所以我不写）：

Some Notes and Comments

• 我在构造曲面和曲线的时候是比较遵从直角坐标系的，交线的方向就是vecI
• 我采用ld.rotate3d来处理这里的旋转，这样用到的点比较少
• 从原图来看，我不知道你为何要用中心投影，更合理的方法应该是斜投影viewdir = {"yz",0.75,40}（类似斜二测画法）
• 要注意的是，如果一个面是透明的，那么在它后面的线都是可见的，不会变成虚线，只是颜色稍微变暗；相反只有当一个面是不透明的，那么它后面的线才会被ld.Hiddenlines = true; ld.Hiddenlinestyle="dashed"控制。如果一个面是半透明的，那么不会（也不应该）出现虚线

可以对比一下思路 :-)

比较取巧的方法：

由于这里遮挡关系并不算太复杂，可以通过画两次（先遮挡再画虚线）的方法来手动维护遮挡关系：

\documentclass{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}
\usepackage{fourier-otf}
\begin{document}
    \begin{luadraw}{name=intersected_surface}
    local ld = luadraw
    local pt3d, cpx = ld.pt3d, ld.cpx
    local M, Mc, Ms = pt3d.M, pt3d.Mc, pt3d.Ms
    local Origin, vecI, vecJ, vecK = pt3d.Origin, pt3d.vecI, pt3d.vecJ, pt3d.vecK
    local g = ld.graph3d:new{
        window3d = {-5,5,-5,5,-5,5},
        window = {-3,3,-2,2},
        -- adjust2d = true, 
        -- viewdir = {30,75}
        viewdir = {"yz",0.75,40}
    }
    -- g:Dpolyline3d({{Origin,3*vecI},{Origin,3*vecJ},{Origin,3*vecK}}, "thick,-Stealth")
    g:Dplane({Origin, vecK}, vecI, 2, 4 , ld.all, "DarkBlue,fill=white")
    local vecN = ld.rotate3d(vecK, -70, {Origin,vecI}) -- rotate vecK 30 degree around vecI
    g:Dplane({Origin, vecN}, vecI, 2, 2.5, ld.all, "Orange,fill=white")
    g:Dplane({Origin, vecK}, vecI, 2, 4 , ld.all, "DarkBlue,dashed")
    g:Dplane({vecJ, vecK}, vecI, 2, 2 , ld.all, "DarkBlue,fill=white") -- trick, fill half plane... not elegant actually...
    g:Dplane({Origin, vecN}, vecI, 2, 2.5, ld.all, "Orange,dashed")
    g:Dpolyline3d({-2.5*vecI,2.5*vecI}, "thick")
    g:Ddots3d({Origin})
    g:Dlabel3d(
        "$\\ell$",2.5*vecI,{pos="SW"},
        "$P$",Origin,{pos="E",dist=0.1},
        "$\\alpha$",vecI-2*vecJ,{pos="NE",dir={vecJ,-vecI},node_options="text=DarkBlue",dist=0},
        "$\\beta$",ld.rotate3d(vecI-vecJ, -70, {Origin,vecI}),{pos="E",dir={-vecI,pt3d.prod(vecN,-vecI)},node_options="text=Orange"}
    )
    g:Show()
\end{luadraw}
\end{document}

简单地说：

• window3d控制的是三维空间的边界盒
• window控制的是三维空间边界盒里的内容按照viewdir投影到纸张平面后，2D平面的边界

luadraw在绘制的过程中，_大致_经历了下面的过程：

• 根据3D盒子的边界确定3D空间盒子的边界，一些在盒子外的内容可能会被某些方法切去
• 基于viewdir确定3D物体投影到屏幕投影面的方法和角度
• 基于adjust2d,window和bbox以及size等参数调整屏幕投影面的最终显示效果

(AI总结，仅供参考）

在你的例子中，3D盒子大小是不影响的最终投影结果的，因此：

    local g = ld.graph3d:new{
        window3d = {0,1.5,0,5.5,0,0.6},
        -- window3d = {0,3,0,11,0,1.2},
        window={-1.1,5.5,-1.1,1.5},
        bbox=false,size={9,9},
        viewdir={"yz",0.5,45}
    }

和

    local g = ld.graph3d:new{
        -- window3d = {0,1.5,0,5.5,0,0.6},
        window3d = {0,3,0,11,0,1.2},
        window={-1.1,5.5,-1.1,1.5},
        bbox=false,size={9,9},
        viewdir={"yz",0.5,45}
    }

的表现都是相同的：

我们可以通过 Dboxaxes3d 方法来更好地可视化这种差距：

\documentclass[border=8pt]{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}
\usepackage{fourier-otf}
\begin{document}
\begin{tabular}{cc}
    \begin{luadraw}{name=cube_small} 
        local ld = luadraw
        local pt3d = ld.pt3d
        local M = pt3d.M
        local g = ld.graph3d:new{
            window3d = {0, 1, 0, 1, 0, 1},
            bbox = false, size = {10, 10},
            viewdir = {"yz", 0.5, 45}
        }
        local Origin, vecI, vecJ, vecK = pt3d.Origin, pt3d.vecI, pt3d.vecJ, pt3d.vecK
        local C = ld.parallelep(Origin, 1.5*vecI, 1.5*vecJ, 1.5*vecK)
        g:Dboxaxes3d({grid = true, gridcolor = "cyan!50", fillcolor = "cyan!20", drawbox = true})
        g:Dpoly(C, {mode = ld.mWireframe})
        g:Show()
    \end{luadraw}
    &
    \begin{luadraw}{name=cube_large} 
        local ld = luadraw
        local pt3d = ld.pt3d
        local M = pt3d.M
        local g = ld.graph3d:new{
            window3d = {0, 2, 0, 2, 0, 2},
            bbox = false, size = {10, 10},
            viewdir = {"yz", 0.5, 45}
        }
        local Origin, vecI, vecJ, vecK = pt3d.Origin, pt3d.vecI, pt3d.vecJ, pt3d.vecK
        local C = ld.parallelep(Origin, 1.5*vecI, 1.5*vecJ, 1.5*vecK)
        g:Dboxaxes3d({grid = true, gridcolor = "cyan!50", fillcolor = "cyan!20", drawbox = true})
        g:Dpoly(C, {mode = ld.mWireframe})
        g:Show()
    \end{luadraw}
    \\
    \texttt{window3d = \{0, 1, 0, 1, 0, 1\}} &
    \texttt{window3d = \{0, 2, 0, 2, 0, 2\}}
\end{tabular}
\end{document}

如果不左右对比，同时不画盒子，只能感受到缩放的差距，因为立方体和2D画布的相对关系是相同的。因为我不怎么会用到那种「盒子外的内容可能会被切去」的方法，所以其实保持默认的 window3d={-5,5,-5,5,-5,5,1,1,1}，只调整window和size其实也是够的...

无论我怎么改变window3d里面的6个参数值，都没有影响到输出的图形,不知道window3d的设置还有什么价值?

是因为只改变「window3d里面的6个参数值」，只会影响3D盒子边界和真实物体的相对大小关系，但是绘制结果是经过上述流程2D投影后得到的，投影后的纸张和物体的投影相对大小关系还是相同

而window的设置可以截取部分图形。我将设置window的代码那行删除,结果右侧显示不完全。

这是因为不写window，就相当使用的是默认值window={-5,5,-5,5}，如果3D投影到2D后还超出了window，那么就会「显示不完全」

我希望是取消留白，只有图形区域,类似border=0pt

那么也许应该使用类似bbox=false的方法....

上面的答案是正确的，我们可以考虑调整角度来说明这一点，下面的代码使用与z轴成90度的「平行方向」来看这个几何体，我们可以看到他们是「对称等高」的：

\documentclass[border=5pt]{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}
\begin{document}
\begin{luadraw}{name=prism}
    local ld = luadraw
    local pt3d = ld.pt3d
    local Origin, vecK = pt3d.Origin, pt3d.vecK
    local M,Mc=pt3d.M,pt3d.Mc
    local g = ld.graph3d:new{
        window3d={-5,5,-5,5,-5,5},
        window={-5,5,-5,5},
        bbox=false,
        -- adjust2d = true,
        viewdir={0,90} -- <- here
    }
    local A1,B1,C1=Mc(1,-120*ld.deg,0),Mc(1,0,0), Mc(1,120*ld.deg,0)
    local prism1 = ld.prism({A1,B1,C1}, M(-2,0,3))
    g:Dpoly(prism1,{mode=ld.mWireframe,hiddenstyle="dashed"})
    local prism = ld.sym3d(prism1, {Origin,vecK})
    g:Dpoly(prism,{mode=ld.mWireframe,hiddenstyle="dashed"})
    local V = prism.vertices
    g:Dlabel3d(
        "$A$",V[1],{pos="W"},
        "$B$",V[2],{pos="S"},
        "$C$",V[3],{pos="E"},
        "$A_1$",V[4],{pos="W"},
        "$B_1$", V[5],{pos="SE"},
        "$C_1$", V[6],{pos="E"}
    )
    g:Show()
\end{luadraw} 
\end{document}

另外，如果我是你，我可能不会考虑这样的构造，我会想办法让「两个棱柱相交的那个面」在「xoy平面」上：

\documentclass[border=5pt]{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}
\usepackage{fourier-otf}
\begin{document}
\begin{luadraw}{name=prism}
    local ld = luadraw
    local pt3d = ld.pt3d
    local Origin, vecK = pt3d.Origin, pt3d.vecK
    local M,Mc=pt3d.M,pt3d.Mc
    local g = ld.graph3d:new{
        window3d={-5,5,-5,5,-5,5},
        window={-5,5,-5,5},
        bbox=false,
        -- adjust2d = true,
        viewdir={"yz",0.5,45}
    }
    local A1,B1,C1=Mc(1,-120*ld.deg,0),Mc(1,0,0), Mc(1,120*ld.deg,0)
    local prism1 = ld.prism({A1,B1,C1}, M(-2,0,3))
    g:Dpoly(prism1,{mode=ld.mWireframe,hiddenstyle="dashed"})
    local prism = ld.sym3d(prism1, {Origin,vecK})
    g:Dpoly(prism,{mode=ld.mWireframe,hiddenstyle="dashed"})
    local V = prism.vertices
    g:Dlabel3d(
        "$A_1$",V[1],{pos="W"},
        "$B_1$", V[2],{pos="SE"},
        "$C_1$", V[3],{pos="E"},
        "$A$",V[4],{pos="W"},
        "$B$",V[5],{pos="S"},
        "$C$",V[6],{pos="E"}
    )
    g:Show()
\end{luadraw} 
\end{document}

我对你的代码做了一下简化：

• 0*ld.deg 这里的0就没必要乘了
• node_options="xshift=-0cm" 默认就不移动，不需要这么改，一般改距离可以直接用dist=0.1
• g:Dlabel3d一个命令可以同时画六个点的坐标
• 在这个例子里，完全可以构造一个对称的棱柱，使用ld.sym3d，这可以简化你的不少坐标...

\documentclass{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}
\usepackage{fourier-otf}
\begin{document}
    \begin{luadraw}{name=prism_sphere}
    local ld = luadraw
    local pt3d, cpx = ld.pt3d, ld.cpx
    local M, Mc, Ms = pt3d.M, pt3d.Mc, pt3d.Ms
    local Origin, vecI, vecJ, vecK = pt3d.Origin, pt3d.vecI, pt3d.vecJ, pt3d.vecK
    local g = ld.graph3d:new{
        window3d = {-5,5,-5,5,-5,5},
        window = {-5,5,-5,5},
        -- adjust2d = true, 
        viewdir = {0,75}
    }
    require 'luadraw_spherical'
    local sM = ld.sM
    ld.Hiddenlines = true;ld.Hiddenlinestyle = "dashed"
    g:Define_sphere{ color = "", edgewidth=10, edgecolor="black" }
    local h = 2.25
    local O1,O2 = M(0,0,h),M(0,0,-h)
    local r = math.sqrt(3^2-h^2)
    local t1, t2, t3 = -80*ld.deg, 45*ld.deg, 150*ld.deg
    local A,B,C = Mc(r,t1,-h), Mc(r,t2,-h), Mc(r,t3,-h)
    local A1,B1,C1 = Mc(r,t1,h), Mc(r,t2,h), Mc(r,t3,h)
    g:DScircle({Origin,vecK},{width=6})
    local P = ld.facet2poly( ld.cvx_hull3d({A,B,C,A1,B1,C1}) )
    g:Dcircle3d(O2,r,vecK,"draw=none,fill=blue!25, fill opacity=0.4") -- lower circle
    g:DScircle({O2,vecK},{width=8})
    g:Dpoly(P,{color="violet!30!white",mode=ld.mFlatHidden,edgewidth=8,hiddenstyle="dashed"})
    g:Dcircle3d(O2,r,vecK,"draw=none,fill=blue!25, fill opacity=0.4") -- upper circle
    g:DScircle({O1,vecK},{width=8})
    g:Dspherical()
    g:Dballdots3d({Origin,O1,O2},"red")
    g:Dpolyline3d({C1,O1,O2,A},"blue!75,dashed,semithick")
    g:Dpolyline3d({C1,Origin,A},"red,dashed,semithick")
    g:Dlabel3d(
        "$O$",Origin,{pos="E"},
        "$O_1$",O1,{pos="W"},
        "$O_2$",O2,{pos="E"},
        "$A$",A,{pos="SW"},
        "$B$",B,{pos="SE"},
        "$C$",C,{pos="NE"},
        "$A_1$",A1,{pos="NW"},
        "$B_1$",B1,{pos="SE"},
        "$C_1$",C1,{pos="NE"},
        "$r$",(O1+C1)/2,{pos="",node_options="text=blue"},
        "$r$",A/6+5*O2/6,{},
        "$h$",O1/4,{pos="W"},
        "$R$",C1/2,{pos="E",node_options="text=red"},
        "$R$",2*A/5,{pos="N",dist=0.15}
    )
    g:Show()
\end{luadraw}
\end{document}

% https://ask.latexstudio.net/ask/question/18066.html
\documentclass{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}
\usepackage{fourier-otf}
\begin{document}
    \begin{luadraw}{name=cylinder_sphere}
    local ld = luadraw
    local pt3d = ld.pt3d
    local M, Mc, Origin, vecK = pt3d.M, pt3d.Mc, pt3d.Origin, pt3d.vecK
    local g = ld.graph3d:new{
        window3d = {-5,5,-5,5,-5,5},
        window = {-5,5,-5,5},
        -- adjust2d = true, 
        viewdir = {0,80} -- should be fixed
    }
    require 'luadraw_spherical'
    local sM = ld.sM
    ld.Hiddenlines = true;ld.Hiddenlinestyle = "dashed"
    g:Define_sphere{ color = "", edgewidth=10, edgecolor="black" }
    local h = 2
    local O1,O2 = M(0,0,h),M(0,0,-h)
    local r = math.sqrt(3^2-h^2)
    local A,B = Mc(r,140*ld.deg,h), Mc(r,45*ld.deg,-h)
    g:DScircle({Origin,vecK});g:DScircle({O1,vecK},{width=8});g:DScircle({O2,vecK},{width=8})
    g:Dcylinder(O1,r,O2,{color="violet!75", edgestyle="noline",hiddenstyle="noline"}) -- hide the frustum border and hidden lines
    local M1, N1, M2, N2 = table.unpack(g:Cylinder_tangency(O1, r, O2)) -- get the tangency points of the cylinder
    g:DSpolyline({{M1,M2},{N1,N2}},{width=8}) -- draw the 
    --
    g:Dspherical()
    g:Dangle3d(A,O1,O2,0.3,"very thick")
    g:Dpolyline3d({B,Origin,A,O1,O2},"blue,dashed,thick")
    g:Dballdots3d({Origin,O1,O2,A,B},"red",0.75)
    g:Dlabel3d(
        "$O$",Origin,{pos="W"},
        "$h$",O2/4,{},
        "$O_1$",O1,{},
        "$O_2$",O2,{},
        "$A$",A,{pos="NE"},
        "$B$",B,{pos="SE"},
        "$r$",(O1+A)/2,{pos="S",node_options="text=red"},
        "$h/2$",O1/2,{pos="W"},
        "$R$",3*A/5,{pos="E"}
    )
    g:Show()
\end{luadraw}
\end{document}

注意右手定则，看注释...

\documentclass[border=5pt]{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}

\begin{document}
\begin{luadraw}{name=prism}
    local ld = luadraw
    local pt3d = ld.pt3d
    local M = pt3d.M
    local g = ld.graph3d:new{
        window3d={-5,5,-5,5,-5,5},
        windows={-5,5,-5,5},
        size={8,8},
        bbox=false,
        viewdir={"yz",0.5,45}
        -- viewdir={30,70}
    }

    -- local base = { M(0,0,0), M(2,0,0), M(0,0,2) }
    local base = { M(0,0,0), M(0,0,2), M(2,0,0) } -- 右手定则
    -- M(0,0,2), M(2,0,0)在右手法则下应该与 (0,3,0) 通向
    local prism = ld.prism(base, M(0,3,0))
    
    g:Dpoly(prism,{
        mode=ld.mWireframe,
        hiddenstyle="dashed",
        -- edgestyle="solid"
    })
    g:Ddots3d(M(0,0,2),"blue")

    g:Show()
\end{luadraw}
\end{document}

更新你的 latex3-backend:

在终端运行：

tlmgr update --all --self

应该就行了...

https://tex.stackexchange.com/q/765234/322482

原图又在骗

如果「D」点确实是截面圆与视界圆的切点的话，那么根本不可能看到如上角度的蓝色三角形：

\documentclass{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}
\usepackage{fourier-otf}
\begin{document}
    \begin{luadraw}{name=cuboid_sphere}
    local ld = luadraw
    local pt3d = ld.pt3d
    local M, Origin, vecK = pt3d.M, pt3d.Origin, pt3d.vecK
    local g = ld.graph3d:new{
        window3d = {-5,5,-5,5,-5,5},
        window = {-5,5,-5,5},
        -- adjust2d = true, 
        viewdir = {20,75} 
    }
   require 'luadraw_spherical'
    local sM = ld.sM
    ld.Hiddenlines = true;ld.Hiddenlinestyle = "dashed"
    g:Define_sphere{ color = "", edgewidth=8, edgecolor="black" }
    local sqrt = math.sqrt
    local h = 1.75
    local O1, P = M(0,0,-h), sM(0,0)
    local Out = {}
    g:DScircle({O1,vecK},{out=Out});
    local D, _ = table.unpack(Out)
    local A = ld.rotate3d(D, 90, {O1, vecK})
    local B = ld.rotate3d(A, 90, {O1, vecK})
    local C = ld.rotate3d(B, 90, {O1, vecK})
    local vecn = pt3d.prod(A-P,D-P)
    local O2 = ld.proj3d(Origin,{P,vecn})
    local h1,h2 = pt3d.abs(Origin-O1),pt3d.abs(Origin-O2)
    local r1,r2 = sqrt(3^2-h1^2),sqrt(3^2-h2^2)
    g:DScircle({Origin,vecK})
    local T = ld.facet2poly( ld.cvx_hull3d({A,B,C,D,P}) )
    g:Dcircle3d(O1,r1,vecK,"draw=none,fill=blue!25, fill opacity=0.4") 
    g:Dcircle3d(O2,r2,vecn,"draw=none,fill=blue!25, fill opacity=0.4") 
    g:DScircle({O1,vecK}); g:DScircle({O2,vecn})
    g:Dpoly(T,{color="violet!30!white",mode=ld.mFlatHidden,edgewidth=8,hiddenstyle="dashed"})
    g:Dspherical()
    g:Dpolyline3d({P,O1,A,Origin},"blue,dashed,semithick")
    g:Dballdots3d({Origin,O1,A,B,C,D,P},"red",0.75)
    g:Dlabel3d(
        "$O$",Origin,{pos="W"},
        "$O_1$",O1,{},
        "$P$",P,{pos="N"},
        "$A$",A,{pos="NE"},
        "$B$",B,{pos="W"},
        "$C$",C,{pos="S"},
        "$D$",D,{pos="E"},
        "$h$",O1/2,{pos="W"},
        "$R$",P/2,{pos="",node_options="text=blue"},
        "$R$",A/2,{},
        "$r$",(O1+A)/2,{pos="E"}
    )
    g:Show()
\end{luadraw}
\end{document}

把底面微调成长方形，修改这里的旋转角度：

    local D, _ = table.unpack(Out)
    local A = ld.rotate3d(D, 75, {O1, vecK}) -- <- 75 degree?
    local B = ld.rotate3d(A, 115, {O1, vecK})
    local C = ld.rotate3d(B, 75, {O1, vecK})

完整代码如下：

\documentclass{standalone}
\usepackage[3d]{luadraw}
\usepackage[svgnames]{xcolor}
\usepackage{fourier-otf}
\begin{document}
    \begin{luadraw}{name=cuboid_sphere}
    local ld = luadraw
    local pt3d = ld.pt3d
    local M, Origin, vecK = pt3d.M, pt3d.Origin, pt3d.vecK
    local g = ld.graph3d:new{
        window3d = {-5,5,-5,5,-5,5},
        window = {-5,5,-5,5},
        -- adjust2d = true, 
        viewdir = {20,75} 
    }
   require 'luadraw_spherical'
    local sM = ld.sM
    ld.Hiddenlines = true;ld.Hiddenlinestyle = "dashed"
    g:Define_sphere{ color = "", edgewidth=8, edgecolor="black" }
    local sqrt = math.sqrt
    local h = 1.75
    local O1, P = M(0,0,-h), sM(0,0)
    local Out = {}
    g:DScircle({O1,vecK},{out=Out});
    local D, _ = table.unpack(Out)
    local A = ld.rotate3d(D, 75, {O1, vecK}) -- <- 75 degree?
    local B = ld.rotate3d(A, 115, {O1, vecK})
    local C = ld.rotate3d(B, 75, {O1, vecK})
    local vecn = pt3d.prod(A-P,D-P)
    local O2 = ld.proj3d(Origin,{P,vecn})
    local h1,h2 = pt3d.abs(Origin-O1),pt3d.abs(Origin-O2)
    local r1,r2 = sqrt(3^2-h1^2),sqrt(3^2-h2^2)
    g:DScircle({Origin,vecK})
    local T = ld.facet2poly( ld.cvx_hull3d({A,B,C,D,P}) )
    g:Dcircle3d(O1,r1,vecK,"draw=none,fill=blue!25, fill opacity=0.4") 
    g:Dcircle3d(O2,r2,vecn,"draw=none,fill=blue!25, fill opacity=0.4") 
    g:DScircle({O1,vecK}); g:DScircle({O2,vecn})
    g:Dpoly(T,{color="violet!30!white",mode=ld.mFlatHidden,edgewidth=8,hiddenstyle="dashed"})
    g:Dspherical()
    g:Dpolyline3d({P,O1,A,Origin},"blue,dashed,semithick")
    g:Dballdots3d({Origin,O1,A,B,C,D,P},"red",0.75)
    g:Dlabel3d(
        "$O$",Origin,{pos="W"},
        "$O_1$",O1,{},
        "$P$",P,{pos="N"},
        "$A$",A,{pos="NE"},
        "$B$",B,{pos="W"},
        "$C$",C,{pos="S"},
        "$D$",D,{pos="E"},
        "$h$",O1/2,{pos="W",dist=0.1},
        "$R$",P/2,{pos="",dist=0,node_options="text=blue"},
        "$R$",A/2,{},
        "$r$",O1/3+2*A/3,{pos="S"}
    )
    g:Show()
\end{luadraw}
\end{document}

我们可以获得比较贴近原图的效果...

真的别用伪3D来骗了😡😡😡😡😡修改和分析成本很大，打咩！