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如何：3d 中 5 個座標的圖片

tag-icon • tag-icon tikz-pgf tikz-graphdrawing
如何：3d 中 5 個座標的圖片

我一直在嘗試製作一張 tikzpicture 來說明 3d 中的 5 個點，其中從一個點它們是 4 個向量到其他 4 個點。我想知道有沒有辦法讓這張圖片看起來更像3d。

\documentclass[landscape]{article}
\usepackage{tikz}
\usepackage{verbatim}
\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}
\setlength\PreviewBorder{10pt}%
\usetikzlibrary{calc}


\begin{document}
\begin{tikzpicture}
[
scale=3,
>=stealth,
point/.style = {draw, circle,  fill = black, inner sep = 1pt},
dot/.style   = {draw, circle,  fill = black, inner sep = .2pt},
]

%% Vanishing points for perspective handling
\coordinate (P1) at (-9cm,1.5cm); % left vanishing point (To pick)
\coordinate (P2) at (9cm,1.5cm); % right vanishing point (To pick)

%% (A1) and (A2) defines the 2 central points of the cuboid
\coordinate (A1) at (0em,3cm); % central top point (To pick)
\coordinate (A2) at (0em,-3cm); % central bottom point (To pick)

%% (A3) to (A8) are computed given a unique parameter (or 2) .8
% You can vary .8 from 0 to 1 to change perspective on left side
\coordinate (A3) at ($(P1)!.8!(A2)$); % To pick for perspective 
\coordinate (A4) at ($(P1)!.8!(A1)$);

% You can vary .8 from 0 to 1 to change perspective on right side
\coordinate (A7) at ($(P2)!.7!(A2)$);
\coordinate (A8) at ($(P2)!.7!(A1)$);

%% Automatically compute the last 2 points with intersections
\coordinate (A5) at
  (intersection cs: first line={(A8) -- (P1)},
            second line={(A4) -- (P2)});
\coordinate (A6) at
  (intersection cs: first line={(A7) -- (P1)}, 
            second line={(A3) -- (P2)});



%% Possibly draw back faces
\fill[gray!90] (A2) -- (A3) -- (A6) -- (A7) -- cycle; % face 6
\node at (barycentric cs:A2=1,A3=1,A6=1,A7=1) {\tiny };
\fill[gray!50] (A3) -- (A4) -- (A5) -- (A6) -- cycle; % face 3
\node at (barycentric cs:A3=1,A4=1,A5=1,A6=1) {\tiny };
\fill[gray!30] (A5) -- (A6) -- (A7) -- (A8) -- cycle; % face 4
\node at (barycentric cs:A5=1,A6=1,A7=1,A8=1) {\tiny };

\draw[thick] (A5) -- (A6);
\draw[thick] (A3) -- (A6);
\draw[thick] (A7) -- (A6);

\node (n0) at (0,0) [point, label = left:$S$] {};
\node (n1) at (1.2,1.7) [point, label = left:$R_{1}$] {};
\node (n2) at (-0.5, 1.2) [point, label = left:$R_{2}$] {};
\node (n3) at (1.35,-1.1) [point, label = left:$R_{3}$] {};
\node (n4) at (-0.2, -1.9) [point, label = left:$R_{4}$] {};

\draw[<-] (n1) -- node (a) [label = {below right:$d_{1}$}] {} (n0);
\draw[<-] (n2) -- node (b) [label = {above right:$d_{2}$}] {} (n0);
\draw[<-] (n3) -- node (c) [label = {above right:$d_{3}$}] {} (n0);
\draw[<-] (n4) -- node (d) [label = {above right:$d_{4}$}] {} (n0);

\end{tikzpicture}
\end{document}

tikz圖片

答案1

如果你不想實際繪製 3D 影像，可以使用tikz-3dplot包，那麼一個非常簡單的解決方案就是用一些非常淺的顏色填滿平面。參見下圖。

在此輸入影像描述

這是您的案例的程式碼

\documentclass[landscape]{article}
\usepackage{tikz}
\usepackage{verbatim}
\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}
\setlength\PreviewBorder{10pt}%
\usetikzlibrary{calc}


\begin{document}
\begin{tikzpicture}
[
scale=3,
>=stealth,
point/.style = {draw, circle,  fill = black, inner sep = 1pt},
dot/.style   = {draw, circle,  fill = black, inner sep = .2pt},
]

%% Vanishing points for perspective handling
\coordinate (P1) at (-9cm,1.5cm); % left vanishing point (To pick)
\coordinate (P2) at (9cm,1.5cm); % right vanishing point (To pick)

%% (A1) and (A2) defines the 2 central points of the cuboid
\coordinate (A1) at (0em,3cm); % central top point (To pick)
\coordinate (A2) at (0em,-3cm); % central bottom point (To pick)

%% (A3) to (A8) are computed given a unique parameter (or 2) .8
% You can vary .8 from 0 to 1 to change perspective on left side
\coordinate (A3) at ($(P1)!.8!(A2)$); % To pick for perspective 
\coordinate (A4) at ($(P1)!.8!(A1)$);

% You can vary .8 from 0 to 1 to change perspective on right side
\coordinate (A7) at ($(P2)!.7!(A2)$);
\coordinate (A8) at ($(P2)!.7!(A1)$);

%% Automatically compute the last 2 points with intersections
\coordinate (A5) at
  (intersection cs: first line={(A8) -- (P1)},
            second line={(A4) -- (P2)});
\coordinate (A6) at
  (intersection cs: first line={(A7) -- (P1)}, 
            second line={(A3) -- (P2)});



%% Possibly draw back faces
\fill[gray!90] (A2) -- (A3) -- (A6) -- (A7) -- cycle; % face 6
\node at (barycentric cs:A2=1,A3=1,A6=1,A7=1) {\tiny };
\fill[gray!50] (A3) -- (A4) -- (A5) -- (A6) -- cycle; % face 3
\node at (barycentric cs:A3=1,A4=1,A5=1,A6=1) {\tiny };
\fill[gray!30] (A5) -- (A6) -- (A7) -- (A8) -- cycle; % face 4
\node at (barycentric cs:A5=1,A6=1,A7=1,A8=1) {\tiny };

\draw[thick] (A5) -- (A6);
\draw[thick] (A3) -- (A6);
\draw[thick] (A7) -- (A6);

\node (n0) at (0,0) [point, label = left:$S$] {};
\node (n1) at (1.2,1.7) [point, label = left:$R_{1}$] {};
\node (n2) at (-0.5, 1.2) [point, label = left:$R_{2}$] {};
\node (n3) at (1.35,-1.1) [point, label = left:$R_{3}$] {};
\node (n4) at (-0.2, -1.9) [point, label = left:$R_{4}$] {};



\draw[fill, green, opacity=.1] (0,0) -- (1.2,1.7) -- (-0.5, 1.2);
\draw[fill, green, opacity=.1] (0,0) -- (-0.2, -1.9) -- (1.35,-1.1);
\draw[fill, red, opacity=.05] (0,0) -- (1.2,1.7) -- (1.35,-1.1);


\draw[<-,very thick] (n1) -- node (a) [label = {below right:$d_{1}$}] {} (n0);
\draw[<-,very thick] (n2) -- node (b) [label = {above right:$d_{2}$}] {} (n0);
\draw[<-,very thick] (n3) -- node (c) [label = {above right:$d_{3}$}] {} (n0);
\draw[<-,very thick] (n4) -- node (d) [label = {above right:$d_{4}$}] {} (n0);


\end{tikzpicture}
\end{document}

答案2

我會使用 Ti 中提供的座標轉換宏kZ 基本上允許定義一組新的座標 x、y 和 z 作為「舊」系統 u、v 的函數，如下所示：

在此輸入影像描述

我發現這段程式碼不那麼棘手且計算量也更少：

\documentclass[tikz,convert]{standalone}
\begin{document}
\def\rot{30}
\begin{tikzpicture}[x={(180+\rot:1cm)},y={(-\rot:1cm)},z={(90:1cm)},scale=.7,>=stealth,thick]


\draw [fill=gray,draw=none] (0,0,0)--(10,0,0)--(10,5,0)--(0,5,0)--cycle;
\draw [fill=gray!50!white,draw=none] (0,0,0)--(10,0,0)--(10,0,15)--(0,0,15)--cycle;
\draw [fill=gray!30!white,draw=none] (0,0,0)--(0,5,0)--(0,5,15)--(0,0,15)--cycle;
\draw (0,0,0)--(10,0,0) (0,0,0)--(0,5,0) (0,0,0)--(0,0,15);


\node [fill=black,circle,inner sep=1pt] (pt0) at (5,0.5,5) {};
\node [fill=black,circle,inner sep=1pt] (pt1) at (0,1,10) {};
\node [fill=black,circle,inner sep=1pt] (pt2) at (9,0,9) {};
\node [fill=black,circle,inner sep=1pt] (pt3) at (0,2,1) {};
\node [fill=black,circle,inner sep=1pt] (pt4) at (8,2,0) {};

\draw [->] (pt0)--(pt1) node [pos=.5,right] {$d_1$};
\draw [->] (pt0)--(pt2) node [pos=.5,right] {$d_2$};
\draw [->] (pt0)--(pt3) node [pos=.5,below] {$d_3$};
\draw [->] (pt0)--(pt4) node [pos=.5,right] {$d_4$};

\node at (pt0) [left] {$S$}; 
\node at (pt1) [above] {$R1$}; 
\node at (pt2) [left] {$R2$}; 
\node at (pt3) [right] {$R3$}; 
\node at (pt4) [below] {$R4$}; 
\end{tikzpicture}
\end{document}

這是輸出：

具有 3D 變換且角度=20 的圖像

請注意，我定義了一個名為 的數字\rot。如果將其值變更為 30（例如），您將自動獲得另一個視角。此外，如果你想要更多的深度，你也可以畫虛線。您可以做的另一件事是，由於深度缺乏主要是由於 y 座標的微小變化造成的，因此您可能需要更改該方向的比例，例如y={((-\rot:3cm))}

透過虛線和不同的視角獲得更多深度

只需在上面的程式碼中添加這幾行（請注意\def\rot{30}序言）：

\begin{tikzpicture}[x={(180+\rot:1cm)},y={(-\rot:3cm)},z={(90:1cm)},...]
...
\draw [dashed] (pt1)--+(0,-1,0) (pt1)--+(0,0,-10);
\draw [dashed] (pt2)--+(0,0,-9) (pt2)--+(-9,0,0);
\draw [dashed] (pt3)--+(0,-2,0) (pt3)--+(0,0,-1);
\draw [dashed] (pt4)--+(-8,0,0) (pt4)--+(0,-2,0);
...
\end{tikzpicture}

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