4

I am adapting code from a response offered by Jan Hlavacek years ago. How do the following commands put perspective into the drawing?

\coordinate (front_right) at ($(top_front)!5!(top_right)$);
\coordinate (front_left) at ($(top_front)!5!(top_left)$);
\coordinate (front_bottom) at ($(top_front)!15!(bottom_front)$);

How would I label three edges so that the label is at the midpoint, 0.1cm from the edge, and sloped? If one of the diagonals of the box were to be drawn, how would it be labeled similarly?

\documentclass{amsart}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}



\begin{document}


\begin{tikzpicture}

%\clip (-3,-3) rectangle (3,3);
\coordinate (top_front) at (0,0);
\coordinate (bottom_front) at (0,-3);
\coordinate (top_right) at (15:2.5cm);
\coordinate (top_left) at (165:2.5cm);

%You can change the perspective by playing with the 5, 5, 15:
\coordinate (front_right) at ($(top_front)!5!(top_right)$);
\coordinate (front_left) at ($(top_front)!5!(top_left)$);
\coordinate (front_bottom) at ($(top_front)!15!(bottom_front)$);

\path[name path=bottom_right_path] (bottom_front) -- (front_right);   
\path[name path=right_back_path] (top_right) -- (front_bottom);
\path[name path=back_left_path] (bottom_front) -- (front_left);
\path[name path=left_back_path] (top_left) -- (front_bottom);
\path[name path=top_right_path] (top_left) -- (front_right);
\path[name path=top_left_path] (top_right) -- (front_left);



\coordinate[name intersections={of=bottom_right_path and right_back_path, by=back_right}];
\coordinate[name intersections={of=back_left_path and left_back_path, by=back_left}];
\coordinate[name intersections={of=top_right_path and top_left_path, by=top_back}];

%\shade[right color=gray!10, left color=black!50, shading angle=105] (top_front) -- (bottom_front) -- (back_left) -- (top_left) -- cycle;
%\shade[left color=gray!10, right color=black!50, shading angle=75] (top_front) -- (bottom_front) -- (back_right) -- (top_right) -- cycle;

\begin{scope}
\clip (top_front) -- (top_right) -- (top_back) -- (top_left) -- cycle;
\shade[inner color = gray!5, outer color=black!50, shading=radial] (top_front) ellipse (3cm and 1.5cm);
\end{scope}

\draw (top_front) -- (bottom_front);
\draw (top_front) -- (top_right);
\draw (top_front) -- (top_left);
\draw (top_right) -- (back_right);
\draw (bottom_front) -- (back_right);
\draw (top_left) -- (back_left);
\draw (bottom_front) -- (back_left);
\draw (top_back) -- (top_right);
\draw (top_back) -- (top_left);



\end{tikzpicture}

\end{document}
A gal named Desire
  • 1,533
  • Could you please provide a link to the original code and explain in more detail what you mean by "How do the following commands put perspective into the drawing?", "How would I label three edges so that the label is at the midpoint, 0.1cm from the edge, and sloped?" and " If one of the diagonals of the box were to be drawn, how would it be labeled similarly?". Perhaps you could add a sketch of what you want? (And why is the bounding box so huge?) –  Nov 18 '18 at 04:35
  • Here is the link to the response from Jan Hlavacek. https://tex.stackexchange.com/questions/12020/what-is-the-easiest-way-to-draw-a-3d-cube-with-tikz – A gal named Desire Nov 18 '18 at 11:26

2 Answers2

4

To answer this problem in the mind of the person who wrote it, I started by visualizing the different points used for construction by drawing the paths that have not been drawn and naming these points, we get this figure with a reduction of 0.4 scale=.4:

totalite

%\clip (-3,-3) rectangle (3,3);
\coordinate[label={[blue]left:top-front}] (top_front) at (0,0);
\coordinate[label={[blue]left:bottom-front}] (bottom_front) at (0,-3);
\coordinate[label={[blue]right:top-right}] (top_right) at (15:2.5cm);
\coordinate[label={[blue]left:top-left}] (top_left) at (165:2.5cm);

%You can change the perspective by playing with the 5, 5, 15:
\coordinate[label={right:front-right}]  (front_right) at ($(top_front)!5!(top_right)$);
\coordinate [label={left:front-left}] (front_left) at ($(top_front)!5!(top_left)$);
\coordinate [label={below:front-bottom}](front_bottom) at ($(top_front)!15!(bottom_front)$);

With the normal size scale=1 and with \clip (-6,-4) rectangle (6,4);, we get this:

normal size and clip

\begin{tikzpicture}[scale=1]

\clip (-6,-4) rectangle (6,4);
\coordinate[label={[blue]left:top-front}] (top_front) at (0,0);
\coordinate[label={[blue]left:bottom-front}] (bottom_front) at (0,-3);
\coordinate[label={[blue]right:top-right}] (top_right) at (15:2.5cm);
\coordinate[label={[blue]left:top-left}] (top_left) at (165:2.5cm);

To place labels on the edges, simply place a node when building the edge cases and you get this:

labels on edge

\draw (top_front) -- (bottom_front)node[midway,above,sloped]{label 1};
\draw (top_front) -- (top_right)node[midway,above,sloped]{label 2};
\draw (top_front) -- (top_left)node[midway,above,sloped]{label 3};
\draw (top_right) -- (back_right)node[midway,above,sloped]{label 4};
\draw (bottom_front) -- (back_right)node[midway,above,sloped]{label 5};
\draw (top_left) -- (back_left)node[midway,above,sloped]{label 6};
\draw (bottom_front) -- (back_left)node[midway,above,sloped]{label 7};
\draw (top_back) -- (top_right)node[midway,above,sloped]{label 8};
\draw (top_back) -- (top_left)node[midway,above,sloped]{label 9};

For a diagonal, it is enough to create a path between two of its vertices. For example, for the diagonal (top_left)--(bottom_front), we may not draw this diagonal and mark the text as follows:

diagonal

\path (top_left)--(bottom_front)node[midway,above,sloped]{diagonal 1};

By drawing the diagonal in dotted lines:

diagonal -draw

we write this:

\path[draw,dashed] (top_left)--(bottom_front)node[midway,above,sloped]{diagonal 1};

or just as well this:

\draw[dashed] (top_left)--(bottom_front)node[midway,above,sloped]{diagonal 1};

complete code :

\documentclass{amsart}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\begin{document}
\begin{tikzpicture}[scale=1]

\clip (-6,-4) rectangle (6,4);
\coordinate[label={[blue]left:top-front}] (top_front) at (0,0);
\coordinate[label={[blue]left:bottom-front}] (bottom_front) at (0,-3);
\coordinate[label={[blue]right:top-right}] (top_right) at (15:2.5cm);
\coordinate[label={[blue]left:top-left}] (top_left) at (165:2.5cm);

%You can change the perspective by playing with the 5, 5, 15:
\coordinate (front_right) at ($(top_front)!5!(top_right)$);
\coordinate (front_left) at ($(top_front)!5!(top_left)$);
\coordinate (front_bottom) at ($(top_front)!15!(bottom_front)$);

\path[name path=bottom_right_path] (bottom_front) -- (front_right);   
\path[name path=right_back_path] (top_right) -- (front_bottom);
\path[name path=back_left_path] (bottom_front) -- (front_left);
\path[name path=left_back_path] (top_left) -- (front_bottom);
\path[name path=top_right_path] (top_left) -- (front_right);
\path[name path=top_left_path] (top_right) -- (front_left);

\coordinate[name intersections={of=bottom_right_path and right_back_path, by=back_right}];
\coordinate[name intersections={of=back_left_path and left_back_path, by=back_left}];
\coordinate[name intersections={of=top_right_path and top_left_path, by=top_back}];

%\shade[right color=gray!10, left color=black!50, shading angle=105] (top_front) -- (bottom_front) -- (back_left) -- (top_left) -- cycle;
%\shade[left color=gray!10, right color=black!50, shading angle=75] (top_front) -- (bottom_front) -- (back_right) -- (top_right) -- cycle;

\begin{scope}
\clip (top_front) -- (top_right) -- (top_back) -- (top_left) -- cycle;
\shade[inner color = gray!5, outer color=black!50, shading=radial] (top_front) ellipse (3cm and 1.5cm);
\end{scope}

\draw (top_front) -- (bottom_front)node[midway,above,sloped]{label 1};
\draw (top_front) -- (top_right)node[midway,above,sloped]{label 2};
\draw (top_front) -- (top_left)node[midway,above,sloped]{label 3};
\draw (top_right) -- (back_right)node[midway,above,sloped]{label 4};
\draw (bottom_front) -- (back_right)node[midway,above,sloped]{label 5};
\draw (top_left) -- (back_left)node[midway,above,sloped]{label 6};
\draw (bottom_front) -- (back_left)node[midway,above,sloped]{label 7};
\draw (top_back) -- (top_right)node[midway,above,sloped]{label 8};
\draw (top_back) -- (top_left)node[midway,above,sloped]{label 9};

\path (top_left)--(bottom_front)node[midway,above,sloped]{diagonal 1};

\end{tikzpicture}

\end{document}

Update: Creation of the bottom_back vertex of the cube.

% Creation of the bottom_back vertex of the cube.
% definition of 2 new paths
\path[name path=bottom_left_path] (back_left) -- (front_right);
\path[name path=back_right_path] (back_right) -- (front_left);
% vertex bottom_back definition
\coordinate[name intersections={of=bottom_left_path and back_right_path, by=bottom_back}];

%\node[red] at (back_right){back-right};
%\node[red] at (back_left){back-left};
%\node[red] at (top_back){top-back};
%\node[red] at (bottom_back){bottom-back};

% hidden sides
\draw[dashed] (back_left) -- (bottom_back);
\draw [dashed](back_right) -- (bottom_back);
\draw[dashed] (top_back) -- (bottom_back);

\draw[dashed,blue] (back_right) -- (top_left)node[midway, sloped, above]{diagonal};

I drew the hidden sides were drawn in dotted lines.

vertex

You will notice that the vertex (top_back), (top_front), (bottom_back) and (bottom_back) are aligned, the diagonal (bottom_back)--(top_front) will not be visible.

To do this, you must change the perspective. Which I'll leave it to you to do.

So I drew the visible diagonal (back_right) -- (top_left)

diagonals

Complete code updated:

\documentclass{amsart}

\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\begin{document}
\begin{tikzpicture}[scale=1]

\clip (-6,-4) rectangle (6,4);
\coordinate[label={[blue]left:top-front}] 
(top_front) at (0,0);
\coordinate[label={[blue]left:bottom-front}] 
(bottom_front) at (0,-3);
\coordinate[label={[blue]right:top-right}] 
(top_right) at (15:2.5cm);
\coordinate[label={[blue]left:top-left}]
 (top_left) at (165:2.5cm);

%You can change the perspective by playing with the 5, 5, 15:
\coordinate%[label={right:front-right}]  
(front_right) at ($(top_front)!5!(top_right)$);
\coordinate% [label={left:front-left}] 
(front_left) at ($(top_front)!5!(top_left)$);
\coordinate %[label={below:front-bottom}]
(front_bottom) at ($(top_front)!15!(bottom_front)$);

\path[name path=bottom_right_path] (bottom_front) -- (front_right);   
\path[name path=right_back_path] (top_right) -- (front_bottom);
\path[name path=back_left_path] (bottom_front) -- (front_left);
\path[name path=left_back_path] (top_left) -- (front_bottom);
\path[name path=top_right_path] (top_left) -- (front_right);
\path[name path=top_left_path] (top_right) -- (front_left);

\coordinate[name intersections={of=bottom_right_path and right_back_path, by=back_right}];
\coordinate[name intersections={of=back_left_path and left_back_path, by=back_left}];
\coordinate[name intersections={of=top_right_path and top_left_path, by=top_back}];

%\shade[right color=gray!10, left color=black!50, shading angle=105] (top_front) -- (bottom_front) -- (back_left) -- (top_left) -- cycle;
%\shade[left color=gray!10, right color=black!50, shading angle=75] (top_front) -- (bottom_front) -- (back_right) -- (top_right) -- cycle;

\begin{scope}
\clip (top_front) -- (top_right) -- (top_back) -- (top_left) -- cycle;
%\shade[inner color = gray!5, outer color=black!50, shading=radial] (top_front) ellipse (3cm and 1.5cm);
\end{scope}

\draw (top_front) -- (bottom_front);
\draw (top_front) -- (top_right);
\draw (top_front) -- (top_left);
\draw (top_right) -- (back_right);
\draw (bottom_front) -- (back_right);
\draw (top_left) -- (back_left);
\draw (bottom_front) -- (back_left);
\draw (top_back) -- (top_right);
\draw (top_back) -- (top_left);


\path[draw,dashed,red] (top_left)--(bottom_front)node[midway,above,sloped]{diagonal 1};

% Creation of the bottom_back vertex of the cube.
% definition of 2 new paths
\path[name path=bottom_left_path] (back_left) -- (front_right);
\path[name path=back_right_path] (back_right) -- (front_left);
% vertex bottom_back definition
\coordinate[name intersections={of=bottom_left_path and back_right_path, by=bottom_back}];

% vertex names
\node[red] at (back_right){back-right};
\node[red] at (back_left){back-left};
\node[red] at (top_back){top-back};
\node[red] at (bottom_back){bottom-back};

% hidden sides
\draw[dashed] (back_left) -- (bottom_back);
\draw [dashed](back_right) -- (bottom_back);
\draw[dashed] (top_back) -- (bottom_back);

\draw[dashed,blue] (back_right) -- (top_left)node[midway, sloped, above]{diagonal 2};
\end{tikzpicture}

\end{document}
AndréC
  • 24,137
  • May you include one more feature in one of your drawings? I see that you drew a diagonal along a face of the box. I do want that; please keep it in your code. I was asking for the code to draw a diagonal of the box. For example, the diagonal from the bottom_back vertex to the top_front vertex. I forgot to mention that I would like the bottom_back_vertex to be located and dashed lines drawn from it to top_back vertex, front_left vertex, and front_right vertex. – A gal named Desire Nov 18 '18 at 11:23
  • 2
    @AgalnamedDesire answer updated – AndréC Nov 18 '18 at 13:13
  • 1
    I have an additional request. I would like the dimensions of the box to be in the ratio 3:4:12. In the code, the height is specified to be 3. The length and width are calculated with commands like \coordinate (front_bottom) at ($(top_front)!15!(bottom_front)$);. (Would you like me to post another question?) – A gal named Desire Nov 21 '18 at 14:24
  • @AgalnamedDesire Yes, ask another question. – AndréC Nov 21 '18 at 15:26
  • "Drawing a box in TikZ - one edge closer to the observer than other edges - with perspective (Version 2)" – A gal named Desire Nov 27 '18 at 19:38
2

It would be great if you could try to reword your question a bit. In the meantime, I'd like to draw your attention to this great answer which allows you to draw 3d objects with perspective. All I did was to use the preamble of this answer to draw a cube in perspective, and to add a few elements. No intersections etc. are needed here.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{tikz-3dplot}
\usepgfmodule{nonlineartransformations}
% Max magic https://tex.stackexchange.com/a/447120/121799
\makeatletter 
% the first part is not in use here
\def\tikz@scan@transform@one@point#1{%
  \tikz@scan@one@point\pgf@process#1%
  \pgf@pos@transform{\pgf@x}{\pgf@y}}
\tikzset{%
  grid source opposite corners/.code args={#1and#2}{%
   \pgfextract@process\tikz@transform@source@southwest{%
     \tikz@scan@transform@one@point{#1}}%
   \pgfextract@process\tikz@transform@source@northeast{%
     \tikz@scan@transform@one@point{#2}}%
  },
  grid target corners/.code args={#1--#2--#3--#4}{%
   \pgfextract@process\tikz@transform@target@southwest{%
     \tikz@scan@transform@one@point{#1}}%
   \pgfextract@process\tikz@transform@target@southeast{%
     \tikz@scan@transform@one@point{#2}}%
   \pgfextract@process\tikz@transform@target@northeast{%
     \tikz@scan@transform@one@point{#3}}%
   \pgfextract@process\tikz@transform@target@northwest{%
     \tikz@scan@transform@one@point{#4}}%
  }
}

\def\tikzgridtransform{%
  \pgfextract@process\tikz@current@point{}%
  \pgf@process{%
    \pgfpointdiff{\tikz@transform@source@southwest}%
      {\tikz@transform@source@northeast}%
  }%
  \pgf@xc=\pgf@x\pgf@yc=\pgf@y%
  \pgf@process{%
    \pgfpointdiff{\tikz@transform@source@southwest}{\tikz@current@point}%
  }%
  \pgfmathparse{\pgf@x/\pgf@xc}\let\tikz@tx=\pgfmathresult%
  \pgfmathparse{\pgf@y/\pgf@yc}\let\tikz@ty=\pgfmathresult%
  %
  \pgfpointlineattime{\tikz@ty}{%
    \pgfpointlineattime{\tikz@tx}{\tikz@transform@target@southwest}%
      {\tikz@transform@target@southeast}}{%
    \pgfpointlineattime{\tikz@tx}{\tikz@transform@target@northwest}%
      {\tikz@transform@target@northeast}}%
}

% Initialize H matrix for perspective view
\pgfmathsetmacro\H@tpp@aa{1}\pgfmathsetmacro\H@tpp@ab{0}\pgfmathsetmacro\H@tpp@ac{0}%\pgfmathsetmacro\H@tpp@ad{0}
\pgfmathsetmacro\H@tpp@ba{0}\pgfmathsetmacro\H@tpp@bb{1}\pgfmathsetmacro\H@tpp@bc{0}%\pgfmathsetmacro\H@tpp@bd{0}
\pgfmathsetmacro\H@tpp@ca{0}\pgfmathsetmacro\H@tpp@cb{0}\pgfmathsetmacro\H@tpp@cc{1}%\pgfmathsetmacro\H@tpp@cd{0}
\pgfmathsetmacro\H@tpp@da{0}\pgfmathsetmacro\H@tpp@db{0}\pgfmathsetmacro\H@tpp@dc{0}%\pgfmathsetmacro\H@tpp@dd{1}

%Initialize H matrix for main rotation
\pgfmathsetmacro\H@rot@aa{1}\pgfmathsetmacro\H@rot@ab{0}\pgfmathsetmacro\H@rot@ac{0}%\pgfmathsetmacro\H@rot@ad{0}
\pgfmathsetmacro\H@rot@ba{0}\pgfmathsetmacro\H@rot@bb{1}\pgfmathsetmacro\H@rot@bc{0}%\pgfmathsetmacro\H@rot@bd{0}
\pgfmathsetmacro\H@rot@ca{0}\pgfmathsetmacro\H@rot@cb{0}\pgfmathsetmacro\H@rot@cc{1}%\pgfmathsetmacro\H@rot@cd{0}
%\pgfmathsetmacro\H@rot@da{0}\pgfmathsetmacro\H@rot@db{0}\pgfmathsetmacro\H@rot@dc{0}\pgfmathsetmacro\H@rot@dd{1}

\pgfkeys{
    /three point perspective/.cd,
        p/.code args={(#1,#2,#3)}{
            \pgfmathparse{int(round(#1))}
            \ifnum\pgfmathresult=0\else
                \pgfmathsetmacro\H@tpp@ba{#2/#1}
                \pgfmathsetmacro\H@tpp@ca{#3/#1}
                \pgfmathsetmacro\H@tpp@da{ 1/#1}
                \coordinate (vp-p) at (#1,#2,#3);
            \fi
        },
        q/.code args={(#1,#2,#3)}{
            \pgfmathparse{int(round(#2))}
            \ifnum\pgfmathresult=0\else
                \pgfmathsetmacro\H@tpp@ab{#1/#2}
                \pgfmathsetmacro\H@tpp@cb{#3/#2}
                \pgfmathsetmacro\H@tpp@db{ 1/#2}
                \coordinate (vp-q) at (#1,#2,#3);
            \fi
        },
        r/.code args={(#1,#2,#3)}{
            \pgfmathparse{int(round(#3))}
            \ifnum\pgfmathresult=0\else
                \pgfmathsetmacro\H@tpp@ac{#1/#3}
                \pgfmathsetmacro\H@tpp@bc{#2/#3}
                \pgfmathsetmacro\H@tpp@dc{ 1/#3}
                \coordinate (vp-r) at (#1,#2,#3);
            \fi
        },
        coordinate/.code args={#1,#2,#3}{
           \pgfmathsetmacro\tpp@x{#1} %<- Max' fix
            \pgfmathsetmacro\tpp@y{#2}
            \pgfmathsetmacro\tpp@z{#3}
        },
}

\tikzset{
    view/.code 2 args={
        \pgfmathsetmacro\rot@main@theta{#1}
        \pgfmathsetmacro\rot@main@phi{#2}
        % Row 1
        \pgfmathsetmacro\H@rot@aa{cos(\rot@main@phi)}
        \pgfmathsetmacro\H@rot@ab{sin(\rot@main@phi)}
        \pgfmathsetmacro\H@rot@ac{0}
        % Row 2
        \pgfmathsetmacro\H@rot@ba{-cos(\rot@main@theta)*sin(\rot@main@phi)}
        \pgfmathsetmacro\H@rot@bb{cos(\rot@main@phi)*cos(\rot@main@theta)}
        \pgfmathsetmacro\H@rot@bc{sin(\rot@main@theta)}
        % Row 3
        \pgfmathsetmacro\H@m@ca{sin(\rot@main@phi)*sin(\rot@main@theta)}
        \pgfmathsetmacro\H@m@cb{-cos(\rot@main@phi)*sin(\rot@main@theta)}
        \pgfmathsetmacro\H@m@cc{cos(\rot@main@theta)}
        % Set vector values
        \pgfmathsetmacro\vec@x@x{\H@rot@aa}
        \pgfmathsetmacro\vec@y@x{\H@rot@ab}
        \pgfmathsetmacro\vec@z@x{\H@rot@ac}
        \pgfmathsetmacro\vec@x@y{\H@rot@ba}
        \pgfmathsetmacro\vec@y@y{\H@rot@bb}
        \pgfmathsetmacro\vec@z@y{\H@rot@bc}
        % Set pgf vectors
        \pgfsetxvec{\pgfpoint{\vec@x@x cm}{\vec@x@y cm}}
        \pgfsetyvec{\pgfpoint{\vec@y@x cm}{\vec@y@y cm}}
        \pgfsetzvec{\pgfpoint{\vec@z@x cm}{\vec@z@y cm}}
    },
}

\tikzset{
    perspective/.code={\pgfkeys{/three point perspective/.cd,#1}},
    perspective/.default={p={(15,0,0)},q={(0,15,0)},r={(0,0,50)}},
}

\tikzdeclarecoordinatesystem{three point perspective}{
    \pgfkeys{/three point perspective/.cd,coordinate={#1}}
    \pgfmathsetmacro\temp@p@w{\H@tpp@da*\tpp@x + \H@tpp@db*\tpp@y + \H@tpp@dc*\tpp@z + 1}
    \pgfmathsetmacro\temp@p@x{(\H@tpp@aa*\tpp@x + \H@tpp@ab*\tpp@y + \H@tpp@ac*\tpp@z)/\temp@p@w}
    \pgfmathsetmacro\temp@p@y{(\H@tpp@ba*\tpp@x + \H@tpp@bb*\tpp@y + \H@tpp@bc*\tpp@z)/\temp@p@w}
    \pgfmathsetmacro\temp@p@z{(\H@tpp@ca*\tpp@x + \H@tpp@cb*\tpp@y + \H@tpp@cc*\tpp@z)/\temp@p@w}
    \pgfpointxyz{\temp@p@x}{\temp@p@y}{\temp@p@z}
}
\tikzaliascoordinatesystem{tpp}{three point perspective}

\makeatother

\begin{document}
  \tdplotsetmaincoords{70}{-22.5}
  \begin{tikzpicture}[scale=6,font=\sffamily,
  view={\tdplotmaintheta}{\tdplotmainphi},
            perspective={
                p = {(4,0,1.5)},
                q = {(0,4,1.5)},
            }
  ]
  \begin{scope} % top face
   \draw[clip]   (tpp cs:0,0,1) coordinate (top_front)
  -- (tpp cs:1,0,1) coordinate (top_right)
   -- (tpp cs:1,1,1) coordinate (top_back)
   -- (tpp cs:0,1,1) coordinate (top_left) --
    cycle;
    \shade[inner color = gray!5, outer color=black!50, shading=radial] (top_front) ellipse (3cm and 1.5cm);
  \end{scope}
  % right face  
  \draw   (top_front) -- (top_right) -- (tpp cs:1,0,0) coordinate (bottom_right)
  -- (tpp cs:0,0,0)  coordinate (bottom_front) --   cycle;
  % top face  
  \draw   (top_front) -- (top_left) -- (tpp cs:0,1,0) coordinate (bottom_left)
  -- (bottom_front) --  cycle;
  % your midway coordinates with Max perspective coordinate system
  \coordinate (front_right) at (tpp cs:0.5,0,1);
  \coordinate (front_left) at (tpp cs:0,0.5,1);
  \coordinate (front_bottom) at (tpp cs:0,0,0.5);
  \path (0,0,0.5) -- (1,0,0.5) node[pos=0.1,sloped] {a label};
  \draw (top_left) -- (bottom_front) node[midway,sloped,above] {a sloped label};
  \end{tikzpicture}
\end{document}

enter image description here

In principle, this great answer also provides tools to write text in perspective. Is this what you are asking here?

  • I was considering typesetting text in perspective. I think that I am going to avoid that, though. – A gal named Desire Nov 27 '18 at 18:53
  • Keep the link for other members of tex.stackexchange. – A gal named Desire Nov 27 '18 at 18:54
  • @AgalnamedDesire Do what you think is right. However, please consider asking a clearer question. In the present situation, users will get dragged away from Max' great answer to some post in which the coordinates are guessed. I do not care about reputation points, but personally I would find it here very unfortunate if the impression was that there is no better way than guessing the coordinates. You should at least explicitly say that you do not care about computing the coordinates but are fine if they are guessed. –  Nov 27 '18 at 19:10
  • I asked how to draw a box with perspective. I got a response that prompted TikZ to compute the coordinates of a box. There is no guessing of coordinates. – A gal named Desire Nov 27 '18 at 19:28
  • @AgalnamedDesire Just for the records: I disagree with that statement. Max great answer computes the screen coordinates from their 3D coordinates, see here for more details. AndréC's answer definitely has an appealing output, but it does not use these techniques as far as I can see. It is fine if we disagree, but I will keep advertizing Max great answer for these reasons. –  Nov 27 '18 at 19:44