How to draw projection a circle on a plane with 3dtools? - TeX

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user 3.14159

You do not necessarily need `3dtools` for that, but in order to have a sloped xy plane it may be a convenient tool. Yet, the main players are still keys like `canvas is xy plane at z=0`.

```
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{3dtools}
\begin{document}
\begin{tikzpicture}[pics/my circle/.style={code={%
   \def\pv##1{\pgfkeysvalueof{/tikz/my circle/##1}}	
   \tikzset{my circle/.cd,#1}	
   \draw circle[radius=\pv{r}] foreach \XX/\YY in {A/-135,B/45,C/135,D/-45}
   {(\YY:\pv{r})coordinate (-\XX) (\YY:\pv{r}+\pv{b})node{$\XX\pv{label suffix}$}}
   (-A) -- pic[midway,sloped]{right angle} (-B) (-C) -- (-D);
 }},my circle/.cd,r/.initial=1,a/.initial=0.25,b/.initial=0.2,label suffix/.initial={},
 /tikz/.cd,pics/right angle/.style={code={
   \def\pv##1{\pgfkeysvalueof{/tikz/right angle/##1}}	
   \tikzset{right angle/.cd,#1}	
  \draw (\pv{a},0) -- (\pv{a},\pv{a}) -- (0,\pv{a})
  (\pv{b},\pv{b}) coordinate (-l);
 }},right angle/.cd,a/.initial=0.2,l/.initial={},b/.initial=0.4,
 /tikz/.cd,pics/arrow/.style={code={
 	\draw[semithick,-stealth] (-0.3,0) -- (0,0);}}]
 \begin{scope}[3d/install view={phi=20,theta=70}]
  \begin{scope}[canvas is xz plane at y=0]
   \draw[right angle/l={$O$}] (-2,0) rectangle (2,4) 
   	pic[midway,transform shape](xz){my circle} (xz-l) node[transform shape]{$O$};
  \end{scope}
  \path (0,0,0) coordinate (O) (1,0,0) coordinate (B) 
   (0,{cos(30)},{sin(30)}) coordinate (C);
  \tikzset{3d/define orthonormal dreibein={A={(O)}}}
  \begin{scope}[x={(ex)},y={(ey)},z={(ez)},canvas is xy plane at z=0]
   \draw[right angle/l={$O$}] (-2,-4) rectangle (2,0) 
   	pic[midway,transform shape,yscale=-1,nodes={yscale=-1}](xy){my circle={label suffix='}} 
	(xy-l) node[transform shape]{$O'$};
  \end{scope}
 \end{scope}
 \draw[semithick] foreach \X in {A,B,C,D} 
 {(xz-\X) -- (xy-\X) pic[pos=0.5,sloped,allow upside down]{arrow}};
\end{tikzpicture}
\end{document}
```
![Screen Shot 2020-12-19 at 8.06.46 AM.png](/image?hash=d1fcc0247d3dd9a6fe61b0737a922f577e09d266133704092bf30eed41f864d5)

I was wrong, there was a commented out code for the projected parabola. So one could use ``` \documentclass{standalone} \usepackage{pgfplots} \pgfplotsset{compat=1.18} \usetikzlibrary{3d,calc} \makeatletter \pgfplotsset{get xy window/.code={\pgfkeys{/pgf/fpu}% \pgfmathfloattofixed{\pgfplots@xmax}\let\myxmax\pgfmathresult \pgfmathfloattofixed{\pgfplots@xmin}\let\myxmin\pgfmathresult \pgfmathfloattofixed{\pgfplots@ymax}\let\myymax\pgfmathresult \pgfmathfloattofixed{\pgfplots@ymin}\let\myymin\pgfmathresult \pgfkeys{/pgf/fpu=false}}, window/ds/.initial=1} \makeatother \pgfplotsset{draw xy background/.style={% execute at end axis={\pgfplotsset{get xy window}% \pgfmathsetmacro{\myxmineff}{\myxmin-\myymax+\myymin}% \pgfmathsetmacro{\myxminnext}{\myxmineff+\pgfkeysvalueof{/pgfplots/window/ds}}% \pgfmathsetmacro{\myxmaxeff}{\myxmax+\myymax-\myymin}% \draw[window background] foreach \X in {\myxmineff,\myxminnext,...,\myxmax} {(axis cs:\X,\myymin) -- ++ (axis direction cs:\myymax-\myymin,\myymax-\myymin)}; }}} \tikzset{window background/.style={thin}} \begin{document} \pgfdeclarelayer{background} \pgfdeclarelayer{main} \pgfsetlayers{background,main} \begin{tikzpicture}[line join=round,line cap=round] \begin{axis}[axis equal, xmin=-.5, xmax=4.5, ymin=-.5, ymax=4.5, draw xy background] \addplot[black, opacity=0.5,smooth] {(x-3)^2}; \end{axis} \end{tikzpicture} \begin{tikzpicture}[line join=round,line cap=round] \begin{scope}[ x={(-0.7cm,0.4cm)}, y={(.9cm,.2cm)}, z={(0cm,1cm)}, canvas is yx plane at z=0 ] \begin{axis}[axis equal, xmin=-.5, xmax=4.5, ymin=-.5, ymax=4.5, draw xy background] \addplot[black, opacity=0.5,smooth] {(x-3)^2}; \end{axis} \end{scope} \end{tikzpicture} \end{document} ```

One cannot use a pattern for that because patterns cannot be projected, at least not without major efforts. One can definitely install a background pattern using e.g. [these methods](https://topanswers.xyz/tex?q=1741#a1971) to determine the dimensions of the plot (rather than what the OP is doing). However, the question is not very well written, for instance the code for the 3d plot is not disclosed, so it is not too surprising that there is no answer.

@marmot, re: [your answer](#a1816), This question have not yet the answer [here](https://tex.stackexchange.com/questions/600723/replicate-axis-region-reference-frame-and-drawing-to-it-saving-coordinates-2d).

My code ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools \begin{document} \begin{tikzpicture}[3d/install view={phi=70,theta=70},line cap=butt,line join=round,c/.style={circle,fill,inner sep=1pt}, declare function={a=3;}] \path (a,-a,-a) coordinate (A) (a,a,-a) coordinate (B) (-a,a,-a) coordinate (C) (-a,-a,-a) coordinate (D) (a,-a,a) coordinate (E) (a,a,a) coordinate (F) (-a,a,a) coordinate (G) (-a,-a,a) coordinate (H) (0,0,0) coordinate (O) ; \tikzset{3d/polyhedron/.cd,O={(O)}, back/.style={3d/polyhedron/complete dashes,fill=none}, fore/.style={3d/visible,fill=none},%3d/polyhedron/edges have complete dashes=false draw face with corners={{(B)},{(C)},{(G)},{(F)}}, draw face with corners={{(D)},{(C)},{(G)},{(H)}}, draw face with corners={{(E)},{(F)},{(G)},{(H)}}, draw face with corners={{(A)},{(B)},{(C)},{(D)}}, draw face with corners={{(A)},{(B)},{(F)},{(E)}}, draw face with corners={{(A)},{(E)},{(H)},{(D)}} } \path foreach \p/\g in {A/-90,B/90,C/0,D/0,E/0,F/0,G/0,H/0} {(\p)node[c]{}+(\g:2.5mm) node{$\p$}}; \end{tikzpicture} \end{document} ```

Yes, this is not an orthonormal projection of a cube. [Here](https://tex.stackexchange.com/a/448048) you can find the proof. This does not prevent you from trying your luck and to use a non-orthographic projection. In some cases this still works by accident. ``` \documentclass[border=3mm,tikz]{standalone} \usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools \begin{document} \begin{tikzpicture} \path foreach \Xa/\Ya in {-1/A,1/B} {foreach \Xb/\Yb in {-1/A,1/B} {foreach \Xc/\Yc in {-1/A,1/B} {(\Xa,\Xb,\Xc) coordinate (\Ya\Yb\Yc)}}} ; \tikzset{3d/polyhedron/.cd, fore/.append style={fill opacity=0.5}, edges have complete dashes, draw face with corners={{(AAA)},{(AAB)},{(ABB)},{(ABA)}}, draw face with corners={{(AAA)},{(AAB)},{(BAB)},{(BAA)}}, draw face with corners={{(AAA)},{(BAA)},{(BBA)},{(ABA)}}, draw face with corners={{(BAA)},{(BAB)},{(BBB)},{(BBA)}}, draw face with corners={{(ABA)},{(ABB)},{(BBB)},{(BBA)}}, draw face with corners={{(AAB)},{(BAB)},{(BBB)},{(ABB)}}} \end{tikzpicture} \end{document} ``` ![Screen Shot 2021-05-11 at 7.25.21 PM.png](/image?hash=1a5feeee65d7e3023974bc43ea00ed8949ff34b16ecd69b5ab8de1ff6bb504db) However, when doing this is there is no well-defined notion of rotating the cube, and so on.

Thank. I see this picture about a cube. I tried `3dtools` and GeoSpace, I do not know how to set up `3d/install view`. ![ScreenHunter 147.png](/image?hash=b240947c8a469f48382ba7ad4156be5fbb9ba392c2ca8a999532e162347c2cfd)

Not sure id [this](https://en.wikipedia.org/wiki/Projection_(linear_algebra)) answers your question. Maybe you mean [this](https://en.wikipedia.org/wiki/Perspective_(graphical)) by non-orthogonal projection, but I do not know if "non-orthogonal projection" is a term.

1 Answer