Level of water in a cylinder with 3dtools - TeX

Anonymous 1123 replying to marmot

I have just posted my answer and thank to you.

Anonymous 1123

Very nice. Thanks.

marmot replying to Anonymous 1123

This does such clips.  
```  
\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}   
```

Anonymous 1123 replying to marmot

I try to find it. Thank you very much.

marmot replying to Anonymous 1123

Well, yes, but then you need to compute the intersection of the circle with the sphere. I was assuming that this is `(C)` because the sphere runs throug `(C)` but what I did not check is if the intersection is on the back side, which it seems to be. If you know that this is the case, you can just use clips.

Anonymous 1123 replying to marmot

Another view  
![ScreenHunter 102.png](/image?hash=35dcdbdf1f1dec3b3c3e8395948bf72a859dcda00c6991039f531ad96a47f3ca)

Anonymous 1123 replying to marmot

I use GeoSpace, like this  
![ScreenHunter 101.png](/image?hash=19967c22ebdb45cbd40d3446fe038f8cb1d619ebaa975adb5fd1886c8d0bdef6)

marmot replying to Anonymous 1123

Maybe  
```  
\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={(A)},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)")}%	 
		\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}

		\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}   
```

Anonymous 1123 replying to marmot

The dashed line of the circle `BEFC` out side of the sphere incorrect. How can I fix it?  
![ScreenHunter 100.png](/image?hash=d571d8c7ca57b634579b8e072eaec4f785fb4b639dedf846959d3ec09bca9429)

Anonymous 1123 replying to marmot

I am trying to write answer [this question](https://tex.stackexchange.com/questions/470177/how-to-draw-a-circle-sphere-passing-through-four-points/470182#470182) by using `3dtools`.   
My code  
```  
\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={(A)},B={(B)},C={(C)}}] coordinate (M);

\path pic[3d/hidden]{3d circle through 3 points={%
		A={(B)},B={(E)},C={(F)},center name=M}};
	
\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}   
```

marmot replying to Anonymous 1123

You can also dash the line in the background.  
```  
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools
\begin{document}
\begin{tikzpicture}[line cap=round,line join=round,
	3d/install view={phi=0,theta=70},
	declare function={H=4;h=2.5;R=2;r=h*R/H;alpha=atan2(R,H);}]
 \path (0,0,-H+h) coordinate (P) 
 	pic[3d/visible/.style={draw=none},3d/hidden/.style={draw=none},
 		3d/cone/inner/.style={fill=blue!50},
		3d/cone/outer/.style={fill=blue!80}]{3d/cone={h=-h,r=r}}
	(0,0,0) pic{3d/cone={h=-H,r=R}}
	pic{3d/circle on
	sphere={C={(0,0,{-H+h+r*tan(alpha)})},P={(P)},R/.evaluated={r/cos(alpha)}}};
\end{tikzpicture}
\end{document}   
```

Anonymous 1123 replying to marmot

Thanks.

marmot replying to Anonymous 1123

Maybe like this?  
```  
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools
\begin{document}
\begin{tikzpicture}[line cap=round,line join=round,
	3d/install view={phi=0,theta=70},
	declare function={H=4;h=2.5;R=2;r=h*R/H;}]
 \path (0,0,-H+h) pic[%3d/visible={draw=none},3d/hidden={draw=none},
 	3d/cone/inner/.style={fill=blue!50},
	3d/cone/outer/.style={fill=blue!80}]{3d/cone={h=-h,r=r}}
	(0,0,0) pic{3d/cone={h=-H,r=R}};
\end{tikzpicture}
\end{document}  
```

Anonymous 1123

@marmot, re: [your answer](#a1974),  How about water in cone, e.g. [here](https://tex.stackexchange.com/questions/290797/draw-background-of-water-in-cone)?

marmot replying to Anonymous 1123

It is there for a long time, just not documented.  
```  
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{3dtools}% https://github.com/marmotghost/tikz-3dtools
\begin{document}
\begin{tikzpicture}[line cap=round,line join=round,
	3d/install view={phi=0,theta=70}]
 \path (1,2,3) coordinate (A) (4,5,6) coordinate (B)  (7,8,9) coordinate (C);	
 \pgfmathsetmacro{\mybarycenter}{barycenter("(A),(B),(C)")}
 \path (\mybarycenter) coordinate (bary);
\end{tikzpicture}
\end{document}  
```

Anonymous 1123 replying to marmot

Have you just added `barycenter of a list of coordinates` into `3dtools`?

Anonymous 1123 replying to marmot

Thanks. Now it works.

marmot replying to Anonymous 1123

Yes, there is an issue. I did not use `fpu` for one computation, so there was a `dimension too large` error. I updated the library https://github.com/marmotghost/tikz-3dtools/blob/master/tikzlibrary3dtools.code.tex.

Anonymous 1123 replying to marmot

This code does not work. I get the same result  
```  
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools
\begin{document}
	\begin{tikzpicture}[line cap=round,line join=round,
		3d/install view={phi=0,theta=70},
		declare function={R=3;}]
	\pic{3d/frustum={r=R,R=R,h=8}};  \end{tikzpicture}
\end{document}  
```

Anonymous 1123 replying to marmot

I tried, and cannot the result  
```  
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools
\begin{document}
	\begin{tikzpicture}[line cap=round,line join=round,
		3d/install view={phi=0,theta=70},
		declare function={R=3;}]
		\pic{3d/frustum={r=3,R=3,h/.evaluated=2*R}};
	\end{tikzpicture}
\end{document}  
```    
I get
![ScreenHunter 99.png](/image?hash=0b00a3c80a5a1143fff2703a6fec9e73eda8cbf288cfc7553f23d49f4630a533)

marmot replying to Anonymous 1123

I think it should be   
```  
\pic{3d/frustum={r=R,R=R,h/.evaluated=2*R}};  
```

Anonymous 1123

@marmot, re: [your answer](#a1974),  My friend told me that syntax `\pic{3d/frustum={r=R,R=R,h=2*R}};` get a bug when `h = 2*R`.

1 Answer