How to draw a Spherical Segment? - TeX

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user 3.14159

Here is a possible way to do that with the [experimental 3dtools library](https://github.com/marmotghost/tikz-3dtools). Please note that this version assumes that both circles have a hidden and a visible stretch. One *can* generalize this to the general case, but I am not sure if this is needed, and this will definitely make the code longer, so I let it be.
```
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,decorations.pathreplacing,3dtools} % https://github.com/marmotghost/tikz-3dtools
\pgfdeclarelayer{background} 
\pgfdeclarelayer{foreground}
\pgfsetlayers{background,main,foreground}
\begin{document}
\begin{tikzpicture}[3d/install view={phi=0,theta=60},
	declare function={R=3;a=2.9;b=2;h1=sqrt(R*R-a*a);h2=sqrt(R*R-b*b;},
	3d/hidden/.append style={on layer=background},
	3d/visible/.append style={on layer=foreground}]
 \path[3d/hidden/.append style={postaction={store path=H1}},
 	3d/visible/.append style={postaction={store path=V1}}] 
		pic{3d/circle on sphere={R=R,P={(0,0,h1)}}};
 \path[3d/hidden/.append style={postaction={store path=H2}},
 	3d/visible/.append style={postaction={store path=V2}}] 
		pic{3d/circle on sphere={R=R,P={(0,0,h2)}}};
 \draw[3d/hidden] [stored path/first coordinate of=V2] coordinate (aux2)
  		[stored path/first coordinate of=V1] coordinate (aux1)
		(b,0,h2) -- node[below] {$b$}(0,0,h2) -- (0,0,h1)
			 --node[below] {$a$}(a,0,h1);		
 \path[3d/screen coords,fill=cyan,opacity=0.6,stored path/restore reversed path=V2]
   let \p1=(aux1),\p2=(aux2),\n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)} in
   -- (aux2) arc[start angle=\n2,end angle=\n1,radius=R]
   [stored path/append path=V1]
   arc[start angle=180-\n1,end angle=180-\n2,radius=R];		
 \path[fill=blue,opacity=0.6,stored path/restore reversed path=V2,
 	stored path/append path=H2] --cycle;
 \draw[3d/screen coords,3d/visible] (0,0) circle[radius=R];
\end{tikzpicture}
\end{document}
```
![Screen Shot 2021-01-15 at 5.27.26 PM.png](/image?hash=1efe461394ecc196e20d2cea96b5135d366629f08a68c7095aaae206b260df18)

If `I` does not coincide with `O`, the circle center, then it will only be a circle if the normal `n` and `I-O` are parallel. Then you can do something like ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools} % https://github.com/marmotghost/tikz-3dtools \begin{document} \begin{tikzpicture} \begin{scope}[3d/install view={phi=110,theta=70},declare function={R=4;}] \path (0,0,0) coordinate (O) (0.5,-1.2,1) coordinate (I); \shade[ball color=white,3d/screen coords] (I) circle[radius=R]; \path (I) pic{3d/circle on sphere={R=R,C={(O)},P={(I)}}}; \end{scope} \end{tikzpicture} \end{document} ```

@marmot, re: [your answer](#a1849), I use tikz-3dplot-circleofsphere, in the syntax `\tdplotCsDrawCircle[style]{r}{alpha}{beta}{epsilon}`, if the circle has center `O`and is intersection of the plane has normal vector `n= (a,b,c)` and the sphere has center `I` and radius `r`. How can I find `{alpha}{beta}{epsilon}` from `a, b, c` and `r`? Please help me.

You can define a growing radius. However, in order to keep the length of the segments fixed, you need to compute the length of the curve. I did this quickly with Mathematica, and found a value that works for these precise parameters. ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{3dtools} % https://github.com/marmotghost/tikz-3dtools \begin{document} \begin{tikzpicture}[3d/install view={phi=0,theta=70}, line cap=round,line join=round, declare function={R=2;h=4;a=0.48;%<-numerical solution by hand reff(\t,\a)=R*(1+abs(\t-\a)/180*abs(\t-\a)/180);}] \draw[fill=gray!20] (-R,0,h) coordinate (D') arc[start angle=180,end angle=0,radius=R] -- (R,0,0) arc[start angle=0,end angle=180,radius=R] -- cycle; \draw[smooth,fill=white,fill opacity=0.6] plot[variable=\t,domain=180:180+a*180] ({reff(\t,180)*cos(\t)},{reff(\t,180)*sin(\t)},h) -- plot[variable=\t,domain=180+a*180:180] ({reff(\t,180)*cos(\t)},{reff(\t,180)*sin(\t)},0) -- cycle; \draw[smooth,fill=white,fill opacity=0.6] plot[variable=\t,domain=0:-a*180] ({reff(\t,0)*cos(\t)},{reff(\t,0)*sin(\t)},h) -- plot[variable=\t,domain=-a*180:0] ({reff(\t,0)*cos(\t)},{reff(\t,0)*sin(\t)},0) -- cycle; \end{tikzpicture} \end{document} ``` ![Screen Shot 2021-01-17 at 8.49.44 PM.png](/image?hash=1248e855c55ce4fe0ec273a7126b650dce6efc4f74b5a44693930cc912ac925f)

@marmot, re: [your answer](#a1849), How to draw this picture (cut a rectangle to make a cylinder)? ![ScreenHunter 1090.png](/image?hash=b7343a573d469499ef9bd9c6df57beccdc3734500cf0dbab6feb952fadb655ff)

For instance: ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{calc,decorations.pathreplacing,3dtools} % https://github.com/marmotghost/tikz-3dtools \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground} \begin{document} \begin{tikzpicture}[3d/install view={phi=0,theta=80}, declare function={R=5;a=R-.5;b=R-1.2;h1=sqrt(R*R-a*a);h2=sqrt(R*R-b*b;}] \path[3d/hidden/.append style={postaction={store path=H1}}, 3d/visible/.append style={postaction={store path=V1}}] pic{3d/circle on sphere={R=R,P={(0,0,h1)}}}; \path[3d/hidden/.append style={postaction={store path=H2}}, 3d/visible/.append style={postaction={store path=V2}}] pic{3d/circle on sphere={R=R,P={(0,0,h2)}}}; \draw[3d/hidden] [stored path/first coordinate of=V2] coordinate (aux2) [stored path/first coordinate of=V1] coordinate (aux1); \path[3d/screen coords,fill=cyan,opacity=0.6,stored path/restore reversed path=V2] let \p1=(aux1),\p2=(aux2),\n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)} in -- (aux2) arc[start angle=\n2,end angle=\n1,radius=R] [stored path/append path=V1] arc[start angle=180-\n1,end angle=180-\n2,radius=R]; \draw[3d/screen coords,3d/visible] (0,0,0) circle[radius=R]; \clip[3d/screen coords] let \p1=(aux1) in (-R,\y1) rectangle (R,-0.1); \path[3d/cone/outer/.style={fill=blue}] (0,0,h1) pic{3d/cone={r=a,h=-h1}}; \end{tikzpicture} \end{document} ``` ![Screen Shot 2021-01-15 at 8.43.02 PM.png](/image?hash=d89f984328790ea4a59d8ae6f51de50aa4f22a0cd7fcd3f2741bbb3de3cdce0b)

With ``` declare function={R=5;a=R-.3;b=R-1;h1=sqrt(R*R-a*a);h2=sqrt(R*R-b*b;} ``` I get ![Screen Shot 2021-01-15 at 6.53.42 PM.png](/image?hash=142cff4c343bb74d4de3e9a42e528593c1efe3acbb7efe39da65324e72c0b6d9) Possibly the problem is that, as mentioned, the code assumes that both circles have both a hidden and visible stretch.

Sure. ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{calc,decorations.pathreplacing,3dtools} % https://github.com/marmotghost/tikz-3dtools \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground} \begin{document} \begin{tikzpicture}[3d/install view={phi=0,theta=60}, declare function={R=3;a=2.9;b=2;h1=sqrt(R*R-a*a);h2=sqrt(R*R-b*b;}, 3d/hidden/.append style={on layer=background}, 3d/visible/.append style={on layer=foreground}] \path[3d/hidden/.append style={postaction={store path=H1}}, 3d/visible/.append style={postaction={store path=V1}}] pic{3d/circle on sphere={R=R,P={(0,0,h1)}}}; \path[3d/hidden/.append style={solid,postaction={store path=H2}}, 3d/visible/.append style={postaction={store path=V2}}] pic{3d/circle on sphere={R=R,P={(0,0,h2)}}}; \draw[3d/hidden] [stored path/first coordinate of=V2] coordinate (aux2) [stored path/first coordinate of=V1] coordinate (aux1) (b,0,h2) -- node[below] {$b$}(0,0,h2) -- (0,0,h1) --node[below] {$a$}(a,0,h1); \path[3d/screen coords,fill=cyan,opacity=0.6,stored path/restore reversed path=V2] let \p1=(aux1),\p2=(aux2),\n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)} in -- (aux2) arc[start angle=\n2,end angle=\n1,radius=R] [stored path/append path=V1] arc[start angle=180-\n1,end angle=180-\n2,radius=R]; \path[fill=blue,opacity=0.6,stored path/restore reversed path=V2, stored path/append path=H2] --cycle; \draw[3d/screen coords,3d/visible] let \p1=(aux2),\n1={atan2(\y1,\x1)} in (\n1:R) arc[start angle=\n1,end angle=-180-\n1,radius=R]; \end{tikzpicture} \end{document} ``` ![Screen Shot 2021-01-15 at 6.12.15 PM.png](/image?hash=8f31c71650bed57eb2b1e83d6887fc2584a2f0040b7384d8e5b3b7df6b69949d)

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