How to draw intersection of a plane and a cone with 3dtool? - TeX

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user 3.14159

Here is a first version. Given a plane that has a slope `alpha` and a cone of fixed base radius and height, you can specify the position of the highest (for `alpha<0`) or lowest (for `alpha>0`) point `I` of the intersection. This point is fixed by its z component `Iz` and a polar angle `gamma`. The intersection curve is then specified. The contour will be terminated when it starts to cross the base circle.

```
\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools
\tikzset{pics/3d/intersection of cone with plane/.style={code={
   \def\pv##1{\pgfkeysvalueof{/tikz/3d/intersection of cone with plane/##1}}		
   \tikzset{declare function={%	
    iconeplaneradius(\x,\y,\z)=\z/(1+\y*cos(\x));% https://tex.stackexchange.com/a/457458
	iconeplaneh(\x)=\pv{h}*(\pv{r}-\x)/\pv{r};
   },
    3d/intersection of cone with plane/.cd,#1,
   	e/.evaluated={cos(90+\pv{alpha})/cos(\pv{beta})},
	s/.evaluated={(1+\pv{e})*\pv{r}*(\pv{h}-\pv{Iz})/(\pv{h}*sin(90+\pv{alpha}))},
	beta/.evaluated={atan2(\pv{r},\pv{h})}}
   \pgfmathtruncatemacro{\itest}{(\pv{e}<0.1?0:1)}	
   \ifnum\itest=0\relax
    \pgfmathsetmacro{\tcrit}{180}
   \else
    \pgfmathsetmacro{\tmp}{(\pv{s}-\pv{r})/(\pv{r}*\pv{e})}
	\pgfmathtruncatemacro{\itest}{(abs(\tmp)<1?1:0)}	
	\ifnum\itest=1
	 \pgfmathsetmacro{\tcrit}{abs(acos(\tmp))}
	 %\typeout{tcrit=\tcrit}
	\else
     \pgfmathsetmacro{\tcrit}{180}
	\fi
   \fi
%    \path ({\pv{r}*cos(\pv{gamma})*(\pv{h}-\pv{Iz})/\pv{h}},
%    	{\pv{r}*sin(\pv{gamma})*(\pv{h}-\pv{Iz})/\pv{h}},\pv{Iz}) coordinate (I);
%    \typeout{alpha=\pv{alpha}, beta=\pv{beta}, e=\pv{e}, s=\pv{s}}
%    \pgfmathsetmacro{\Itest}{TD("({iconeplaneradius(0,\pv{e},\pv{s})*cos(0-\pv{gamma})},
%    	{iconeplaneradius(0,\pv{e},\pv{s}))*sin(0-\pv{gamma})},{iconeplaneh(iconeplaneradius(0,\pv{e},\pv{s}))})")}	
%    \path (\Itest) node[dot]{};	
%    \typeout{(Itest)=(\Itest)}	
   \draw[pic actions] plot[variable=\t,domain=-\tcrit:\tcrit,samples=101,smooth]
     ({iconeplaneradius(\t,\pv{e},\pv{s})*cos(\t-\pv{gamma})},
  	{iconeplaneradius(\t,\pv{e},\pv{s}))*sin(\t-\pv{gamma})},
	{iconeplaneh(iconeplaneradius(\t,\pv{e},\pv{s}))});
}},/tikz/3d/intersection of cone with plane/.cd,
		h/.initial=5,% height of the cone
		r/.initial=3,% radius of the base
  		gamma/.initial=0,% angle of intersection
		Iz/.initial=3,% z coordinate of the intersection
		alpha/.initial=0,% slope of the plane
		beta/.initial=0,% will be computed
		e/.initial=1,
		s/.initial=1}
\begin{document}
\begin{tikzpicture}
 \begin{scope}[3d/install view={phi=30,theta=70},
 	declare function={R=3;% radius of the base of the cone
	H=5;% height of the cone
	},
	dot/.style={circle,fill,inner sep=1.2pt,label=above:{#1}}]
  \path 
   (0,0,0) coordinate (O) % center of base of the cone (fixed)
   (0,0,H) coordinate (T); % tip of the cone
%   
  \path (O) pic{3d/cone={r=R,h=H}};
% 
  \begin{scope}[3d/intersection of cone with plane/.cd,r=R,h=H,gamma=-20]
  \path pic[fill=blue]{3d/intersection of cone with plane={Iz=2.75,alpha=-45}};
  \path pic[fill=green!70!black]{3d/intersection of cone with plane={Iz=3,alpha=-30}};
  \path pic[fill=red]{3d/intersection of cone with plane={Iz=3.25,alpha=-15}};
  \path pic[fill=orange]{3d/intersection of cone with plane={Iz=3.5,alpha=0}};
  \end{scope}
 \end{scope}
\end{tikzpicture}
\end{document}
```

![Screen Shot 2020-12-09 at 11.35.15 PM.png](/image?hash=e01da05f5911c0f583baec25e664f78ba9f9860fc95facf7f9b01fa861090938)

This is a (not thoroughly tested) version that discriminates between visible and hidden contour parts. However, I do not know what you mean by "add plane to the picture". A plane is usually an infinitely large object, without further specification it is not clear what you want to see. ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools \makeatletter \tikzset{pics/3d/intersection of cone with plane/.style={code={ \def\pv##1{\pgfkeysvalueof{/tikz/3d/intersection of cone with plane/##1}} \tikzset{declare function={% iconeplaneradius(\x,\y,\z)=\z/(1+\y*cos(\x));% https://tex.stackexchange.com/a/457458 iconeplaneh(\x)=\pv{h}*(\pv{r}-\x)/\pv{r}; }, 3d/intersection of cone with plane/.cd,#1, e/.evaluated={cos(90+\pv{alpha})/cos(\pv{beta})}, s/.evaluated={(1+\pv{e})*\pv{r}*(\pv{h}-\pv{Iz})/(\pv{h}*sin(90+\pv{alpha}))}, beta/.evaluated={atan2(\pv{r},\pv{h})}} \pgfmathtruncatemacro{\itest}{(\pv{e}<0.1?0:1)} \ifnum\itest=0\relax \pgfmathsetmacro{\tcrit}{180} \else \pgfmathsetmacro{\tmp}{(\pv{s}-\pv{r})/(\pv{r}*\pv{e})} \pgfmathtruncatemacro{\itest}{(abs(\tmp)<1?1:0)} \ifnum\itest=1 \pgfmathsetmacro{\tcrit}{abs(acos(\tmp))} %\typeout{tcrit=\tcrit} \else \pgfmathsetmacro{\tcrit}{180} \fi \fi % warning: this may only work for psi=0 \pgfmathsetmacro{\myphi}{atan2(\pgf@yx,\pgf@xx)} \pgfmathsetmacro{\sdtip}{screendepth(0,0,\pv{h})} \pgfmathsetmacro{\aspectangle}{atan2(\sdtip,sqrt(\pv{h}*\pv{h}-\sdtip*\sdtip))} \pgfmathsetmacro{\alphacrit}{acos(tan(\aspectangle)*\pv{r}/\pv{h})} \pgfmathsetmacro{\tvis}{90-\alphacrit+\myphi+\pv{gamma}} %\typeout{alpha_crit=\alphacrit, phi=\myphi, t_vis=\tvis} \path[pic actions] plot[variable=\t,domain=-\tcrit:\tcrit,samples=101,smooth] ({iconeplaneradius(\t,\pv{e},\pv{s})*cos(\t-\pv{gamma})}, {iconeplaneradius(\t,\pv{e},\pv{s}))*sin(\t-\pv{gamma})}, {iconeplaneh(iconeplaneradius(\t,\pv{e},\pv{s}))}); \pgfmathtruncatemacro{\isamples}{50*abs(\tcrit-\tvis)/abs(\tcrit)+1} \draw[3d/hidden] plot[variable=\t,domain=\tcrit:\tvis,samples=\isamples,smooth] ({iconeplaneradius(\t,\pv{e},\pv{s})*cos(\t-\pv{gamma})}, {iconeplaneradius(\t,\pv{e},\pv{s}))*sin(\t-\pv{gamma})}, {iconeplaneh(iconeplaneradius(\t,\pv{e},\pv{s}))}); \pgfmathtruncatemacro{\isamples}{50*abs(\tcrit+\tvis)/abs(\tcrit)+1} \draw[3d/visible] plot[variable=\t,domain=-\tcrit:\tvis,samples=\isamples,smooth] ({iconeplaneradius(\t,\pv{e},\pv{s})*cos(\t-\pv{gamma})}, {iconeplaneradius(\t,\pv{e},\pv{s}))*sin(\t-\pv{gamma})}, {iconeplaneh(iconeplaneradius(\t,\pv{e},\pv{s}))}); }},/tikz/3d/intersection of cone with plane/.cd, h/.initial=5,% height of the cone r/.initial=3,% radius of the base gamma/.initial=0,% angle of intersection Iz/.initial=3,% z coordinate of the intersection alpha/.initial=0,% slope of the plane beta/.initial=0,% will be computed e/.initial=1, s/.initial=1} \makeatother \begin{document} \begin{tikzpicture} \begin{scope}[3d/install view={phi=30,theta=70}, declare function={R=3;% radius of the base of the cone H=5;% height of the cone }, dot/.style={circle,fill,inner sep=1.2pt,label=above:{#1}}] \path (0,0,0) coordinate (O) % center of base of the cone (fixed) (0,0,H) coordinate (T); % tip of the cone % \path (O) pic{3d/cone={r=R,h=H}}; % \begin{scope}[3d/intersection of cone with plane/.cd,r=R,h=H,gamma=-20, /tikz/fill opacity=0.3] \path pic[fill=blue]{3d/intersection of cone with plane={Iz=2.75,alpha=-45}}; \path pic[fill=green!70!black]{3d/intersection of cone with plane={Iz=3,alpha=-30}}; \path pic[fill=red]{3d/intersection of cone with plane={Iz=3.25,alpha=-15}}; \path pic[fill=orange]{3d/intersection of cone with plane={Iz=3.5,alpha=0}}; \end{scope} \end{scope} \end{tikzpicture} \end{document} ```

Let given a cone with tip of the cone `T`; `A` and `B` are two points on the circle of base of the cone; `O` center of base of the cone. we consider four cases. 1) The plane perpendicular to `TO`; 2) The plane is parallel to `TA` and `TB`; 3) The plane is parallel to one and only one `TM`, where `M` line on base of the cone; 4) The plane cuts all the segments `TM`, where `M` line on base of the cone and does not perpendiuclar to `TO`;

On the other hand, if one was given the point at which the plane intersects with the cone and the normal of the plane, this could lead to a much shorter code, which also spares the user from disappointments because the cone does not intersect with the plane. The intersection point could e.g. be specified by an angle and height or an angle and the radius.

Well, let me try to reiterate my question. Below is a start of a code that computes the intersection of a cone with a plane. (The code is **not** complete, it is just supposed to serve as an illustration.) As you can see, there are many many cases. The plane may not intersect at all, it may run through the tip (something the Wikipedia article fails to mention clearly), run through the center of the base, etc. etc. As long as I do not know which data you want to use, I have the choice of producing a code monster (which will take a while) or just let it be. ``` \documentclass[tikz,border=3mm]{standalone} \usetikzlibrary{calc,3dtools}% https://github.com/marmotghost/tikz-3dtools \begin{document} \begin{tikzpicture} \begin{scope}[3d/install view={phi=110,theta=70}, declare function={R=3;}, dot/.style={circle,fill,inner sep=1.2pt,label=above:{#1}}] \path (0,0,0) coordinate (O) % center of base of the cone (fixed) (0,0,5) coordinate (T) % tip of the cone (1,0,1) coordinate (A) % point on the plane (1,0,0) coordinate (N) % normal of the plane ; \tikzset{3d/.cd, plane with normal={(N) through (A) named pA}, plane with normal={(T) through (O) named pO}, plane with normal={(T) through (T) named pT}, line through={(O) and (T) named lOT}} \path[3d/project={(N) on pO}] coordinate (N'); \pgfmathsetmacro{\myh}{tddistance("(T)","(O)")} \path (O) pic{3d/cone={r=R,h=\myh}}; \pgfmathsetmacro{\myd}{tddistance("(N')","(O)")} \typeout{\myd} \ifdim\myd pt<0.01pt\relax % circle \typeout{Normal of the plane is more or less collinear with the axis of the cone.} \pgfmathsetmacro{\myAz}{TD("(T)-(O)o(A)")/\myh} \pgfmathtruncatemacro{\itest}{(\myAz>=0&&\myAz<=\myh?1:0)} \ifnum\itest=0 \typeout{Plane does not intersect with the cone.} \else \pgfmathsetmacro{\myr}{R-R*\myAz/\myh} \draw (0,0,\myAz) circle[radius=\myr]; \fi \else \path[3d/project={(O) on pA}] coordinate[dot=$O'$] (O'); \path[3d/project={(T) on pA}] coordinate[dot=$T'$] (T'); \tikzset{3d/.cd,line through={(O') and (T') named lO'T'}} \path[3d/intersection of={lO'T' with pO}] coordinate[dot=$i_1$] (i1) [3d/intersection of={lO'T' with pT}] coordinate[dot=$i_2$] (i2); \pgfmathsetmacro{\mydi}{tddistance("(i1)","(O)")} \pgfmathtruncatemacro{\itest}{(\mydi<R?1:0)} \typeout{first itest=\itest} \ifnum\itest=0 \typeout{The intersection (i1) of the relevant line on the plane with the plane containing the base of the plane has distance larger than the base radius of the cone.} \pgfmathtruncatemacro{\itest}{(TD("(T')-(T)o(O')-(O)")>0?0:1)} \typeout{second itest=\itest} \else \pgfmathtruncatemacro{\itest}{(tddistance("(i1)","(T')")+tddistance("(O)","(O')"))>0.1?1:0)} \typeout{third itest=\itest} \fi \ifnum\itest=0 \typeout{Plane does not intersect with the cone.} \else % plane and cone intersect \tikzset{3d/.cd,line through={(T') and (i1) named lT'i1}} \pgfmathsetmacro{\mys}{R/\myd} \path[overlay,3d coordinate={(i3)=(O)+\mys*(N')}]; \tikzset{3d/.cd,line through={(i3) and (T) named li3T}} \path (i3) node[dot={$i_3$}]{}; \path[3d/intersection of={lT'i1 with li3T}] coordinate[dot={$i_4$}] (i4); \pgfmathsetmacro{\myproj}{TD("(i4)o(T)-(O)")/TD("(T)-(O)o(T)-(O)")} \ifdim\myproj pt>1pt \typeout{Plane does not intersect with the visible part of the cone because the intersection is "above" the tip of the cone.} \else \ifdim\myproj pt<0pt \typeout{Plane does not intersect with the visible part of the cone.} \else % we have found an intersection in the visible range: (i4) \fi \fi \fi % todo: special cases in which the plane runs through O and/or T %\pgfmathsetmacro{\mys}{R/\mydi} %\path[3d/intersection of={lT'i1 with lO}] coordinate (i1); \fi \end{scope} \end{tikzpicture} \end{document} ```

@Anonymous, re: [your question](#question), it is possible to compute the intersections along the lines of the answer you link to, and simplify and generalize this using `3dtools`. Yet I wonder what the input and the output will be. The cone is easy to specify, but what are your assumptions on the plane?

1 Answer