Edge perpendicular to line.
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Edge perpendicular to line.
by Paulo J. Matos
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Hello,
I have two nodes (n1 and n2) with an edge (e1) between then. Now I have a third node (n3) and I would like to have an edge (e2) from node n3 to the edge e1, such that e2 is perpendicular to e1. Is there any way to do this automatically, without precomputing ? Thanks, -- Paulo Jorge Matos - pocmatos at gmail.com http://www.pmatos.net ------------------------------------------------------------------------------ Come build with us! The BlackBerry(R) Developer Conference in SF, CA is the only developer event you need to attend this year. Jumpstart your developing skills, take BlackBerry mobile applications to market and stay ahead of the curve. Join us from November 9 - 12, 2009. Register now! http://p.sf.net/sfu/devconference _______________________________________________ pgf-users mailing list pgf-users@... https://lists.sourceforge.net/lists/listinfo/pgf-users |
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Re: Edge perpendicular to line.
by Rouben Rostamian-3
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On Sun, Oct 25, 2009 at 02:42:57PM +0000, Paulo J. Matos wrote:
> > I have two nodes (n1 and n2) with an edge (e1) between then. > Now I have a third node (n3) and I would like to have an edge (e2) > from node n3 to the edge e1, such that e2 is perpendicular to e1. Is > there any way to do this automatically, without precomputing ? Will this do? \documentclass{article} \usepackage{tikz} \begin{document} \begin{tikzpicture} \tikzstyle{mynode}=[draw=black, minimum width=2cm, minimum height=1cm] \node[mynode] (A) at (0,0) {Node A}; \node[mynode] (B) at (6,0) {Node B}; \node[mynode] (C) at (4,3) {Node C}; \draw (A.east) -- (B.west); \draw (A.east) -| (C.south); \end{tikzpicture} \end{document} -- Rouben Rostamian ------------------------------------------------------------------------------ Come build with us! The BlackBerry(R) Developer Conference in SF, CA is the only developer event you need to attend this year. Jumpstart your developing skills, take BlackBerry mobile applications to market and stay ahead of the curve. Join us from November 9 - 12, 2009. Register now! http://p.sf.net/sfu/devconference _______________________________________________ pgf-users mailing list pgf-users@... https://lists.sourceforge.net/lists/listinfo/pgf-users |
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Re: Edge perpendicular to line.
by Paulo J. Matos
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On Sun, 2009-10-25 at 11:59 -0400, Rouben Rostamian wrote:
> On Sun, Oct 25, 2009 at 02:42:57PM +0000, Paulo J. Matos wrote: > > > > I have two nodes (n1 and n2) with an edge (e1) between then. > > Now I have a third node (n3) and I would like to have an edge (e2) > > from node n3 to the edge e1, such that e2 is perpendicular to e1. Is > > there any way to do this automatically, without precomputing ? > > Will this do? > > \documentclass{article} > \usepackage{tikz} > \begin{document} > \begin{tikzpicture} > \tikzstyle{mynode}=[draw=black, minimum width=2cm, minimum height=1cm] > \node[mynode] (A) at (0,0) {Node A}; > \node[mynode] (B) at (6,0) {Node B}; > \node[mynode] (C) at (4,3) {Node C}; > \draw (A.east) -- (B.west); > \draw (A.east) -| (C.south); > \end{tikzpicture} > \end{document} > Not really. This is maybe an ugly hack but not a solution and it would work on my case that I have the connection from A to B decorated with snake. What I want is something line: A ^ | | C ------------>| | B Unfortunately the point that C is pointing to is not the mid point of B->A. I really need to say. I want to point from C to the point between B->A that makes the edge from C perpendicular to B->A. -- Paulo Jorge Matos - pocmatos at gmail.com Webpage: http://www.pmatos.net ------------------------------------------------------------------------------ Come build with us! The BlackBerry(R) Developer Conference in SF, CA is the only developer event you need to attend this year. Jumpstart your developing skills, take BlackBerry mobile applications to market and stay ahead of the curve. Join us from November 9 - 12, 2009. Register now! http://p.sf.net/sfu/devconference _______________________________________________ pgf-users mailing list pgf-users@... https://lists.sourceforge.net/lists/listinfo/pgf-users |
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Re: Edge perpendicular to line.
by Bugzilla from avatar@hot.ee
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> What I want is something line: > A > ^ > | > | > C ------------>| > | > B > Did you look at projection modifiers - 12.4.5 in the PGF manual? Misha ------------------------------------------------------------------------------ Come build with us! The BlackBerry(R) Developer Conference in SF, CA is the only developer event you need to attend this year. Jumpstart your developing skills, take BlackBerry mobile applications to market and stay ahead of the curve. Join us from November 9 - 12, 2009. Register now! http://p.sf.net/sfu/devconference _______________________________________________ pgf-users mailing list pgf-users@... https://lists.sourceforge.net/lists/listinfo/pgf-users |
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Re: Edge perpendicular to line.
by Mark Wibrow-2
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Hi,
Try using the perpendicular path specification in the coordinate, e.g., \draw (C) -- (C -| A); Mark 2009/10/25 Paulo J. Matos <pocmatos@...>: > On Sun, 2009-10-25 at 11:59 -0400, Rouben Rostamian wrote: >> On Sun, Oct 25, 2009 at 02:42:57PM +0000, Paulo J. Matos wrote: >> > >> > I have two nodes (n1 and n2) with an edge (e1) between then. >> > Now I have a third node (n3) and I would like to have an edge (e2) >> > from node n3 to the edge e1, such that e2 is perpendicular to e1. Is >> > there any way to do this automatically, without precomputing ? >> >> Will this do? >> >> \documentclass{article} >> \usepackage{tikz} >> \begin{document} >> \begin{tikzpicture} >> \tikzstyle{mynode}=[draw=black, minimum width=2cm, minimum height=1cm] >> \node[mynode] (A) at (0,0) {Node A}; >> \node[mynode] (B) at (6,0) {Node B}; >> \node[mynode] (C) at (4,3) {Node C}; >> \draw (A.east) -- (B.west); >> \draw (A.east) -| (C.south); >> \end{tikzpicture} >> \end{document} >> > > Not really. This is maybe an ugly hack but not a solution and it would > work on my case that I have the connection from A to B decorated with > snake. > > What I want is something line: > A > ^ > | > | > C ------------>| > | > B > > Unfortunately the point that C is pointing to is not the mid point of > B->A. I really need to say. I want to point from C to the point between > B->A that makes the edge from C perpendicular to B->A. > -- > Paulo Jorge Matos - pocmatos at gmail.com > Webpage: http://www.pmatos.net > > > ------------------------------------------------------------------------------ > Come build with us! The BlackBerry(R) Developer Conference in SF, CA > is the only developer event you need to attend this year. Jumpstart your > developing skills, take BlackBerry mobile applications to market and stay > ahead of the curve. Join us from November 9 - 12, 2009. Register now! > http://p.sf.net/sfu/devconference > _______________________________________________ > pgf-users mailing list > pgf-users@... > https://lists.sourceforge.net/lists/listinfo/pgf-users > ------------------------------------------------------------------------------ Come build with us! The BlackBerry(R) Developer Conference in SF, CA is the only developer event you need to attend this year. Jumpstart your developing skills, take BlackBerry mobile applications to market and stay ahead of the curve. Join us from November 9 - 12, 2009. Register now! http://p.sf.net/sfu/devconference _______________________________________________ pgf-users mailing list pgf-users@... https://lists.sourceforge.net/lists/listinfo/pgf-users |
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