# Mixing relative and absolute coordinates in tikz

The following example uses circuitikz, but the question is about expressing coordinates in generic tikz.

Look at the following code:

\documentclass[border=10pt]{standalone}
\usepackage{circuitikz}
\begin{document}
\begin{circuitikz}[american]
\draw (0,0) node[npn](Q1){};
\draw (Q1.E) to[R=$R_E$] (0,-2.5) node[ground]{};
\draw [gray] (Q1.E) -- ++(2,0)
coordinate(tmp) to[C, color=gray] (tmp |- 0, -2.5)
node[ground, color=gray]{};
\end{circuitikz}
\end{document}


where all the grounds are supposed to be at (x, -2.5). The tricky thing is the coordinate (tmp |- 0, -2.5) (hint from here): what I want to express is the concept of a coordinate which is

(the current path x, the absolute y -2.5)

and the solution I found is, as you can see, using a temporary node. Is there an easier way to write that coordinate?

BTW, for easier way I mean something to directly say "move to the coordinate that has the current value of x, and y=-2.5", like for example

\draw [gray] (Q1.E) -- ++(2,0)
to[C, color=gray] (\magiccurrentx, -2.5)
node[ground, color=gray]{};


...or a similar thing. • @phelype-oleinik, thanks for the edit. – Rmano Oct 3 '18 at 11:10
• if you know distance from (Q1.E) to (0,-2.5) than you can use only relative coordinates: \draw [gray] (Q1.E) -- ++(2,0) to[C, color=gray] ++ (0,-1.73) node[ground, color=gray]{};. as i see your mwe, this distance is not known, so you determine end of this path by coordinate defined by (tmp |- 0, -2.5) . as far as i know, for this is no better way. – Zarko Oct 3 '18 at 11:42
• I am sorry, I do not understand the question. Isn't (tmp |- 0, -2.5) already what you're looking for? If not, please consider editing (the current path x, the absolute y -2.5) in such a way that it is a bit clearer. – user121799 Oct 3 '18 at 14:55
• @marmot sorry --- for simpler I mean not having to use a temporary node; some magic to say "move to a coordinate that has the current value of x, and y=-2.5". I have made myself clearer in the question (I hope). – Rmano Oct 3 '18 at 15:26

If you want to save code characters you simply change coordinate(coordinate_name) for node(node_name){} and use one \draw statement, relative coordinates, and you can use the node ground coordinate named as G to find the coordinate intersection of the named coordinates from the short of the capacitor and the ground; to change the color including the nodes use color=your_fav_color, in my case red.

RESULT: MWE:

\documentclass[border=10pt]{standalone}
\usepackage{circuitikz}
\begin{document}
\begin{tikzpicture}[american]
\draw
(0,0)
node[npn](Q1){}
(Q1.E)
to [R=$R_E$] ++(0,-2.5)
node[ground](G){}
;
\draw[color=red]
(Q1.E)
to[short,*-] ++ (2,0) node(C){}
to[C] (C |- G)
node[ground]{}
;
\end{tikzpicture}
\end{document}


OPTIONAL:

Using the two nodenames available from the first ground and the second, and drawing from the second ground to the emiter...

\documentclass[border=10pt]{standalone}
\usepackage{circuitikz}
\begin{document}
\begin{tikzpicture}[american]
\draw
(0,0)
node[npn](Q1){}
(Q1.E)
to [R=$R_E$] ++(0,-2.5)
node[ground](G1){}
;
\draw[color=red]
(G1)++(2,0)
node[ground](G2){}
to[C] (G2 |- Q1.E)
to[short,-*] (Q1.E)
;
\end{tikzpicture}
\end{document}

• This is more elegant than my solution because the ground level is labelled (and I'll use it), but still need to use a coordinate (node(C){} and \coordinate(C) I think are the same thing). Thanks anyway, this is a good idea. – Rmano Oct 3 '18 at 18:01
• This coordinate is needed to calculate the intersection to put the ground at the same x position of the first, the other option is using shftings, but writing shiftx=Some_value in a coordinate is not elegant, but I'll try another method to avoid declare the node(C){}... – J Leon V. Oct 3 '18 at 18:13
• I was toying with the [turn] coordinate system, but no joy, still relative... ;-) – Rmano Oct 3 '18 at 18:20
• Check the optional code... – J Leon V. Oct 3 '18 at 18:22
• Yes, it works... although in the end I prefer your first code. Thanks! I'll wait a bit to see if something magic happens... ;-) – Rmano Oct 3 '18 at 18:31

Here is an alternative, but I would not call it simpler.

\documentclass[border=10pt]{standalone}
\usepackage{circuitikz}

\begin{document}
\begin{circuitikz}[american]
\draw (0,0) node[npn](Q1){};
\draw (Q1.E) to[R=$R_E$] (0,-2.5) node[ground]{};
\draw [gray] (Q1.E) -- ++(2,0) \pgfextra{\pgfgetlastxy{\xlast}{\ylast}}
to[C, color=gray]  (\xlast,-2.5)
node[ground, color=gray]{};
\end{circuitikz}
\end{document}


Very similar to John Kormylo's answer but without \pgfextra (which may cause serious trouble). Similar to John, I am not sure this is more elegant.

\documentclass[border=10pt]{standalone}
\usepackage{circuitikz}
\tikzset{magic xy/.code n args={2}{\pgfgetlastxy{#1}{#2}}}
\begin{document}
\begin{circuitikz}[american]
\draw (0,0) node[npn](Q1){};
\draw (Q1.E) to[R=$R_E$] (0,-2.5) node[ground]{};
\draw [gray] (Q1.E) -- ++(2,0) [magic xy={\magicx}{\magicy}]
to[C, color=gray]  (\magicx,-2.5)
node[ground, color=gray]{};
\end{circuitikz}
\end{document}


Or with calc. Both reproduce your picture.

\documentclass[border=10pt]{standalone}
\usepackage{circuitikz}
\usetikzlibrary{calc}
\begin{document}
\begin{circuitikz}[american]
\draw (0,0) node[npn](Q1){};
\draw (Q1.E) to[R=$R_E$] (0,-2.5) node[ground]{};
\draw[gray]  let \p1=(Q1.E) in (Q1.E)  -- ++ (2,0)
to[C, color=gray]  (\x1+2cm,-2.5)  node[ground, color=gray]{};
node[ground, color=gray]{};
\end{circuitikz}
\end{document}