Mechanics on arcs (spring, damper)

Question

I am used to drawing basic mechanical components (spring, damper) using the tikz and circuitikz this way:

\coordinate (A) at (0,0);
\coordinate (B) at (3,0);

[..]

\draw (A) to[spring] (B);
\draw (C) to[damper] (D);

Now I am in a configuration where I want to draw the same component but on an arc, not a line:

\draw (0,0) + (30:3cm) arc (30:80:3cm);

I've tried without success using to[spring] positioned before and after the arc command (see figure) and also:

\draw[damper] (0,0) + (30:3cm) arc (30:80:3cm);

The last command just draw this arc, without the mechanical component.

I've seen here, here and here that it is possible with coils using decorations, but I wish to keep the same spring representation, and the solution for the dampers seems quite complicated, is there a simpler way to achieve that ?

EDIT: working example

\usepackage{tikz}
\usepackage{circuitikz}
%
\begin{figure} [h]
\begin{tikzpicture}         
    \coordinate (elbow) at (0,0);
    \coordinate (shoulder) at ($(elbow) + (0, 3.25)$);
    %
    \draw[rounded corners=1ex, rotate around={-60:(elbow)}] (0.175, 0) rectangle (-0.175, 3.25); 
    \draw[rounded corners=0.5ex] ($(elbow) + (0.175, 0)$) rectangle ($(shoulder) + (-0.175, 0)$) node (wrist){};
\node[circle, draw, minimum size=1.25cm, fill=white] (joint) at (elbow) {};
\node[circle, draw, minimum size=0.15cm, inner sep=0, fill=white] (jointBis) at (elbow) {};
%
\draw[white, line width=1.5pt] (0,0) + (15:0.625cm) arc (15:45:0.625cm);
%
\draw (0,0) + (33:3cm) arc (33:87:3cm); %to[spring]
\draw[damper] (0,0) + (35:2cm) arc (35:85:2cm);
%
\draw[gray, dashed] (0.625,0) -- (3,0);
\draw[-{Latex[length=2mm]}] (0,0) + (0:2cm) arc (0:25:2cm) node[midway, right]{$\theta$};

\end{tikzpicture}
\end{figure}

Hello and welcome to TeX-SE. Please, add a complete Minimal Working Example to let users copy/paste and look for improvements to help you out. — SebGlav, Jul 28 '21 at 13:26
You can position a spring in the middle of the arc, but the element will not be bent; circuitikz elements are rigid. The only way would be to use a non-linear coordinate transformation, but that's quite complex to do. — Rmano, Jul 28 '21 at 13:39
@Rmano it could be a back up solution (the element that do not bent), but I don't even manage to achieve this. — fo fu, Jul 28 '21 at 14:02

Rmano · Accepted Answer · 2021-07-28T15:00:21.767

Positioning a node along an arc is not so difficult --- the problem is that circuitikz elements can only be automatically positioned along a straight line, with the to statement.

So to position an element on an arc we must rely on using the "naked" node and position it. For example

\path (0,0) +(30:3cm) arc (30:80:3cm) 
   node[draw, springshape, pos=0.5, sloped](S){};

will give:

Now, the big problem is to connect the element and remove the line underneath. I used a bit of trigonometry here (and for sure it could be automated in some macro...); for example --- I let in red the help lines that guided me in the construction:

\begin{tikzpicture}[]
    % these are to show the construction
    \draw[red,thin]  (0,0) + (30:3cm) coordinate(B) 
         arc (30:80:3cm) coordinate(E);
    \node[red] at (B){B}; \node[red] at (E){E};
    \path (B) arc (30:80:3cm) node[draw, springshape, pos=0.5, sloped](S){};
    % from B the circle starts at 120 degree, let's se e the landing angle and use a spline
    \draw let \p1=(S.right), \n1={-90+acos(\x1/3cm)} 
          in (B) to[out=120,in=\n1] (S.right);
    % from E it starts at -10, let's caluclate the landing angle
    \draw let \p1=(S.left), \n1={90+acos(\x1/3cm)} 
          in (E) to[out=-10,in=\n1] (S.left);
\end{tikzpicture}

So the arcs are not really arcs, but the final result (removing the red parts) seems acceptable to me:

\path (0,0) +(30:3cm) coordinate(B)
        arc (30:80:3cm) coordinate(E)
        node[draw, springshape, pos=0.5, sloped](S){};
    % from B the circle starts at 120 degree, let's se e the landing angle and use a spline
    \draw let \p1=(S.right), \n1={-90+acos(\x1/3cm)}
          in (B) to[out=120,in=\n1] (S.right);
    % from E it starts at -10, let's caluclate the landing angle
    \draw let \p1=(S.left), \n1={90+acos(\x1/3cm)}
          in (E) to[out=-10,in=\n1] (S.left);

with a damper, use the nodename for the element you find in the manual:

to obtain:

Yes --- and every component in circuitikz. The "nodename" is normally the path name with a shape appended --- it's marked in the manual anyway for each element. — Rmano, Jul 28 '21 at 14:58
Notice that the code is supersimplyfied and works only for arcs centered in 0 and with a 3cm diameter... I've been too lazy to generalize it. The trick is getting the orientation of the arc near the points where the component left and right anchors are. — Rmano, Jul 29 '21 at 08:07

Mechanics on arcs (spring, damper)

1 Answers1