Imitating a conveyor belt in TikZ

I have two circles of radii 4 and 5 drawn in blue and line segments drawn in purple between them so that the endpoints of them are perpendicular to radii of them. To give the impression that this depiction represents a conveyor belt, I will add two arc commands drawn in purple and will reduce the radii of the circles from 5cm and 4cm to 5cm-0.2pt and 4cm-0.2pt. Why is the command for drawing the circle 2pt less than 5cm not being implemented? This is the command that was misinterpreted.

\draw[blue] (center_of_first_circle) circle ({5cm - 0.2pt});


In my code, I used the following command to render a diagram that I am looking to modify.

\draw[blue] (center_of_first_circle) circle (5);


Here is my code.

\documentclass[10pt]{amsart}
\usepackage{mathtools,array}

\usepackage{tikz}
\usetikzlibrary{calc,positioning,intersections}

\begin{document}

\begin{tikzpicture}[x=0.25cm, y=0.25cm]

%The centers of a circle of radius 4 and a circle of radius 5 are at a distance of 16 from each other. Two line segments
%are to be drawn between them so that they are tangent to the circles. (It is a sketch of a conveyor belt.) The radii to
%the point of tangency of the same line segment are parallel to each other. The centers of the circles are located on the
%x-axis - the center of the circle of radius 5 is centered at the origin, and the center of the circle of radius 4 is
%centered at (16,0). If \theta is the measure of the angle between the x-axis and the two radii, the slope of
%the line segment is sin\theta/(16 - cos\theta). So, the slope of the radii is (cos\theta - 16)/sin\theta. The slope is
%also tan\theta.
%x = 5cos\theta.
%x^2 + ((cos\theta - 16)/sin\theta)^2*x^2 = 5^2.
%tan\theta=(cos\theta - 16)/sin\theta.
%This is a quartic equation in the variable x. The solution is 5*sqrt(65/2 - 4*sqrt(66)). The point of tangency on the circle
%(5*sqrt(65/2 - 4*sqrt(66)), 5*sqrt(4*sqrt(66) - 63/2)).
%
%
\path (0,0) coordinate (center_of_first_circle) (16,0) coordinate (center_of_second_circle);
%
%
%
\path let \n1={5*sqrt(65/2 - 4*sqrt(66))}, \n2={5*sqrt(4*sqrt(66) - 63/2)} in coordinate (a_point_of_tangency_on_bigger_circle) at (\n1,\n2);
\path let \n1={5*sqrt(65/2 - 4*sqrt(66))}, \n2={5*sqrt(4*sqrt(66) - 63/2)} in coordinate (another_point_of_tangency_on_bigger_circle) at (\n1,-\n2);
%
\path let \n1={2*sqrt(66)}, \n2={sqrt(64*sqrt(66)-504)} in coordinate (a_point_of_tangency_on_smaller_circle) at (\n1,\n2);
\path let \n1={2*sqrt(66)}, \n2={sqrt(64*sqrt(66)-504)} in coordinate (another_point_of_tangency_on_smaller_circle) at (\n1,-\n2);
%
%
\draw (a_point_of_tangency_on_bigger_circle) -- (a_point_of_tangency_on_smaller_circle);
\draw (another_point_of_tangency_on_bigger_circle) -- (another_point_of_tangency_on_smaller_circle);
%
%
%
\draw[blue] (center_of_first_circle) circle (5);
\draw[blue]  (center_of_second_circle) circle (4);

\end{tikzpicture}

\end{document}


Explanation for why your command does not work:

Your transformation (x=.25cm,y=.25cm) kills it. The transformation has no effect in numbers which have units, so 5cm-2pt is still 5cm-2pt, it is not scaled.

To prevent it, you will need to scale 5cm-2pt yourself. In the following code I use 1.25cm-2pt and a similar number for the smaller circle.

If I understand your question, you can achieve that with tkz-euclide

\documentclass[tikz]{standalone}
\usepackage{mathtools,array}
\usetikzlibrary{calc,positioning,intersections}
\usepackage{tkz-euclide}
\usetkzobj{all}
\begin{document}
\begin{tikzpicture}[x=0.25cm, y=0.25cm]
%The centers of a circle of radius 4 and a circle of radius 5 are at a distance of 16 from each other. Two line segments
%are to be drawn between them so that they are tangent to the circles. (It is a sketch of a conveyor belt.) The radii to
%the point of tangency of the same line segment are parallel to each other. The centers of the circles are located on the
%x-axis - the center of the circle of radius 5 is centered at the origin, and the center of the circle of radius 4 is
%centered at (16,0). If \theta is the measure of the angle between the x-axis and the two radii, the slope of
%the line segment is sin\theta/(16 - cos\theta). So, the slope of the radii is (cos\theta - 16)/sin\theta. The slope is
%also tan\theta.
%x = 5cos\theta.
%x^2 + ((cos\theta - 16)/sin\theta)^2*x^2 = 5^2.
%tan\theta=(cos\theta - 16)/sin\theta.
%This is a quartic equation in the variable x. The solution is 5*sqrt(65/2 - 4*sqrt(66)). The point of tangency on the circle
%(5*sqrt(65/2 - 4*sqrt(66)), 5*sqrt(4*sqrt(66) - 63/2)).
%
%
\path (0,0) coordinate (center_of_first_circle) (16,0) coordinate (center_of_second_circle);
%
\path let \n1={5*sqrt(65/2 - 4*sqrt(66))}, \n2={5*sqrt(4*sqrt(66) - 63/2)} in coordinate (a_point_of_tangency_on_bigger_circle) at (\n1,\n2);
\path let \n1={5*sqrt(65/2 - 4*sqrt(66))}, \n2={5*sqrt(4*sqrt(66) - 63/2)} in coordinate (another_point_of_tangency_on_bigger_circle) at (\n1,-\n2);
%
\path let \n1={2*sqrt(66)}, \n2={sqrt(64*sqrt(66)-504)} in coordinate (a_point_of_tangency_on_smaller_circle) at (\n1,\n2);
\path let \n1={2*sqrt(66)}, \n2={sqrt(64*sqrt(66)-504)} in coordinate (another_point_of_tangency_on_smaller_circle) at (\n1,-\n2);
%
\draw[violet] (a_point_of_tangency_on_bigger_circle) -- (a_point_of_tangency_on_smaller_circle);
\draw[violet] (another_point_of_tangency_on_bigger_circle) -- (another_point_of_tangency_on_smaller_circle);
%
\draw[blue] (center_of_first_circle) circle (1.25cm-2pt);
\draw[blue]  (center_of_second_circle) circle (1cm-2pt);

\tkzDrawArc[color=violet](center_of_first_circle,a_point_of_tangency_on_bigger_circle)(another_point_of_tangency_on_bigger_circle)
\tkzDrawArc[color=violet](center_of_second_circle,another_point_of_tangency_on_smaller_circle)(a_point_of_tangency_on_smaller_circle)
\end{tikzpicture}
\end{document} Btw you should never name your coordinates with such looooooong names... typing the names takes me considerable time :)

• I want to keep this in TikZ. – A gal named Desire May 24 at 13:10
• @AgalnamedDesire It is still TikZ. – user156344 May 24 at 13:11
• Why is TikZ misinterpreting the command I issued? – A gal named Desire May 24 at 13:11
• @AgalnamedDesire Can you tell me what that command is? Both circle commands are not the commands which are "ignored" – user156344 May 24 at 13:12
• @AgalnamedDesire I added an explanation. Shortly: the transformation kills it. – user156344 May 24 at 13:21

A PSTricks solution only for fun purposes! \documentclass[pstricks]{standalone}
\begin{document}
\pspicture[linewidth=1.2pt](15,8)
\pnodes{P}(4,4)(11,4)
\psCircleTangents(P0){3.2}(P1){2.7}
\psset{linecolor=magenta}
\psarc[origin=P0](P0){3.2}{(CircleTO2)}{(CircleTO4)}
\psarc[origin=P1](P1){2.7}{(CircleTO3)}{(CircleTO1)}
\psline(CircleTO2)(CircleTO1)
\psline(CircleTO4)(CircleTO3)
\psset{linecolor=blue}
\pscircle(P0){3}
\pscircle(P1){2.5}
\endpspicture
\end{document}


Unrealistic animation \documentclass[pstricks]{standalone}
\begin{document}
\multido{\r=0.1+0.1}{29}{
\pspicture[linewidth=1.2pt](15,8)
\pnodes{P}(4,4)(11,4)
\psCircleTangents(P0){3.2}(P1){\pscalculate{\r+0.2}}
\psset{linecolor=magenta}
\psarc[origin=P0](P0){3.2}{(CircleTO2)}{(CircleTO4)}
\psline(CircleTO2)(CircleTO1)
\psline(CircleTO4)(CircleTO3)
\psset{linecolor=blue}
\pscircle(P0){3}
\pscircle(P1){\r}
\endpspicture}
\end{document}

• It would be very nice to see some animations! – user156344 May 24 at 13:31
• Why do we need \space following by \r ? – Trong Vuong May 24 at 13:44
• @TrongVuong: Without \space any characters that follows will be eaten. :-) – Money Oriented Programmer May 24 at 13:57
• By "animation" I mean it is something like the rotation of a chain wheel in a bicycle. Hope to see it soon! – user156344 May 24 at 14:04

With the style from this answer. Earlier TIkZ only solutions can be found here.

\documentclass[10pt]{amsart}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}

\begin{tikzpicture}[tangent of circles/.style args={%
at #1 and #2 with radii #3 and #4}{insert path={%
let \p1=($(#2)-(#1)$),\n1={atan2(\y1,\x1)},\n2={veclen(\y1,\x1)*1pt/1cm},
\n3={atan2(#4-#3,\n2)}
in ($(#1)+(\n3+\n1+90:#3)$) -- ($(#2)+(\n3+\n1+90:#4)$)}}]
\pgfmathsetmacro{\offscale}{1.04}
%
\draw[red,tangent of circles={at c1 and c2 with radii \offscale*1.25 and \offscale*1}]
coordinate[pos=0] (tl) coordinate[pos=1] (tr)
[tangent of circles={at c2 and c1 with radii \offscale*1 and \offscale*1.25}]
coordinate[pos=0] (br) coordinate[pos=1] (bl);
\draw[red] let \p1=($(tl)-(c1)$),\p2=($(bl)-(c1)$),\n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)}
in (tl) arc(\n1:\n2+360:\offscale*1.25);
\draw[red] let \p1=($(tr)-(c2)$),\p2=($(br)-(c2)$),\n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)}
in (tr) arc(\n1:\n2:\offscale*1);
\end{tikzpicture}
\end{document} Animation:

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{calc}
\tikzset{tangent of circles/.style args={%
at #1 and #2 with radii #3 and #4}{insert path={%
let \p1=($(#2)-(#1)$),\n1={atan2(\y1,\x1)},\n2={veclen(\y1,\x1)*1pt/1cm},
\n3={atan2(#4-#3,\n2)}
in ($(#1)+(\n3+\n1+90:#3)$) -- ($(#2)+(\n3+\n1+90:#4)$)}}}
\begin{document}
\foreach \X in {1.25,1.2,...,0.25,0.3,0.35,...,1.2}
{\begin{tikzpicture}
\pgfmathsetmacro{\rOne}{1.25}
\pgfmathsetmacro{\rTwo}{\X}
\pgfmathsetmacro{\rPlus}{0.05}
%
\path[use as bounding box] (-1.5,-1.5) rectangle (5.5,1.5);
\draw[red,tangent of circles={at c1 and c2 with radii {\rOne+\rPlus} and {\rTwo+\rPlus}}]
coordinate[pos=0] (tl) coordinate[pos=1] (tr)
[tangent of circles={at c2 and c1 with radii {\rTwo+\rPlus} and {\rOne+\rPlus}}]
coordinate[pos=0] (br) coordinate[pos=1] (bl);
\draw[red] let \p1=($(tl)-(c1)$),\p2=($(bl)-(c1)$),\n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)}
in (tl) arc(\n1:\n2+360:\rOne+\rPlus);
\draw[red] let \p1=($(tr)-(c2)$),\p2=($(br)-(c2)$),\n1={atan2(\y1,\x1)},\n2={atan2(\y2,\x2)}
in (tr) arc(\n1:\n2:\rTwo+\rPlus);
\end{tikzpicture}}
\end{document} • For a little fun, would you please add an animation which looks like a rotating chain wheel of a bicycle? :) – user156344 May 24 at 14:05
• @JouleV Like tex.stackexchange.com/a/458120/121799 ? – marmot May 24 at 14:09
• Yes! (of course with the addition of the two wheels) – user156344 May 24 at 14:10
• @JouleV You can ask a question for that... – marmot May 24 at 14:11
• Or maybe like this answer. – Mark Wibrow May 24 at 14:11