# How to make the length of a car moving on a roller coaster track remain unchanged?

For any arbitrary parametric smooth curve, how can I make the length of a car remain unchanged?

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-plot,pst-node}

\pstVerb
{
tx@Derive begin
/EvalVariable { 2 index (t) eq { (1) } { (0) } ifelse 4 -1 roll exch 6 2 roll } def
end
}

\def\x{t}
\def\y{2*cos(t)}

\def\xx{Derive(1,\x)}
\def\yy{Derive(1,\y)}
\def\rr{sqrt((\xx)^2+(\yy)^2)}

\def\Nx{-\yy/\rr}
\def\Ny{\xx/\rr}

\def\R{.2}
\def\xP{\x+\R*\Nx}
\def\yP{\y+\R*\Ny}

\begin{document}

\multido{\rx=.0+.3,\ry=.4+.3}{41}{%
\begin{pspicture}[algebraic](-.2,-2.2)(\dimexpr\psPiFour cm+.2cm,2.2)
\psparametricplot[linecolor=gray,plotpoints=100]{0}{TwoPi 2 mul}{\x|\y}
\curvepnodes[plotpoints=2]{\rx}{\ry}{\xP|\yP}{P}
\pspolygon*[linecolor=gray](P0)(P1)([offset=-.5]{P0}P1)([offset=.5]{P1}P0)
\qdisk(P0){\R}\qdisk(P1){\R}
\psparametricplot[linecolor=red,linewidth=2pt,plotpoints=100]{0}{\rx}{\xP|\yP}
\end{pspicture}}

\end{document}


## Commenting Herbert's solution

It is too long as comment. Herbert's solution apparently keeps the length of curve segment bounded by the wheels constant rather than the length of the car. The following animation shows it.

\documentclass[pstricks,border=12pt]{standalone}

\pstVerb
{
tx@Derive begin
/EvalVariable { 2 index (t) eq { (1) } { (0) } ifelse 4 -1 roll exch 6 2 roll } def
end
}

\def\x{t}
\def\y{2*cos(t)}

\def\xx{Derive(1,\x)}
\def\yy{Derive(1,\y)}
\def\rr{sqrt((\xx)^2+(\yy)^2)}

\def\Nx{-\yy/\rr}
\def\Ny{\xx/\rr}

\def\R{.2}
\def\xP{\x+\R*\Nx}
\def\yP{\y+\R*\Ny}

\begin{document}

\multido{\iA=1+1,\iB=10+1}{40}{%
\begin{pspicture}[algebraic](-.2,-2.2)(\dimexpr\psPiFour cm+.2cm,2.2)
\psparametricplot[linecolor=gray,plotpoints=100]{0}{TwoPi 2 mul}{ \x | \y }
\pscurvepoints[plotpoints=50]{0}{TwoPi 2 mul}{ \xP | \yP }{P}
\pspolylineticks[ticksize=0 0,metricInitValue=1,Os=1,Ds=.3]{P}{ ds }{1}{50}%
\pspolygon*[linecolor=gray](PTick\iA)(PTick\iB)([offset=-.05]{PTick\iA}PTick\iB)([offset=.05]{PTick\iB}PTick\iA)
\qdisk(PTick\iA){\R}\qdisk(PTick\iB){\R}
\end{pspicture}%
}

\end{document}


-
you could have the body of the car a fixed size rectangle centered mid-way between the wheels, then draw springs from the wheel to the body of the car. The wheels will thus be allowed to be at various distances, but the car itself will keep fixed shape. At least, I would like to see such an animation. – jfbu May 7 '14 at 16:36
I just posted a MP solution to this problem in my answer to this question. – Thruston May 7 '14 at 18:08
Note that my earlier Asymptote answer to the same question Thruston linked to keeps the length of the car constant, using essentially the same approach. – Charles Staats May 7 '14 at 21:41
I'm sorry, but I consider "This question has not received enough attention." to be a bit rude considering the amount of attention it has had and am consequently deleting my answer. – Loop Space May 10 '14 at 19:19
@HarishKumar: More precisely, I meant attention from other PSTricks experts who can fix the shortcoming in the existing PSTricks answers. :-) TikZ, Metapost, and Asymptote answers are not the focal point for me but just for comparison purpose. I appreciate those answers, of course. – kiss my armpit May 11 '14 at 5:27

Ok, here you go :) This code works similar to the Asymptote, Metapost and TikZ answers. First a path parallel to the original track path is created, the rear wheel is advanced with constant velocity and the position of the front wheel is calculated as intersection of a circle centered around the rear wheel with the parallel track path.

The code is based on the pst-intersect package. The largest part is required to create a path which is parallel to an other path:

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-plot,pst-node,pst-intersect}
\def\x{t}
\def\y{2*cos(t)}

\makeatletter
\pstVerb{ \pst@intersectdict
/ExtrudePath {
/@myshift exch def
[ exch
{
{ aload pop } forall
4 copy VecSub
2 copy tx@Dict begin Pyth end dup 3 1 roll div 3 1 roll div exch % normalized
@myshift VecScale
-90 matrix rotate dtransform 2 copy 8 -2 roll VecAdd 6 2 roll VecAdd
[ 5 1 roll ] ArrayToPointArray
counttomark 1 roll
} forall ]
} bind def
/CleanupPath {
[ exch
5 dict begin
dup length dup /N exch def 1 gt {
/i 1 def
dup 0 get /A exch def
{
dup i get /B exch def
A B IntersectLines pop /tA exch def pop pop
{
i N 1 sub eq { exit } if
/i i 1 add def
dup i get A exch IntersectLines pop
dup 0 get tA 0 get lt { %(skip current line) ==
/tA exch def /B exch def pop
} { % (use current line) ==
pop pop pop /i i 1 sub def exit
} ifelse
} loop
A tA LoadLineIntersectionPoints dup
[ A 0 get 3 -1 roll ] counttomark 1 roll
/A [ 3 -1 roll B 1 get ] def
/i i 1 add def
i N eq { A counttomark 1 roll exit } if
} loop
pop ]
} if
end
} bind def
/GetCurvePointAtLength {
3 dict begin
/l_to exch def
/L 0 def
/segm 0 def
PreparePath [ exch aload pop counttomark -1 2 { 1 roll } for ]
dup 0 get L 3 -1 roll
{
dup { aload pop } forall
tx@Dict begin Pyth2 end
dup L add /L exch def
L l_to gt { 4 2 roll pop pop exit } { /segm segm 1 add def pop pop } ifelse
} forall
L l_to sub exch div neg 1 add dup segm add 3 1 roll [ exch ]
end
} bind def
end
}%
\def\psGetCurveCoorAtLength#1#2{%
\pst@intersectdict
currentdict /\PIT@name{#1} known not {
(You haven't defined the curve or path '#1') ==
} if
\PIT@name{#1} #2 \pst@number\psxunit\space mul
GetCurvePointAtLength 3 -1 roll pop
\tx@UserCoor
end
}%
\def\pssaveparallelpath#1#2#3{%
\pstVerb{ \pst@intersectdict
/\PIT@name{#1} load PreparePath #3 \pst@number\psxunit mul
ExtrudePath
CleanupPath
[ exch dup dup length 1 sub get 0 get aload pop /movetype 4 -1 roll
{ 1 get aload pop /linetype } forall counttomark -3 roll ]
/\PIT@name{#2} exch def
end }%
}
\def\psGetFrontWheelCoor#1#2#3{%
\pst@intersectdict
\PIT@name{#1} /\PIT@name{#2} get #3 \pst@number\psxunit\space mul
GetCurvePointAtLength pop pop /t_val exch def
\PIT@name{#1} /\PIT@name{#2@t} get
{ dup t_val gt { exit }{ pop } ifelse } forall
\PIT@name{#1} /\PIT@name{#2} get
PreparePath dup length 1 sub
3 -1 roll dup dup
cvi sub 4 1 roll
cvi sub get
PointArrayToArray
tx@FuncDict begin 2 dict begin
dup length 2 idiv 1 sub /BezierType exch def /Points exch def GetBezierCoor
end end end
\tx@UserCoor
}
\makeatother
\def\Car{%
\ncline[linestyle=none]{RearWheel}{FrontWheel}
\ncput[nrot=:U]{\psline[linearc=0.05,fillstyle=solid,fillcolor=red](0,0)(0.5,0)(0.45,0.25)(0.1,0.25)(-0.1,0.4)(-0.45,0.4)(-0.45,0)(0,0)}
\pstracecurve[fillstyle=solid,fillcolor=black]{RearWheel}}
\begin{document}

\multido{\r=0+0.45}{95}{%
\def\CarLength{0.4}
\begin{pspicture}[algebraic](-.2,-2.2)(\dimexpr\psPiFour cm+.2cm,2.2)
\pssavepath[linestyle=none]{A}{\psparametricplot[plotpoints=100]{0}{TwoPi 2 mul}{\x|\y}}
\pstracecurve[linecolor=gray]{A}
\pssavepath[linestyle=none]{B}{%
\pstracecurve{B1}
\psparametricplot[plotpoints=10]{Pi -0.5 mul}{Pi 0.5 mul}{12.56637+0.2*cos(t)|2-0.2*sin(t)}
\pstracecurve[tstart=99,tstop=0]{B2}
\psparametricplot[plotpoints=10]{Pi 0.5 mul}{Pi 1.5 mul}{0.2*cos(t)|2-0.2*sin(t)}}%
\pnode(!\psGetCurveCoorAtLength{B}{\r}){RearWheel}
\pssavepath[linestyle=none]{CircIsect}{\pscircle(RearWheel){\CarLength}}
\psintersect[name=I]{B}{CircIsect}
\pnode(!\psGetFrontWheelCoor{I}{B}{\r}){FrontWheel}
\Car
\end{pspicture}}
\end{document}


In your case I should mention, that the code doesn't yet account for all possible or impossible path windings :)

-

A bit slow to compile and a bit inaccurate in places.

\documentclass[tikz,border=5]{standalone}
\usetikzlibrary{decorations}
\makeatletter
\pgfdeclaredecoration{cart}{start}{
\state{start}[width=0pt, next state=move,
persistent precomputation={
\ifx\pgfdecorationcartdistance\pgfutil@empty%
\else%
\pgfmathparse{\pgfdecorationcartdistance/\pgfdecoratedpathlength}%
\let\pgfdecorationcarttime=\pgfmathresult%
\fi%
}]{}
\state{move}[width=\pgfdecorationcarttime*\pgfdecoratedpathlength,
next state=pre-calculate]{}
\state{pre-calculate}[width=1pt, next state=calculate]{
\pgfcoordinate{cart-start}{\pgfpointorigin}%
}
\state{calculate}[width=1pt,
persistent postcomputation={
\pgfpointdiff{\pgfpointanchor{cart-start}{center}}%
{\pgfpointanchor{cart-end}{center}}%
\pgfmathveclen{\the\pgf@x}{\the\pgf@y}%
\ifdim\pgfmathresult pt>\pgfdecorationcartlength\relax
\def\pgf@decorate@next@state{final}
\fi
}
]{ \pgfcoordinate{cart-end}{\pgfpointorigin} }
\state{final}{%
\pgftransformreset
\pgftransformshift{\pgfpointanchor{cart-start}{center}}%
\pgfmathanglebetweenpoints{\pgfpointanchor{cart-start}{center}}{\pgfpointanchor{cart-end}{center}}%
\pgftransformrotate{\pgfmathresult}%
\path pic [transform shape] {cart};
}
}
\begin{document}

\pgfkeys{/pgf/decoration/.cd,
cart length/.store in=\pgfdecorationcartlength,
cart height/.store in=\pgfdecorationcartheight,
cart time/.store in=\pgfdecorationcarttime,
cart distance/.store in=\pgfdecorationcartdistance,
cart length=0.375cm,
cart height=0.25cm,
cart time=0.5,
cart distance=,
}

\tikzset{cart/.pic={
\fill [gray]  (0cm, \pgfdecorationcartwheelradius)
rectangle (\pgfdecorationcartlength, \pgfdecorationcartheight);
\fill [black] (0cm, \pgfdecorationcartwheelradius)
\fill [black] (\pgfdecorationcartlength, \pgfdecorationcartwheelradius)
}}

\foreach \p in {0,...,49}{%
\begin{tikzpicture}
\useasboundingbox (-1, -2) rectangle (6, 3);
\draw [postaction={decoration={cart, cart time=\p/50}, decorate},
postaction={decoration={cart, cart time=\p/50, reverse path}, decorate}]
(0,0) .. controls ++(90:1) and ++(240:2) .. (3,2)
.. controls ++(60:2) and ++(90:2) .. (5,0) .. controls ++(270:2)
and ++(270:2) .. cycle;
\end{tikzpicture}
}
\end{document}


-
These cars defy gravity! good work :) – Harish Kumar May 8 '14 at 8:24
One spot of inaccuracy appears to be when both cars are rounding the sharp curve on the left. But you I had to pause the animation there to see it. – Charles Staats May 9 '14 at 2:09
@CharlesStaats It is pulling a wheelie (which is more spectacular the nearer the position/time gets to 1). Or it is because it is at the end of the path and the decoration does not recycle the path. – Mark Wibrow May 9 '14 at 4:35
ooh this is very neat! :) I'm dreaming about these cars in a Möbius strip. :) – Paulo Cereda May 10 '14 at 19:50

Here's an Asymptote solution, adapted from my answer here:

unitsize(5cm);
import graph;
import animation;
real wheelradius = 0.1, wheeldistance = 1.0;

pair torusknot(real t) {
int p = 3, q=5;
real r = cos(q*t) + 2;
return (r*cos(p*t), r*sin(p*t));
}

path loop = graph(torusknot, 0, 2pi, operator..) & cycle;

//Where will a wheel center be when it's tangent to the loop at path time t?
pair wheelcenter(real t) {
return point(loop, t) + wheelradius*(rotate(90)*dir(loop,t));
}
//This path is for computation, not drawing:
path wheelpath = graph(wheelcenter, 0, length(loop), operator ..) & cycle;

void drawcart(pair trailingwheel, pair leadingwheel = trailingwheel + (wheeldistance, 0)) {
draw(trailingwheel -- leadingwheel, gray);
}

//t is specified in arclength
void drawcart(real t) {
pair trailingwheel = arcpoint(wheelpath, t);
pair estimateleading = arcpoint(wheelpath, t + wheeldistance);
path samedist = circle(c=trailingwheel, r=wheeldistance);
pair[] intersections = intersectionpoints(samedist, wheelpath);
pair leadingwheel = intersections[0];
for (pair candidate : intersections) {
}
}

//Draw the loop:
draw(loop);

animation A;

int n = 200;

real length = arclength(wheelpath);

for (int i = 0; i < n; ++i) {
save();
drawcart(i*length/n);
restore();
}

A.movie(delay=300);


The result:

-

Here is a method which is only approximate. There is an underlying fact which is neat: the path followed by the center of the wheels has the same centers of curvatures as the original path. Thus to find the location of the front wheel, one may proceed approximately like this

1. consider a point M=(f(t),g(t)) on the path, its unit tangent vector, and the normal vector.
2. The center of the rear wheel is at the distance r along the normal vector.
3. compute the algebraic curvature k, hence the center of curvature which is at algebraic distance 1/k from M on the normal line,
4. and find then point which is simultaneously at distance d of the center of the rear wheel and at distance |1/k-r| from the center of curvature, and has positive coordinate along the tangent vector: this gives the approximate location of the center of the front wheel.

For this to work, d must be small compared to the radius of curvatures, and r small compared to d. Here are two implementations (where the condition that r is small compared to d is not satisfied):

code for torus knot:

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-plot,pst-node}

\def\p {3}
\def\q {5}
\def\x{(2+cos(\q*t))*cos(\p*t)}
\def\y{(2+cos(\q*t))*sin(\p*t)}

% derivative vector
\def\xx {(-sin(\q*t)*\q*cos(\p*t)-(2+cos(\q*t))*sin(\p*t)*\p)}
\def\yy {(-sin(\q*t)*\q*sin(\p*t)+(2+cos(\q*t))*cos(\p*t)*\p)}

\def\rr{sqrt(4*\p^2+4*\p^2*cos(\q*t)+\p^2*cos(\q*t)^2+\q^2-cos(\q*t)^2*\q^2)}

\def\Tx {\xx/\rr}  % unit tangent vector
\def\Ty {\yy/\rr}
\def\Nx {-\yy/\rr} % normal vector= tangent rotated +90 degrees
\def\Ny {\xx/\rr}

\def\R{.2}         % wheel radius

\def\xP{\x+\R*\Nx} % wheel center
\def\yP{\y+\R*\Ny}

\def\gg{(\p*(2*cos(\q*t)*\q^2+4*\p^2+4*\p^2*cos(\q*t)-cos(\q*t)^2*\q^2+\p^2*cos(\q*t)^2+2*\q^2)/\rr^3)}% algebraic curvature (thank you maple)

\def\GG{\gg/(1-\R*\gg)}% algebraic curvature of wheel center path

\def\D{.4}   % intra-wheel distance

% approximate location of front wheel center (exact when path is a circle)
\def\xQ {\x + (\R+\GG*(\D)^2/2)*\Nx + \D*sqrt(1-(\GG*\D/2)^2)*\Tx }
\def\yQ {\y + (\R+\GG*(\D)^2/2)*\Ny + \D*sqrt(1-(\GG*\D/2)^2)*\Ty }

% location of centers of curvatures (for testing)
% \def\xG {\x + \Nx/\gg }
% \def\yG {\y + \Ny/\gg }

\begin{document}

\multido{\rx=.0+.1}{63}{%
\begin{pspicture}[algebraic](-3,-3)(3,3)
\psparametricplot[linecolor=gray,plotpoints=300]{0}{TwoPi}{\x|\y}
\curvepnode {\rx}{\xP|\yP}{P} % is this efficient way to set one node?
\curvepnode {\rx}{\xQ|\yQ}{Q} %  (I don't know pstricks!)
\pspolygon*[linecolor=gray](P)(Q)([offset=-\D]{P}Q)([offset=\D]{Q}P)
\qdisk(P){\R}\qdisk(Q){\R}
\end{pspicture}}

\end{document}


and for roller coaster:

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-plot,pst-node}

\def\x{t}
\def\y{2*cos(t)}

\def\xx{1}         % derivative vector
\def\yy{-2*sin(t)}
\def\rr{sqrt(1+4*(sin(t))^2)}

\def\Tx {\xx/\rr}  % unit tangent vector
\def\Ty {\yy/\rr}
\def\Nx {-\yy/\rr} % normal vector= tangent rotated +90 degrees
\def\Ny {\xx/\rr}

\def\R{.2}         % wheel radius

\def\xP{\x+\R*\Nx} % wheel center
\def\yP{\y+\R*\Ny}

\def\gg{(-2*cos(t)/(\rr)^3)}% algebraic curvature

\def\GG{\gg/(1-\R*\gg)}   % algebraic curvature of wheel center path

\def\D{.4}   % intra-wheel distance

% approximate location of front wheel center (exact when path is a circle)
\def\xQ {\x + (\R+\GG*(\D)^2/2)*\Nx + \D*sqrt(1-(\GG*\D/2)^2)*\Tx }
\def\yQ {\y + (\R+\GG*(\D)^2/2)*\Ny + \D*sqrt(1-(\GG*\D/2)^2)*\Ty }

% location of centers of curvatures (for testing)
% \def\xG {\x + \Nx/\gg }
% \def\yG {\y + \Ny/\gg }

\begin{document}

\multido{\rx=.0+.3}{40}{%
\begin{pspicture}[algebraic](-.2,-2.2)(\dimexpr\psPiFour cm+.2cm,2.2)
\psparametricplot[linecolor=gray,plotpoints=100]{0}{TwoPi 2 mul}{\x|\y}
\curvepnode {\rx}{\xP|\yP}{P} % is this efficient way to set one node?
\curvepnode {\rx}{\xQ|\yQ}{Q} %  (I don't know pstricks!)
\pspolygon*[linecolor=gray](P)(Q)([offset=-\D]{P}Q)([offset=\D]{Q}P)
\qdisk(P){\R}\qdisk(Q){\R}
\end{pspicture}}

\end{document}

-
Your approximation produces some defects. – kiss my armpit May 8 '14 at 15:00
@IamwhoIsayIam yes, the car is too big compared to the radius of curvature, at some places. On the other hand, one may say it creates a realistic effect of the car bouncing off the road... – jfbu May 8 '14 at 17:02
Bouncing off the road, or in some cases, rearing up for a climb. – Charles Staats May 8 '14 at 19:41
@CharlesStaats reminds me of my first car, needed full gas at the bottom of such climbs... – jfbu May 8 '14 at 19:45
@IamwhoIsayIam yes, the Derive from your updated question and this answer to another question is very neat. Nice bounty; if somebody reads this: my approximate method is exact when the path is a circle; this means that it is not so much the size of the car and wheels versus curvature which counts for accuracy, but the sizes versus rate of change of curvature, although I have not tried to make this quantitative. – jfbu May 12 '14 at 17:25
\documentclass{beamer}
%\url{http://tex.stackexchange.com/q/175874/86}
\usepackage{tikz}
\usetikzlibrary{%
% Intersections is needed to work out where the front of the car will be
intersections,%
% Hobby is just to get a track that doesn't have a nice'' function
hobby,%
% Calc is to make it easy to draw the car, and to make the clipping path easy to compute
calc,%
% Decorations makes it easy to locate the car on the track
decorations.markings%
}

% Converts the beamer@slideinframe'' to a LaTeX counter to make it easier to animate
\makeatletter
\def\c@slideinframe{\beamer@slideinframe}
\makeatother
\begin{document}
% The track appears to be 125mm long, give or take the length of the car, so we animate over that length
\begin{frame}<1-125>
\begin{tikzpicture}
% Set the position of the car dependent on the slide number
\pgfmathsetmacro\xpos{\the\value{slideinframe}}
% Length of the car
\def\clen{10mm}
% Radius of car wheel
% We're going to use this path a few times, so we save it for easy restoration.
% This also sets the bounding box and marks the position of the back wheel of the car.
\path[
use as bounding box,
use Hobby shortcut,
save Hobby path={track},
decoration={
markings,
mark=at position {\xpos mm} with {\coordinate (bwheel);}
},
decorate
]
([out angle=70]0,0) .. (2,2) .. (4,-2) .. (6,0) .. ([in angle=120]8,-2);
\begin{scope}
% Now we draw the track.
% The actual track is offset from the track path by the radius of the car wheels.
% Since an offset path is unlikely to be a bezier curve, we can't draw it directly.
% So we cheat: we draw a thick path of the right width but clip it against the original path to only draw one side of it.
% Then we overlay with a narrower white path (this is how the double'' key works, except that we only want one of the sides not both, hence the clip).
% The clip path consists of the track path and a lower box, large enough to encompass the thickened track.
\clip (8,-2) |- ($(current bounding box.south)+(0,-2*\crad)$) -| (0,0) [restore and use Hobby path={track}{}];
\draw[restore and use Hobby path={track}{disjoint}, name path global=track,line width=2*\crad+1pt];
\end{scope}
\draw[restore and use Hobby path={track}{disjoint}, name path global=track,line width=2*\crad,white];
% This path is a circle of radius the length of the car.
% We use this to intersect with the track path to find out where the front wheel will be.
\path[name path=base] (bwheel) circle[radius=\clen];
% The intersections library gives us one or two intersections.
% We want the front one, so we order the intersections by the track path and then take the last one.
\path [name intersections={of=track and base,total=\carint,sort by=track}]  coordinate (fwheel) at (intersection-\carint);
% We draw the wheels.
% And lastly the car, using the calc library to draw the sides perpendicular to the base.
\draw[fill=white] (bwheel) -- (fwheel) -- ($(fwheel)!15pt!-90:(bwheel)$) -- ($(bwheel)!15pt!90:(fwheel)$) -- cycle ;
\end{tikzpicture}
\end{frame}
\end{document}


-
\documentclass[pstricks,border=12pt]{standalone}
\begin{document}

\multido{\iA=1+1,\iB=4+1}{40}{%
\begin{pspicture}[algebraic](-.2,-2.2)(\dimexpr\psPiFour cm+.2cm,2.2)
\psparametricplot[linecolor=gray,plotpoints=200]{0}{TwoPi 2 mul}{ t | 2*cos(t) }
\pscurvepoints[plotpoints=100]{0}{4 Pi mul}{ t | 2*cos(t) }{P}
\pspolylineticks[ticksize=0 0,metricInitValue=1,Os=1,Ds=.2]{P}{ ds }{1}{100}%
\pspolygon*[linecolor=gray](PTick\iA)(PTick\iB)([offset=-.5]{PTick\iA}PTick\iB)([offset=.5]{PTick\iB}PTick\iA)
\qdisk(PTick\iA){.1}\qdisk(PTick\iB){.1}
\end{pspicture}%
}

\end{document}


-
Sorry. I noticed the length of the curve is kept constant rather than the length of the car. :-) – kiss my armpit May 7 '14 at 19:10