Wie man mit 10 Bällen jongliert



Und so macht es Mathematica


n = 5;
T = 1;
b = 20;
tw = t /.FindRoot[1 == Pi / T Tan[2 Pi t/T] ((2n-1)/2 T - 2 t),{t,0.9 T/4}];
r = 5( (2n-1) /2 T - 2 tw) T / (2 Pi Cos[2 Pi tw/T]) //N;

SHL[t_] := r { Cos[2 Pi t /T ], Sin[2 Pi t/ T]} //N ;
SHR[t_] := {-1,1} SHL[t+T/2];
VH[t_] := r {- 2 Pi Sin[(2*Pi*t)/T] /T, 2 Pi Cos[(2 Pi t)/T] /T} //N ; 
SB[t_] := SHL[tw] + VH[tw] (t-tw) - {0, 5 (t-tw)^2} //N;

hw = SHL[tw][[2]];
h  = VH[tw][[2]]^2 /20 + hw;
la = n/2;
Do[
Bll = Table[ If[SB[t+(b-1)T ][[2]] > hw, 
  Point[SB[t+(b-1) T]-{la,0}],Point[SHL[t]-{la,0}]] ,{b,n}];
Blr = Table[ If[SB[t+(b-1.5)T ][[2]] > hw, 
  Point[{-1,1} SB[t+(b-1.5) T]+{la,0}],Point[SHR[t]+{la,0}]] ,{b,n}];
Show[Graphics[ Join[
    Bll,
    Blr,
    {Line[{{la,0},SHR[t]-{-la,0}}]},
    {Line[{{-la,0},SHL[t]-{la,0}}]},
    {Line[{{la,0},{0,1}}]},
    {Line[{{-la,0},{0,1}}]},
    {Line[{{-la,-2r},{-2/3 la,-2r},{0,1},{2/3 la,-2r},{la,-2r}}]},
    {Circle[{0,3},1]},
    {Text["by F.Holderied",{10,-7}]}
    ]],
  PlotRange->1.1{{-(h+2r)/2, (h+2r)/2},{-2r,h}},
  AspectRatio -> 1,
  Prolog->AbsolutePointSize[7]
    ],
  {t,0,T (1-1/b),T/b}]

Felix Holderied, (felix@holderied.de)