Animating wave motion in water












19












$begingroup$


Further to this question I found on MSE, I tried to replicate



enter image description here



from here



this is as far as I got:



fun[a_, b_, c_, x_, y_] := 

  Point[{#[[1]] + x, #[[2]] + y} &[

    Part[CirclePoints[360] c, 

     If[a + b == 360, 360, Mod[a + b, 360]]]]];

tab = With[{a = #}, 

     Flatten[Table[

       Table[fun[a, 90 + 15 n, 1 - .15 m, -1 + .5 n, -.35 m], {m, 0, 

         10}], {n, 0, 24}], 1]] & /@ Range[1, 360, 15];



Module[{t, x, y, fun, xf, yf, a}, x = -.5; y = 1;

 fun[a_, b_, c_, x_, y_] := 

  Point[{#[[1]] + x, #[[2]] + y} &[

    Part[CirclePoints[360] c, 

     If[a + b == 360, 360, Mod[a + b, 360]]]]];

 xf[t_, a_, b_] := a t - b Sin[t]; yf[t_, a_, b_] := a - b Cos[t];

 Animate[

  Show[

   Graphics[

    {PointSize[.01], tab[[a]]},

    PlotRange -> {{-1 - x, 10 + x}, {-1 - y, 1}}

    ],

   ParametricPlot[

    {(Pi/2) xf[t + 2 Pi a/24, 1.25, .6] - 4 Pi a/24 - Pi^2 + .05,  

     2.05 - 1.65 yf[t + 2 Pi a/24, 1.25, .6]},

    {t, -4 Pi, 4 Pi}, Axes -> False

    ]

   ],

  {a, 1, 24, 1}, ControlPlacement -> Top, AnimationRate -> 5, 

  AnimationDirection -> Backward

  ]

 ]


which is not very efficient (I'm sure Part could be applied more efficiently), and despite various tweeks, I couldn't quite manage to get the cycloid to line up with the points.



What is a better way to approach this?






performance-tuning animation







share|improve this question











$endgroup$












  • $begingroup$
    See this: mathematica.stackexchange.com/questions/123127/…
    $endgroup$
    – LCarvalho
    2 days ago
















19












$begingroup$


Further to this question I found on MSE, I tried to replicate



enter image description here



from here



this is as far as I got:



fun[a_, b_, c_, x_, y_] := 

  Point[{#[[1]] + x, #[[2]] + y} &[

    Part[CirclePoints[360] c, 

     If[a + b == 360, 360, Mod[a + b, 360]]]]];

tab = With[{a = #}, 

     Flatten[Table[

       Table[fun[a, 90 + 15 n, 1 - .15 m, -1 + .5 n, -.35 m], {m, 0, 

         10}], {n, 0, 24}], 1]] & /@ Range[1, 360, 15];



Module[{t, x, y, fun, xf, yf, a}, x = -.5; y = 1;

 fun[a_, b_, c_, x_, y_] := 

  Point[{#[[1]] + x, #[[2]] + y} &[

    Part[CirclePoints[360] c, 

     If[a + b == 360, 360, Mod[a + b, 360]]]]];

 xf[t_, a_, b_] := a t - b Sin[t]; yf[t_, a_, b_] := a - b Cos[t];

 Animate[

  Show[

   Graphics[

    {PointSize[.01], tab[[a]]},

    PlotRange -> {{-1 - x, 10 + x}, {-1 - y, 1}}

    ],

   ParametricPlot[

    {(Pi/2) xf[t + 2 Pi a/24, 1.25, .6] - 4 Pi a/24 - Pi^2 + .05,  

     2.05 - 1.65 yf[t + 2 Pi a/24, 1.25, .6]},

    {t, -4 Pi, 4 Pi}, Axes -> False

    ]

   ],

  {a, 1, 24, 1}, ControlPlacement -> Top, AnimationRate -> 5, 

  AnimationDirection -> Backward

  ]

 ]


which is not very efficient (I'm sure Part could be applied more efficiently), and despite various tweeks, I couldn't quite manage to get the cycloid to line up with the points.



What is a better way to approach this?






performance-tuning animation







share|improve this question











$endgroup$












  • $begingroup$
    See this: mathematica.stackexchange.com/questions/123127/…
    $endgroup$
    – LCarvalho
    2 days ago














19












19








19


8



$begingroup$


Further to this question I found on MSE, I tried to replicate



enter image description here



from here



this is as far as I got:



fun[a_, b_, c_, x_, y_] := 

  Point[{#[[1]] + x, #[[2]] + y} &[

    Part[CirclePoints[360] c, 

     If[a + b == 360, 360, Mod[a + b, 360]]]]];

tab = With[{a = #}, 

     Flatten[Table[

       Table[fun[a, 90 + 15 n, 1 - .15 m, -1 + .5 n, -.35 m], {m, 0, 

         10}], {n, 0, 24}], 1]] & /@ Range[1, 360, 15];



Module[{t, x, y, fun, xf, yf, a}, x = -.5; y = 1;

 fun[a_, b_, c_, x_, y_] := 

  Point[{#[[1]] + x, #[[2]] + y} &[

    Part[CirclePoints[360] c, 

     If[a + b == 360, 360, Mod[a + b, 360]]]]];

 xf[t_, a_, b_] := a t - b Sin[t]; yf[t_, a_, b_] := a - b Cos[t];

 Animate[

  Show[

   Graphics[

    {PointSize[.01], tab[[a]]},

    PlotRange -> {{-1 - x, 10 + x}, {-1 - y, 1}}

    ],

   ParametricPlot[

    {(Pi/2) xf[t + 2 Pi a/24, 1.25, .6] - 4 Pi a/24 - Pi^2 + .05,  

     2.05 - 1.65 yf[t + 2 Pi a/24, 1.25, .6]},

    {t, -4 Pi, 4 Pi}, Axes -> False

    ]

   ],

  {a, 1, 24, 1}, ControlPlacement -> Top, AnimationRate -> 5, 

  AnimationDirection -> Backward

  ]

 ]


which is not very efficient (I'm sure Part could be applied more efficiently), and despite various tweeks, I couldn't quite manage to get the cycloid to line up with the points.



What is a better way to approach this?






performance-tuning animation







share|improve this question











$endgroup$




Further to this question I found on MSE, I tried to replicate



enter image description here



from here



this is as far as I got:



fun[a_, b_, c_, x_, y_] := 

  Point[{#[[1]] + x, #[[2]] + y} &[

    Part[CirclePoints[360] c, 

     If[a + b == 360, 360, Mod[a + b, 360]]]]];

tab = With[{a = #}, 

     Flatten[Table[

       Table[fun[a, 90 + 15 n, 1 - .15 m, -1 + .5 n, -.35 m], {m, 0, 

         10}], {n, 0, 24}], 1]] & /@ Range[1, 360, 15];



Module[{t, x, y, fun, xf, yf, a}, x = -.5; y = 1;

 fun[a_, b_, c_, x_, y_] := 

  Point[{#[[1]] + x, #[[2]] + y} &[

    Part[CirclePoints[360] c, 

     If[a + b == 360, 360, Mod[a + b, 360]]]]];

 xf[t_, a_, b_] := a t - b Sin[t]; yf[t_, a_, b_] := a - b Cos[t];

 Animate[

  Show[

   Graphics[

    {PointSize[.01], tab[[a]]},

    PlotRange -> {{-1 - x, 10 + x}, {-1 - y, 1}}

    ],

   ParametricPlot[

    {(Pi/2) xf[t + 2 Pi a/24, 1.25, .6] - 4 Pi a/24 - Pi^2 + .05,  

     2.05 - 1.65 yf[t + 2 Pi a/24, 1.25, .6]},

    {t, -4 Pi, 4 Pi}, Axes -> False

    ]

   ],

  {a, 1, 24, 1}, ControlPlacement -> Top, AnimationRate -> 5, 

  AnimationDirection -> Backward

  ]

 ]


which is not very efficient (I'm sure Part could be applied more efficiently), and despite various tweeks, I couldn't quite manage to get the cycloid to line up with the points.



What is a better way to approach this?






performance-tuning animation




performance-tuning animation






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 days ago









Kuba♦

107k12210530




107k12210530










asked 2 days ago









martinmartin

4,01821249




4,01821249












  • $begingroup$
    See this: mathematica.stackexchange.com/questions/123127/…
    $endgroup$
    – LCarvalho
    2 days ago


















  • $begingroup$
    See this: mathematica.stackexchange.com/questions/123127/…
    $endgroup$
    – LCarvalho
    2 days ago
















$begingroup$
See this: mathematica.stackexchange.com/questions/123127/…
$endgroup$
– LCarvalho
2 days ago




$begingroup$
See this: mathematica.stackexchange.com/questions/123127/…
$endgroup$
– LCarvalho
2 days ago










1 Answer
1






active

oldest

votes


















27












$begingroup$

DynamicModule[{t = 0, d = 5, a = .08, base, distortion, pts, r, f, n = 10},



 r[y_] := .08 y^4;

 f[x_] := -2 Pi Dynamic[t] + d x; 

 (*f does not evaluate to a number but FE will take care of that later*)



 base = Array[List, n {3, 1}, {{0, Pi}, {0, 1}} ];



 distortion = Array[ 

   Function[{x, y}, r[y] {Cos @ f @ x, Sin @ f @ x}], n {3, 1}, {{0, Pi}, {0, 1}} 

 ];



 pts = base + distortion;



 Row[{

   Animator[Dynamic @ t, AnimationRate -> .8, AppearanceElements -> {}],

   Graphics[{   

     LightBlue,

     Polygon @ Join[ pts[[;; , -1]], {Scaled[{1, 0}], Scaled[{0, 0}]}],



     Darker @ Blue, AbsolutePointSize @ 5, Point @ Catenate @ pts,



     AbsolutePointSize @ 7, Orange, Thick,

     Point @ pts[[15, -1]],  Circle[base[[15, -1]], r @ base[[15, -1, 2]]],

     Point @ pts[[15, 7]],  Circle[base[[15, 7]], r @ base[[15, 7, 2]]]     

     },

    PlotRange -> {{0 + .1, Pi - .1}, {0, 1.2}}, 

    PlotRangePadding -> 0,

    PlotRangeClipping -> True, ImageSize -> 800]

   }]

 ]


enter image description here






share|improve this answer











$endgroup$













  • $begingroup$
    thanks - a vastly better approach!
    $endgroup$
    – martin
    2 days ago










  • $begingroup$
    @martin thanks, let me know if anything is not clear.
    $endgroup$
    – Kuba♦
    yesterday











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









27












$begingroup$

DynamicModule[{t = 0, d = 5, a = .08, base, distortion, pts, r, f, n = 10},



 r[y_] := .08 y^4;

 f[x_] := -2 Pi Dynamic[t] + d x; 

 (*f does not evaluate to a number but FE will take care of that later*)



 base = Array[List, n {3, 1}, {{0, Pi}, {0, 1}} ];



 distortion = Array[ 

   Function[{x, y}, r[y] {Cos @ f @ x, Sin @ f @ x}], n {3, 1}, {{0, Pi}, {0, 1}} 

 ];



 pts = base + distortion;



 Row[{

   Animator[Dynamic @ t, AnimationRate -> .8, AppearanceElements -> {}],

   Graphics[{   

     LightBlue,

     Polygon @ Join[ pts[[;; , -1]], {Scaled[{1, 0}], Scaled[{0, 0}]}],



     Darker @ Blue, AbsolutePointSize @ 5, Point @ Catenate @ pts,



     AbsolutePointSize @ 7, Orange, Thick,

     Point @ pts[[15, -1]],  Circle[base[[15, -1]], r @ base[[15, -1, 2]]],

     Point @ pts[[15, 7]],  Circle[base[[15, 7]], r @ base[[15, 7, 2]]]     

     },

    PlotRange -> {{0 + .1, Pi - .1}, {0, 1.2}}, 

    PlotRangePadding -> 0,

    PlotRangeClipping -> True, ImageSize -> 800]

   }]

 ]


enter image description here






share|improve this answer











$endgroup$













  • $begingroup$
    thanks - a vastly better approach!
    $endgroup$
    – martin
    2 days ago










  • $begingroup$
    @martin thanks, let me know if anything is not clear.
    $endgroup$
    – Kuba♦
    yesterday
















27












$begingroup$

DynamicModule[{t = 0, d = 5, a = .08, base, distortion, pts, r, f, n = 10},



 r[y_] := .08 y^4;

 f[x_] := -2 Pi Dynamic[t] + d x; 

 (*f does not evaluate to a number but FE will take care of that later*)



 base = Array[List, n {3, 1}, {{0, Pi}, {0, 1}} ];



 distortion = Array[ 

   Function[{x, y}, r[y] {Cos @ f @ x, Sin @ f @ x}], n {3, 1}, {{0, Pi}, {0, 1}} 

 ];



 pts = base + distortion;



 Row[{

   Animator[Dynamic @ t, AnimationRate -> .8, AppearanceElements -> {}],

   Graphics[{   

     LightBlue,

     Polygon @ Join[ pts[[;; , -1]], {Scaled[{1, 0}], Scaled[{0, 0}]}],



     Darker @ Blue, AbsolutePointSize @ 5, Point @ Catenate @ pts,



     AbsolutePointSize @ 7, Orange, Thick,

     Point @ pts[[15, -1]],  Circle[base[[15, -1]], r @ base[[15, -1, 2]]],

     Point @ pts[[15, 7]],  Circle[base[[15, 7]], r @ base[[15, 7, 2]]]     

     },

    PlotRange -> {{0 + .1, Pi - .1}, {0, 1.2}}, 

    PlotRangePadding -> 0,

    PlotRangeClipping -> True, ImageSize -> 800]

   }]

 ]


enter image description here






share|improve this answer











$endgroup$













  • $begingroup$
    thanks - a vastly better approach!
    $endgroup$
    – martin
    2 days ago










  • $begingroup$
    @martin thanks, let me know if anything is not clear.
    $endgroup$
    – Kuba♦
    yesterday














27












27








27





$begingroup$

DynamicModule[{t = 0, d = 5, a = .08, base, distortion, pts, r, f, n = 10},



 r[y_] := .08 y^4;

 f[x_] := -2 Pi Dynamic[t] + d x; 

 (*f does not evaluate to a number but FE will take care of that later*)



 base = Array[List, n {3, 1}, {{0, Pi}, {0, 1}} ];



 distortion = Array[ 

   Function[{x, y}, r[y] {Cos @ f @ x, Sin @ f @ x}], n {3, 1}, {{0, Pi}, {0, 1}} 

 ];



 pts = base + distortion;



 Row[{

   Animator[Dynamic @ t, AnimationRate -> .8, AppearanceElements -> {}],

   Graphics[{   

     LightBlue,

     Polygon @ Join[ pts[[;; , -1]], {Scaled[{1, 0}], Scaled[{0, 0}]}],



     Darker @ Blue, AbsolutePointSize @ 5, Point @ Catenate @ pts,



     AbsolutePointSize @ 7, Orange, Thick,

     Point @ pts[[15, -1]],  Circle[base[[15, -1]], r @ base[[15, -1, 2]]],

     Point @ pts[[15, 7]],  Circle[base[[15, 7]], r @ base[[15, 7, 2]]]     

     },

    PlotRange -> {{0 + .1, Pi - .1}, {0, 1.2}}, 

    PlotRangePadding -> 0,

    PlotRangeClipping -> True, ImageSize -> 800]

   }]

 ]


enter image description here






share|improve this answer











$endgroup$



DynamicModule[{t = 0, d = 5, a = .08, base, distortion, pts, r, f, n = 10},



 r[y_] := .08 y^4;

 f[x_] := -2 Pi Dynamic[t] + d x; 

 (*f does not evaluate to a number but FE will take care of that later*)



 base = Array[List, n {3, 1}, {{0, Pi}, {0, 1}} ];



 distortion = Array[ 

   Function[{x, y}, r[y] {Cos @ f @ x, Sin @ f @ x}], n {3, 1}, {{0, Pi}, {0, 1}} 

 ];



 pts = base + distortion;



 Row[{

   Animator[Dynamic @ t, AnimationRate -> .8, AppearanceElements -> {}],

   Graphics[{   

     LightBlue,

     Polygon @ Join[ pts[[;; , -1]], {Scaled[{1, 0}], Scaled[{0, 0}]}],



     Darker @ Blue, AbsolutePointSize @ 5, Point @ Catenate @ pts,



     AbsolutePointSize @ 7, Orange, Thick,

     Point @ pts[[15, -1]],  Circle[base[[15, -1]], r @ base[[15, -1, 2]]],

     Point @ pts[[15, 7]],  Circle[base[[15, 7]], r @ base[[15, 7, 2]]]     

     },

    PlotRange -> {{0 + .1, Pi - .1}, {0, 1.2}}, 

    PlotRangePadding -> 0,

    PlotRangeClipping -> True, ImageSize -> 800]

   }]

 ]


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago

























answered 2 days ago









Kuba♦Kuba

107k12210530




107k12210530












  • $begingroup$
    thanks - a vastly better approach!
    $endgroup$
    – martin
    2 days ago










  • $begingroup$
    @martin thanks, let me know if anything is not clear.
    $endgroup$
    – Kuba♦
    yesterday


















  • $begingroup$
    thanks - a vastly better approach!
    $endgroup$
    – martin
    2 days ago










  • $begingroup$
    @martin thanks, let me know if anything is not clear.
    $endgroup$
    – Kuba♦
    yesterday
















$begingroup$
thanks - a vastly better approach!
$endgroup$
– martin
2 days ago




$begingroup$
thanks - a vastly better approach!
$endgroup$
– martin
2 days ago












$begingroup$
@martin thanks, let me know if anything is not clear.
$endgroup$
– Kuba♦
yesterday




$begingroup$
@martin thanks, let me know if anything is not clear.
$endgroup$
– Kuba♦
yesterday


















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