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Animating wave motion in water
Obtaining a 3D animation as a drop in a liquid surfaceChanging a wave simulationAnimating arrowsAnimating labelsAnimating a parametric plotHow to solve the tsunami model and animate the shallow water wave?Solve and design “water flow”Not getting smooth motion when animating a 3D plot (with jumps in rotation)Animating pointAnimating a 3D plot of electromagnetic wavesPlotting and Animating Spherical Spring Pendulum
22
$begingroup$
Further to this question I found on MSE, I tried to replicate
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 physics simulation
edited 21 hours ago
J. M. is slightly pensive♦
98.2k10306466
asked Mar 18 at 11:08
martinmartin
4,03321249
$endgroup$
add a comment |
22
$begingroup$
Further to this question I found on MSE, I tried to replicate
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 physics simulation
edited 21 hours ago
J. M. is slightly pensive♦
98.2k10306466
asked Mar 18 at 11:08
martinmartin
4,03321249
$endgroup$
$begingroup$
See this: mathematica.stackexchange.com/questions/123127/…
$endgroup$
– LCarvalho
Mar 18 at 11:23
$begingroup$
For searching purposes: this is a trochoidal wave.
$endgroup$
– J. M. is slightly pensive♦
21 hours ago
$begingroup$
@J.M.isslightlypensive thanks - looks like next step will be stokes drift!
$endgroup$
– martin
20 hours ago
add a comment |
22
22
22
9
$begingroup$
Further to this question I found on MSE, I tried to replicate
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 physics simulation
edited 21 hours ago
J. M. is slightly pensive♦
98.2k10306466
asked Mar 18 at 11:08
martinmartin
4,03321249
$endgroup$
Further to this question I found on MSE, I tried to replicate
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 physics simulation
performance-tuning animation physics simulation
edited 21 hours ago
J. M. is slightly pensive♦
98.2k10306466
asked Mar 18 at 11:08
martinmartin
4,03321249
edited 21 hours ago
J. M. is slightly pensive♦
98.2k10306466
asked Mar 18 at 11:08
martinmartin
4,03321249
edited 21 hours ago
J. M. is slightly pensive♦
98.2k10306466
edited 21 hours ago
J. M. is slightly pensive♦
98.2k10306466
edited 21 hours ago
J. M. is slightly pensive♦
98.2k10306466
98.2k10306466
asked Mar 18 at 11:08
martinmartin
4,03321249
asked Mar 18 at 11:08
martinmartin
4,03321249
asked Mar 18 at 11:08
martinmartin
4,03321249
4,03321249
$begingroup$
See this: mathematica.stackexchange.com/questions/123127/…
$endgroup$
– LCarvalho
Mar 18 at 11:23
$begingroup$
For searching purposes: this is a trochoidal wave.
$endgroup$
– J. M. is slightly pensive♦
21 hours ago
$begingroup$
@J.M.isslightlypensive thanks - looks like next step will be stokes drift!
$endgroup$
– martin
20 hours ago
add a comment |
$begingroup$
See this: mathematica.stackexchange.com/questions/123127/…
$endgroup$
– LCarvalho
Mar 18 at 11:23
$begingroup$
For searching purposes: this is a trochoidal wave.
$endgroup$
– J. M. is slightly pensive♦
21 hours ago
$begingroup$
@J.M.isslightlypensive thanks - looks like next step will be stokes drift!
$endgroup$
– martin
20 hours ago
$begingroup$
See this: mathematica.stackexchange.com/questions/123127/…
$endgroup$
– LCarvalho
Mar 18 at 11:23
$begingroup$
See this: mathematica.stackexchange.com/questions/123127/…
$endgroup$
– LCarvalho
Mar 18 at 11:23
$begingroup$
For searching purposes: this is a trochoidal wave.
$endgroup$
– J. M. is slightly pensive♦
21 hours ago
$begingroup$
For searching purposes: this is a trochoidal wave.
$endgroup$
– J. M. is slightly pensive♦
21 hours ago
$begingroup$
@J.M.isslightlypensive thanks - looks like next step will be stokes drift!
$endgroup$
– martin
20 hours ago
$begingroup$
@J.M.isslightlypensive thanks - looks like next step will be stokes drift!
$endgroup$
– martin
20 hours ago
add a comment |
1 Answer
1
active
oldest
votes
31
$begingroup$
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]
]
]
answered Mar 18 at 13:16
Kuba♦Kuba
107k12210530
$endgroup$
$begingroup$
thanks - a vastly better approach!
$endgroup$
– martin
Mar 18 at 13:34
$begingroup$
@martin thanks, let me know if anything is not clear.
$endgroup$
– Kuba♦
Mar 19 at 21:31
$begingroup$
a = .08doesn't get used; maybe you meantr[y_] := a y^4? (Also, how did you pick the form ofr[y]?)
$endgroup$
– J. M. is slightly pensive♦
14 hours ago
1
$begingroup$
@J.M.isslightlypensive right, should be deleted. About formulas, I got them from looking at the original .gif :-) I was too lazy to think about physics etc so I assumed something like that should work. I also ignored OP's code :(
$endgroup$
– Kuba♦
14 hours ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
31
$begingroup$
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]
]
]
answered Mar 18 at 13:16
Kuba♦Kuba
107k12210530
$endgroup$
$begingroup$
thanks - a vastly better approach!
$endgroup$
– martin
Mar 18 at 13:34
$begingroup$
@martin thanks, let me know if anything is not clear.
$endgroup$
– Kuba♦
Mar 19 at 21:31
$begingroup$
a = .08doesn't get used; maybe you meantr[y_] := a y^4? (Also, how did you pick the form ofr[y]?)
$endgroup$
– J. M. is slightly pensive♦
14 hours ago
1
$begingroup$
@J.M.isslightlypensive right, should be deleted. About formulas, I got them from looking at the original .gif :-) I was too lazy to think about physics etc so I assumed something like that should work. I also ignored OP's code :(
$endgroup$
– Kuba♦
14 hours ago
add a comment |
31
$begingroup$
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]
]
]
answered Mar 18 at 13:16
Kuba♦Kuba
107k12210530
$endgroup$
$begingroup$
thanks - a vastly better approach!
$endgroup$
– martin
Mar 18 at 13:34
$begingroup$
@martin thanks, let me know if anything is not clear.
$endgroup$
– Kuba♦
Mar 19 at 21:31
$begingroup$
a = .08doesn't get used; maybe you meantr[y_] := a y^4? (Also, how did you pick the form ofr[y]?)
$endgroup$
– J. M. is slightly pensive♦
14 hours ago
1
$begingroup$
@J.M.isslightlypensive right, should be deleted. About formulas, I got them from looking at the original .gif :-) I was too lazy to think about physics etc so I assumed something like that should work. I also ignored OP's code :(
$endgroup$
– Kuba♦
14 hours ago
add a comment |
31
31
31
$begingroup$
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]
]
]
answered Mar 18 at 13:16
Kuba♦Kuba
107k12210530
$endgroup$
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]
]
]
answered Mar 18 at 13:16
Kuba♦Kuba
107k12210530
edited Mar 18 at 21:35
answered Mar 18 at 13:16
Kuba♦Kuba
107k12210530
answered Mar 18 at 13:16
Kuba♦Kuba
107k12210530
answered Mar 18 at 13:16
Kuba♦Kuba
107k12210530
107k12210530
$begingroup$
thanks - a vastly better approach!
$endgroup$
– martin
Mar 18 at 13:34
$begingroup$
@martin thanks, let me know if anything is not clear.
$endgroup$
– Kuba♦
Mar 19 at 21:31
$begingroup$
a = .08doesn't get used; maybe you meantr[y_] := a y^4? (Also, how did you pick the form ofr[y]?)
$endgroup$
– J. M. is slightly pensive♦
14 hours ago
1
$begingroup$
@J.M.isslightlypensive right, should be deleted. About formulas, I got them from looking at the original .gif :-) I was too lazy to think about physics etc so I assumed something like that should work. I also ignored OP's code :(
$endgroup$
– Kuba♦
14 hours ago
add a comment |
$begingroup$
thanks - a vastly better approach!
$endgroup$
– martin
Mar 18 at 13:34
$begingroup$
@martin thanks, let me know if anything is not clear.
$endgroup$
– Kuba♦
Mar 19 at 21:31
$begingroup$
a = .08doesn't get used; maybe you meantr[y_] := a y^4? (Also, how did you pick the form ofr[y]?)
$endgroup$
– J. M. is slightly pensive♦
14 hours ago
1
$begingroup$
@J.M.isslightlypensive right, should be deleted. About formulas, I got them from looking at the original .gif :-) I was too lazy to think about physics etc so I assumed something like that should work. I also ignored OP's code :(
$endgroup$
– Kuba♦
14 hours ago
$begingroup$
thanks - a vastly better approach!
$endgroup$
– martin
Mar 18 at 13:34
$begingroup$
thanks - a vastly better approach!
$endgroup$
– martin
Mar 18 at 13:34
$begingroup$
@martin thanks, let me know if anything is not clear.
$endgroup$
– Kuba♦
Mar 19 at 21:31
$begingroup$
@martin thanks, let me know if anything is not clear.
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– Kuba♦
Mar 19 at 21:31
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a = .08 doesn't get used; maybe you meant r[y_] := a y^4? (Also, how did you pick the form of r[y]?)$endgroup$
– J. M. is slightly pensive♦
14 hours ago
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a = .08 doesn't get used; maybe you meant r[y_] := a y^4? (Also, how did you pick the form of r[y]?)$endgroup$
– J. M. is slightly pensive♦
14 hours ago
1
1
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@J.M.isslightlypensive right, should be deleted. About formulas, I got them from looking at the original .gif :-) I was too lazy to think about physics etc so I assumed something like that should work. I also ignored OP's code :(
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– Kuba♦
14 hours ago
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@J.M.isslightlypensive right, should be deleted. About formulas, I got them from looking at the original .gif :-) I was too lazy to think about physics etc so I assumed something like that should work. I also ignored OP's code :(
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– Kuba♦
14 hours ago
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$begingroup$
See this: mathematica.stackexchange.com/questions/123127/…
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– LCarvalho
Mar 18 at 11:23
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For searching purposes: this is a trochoidal wave.
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– J. M. is slightly pensive♦
21 hours ago
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@J.M.isslightlypensive thanks - looks like next step will be stokes drift!
$endgroup$
– martin
20 hours ago