Per the documentation for OpenModelica , DynamicSelect,
Any value (coordinates, color, text, etc.) in graphical annotations can be dependent on class variables using the DynamicSelect expression. DynamicSelect has the syntax of a function call with two arguments, where the first argument specifies the value of the editing state and the second argument the value of the non-editing state. The first argument must be a literal expression and this value is used for the annotation when editing and/or browsing the diagram layer. The second argument may contain references to variables to enable a dynamic behavior and the actual value is used for the annotation for schematic animation of the diagram layer, e.g., after a simulation.
To test, a model with DynamicSelect was created to change colors and coordinates during a simulation.
model BarGraph_v001
Modelica.Blocks.Interfaces.RealInput u annotation(
Placement(visible = true, transformation(origin = {-104, 0}, extent = {{-20, -20}, {20, 20}}, rotation = 0), iconTransformation(origin = {-62, 0}, extent = {{-20, -20}, {20, 20}}, rotation = 0)));
equation
annotation(
Icon(graphics = {
Rectangle(lineThickness = 2, extent = DynamicSelect({{-40, 100}, {40, -100}},{{-40, 100}, {u, -100}})),
Rectangle(visible=true, origin = DynamicSelect({0, 27},{0,u}), fillColor=DynamicSelect({192,192,192}, {125+u*5,125-u*5,125+u*5}), fillPattern = FillPattern.Solid, extent = {{-40, 13}, {40, -13}})}));
end BarGraph_v001;
Note that DynamicSelect is used to change the coordinates of the origin and the color of the rectangle in the icon.
The static icon for this model is:
Then this was integrated into a model with a stimulus and several other animated Modelica library elements.
model BarGraph_test_001
Anim.BarGraph_v001 barGraph_v001 annotation(
Placement(visible = true, transformation(origin = {22, 0}, extent = {{-10, -10}, {10, 10}}, rotation = 0)));
Modelica.Blocks.Sources.Sine sine(f = 4) annotation(
Placement(visible = true, transformation(origin = {-54, 0}, extent = {{-10, -10}, {10, 10}}, rotation = 0)));
Modelica.Blocks.Math.Gain gain(k = 25) annotation(
Placement(visible = true, transformation(origin = {-14, 0}, extent = {{-10, -10}, {10, 10}}, rotation = 0)));
Modelica.Blocks.Interaction.Show.RealValue realValue annotation(
Placement(visible = true, transformation(origin = {28, 24}, extent = {{-10, -10}, {10, 10}}, rotation = 0)));
Modelica.Blocks.Interaction.Show.BooleanValue booleanValue annotation(
Placement(visible = true, transformation(origin = {56, -36}, extent = {{-10, -10}, {10, 10}}, rotation = 0)));
Modelica.Blocks.Logical.GreaterThreshold greaterThreshold annotation(
Placement(visible = true, transformation(origin = {18, -36}, extent = {{-10, -10}, {10, 10}}, rotation = 0)));
equation
connect(gain.y, barGraph_v001.u) annotation(
Line(points = {{-2, 0}, {16, 0}}, color = {0, 0, 127}));
connect(sine.y, gain.u) annotation(
Line(points = {{-42, 0}, {-26, 0}}, color = {0, 0, 127}));
connect(realValue.numberPort, gain.y) annotation(
Line(points = {{16, 24}, {8, 24}, {8, 0}, {-2, 0}}, color = {0, 0, 127}));
connect(greaterThreshold.y, booleanValue.activePort) annotation(
Line(points = {{30, -36}, {44, -36}}, color = {255, 0, 255}));
connect(greaterThreshold.u, gain.y) annotation(
Line(points = {{6, -36}, {-2, -36}, {-2, -12}, {10, -12}, {10, 0}, {-2, 0}}, color = {0, 0, 127}));
end BarGraph_test_001;
The model looks like this in the editor:
And after simulation, it is animated like this. Note that colors change, the library parts are animated, but the DynamicSelect used to change coordinates is not working.
Questions:
Is DynamicSelect being used correctly in this example?
Is DynamicSelect changes in coordinates supported in OpenModelica Connection Editor 2.19.0?
Changing the size of the rectangle works in Dymola:
But it seems like the current version of OMEdit (v1.19.2) does not fully support DynamicSelect.
There is an issue on github about DynamicSelect support in OpenModelica: #3675: Add interactive simulation support (e.g., DynamicSelect display)
Related
Hello! Help me please to make a procedure from my program code.
I've got code, which includs 2 parts (they're Graphics[...]), it plots functions with rectengles.
But it seems to be too much code, so I have to write a procedure with an argument of a function and then when I call it, this procedure would do the same (plots functions with rectengles). (can somehows to use spx1,spx2.)
This is whole code:
(*Sin[x]*)
data1 = Table[{x, Sin[x]}, {x, -1, 6, 0.1}];
data11 = data1[[;; , 2]];(*отримуємо набір ординат*)localmins1 =
Pick[data1, MinDetect[data11, 10^-6], 1];
localmaxs1 = Pick[data1, MaxDetect[data11, 10^-6], 1];
Graphics[{Thick, Blue, Line[data1], Red, PointSize[0.01],
Point[localmins1], Point[localmaxs1]}, Axes -> True];
spx1 = {-1, -0.35, 0.3, 0.95, 1.6, 2.375, 3.15, 3.925, 4.7, 6};
Graphics[{Thick, Blue, Line[data1], Red, PointSize[0.01],
Point[localmins1], Point[localmaxs1],
Point[{{-1, 0}, {1.6, 0}, {-0.35, 0}, {0.3, 0}, {0.95, 0}}], Green,
Point[{{-0.35, -0.1342898}, {0.3, 0.29552}, {0.95, 0.813416}}], Red,
Point[{{2.375, 0}, {3.15, 0}, {3.925, 0}, {4.7, 0}}], Green,
Point[{{2.375, 0.693685}, {3.15, -0.0084}, {3.925, -0.70569766}}],
Pink, Opacity[.7], EdgeForm[Directive[Dashed, Pink]],
Rectangle[{-1, -0.84147}, {-0.35, -0.342898}],
Rectangle[{-0.35, -0.342898}, {0.3, 0.29552}],
Rectangle[{0.3, 0.29552}, {0.95, 0.813416}],
Rectangle[{0.95, 0.813416}, {1.6`, 0.9995736030415051`}],
Rectangle[{1.6, 0.99957}, {2.375, 0.693685}],
Rectangle[{2.375, 0.693685}, {3.15, -0.0084}],
Rectangle[{3.15, -0.0084}, {3.925, -0.70569766}],
Rectangle[{3.925, -0.70569766}, {4.7, -0.9999}],}, Axes -> True]
(*Cos[x]*)
data2 = Table[{x, Cos[x]}, {x, -3, 4, 0.1}]; data22 =
data2[[;; , 2]];(*отримуємо набір ординат*)localmins2 =
Pick[data2, MinDetect[data22, 10^-6], 1];
localmaxs2 = Pick[data2, MaxDetect[data22, 10^-6], 1];
Graphics[{Thick, Blue, Line[data2], Red, PointSize[0.01],
Point[localmins2], Point[localmaxs2]}, Axes -> True];
spx2 = {-3,-2.25, -1.5, -0.75,0, 0.75, 1.5, 2.25, 3};
spxrozb11 = {-3, -2.25, -1.5, -0.75, 0};
spxrozb22 = {0, 0.75, 1.5, 2.25, 3};
Graphics[{Thick, Blue, Line[data2], Red, PointSize[0.01],
Point[localmins2], Point[localmaxs2],
Point[{{-2.25, 0}, {-1.5, 0}, {-0.75, 0}, {0, 0}}], Green,
Point[{{-2.25, -0.628}, {-1.5, 0.07}, {-0.75, 0.73}}], Red,
Point[{{0.75, 0}, {1.5, 0}, {2.25, 0}, {3, 0}}], Green,
Point[{{0.75, 0.7316}, {1.5, 0.07}, {2.25, -0.628}}], Pink,
Opacity[.7], EdgeForm[Directive[Dashed, Pink]],
Rectangle[{-3, -0.989992}, {-2.25, -0.628}],
Rectangle[{-2.25, -0.628}, {-1.5, 0.07}],
Rectangle[{-1.5, 0.07}, {-0.75, 0.7316}],
Rectangle[{-0.75, 0.7316}, {1.6653345369377348`*^-16, 1.`}],
Rectangle[{1.6653345369377348`*^-16, 1.`}, {0.75, 0.7316}],
Rectangle[{0.75, 0.7316}, {1.5, 0.07}],
Rectangle[{1.5, 0.07}, {2.25, -0.628}],
Rectangle[{2.25, -0.628}, {3.1000000000000005`, \
-0.9991351502732795`}]}, Axes -> True]
I tried to write the first step in procedure, and even it doesn't work;(
f1 = Table[{x, Sin[x]}, {x, -1, 6, 0.1}];
f2 = Table[{x, Cos[x]}, {x, -3, 4, 0.1}];
data = {f1, f2};
spx1 = {-1, -0.35, 0.3, 0.95, 1.6, 2.375, 3.15, 3.925, 4.7, 6};
graph[i_] :=
Graphics[For[i = 0, i < 3,
i++, {Thick, Blue, Line[data[[i]]], Red, Point[{spx1[[i]], 0}]}]]
graph[1]
******How to write For in procedure correctly?******
Help me, please!
There's the procedure operation[f_].
It works correctly and plot for functions:Cos,Sin. But, Unfortunately, it doesn't work for E^x and Log[E,x] and outputs errors, maybe because inputting not correct name of function or something else;(( What's the problem?
spxsin = {-1, -0.35, 0.3, 0.95, 1.6, 2.375, 3.15, 3.925, 4.7, 5.025,
5.35, 5.675, 6};
spxcos = {-1, -0.75, -0.5, -0.25, 0, 0.775, 1.55, 2.325, 3.1, 3.825,
4.55, 5.275, 6};
spxlny = {-1, 0.75, 2.5, 4.25, 6};
spxey = {-1, 0.75, 2.5, 4.25, 6};
operation[f_] := Block[{data},
data = Table[{x, f[x]}, {x, -1, 6, 0.1}];
Graphics[{Thick, Blue, Line[data],
Green, Table[Point[{spx[[i]], f[spx[[i]]]}], {i, 1, Length[spx]}],
Pink, Opacity[.7],
Table[Rectangle[{spx[[i]], f[spx[[i]]]}, {spx[[i + 1]],
f[spx[[i + 1]]]}], {i, 1, Length[spx] - 1}]
}, Axes -> True]]
Which[ f == Sin, spx := spxsin, f == Cos, spx := spxcos, f == E^x ,
spx := spxlny, f == Log, spx := spxey]
operation[Sin]
operation[Cos]
operation[E^x]
operation[Log]
Edit now tested: you can pass pure functions to your operation, so instead of: operation[E^x] try
operation[E^# &]
or for example if you wanted a base 2 log it would be
operation[Log[2,#]&]
A few other things to point out: Log fails simply because your x table range is negative.
Also, the Which statement you have doesn't do anything. Being outside your function, f is not defined so none of the conditionals are True. Moving Which inside the function, this works:
spxsin = {-1, -0.35, 0.3, 0.95, 1.6, 2.375, 3.15, 3.925, 4.7, 5.025,
5.35, 5.675, 6};
spxcos = {-1, -0.75, -0.5, -0.25, 0, 0.775, 1.55, 2.325, 3.1, 3.825,
4.55, 5.275, 6};
spxlny = {-1, 0.75, 2.5, 4.25, 6};
spxey = {-1, 0.75, 2.5, 4.25, 6};
operation[f_] :=
Block[{data}, data = Table[{x, f[x]}, {x, -1, 6, 0.1}];
Clear[spx];
Which[
TrueQ[f == Sin], spx := spxsin,
TrueQ[f == Cos], spx := spxcos ,
TrueQ[f == (E^# &)], spx := spxey ];
Graphics[{Thick, Blue, Line[data], Green,
Table[{PointSize[.1], Point[{spx[[i]], f[spx[[i]]]}]}, {i, 1, Length[spx]}],
Pink, Opacity[.7],
Table[Rectangle[{spx[[i]], f[spx[[i]]]}, {spx[[i + 1]],
f[spx[[i + 1]]]}], {i, 1, Length[spx] - 1}]}, Axes -> True,
AspectRatio -> 1/GoldenRatio]]
Note each which test is wrapped in TrueQ to ensure it is either True or False ( the test Sin==Cos is not false for all values and so does not return False )
operation[Sin]
operation[Cos]
operation[E^# &]
Now if you want Exp to also work you need to explicitly put that form in your Which statement. ( f==(E^#&) || f==Exp )
Euler's E needs to be entered as Esc ee Esc.
It looks to me at you entered is a standard E.
Note also that Exp is the exponential function in Mathematica.
here's a simple Manipulate Menu created with Mathematica:
SphereSection[d_, h_] :=
RegionPlot3D[(x^2 / d) + (y^2 / d) + (z^2 / h) <= 1,
{x, -5, 5}, {y, -5, 5}, {z, 0, 5},
AxesLabel->{"Lunghezza Km" , "Larghezza Km", "Profondità Km"},
PlotLabel->Style[Framed["Semisfera di riferimento"],25]]
Manipulate[
{diameter,highness,estimatedPorosity,waterSaturation,netOnGross,bg},
{diameter, 0, 200},{highness,0,200},{estimatedPorosity,0,100},
{waterSaturation, 0, 100}, {netOnGross, 0, 100}, {bg, 0, 1},
Button["GO", SphereSection[diameter, highness]]]
When I set the diameter and highness with the graphic menu, I expect that the function SphereSection is called after I've pressed the button but nothing happens...instead if I call directly
SphereSection[2,3]
then it draws the section of the sphere correctly...how can I link the two?
Does this do what you want?
Manipulate[SphereSection[diameter, highness],
{{diameter, 100}, 0, 200},
{{highness, 100}, 0, 200},
{estimatedPorosity, 0, 100},
{waterSaturation, 0, 100},
{netOnGross, 0, 100},
{bg, 0, 1}]
If you actually want the result held until you press a button that's a little harder.
This is quick and dirty approach. Note you need to "use" the dummy variable update in order for it to wrok as a TrackedSymbol
Manipulate[update; SphereSection[diameter, highness],
{{diameter, 100}, 0, 200},
{{highness, 100}, 0, 200},
{estimatedPorosity, 0, 100},
{waterSaturation, 0, 100},
{netOnGross, 0, 100},
{bg, 0, 1}, {update, {True, False}}, TrackedSymbols :> {update}]
I have an issue with the reconstrution of a affine transformation matrix.
The translation matrix reconstructions works just fine, but not the rotation.
Thank you guys!
(*Works just fine*)
Clear["Global`*"]
data = RandomReal[10, {100, 3}];
data0 = TranslationTransform[{1, -1, 1}]#data;
{dX0, dY0, dZ0} /.
Solve[data0 == TranslationTransform[{dX0, dY0, dZ0}]#data, {dX0, dY0,
dZ0}]
(*Yields {} ????*)
Clear["Global`*"]
data = RandomReal[10, {10, 3}];
data0 = RotationTransform[10 , {1, 0, 0}]#data;
Solve[data0 == RotationTransform[aZ0 Degree, {0, 0, 1}]#data, {aZ0}]
You have too many equations for only one var.
data = RandomReal[1, {10, 3}];
data0 = RotationMatrix[1/2, {1, 0, 0}].# & /# data;
Solve[Thread[data[[1]] == RotationMatrix[aZ0, {1, 0, 0}].data0[[1]]][[2]], {aZ0}]
(*
-> -0.5
*)
I have an ODE and I solve it with NDSolve, then I plot the solution on a simplex in 2D.
Valid XHTML http://ompldr.org/vY2c5ag/simplex.jpg
Then I need to transform (align or just plot) this simplex in 3D at coordinates (1,0,0),(0,1,0),(0,0,1), so it looks like this scheme:
Valid XHTML http://ompldr.org/vY2dhMg/simps.png
I use ParametricPlot to do my plot so far. Maybe all I need is ParametricPlot3D, but I don't know how to call it properly.
Here is my code so far:
Remove["Global`*"];
phi[x_, y_] = (1*x*y)/(beta*x + (1 - beta)*y);
betam = 0.5;
betaf = 0.5;
betam = s;
betaf = 0.1;
sigma = 0.25;
beta = 0.3;
i = 1;
Which[i == 1, {betam = 0.40, betaf = 0.60, betam = 0.1,
betaf = 0.1, sigma = 0.25 , tmax = 10} ];
eta[x2_, y2_, p2_] = (betam + betaf + sigma)*p2 - betam*x2 -
betaf*y2 - phi[x2, y2];
syshelp = {x2'[t] == (betam + betaf + sigma)*p2[t] - betam*x2[t] -
phi[x2[t], y2[t]] - eta[x2[t], y2[t], p2[t]]*x2[t],
y2'[t] == (betaf + betam + sigma)*p2[t] - betaf*y2[t] -
phi[x2[t], y2[t]] - eta[x2[t], y2[t], p2[t]]*y2[t],
p2'[t] == -(betam + betaf + sigma)*p2[t] + phi[x2[t], y2[t]] -
eta[x2[t], y2[t], p2[t]]*p2[t]};
initialcond = {x2[0] == a, y2[0] == b, p2[0] == 1 - a - b};
tmax = 50;
solhelp =
Table[
NDSolve[
Join[initialcond, syshelp], {x2, y2, p2} , {t, 0, tmax},
AccuracyGoal -> 10, PrecisionGoal -> 15],
{a, 0.01, 1, 0.15}, {b, 0.01, 1 - a, 0.15}];
functions =
Map[{y2[t] + p2[t]/2, p2[t]*Sqrt[3]/2} /. # &, Flatten[solhelp, 2]];
ParametricPlot[Evaluate[functions], {t, 0, tmax},
PlotRange -> {{0, 1}, {0, 1}}, AspectRatio -> Automatic]
Third day with Mathematica...
You could find a map from the triangle in the 2D plot to the one in 3D using FindGeometricTransformation and use that in ParametricPlot3D to plot your function, e.g.
corners2D = {{0, 0}, {1, 0}, {1/2, 1}};
corners3D = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}};
fun[pts1_, pts2_] := FindGeometricTransform[Append[pts2, Mean[pts2]],
PadRight[#, 3] & /# Append[pts1, Mean[pts1]],
"Transformation" -> "Affine"][[2]]
ParametricPlot3D[Evaluate[fun[corners2D, corners3D][{##, 0}] & ### functions],
{t, 0, tmax}, PlotRange -> {{0, 1}, {0, 1}, {0, 1}}]
Since your solution has the property that x2[t]+y2[t]+p2[t]==1 it should be enough to plot something like:
functions3D = Map[{x2[t], y2[t], p2[t]} /. # &, Flatten[solhelp, 2]];
ParametricPlot3D[Evaluate[functions3D], {t, 0, tmax},
PlotRange -> {{0, 1}, {0, 1}, {0, 1}}]