Skip to main content

gui construction - How to set the slider width inside Manipulate


I use the code below to visualise the connection between a sine curve and the unit circle, but would like to make the slider wider, so it has at least almost the same with as the image. I have tried to replace ControlType->Slider[] with ControlType -> Slider[ImageSize->800], but this seems to have no effect at all och the resulting graphics. The question is thus how to set the width of the slider inside Manipulate?


Manipulate[
 With[{ar = 1/(2*Pi), o = v - Cos[v]}, 
  Show[Plot[Sin[x], {x, -6*Pi, 6*Pi}, 

    PlotRange -> {{-2*Pi, 2*Pi}, {-1, 1}}, PlotStyle -> {Black}, 
    AxesLabel -> {"v", Sin["v"]}, AspectRatio -> ar],
   ListLinePlot[{{o + Cos[v], 0}, {o + Cos[v], Sin[v]}}, 
    PlotStyle -> {Thick, Black}, AspectRatio -> ar],
   ListLinePlot[{{o + Cos[v], Sin[v]}, {o, 0}}, 
    PlotStyle -> {Dashed, Black}, AspectRatio -> ar],
   Graphics[{AbsolutePointSize[8], Point[{o, 0}]}, AspectRatio -> ar],
   Plot[{-Sqrt[1 - (x - o)^2], Sqrt[1 - (x - o)^2]}, {x, o - 1, 
     o + 1}, PlotRange -> {{-2*Pi, 2*Pi}, {-1, 1}}, 
    PlotStyle -> {{Thick, Orange}, {Thick, Orange}}, 

    AspectRatio -> ar], ImageSize -> 800]], {v, -3/2*Pi, 3/2*Pi}, 
 ControlType -> Slider[]]

The resulting graphics



Answer



You have to study the documentation carefully, but I agree that help-pages like the one of Manipulate are very densely packed with information. In the Details and Options section you find how to set options for controls:



{{u,...},...,opts}    control with particular options

The non-obvious part is, that you have to set the ControlType as well to make this work. Therefore, you can use



{v, -3/2*Pi, 3/2*Pi, ControlType -> Slider, ImageSize -> 800}

to achieve the wanted behavior. Another way is to replace Manipulate by a full DynamicModule which is a bit more code but gives you some more flexibility


DynamicModule[{v = -3/2 Pi, o},
 o = v - Cos[v];
 Panel@
  With[{ar = 1/(2*Pi)},
   Column[{
     Slider[Dynamic[v, (v = #; o = v - Cos[v]; &)], {-2 Pi, 2 Pi}, 
      ImageSize -> 800],

     Dynamic@
      Show[Plot[Sin[x], {x, -6*Pi, 6*Pi}, 
        PlotRange -> {{-2*Pi, 2*Pi}, {-1, 1}}, PlotStyle -> {Black}, 
        AxesLabel -> {"v", Sin["v"]}, AspectRatio -> ar], 
       ListLinePlot[{{o + Cos[v], 0}, {o + Cos[v], Sin[v]}}, 
        PlotStyle -> {Thick, Black}, AspectRatio -> ar],
       ListLinePlot[{{o + Cos[v], Sin[v]}, {o, 0}}, 
        PlotStyle -> {Dashed, Black}, AspectRatio -> ar], 
       Graphics[{AbsolutePointSize[8], Point[{o, 0}]}, 
        AspectRatio -> ar], 

       Plot[{-Sqrt[1 - (x - o)^2], Sqrt[1 - (x - o)^2]}, {x, o - 1, 
         o + 1}, PlotRange -> {{-2*Pi, 2*Pi}, {-1, 1}}, 
        PlotStyle -> {{Thick, Orange}, {Thick, Orange}}, 
        AspectRatio -> ar], ImageSize -> 800, Background -> White]
     }]
   ]]

enter image description here


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more