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319

I have to confess that I see this as a proper challenge, as I am usually quite creative in finding/combining functions to provide a desired behavior. So I will give it another try. which is generated using box[x_, x1_, x2_, a_, b_] := Tanh[a (x - x1)] + Tanh[-b (x - x2)]; ex[z_, z0_, s_] := Exp[-(z - z0)^2/s] and r[z_, x_] := (*body*).4 (1.0 - .4 ...


183

This might get me suspended from the site butt I cannot resist. The shape you are looking for can probably be approximated (depending how anal you want to be about the outcome) by two assymetric probability distributions. The obvious choices would be a Poasson or a log normal distribution. I will use the latter as it is continuous. Now the bummer is that ...


90

Parametric Buttocks Manipulator Manipulate[ ParametricPlot3D[{ (e u^p + (1 + (c - a u) (u - 1)) Cos[t]^2) Sin[t], (e u^p + (1 + (d - b u) (u - 1)) Cos[t]^2) Cos[t], 2 u}, {t, -0.2, Pi + 0.2}, {u, 0, 1.1}, Lighting -> "Neutral", Mesh -> None, PlotStyle -> Directive[Specularity[0], RGBColor[0.92, 0.85, 0.73]], Axes -> False], {{a, ...


52

Yes we can. The following DashedGraphics3D[ ] function is designed to convert ordinary Graphics3D object to the "line-drawing" style raster image. Clear[DashedGraphics3D] DashedGraphics3D::optx = "Invalid options for Graphics3D are omitted: `1`."; Off[OptionValue::nodef]; Options[DashedGraphics3D] = {ViewAngle -> 0.4, ViewPoint ...


37

Basic method There appears to be a mechanism for doing just that, though I have yet to map its capabilities. As a basic example for the time being: Themes`AddThemeRules["wizard", DefaultPlotStyle -> Thread@Directive[{Purple, Orange, Hue[0.6]}, Thick], LabelStyle -> 18, AxesStyle -> White, TicksStyle -> LightGray, Background -> ...


29

The triangle plot markers It is natural to expect that the triangle marker is placed in such a way that its center of mass (center of circumcircle) coincides with the point it marks. That's how it is implemented in all major scientific plotting software, for example Origin: Some time ago I published my own implementation of triangle-based plot markers. ...


28

Well, an unusual question to answer, what about something like this Plot3D[.7*(1 + Tanh[1 - (2*Y^2 + X^2 + X^4)]) - .3*Exp[-X^2/.0025]* Exp[-(Y - .1)^2/.15] - .2*(Exp[-(X - .7)^2/.02]*Exp[-(Y - .0)^2/.08] + Exp[-(X + .7)^2/.02]*Exp[-(Y - .0)^2/.08]), {X, -1, 1}, {Y, -1, 1}]


24

First define the Morse code (from rosettacode.org with corrections by @evanb) morsecode = (#1 -> Characters[#2]) & @@@ { {"a", ".-"}, {"b", "-..."}, {"c", "-.-."}, {"d", "-.."}, {"e", "."}, {"f", "..-."}, {"g", "--."}, {"h", "...."}, {"i", ".."}, {"j", ".---"}, {"k", "-.-"}, {"l", ".-.."}, {"m", "--"}, {"n", "-."}, {"o", "---"}, ...


22

i = Import@"http://i.stack.imgur.com/8I3B1.jpg"; f[{{tmin_, tmax_}, {rmin_, rmax_}}, ___] := Module[{l = Join[{{0, 0}}, Table[{Cos@t, Sin@t}, {t, tmin, tmax, (tmax-tmin)/100}]]}, {Texture[i], EdgeForm[], Polygon[l, VertexTextureCoordinates -> 1/2 Transpose[Transpose[l] + {1, 1}]]}] Framed@PieChart[{1, 2, 3, 4, 5, 6}, ChartElementFunction -> f] ...


22

This is how I would do this. Define frequencies and sampling rate precisely. Then use Periodogram because it takes SampleRate as an option and rescales frequency axis automatically. Read up Docs on Periodogram - see examples there. data = Table[{t, Sin[2 Pi 697 t] + Sin[2 Pi 1209 t]}, {t, 0., 0.1, 1/8000.}]; ListLinePlot[data, AspectRatio -> 1/4, ...


21

Here is another approach. It could be improved (I am sure) to properly determined the principle axes and translation (if I get time I will aim to update): lin = {#1^2, #1, #2, 2 #1 #2, #2^2} & @@@ points; lm = LinearModelFit[lin, {1, a, b, c, d}, {a, b, c, d}] Exploring model: lm["ParameterTable"] Determining quadric formula: pa = ...


20

Something like this perhaps. Thanks to halmir for pointing out that multiple highlighted elements might be useful. pieHighlight[chart_, n_] := chart /. DynamicModule[init_, body__] :> (ReplacePart[Hold[init], Thread @ {1, n, 2} -> True] /. Hold[x_] :> DynamicModule[x, body]) e.g. PieChart[{1, 2, 3}] ~pieHighlight~ 2 PieChart[{1, 2, ...


20

Taking the cube root on both sides fixes the problem, and then you don't need lots of PlotPoints any more. ContourPlot3D[(x^2 + 9/4 y^2 + z^2 - 1) == CubeRoot[x^2 z^3 + 9/80 y^2 z^3], {x, -6/5, 6/5}, {y, -6/5, 6/5}, {z, -6/5, 3/2}, Mesh -> None, Boxed -> False, AxesLabel -> {"x", "y", "z"}, Axes -> False, ContourStyle -> Directive[Red, ...


20

Your comment raises interesting question: I got everything simply faded and flat, while in the example - its all bright and sharp I think the problem is related to the fact that Mathematica's graphical functions always work in linear colorspace. Starting from Pickett's solution with better resampling method, in version 10.0.1 we obtain: ...


20

Some function definitions first. AkimaInterpolation[] stolen from here (Thanks JM, wherever you are!): AkimaInterpolation[data_] := Module[{dy}, dy = #2/#1 & @@@ Differences[data]; Interpolation[Transpose[{List /@ data[[All, 1]], data[[All, -1]], With[{wp = Abs[#4 - #3], wm = Abs[#2 - #1]}, If[wp + wm == 0, (#2 + #3)/2, (wp #2 + wm ...


18

body[t_] = Integrate[#[u^2], {u, 0, t}] & /@ {Cos, Sin} ParametricPlot3D[body[t]~Join~{t}, {t, -2 Pi, 2 Pi}, BoxRatios -> 1, SphericalRegion -> True]


18

In an presentation by Markus van Almsick, he gives an solution to visualize atomic orbitals using Image3D. Radius wave function (hydrogen): R[n_Integer?Positive, l_Integer?NonNegative, r_] := Block[{ρ = (2 r)/n}, Sqrt[(2/n)^3 (n - l - 1)!/(2 n (n + l)!)] E^(-ρ/2) ρ^l LaguerreL[n - l - 1, 2 l + 1, ρ]] /; l < n full wave function: ψ[n_, l_, m_, ...


18

The colors alone are indexed color scheme #97: ColorData[97, "ColorList"] Update: further digging in reveals these PlotTheme indexed color relationships: {"Default" -> 97, "Earth" -> 98, "Garnet" -> 99, "Opal" -> 100, "Sapphire" -> 101, "Steel" -> 102, "Sunrise" -> 103, "Textbook" -> 104, ...


18

Update ticks[x1_, x2_] := {#/10 + π/2, #} & /@ FindDivisions[{10 (x1 - π), 10 (x2 - π)}, 20] funcs = Table[3 + BesselJ[i, 10 (x -π/2)], {i, 0, 3}]; PolarPlot[funcs // Evaluate, {x, -π/2, 3π/2}, PolarAxes -> Automatic, PolarTicks -> {ticks[0, 2 π][[2 ;; -2]], Automatic} ] (*thanks @kguler 's and @rm-rf 's advice*) Manipulate ...


17

This is based on Rahul's ideas, but a different implementation: contourRegionPlot3D[region_, {x_, x0_, x1_}, {y_, y0_, y1_}, {z_, z0_, z1_}, opts : OptionsPattern[]] := Module[{reg, preds}, reg = LogicalExpand[region && x0 <= x <= x1 && y0 <= y <= y1 && z0 <= z <= z1]; preds = Union@Cases[reg, _Greater | ...


17

data = Table[{RandomReal[{-10, 10}], RandomReal[{-10, 10}]}, {i, 1, 300}]; L0 = ListPlot[data, Frame -> True, Axes -> False, AspectRatio -> 1, ImageSize -> 400, BaseStyle -> PointSize[.02]]; Using @rm-rf's function inPolyQ inPolyQ[poly_, pt_] := Graphics`Mesh`PointWindingNumber[poly, pt] =!= 0 from this Q/A: Deploy@ ...


16

I wrote the ColorBar package exactly for this purpose and it makes such modifications easy. The README.m should give you all the instructions you need, but I'll summarize it here. After installing the package (copy ColorBar.m to FileNameJoin[{$UserBaseDirectory, "Applications"}]), do the following: ColorBar["TemperatureMap"] Now you can click on the ...


15

The default colours for charts are defined by Charting`CommonDump`rogerStyles. For up to five different colours the values are hard-coded: Charting`CommonDump`rogerStyles[5] (* {RGBColor[0.798413, 0.82472, 0.968322], RGBColor[0.733333, 1., 0.833722], RGBColor[1, 0.986999, 0.742123], RGBColor[1, 0.860624, 0.662562], RGBColor[1, 0.696086, 0.721935]} *) ...


15

This is a solution based on interval operations. Usage and examples First, let's look at how to use the function. The code is at the end. Let's generate some sample data and plot it: data1 = RandomVariate[ExponentialDistribution[1], 200]; data2 = RandomVariate[NormalDistribution[2, 1], 200]; beeswarmPlot[data1] Now let's plot two together: ...


15

The order of arguments in Show makes a difference in two ways: The Graphics expression produced by Show will inherit options from the first argument The first argument will appear in the bottom layer, the last one in the top layer. The most common problem (1) causes is that the plot range is inherited from the first argument, causing parts of the second ...


15

Seeing Silvia's phenomenal answer I've been inspired to take a crack at this. My method requires the use of ColorFunction so it only works for plots rather than general Graphics3D geometry. However, it does find silhouette edges in the interior of the image, as well as those hidden behind other surfaces (such as the missing side walls of the internal ...


15

I think that this question is too localized as it concerns the physics of a specific scientific instrument. Nonetheless, it is upvoted, so here I provide an answer for the benefit of the voters. I would still be happy to discuss this in the chat. The mathematics of the quadrupole mass filter is more complicated than you might think. Basically, your ...


15

In analytic geometry, the ellipse is defined as the set of points (X,Y) of the Cartesian plane that, in non-degenerate cases, satisfy the implicit equation with and where Lets fit points with second-order curve (which include ellipse). elipse = a11*x^2 + a22*y^2 + 2*a12*x*y + 2*a13*x + 2*a23*y + a33; coeff = {a11, a22, a12, a13, a23, a33}; ...


15

To long for a comment, but here's one approach, using information readily available in the docs and on this site: First, make a map that wraps a globe changing the Geoprojection to something a bit more useful. img = With[{Δ = 30}, Row[Table[ GeoGraphics[GeoBackground -> GeoStyling["ReliefMap"], GeoRange -> {{-90, 90}, {λ, λ + Δ}}, ...


14

Here is a standard direct way to get the principal exes and other transformation data. Find the mean of the points, subtract it to center them, and take the singular value decomposition. The third and second components thereof give the rotation and scaling data necessary to form a circle on which the first component, viewed as a point set, roughly lies. The ...



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