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335

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 ...


186

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, ...


39

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}]


25

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", "---"}, ...


24

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 ...


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

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_, ...


22

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 = ...


21

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, ...


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: ...


19

maybe this will provide a little insight: first look at the evaluation points used by ContourPlot: f[x_?NumericQ, y_?NumericQ] := (Sow[{x, y}]; Sin[3.2 x]*Sin[1.3*y] - 2.1*Sin[1.3*x]*Sin[3.2*y]); {plot, dat} = Reap[ContourPlot[f[x, y] == 0, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, ContourStyle -> Red]]; Row[{ ...


18

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 | ...


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

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@ ...


17

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 ...


16

RegionFunction is the option you are looking for. ContourPlot[ Evaluate[Sum[Sin[RandomReal[5, 2].{x, y}], {5}]], {x, -3, 3}, {y, -3, 3}, BoundaryStyle -> {Thick, Black}, RegionFunction -> Function[{x, y, z}, x^2 + y^2 < 9] ]


16

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}}, ...


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

Edit Here is a version that avoids the use of Inset and instead uses Overlay. I think this version covers all of the OPs requests. I have not tried to functionalize the code at this point since there will likely be some tweaking of parameters based on the actual functions plotted. optsall = {Axes -> False, Frame -> True, ImageSize -> 600, ...


15

A crude attempt This is for Mathematica 10+ only. To construct each face, I use an intersection between a unit 3-ball centred at the origin and a pyramid whose base is at infinity and apex is at the origin. Each edge of the pyramid passes through each vertex of the spherical face. The pyramid is given by ConicHullRegion[{origin}, {vertices}]. The ...


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 ...


14

In Version 10 you can use the PlotTheme "OpenMarkersThick": data = Table[{x, x^k}, {k, 1, 4}, {x, 0, 1, 0.1}] ListLinePlot[data, PlotTheme -> {"OpenMarkersThick", "LargeLabels"}, PlotLegends -> {x, x^2, x^3, x^4}]


14

Here is another approach which is based on converting the plot to PDF format first. It makes the tick marks accessible as regular Graphics objects. Specifically, they (and the frame) show up as open JoinedCurve that can be identified by pattern matching. That leads to the following: p = Plot[{Sin[x], Cos[x]}, {x, 0, 3 Pi}, Frame -> True, FrameStyle ...


14

Bob Hanlon's answer works very well, but in some ways it is the hard way of doing things. If you have v9 or v10, then it is arguably easier to use the legend constructs within it. Similar to his answer, we get the image and element names: img = Import["ExampleData/1PPT.pdb", "Rendering" -> "BallAndStick", ImageSize -> 500]; elements = ...


14

Update: With the function top defined in the original post you can replicate all the cool things you see in rm-rf's answer in the linked Q/A. For example, with a slight modification of gr1, i.e., Graphics3D[hexTile[20, 20] /. Polygon[l_] :> {Directive[Orange, Opacity[0.8], Specularity[White, 30]], Polygon[l], Polygon[{Pi/5, 0} + {-1, 1} # & ...



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