Skip to main content

plotting - How to ContourPlot a function of the coordinates on the Earth's surface on a map projection


There are two similar questions (this and this), but one is too specific and the other is apparently not clear and has no answers. I'll try to make the question more complete.


Suppose you have a vector field depending on the coordinates x,y,z of a point on the Earth's surface [Edit: I found a mistake on this expression but to keep discussions consistent I'll leave as is.]


field = With[{rl$ = 
Transpose[{{1.45031*10^7, -1.46446*10^11}, {-3.86086*10^8,
2.25113*10^8}, {1.00319*10^8, 3.47297*10^10}} - 1. {##1}]},
Total[({4.90254*10^12, 1.32705*10^20} rl$)/rl$^3] -
1. {-0.00569904, -0.0000209977, 0.00135953}] &


With some trick I can represent the Norm of this vector field on a map with the equirectangular projection. If I well understood the topic:


[Edit: I found another supposed mistake in the original code following. From what I now understand on a map usaully we represent the geodetic latitude. But the well known parametrization of an oblate spheroid (the one I originally used) refers to the geocentric latitude. The difference between the two expression is not more than about 0.2° for GRS80, but to be rigorous we need to take into account this difference. This is the reason for the function toGeocentric I added. Feel free to comment on this to confirm or to contradict my understanding of this topic.]


{a, b} = GeodesyData["GRS80", #] & /@ {"SemimajorAxis", 
"SemiminorAxis"} // UnitConvert // QuantityMagnitude

toGeocentric = λ \[Function]
Evaluate[GeodesyData[
"GRS80", {"GeocentricLatitude", λ}] /. λ -> λ °]


gg = GeoGraphics[GeoRange -> "World",
GeoProjection -> "Equirectangular",
GeoGridLines -> Quantity[15, "AngularDegrees"], ImageSize -> 800]

cp = ContourPlot[
Norm[field[a Cos[toGeocentric @θ °] Cos[φ °],
a Cos[toGeocentric @θ °] Sin[φ °],
b Sin[toGeocentric @θ °]]], {φ, -180,
180}, {θ, -90, 90},
ColorFunction -> (Opacity[0.6, ColorData["TemperatureMap"][#]] &)]


Show[Graphics @@ gg, cp]

Mathematica graphics


It is the "right" way to do that (I'm a bit concerned by Graphics@@gg for example, I read that on another question)?


It is possible to use many projections (ideally all map projections supported by Mathematica) in a consistent and general way without having to know and to input all the necessary projection transformations but taking advantage of the cartography framework in Mathematica?


I'm interested for example to "LambertAzimuthal", "Bonne", "Robinson"...




Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.