functions - How can you use and convert Gauss-Krüger-Coordinates in Mathematica?

I am given geographic data in the form of Gauss-Krüger-Coordinates and would like to calculate distances with them and convert them to longitude/latitude coordinates in another system for plotting (e.g. WGS84).

Gauss-Krüger-Coordinates are essentially like UTM-coordinates based upon a transversal Mercator projection where positions are indicated by a Right-value (East-value in UTM; y-coordinate in the geodetic coordinate system) and a High-value (North-value in UTM; x-coordinate int the geodetic coordinate system)

Far from being an expert in geodesy the way I understand it Mathematica's geodetic functions allow entering coordinates as GeoPosition, as GeoPositionENU or as GeoGridPosition. GeoGridPosition essentially will reference to a position on a projection so probably should be the way the Gauss-Krüger-Coordinates are entered, but I am not so sure. Mathematica does not seem to know Gauss-Krüger-Projection (not listed among GeoProjectionData[]) and I do not know which parameters to use for Gauss-Krüger and how best to do this. Thus:

How can I enter the Gauss-Krüger-Coordinates into Mathematica so they can be used for distance-calculations and conversions using Mathematica's geodetic functions?

Answer

With a little help from other sources and the links given as a comment to my question I came up with a solution. Essentially Gauss-Krüger is a variant of the Transversal Mercator projection so this projection can be used and made to fit the Gauss-Krüger specialties:

gaussKruegerPosition[{right_Integer,high_Integer},centralMeridian_Integer,

falseEasting_Integer]:=

  GeoGridPosition[
    {right-falseEasting-centralMeridian/3*10^6,high,0},
    {"TransverseMercator","Centering"-> {0,centralMeridian},"ReferenceModel"->"Bessel1841"}
  ];

This short function will convert Gauss-Krüger-Positions into a valid Mathematica position which can then be converted using GeoPosition or referenced for calculation of distances using GeoDistance[pos1, pos2].

Here in Germany the central Meridians will be spaced 3° apart and one can recognize this in the first digit of the Right-value of the coordinates which has to be multiplied by 3 to find the central Meridian.

In order to avoid negative Right-values a False Easting is usually given; in Germany it is 500 000.

Comments

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.

Blog

Search This Blog