Skip to main content

functions - FunctionInterpolation Errors / Question re Evaluation Order and Options


I have using Mathematica functions that takes a Cartesian coordinate relative to the Earth (xyz) and converts it to a latitude, longitude, and altitude (lla). And here it is:


xyz2lla = First@GeoPosition@GeoPositionXYZ[#, "WGS84"] &

I'm using it to convert a satellite viewing line (line of sight or los) from xyz to lla. For instance, if I have a satellite at the following observation point (obs, meters) looking in the direction of look:


obs = {2.560453600382259, 5.245110323032143, -3.819772142191310} 1*^6;
look = {-0.233218833096895, -0.814561997858160, -0.531128729720249};


It has the line of sight:


obs+d look

The transformation xyz2lla is smooth, so I was hoping to use function interpolation:


f = FunctionInterpolation[{xyz2lla[obs + # look]} &[d], 
{d, 2000000, 3700000}
]

And while this works, and I get the following lat, lon, alt functions from it:


enter image description here:



enter image description here


enter image description here


I also get the following errors that I'm wondering why I get:


GeoPositionXYZ::invcoord: "\!\(\"{2.560453600382259*^6 - 0.233218833096895*d, 5.245110323032143*^6 - 0.81456199785816*d, -3.81977214219131*^6 - 0.531128729720249*d}\"\) is not a valid coordinate specification."

Thread::tdlen: Objects of unequal length in {-212500.,-70833.3,70833.3,212500.}^{} cannot be combined. >>

Thread::tdlen: "Objects of unequal length in {12.2667 +3.14159\ I,11.1681 +3.14159\ I,11.1681,12.2667}\ {}\\n cannot be combined."

FunctionInterpolation::nreal: Near d = 2.2125`*^6, the function did not evaluate to a real number.


Does anyone have any insight into these errors?


Also, how can I adjust the quality of the interpolation / how many points are investigated?



Answer



I don't think InterpolatingFunction is intended to work on vector functions. The doc page doesn't say anything about it. Try for instance


f = FunctionInterpolation[{x, x}, {x, 0, 6}]


Mathematica graphics


And


Table[f[x], {x, 0, 6, 1}]


returns


(*
==> {0., 0.9999652778, 1.998576389, 3., 4.004131944, \
4.994409722, 6.}
*)

So, no two dimensional output.


It's probably better to come up with three separate interpolating functions for each of the three components of the output.




Having said that, it doesn't look like this is the end of the problems.



xyz2llaPhi = GeoPosition[GeoPositionXYZ[obs + # look, "WGS84"]][[1, 1]] &;
fPhi = FunctionInterpolation[xyz2llaPhi[d], {d, 2000000, 3700000}]

yields


Mathematica graphics


Whereas


xyz2llaPhi[200000]

yields a nice numerical result:


(*

==> -34.86629487
*)

It looks like the combination of FunctionInterpolation and GeoPosition isn't a healthy one. Somehow, the index d is held unevaluated.


A workaround would be to generate a table of values and then use ListInterpolation.


Comments

Popular posts from this blog

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 - 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 - 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....