Skip to main content

plotting - Seeing coordinates in Graphics3D


I'm writing some structures built up from different strings that specify the coordinates of the lines. The code used to generate these is built up from a ton of user defined functions from my library, so posting them here doesn't work very well. However, I can provide an example here that'll suffice.


What my question is about is how to show the coordinates in the structures that I draw, when looking at them with Graphics3D. When plotting functions, there's always this right mouse button option that has get coordinates, which will tell me exactly where I am on the plot. For Graphics3D this doesn't appear to be there. The underlying grid however definitely exists, as the lines are drawn at exact coordinates.


So lets make an example. An example string based structure is


x = {{Line[{{-273.5, 180., 0.}, {-273.5, 250., 0.}, {-237.5, 250., 
0.}, {-237.5, 180., 0.}, {-242.5, 180., 0.}, {-242.5, 245.,

0.}, {-268.5, 245., 0.}, {-268.5, 180., 0.}, {-273.5, 180.,
0.}}], Line[{{-242.5, 250., 0.}, {-242.5, 180., 0.}, {-206.5,
180., 0.}, {-206.5, 250., 0.}, {-211.5, 250., 0.}, {-211.5,
185., 0.}, {-237.5, 185., 0.}, {-237.5, 250., 0.}, {-242.5,
250., 0.}}],
Line[{{-211.5, 180., 0.}, {-211.5, 250., 0.}, {-175.5, 250.,
0.}, {-175.5, 180., 0.}, {-180.5, 180., 0.}, {-180.5, 245.,
0.}, {-206.5, 245., 0.}, {-206.5, 180., 0.}, {-211.5, 180.,
0.}}], Line[{{-180.5, 250., 0.}, {-180.5, 180., 0.}, {-144.5,
180., 0.}, {-144.5, 250., 0.}, {-149.5, 250., 0.}, {-149.5,

185., 0.}, {-175.5, 185., 0.}, {-175.5, 250., 0.}, {-180.5,
250., 0.}}],
Line[{{-149.5, 180., 0.}, {-149.5, 250., 0.}, {-113.5, 250.,
0.}, {-113.5, 180., 0.}, {-118.5, 180., 0.}, {-118.5, 245.,
0.}, {-144.5, 245., 0.}, {-144.5, 180., 0.}, {-149.5, 180.,
0.}}],
Line[{{-118.5, 250., 0.}, {-118.5, 180., 0.}, {-82.5, 180.,
0.}, {-82.5, 250., 0.}, {-87.5, 250., 0.}, {-87.5, 185.,
0.}, {-113.5, 185., 0.}, {-113.5, 250., 0.}, {-118.5, 250.,
0.}}], Line[{{-87.5, 180., 0.}, {-87.5, 250., 0.}, {-51.5, 250.,

0.}, {-51.5, 180., 0.}, {-56.5, 180., 0.}, {-56.5, 245.,
0.}, {-82.5, 245., 0.}, {-82.5, 180., 0.}, {-87.5, 180., 0.}}]},
222.}

I plot this using my plotting function


CoolView[x_] := 
Graphics3D[x /. {Line -> Polygon}, ViewPoint -> {0, 0, 1},
ImageSize -> Large]

And you can then see the structure with



CoolView[x[[1]]]

It looks like this enter image description here


While I can find the exact distance from my generated code if I try a little bit, preferably I'd like to find a way to for example read off the distance between the two fingers plotted above. That's my question, how I can do this. It should be precicely defined as the lines are drawn onto specific coordinates.



Answer



I am not sure how well the following is answering your question.


Here is the general idea:




  1. Write a function that finds distances from an arbitrary point to each of the lines defined by the polygons sides (line segments).





  2. Use some sort of dynamic manipulation to plot the most interesting point-segment distances.




I assume it is important to stay in 3D, but I was able to make a dynamic interface working only for 2D.


Code 3D


Here is a function for calculating distances from a point to polygon lines:


Clear[DistanceToAllPolygonLines]
DistanceToAllPolygonLines[rpoint_, polys : {_Polygon ..}] :=

Block[{segments, distances},
segments = Join @@ Map[Partition[#[[1]], 2, 1] &, polys];
distances =
Map[Norm[Cross[rpoint - #[[1]], rpoint - #[[2]]]]/
Norm[#[[2]] - #[[1]]] &, segments];
Transpose[{segments, distances}]
];

See this article for the point-line distance formula.


Here is how we use the function above (in 3D):



rpoint = Flatten[RandomReal[#, 1] & /@ pointsRange]
res = DistanceToAllPolygonLines[rpoint, Cases[gr[[1]], _Polygon, \[Infinity]]];
Graphics3D[{gr[[1]], Map[{Blue, Tooltip[Line[#[[1]]], #[[2]]]} &, res],
Arrowheads[Medium],
Map[{Black, Arrow[{rpoint, Mean[#[[1]]]}], Text[#[[2]], Mean[Append[#[[1]], rpoint]]]} &,
Take[SortBy[res, Last], UpTo[6]]],
Red, PointSize[0.01], Point[rpoint]
}, ViewPoint -> {0, 0, 1.8}, ImageSize -> 900]

enter image description here



Note the tooltip -- it is given to any line segment in blue.


Interactive interface 2D


Here is an interactive interface in 2D:


DynamicModule[{p},
polys = gr[[1]] /. (x : {_?NumberQ, _?NumberQ, _?NumberQ}) :> Most[x];
Column[{
Slider2D[Dynamic[p], Transpose[Most@pointsRange]],
Row[{"point:", Dynamic[p]}],
Dynamic[
(res =

DistanceToAllPolygonLines[Append[p, 0],
Cases[gr[[1]], _Polygon, \[Infinity]]];
res = res /. (x : {_?NumberQ, _?NumberQ, _?NumberQ}) :> Most[x];
Graphics[{GrayLevel[0.7], polys,
Map[{Blue, Tooltip[Line[#[[1]]], #[[2]]]} &, res],
Arrowheads[Medium],
Map[{Black, Arrow[{p, Mean[#[[1]]]}],
Text[#[[2]], Mean[Append[#[[1]], p]]]} &,
Take[SortBy[res, Last], UpTo[6]]],
Red, PointSize[0.01], Point[p]

}, ImageSize -> 900, GridLines -> Automatic, Frame -> True])]
}]]

enter image description here


Again notice the tooltip -- it is given to any segment in blue.


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....