Skip to main content

functions - Make mathematica create equations and put them into NSolve


I want to be able to define n shapes and find the intersection between them and a line. I'm treating the shapes as "one" shape, i.e. I do not want any internal intersections (hence the volume equations).


I can do this by typing out each shape and equation as follows:


line = InfiniteLine[{{0.5, 0.5, 0.5}, {0.5, 0.5, 1}}];



(*shapes - I can define these*) 
shape1 = Ball[];
shape2 = Cone[{{0, 0, 0}, {1, 1, 1}}, 1/2];


(*------------ I want from here  automated ---------*) 
(*surface equations *) 
surfaceequations1 = RegionMember[RegionBoundary[shape1], {x, y, z}];
surfaceequations2 = RegionMember[RegionBoundary[shape2], {x, y, z}];


volumeequation1 = RegionMember[shape1, {x, y, z}];
volumeequation2 = RegionMember[shape2, {x, y, z}];

intersection = 
 NSolve[{x, y, z} \[Element] 
    line && (surfaceequations1 || 
     surfaceequations2) && ! (volumeequation1 && volumeequation2), {x,
    y, z}];


(* ---------- automation can stop here ---------- *) 


points = Point[{x, y, z}] /. intersection

Graphics3D[{{Opacity[0.5], shape1}, {Opacity[0.7], shape2}, 
  line, {Red, PointSize[0.015], points}}]

enter image description here


But this method means I have to type out each equation for each shape. If I change the number of shapes, I have to go through code and delete some.



Edit:


The correct logic to find the intersection with the shape as whole (rather than the intersections with the individual n shapes) is actually:


 intersection = 
 NSolve[{x, y, z} \[Element] 
    line && (surfaceequations1 || surfaceequations2 || 
     surfaceequations3) && ! (volumeequation1 && 
      volumeequation2) && ! (volumeequation2 && 
      volumeequation3) && ! (volumeequation1 && volumeequation3), {x, 
   y, z}]


i.e. for the intersection with the line to not to be within the shape, it must not be within any two shapes. This means its properly treated as one shape.


So, now shape1 = Ball[]; shape2 = Cone[]; shape3 = Cuboid[]; works as expected.


The Problem


I want to be able to define n shapes, then mathematica defines the n surface equations, and the n volume equations and puts them into NSolve.


Also; apologies for the bad question title - I'm not sure what to call this problem.



Answer



ClearAll[constraints, intersections]
constraints[shapes__]:= And[##& @@ (Not /@ 
  Through[(RegionMember[RegionIntersection @ ##] & @@@ Subsets[{shapes}, {2}])@#]), 
 RegionMember[RegionUnion @@ (RegionBoundary /@ {shapes})]@#] &


intersections[l_, s__]:=NSolve[#\[Element] l && constraints[s][#], #]& @
 ({x, y, z}[[;; RegionEmbeddingDimension[l]]])

Examples:


line = InfiniteLine[{{0.5, 0.5, 0.5}, {0.5, 0.5, 1}}];
shape1 = Ball[]; 
shape2 = Cone[{{0, 0, 0}, {1, 1, 1}}, 1/2];
shape3 = Cuboid[];


intersections[line, shape1, shape2]


{{x -> 0.5 ,y -> 0.5, z -> -0.7071067811865475},
{x -> 0.5, y -> 0.5, z -> 0.7542812707978059}}



 Graphics3D @ {line, Opacity[.5, Green], shape1, Opacity[.5, Blue], shape2, 
    PointSize[.05], Opacity[1, Red], 
   Point[{x, y,z} /. intersections[line, shape1,shape2]]}


enter image description here


intersections[line, shape1, shape2, shape3]


 {{x -> 0.5, y -> 0.5, z -> 1.}, {x -> 0.5, y -> 0.5, z -> 1.}, {x -> 0.5, y -> 0.5, z -> -1.}}



Graphics3D @ {line, Opacity[.5, Green], shape1, Opacity[.5, Blue], shape2, 
   Opacity[.5, Yellow], shape3, PointSize[.05], Opacity[1, Red], 
   Point[{x, y,z} /. intersections[line, shape1,shape2, shape3]]}


enter image description here


Replace shape3 with Cuboid[{0, 0, -3/2}] to get


enter image description here


2D examples


 DeleteDuplicates @ intersections[InfiniteLine[{-1, -1}, {1, 1}], 
    Disk[{0, 0}, .5], Triangle[]]


  {{x -> -0.353553, y -> -0.353553}, {x -> 0.5, y -> 0.5}}




Graphics @ {#, Opacity[.5, Green], #2, Opacity[.5, Blue], #3,
   PointSize[.05], Opacity[1, Red], Point[{x, y} /. intersections[##]]}& @@ 
 {InfiniteLine[{-1, -1}, {1, 1}],Disk[{0, 0}, .5], Triangle[]}

enter image description here


 DeleteDuplicates @ intersections[InfiniteLine[{-1, -1}, {1, 1}], 
    Disk[{0, 0}, .5], Triangle[], Disk[{1, 1}, .5]]


  {{x -> 1.35355, y -> 1.35355}, {x -> 0.646447, y -> 0.646447}, {x -> -0.353553, y -> -0.353553}, {x -> 0.5, y -> 0.5}}




Graphics@{#, Opacity[.5, Green], #2, Opacity[.5, Blue], #3, 
    Opacity[.5, Magenta], #4, PointSize[.05], Opacity[1, Red], 
    Point[{x, y} /. intersections[##]]} & @@ {InfiniteLine[{-1, -1}, {1, 1}], 
  Disk[{0, 0}, .5], Triangle[], Disk[{1,1}, .5]}

enter image description here


DeleteDuplicates @ intersections[InfiniteLine[{-1, -1}, {1, 1}], 
   Disk[{0, 0}, .5], Triangle[], Disk[{1, 1}, .75]]



 {{x -> 1.53033, y -> 1.53033}, {x -> -0.353553, y -> -0.353553}}



Graphics@{#, Opacity[.5, Green], #2, Opacity[.5, Blue], #3, 
    Opacity[.5, Magenta], #4,PointSize[.05], Opacity[1, Red], 
    Point[{x, y} /. intersections[##]]} & @@ {InfiniteLine[{-1, -1}, {1, 1}], 
  Disk[{0, 0}, .5], Triangle[], Disk[{1,1}, .75]}

enter image description here


Comments

Post a Comment

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}]
Post a Comment
Read more

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}]
Post a Comment
Read more

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....
Post a Comment
Read more