Skip to main content

calculus and analysis - Finding surface normal for 3D region at a specific point


I would like to find the surface normal for a point on a 3D filled shape in Mathematica.


I know how to calculate the normal of a parametric surface using the cross product but this method will not work for a shape like Cone[] or Ball[].




  1. Is there some sort of RegionNormal option? There is an option to find VertexNormals here, but this is something to with shading and seems unhelpful.

  2. Is there a method I can use to convert the region into a parametric expression and use the normal cross product method?


The plan is to take an arbitrary line and find the angle of intersection between the line and the surface of the shape.



Answer



(* put inequality into u ≤ 0 form, return just u *)

standardize[a_ <= b_] := a - b;
standardize[a_ >= b_] := b - a;
regnormal[reg_, {x_, y_, z_}] := Module[{impl},

   impl = LogicalExpand@ Simplify[RegionMember[reg, {x, y, z}], {x, y, z} ∈ Reals];
   If[Head@impl === Or,
    impl = List @@ impl,
    impl = List@impl];
   impl = Replace[impl, {Verbatim[And][a___] :> {a}, e_ :> {e}}, 1];
   Piecewise[
    Flatten[
     Function[{component},
       Table[{
         D[standardize[component[[i]]], {{x, y, z}}],

         Simplify[
          (And @@ Drop[component, {i}] /. {LessEqual -> Less, GreaterEqual -> Greater}) &&
           (component[[i]] /. {LessEqual -> Equal, GreaterEqual -> Equal}),
          TransformationFunctions -> {Automatic, 
            Reduce[#, {}, Reals] &}]
         }, {i, Length@component}]
       ] /@ impl,
     1],
    Indeterminate]
   ];


Examples:


regnormal[Cone[{{0, 0, 0}, {1, 1, 1}}, 1/2], {x, y, z}]

Mathematica graphics


regnormal[Ball[{1, 2, 3}, 4], {x, y, z}]

Mathematica graphics


regnormal[RegionUnion[Ball[], Cone[{{0, 0, 0}, {1, 1, 1}}, 1/2]], {x, y, z}]


Mathematica graphics


regnormal[Cylinder[{{1, 1, 1}, {2, 3, 1}}], {x, y, z}]

Mathematica graphics


It assumes that the RegionMember expression can be computed (which is not always the case) and that it will be a union (via Or) of intersections (via And). It also assumes that the RegionMember expression includes the boundary. Thus, it is probably not very robust, but it handles the OP's examples.


Also, if this is used in numerical applications, which seems to be the case for the OP, one should worry about the exact conditions in the Piecewise expressions returned. It's unlikely the numerical calculations will be accurate enough to satisfy Equal. So either change the conditions or possible change Internal`$EqualTolerance:


Block[{Internal`$EqualTolerance = Log10[2.^28]}, (* ~single-precision FP equality *)
 
 ]

Comments

Post a Comment

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}]
Post a Comment
Read more

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 - 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