Skip to main content

graphics3d - Draw an arbitrary convex polyhedron without excess diagonals drawn


I want to draw a set of convex polyhedrons whose vertices are defined by spherical coordinates on the surface of a unit sphere.


Currently I followed the advice from here: https://mathematica.stackexchange.com/a/21842/651 But it draws diagonals on any tetragonal face. Each face is composed of triangles. Is it possible to accomplish the same but without tetragonal faces diagonaled?


The current code:


Needs["TetGenLink`"]
mapping[{1,0,0}]={0,0,1}
mapping[{1,Pi,0}]={0,0,-1}
mapping:=CoordinateTransformData[{"Spherical"->"Cartesian"},"Mapping"]

verteces[n_]:=Flatten[Table [{1,Pi k/n, i 2Pi/Binomial[n,k]},{k,0,n},{i,0,Binomial[n,k]-1}],1]
Convex[n_]:=TetGenConvexHull[mapping/@ verteces[n]]
pts[n_]:=First[Convex[n]]
surface[n_] :=Last[Convex[n]]
b:=5
Graphics3D[{Yellow,Opacity[.9],GraphicsComplex[pts[b],Polygon[surface[b]]], Black, Line[{{0,0,-1.1},{0,0,1.1}}]}]

Answer



<< ComputationalGeometry`
ComputationalGeometry`Methods`ConvexHull3D[mapping /@ verteces[5], 
                                           Axes -> None, Graphics`Mesh`FlatFaces -> False]


Mathematica graphics



Mapping over n (well, with a trick because it fails with more than three calculations in a row):


Mathematica graphics



Edit


merging with your code:


<< ComputationalGeometry`
mapping[{1, 0, 0}] = {0, 0, 1}

mapping[{1, Pi, 0}] = {0, 0, -1}
mapping := CoordinateTransformData[{"Spherical" -> "Cartesian"}, "Mapping"]
verteces[n_] := Flatten[Table[{1, Pi k/n, i 2 Pi/Binomial[n,k]}, {k,0,n}, {i, 0,Binomial[n,k]-1}], 1]

Graphics3D[{Yellow, Opacity[.9], 
           Sequence @@ ComputationalGeometry`Methods`ConvexHull3D[mapping /@ verteces[5], 
           Axes -> None, Graphics`Mesh`FlatFaces -> False],Black,Line[{{0, 0, -1.1}, {0, 0, 1.1}}]}]

Mathematica graphics


Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more