Skip to main content

equation solving - Finding Intersections Between Arbitrary Surface and A Line


I have a self-intersecting surface H defined as follows:


M = 0; phi = Pi/2;


b1 = {{-Sqrt[3]/2}, {3/2}};
b2 = {{-Sqrt[3]/2}, {-3/2}};
b3 = {{Sqrt[3]}, {0}};

hx[kx_, ky_] := 1 + Cos[{kx, ky}.b1] + Cos[{kx, ky}.b2]
hy[kx_, ky_] := Sin[{kx, ky}.b1] - Sin[{kx, ky}.b2]
hz[kx_, ky_] :=
M - 2 Sin[
phi] (Sin[{kx, ky}.b1] + Sin[{kx, ky}.b2] + Sin[{kx, ky}.b3])
H[kx_, ky_] =

Flatten[{hx[kx, ky], hy[kx, ky], hz[kx, ky]}, 1] // Simplify;

and I wish to find the number of times a line from the origin to some point {10, 10, 10} intersects with this surface.


What I have tried:




  1. Going off this reference - Graphics3D: Finding intersection of 3d objects and lines - I attempted to define a Region and line as follows:


    R = ParametricRegion[{H[u, v], 1.8 <= u <= 5.5 && -2.2 <= v <= 2.2}, {u,v}]; 
    TheLine = Line[{{0,0,0}, {10,10,10}}];


    and continue with the method given. However, Mathematica does not correctly give me the number of intersections, as it does not identify H as a properly defined Region. Additionally, my arbitrary surface is apparently "not a proper Graphics3D primitive or directive".




  2. To tackle this numerically, I tried populating a table of coordinates for the line and comparing it with values of my surface to see where the difference between coordinates for rows would be least (after ordering rows in ascending order). The line now goes as:


    p1 = {0, 0, 0};
    p2 = {10, 0, 10};
    v = p2 - p1;
    TheLine[t_] := p1 + t v

    However, this method is problematic because I miss several potential points. I read online that this could have worked if the third component of H is a function of the first two: H = {x,y,z(x,y)}. However, this is not the case.





  3. I tried solving the following system of 'equations':


    Solve[{x, y, z} \[Element] R && {x, y, z} \[Element] theLine, {x, y, z}, Reals]

    but this takes forever to run. I doubt this would work.


    Additionally, I don't think that I can solve for a system of equations like https://www.wolfram.com/mathematica/new-in-10/basic-and-formula-regions/surface-intersection.html in my case because a line cannot have a stand-alone equation of the form l(x,y,z) because a line is the intersection of two planes.




Perhaps I am missing something trivial, but none of the other potential solutions on this website seem to work. Therefore I would appreciate insight on any approach that would work! Thanks!




Answer



(I managed to borrow someone else's computer for a few minutes...)


Before anything else: using actual vectors instead of $2\times 1$ matrices will make constructing your parametric equations much easier. Thus:


M = 0; ϕ = π/2;
b1 = {-Sqrt[3]/2, 3/2}; b2 = {-Sqrt[3]/2, -3/2}; b3 = {Sqrt[3], 0};

Set up the parametric equations:


eqs = Thread[{x, y, z} == {1 + Cos[{kx, ky}.b1] + Cos[{kx, ky}.b2], 
Sin[{kx, ky}.b1] - Sin[{kx, ky}.b2],
M - 2 Sin[Ï•] (Sin[{kx, ky}.b1] + Sin[{kx, ky}.b2] +

Sin[{kx, ky}.b3])}];

Now, we can use GroebnerBasis[] to derive the implicit Cartesian equation:


impl = GroebnerBasis[Join[TrigExpand[eqs],
{Cos[(Sqrt[3] kx)/2]^2 + Sin[(Sqrt[3] kx)/2]^2 == 1,
Cos[ky/2]^2 + Sin[ky/2]^2 == 1}], {x, y, z},
{Cos[(Sqrt[3] kx)/2], Sin[(Sqrt[3] kx)/2],
Cos[ky/2], Sin[ky/2]}][[1]] // FullSimplify
(3 + (-4 + x) x + y^2)^2 (-3 + (-2 + x) x + y^2) + ((-1 + x)^2 + y^2) z^2


Deriving the intersection points is then as easy as


Solve[impl == 0, {x, y, z} ∈ InfiniteLine[{{0, 0, 0}, {10, 10, 10}}]] // FullSimplify
{{x -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 1],
y -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 1],
z -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 1]},
{x -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 2],
y -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 2],
z -> Root[-27 + 54 #1 - 17 #1^2 - 58 #1^3 + 78 #1^4 - 40 #1^5 + 8 #1^6 &, 2]}}

N[%]

{{x -> -0.814231, y -> -0.814231, z -> -0.814231},
{x -> 1.30996, y -> 1.30996, z -> 1.30996}}

Show the geometry:


Show[ParametricPlot3D[{1 + Cos[{kx, ky}.b1] + Cos[{kx, ky}.b2], 
Sin[{kx, ky}.b1] - Sin[{kx, ky}.b2],
M - 2 Sin[Ï•] (Sin[{kx, ky}.b1] + Sin[{kx, ky}.b2] +
Sin[{kx, ky}.b3])},
{kx, 1.8, 5.5}, {ky, -2.2, 2.2},
Mesh -> False, PlotStyle -> Opacity[1/2]],

Graphics3D[{Tube[Line[{{0, 0, 0}, {10, 10, 10}}]],
{Red, Sphere[{x, y, z}, 1/12] /. %}}]]

surface and intersection points with a line


Comments

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...

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...