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Re: Intersection of surfaces

  • To: mathgroup at smc.vnet.net
  • Subject: [mg88546] Re: Intersection of surfaces
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Thu, 8 May 2008 04:12:45 -0400 (EDT)
  • Organization: Uni Leipzig
  • References: <fvs2jg$en2$1@smc.vnet.net>
  • Reply-to: kuska at informatik.uni-leipzig.de

Hi,

TUBE = {.6 Cos[V], 2 U + 3, .6 Sin[V] + 2};
tube = ParametricPlot3D[TUBE, {U, -1.2, .2}, {V, 0, 2 Pi},
    PlotPoints -> {10, 25}];
BOWL = {p Cos[q], p^2/2, p Sin[q]};
bowl = ParametricPlot3D[BOWL, {p, 1, 2.75}, {q, 0, 2 Pi},
    PlotPoints -> {20, 35}];

and

eqn = Eliminate[{TUBE == {x, y, z} // Thread,
     BOWL == {x, y, z} // Thread} // Flatten, {p, q, U}];

and
sol = Solve[eqn, {x, y, z}];

gives

{{y -> 0.02 (100.+ 9. Cos[V]^2 + 60. Sin[V] + 9. Sin[V]^2),
   x -> 0.6 Cos[V], z -> 0.2 (10.+ 3. Sin[V])}}

and

ll = ParametricPlot3D[{x, y, z} /. sol[[1]], {V, 0, 2 Pi}];


Show[bowl, tube,
  ll /. l_Line :> {AbsoluteThickness[4], RGBColor[1, 0, 0], l}]

show that the result is right.

Regards
   Jens

Narasimham wrote:
> How to find the space curve formed by intersecting 3D patches in
> simple cases like:
> 
> TUBE = {.6  Cos[V], 2 U + 3, .6 Sin[V] + 2};
> tube = ParametricPlot3D[TUBE, {U, -1.2, .2}, {V, 0, 2 Pi}, PlotPoints -
>> {10, 25}]
> BOWL = {p Cos[q], p^2/2, p Sin[q]};
> bowl = ParametricPlot3D [ BOWL, {p, 1, 2.75}, {q, 0, 2 Pi}, PlotPoints
> -> {20, 35}]
> Show[bowl, tube]
> 
> or in slightly more complicated surface cases like:
> 
> terr = ParametricPlot3D[{Cos[u + 1] Cos[v + 2.1], 0.6 + u^2/3,Exp[-v/
> 4] }, {v, -3, 3}, {u, -3, 3}, PlotPoints -> {45, 30}]
> Show[terr, tube]
> 
> How to solve for x,y and z from {0.6 Cos[V] == p Cos[q], 3 + 2 U ==
> p^2/2, 2 + 0.6 Sin[V] == p Sin[q]} obtaining t as a function of (U,V,p
> and q) so as to be able to Show with
> 
> ParametricPlot3D[{x[t], y[t], z[t]},{t,tmin,tmax}]?
> 
> Regards
> Narasimham
> 
> 


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