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Re: Intersection of surfaces

  • To: mathgroup at smc.vnet.net
  • Subject: [mg88536] Re: Intersection of surfaces
  • From: dh <dh at metrohm.ch>
  • Date: Thu, 8 May 2008 04:10:55 -0400 (EDT)
  • References: <fvs2jg$en2$1@smc.vnet.net>


Hi Narasimham,

here is one way of doing it:

we first set the components of TUBE == BOWL, this gives 3 equations. We 

eliminate one sets of parameters (e.g. U,V). This leaves one equation 

with 2 parameters. We solve for one of these (e.g. p). This gives p as a 

function of q. We replace p in e.g. BOWL by this function. This gives a 

parametric representation of the space curve:

res1=Eliminate[{TUBE==BOWL},{U,V}];

res2=Solve[res1,p];

intersection=ParametricPlot3D[BOWL/.res2[[1]],{q,0,2Pi},PlotStyle->{Thickness[0.02]}]

To check, we may show everything together:

Show[bowl,tube,intersection]

hope this helps, Daniel





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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