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 > >