Re: Intersection of surfaces
- To: mathgroup at smc.vnet.net
- Subject: [mg88651] Re: Intersection of surfaces
- From: "ahallam at iastate.edu" <ahallam at iastate.edu>
- Date: Sun, 11 May 2008 15:15:16 -0400 (EDT)
- References: <fvs2jg$en2$1@smc.vnet.net> <fvudl0$eak$1@smc.vnet.net>
On May 8, 3:29 am, Jens-Peer Kuska <ku... at informatik.uni-leipzig.de>
wrote:
> 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
So saw this post and found if useful for something I was doing.
But as I looked at the suggested code, I was not sure why thread and
flatten were used in this particular case.
So I deleted them one at a time and still got the same answer.
What is the difference then between.
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}];
eqn = Eliminate[{TUBE == {x, y, z} // Thread,
BOWL == {x, y, z} // Thread} // Flatten, {p, q, U}]
eqn1 = Eliminate[{TUBE == {x, y, z}, BOWL == {x, y, z}} //
Flatten, {p, q, U}]
eqn2 = Eliminate[{TUBE == {x, y, z}, BOWL == {x, y, z}}, {p, q, U}]
sol = Solve[eqn, {x, y, z}]
sol1 = Solve[eqn1, {x, y, z}]
sol2 = Solve[eqn2, {x, y, z}]
ll = ParametricPlot3D[{x, y, z} /. sol[[1]], {V, 0, 2 Pi}];
l11 = ParametricPlot3D[{x, y, z} /. sol1[[1]], {V, 0, 2 Pi}];
l12 = ParametricPlot3D[{x, y, z} /. sol2[[1]], {V, 0, 2 Pi}];
Show[bowl, tube,
ll /. l_Line :> {AbsoluteThickness[4], RGBColor[1, 0, 0], l}]
Show[bowl, tube,
l11 /. l_Line :> {AbsoluteThickness[4], RGBColor[1, 0, 0], l}]
Show[bowl, tube,
l12 /. l_Line :> {AbsoluteThickness[4], RGBColor[1, 0, 0], l}]