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MathGroup Archive 2004

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Re: Intersection of two surfaces in 3D

  • To: mathgroup at smc.vnet.net
  • Subject: [mg52863] Re: [mg52822] Intersection of two surfaces in 3D
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Wed, 15 Dec 2004 04:26:25 -0500 (EST)
  • References: <200412141059.FAA24571@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 14 Dec 2004, at 19:59, Narasimham wrote:

> *This message was transferred with a trial version of CommuniGate(tm) 
> Pro*
> There are threads currently on sci.math  on this topic. How do we find
> space intersection curve of two  parameterized surfaces? One needs to
> solve for two unknown functions f1(t1,t2)=0 and f2(s1,s2)=0 to print
> out/output coordinates of intersection. I do believe it is within the
> capability of Mathematica, at least when surfaces are algebraically
> generatable. An example/approach considered is:
>
> Clear[x,y,z,t1,t2,s1,s2];
> x1=4*t2* Cos[t1]; y1=4Sin[t1]; z1=3t2;
> x2=s2 Sin[s1];y2=s2 Cos[s1];z2=(s2^2/4);
> pp1=ParametricPlot3D[{x1,y1,z1},{t1,0,2 Pi},{t2,0,1}];
> pp2=ParametricPlot3D[{x2,y2,z2},{s1,0,2 Pi},{s2,0,4}];
> Show[pp1,pp2];
> S1={x-x1,y-y1,z-z1}; S2={x-x2,y-y2,z-z2};
> NSolve[Join[S1,S2],{x,y,z},{t1,t2,s1,s2}];
>
>

In fact in principle Mathematica can fully solve this problem without 
the need for numerical methods.

f = GroebnerBasis[{x - 4*t*Cos[s], y - 4*Sin[s], z - 3*t, Sin[s]^2 + 
Cos[s]^2 - 1}, {x, y, z}, {t, Sin[s], Cos[s]}]

{9*x^2 - 16*z^2 + y^2*z^2}

So the space of zeros of the first equation is a subset of the space of 
solutions of the Cartesian equation:

9*x^2 - 16*z^2 + y^2*z^2==0


  (They may actually be the same, I have not tried to check.)


g = GroebnerBasis[{x - t*Sin[s], y - t*Cos[s], z - t^2/4, Sin[s]^2 + 
Cos[s]^2 - 1}, {x, y, z}, {t, Sin[s], Cos[s]}]


{x^2 + y^2 - 4*z}

So again all points on the surface satisfy

x^2 + y^2 - 4*z==0


Let's now try to find the solutions of this



sols=Reduce[Join[f, g] == 0, {x, y}, Reals]


(0 <= z <= 9/4 && ((x == -Sqrt[(-16*z^2 + 4*z^3)/(-9 + z^2)] && (y == 
-Sqrt[-x^2 + 4*z] || y == Sqrt[-x^2 + 4*z])) ||
     (x == Sqrt[(-16*z^2 + 4*z^3)/(-9 + z^2)] && (y == -Sqrt[-x^2 + 4*z] 
|| y == Sqrt[-x^2 + 4*z])))) ||
   (z >= 4 && ((x == -Sqrt[(-16*z^2 + 4*z^3)/(-9 + z^2)] && (y == 
-Sqrt[-x^2 + 4*z] || y == Sqrt[-x^2 + 4*z])) ||
     (x == Sqrt[(-16*z^2 + 4*z^3)/(-9 + z^2)] && (y == -Sqrt[-x^2 + 4*z] 
|| y == Sqrt[-x^2 + 4*z]))))


We can think of these answers as representing several parametric curves 
in space. The interesection points of the original surfaces should be 
contained among them. Unfortunately it takes a bit of work to get it 
all into the right form. We want to express x and y in terms of z and 
using z as the parameter (using only the real values of z returned by 
Reduce) plot the curves. Note that there seem to be two pieces, for 
0<z<9/4 and for z>4. One can plot them as follows:


g1=ParametricPlot3D[Evaluate[{x, y, z} /. Solve[{x == -Sqrt[(-16*z^2 + 
4*z^3)/(-9 + z^2)], y == -Sqrt[-x^2 + 4*z]}, {x, y}]], {z, 0, 9/4}]

and


g2=ParametricPlot3D[Evaluate[{x, y, z} /. Solve[{x == -Sqrt[(-16*z^2 + 
4*z^3)/(-9 + z^2)], y == -Sqrt[-x^2 + 4*z]}, {x, y}]], {z, 4, 8}]

In general you expect them to be a superset of the intersection curve. 
We can try to check this graphically. Doing so is a beautiful 
application of Mathematica's interactive graphics.

We simply turn on RealTime graphics with

<< RealTime3D`

Now look at

Show[g1, pp1]

you can clearly see the curve g1 lying on the surface pp1 (defined by 
you).

Now do the same thing with pp2:

Show[g1, pp2]

again by rotating the graphic into a suitable position we can see 
clearly that the curve also lies on pp2.

Replacing g1 by g2 is slightly less convincing. We need to change the 
parameters in both surface plots because the curve is actually in a 
different region. But by changing  the plots of the surfaces to

pp1 = ParametricPlot3D[{x1, y1, z1}, {t1, 0, 2 Pi}, {t2, 0, 3}];
pp2 = ParametricPlot3D[{x2, y2, z2}, {s1, 0, 2 Pi}, {s2, 0, 10}];

we see that this curve also lies on both surfaces. So the problem seems 
to have been solved.


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