Tags: curves, defined, dimensions, functions, intersection, matlab, parametric, programming, sox1, variable

Intersection of two 3D parametric curves

On Programmer » Matlab

2,007 words with 3 Comments; publish: Tue, 29 Apr 2008 14:40:00 GMT; (200234.38, « »)

Hi, I have two parametric curves defined in three dimensions, which are func

tions of a variable t, like so:

x1 = f1(t)

y1 = f2(t)

z1 = f3(t)

x2 = f4(t)

y2 = f5(t)

z2 = f6(t)

I am trying to find the intersection of these two curves, but I am having so

me difficulty with the mathematics. In two dimensions, I simply solve for t

as a function of x, and then plug that value of t into my y function to obta

in y as a function of x. Wi

th three dimensions, I cannot do this.

Any idea of how I should approach this problem? Thanks!

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  • 3 Comments
    • JFG wrote:

      > Hi, I have two parametric curves defined in three dimensions, which are fu

      nctions of a variable t, like so:

      > x1 = f1(t)

      > y1 = f2(t)

      > z1 = f3(t)

      > x2 = f4(t)

      > y2 = f5(t)

      > z2 = f6(t)

      > I am trying to find the intersection of these two curves, but I am having some dif

      ficulty with the mathematics. In two dimensions, I simply solve for t as a function

      of x, and then plug that value of t into my y function to obtain y as a function of

      x.

      With three dimensions, I cannot do this.

      > Any idea of how I should approach this problem? Thanks!

      %x1,x2..already defined as functions

      guess = whatever;

      dist=.matlab.todaysummary.com.(t) norm( x2(t)-x1(t), y2(t)-y1(t),z2(t)-z1(t));

      fminsearch(dist, guess)

      I'll let you at least figure out how to tell if they intersect -

      hope it's not homework.

      Cheers,

      Eric Carlson

      #1; Tue, 29 Apr 2008 14:41:00 GMT
    • I guess I should mention that I am interested in an analytical solution. Sor

      ry, I should have mentioned that in the original post.

      #2; Tue, 29 Apr 2008 14:42:00 GMT
    • JFG wrote:

      > I guess I should mention that I am interested in an analytical solution. Sorry, I

      should have mentioned that in the original post.

      norm(vx,vy,vz)=sqrt(vx^2+vy^2+vz^2)

      presuming x1, x2, y1,y2,z1,and z2 are all defined, you can take

      derivative of norm and...

      #3; Tue, 29 Apr 2008 14:43:00 GMT