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Re: Intersection between two cones
Posted:
Feb 7, 2013 4:36 AM
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"Doctor61" wrote in message <kevmfe$gku$1@newscl01ah.mathworks.com>...
> "Bill Whiten" <W.Whiten@uq.edu.au> wrote in message <keteb4$gmc$1@newscl01ah.mathworks.com>...
> > "Doctor61" wrote in message <ket0to$214$1@newscl01ah.mathworks.com>...
> > > I have two 3d circles (centre coordiantes and radii) with their normals passing through origin. If you consider the origin to be the vertex of a right cone having the circle as the base, how can I determine if there is an intersection between these cones?
> >
> > if centre is x1,y1,z1 and radius r set d^2=x1^2+y1^2+z1^2
> > Cone is intersection of spheres a^2 (d^2+r^2)=x^2+y^2+z^2 and
> > a^2 r^2 = (a x1-x)^2+(a y1-y)^2+(a z1-z)^2
> >
> > Eliminate a to get equation of cone:
> > x^2+y^2+z^2 = (r^2+d^2) (x1 x+y1 y+z1 z)^2/d^4
> > (check this)
> >
> > Set x^2+y^2+z^2=1 and solve for points on two cones and this sphere.
> > E.g. Solve linear equations (1= (r^2+d^2) (x1 x+y1 y+z1 z)^2/d^4 etc)
> > for x and y then substitute into x^2+y^2+z^2=1 to get quadratic for z.
> >
> > Regards
>
> Thanks Bill, but I do not get your method. You said "Cone is intersection of spheres a^2 (d^2+r^2)=x^2+y^2+z^2 and
> > a^2 r^2 = (a x1-x)^2+(a y1-y)^2+(a z1-z)^2". But intersection of two spheres is a circle. Could you please elaborate a little?
The free parameter a moves the circle thus forming the cone. Eliminating a then gives the equation of the cone.
You can generate a simpler case with circle origin [0,0,a] and diameter a*r, giving z=a and (a r)^2=x^2+y^2 as the circle, eliminate a to get formula of cone.
Regards
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