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Roja
Posts:
6
Registered:
4/16/12


Re: Intersection between two cones
Posted:
Feb 7, 2013 2:52 AM


"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 x1x)^2+(a y1y)^2+(a z1z)^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 x1x)^2+(a y1y)^2+(a z1z)^2". But intersection of two spheres is a circle. Could you please elaborate a little?



