Rick Gutleber wrote:
>If the points are _tangential_ to the sphere, wouldn't there only be one
>solution?

Rick,
I don't think one can say about points
that they can be tangential to anything.

You probably mean that they are on the
surface of the sphere.

If you have four points in 3D space that
are not coplanar (*), then yes; there
are only one specific sphere that have
them all on its surface. (And the macro
I mentioned in my other post finds that
very sphere.)

This sphere will, of coarse, enclose a
tetrahedron that has these four points
as its vertices.

But as Peter and I pointed out, this is
not always the most optimal sphere to
choose for bounding of such a tetrahedron.

It will in some cases be possible to
find spheres with smaller radii, that
encloses the tetrahedron.

And in these cases only 2 or 3 of the
4 points (vertices) will be on the surface
of the sphere. The other 2 or 1 will be
inside it.

The problem is now to find the bounding
sphere with the smallest radius.


(*) More precisely I mean:
vdot(p1 - p0, vcross(p2 - p0, p3 - p0)) != 0


Tor Olav


>"Peter Popov" <pet### [at] vipbg> wrote in message
>news:6ldgvu8m3haqih3b2cud5narcqn8p82pvi[at]4ax.com...
>> On Wed, 11 Dec 2002 11:45:10 -0500, "Rick Gutleber" <ric### [at] hiscom>
>> wrote:
>>
>> >I bet the Graphics Gems books have code that will allow you to determine
>a
>> >sphere tangential to 4 points in 3-space.  The code for those books can
>be
>> >found on-line.
>>
>> There sure is, but keep in mind that the circumscribed sphere is
>> usually not the smallest sphere containing four points.

Post a reply to this message