Let
be the maximum curvature of the
spine curve, and
,
be the maximum possible
upper limit radius of the pipe surface such that it does not globally
self-intersect between end circle to end circle, body to body and end
circle to body of the pipe surface, respectively.
Then we have [256]:
Theorem 11.6.1. Let
be the pipe surface with spine curve
and radius
. Then,
is nonsingular if and
only if
=
,
,
,
.
Proof (if): It is apparent from the discussion in Sects.
11.6.2 and 11.6.3 that
if
then the pipe surface
is nonsingular.
(only if): Suppose now that
is nonsingular. It is enough to show
that for all
,
is singular. But this is
obvious since if
is as indicated, the pipe surface will either
have a singularity due to local self-intersection or one
due to global self-intersection, or both.
Remark 11.6.1 When the spine curve is planar, Theorem 11.6.1 can be used to find the maximum offset distance such that the offset of the planar spine curve will not self-intersect.
Example 11.6.5.
(2-D spine curve)
The quartic spine curve with control points (-0.3, 0.8, 0), (0.6, 0.3,
0), (0,0,0), (-0.3, 0.2, 0) and (-0.15, 0.6, 0) and weights
1, 1, 2, 3, 1 respectively, has minimum distance 0.0777421 between two
points
and
. By using Newton's method we obtain the
touching radius as
. This distance is the maximum offset
distance such that the offset of the planar spine curve will not
self-interset. Figure 11.38 shows the offset curves when
.