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 .