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6.3.3 Torsion and third order derivative vector
Since
and
lie on the
normal plane, the terms
in
(6.5) can be replaced by
. Thus
(6.35)
Now, if we project
onto the unit surface normal vector
at
and denote by
, we have
(6.36)
Solving the linear system for
and
, and
substituting them into (6.35) yields
(6.37)
Similar to the curvature vector case in Sect. 6.3.2, we
need to provide
and
to evaluate
.
For a parametric surface,
can be obtained by projecting
, which is the third order derivative of the intersection
curve as a curve on a parametric surface, i.e. (6.19),
onto the unit surface normal vector
, resulting in
(6.38)
where
(6.39)
and
and
in (6.38) are evaluated by
taking the dot product on the both sides of
(6.18) with
and
. Noting that
leads to a linear system
(6.40)
(6.41)
which can be solved for
and
.
For an implicit surface, the projection of
onto the unit normal vector of the surface
can be obtained
from (6.22) as
(6.42)
where
(6.43)
(6.44)
(6.45)
and
are given by (6.28).
Finally, the torsion can be obtained from (6.6) and
(6.37) as
follows
(6.46)
where the binormal vector and curvature are evaluated by
(2.40) and (6.34).
Next: 6.3.4 Higher order derivative
Up: 6.3 Transversal intersection curve
Previous: 6.3.2 Curvature and curvature
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December 2009