The algorithm introduced in Sect. 6.3.3 to
compute the third order derivative vector of the intersection curve can
be generalized to compute the higher order derivative vectors
, under the assumption that we have evaluated
for
,
and
for
. The
algorithm is as follows:
Evaluate the
-
(
) order derivative vector
by successively differentiating (6.5). At
each differentiation step replace
,
, and
by
,
, and
using the Frenet-Serret formulae
(2.56), which leads to the equation
(6.47)
where
,
and
are the coefficients that depend
exclusively on
and
and their derivatives (see
(6.10), (6.13) for reference). As we
will see in step 2, it is not necessary to evaluate
and
. The coefficient
consists of
,
and their
derivatives of order up to
and
,
respectively, which have already been evaluated in the
earlier stages of the computation. For example
,
,
,
can be obtained by taking the dot product between
the curve derivative vectors with
or
, thus from
(6.5), (6.10) and
(6.13):
(6.48)
(6.49)
(6.50)
(6.51)
Replace the terms
in
(6.47) by
since
they both lie on the normal plane:
(6.52)
Evaluate
by projecting
, which is the
-
derivative of the intersection curve evaluated as a curve on
a surface, onto the unit surface normal.
Parametric surface:
Differentiate (6.19) with respect to the arc
length using the chain rule to evaluate
as a curve on
the parametric surface. To compute
we also need to
obtain
and
, which can be done by taking the
dot product on the both sides of
with
and
and solving the linear system. Project
onto two surface normal to obtain
.
Implicit surface: Differentiate (6.22)
with respect to the arc length successively. The resulting expression
always involves the terms of the form
, which is the projection of
onto the surface
normal
. Thus, by moving the rest of the terms to the right
hand side we have
(6.53)
The numerator involves terms
,
,
, for
(not
explicitly expressed here), are obtained by the components of
.
Project (6.52) onto both the unit
surface normal vectors yielding
(6.54)
where
.
Substitute
and
, obtained from Step 3, into
(6.54), and solve the linear system for
and
and substitute into (6.52), resulting
in
(6.55)
Next: 6.4 Intersection curve at
Up: 6.3 Transversal intersection curve
Previous: 6.3.3 Torsion and third
Contents Index
December 2009